GEOMETRIC ANALYSIS (draft)

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GEOMETRIC ANALYSIS

Peter Li Department of Mathematics University of California Irvine, CA 92697-3875 USA [email protected] February 10, 2011

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PETER LI

To my wife Glenna for her love and support

GEOMETRIC ANALYSIS

Table of Contents §0 Introduction §1 First and Second Variational Formulas for Area §2 Volume Comparison Theorem §3 Bochner-Weitzenb¨ ock Formulas §4 Laplacian Comparison Theorem §5 Poincar´e Inequality and the First Eigenvalue §6 Gradient Estimate and Harnack Inequality §7 Mean Value Inequality §8 Reilly’s Formula and Applications §9 Isoperimetric Inequalities and Sobolev Inequalities §10 The Heat Equation §11 Properties and Estimates of the Heat Kernel §12 Gradient Estimate and Harnack Inequality for the Heat Equation §13 Upper and Lower Bounds for the Heat Kernel §14 Sobolev Inequality, Poincar´e Inequality and Parabolic Mean Value Inequality §15 Uniqueness and Maximum Principle for the Heat Equation §16 Large Time Behavior of the Heat Kernel §17 Green’s Function §18 Parabolicity §19 Harmonic Functions and Ends §20 Manifolds with Positive Spectrum §21 Manifolds with Ricci Curvature Bounded from Below §22 Manifolds with Finite Volume §23 Stability of Minimal Hypersurfaces in a 3-Manifold §24 Stability of Minimal Hypersurfaces in a Higher Dimensional Manifold §25 Linear Growth Harmonic Functions §26 Polynomial Growth Harmonic Functions §27 Mean Value Constant, Liouville Property, and Minimal Submanifolds §28 Massive Sets §29 The Structure of Harmonic Maps into a Cartan-Hadamard Manifold §30 Lq Harmonic Functions §31 Measured Neumann Poincar´e Inequality and Measured Sobolev Inequality §32 Parabolic Harnack Inequality and Regularity Theory Appendix A: Computation of Warped Product Metrics Appendix B: Polynomial Growth Harmonic Functions on Euclidean Space References

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§0 Introduction The main goal of this book is to present the basic tools that are necessary for research in geometric analysis. Though the main theme centers around linear theory, i.e., the Laplace equation, the heat equation, and eigenvalues for the Laplacian, the methods in dealing with these problems are quite often useful to the study of nonlinear partial differential equations that arise in geometry. The first part of this book originated from a series of lectures given by the author at a Geometry Summer Program in 1990 at the Mathematical Sciences Research Institute in Berkeley. The lecture notes were revised and expanded when the author taught a regular course in geometric analysis. During the author’s visit with the Global Analysis Research Institute at Seoul National University, he was encouraged to submit these notes, though still in a rather crude form, for publication in their lecture notes series [L6]. Another part of this book on harmonic functions originated from the lecture notes of the author while he gave a sequence of courses on this topic at the University of California, Irvine. A certain part of this material was also used in a series of lectures the author gave in the XIV Escola de Geometria Diferencial in Brazil during the summer of 2006. These notes [L9] were printed for distribution to the participants of the program. After updating the Korea lecture notes with more recent developments and combining with them the harmonic function notes, the author also inserted a few sections on the heat equation. The resulting product now takes the form of an introduction to the subject of geometric analysis on the one hand, with some application to geometric problems via linear theory on the other. Due to the vast literature in geometric analysis, it is prudent not to make any attempt to try to discuss the nonlinear theory. The interested reader is encouraged to consult the excellent book of Schoen and Yau [SY2] in this direction. The readers should be aware that the aim of this book is to address entry level geometric analysts by introducing the basic techniques in geometric analysis in the most economical way. The theorems discussed are chosen sometimes for their fundamental usefulness and sometimes for purpose of demonstrating various techniques. In many cases, they often do not represent the best possible results that are available. The author would like to express his gratitude to Ovidiu Munteanu, Lei Ni and Jiaping Wang for their suggestions to improve this book. He is particular in debt to Munteanu for his detail proof-reading of the draft. Acknowledgement is also due to his graduate students Lihan Wang and Fei He, who were extremely helpful in pointing out necessary corrections to the manuscript.

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§1. First and Second Variational Formulas for Area Let M be a Riemannian manifold of dimension m with metric denoted by ds2 . In terms of local coordinates {x1 , . . . , xm } the metric is written in the form ds2 = gij dxi dxj , where we are adapting the convention that repeated indices are being taken the summation over. If X and Y are tangent vectors at a point p ∈ M, we will also denote their inner product by ds2 (X, Y ) = hX, Y i. If we denote S(T M ) to be the set of smooth vector fields on M, then the Riemannian connection ∇ : S(T M ) × S(T M ) → S(T M ) satisfies the following properties: (1) ∇(f1 X1 +f2 X2 ) Y = f1 ∇X1 Y + f2 ∇X2 Y, for all X1 , X2 , Y ∈ S(T M ) and for all f1 , f2 ∈ C ∞ (M ); (2) ∇X (g1 Y1 + g2 Y2 ) = X(g1 )Y1 + g1 ∇X Y1 + X(g2 )Y2 + g2 ∇X Y2 , for all X, Y1 , Y2 ∈ S(T M ) and for all g1 , g2 ∈ C ∞ (M ); (3) XhY, Zi = h∇X Y, Zi + hY, ∇X Zi, for all X, Y, Z ∈ S(T M ); and (4) ∇X Y − ∇Y X = [X, Y ], for all X, Y ∈ S(T M ). Property (3) says that ∇ is compatible with the Riemannian metric, while property (4) means that ∇ is torsion free. The curvature tensor of the Riemannian metric is then given by RXY Z = ∇X ∇Y Z − ∇Y ∇X Z − ∇[X,Y ] Z, for X, Y, Z ∈ S(T M ), and it satisfies the properties: (1) RXY Z = −RY X Z, for all X, Y, Z ∈ S(T M ); (2) RXY Z + RY Z X + RZX Y = 0, for all X, Y, Z ∈ S(T M ); and (3) hRXY Z, W i = hRZW X, Y i, for all X, Y, Z, W ∈ S(T M ). The sectional curvature of the 2-plane section spanned by a pair of orthonormal vectors X and Y are defined by K(X, Y ) = hRXY Y, Xi. If we take {e1 , . . . , em } to be an orthonormal basis of the tangent space of M , then the Ricci curvature is defined to be the symmetric 2-tensor given by Rij =

m X

hRei ,ek ek , ej i.

k=1

Observe that Rii =

X

K(ei , ek ).

k6=i

Let N be an n-dimensional submanifold in M with n < m. The Riemannian metric ds2M defined on M when restricted to N induces a Riemannian metric ds2N on N. One checks easily that for vector fields X, Y ∈ S(T N ), if we define ∇tX Y = (∇X Y )t to be the tangential component of ∇X Y to N, then ∇t is the Riemannian connection of N with respect to ds2N . The normal component of ∇ yields the second fundamental form of N. In particular, one defines − → − II (X, Y ) = (∇X Y )n

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− → and checks that it is tensorial with respect to X, Y ∈ S(T N ). Taking the trace of the bilinear form II over the tangent space of N yields the mean curvature vector, given by − → − → tr II = H . In the remaining of this section we will derive the first and second variational formulas for the area functional of a submanifold. Let N n ⊂ M m be a n-dimensional submanifold of a m-dimensional manifold M with m > n. Consider a 1-parameter family of deformations of N given by Nt = φ(N, t) for t ∈ (−, ) with N0 = N. Let {x1 , . . . , xn } be a coordinate system around a point p ∈ N. We can consider  {x1 , . . . , xn , t} to

∂ for i = 1, . . . , n be a coordinate system of N × (−, ) near the point (p, 0). Let us denote ei = dφ ∂x i
∂ and T = dφ ∂t . The induced metric on Nt from M is then given by gij = hei , ej i. We may further assume that {x1 , . . . , xn } form a normal coordinate system at p ∈ N. Hence gij (p, 0) = δij and ∇ei ej (p, 0) = 0. Let us define dAt to be the area element of Nt with respect to the induced metric. For t sufficiently close to 0, we can write dAt = J(x, t) dA0 . With respect to the normal coordinate system {x1 , . . . , xn }, the function J(x, t) is given by p g(x, t) J(x, t) = p g(x, 0)

with g(x, t) = det(gij (x, t)). To compute the first variation for the area of N, we compute J 0 (p, t) = By the assumption that gij (p, 0) = δij , we have J 0 (p, 0) = 21 g 0 (p, 0). However,

∂J ∂t (p, t).

g = det(gij ) =

n X

g1j c1j

j=1

where cij are the cofactors of gij . Therefore g 0 (p, 0) =

n X

0 g1j (p, 0) c1j (p, 0) +

j=1

n X

g1j (p, 0) c01j (p, 0)

j=1

0 = g11 (p, 0) + c011 (p, 0).

By induction on the dimension, we conclude that g 0 (p, 0) =

Pn

0 i=1 gii .

On the other hand,

0 gii = T hei , ei i

= 2h∇T ei , ei i = 2h∇ei T, ei i because {x1 , . . . , xn , t} form a coordinate system for N × (−, ). Let us point out that the quantity n X

h∇ei T, ei i

i=1

is now well-defined under orthonormal change of basis and hence is globally defined. If we write T = T t + T n where T t is the tangential component of T on N and T n its normal component, then n X i=1

h∇ei T, ei i =

n n X X h∇ei T t , ei i + h∇ei T n , ei i i=1

i=1

= div(T t ) +

n X i=1

ei hT n , ei i −

− → = div(T ) + hT , H i t

n

n X hT n , ∇ei ei i i=1

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− → where H is the mean curvature vector of N. Hence the first variation for the volume form at the point (p, 0) is given by d − → dAt |(p,0) = (divT t + hT n , H i)dA0 |(p,0) . dt However, the right hand side is intrinsically defined independent on the choice of coordinates and hence this formula is valid at any arbitrary point. If T is a compactly supported variational vector field on N, then using the divergence theorem the first variation of the area of N is given by Z d − → = A(Nt ) hT n , H i. dt N t=0 This shows that the mean curvature of N is identically 0 if and only if N is a critical point of the area functional. Definition 1.1. An immersed submanifold N ,→ M is said to be minimal if its mean curvature vector − → vanishes identically, i.e., H ≡ 0. When N is a curve in M that is parametrized by arc-length with unit tangent vector e, then the first variational formula for length can be written as Z l l d hT n , ∇e ei = hT t , ei 0 − L dt t=0 0 Z l l = hT, ei|0 − hT, ∇e ei. 0

We will now proceed to derive the second variational formula for area. Let φ : N × (−, ) × (−, ) −→ M ∂ be a 2-parameter family of variations of N . Following the similar notation, we denote dφ( ∂x ) = ei for i ∂ ∂ ) = T and dφ( ∂s ) = S. In terms of a general i = 1, . . . , n, and the variational vector fields by dφ( ∂t coordinate system, the first partial derivative of J can be written as n X ∂J (x, t, s) = g ij h∇ei T, ej i J(x, t, s), ∂t i,j=1

where (g ij ) denotes the inverse matrix of (gij ). Differentiating this with respect to s and evaluating at (p, 0, 0) we have (1.1)

n X
∂2J = S g ij h∇ei T, ej iJ ∂s∂t i,j=1

=

n X

(Sg ij )h∇ei T, ej iJ +

i,j=1

n X

g ij (Sh∇ei T, ej i) J

i,j=1 n X

+

g ij h∇ei T, ej iS(J)

i,j=1

=

n X

(Sg ij )h∇ei T, ej i +

i,j=1

+

n X

Sh∇ei T, ei i

i=1

 ! n n X X h∇ei T, ei i  h∇ej S, ej i . i=1

j=1

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PETER LI

However, differentiating the formula

Pn

k=1

n X

g ik gkj = δij , we obtain

(Sg ik )gkj = −

k=1

n X

g ik (Sgkj ),

k=1

hence Sg ij = −

n X

g ik (Sgkl )g lj

k,l=1

= −Sgij = −Shei , ej i = −h∇S ei , ej i − h∇S ej , ei i = −h∇ei S, ej i − h∇ej S, ei i. The first term on the right hand side of (1.1) now becomes n X

(Sg ij )h∇ei T, ej i = −

n X

h∇ei S, ej ih∇ei T, ej i −

h∇ej S, ei ih∇ei T, ej i.

i,j=1

i,j=1

i,j=1

n X

The second term on the right hand side of (1.1) can be written as n X

Sh∇ei T, ei i =

i=1

n n X X h∇S ∇ei T, ei i + h∇ei T, ∇S ei i i=1

i=1

n n n X X X h∇ei T, ∇ei Si h∇ei ∇S T, ei i + hRSei T, ei i + = i=1

i=1

i=1

where the term hRSei T, ei i on the right hand side denotes the curvature tensor of M. Therefore, we have (1.2)

n n n X X X ∂2J hRSei T, ei i h∇ej S, ei ih∇ei T, ej i + h∇ei S, ej ih∇ei T, ej i − =− ∂s∂t i=1 i,j=1 i,j=1  ! n n n n X X X X h∇ei T, ∇ei Si + h∇ei T, ei i  h∇ej S, ej i . h∇ei ∇S T, ei i + + i=1

i=1

i=1

j=1

We will now consider some special cases that will simplify (1.2). Let us first assume that N is a curve parametrized by arc-length in M with unit tangent vector given by e, the second variational formula for the length is given by Z l ∂ 2 L = {−h∇e S, eih∇e T, ei + hRSe T, ei} ∂s∂t (s,t)=(0,0) 0 Z l + {h∇e ∇S T, ei + h∇e T, ∇e Si} . 0

If we further assumed that N is a geodesic, ie. ∇e e ≡ 0, then we have Z l Z l ∂ 2 L = {−(ehS, ei)(ehT, ei) + hR T, ei} + {eh∇S T, ei + h∇e T, ∇e Si} Se ∂s∂t (s,t)=(0,0) 0 0 Z l l = {h∇e T, ∇e Si + hRSe T, ei − (ehS, ei)(ehT, ei)} + h∇S T, ei|0 . 0

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The second special case is when N is a general n-dimensional manifold and if the two variational vector fields are the same and are normal to N , then (1.2) becomes (1.3)

n n n X X X ∂ 2 J 2 = − h∇ T, e i − h∇ T, e ih∇ T, e i + hRT ei T, ei i e j e i e j i j i ∂t2 t=0 i,j=1 i,j=1 i=1 !2 n n n X X X 2 + h∇ei ∇T T, ei i + |∇ei T | + h∇ei T, ei i i=1 n X

=−

i=1

h∇ei T, ej i2 −

i,j=1

n X

i=1

h∇ej T, ei ih∇ei T, ej i −

i,j=1

n X hRei T T, ei i i=1

− → + div(∇T T )t + h(∇T T )n , H i +

n X

− → |∇ei T |2 + hT, H i2 .

i=1

On the other hand, if {en+1 , . . . , em } denotes an orthonormal set of vectors normal to N in M , then n X

h∇ei T, ∇ei T i =

i=1

n X

2

h∇ei T, ej i +

m n X X

h∇ei T, eν i2 .

i=1 ν=n+1

i,j=1

Also − → h∇ei T, ej i = hT, II ij i = h∇ej T, ei i, − → where II ij denotes the second fundamental form with value in the normal bundle of N . Hence, (1.3) becomes n X ∂ 2 J − → 2 X hRei T T, ei i + div(∇T T )t = − hT, II i − ij ∂t2 t=0 i=1 i, j n m X X − → − → + h(∇T T )n , H i + h∇ei T, eν i2 + hT, H i2 . i=1 ν=n+1

Therefore, the second variational formula for area in terms of compactly supported normal variations is given by Z n  X d2 − → − → 2 X A(N ) = − hT, II i − hRei T T, ei i + h(∇T T )n , H i t ij 2 dt N t=0 i, j i=1 Z n m X X  − → + h∇ei T, eν i2 + hT, H i2 . N

i=1 ν=n+1

Definition 1.2. A minimally immersed submanifold N ,→ M is said to be stable if the second variation for area with respect to all compactly supported normal variations are nonnegative. This means that the stability inequality Z X Z X Z X n n m X − → 0≤− hT, II ij i2 − hRei T T, ei i + h∇ei T, eν i2 N i, j

N i=1

is valid for any compactly supported normal vector field T.

N i=1 ν=n+1

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PETER LI

If we further restrict N to be an orientable codimension-1 minimal submanifold of an orientable manifold M , we can write any normal variation in the form T = ψem , where ψ is a differentiable function on N and em is a unit normal vector field to N. Then the second variational formula can be written as Z n X  X d2 − → 2 = A(N ) − hT, II i − R(T, T ) + h∇ei T, em i2 t ij 2 dt N t=0 i, j i=1 Z  = − ψ 2 h2ij − ψ 2 R(em , em ) + |∇ψ|2 N

− → where II ij = hij em with hij being the component of the second fundamental form and R(T, T ) denotes the Ricci curvature of M in the direction of T. Here we have also used the fact that h∇ei T, em i = ψh∇ei em , em i + ei (ψ)hem , em i = ei (ψ). In particular, the stability inequality in this case is given by Z Z Z (1.4) |∇ψ|2 ≥ ψ 2 h2ij + ψ 2 R(em , em ). N

N

N

The last special case is again to assume that N is an oriented hypersurface in an oriented manifold M and we restrict the variation to be given by hypersurfaces which are constant distant from N, the variational vector field is then given by em with ∇em em ≡ 0. This situation is particularly useful for the purpose of − → controlling the growth of the volume of geodesic balls of radius r. In this case, if we write H = H em , the first variational formula for the area element becomes (1.5)

∂J (x, 0) = H(x) J(x, 0), ∂t

and the second variational formula can be written as (1.6)

m−1 X ∂2J (x, 0) = − h2ij (x) J(x, 0) − R(em , em )(x) J(x, 0) + H 2 (x) J(x, 0). 2 ∂t i,j=1

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§2 Volume Comparison Theorem In this section, we will develop a volume comparison theorem originally proved by Bishop (see [BC]). Let p ∈ M be a point in a complete Riemannian manifold of dimension m. In terms of polar normal coordinates at p, we can write the volume element as J(θ, r)dr ∧ dθ where dθ is the area element of the unit (m − 1)-sphere. The Gauss lemma asserts that the area element of submanifold ∂Bp (r) which is the boundary of the geodesic ball of radius r is given by J(θ, r)dθ. By the first and second variational formulas (1.5) and (1.6), if x = (θ, r) is not in the cut-locus of p, we have ∂J (θ, r) ∂r = H(θ, r) J(θ, r)

J 0 (θ, r) =

(2.1)

and (2.2)

∂2J (θ, r) ∂r2 m−1 X h2ij (θ, r) J(θ, r) − Rrr (θ, r) J(θ, r) + H 2 (θ, r) J(θ, r) =−

J 00 (θ, r) =

i,j=1 ∂ ∂ where Rrr = R( ∂r , ∂r ), H(θ, r), and (hij (θ, r)) denote the Ricci curvature in the radius direction, the mean curvature and the second fundamental form of ∂Bp (r) at the point x = (θ, r) with respect to the unit normal ∂ vector ∂r , respectively. Using the inequalities m−1 X

(2.3)

h2ij ≥

i,j=1

m−1 X

h2ii

i=1

m−1 P ≥

2 hii

i=1

m−1 H2 = m−1 and (2.1), we can estimate (2.2) by (2.4)

m−2 2 H J − Rrr J m−1 m − 2 0 2 −1 = (J ) J − Rrr J. m−1

J 00 ≤

Since any smooth metric is locally Euclidean, we have the initial conditions J(θ, r) ∼ rm−1 and J 0 (θ, r) ∼ (m − 1)rm−2 , as r → 0. Let us point out that if M is a simply connected constant curvature space form with constant sectional curvature K, then all the above inequalities become equalities. In particular (2.4) becomes J 00 =

m − 2 0 2 −1 (J ) J − (m − 1)K J. m−1

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Theorem 2.1. (Bishop [BC]) Let M be a complete Riemannian manifold of dimension m, and p a fixed point of M. Let us assume that the Ricci curvature tensor of M at any point x is bounded below by (m−1)K(r(p, x)) for some function K depending only on the distance from p. If J(θ, r) dθ is the area element of ∂Bp (r) as ¯ defined above and J(r) is the solution of the ordinary differential equation m − 2 ¯0 2 ¯−1 (J ) J − (m − 1)K J¯ J¯00 = m−1 with initial conditions ¯ ∼ rm−1 J(r) and J¯0 (r) ∼ (m − 1)rm−2 , as r → 0, then within the cut-locus of p, the function J(θ,r) is a non-increasing function of r. Also, if ¯ J(r) J¯0 ¯ ¯ H(r) = J¯ , then H(θ, r) ≤ H(r) whenever (θ, r) is within cut-locus of p. In particular, if K is a constant, ¯ corresponds to the area element of the sphere of radius r in the simply connected space form of then Jdθ constant curvature K. 1

Proof. By setting f = J m−1 , (2.1) and (2.4) can be written as f0 =

1 Hf m−1

and −1 Rrr f m−1 ≤ −K f.

f 00 ≤

The initial conditions become f (θ, 0) = 0 and f 0 (θ, 0) = 1. 1 ¯ The function f¯ satisfies Let f¯ = J¯m−1 be the corresponding function defined using J.

f¯00 = −K f¯, f¯(0) = 0, and f¯0 (0) = 1. Observe that when K is a constant, the function f¯ > 0 for all values of r ∈ (0, ∞) when K ≤ 0, and for r ∈ (0, √πK ) when K > 0. In general, f¯ > 0 on an interval (0, a) for some a > 0. At those values of r we can define f (θ, r) F (θ, r) = ¯ . f (r) We have F 0 = f¯−2 (f 0 f¯ − f f¯0 ) and F 00 = f¯−1 f 00 − 2f¯−2 f 0 f¯0 − f¯−2 f f¯00 + 2f¯−3 f (f¯0 )2 ≤ −2f¯−1 f¯0 F 0 ,

GEOMETRIC ANALYSIS

13

hence (f¯2 F 0 )0 = f¯2 (F 00 + 2f¯−1 f¯0 F 0 ) ≤ 0. Integrating from  to r yields F 0 (r) ≤ F 0 () f¯2 () f¯−2 (r) = (f¯() f 0 () − f () f¯0 ()) f¯−2 (r). Letting  → 0, the initial conditions of f and f¯ implies that F 0 (r) ≤ 0. In particular, f¯ f 0 − f¯0 f ≤ 0, implying ¯ H(θ, r) ≤ H(r). Moreover, F is a non-increasing function of r implies that

J(θ,r) ¯ J(r)

is also a non-increasing function of r.



By computing the area element and the mean curvature of the constant curvature space form explicitly, we have the follow corollary. Corollary 2.1. Under the same assumption of the theorem, if K is a constant, then  √ √  (m − 1) K cot( Kr), for K > 0   −1 H≤ (m − 1)r , for K = 0    (m − 1)√−K coth(√−Kr), for K < 0, and

J(θ, r) ¯ J(r)

is a non-increasing function of r, where   m−1 √  1   √ sinm−1 ( Kr),    K  ¯ rm−1 , J(r) =     m−1   √ 1   sinhm−1 ( −Kr),  √ −K

for K > 0 for K = 0 for K < 0.

Let us take this opportunity to point out that this estimate implies that when K > 0, one must encounter a cut point along any geodesic which has length √πK . In particular, this proves Myers’ theorem. Corollary 2.2. (Myers) Let M be an m-dimensional complete Riemannian manifold with Ricci curvature bounded from below by Rij ≥ (m − 1)K for some constant K > 0. Then M must be compact with diameter d bounded from above by π d≤ √ . K

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PETER LI

Corollary 2.3. Let M be an m-dimensional complete Riemannian manifold with Ricci curvature bounded ¯ is an m-dimensional simply connected space form with from below by a constant (m − 1)K. Suppose M constant sectional curvature K. Let us denote Ap (r) to be the area of the boundary of the geodesic ball ¯ ¯ ∂Bp (r) centered at p ∈ M of radius r and A(r) to be the area of the boundary of a geodesic ball ∂ B(r) of ¯ radius r in M . Then for 0 ≤ r1 ≤ r2 < ∞, we have ¯ 2 ) ≥ Ap (r2 ) A(r ¯ 1 ). Ap (r1 ) A(r

(2.5)

¯ If we denote Vp (r) and V¯ (r) to be the volume of Bp (r) and B(r) respectively, then for 0 ≤ r1 ≤ r2 , r3 ≤ r4 < ∞ we have

(2.6) (Vp (r2 ) − Vp (r1 )) V¯ (r4 ) − V¯ (r3 ) ≥ (Vp (r4 ) − Vp (r3 )) V¯ (r2 ) − V¯ (r1 ) . Proof. Let us define C(r) to be a subset of the unit tangent sphere Sp (M ) at p such that for all θ ∈ C(r) the geodesic given by γ(s) = expp (sθ) is minimizing up to s = r. Clearly for r1 ≤ r2 we have C(r2 ) ⊂ C(r1 ). By Theorem 2.1, we have ¯ 2 ) ≥ J(θ, r2 ) J(r ¯ 1) J(θ, r1 ) J(r for θ ∈ C(r2 ). Integrating over C(r2 ) yields Z Z ¯ 2) ≥ J(θ, r1 ) dθ J(r C(r2 )

¯ 1) J(θ, r2 ) dθ J(r

C(r2 )

¯ 1 ). = Ap (r2 ) J(r On the other hand Z Ap (r1 ) =

J(θ, r1 ) dθ C(r1 )

Z ≥

J(θ, r1 ) dθ. C(r2 )

Hence, together with the fact that ¯ ¯ A(r) = αm−1 J(r) with αm−1 being the area of the unit (m − 1)-sphere, we conclude (2.5). To see (2.6), we first assume that r1 ≤ r2 ≤ r3 ≤ r4 and we simply integrate the inequality ¯ 2 ) ≥ Ap (t2 ) A(t ¯ 1) Ap (t1 ) A(t over r1 ≤ t1 ≤ r2 and r3 ≤ t2 ≤ r4 . For the case when r1 ≤ r3 ≤ r2 ≤ r4 , we write
(Vp (r2 ) − Vp (r1 )) V¯ (r4 ) − V¯ (r3 )

= (Vp (r3 ) − Vp (r1 )) V¯ (r2 ) − V¯ (r3 ) + (Vp (r3 ) − Vp (r1 )) V¯ (r4 ) − V¯ (r2 )

+ (Vp (r2 ) − Vp (r3 )) V¯ (r2 ) − V¯ (r3 ) + (Vp (r2 ) − Vp (r3 )) V¯ (r4 ) − V¯ (r2 )

≥ (Vp (r2 ) − Vp (r3 )) V¯ (r3 ) − V¯ (r1 ) + (Vp (r4 ) − Vp (r2 )) V¯ (r3 ) − V¯ (r1 )

+ (Vp (r2 ) − Vp (r3 )) V¯ (r2 ) − V¯ (r3 ) + (Vp (r4 ) − Vp (r2 )) V¯ (r2 ) − V¯ (r3 )
= (Vp (r4 ) − Vp (r3 )) V¯ (r2 ) − V¯ (r1 ) .  ¯ Let us point out that equality in (2.6) holds if and only if C(r1 ) = C(r4 ) and J(θ, r) = J(r) for all ¯ 0 ≤ r ≤ r4 and θ ∈ C(r1 ). In particular, if r1 = 0 then J(θ, r) = J(r) for all r ≤ r4 and θ ∈ Sp (M ). This ¯ 4 ). implies that Bp (r4 ) is isometric to B(r

GEOMETRIC ANALYSIS

15

Theorem 2.2. Let M be an m-dimensional complete Riemannian manifold with nonnegative Ricci curvature. Then the volume growth of M must satisfy the following estimates: (1) (Bishop) If αm−1 is the area of the unit (m − 1)-sphere, then αm−1 m ρ Vp (ρ) ≤ m for all p ∈ M and ρ ≥ 0. (2) (Yau [Y2]) For all p ∈ M, there exists a constant C(m) > 0 depending only on m, such that Vp (ρ) ≥ C Vp (1) ρ for all ρ > 2. Proof. Applying (2.6) to r1 = 0 = r3 and r4 = r, we have Vp (r2 ) V¯ (r) ≥ Vp (r) V¯ (r2 ). Observing that Vp (r2 ) lim ¯ = 1, V (r2 )

r2 →0

and the upper bound follows. To prove the lower bound, let x ∈ ∂Bp (1 + ρ). Then (2.6) and the curvature assumption implies that (2.7)

Vx (2 + ρ) − Vx (ρ) ≤ Vx (ρ)

(2 + ρ)m − ρm . ρm

However, since the distance between p and x is r(p, x) = 1 + ρ, we have Bp (1) ⊂ (Bx (2 + ρ) \ Bx (ρ)), hence (2.8)

Vp (1) ≤ Vx (2 + ρ) − Vx (ρ).

Since Bx (ρ) ⊂ Bp (1 + 2ρ), we have Vx (ρ) ≤ Vp (1 + 2ρ), therefore combining with (2.7) and (2.8), we conclude that Vp (1) ≤ Vp (1 + 2ρ)

(2 + ρ)m − ρm . ρm

The lower bounded follows by observing that (2 + ρ)m − ρm = C ρ−1 ρm for 21 ρ → ∞.  We would like to remark that if we assume that for a sufficiently small  > 0 the Ricci curvature has a lower bound of the form Rij (x) ≥ −(1 + r(p, x))−2 , then one can show that M must have infinite volume. On the other hand, if the Ricci curvature is bounded from below by Rij (x) ≥ −C0 (1 + r(p, x))−2−δ for some constants C0 , δ > 0, then and the upper bound is also valid and is of the form Vp (r) ≤ Crm where C(C0 , δ, m) > 0 is a constant depending on C0 , δ and m. It is also a good exercise to show that if a complete manifold has Ricci curvature bounded from below by Rij ≥ r(p, x)−2 for some constant  > 41 and for all r > 1, then M must be compact. The next theorem shows that when the upper bound of the diameter given by Myers’ theorem is achieved, then the manifold must be isometrically a sphere.

16

PETER LI

Theorem 2.3. (Cheng [Cg]) Let M be a complete m- dimensional Riemannian manifold with Ricci curvature bounded from below by Rij ≥ (m − 1)K for some constant K > 0. If the diameter d of M satisfies π d= √ , K then M is isometric to the standard sphere of radius

√1 . K

Proof. By scaling, we may assume that K = 1. Let p and q be a pair of points in M which realize the diameter. The volume comparison theorem implies that   ¯ d V (d)
. Vp (d) ≤ Vp ¯ 2 V d 2

The assumption that d =

√π K

implies that V¯ (d)
= 2, ¯ V d2

hence Vp (d) ≤ 2Vp

  d . 2

Similarly, we have   d Vq (d) ≤ 2Vq . 2 However, by the triangle inequality and the fact that d = r(p, q), we have Bp ( d2 ) ∩ Bq ( d2 ) = ∅. Therefore, 2V (M ) = Vp (d) + Vq (d)      d d + Vq ≤ 2 Vp 2 2 ≤ 2V (M ) where V (M ) denotes the volume of M. This implies that the inequalities in the volume comparison theorem are in fact equalities. Hence by the remark following Corollary 2.3 M must be the standard sphere. 

GEOMETRIC ANALYSIS

17

¨ ck Formulas §3 Bochner-Weitzenbo The Bochner-Weitzenb¨ ock formulas, sometimes referred to as the Bochner technique, is one of the most important techniques in the theory of geometric analysis. There are many formulas which can be derived for various situations. In this section, we will only derive the formula for differential forms so as to illustrate the flavor of this technique. For convenience sake, we will also introduce the moving frame notation. Let {e1 , . . . , em } be locally defined orthonormal frame field of the tangent bundle. Let us denote the dual coframe field by {ω1 , . . . , ωm }. They have the property that ωi (ej ) = δij . The connection 1-forms ωij are given by exterior differentiating the ωi ’s, and are uniquely defined by the Cartan’s 1st structural equations dωi = ωij ∧ ωj where ωij + ωji = 0. Cartan’s 2nd structural equations yield the curvature tensor dωij = ωik ∧ ωkj + Ωij , with

1 Rijkl ωl ∧ ωk . 2 Now let us consider the case that N is an n-dimensional submanifold of M. Let {e1 , . . . , em } be an adapted orthonormal frame field of M such that {e1 , . . . , en } are orthonormal to N. We will now adopt the indexing convention that 1 ≤ i, j, k ≤ n and n + 1 ≤ ν, µ ≤ m. The second fundamental form of N is given by Ωij =

ωνi = hνij ωj . Relating the two notations, we have the formulas ωij (ek ) = h∇ek ei , ej i, Rijkl = hRei ej el , ek i, and

− → hνij = h II (ei , ej ), eν i.

The sectional curvature of the 2-plane section spanned by ei and ej is given by Rijij and the Ricci curvature is given by m X Rij = Rikjk . k=1 ∞

Let f ∈ C (M ) be a smooth function defined on M. Its exterior derivative is given by (3.1)

df = fi ωi .

The second covariant derivative of f can be defined by (3.2)

fij ωj = dfi + fj ωji .

Exterior differentiating (3.1), and applying the 1st structural equations, we have 0 = dfi ∧ ωi + fi dωi = dfi ∧ ωi + fi ωij ∧ ωj = (dfi + fj ωji ) ∧ ωi = fij ωj ∧ ωi .

18

PETER LI

This implies that fij − fji = 0 for all i and j. The symmetric 2-tensor given by fij ωj ⊗ ωi is called the hessian of f. Taking the trace of the hessian, we define the Laplacian of f by ∆f = fii . The third covariant derivative of f is defined by fijk ωk = dfij + fkj ωki + fik ωkj . Exterior differentiating (3.2) gives dfij ∧ ωj + fij dωj = dfj ∧ ωji + fj dωji . However, the 1st and 2nd structural equations imply that 0 = −dfij ∧ ωj − fij dωj + dfj ∧ ωji + fj dωji = −dfij ∧ ωj − fij ωjk ∧ ωk + dfj ∧ ωji + fj ωjk ∧ ωki + fj Ωji = −(dfij + fik ωkj ) ∧ ωj + (dfj + fk ωkj ) ∧ ωji + fj Ωji = −(dfij + fik ωkj ) ∧ ωj + fjk ωk ∧ ωji + fj Ωji = −(dfij + fik ωkj + fkj ωki ) ∧ ωj + fj Ωji 1 = −fijk ωk ∧ ωj + fj Rjikl ωl ∧ ωk . 2 This yields the commutation formula 1 fl (Rlijk − Rlikj ) 2 = fl Rlijk .

fijk − fikj =

Contracting the indices k and i by setting k = i and summing over 1 ≤ i ≤ m, we have the Ricci identity fiji − fiij = fl Rlj . For p ≤ m, we will now take the convention on the indices so that 1 ≤ i, j, k, l ≤ m, 1 ≤ α, β, γ ≤ p, and 1 ≤ a, b, c, d ≤ p − 1. Let ω ∈ Λp (M ) be an exterior p-form defined on M. Then in terms of the basis, we can write ω = ai1 ...ip ωip ∧ · · · ∧ ωi1 where the summation is being performed over the multi-index I = (i1 , . . . , ip ). With this understanding, we can write ω = aI ωI . Exterior differentiating yields dω = daI ∧ ωI + aI dωI = daI ∧ ωI + aI (−1)p−α ωip ∧ · · · ∧ dωiα ∧ · · · ∧ ωi1 = daI ∧ ωI + aI ωiα jα ∧ ωip ∧ · · · ∧ ωjα ∧ · · · ∧ ωi1 = (daI + ai1 ...jα ...ip ωjα iα ) ∧ ωI . One defines the covariant derivatives ai1 ,...ip ,j by m X j=1

ai1 ...ip ,j ωj = dai1 ...ip +

X 1≤α≤p jα

ai1 ...jα ...ip ωjα iα

GEOMETRIC ANALYSIS

19

for each multi-index I = (i1 , . . . , ip ). Similarly, for (p − 1)-forms, we have ai1 ...ip−1 ,j ωj = dai1 ...ip−1 + ai1 ...ja ...ip−1 ωja ia . Exterior differentiation yields dai1 ...ip−1 ,j ∧ ωj + ai1 ...ip−1 ,k ωkj ∧ ωj = dai1 ...ja ...ip−1 ∧ ωja ia + ai1 ...ja ...ip−1 ωja k ∧ ωkia 1 + ai1 ...ja ...ip−1 Rja ia kl ωl ∧ ωk . 2 The left hand side becomes dai1 ...ip−1 ,j ∧ ωj + ai1 ...ip−1 ,k ωkj ∧ ωj = ai1 ...ip−1 ,jk ωk ∧ ωj − ai1 ...ka ...ip−1 ,j ωka ia ∧ ωj = ai1 ...ip−1 ,jk ωk ∧ ωj + dai1 ...ka ...ip−1 ∧ ωka ia X + ai1 ...jb ...ka ...ip−1 ωjb ib ∧ ωka ia b6=a

+ ai1 ...ja ...ip−1 ωja ka ∧ ωka ia . Equating to the right hand side gives ai1 ...ip−1 ,jk ωk ∧ ωj +

X

ai1 ...jb ...ka ...ip−1 ωjb ib ∧ ωka ia =

b6=a

1 Rj i lk ai1 ...ja ...ip−1 ωk ∧ ωl . 2 aa

We now claim that the second term on the left hand side is identically 0. Indeed, since X

ai1 ...jb ...ka ...ip−1 ωjb ib ∧ ωka ia =

b6=a

X

ai1 ...jb ...ka ...ip−1 ωjb ib ∧ ωka ia

ba

and the claim follows by interchanging the role of ka and jb in the second term. Hence ai1 ...ip−1 ,jk ωk ∧ ωj =

1 Rj i lk ai1 ...ja ...ip−1 ωk ∧ ωl 2 aa

implying that ai1 ...ip−1 ,lk − ai1 ...ip−1 ,kl = Rja ia lk ai1 ...ja ...ip−1 . Similarly, for p-forms, we also have (3.3)

ai1 ...ip ,jk − ai1 ...ip ,kj = Rlα iα jk ai1 ...lα ...ip .

Let us now compute the Laplacian ∆ω = −dδω − δdω for p-forms. First we have dω = aI,j ωj ∧ ωI X =

X

i1 0 for K = 0 for K < 0

in the sense of distribution. Proof. For a smooth function f , let x ∈ M be a point such that ∇f (x) 6= 0. Then locally the level surface N of f through the point x is a smooth hypersurface. Let {e1 , . . . , em−1 } be an orthonormal vector field tangent to N. Let us denote em to be the unit normal vector to N. The Laplacian of f at x is defined to be ∆f (x) =

m−1 X

(ei ei − ∇ei ei )f (x) + (em em − ∇em em )f (x)

i=1

=

m−1 X

m−1 X

i=1

i=1

(ei ei − (∇ei ei )t )f (x) −

(∇ei ei )n f (x) + (em em − ∇em em )f (x)

− → = ∆N f (x) + H f (x) + (em em − ∇em em )f (x) − → = H f (x) + (em em − ∇em em )f (x) − → where ∆N denotes the Laplacian of N with respect to the induced metric and H denotes the mean curvature vector of N. If we take f = r, then for the point x which is not in the cut-locus of p, we have N = ∂Bp (r). ∂ . Hence, we have Moreover, the unit normal vector em = ∂r ∆r(x) = H(x) ∂ where H(x) is the mean curvature of ∂Bp (r) with respect to ∂r . Therefore according to Corollary 2.1 the theorem is true for those points that are not in the cut-locus of p. Using the same notation as in Theorem 2.1, let us denote

¯ H(r) =

   

√ √ (m − 1) K cot( Kr), −1

(m − 1)r ,    (m − 1)√−K coth(√−Kr),

for K > 0 for K = 0 for K < 0.

GEOMETRIC ANALYSIS

27

To show that ∆r has the desired estimate in the sense of distribution, it suffices to show that for any nonnegative compactly supported smooth function φ, we have Z

Z

¯ φ H(r).

(∆φ) r ≤ M

M

In terms of polar normal coordinates at p, we can write Z

∞

Z

¯ φ H(r) =

M

Z

0

¯ φ H(r) J(θ, r) dθ dr.

C(r)

On the other hand, for each θ ∈ Sp (M ) if we denote R(θ) to be the maximum value of r > 0 such that the geodesic γ(s) = expp (sθ) minimizes up to s = r, then by Fubini’s theorem we can write Z

∞

Z

¯ φ H(r) J(θ, r) dθ dr =

C(r)

0

Z

Z

Sp (M )

R(θ)

¯ φ H(r) J(θ, r) dr dθ.

0

However, for r < R(θ), we have ¯ H(r) J(θ, r) ≥ H(θ, r) J(θ, r) ∂ J(θ, r). = ∂r Therefore, Z

¯ φ H(r) ≥

M

Z

R(θ)

Z

φ Sp (M )

0

Z

Z

R(θ)

=− Sp (M )

Z =− M

0

∂φ + ∂r

∂J dr dθ ∂r ∂φ J dr dθ + ∂r

Z

R(θ)

[φ J]0

dθ

Sp (M )

Z φ(θ, R(θ)) J(θ, R(θ)) dθ Sp (M )

Z ≥−

h∇φ, ∇ri, M

where we have used the fact that both φ and J are nonnegative and J(θ, 0) = 0. On the other hand, since r is Lipschitz, we conclude that Z Z − h∇φ, ∇ri = (∆φ)r, M

M

which proves the theorem.  We are now ready to prove a structural theorem for manifolds with nonnegative Ricci curvature. Let us first define the notions of a line and a ray in a Riemannian manifold. Definition 4.1. A line is a normal geodesic γ : (−∞, ∞) −→ M such that any of its finite segment γ|ba is a minimizing geodesic. Definition 4.2. A ray is half-line γ+ : [0, ∞) −→ M, which is a normal minimizing geodesic.

28

PETER LI

Theorem 4.2. (Cheeger-Gromoll [CG2]) Let M be a complete, m-dimensional, manifold with nonnegative Ricci curvature. If there exists a line in M, then M is isometric to R × N, the product of a real line and an (m − 1)-dimensional manifold N with nonnegative Ricci curvature. Proof. Let γ+ : [0, ∞) −→ M be a ray in M. One defines the Buseman function β+ with respect to γ+ by β+ (x) = lim (t − r(γ+ (t), x)). t→∞

We observe that β+ is a Lipschitz function with Lipschitz constant 1. Moreover, by the Laplacian comparison theorem, ∆β+ (x) ≥ − lim (m − 1)r(γ+ (t), x)−1 t→∞

=0 in the sense of distribution. If γ is a line, then γ+ (t) = γ(t) and γ− (t) = γ(−t) for t ≥ 0 are rays. The corresponding Buseman functions β+ and β− are subharmonic in the sense of distribution. In particular, β− + β+ is also subharmonic on M. On the other hand, since γ is a line, the triangle inequality implies that 2t = r(γ(−t), γ(t)) ≤ r(γ(−t), x) + r(γ(t), x). Hence t − r(γ− (t), x) + t − r(γ+ (t), x) ≤ 0, and by taking the limit as t → ∞ we have β− (x) + β+ (x) ≤ 0. Moreover, it is also clear that β− (x) + β+ (x) = 0 for all x on γ. However, by the strong maximum principle, since the subharmonic function β− + β+ has an interior maximum, it must be identically constant. In particular, both β− and β+ are harmonic and β− ≡ −β+ . By regularity theory, β+ is a smooth harmonic function with |∇β+ | ≤ 1, and |∇β+ | ≡ 1 on the geodesic γ. For simplicity, let us write β = β+ . The Bochner formula gives (4.1)

2 ∆|∇β|2 = 2βij + 2Rij βi βj + 2h∇β, ∇∆βi

≥ 0. Hence by the fact that |∇β|2 achieves its maximum in the interior of M, and the maximum principle for subharmonic functions again implies that, |∇β|2 ≡ 1 on M . Substituting this into (4.1) yields βij ≡ 0, and ∇β is a parallel vector field on M. This implies that M must split, which proves the theorem.  Corollary 4.1. Let M be a complete m-dimensional Riemannian manifold with nonnegative Ricci curvature. If M has at least 2 ends, then there exists a compact (m − 1)-dimensional manifold N of nonnegative Ricci curvature such that M = R × N. Proof. The assumption that M has at least 2 ends implies that there exists a compact set D ⊂ M such that M \ D has at least 2 unbounded components. Hence there are 2 unbounded sequences of points {xi }∞ i=1 and {yi }∞ i=1 , such that the minimal geodesics γi joining xi to yi must intersect D. By compactness of D, if pi ∈ γi ∩ D and vi = γi0 (pi ), then by passing through a subsequence we have pi → p and vi → v for some p ∈ D and v ∈ Tp M. We now claim that the geodesic γ : (−∞, ∞) −→ M given by the initial conditions γ(0) = p and γ 0 (0) = v is a line. To see this, let us consider an arbitrary segment γ|[a,b] of γ. By continuity of initial conditions of second order ordinary differential equation, we know that γi |[a,b] → γ|[a,b] because (pi , vi ) → (p, v). However, by the assumption, γi |[a,b] are minimizing geodesics, hence γ|[a,b] is also minimizing, and γ is a line. Theorem 4.2 now implies that M = R × N. The compactness of N follows from the assumption that M has at least 2 ends.  Another application of the Laplacian comparison theorem is the eigenvalue comparison theorem of Cheng.

GEOMETRIC ANALYSIS

29

Theorem 4.3. (Cheng [Cg1]) Let M be a compact Riemannian manifold of dimension m and ∂M is the boundary of M. Assume that the Ricci curvature of M is bounded by Rij ≥ (m − 1)K ¯ to be the simply connected space form with constant curvature K. for some constant K. Let us consider M Let us denote µ1 (M ) to be the first nonzero eigenvalue of the Dirichlet Laplacian on M and i to be the ¯ ¯ with radius i and µ1 (B(i)) ¯ inscribe radius of M. If B(i) is a geodesic ball in M its first Dirichlet eigenvalue, then ¯ µ1 (M ) ≤ µ1 (B(i)). When ∂M = ∅ , let us denote λ1 (M ) to be the first nonzero eigenvalue of the Laplacian and d to be the ¯ with radius d and µ1 (B( ¯ d )) its first Dirichlet eigenvalue, ¯ d ) is a geodesic ball in M diameter of M. If B( 2 2 2 then    ¯ d λ1 (M ) ≤ µ1 B . 2 Proof. Let us first consider the case when M has boundary. Let Bp (i) be an inscribed ball in M. By the monotonicity of eigenvalues, it suffices to show that ¯ µ1 (Bp (i)) ≤ µ1 (B(i)). ¯ Let u ¯ be the first Dirichlet eigenfunction on B(i). By the uniqueness of u ¯, we may assume that u ¯ ≥ 0. If we ¯ denote p¯ ∈ B(i) to be its center and r¯ to be the distance function to p¯, then u ¯(¯ r) must be a function of r¯ alone. By the fact that u ¯ ≥ 0 and u ¯(i) = 0, the strong maximum principle implies that ∂∂ur¯¯ (i) < 0. If there is ¯ has a interior local minimum. However, some value of r¯ < i such that ∂∂ur¯¯ > 0 then this would imply that u this violates the strong maximum principle. Hence, u ¯0 = ∂∂ur¯¯ ≤ 0. Let us define a Lipschitz function on Bp (i) by u(r) = u ¯(r), where r denotes the distance function to p. Clearly, u satisfies the Dirichlet boundary condition. Computing the Laplacian of u gives ∆u = u ¯0 ∆r + u ¯00 |∇r|2 ≥u ¯0 ∆r + u ¯00 . ¯ to be the Laplacian on M ¯ , then On the other hand, if we denote ∆ ¯ ¯u −µ1 (B(i)) u ¯ = ∆¯ ¯r + u =u ¯0 ∆¯ ¯00 . By Theorem 4.1 and the fact that u ¯0 ≤ 0, we conclude that ¯ ∆u ≥ −µ1 (B(i)) u in the sense of distribution. Hence, by the Rayleigh principle for eigenvalues, we conclude that R µ1 (Bp (i)) ≤

Bp (i)

|∇u|2

R Bp (i)

− =

u2

R RBp (i) Bp (i)

u∆u u2

¯ ≤ µ1 (B(i)).

30

PETER LI

To prove the upper bound for the case when M has no boundary, we consider the disjoint balls Bp ( d2 ) and Bq ( d2 ) centered at a pair of points p and q which realize the diameter. By the above estimate,       d ¯ d µ1 Bp ≤ µ1 B 2 2 and

 µ1

     d ¯ d Bq ≤ µ1 B . 2 2

We now claim that λ1 (M ) ≤ max{µ1 (M1 ), µ1 (M2 )} for any disjoint pair of open subsets M1 , M2 ⊂ M. This will establish the theorem by setting M1 = Bp ( d2 ) and M2 = Bq ( d2 ). To see the claim, let u1 and u2 be nonnegativeRfirst Dirichlet eigenfunctions on M1 and M2 respectively. By multiplying a constant, we may assume that Mi ui = 1 for i = 1, 2. Let us define a Lipschitz function on M by  on M1   u1 , on M2 u = −u2 ,   0, on M \ (M ∪ M ). 1

Clearly,

R M

2

u = 0. Hence by the Rayleigh principle, Z

u21 +

λ1 (M ) M1

Z M2

u22



Z = λ1 (M ) u2 M Z 2 ≤ |∇u| ZM Z 2 = |∇u1 | + |∇u2 |2 M1 M2 Z Z 2 u22 u1 + µ1 (M2 ) = µ1 (M1 ) M2 M1 Z Z 2 ≤ max{µ1 (M1 ), µ1 (M2 )} u1 + M1

This establishes the claim.



M2

u22

 .

GEOMETRIC ANALYSIS

31

´ Inequality and the First Eigenvalue §5 Poincare In this section, we will obtain lower estimates for the first eigenvalue for the Laplacian on a compact manifold. For the moment, we will primarily concern with the case when M has no boundary. The following lower bound was proved by Lichnerowicz [Lz] while Obata [O] considered the case when the estimate is achieved. Theorem 5.1. (Lichnerowicz and Obata) Let M be an m-dimensional compact manifold without boundary. Suppose that the Ricci curvature of M is bounded from below by Rij ≥ (m − 1)K for some constant K > 0, then the first nonzero eigenvalue of the Laplacian on M must satisfy λ1 ≥ mK. Moreover, equality holds if and only if M is isometric to a standard sphere of radius

√1 . K

Proof. Let u be a nonconstant eigenfunction satisfying ∆u = −λu, with λ > 0. Consider the smooth function Q = |∇u|2 +

λ 2 u m

defined on M . Computing its Laplacian 2λ uui )i m 2λ 2 2λ u + uuii = 2u2ji + 2uj ujii + m i m 2λ 2λ = 2u2ji + 2Rij ui uj + 2uj (∆u)j + |∇u|2 + u(∆u) m m

∆Q = (2uj uji +

where we have used the Ricci identity and the convention that summation is being performed on repeated indices. On the other hand m X

u2ji ≥

i,j=1

m X

u2ii

i=1

Pm 2 ( i=1 uii ) m (∆u)2 = m λ2 u2 = . m

≥

Hence, by the assumption on the Ricci curvature, we have (5.1)

2λ2 u2 2λ 2λ2 u2 + 2(m − 1)K|∇u|2 − 2λ|∇u|2 + |∇u|2 − m m m   λ = 2(m − 1) K − |∇u|2 . m

∆Q ≥

32

PETER LI

If λ ≤ mK, then Q is a subharmonic function. By the compactness of M and the maximum principle, Q must be identically constant and all the above inequalities are equalities. In particular, the right hand side of (5.1) must be identically 0. Hence λ = mK because u is nonconstant. Moreover, |∇u|2 +

λ 2 λ u = |u|2∞ m m

where |u|∞ = supM |u|. If we normalize u such that |u|∞ = 1, and observe that at the maximum and the minimum points of u its gradient must vanish, then we conclude that max u = 1 = − min u and √ |∇u| √ = K. 1 − u2 Integrating this along a minimal geodesic γ joining the points where u = 1 and u = −1, we have Z √ |∇u| √ d K≥ 1 − u2 γ Z 1 du √ ≥ 1 − u2 −1 =π where d denotes the diameter of M. However, Cheng’s theorem (Theorem 2.3) implies that M must be the standard sphere.  We will now give a sharp lower bound for the first eigenvalue on manifolds with nonnegative Ricci curvature. The estimate of Lichnerowicz becomes trivial in this case, since the Ricci curvature does not have a positive lower bound. However, one could still estimate the first eigenvalue in terms of the diameter of M alone. Let λ1 be the least nontrivial eigenvalue of a compact manifold and let φ be the corresponding eigenfunction. By multiplying with a constant it is possible to arrange that a − 1 = inf φ, M

a + 1 = sup φ M

where 0 ≤ a(φ) < 1 is the median of φ. Lemma 5.1. (Li-Yau [LY1]) Suppose M m is a compact manifold without boundary whose Ricci curvature is nonnegative. Then the first nontrivial eigenvalue satisfies λ1 ≥

π2 (1 + a)d2

where d is the diameter of M . Proof. Setting λ = λ1 and u = φ − a the eigenvalue equation becomes ∆u = −λ(u + a). Let P = |∇u|2 + cu2 where c = λ(1 + a). Let x0 ∈ M be the point where P is maximum. If |∇u(x0 )| = 6 0 we may rotate the frame so that u1 (x0 ) = |∇u(x0 )|. Differentiating in the ei direction yields 1 Pi = uj uji + cuui . 2

GEOMETRIC ANALYSIS

33

At the point x0 , this yields 0 = u1 (u11 + cu) and uji uji ≥ u211 = c2 u2 .

(5.2)

Covariant differentiating with respect to ei again, using the commutation formula, (5.2), the definition of P , and evaluating at x0 , we have 1 ∆P 2 = uji uji + uj ujii + cu2i + cu∆u

0≥

≥ c2 u2 + uj (∆u)j + Rji uj ui + cu2i − cλu(u + a) ≥ c2 u2 − λu2i + cu2i − cλu(u + a)
= (c − λ) u2i + cu2 − acλu ≥ aλP (x0 ) − acλ. Hence, for all x ∈ M
|∇u(x)|2 ≤ λ(1 + a) 1 − u(x)2 .

(5.3)

Note that (5.3) is trivially satisfied if ∇u(x0 ) = 0, hence the assumption ∇u(x0 ) 6= 0 is not necessary. Let γ be the shortest geodesic from the minimizing point of u to the maximizing point. The length of γ is at most d. Integrating the gradient estimate (5.3) along this segment with respect to arclength, we obtain Z Z Z 1 p p |∇u|ds du √ √ ≥ = π.  d λ(1 + a) ≥ λ(1 + a) ds ≥ 2 1−u 1 − u2 γ γ −1 In view of Lemma 5.1 and known examples, Li and Yau conjectured that the sharp estimate λ1 ≥

π2 d2

should hold for compact manifolds with nonnegative Ricci curvature. In fact, if the first eigenspace has multiplicity greater than 1, this was verified in [L3]. This conjecture was finally proved by Zhong and Yang by applying the maximum principle to a judicious choice of test function. We will now present this argument. Lemma 5.2. The function z(u) =

 p 2 arcsin(u) + u 1 − u2 − u π

defined on [−1, 1] satisfies (5.4)

zu ˙ + z¨(1 − u2 ) + u = 0,

(5.5)

z˙ 2 − 2z z¨ + z˙ ≥ 0,

(5.6)

2z − zu ˙ + 1 ≥ 0,

34

PETER LI

and (1 − u2 ) ≥ 2|z|.

(5.7)

Proof. Differentiating the definition of z(u) yields z˙ =

4p 1 − u2 − 1 π

and z¨ =

−4u √ . π 1 − u2

Clearly (5.4) is satisfied. To see (5.5), we have z˙ 2 − 2z z¨ + z˙ =

4 π 1 − u2 √



  4 p 1 − u2 + u arcsin(u) − (1 + u2 ) . π

Since the right hand side is an even function, it suffices to check that  4 p 1 − u2 + u arcsin(u) − (1 + u2 ) ≥ 0 π on [0, 1]. Computing its derivative d du



  4 p 4 2 2 1 − u + u arcsin(u) − (1 + u ) = arcsin(u) − 2u π π

which is nonpositive on [0, 1]. Hence  4 p 1 − u2 + u arcsin(u) − (1 + u2 ) π    4 p 2 2 ≥ 1 − u + u arcsin(u) − (1 + u ) π u=1 = 0. Inequality (5.6) follows easily because 2z − zu ˙ +1=

4 arcsin(u) + 1 − u π

which is obviously nonnegative. To see (5.7), we will consider the cases −1 ≤ u ≤ 0 and 0 ≤ u ≤ 1 separately. It is clearly that the inequality is valid at −1, 0, and 1. Let us set f (u) = 1 − u2 − Then

 p 4 arcsin(u) + u 1 − u2 + 2u. π

4 p f˙ = −2u − (2 1 − u2 ) + 2, π 8u f¨ = −2 + √ , π 1 − u2

GEOMETRIC ANALYSIS

and

... f =

8 3

π(1 − u2 ) 2

35

.

... When −1 ≤ u ≤ 0, f¨ ≤ 0. Hence f (u) ≥ min{f (−1), f (0)} = 0. For the case 0 ≤ u ≤ 1, f ≥ 0. Hence f˙ ≤ max{f˙(0), f˙(1)} 8 = max{2 − , 0} π = 0. Therefore f (u) ≥ f (1) proving (1 − u2 ) ≥ 2z on [−1, 1]. Note that (1 − u2 ) ≥ 2|z| follows from the previous inequality because z is an odd function while (1 − u2 ) is even.  Lemma 5.3. Suppose M is a compact manifold without boundary whose Ricci curvature is nonnegative. Assume that a nontrivial eigenfunction φ corresponding to the eigenvalue λ is normalized so that for 0 ≤ a < 1, a + 1 = sup φ and a − 1 = inf M φ. By setting u = φ − a, its gradient must satisfy the estimate (5.8)


|∇u|2 ≤ λ 1 − u2 + 2aλz(u)

where (5.9)

z(u) =

 p 2 arcsin(u) + u 1 − u2 − u. π

Proof. We will first prove an estimate similar to (5.8) for u = (φ − a) where 0 <  < 1. The lemma will follow by letting  → 1. By the definition of u, we have ∆u = −λ(u + a) with − ≤ u ≤ . By (5.3) we may assume a > 0. Consider the function Q = |∇u|2 − c(1 − u2 ) − 2aλz(u), and because of (5.3) and (5.7) we can choose c large enough so that supM Q = 0. The lemma follows if c ≤ λ, for a sequence of  → 1, hence we may assume that c > λ. Let the maximum point of Q be x0 . We claim that |∇u(x0 )| > 0 since otherwise ∇u(x0 ) = 0 and 0 = Q(x0 ) = −c(1 − u2 )(x0 ) − 2aλz(x0 ) ≤ −(c − aλ)(1 − 2 ) by (5.7), which is a contradiction. Differentiating in the ei direction gives (5.10)

1 Qi = uj uji + cuui − aλzu ˙ i. 2

At x0 , we can rotate the frame so u1 (x0 ) = |∇u(x0 )| and using Qi = 0, we have (5.11)

2

uji uji ≥ u211 = (cu − aλz) ˙ .

36

PETER LI

Differentiating again, using the commutation formula, Q(x0 ) = 0, (5.7), (5.10), and (5.11), we get 1 ∆Q(x0 ) 2 = uji uji + uj ujii + cu2i + cu∆u − aλ¨ z u2i − aλz∆u ˙

0≥

= uji uji + uj (∆u)j + Rji uj ui + (c − aλ¨ z ) u2i + (cu − aλz) ˙ ∆u 
 2 2 ≥ (cu − aλz) ˙ + (c − λ − aλ¨ z ) c 1 − u + 2aλz − λ (cu − aλz) ˙ (u + a)   2 = −acλ (1 − u )¨ z + uz˙ + u + a2 λ2 −2z z¨ + z˙ 2 + z˙ + aλ(c − λ) {−uz˙ + 2z + 1} + (c − λ)(c − aλ). However by (5.4), (5.5), and (5.6),we conclude that 0 ≥ acλ(1 − )u − a2 λ2 (1 − )z˙ + (c − λ)(c − aλ) 4 ≥ −acλ(1 − ) − a2 λ2 (1 − )( − 1) + (c − λ)(c − aλ) π ≥ −(c + λ)λ(1 − ) + (c − λ)2 . This implies that ( c≤λ

2 + (1 − ) +

p 2

(1 − )(9 − )

) .

Clearly, when  → 1 this yields the desired estimate.  Theorem 5.2. (Zhong-Yang [ZY]) Suppose M is a compact manifold without boundary whose Ricci curvature is nonnegative. Let a ≥ 0 be the median of a normalized first eigenfunction with a + 1 = sup φ and a − 1 = inf φ, and let d be the diameter. Then the first nontrivial eigenvalue satisfies d2 λ1 ≥ π 2 +

4 6 π − 1 a2 ≥ π 2 (1 + .02a2 ). π 2

Proof. Arguing with u = φ − a as before, let γ be the shortest geodesic from a minimum point of u to a maximum point with length at most d. Integrating the gradient estimate (5.8) along this segment with respect to arclength and using oddness of the function, we have 1 2

dλ ≥ λ

1 2

Z ds γ

|∇u|ds p 1 − u2 + 2az(u) γ  Z 1 1 1 √ ≥ +√ du 1 − u2 + 2az 1 − u2 − 2az 0   Z 1 1 3a2 z 2 √ ≥ 2+ du 1 − u2 1 − u2 0 Z 1 2 z √ ≥ π + 3a2 1 − u2 0 4 2  3a π =π+ 2 −1 .  π 2 Z

≥

GEOMETRIC ANALYSIS

37

This technique also applies to manifolds with boundary. Let M n be a compact manifold with smooth boundary whose Ricci curvature is nonnegative. Suppose that the second fundamental form of ∂M is nonnegative. Then the first nontrivial eigenvalue of the Laplacian with Neumann boundary conditions also satisfies the inequality (5.8). The proof is the same as Lemma 5.3 except that the possibility of the maximum of the the function Q may occur at the boundary must be handled. In fact, the boundary convexity assumption implies that the maximum of Q cannot occur on the boundary as indicated in the proof of Corollary 5.1. The next theorem gives an estimate of the first eigenvalue for general compact Riemannian manifold without boundary. The estimate depends on the lower bound of the Ricci curvature, the upper bound of the diameter, and the dimension of M alone. An estimate of this form was first conjectured by Yau in [Y3]. It was first proved by the author [L1] for manifolds with nonnegative Ricci curvature with an improvement by Li-Yau [LY1] as stated in Lemma 5.1. The general case was also proved by Li-Yau in [LY1] and will be presented in Theorem 5.3. The following lemma will be useful for the upcoming theorem and also for various estimates in the next section. Lemma 5.4. Let M be a complete m-dimensional Riemannian manifold. Suppose the Ricci curvature of M is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. Let u be a function defined on M satisfying the equation ∆u = −λu, and if we define Q = |∇ log(a + u)|2 then   2(m − 2) 2 λu m |∇Q|2 Q−1 + h∇v, ∇Qi Q−1 Q− 2(m − 1) m−1 m−1 a+u    2 2 4 λu 2λa 2 λu 2 ≥ Q + − − 2(m − 1)R Q + . m−1 m−1a+u a+u m−1 a+u

∆Q −

Proof. If we set v = log(a + u), then a direct computation shows that v satisfies the equation ∆v = −|∇v|2 − λ +

λa . a+u

Using the Bochner formula, we compute (5.12)

2 ∆Q = 2vij + 2Rij vi vj + 2h∇v, ∇∆vi 2 ≥ 2vij − 2(m − 1)R Q − 2h∇v, ∇Qi −

2λa |∇v|2 . (a + u)

Choosing an orthonormal frame {e1 , . . . , em } at a point so that |∇v| e1 = ∇v, we can write !2 m m X X 2 2 (5.13) |∇|∇v| | = 4 vi vij j=1

=

=

i=1 m X 2 4v12 v1j j=1 m X 2 4|∇v|2 v1j . j=1

38

PETER LI

On the other hand, (5.14)

2 2 vij ≥ v11 +2

2 ≥ v11 +2

m X α=2 m X α=2 m X

2 v1α +

m X

2 vαα

α=2 2 v1α +

Pm 2 ( α=2 vαα ) m−1

(∆v − v11 )2 m−1 α=2  2 m X λu 1 2 2 2 |∇v| + + v11 = v11 + 2 v1α + m−1 a+u α=2  2   m 2v11 m X 2 1 λu λu + ≥ v1j + |∇v|2 + |∇v|2 + . m − 1 j=1 m−1 a+u m−1 a+u 2 = v11 +2

2 v1α +

However, using the identity 2v1 v11 = e1 (|∇v|2 ) = h∇|∇v|2 , ∇vi |∇v|−1 we conclude that 2v11 = |∇v|−2 h∇|∇v|2 , ∇vi. Substituting into (5.14), we obtain 2 vij

  2  m 1 λu m X 2 λu 1 2 −2 2 2 v + h∇|∇v| , ∇vi |∇v| |∇v| + ≥ |∇v| + + . m − 1 j=1 1j m − 1 a+u m−1 a+u

Combining with (5.12) and (5.13) yields (5.15)

∆Q ≥

m |∇Q|2 Q−1 − 2(m − 1)R Q 2(m − 1)   2λu 2(m − 2) 2 −1 + h∇v, ∇Qi Q − Q + Q2 (m − 1)(a + u) m−1 m−1  2 4 λu 2 λu 2λa + Q+ − Q.  m−1 a+u m−1 a+u a+u

Theorem 5.3. (Li-Yau [LY1]) Let M be a compact m-dimensional Riemannian manifold without boundary. Suppose that the Ricci curvature of M is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0, and d denotes the diameter of M. Then there exists constants C1 (m), C2 (m) > 0 depending on m alone, such that, the first nonzero eigenvalue of M satisfies λ1 ≥

√ C1 exp(−C2 d R). 2 d

Proof. Let u be a nonconstant eigenfunction satisfying ∆u = −λu.

GEOMETRIC ANALYSIS

39

By the fact that Z

Z

−λ

u= M

∆u = 0, M

u must change sign. Hence we may normalize u to satisfy min u = −1 and max u ≤ 1. Let us consider the function v = log(a + u) for some constant a > 1. If x0 ∈ M is a point where the function Q = |∇v|2 achieved its maximum, then Lemma 5.4 and the maximum principle asserts that   2  2 λu 2λa 2 λu 4 2 0≥ Q + − − 2(m − 1)R Q + m−1 m−1a+u a+u m−1 a+u   2 4λ 2(m + 1) λa ≥ Q2 + − − 2(m − 1)R Q, m−1 m−1 m−1 a+u which implies that Q(x) ≤ Q(x0 ) ≤ (m − 1)2 R +

(m + 1)aλ (a − 1)

1

for all x ∈ M. Integrating Q 2 = |∇ log(a + u)| along a minimal geodesic γ joining the points at which u = −1 and u = max u, we have  log

a a−1



  a + max u ≤ log a−1 Z ≤ |∇ log(a + u)| γ

r ≤d for all a > 1. Setting t =

a−1 a ,

(m + 1)aλ + (m − 1)2 R a−1

we have (m + 1)λ ≥ t

1 d2



1 log t

2

! − (m − 1)2 R

for all 0 < t < 1. Maximizing the right hand side as a function of t by setting t = exp(−1 −

p 1 + (m − 1)2 Rd2 ),

we obtain the estimate p p 2 (1 + 1 + (m − 1)2 Rd2 ) exp(−1 − 1 + (m − 1)2 Rd2 ) (m + 1)d2 √ C1 ≥ 2 exp(−C2 d R) d

λ≥

as claimed.



40

PETER LI

We would like to point out that when M is a compact manifold with boundary, there are corresponding estimates for the first Dirichlet eigenvalue and the first nonzero Neumann eigenvalue using the maximum principle. In fact, Reilly [R] proved the Lichnerowicz-Obata result for the Dirichlet eigenvalue on manifolds whose boundary has nonnegative mean curvature with respect to the outward normal vector. In1990, Escobar [E] established the Lichnerowicz-Obata result for the first nonzero Neumann eigenvalue on manifolds whose boundary is convex with respect to the second fundamental form. There are analogous estimates to that of Theorem 5.3 for both the Dirichlet and Neumann eigenvalues on manifolds with boundary. In general, the estimate for the Dirichlet eigenvalue [LY1] will depend also on the lower bound of the mean curvature of the boundary with respect to the outward normal, and the estimate for the Neumann eigenvalue will depend also on the lower bound of the second fundamental form of the boundary and the -ball condition (see [Cn]). However, when the boundary is convex, the Neumann eigenvalue has an estimate similar to manifolds without boundary. Corollary 5.1. (Li-Yau [LY1]) Let M be a compact m-dimensional Riemannian manifold whose boundary is convex in the sense that the second fundamental form is nonnegative with respect to the outward pointing normal vector. Suppose that the Ricci curvature of M is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0, and d denotes the diameter of M. Then there exists constants C1 (m), C2 (m) > 0 depending on m alone, such that, the first nonzero Neumann eigenvalue of M satisfies λ1 ≥

√ C1 exp(−C2 d R). d2

Proof. In view of the proof of Theorem 5.3, it suffices to show that the maximum value for the functional Q does not occur on the boundary of M. Supposing the contrary that the maximum point for Q is x0 ∈ ∂M. Let us denote the outward pointing unit normal vector by em , and assume that {e1 , . . . , em−1 } are orthonormal tangent vectors to ∂M. Since Q satisfies the differential inequality (5.15), the strong maximum principle implies that em (Q)(x0 ) > 0. Using the Neumann boundary condition on u, we conclude that em v = 0. Moreover, since the second covariant derivative of v is defined by vji = (ei ej − ∇ei ej ) v, we have em (Q) = 2

m X

(ei v)(em ei v)

i=1

=2

m−1 X

(eα v)(vαm + ∇em eα v)

α=1

=2

m−1 X

(eα v) (vmα + ∇em eα v)

α=1

=2

m−1 X

(eα v) (eα em v − ∇eα em v + ∇em eα v) .

α=1

Using em v = 0 again and the fact that the second fundamental form is defined by hαβ = h∇eα em , eβ i,

GEOMETRIC ANALYSIS

41

we have em (Q) = −2

m−1 X

(eα v)hαβ (eβ v) + 2

α,β=1 m−1 X

≤2

m−1 X

(eα v)h∇em eα , eβ i(eβ v)

α,β=1

(eα v)h∇em eα , eβ i(eβ v).

α,β=1

On the other hand, since h∇em eα , eβ i = −h∇em eβ , eα i, we conclude that 2

m−1 X

(eα v)h∇em eα , eβ i(eβ v) = −2

α,β=1

m−1 X

(eα v)h∇em eβ , eα i(eβ v).

α,β=1

Hence em (Q) ≤ 0, which is a contradiction.



When M is a complete manifold, it is often useful to have a lower bound of the first eigenvalue for the Dirichlet Laplacian on a geodesic ball. This is provided by the next theorem. Theorem 5.4. (Li-Schoen [LS]) Let M be a complete manifold of dimension m. Let p ∈ M be a fixed point such that Bp (2ρ) ∩ ∂M = ∅ for 2ρ ≤ d. Assume that the Ricci curvature on Bp (2ρ) satisfies Rij ≥ −(m − 1)R for some constant R ≥ 0. For any α ≥ 1, there exists constants C1 (α), C2 (m, α) > 0, such that for any compactly supported function f on Bp (ρ) Z

Z √ |∇f |α ≥ C1 ρ−α exp(−C2 (1 + ρ R))

|f |α .

Bp (ρ)

Bp (ρ)

In particular, the first Dirichlet eigenvalue of Bp (ρ) satisfies √ µ1 ≥ C1 ρ−2 exp(−C2 (1 + ρ R)). Proof. Let q ∈ ∂Bp (2ρ). By the triangle inequality Bp (ρ) ⊂ (Bq (3ρ) \ Bq (ρ)). Theorem 4.1 implies that √ √ ∆r ≤ (m − 1) R coth(r R) √ ≤ (m − 1)(r−1 + R) for r(x) = r(q, x). For k > m − 2, we have ∆r−k = −kr−k−1 ∆r + k(k + 1)r−k−2 √ ≥ −k(m − 1)r−k−1 (r−1 + R) + k(k + 1)r−k−2 √ = kr−k−1 ((k + 2 − m)r−1 − (m − 1) R) √ ≥ kr−k−1 ((k + 2 − m)(3ρ)−1 − (m − 1) R)

42

PETER LI

√ on Bp (ρ). Choosing k = m − 1 + 3(m − 1)ρ R this becomes ∆r−k ≥ kr−k−1 (3ρ)−1

(5.16)

≥ k(3ρ)−k−2 on Bp (ρ). Let f be a nonnegative function supported on Bp (ρ). Multiplying (5.16) with f and integrating over Bp (ρ) yields Z Z k(3ρ)−k−2 f≤ f ∆r−k Bp (ρ)

Bp (ρ)

Z

h∇f, ∇r−k i

=− Bp (ρ)

Z

|∇f |r−k−1

≤k Bp (ρ) −k−1

Z

≤ kρ

|∇f |. Bp (ρ)

Hence

Z √ |∇f | ≥ C1 ρ−1 exp(−C2 (1 + ρ R))

Z

f.

Bp (ρ)

Bp (ρ)

This shows the case when α = 1 by simply applying the above inequality to |f |. For α > 1, we set f = |g|α . Then we have Z α

! α1 |∇g|α

! α−1 α

Z

Bp (ρ)

|g|α

Z

|g|α−1 |∇g|

≥α

Bp (ρ)

Bp (ρ)

Z

|∇g α |

= Bp (ρ)

≥ C1 ρ

−1

√

Z

exp(−C2 (1 + ρ R)) Bp (ρ)

which implies the desired inequality.



|g|α ,

GEOMETRIC ANALYSIS

43

§6 Gradient Estimate and Harnack Inequality In this section, we discuss an important estimate that is essential to the study of harmonic functions and many elliptic and parabolic problems. In 1975, Yau [Y1] developed a maximum principle method to prove that complete manifolds with nonnegative Ricci curvature must have a Liouville property. His argument was later localized in his joint paper with Cheng [CgY] and resulted in a gradient estimate for a rather general class of elliptic equations. In 1979, the maximum principle method was used by Li [L1] in proving eigenvalue estimates for compact manifolds. This method was then refined and used by many authors ([LY1], [ZY], etc) for obtaining sharp eigenvalue estimates. In 1986, Li and Yau [LY2] used a similar method to prove a parabolic version of the gradient estimate for the parabolic Schr¨odinger equation. This method has since been used by Hamilton, Chow, Cao, and many other authors to yield estimates for various non-linear parabolic equations. In a recent work [LW6] of Li and Wang, they realized that Yau’s gradient estimate is sharp and equality is achieved on a manifold with negative Ricci curvature. The sharpness of this was somewhat surprising since the parabolic gradient estimate is sharp on a manifold with nonnegative Ricci curvature. In the following treatment, we will present a proof for the sharp gradient estimate for positive functions satisfying the equation ∆f = −λ f where λ ≥ 0 is a constant. The argument will give both the local and global estimates. Various immediate consequences of the gradient estimate will also be derived. Theorem 6.1. Let M m be a complete Riemannian manifold of dimension m. Assume that the geodesic ball Bp (2ρ) ∩ ∂M = ∅. Suppose that the Ricci curvature on Bp (2ρ) is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. If u is a positive function defined on Bp (2ρ) ⊂ M satisfying ∆u = −λ u for some constant λ ≥ 0, then there exists a constant C depending on m such that (4(m − 1)2 + 2)R |∇u|2 (x) ≤ + C((1 + −1 )ρ−2 + λ), u2 4 − 2 for all x ∈ Bp (ρ) and for any  < 2. Moreover, if ∂M = ∅ with Rij ≥ −(m − 1)R everywhere and u is defined on M , then |∇u|2 (m − 1)2 R (x) ≤ −λ+ u2 2 and λ≤

r

(m − 1)4 R2 − (m − 1)2 λ R. 4

(m − 1)2 R . 4

Proof. If we set v = log u and Q = |∇v|2 , then Lemma 5.4 asserts that   m 2(m − 2) 2λ 2 −1 −1 ∆Q − |∇Q| Q + h∇v, ∇Qi Q Q− 2(m − 1) m−1 m−1   2 2 4λ 2λ Q2 + − 2(m − 1)R Q + . ≥ m−1 m−1 m−1

44

PETER LI

Let φ be a nonnegative cutoff function and let G = φ Q, then we have (6.1) ∆G = (∆φ) Q + 2h∇φ, ∇Qi + φ ∆Q m m ∆φ G + 2φ−1 h∇φ, ∇Gi − 2|∇φ|2 φ−2 G + |∇G|2 G−1 + |∇φ|2 φ−2 G ≥ φ 2(m − 1) 2(m − 1)   m 2(m − 2) 2λ − φ−1 h∇φ, ∇Gi − h∇v, ∇Gi Q−1 Q− (m − 1) m−1 m−1     2(m − 2) 2λ 2 2λ2 4λ + h∇v, ∇φi Q− + φ−1 G2 + − 2(m − 1)R G + φ. m−1 m−1 m−1 m−1 m−1 Using the inequality 1

1

|h∇v, ∇φi| ≤ |∇φ| φ− 2 G 2 , (6.1) can be written as ∆G ≥

∆φ m − 2 −1 3m − 4 G+ φ h∇φ, ∇Gi − |∇φ|2 φ−2 G φ m−1 2(m − 1)   m 4λ 2(m − 2) 2 −1 + |∇G| G + − 2(m − 1)R G − h∇v, ∇Gi 2(m − 1) m−1 m−1 3 3 1 1 2(m − 2) 2λ 2λ − |∇φ| φ− 2 G 2 + h∇v, ∇Gi − |∇φ| φ− 2 G 2 m−1 (m − 1)Q m−1 2 2λ2 + φ−1 G2 + φ. m−1 m−1

However, at the maximum point x0 of G, the maximum principle asserts that ∆G(x0 ) ≤ 0 and ∇G(x0 ) = 0. Hence at the point x0 we have (6.2)


3m − 4 |∇φ|2 φ−1 G + 4λ − 2(m − 1)2 R φ G 2 3 1 1 − 12 − 2(m − 2)|∇φ| φ G 2 − 2λ |∇φ| φ 2 G 2 + 2G2 + 2λ2 φ2 .

0 ≥ (m − 1)(∆φ) G −

Let us choose φ(x) = φ(r(x)) to be a function of the distance r to the fixed point p with the property that φ=1 φ=0

on on

1

−Cρ−1 φ 2 ≤ φ0 ≤ 0

Bp (ρ), M \ Bp (2ρ), on

Bp (2ρ) \ Bp (ρ),

and |φ00 | ≤ C ρ−2

on

Bp (2ρ) \ Bp (ρ).

Then the Laplacian comparison theorem asserts that ∆φ = φ0 ∆r + φ00 √ ≥ −C1 (ρ−1 R + ρ−2 ),

GEOMETRIC ANALYSIS

45

and also |∇φ|2 φ−1 ≤ C2 ρ−2 . Hence (6.2) yields (6.3)

√ 0 ≥ −(C3 ρ−1 R + C4 ρ−2 ) G + (4 λ φ − 2(m − 1)2 R φ) G 3

1

− C5 ρ−1 G 2 − C6 λ ρ−1 G 2 + 2G2 + 2λ2 φ2 . We observe that since x0 is the maximum point of G and φ = 1 on Bp (ρ), φ(x0 ) |∇v|2 (x0 ) ≥ sup |∇v|2 (x). Bp (ρ)

On the other hand, using the fact that φ(x0 ) |∇v|2 (x0 ) ≤ φ(x0 ) sup |∇v|2 (x), Bp (2ρ)

we conclude that σ(ρ) ≤ φ(x0 ) ≤ 1, where σ(ρ) denotes σ(ρ) =

supBp (ρ) |∇v|2 (x) supBp (2ρ) |∇v|2 (x)

.

Applying this estimate to (6.3), we have √ 0 ≥ −(C3 ρ−1 R + C4 ρ−2 − 4 λ σ(ρ) + 2(m − 1)2 R) G. (6.4) 3

1

− C5 ρ−1 G 2 − C6 λ ρ−1 G 2 + 2G2 + 2λ2 σ 2 (ρ). On the other hand, Schwarz inequality asserts that 3

−C5 ρ−1 G 2 ≥ − G2 − 1

−C6 λ ρ−1 G 2 ≥ − λ2 − and

C52 −1 −2  ρ G, 4 C62 −1 −2  ρ G, 4

√ −C3 ρ−1 R ≥ − R − C7 −1 ρ−2

for  > 0. Hence combining with (6.4) we obtain 0 ≥ −(C8 (1 + −1 ) ρ−2 − 4 λ σ(ρ) + (2(m − 1)2 + )R) G + (2 − )G2 + 2λ2 σ 2 (ρ) −  λ2 . In particular, if  < 2 then for any x ∈ Bp (ρ), we conclude that (6.5)

|∇v|2 (x) ≤ G(x0 ) p B + B 2 − 4(2 − ) λ2 (2σ 2 (ρ) − ) ≤ 4 − 2

where B = C8 (1 + −1 ) ρ−2 − 4 λ σ(ρ) + (2(m − 1)2 + )R.

46

PETER LI

Since 0 ≤ σ(ρ) ≤ 1, we conclude that (4(m − 1)2 + 2)R |∇u|2 (x) ≤ + C((1 + −1 )ρ−2 + λ) u2 4 − 2 for all x ∈ Bp (ρ) and for all  < 2. If u is defined on M , then it follows that |∇v|2 is a bounded function. In particular, if we take ρ → ∞ in (6.5), then σ(ρ) → 1 and inequality (6.5) becomes (2(m − 1)2 + )R − 4λ + |∇v| ≤ 4 − 2 2

p ((2(m − 1)2 + )R − 4λ)2 − 4(2 − )2 λ2 . 4 − 2

Letting  → 0, we obtain (m − 1)2 R |∇v| ≤ −λ+ 2 2

This gives a contradiction if λ >

(m−1)2 R , 4

r

(m − 1)4 R2 − λ (m − 1)2 R. 4

and the second assertion follows. 

The following Harnack type inequality is a direct consequence of the gradient estimate. Corollary 6.1. Let M m be a complete manifold with Ricci curvature bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. If u is a positive function defined on the geodesic ball Bp (2ρ) ⊂ M satisfying ∆u = −λ u for some constant λ ≥ 0, then there exists constants C9 , C10 > 0 depending on m such that √ u(x) ≤ u(y) C9 exp(C10 ρ R + λ) for all x, y ∈ Bp ( ρ2 ). Proof. Let γ be a shortest curve in Bp (ρ) joining y to x. Clearly, the length of γ is at most 2ρ. Integrating the quantity |∇ log u| along γ yields Z (6.6) log u(x) − log u(y) ≤ |∇ log u|. γ

On the other hand, applying the gradient estimate of Theorem 6.1, we obtain Z

(4(m − 1)2 + 2)R + C((1 + −1 )ρ−2 + λ) 4 − 2 γ Z   √ ≤ C10 R + λ + C12 ρ−1 Z 

 12

|∇ log u| ≤ γ

γ

≤ C10 ρ

√

R + λ + 2C12 .

The corollary follows by combining with (6.6).



An upper bound for the infimum of the spectrum that was a consequence of the comparison theorem of Cheng (Theorem 4.3) can also be recovered by the above estimate.

GEOMETRIC ANALYSIS

47

Definition 6.1. Let M m be a complete Riemannian manifold without boundary. We denote the greatest lower bound for the L2 -spectrum of the Laplacian ∆ by µ1 (M ). The notation µ1 (M ) does not necessarily mean that it is an eigenvalue of ∆, but is motivated by the characterization of µ1 (M ) by µ1 (M ) = lim µ1 (Di ) i→∞

for any compact exhaustion {Di } of M . Corollary 6.2. (Cheng [Cg1]) Let M m be a complete Riemannian manifold without boundary. Suppose the Ricci curvature of M is bounded from below by Rij ≥ −(m − 1), then the greatest lower bound of the L2 -spectrum of the Laplacian has an upper bound given by µ1 (M ) ≤

(m − 1)2 . 4

The following lemma was first proved by Fischer-Colbrie and Schoen [FCS] for general operators of the form L = ∆ + V (x), where V (x) is a smooth potential function. We will provide the proof for the case V is a constant λ as an application of Theorem 6.1. However, it is important to point out that a Harnack inequality similar to Theorem 6.1 is valid in a more general setting of L = ∆ + V (x) (see [CgY]) hence the argument provided below will yield the result of Fischer-Colbrie and Schoen. We will also like to point out that using the Nash-Moser’s Harnack inequality of §32, the potential function V is only required to be in some Lp space. Lemma 6.1. (Fischer-Colbrie and Schoen) Let M m be a complete Riemannian manifold, and µ1 (M ) be the greatest lower bound of ∆. Then the value µ satisfies µ ≤ µ1 if and only if the operator L = ∆ + µ admits a positive solution of the equation Lu = 0 on M. Proof. Let us first assume that there exists a positive function u defined on M , such that, Lu = 0. For any smooth, compact, subdomain D ⊂ M, let f be the first Dirichlet eigenfunction for the operator L = ∆ + µ, satisfying Lf = −µ1 (D, L) f on D, with boundary condition f =0

on

∂D.

By multiplication of −1 if necessary, we may assume that f is positive in the interior of D. In particular, Z −µ1 (D, L)

Z

D

Z u Lf −

uf = D

f Lu D

Z =

u fν . ∂D

The positivity of f on D and its boundary condition asserts that fν < 0, hence together with the positivity of u, we conclude that µ1 (D, L) > 0. Since D is arbitrary, this implies that µ1 (M, L) ≥ 0, hence µ1 (M ) ≥ µ.

48

PETER LI

Conversely, let us assume that µ1 (M ) ≥ µ and µ1 (M, L) ≥ 0. In particular, by monotonicity µ1 (D, L) ≥ 0 for all compact subdomain D ⊂ M. Let us consider a compact exhaustion of M by a sequence of subdomains {Di } with Di ⊂ Di+1 and ∪∞ i=1 Di = M. On each Di , since µ1 (Di , L) > 0, one can find a positive solution to the Dirichlet problem Lfi = 0 on Di , with boundary condition fi = 1

on

∂Di .

Note that fi must be nonnegative on Di since the subdomain Ω = {fi < 0} also has µ1 (Ω, L) > 0 will contradict the fact that fi is an eigenfunction with eigenvalue 0 on Ω. We claim that fi must be positive in the interior of Di . If not, then there exists x ∈ Di with fi (x) = 0 which is a local minimum hence a critical point of fi . On the other hand, a regularity result (see [Cg2]) asserts that at x, the function fi is asymptotically a spherical harmonic in Rm , contradicting the fact that fi has a local minimum at x. For a fixed point p ∈ M , after multiplication by a constant, we may assume that fi (p) = 1. Of course, fi still satisfy Lfi = 0 but with a different boundary condition. For any fixed ρ > 0, Corollary 6.1 asserts that 0 ≤ fi ≤ C on Bp (2ρ), for some constant C > 0 independent of i. Hence a subsequence of {fi } converges uniformly on Bp (2ρ). We also consider the nonnegative cutoff function φ supported on Bp (2ρ) with the property that φ=1 on Bp (ρ), φ=0

M \ Bp (2ρ),

on

and |∇φ|2 ≤ C ρ−2

on

Bp (2ρ) \ Bp (ρ)

for some fixed constant C > 0 independent of ρ. For sufficiently large i since Bp (2ρ) ⊂ Di , we have Z

φ2 fi Lfi

0= M

Z M

1 ≤− 2

Z φ fi h∇φ, ∇fi i + µ φ2 fi2 M M Z Z 2 2 2 2 φ |∇fi | + 2 |∇φ| fi + µ φ2 fi2 .

φ2 |∇fi |2 − 2

=− Z

Z

M

M

M

This implies Z

|∇fi |2 ≤

Z

Bp (ρ)

φ2 |∇fi |2

M

Z

|∇φ|2 fi2 + 2µ

≤4

Z

M

φ2 fi2

M

≤ 4C (ρ−2 + µ) Vp (2ρ), hence we conclude that fi is bounded in H 1,2 (Bp (ρ)). In particular, we choose a subsequence of fi converging uniformly on compact subsets and in H 1,2 to a nonnegative function f . In particular, f is a positive weak solution and by regularity is a strong solution to Lf = 0.  Obviously, Corollary 6.2 follows by combining the second part Theorem 6.1 and Lemma 6.1. Corollary 6.3. (Yau [Y1], Cheng [Cg3]) Let M m be a complete manifold with nonnegative Ricci curvature. There exists a constant C(m) > 0, such that, for any harmonic function defined on M if we denote i(ρ) =

inf x∈Bp (ρ)

u(x)

GEOMETRIC ANALYSIS

49

then |∇u|(x) ≤ C ρ−1 (u(x) − i(2ρ)) for all x ∈ Bp (ρ). In particular, M does not admit any nonconstant harmonic function satisfying the growth estimate lim inf r−1 (x) u(x) ≥ 0. x→∞

Proof. Observe that u − i(2ρ) is a positive harmonic function defined on Bp (2ρ). Applying Theorem 6.1 to this function and setting  = 1, we obtain |∇u|2 (x) ≤ 2Cρ−2 (u(x) − i(2ρ))2 for all x ∈ Bp (ρ), proving the first part of the Corollary. To prove the second part, we observe that the maximum principle asserts that i(ρ) is achieve at some point x ∈ ∂Bp (ρ). In particular, ρ−1 i(ρ) = r−1 (x) u(x) and the growth assumption on u implies that lim ρ−1 i(ρ) ≥ 0.

ρ→∞

Hence for any fixed x ∈ M , we apply the estimate |∇u|(x) ≤ lim C ρ−1 (u(x) − i(2ρ)) ρ→∞

≤0 and conclude that u must be a constant function.



The following example will demonstrate the sharpness of Theorem 6.1 even on manifolds with negative curvature. Example 6.1 Let M m = Hm be the hyperbolic m-space with constant curvature −1. Using the upper half-space model, Hm is given by Rm + = {(x1 , x2 , . . . , xm ) | xm > 0} with metric 2 2 ds2 = x−2 m (dx1 + · · · + dxm ).

For x = (x1 , . . . , xm ) let us consider the function u(x) = xα m . Direct computation yields |∇u|2 = x2m

2 m  X ∂u

i=1 2 2

∂xi

=α u . and ∆u =

x2m

m X ∂2u i=1

∂x2i

− (m − 2)xm

∂u ∂xm

= α(α + 1 − m)u. In particular, for m−1 ≤ α ≤ m − 1, we see that ∆u = −λ u with λ = α(m − 1 − α). This implies that the 2 gradient estimate of Theorem 6.1 is sharp since (m − 1)2 −λ+ 2

r

(m − 1)4 − (m − 1)2 λ = α2 . 4

50

PETER LI

§7 Mean Value Inequality We will prove a version of mean value inequality which is adapted to the theory of subharmonic functions on a Riemannian manifold. Let us begin by proving a theorem of Yau [Y2]. Lemma 7.1. (Yau) Let M be a complete Riemannian manifold. Suppose p ∈ M and ρ4 > 0 be such that the geodesic ball Bp (ρ4 ) centered at p of radius ρ4 satisfies Bp (ρ4 )∩∂M = ∅. Let f be a nonnegative subharmonic function defined on Bp (ρ4 ). Then for any constant α > 1 and for any 0 ≤ ρ1 < ρ2 < ρ3 < ρ4 , we have the estimates ! Z Z Z 4 −2 α −2 α α−2 2 (ρ2 − ρ1 ) f + (ρ4 − ρ3 ) f f |∇f | ≤ (α − 1)2 Bp (ρ2 )\Bp (ρ1 ) Bp (ρ4 )\Bp (ρ3 ) Bp (ρ3 )\Bp (ρ2 ) and

Z

f α−2 |∇f |2 ≤

Bp (ρ3 )

4 (ρ4 − ρ3 )−2 (α − 1)2

Z

f α.

Bp (ρ4 )\Bp (ρ3 )

In particular, if M has no boundary, then there does not exist any nonconstant, nonnegative, Lα subharmonic function. Proof. Let φ(r(x)) be a cutoff function given by a function of r(x) = r(p, x) alone. We may take φ to satisfy  0 for r ≤ ρ1     r − ρ  1  , for ρ1 ≤ r ≤ ρ2     ρ2 − ρ1 1 for ρ2 ≤ r ≤ ρ3 φ(r) =    ρ4 − r   for ρ3 ≤ r ≤ ρ4   ρ  4 − ρ3   0 for ρ4 ≤ r. Let us now consider the integral Z 0≤ φ2 f α−1 ∆f Bp (ρ4 )

Z = −(α − 1)

2 α−2

φ f

Z

2

φf α−1 h∇φ, ∇f i.

|∇f | − 2

Bp (ρ4 )

Bp (ρ4 )

However, an algebraic inequality implies that Z Z α−1Z 2 2 α−2 2 α−1 2 φ f |∇f | + |∇φ|2 f α . φf h∇φ, ∇f i ≤ Bp (ρ4 ) 2 α − 1 Bp (ρ4 ) Bp (ρ4 ) Hence, we have α−1 2

Z

f α−2 |∇f |2 ≤

Bp (ρ3 )\Bp (ρ2 )

≤ ≤

α−1 2

Z

2 α−1

Z

φ2 f α−2 |∇f |2

Bp (ρ4 )

|∇φ|2 f α

Bp (ρ4 )

2 (ρ2 − ρ1 )2 (α − 1)

Z

fα

Bp (ρ2 )\Bp (ρ1 )

2 + (ρ4 − ρ3 )2 (α − 1)

Z Bp (ρ4 )\Bp (ρ3 )

f α.

GEOMETRIC ANALYSIS

51

The second inequality follow similarly by taking φ to be 1 ρ4 − r φ(r) =   ρ4 − ρ3  0    

for

r ≤ ρ3

for

ρ3 ≤ r ≤ ρ4

for

ρ4 ≤ r.

If M has no boundary, we simply set ρ3 = ρ, ρ4 = 2ρ and let ρ → ∞. The fact that f is Lα implies that the integral Z f α−2 |∇f |2 = 0, M

which implies that |∇f | ≡ 0. Hence f must be a constant function.



Theorem 7.1. (Li-Schoen [LS]) Let M be a complete Riemannian manifold of dimension m. Let p ∈ M be a fixed point such that the geodesic ball Bp (4ρ) centered at p of radius 4ρ satisfy Bp (4ρ) ∩ ∂M = ∅. Suppose f is a nonnegative subharmonic function defined on Bp (4ρ). Assume that the Ricci curvature on Bp (4ρ) is bounded by Rij ≥ −(m − 1)R for some constant R ≥ 0. Then there exists constants C3 , C4 (m) > 0 with C4 depending only on m such that Z √ f 2. sup f 2 ≤ C3 (1 + exp(C4 ρ R))Vp (4ρ)−1 x∈Bp (ρ)

Bp (4ρ)

Proof. Let h be a harmonic function on Bp (2ρ) obtained by the solving the Dirichlet boundary problem ∆h = 0

on Bp (2ρ),

and h=f

on ∂Bp (2ρ).

Since f is nonnegative, by the maximum principle h is positive on the ball Bp (ρ). Moreover, f ≤ h on Bp (ρ). The Harnack inequality (Corollary 6.1) implies that   √ sup h ≤ inf h exp(C(1 + ρ R)). Bp (ρ)

Bp (ρ)

Hence, in particular, we have (7.1)

sup f 2 ≤ sup h2 Bp (ρ)

Bp (ρ)

Z √ ≤ exp(2C(1 + ρ R))Vp (ρ)−1

h2 .

Bp (ρ)

We will now estimate the L2 -norm of h in terms of the L2 -norm of f . By triangle inequality, we observe that Z Z Z (7.2) h2 ≤ 2 (h − f )2 + 2 f2 Bp (ρ)

Bp (ρ)

Z ≤2

B (ρ)

2

Zp

(h − f ) + 2 Bp (2ρ)

Bp (4ρ)

f 2.

52

PETER LI

However, since the functions h − f vanishes on ∂Bp (2ρ), the Poincar´e inequality (Theorem 5.4) implies that Z √ (h − f )2 ≤ C1 ρ2 exp(C2 (1 + ρ R))

Z (7.3) Bp (2ρ)

|∇(h − f )|2

Bp (2ρ)

for some constants C1 > 0 and C2 (m) > 0. By triangle inequality again, we have Z

|∇(h − f )|2 ≤ 2

Bp (2ρ)

Z

|∇h|2 + 2

Bp (2ρ)

Z

|∇f |2 .

Bp (2ρ)

The fact that a harmonic function minimizes Dirichlet integral among functions with the same boundary data asserts that Z Z |∇(h − f )|2 ≤ 4 |∇f |2 . Bp (2ρ)

Bp (2ρ)

Now the argument in Lemma (7.1) implies that Z

|∇f |2 ≤ Cρ−2

Bp (2ρ)

Z

f2

Bp (4ρ)

for some constant C > 0. Hence together with (7.1), (7.2), (7.3), and the volume comparison (Corollary 2.3), the theorem follows.  Let us point out that the fact that the constant in the mean value inequality depends only on the lower bound of the Ricci curvature and the radius of the ball is essential in some of the geometric application. In fact, it is well known that one can prove another version of the mean value inequality by using an iteration method of Moser. This will be presented in a later section and the difference of these two methods will also be pointed out later. We will now give an application of this mean value inequality to study the space of harmonic functions on a certain class of manifolds. This result can be viewed as a generalization of Yau’s Liouville theorem. Let us first prove a lemma. Lemma 7.2. (Li [L2]) Let H be a finite dimensional space of L2 functions defined over a set D. If V (D) denotes the volume of the set D, then there exists a function f0 in H such that Z dim H f02 ≤ V (D) sup f02 . D

D

Proof. Let f1 , . . . , fk be an orthonormal basis for H with respect to the L2 inner product. Let us consider the function k X F (x) = fi2 (x) i=1

which is well-defined under orthonormal change of basis. Clearly Z dim H = F (x). D

Now let us consider the subspace Hp of H which consists of functions f vanishing at p ∈ D. The space is clearly of at most codimension 1. Otherwise, there are f1 and f2 in the compliment of Hp which are linearly independent. This implies that both f1 (p) 6= 0 and f2 (p) 6= 0. However, clearly the linearly combination f1 (p)f2 − f2 (p)f1

GEOMETRIC ANALYSIS

53

is a function in Hp , which is a contradiction. This implies that by a change of orthonormal bases, there exist f0 in the orthogonal compliment of Hp and has unit L2 -norm, such that F (p) = f02 (p). Hence, in particular, if we choose p ∈ D such that F achieves its maximum then Z dim H = F D

≤ V (D)F (p) = V (D)f02 (p) = V (D) sup f02 . D

This proves that lemma.  Theorem 7.2. (Li-Tam [LT6]) Let M be an m-dimensional complete noncompact Riemannian manifold without boundary. Suppose that the Ricci curvature of M is nonnegative on M \ Bp (1) for some unit geodesic ball centered at p ∈ M . Let us assume that the lower bound of the Ricci curvature on Bp (1) is given by Rij ≥ −(m − 1)R for some constant R ≥ 0. If we denote H0 (M ) to be the space of functions spanned by the set of harmonic functions f which has the property that when restricted to each unbounded component of M \ D is either bounded from above or from below for some compact subset D ⊂ M , then H0 (M ) is of finite dimension. Moreover, there exists a constant C(m, R) > 0 depending only on m and R, such that, dim H0 (M ) ≤ C(m, R). Proof. By the definition of H0 (M ), there exists R0 > 1 such that f=

k X

vi ,

i=1

where each vi is bounded from one side on each end of M \ Bp (R0 ). Let E be an end of M \ Bp (R0 ). If v is a harmonic function defined on M which is positive on E and if x is a point in E with r(p, x) ≥ 2R0 , then by applying Theorem 6.1 to the ball Bx ( r(p,x) 2 ) and using the curvature assumption, there is a constant C > 0 independent of v, such that (7.4)

|∇v|(x) ≤ Cr−1 (p, x)v(x).

Since all the vi0 s are bounded on one side on E, there are constants a1 , · · · , ak and i = ±1 such that the harmonic functions ui = ai + i vi are positive on E. Hence, by applying (7.4) to u1 , · · · , um , we can estimate the gradient of f by (7.5)

|∇f |(x) ≤

k X

|∇vi |(x)

i=1

=

k X

|∇ui |(x)

i=1

≤ Cr−1 (p, x)

k X i=1

ui (x).

54

PETER LI

Using the fact that |∇f | is a subharmonic function on M \ Bp (1), the maximum principle implies that for any given δ > 0, ! m X |∇f |(x) − sup |∇f | ≤ Cδ ui (x) + 1 ∂E

i=1

for all x ∈ E. Letting δ → 0, we conclude that sup |∇f | ≤ sup |∇f |. E

∂E

Since E is an arbitrary end of M \ Bp (R0 ), we have (7.6)

|∇f | ≤

sup M \Bp (R0 )

sup |∇f | ∂Bp (R0 )

In fact, sup

|∇f | ≤ sup |∇f |

M \Bp (1)

∂Bp (1)

after applying the maximum principle to the subharmonic function |∇f | on the set M \ Bp (1) and (7.6). In particular, this implies that sup |∇f | ≤ sup |∇f |.

(7.7)

M

Bp (1)

Let us now consider the codimension-1 subspace Hp0 (M ) of H0 (M ) defined by Hp0 (M ) = {f ∈ H0 (M )|f (p) = 0}. For any f ∈ Hp0 (M ), the fundamental theorem of calculus implies that sup f 2 ≤ 16 sup |∇f |2 . Bp (4)

Bp (4)

Together with (7.7), we have sup f 2 ≤ 16 sup |∇f |2 . Bp (4)

Bp (1)

Applying the gradient estimate (Theorem 6.1) to the function f + supBp (2) |f | yields sup f 2 ≤ 16 sup |∇f |2 Bp (4)

Bp (1)

!2 ≤ C(R + 1) sup

f + sup |f |

Bp (1)

Bp (2) 2

≤ C(R + 1) sup f . Bp (2)

However, this together with the mean value inequality (Theorem 7.1) when applied to the nonnegative subharmonic function |f | asserts that there exists constants C3 , C4 (n) > 0 such that (7.8)

√ Z Vp (4) sup f 2 ≤ C3 exp(C4 R) Bp (4)

Bp (4)

f 2.

GEOMETRIC ANALYSIS

55

On the other hand Lemma 7.2 implies that for any finite dimensional subspace H of Hp0 (M ), there exists a function f0 such that Z dim H f02 ≤ Vp (4) sup f02 . Bp (4)

Bp (4)

Hence applying (7.8) to f0 yields the estimate √ dim H ≤ C3 exp(C4 R). Since this estimate holds for any finite dimensional subspace H, this implies that √ dim Hp0 (M ) ≤ C3 exp(C4 R). Therefore, as to be proven.

√ dim H0 (M ) ≤ C3 exp(C4 R) + 1 

Let us remark that if M has nonnegative Ricci curvature, then (7.7) can be written as sup |∇f | ≤ |∇f |(p). M

However, since |∇f | is a subharmonic function on M , the maximum principle implies that |∇f | must be identically constant. If f is not a constant function, we can apply the Bochner formula to |∇f | again, and conclude that ∇f is a parallel vector field. This implies that M must split and that f is a linear growth harmonic function. In particular, f cannot be a positive harmonic function, hence recovering Yau’s theorem.

56

PETER LI

§8 Reilly’s Formula and Applications We will discuss a few applications of the integral version of Bochner’s formula derived by Reilly [R]. In particular, this formula is useful in the studying of embedded minimal surfaces and surfaces with constant mean curvature. Let us first point out some standard formulas about submanifolds in Rm+1 and Sm+1 . Lemma 8.1. Let {x1 , . . . , xm+1 } be rectangular coordinates of Rm+1 , and let us denote the position vector − → − → by X = (x1 , . . . , xm+1 ). If M is a submanifold of Rm+1 with the induced metric and if II and H denotes the second fundamental form and the mean curvature vector of M , then − → HessM X = − II and

− → ∆M X = − H ,

where HessM (X) and ∆M (X) are the Hessian of X and the Laplacian of X computed on M . Proof. Let us first assume that M m ⊂ N n is a submanifold of an arbitrary manifold N . Let us choose n adapted orthonormal frame with {ei }m i=1 tangential to M and {eν }ν=m+1 normal to M . Then the Hessian of M and the Hessian of N are related by (HessN f )ij = (ei ej − ∇ei ej )f

(8.1)

= (HessM f )ij − = (HessM f )ij +

n X

h∇ei ej , eν i fν

ν=m+1 n X

− → h II ij , νi fν ,

ν=m+1

for any 1 ≤ i, j ≤ m and for any function f defined on a neighborhood of M. For N = Rm+1 , we apply (8.1) to the position vector X = (x1 , . . . , xm+1 ) and observe that HessRm+1 X = 0, we conclude that − → (HessM X)ij = −h II ij , em+1 i Xm+1 − → = −h II ij , em+1 i em+1 − → = − II ij , and the lemma follows.



Corollary 8.1. A submanifold M of Rm+1 is minimal if and only if the coordinate functions are harmonic. In particular, there are no compact minimal submanifolds in Rm+1 other than points. Lemma 8.2. Let M be an m-dimensional submanifold of the standard unit sphere Sn , then M is minimal if and only if all the coordinate functions of Sn ⊂ Rn+1 are eigenfunctions of M satisfying ∆M X = −mX. Proof. By Lemma 8.1, and using the fact that the position vector X is also the unit normal vector on Sn , we have − → (HessSn X)ij = − II ij = −δij X.

GEOMETRIC ANALYSIS

57

Applying (8.1) to the pair M ⊂ Sn yields − → (HessSn X)αβ = (HessM X)αβ + ( II M )αβ for tangent vectors eα and eβ which are tangential to M . Taking the trace of both sides gives − → −mX = ∆M X + H M , which proves the lemma.  The following integral formula was proved by Reilly in his work of reproving Aleksandrov’s theorem. Theorem 8.1. (Reilly) Let D be a manifold of dimension m + 1 whose boundary is given by a smooth m-dimensional manifold M. Suppose f is function defined on D satisfying ∆f = g

on D

and f =u

on M,

then m m+1

Z D

g2 ≥

Z

H fν2 + 2

Z

m X

Z fν ∆M u +

M

M

Z Rij fi fj

hαβ uα uβ +

M α,β=1

D

where H and hαβ denote the mean curvature and the second fundamental form of M with respect to the outward unit normal ν, ∆M is the Laplacian on M, and Rij is the Ricci curvature of D. Moreover, equality holds if and only if g δij fij = m+1 on D. Proof. Let us consider the Bochner formula 1 2 ∆|∇f |2 = fij + fi fijj 2 2 = fij + fi (∆f )i + Rij fi fj 2 = fij + h∇f, ∇gi + Rij fi fj .

Using the inequality m+1 X

P 2 fij ≥

i,j=1

= we have

m+1 i=1

fii

2

m+1 g2 , m+1

1 g2 ∆|∇f |2 ≥ + h∇f, ∇gi + Rij fi fj . 2 m+1

Integrating this over D yields Z Z Z Z 1 1 (8.2) ∆|∇f |2 ≥ g2 + h∇f, ∇gi + Rij fi fj . 2 D m+1 D D D

58

PETER LI

On the other hand, after integration by parts, the second term on the right hand side becomes Z

Z

g2 +

h∇f, ∇gi = − D

Z gfν

D

M

where ν is the outward unit normal to M, hence (8.2) becomes 1 2

(8.3)

Z

∆|∇f |2 ≥ −

D

m m+1

Z

g2 +

Z

D

Z Rij fi fj .

gfν + M

D

If we pick orthonormal frame {e1 , . . . , em+1 } near the boundary of D such that {e1 , . . . , em } are tangential to M , and ν = em+1 is the outward unit normal vector, then divergence theorem implies that 1 2

Z

∆|∇f |2 =

D

Z

m+1 X

(ei f )(em+1 ei f ).

M i=1

Using the boundary data of f , and choosing ∇em+1 em+1 = 0 at a point, we conclude that (8.4)

m+1 X

m X

i=1

α=1

(ei f )(em+1 ei f ) = (em+1 f )(em+1 em+1 f ) + = (em+1 f ) ∆f −

m X

(eα f )(em+1 eα f )

! fαα

+

α=1

= fν (g − Hfν − ∆M u) +

m X

(eα f )(em+1 eα f )

α=1 m X

(eα f )(em+1 eα f )

α=1

where ∆M is the Laplacian on M and H is the mean curvature of M with respect to the unit normal ν. However, (8.5)

em+1 eα f = eα em+1 f + ∇em+1 eα f − ∇eα em+1 f = eα em+1 f +

m X

h∇em+1 eα , eβ ifβ −

β=1

m X

h∇eα em+1 , eβ ifβ ,

β=1

because h∇em+1 eα , em+1 i = −heα , ∇em+1 em+1 i =0 and 1 eα |em+1 |2 2 = 0.

h∇eα em+1 , em+1 i =

Using (8.4), (8.5), and the fact that h∇eα em+1 , eβ i = −hem+1 , ∇eα eβ i = hαβ ,

GEOMETRIC ANALYSIS

59

we can write 1 2

(8.6)

Z

∆|∇f |2

D

Z

Z gfν −

= M

M

m Z X

+

H fν2 −

Z M

(eα f )(eα em+1 f )

M α=1 m X

Z h∇em+1 eα , eβ ifα fβ −

M

α,β=1

m X

Z fν ∆M u +

hαβ uα uβ .

M α,β=1

On the other hand, Z

m X

Z h∇em+1 eα , eβ ifα fβ = −

M α,β=1

=−

m X

heα , ∇em+1 eβ ifα fβ

M α,β=1 Z X m

h∇em+1 eα , eβ ifα fβ

M α,β=1

implies that it must be identically 0. Also integrating by parts yields Z

m X

Z (eα f )(eα em+1 f ) = −

fν ∆M u. M

M α=1

Combining this with (8.3) and (8.6), we have Z − M

H fν2 − 2

Z

Z fν ∆ M u −

M

m X

hαβ uα uβ ≥ −

M α,β=1

m m+1

which was to be proved. Equality case is clear from the above argument.

Z

g2 +

D

Z Rij fi fj D



Theorem 8.2. (Aleksandrov-Reilly) Any compact embedded hypersurface of constant mean curvature in Rm+1 is a standard sphere. Proof. Let M m be a compact, embedded hypersurface in Rm+1 with constant mean curvature H. By compactness, we claim that H > 0. Indeed, if p ∈ M is the maximum point for the function |X|2 on M , then the identity ∆M |X|2 = 2h∆M X, Xi + 2|∇X|2 and the maximum principle implies that 0 ≥ h∆M X, Xi − → = −h H , Xi at the point p. On the other hand, it is clear that X must be a multiple of the unit normal vector ν at p, hence 0 ≥ −H |X| and H ≥ 0. On the other hand, Corollary 8.1 asserts that H 6= 0, confirming the claim that H > 0. After scaling, we may assume that H = m. The assumption that M is embedded implies that M must enclose a bounded domain D in Rm+1 . Let us now consider the solution f of the boundary value problem ∆f = −1

on D

60

PETER LI

and f =0

on M = ∂D.

Applying Theorem 8.1 to f, we have V (D) ≥ m+1

(8.7)

Z

fν2 .

M

Schwarz inequality now implies that Z

fν2

A(M ) M

Z ≥( fν )2 M Z = ( ∆f )2 D

= V 2 (D), where A(M ) is the area of M . Therefore, together with (8.7), we obtain the inequality A(M ) ≥ (m + 1)V (D).

(8.8)

On the other hand, Lemma 8.1 asserts that ∆M X = −m ν, hence Z hX, ∆Xi Z Z 2 =− |∇X| + hX, Xν i D M Z 1 hX, ∆M Xi = −(m + 1)V (D) − m M Z 1 = −(m + 1)V (D) + |∇M X|2 m M = −(m + 1)V (D) + A(M ),

0=

D

where we have used the fact that |∇X|2 = m + 1 and |∇M X|2 = m. Together with (8.8), the inequalities which were used to derive (8.8) are all equalities. In particular, (8.9)

fij = −

δij m+1

on D,

and fm+1 must be identically constant on M. Let us denote the maximum point of f in D by p. Since p is a critical point and (8.9) implies that it is an isolated maximum. On other than hand, by choosing p to be the origin of Rm+1 , the function |X|2 g=− 2(m + 1) has a maximum at p and its hessian is given by gij = −

δij . m+1

In particular, this implies that f − g must be a constant function on D. Since f = 0 on M , we conclude M must be the sphere given by the level set of |X|2 . In particular, M must be the standard sphere since the mean curvature of M is m. 

GEOMETRIC ANALYSIS

61

Theorem 8.3. (Choi-Wang [CW]) Let M m be a compact embedded oriented minimal hypersurface in a compact oriented Riemannian manifold N m+1 . Suppose that the Ricci curvature of N is bounded from below by Rij ≥ mR for some positive constant R. Then the first non-zero eigenvalue of M has a lower bound given by λ1 (M ) ≥

mR . 2

Proof. The assumption that N has positive Ricci curvature implies that its first homology group H 1 (N, R) is trivial. By an exact sequence argument, we conclude that M divides N into 2 connected components N1 and N2 with ∂N1 = M = ∂N2 . Let us denote D to be one of the component to be chosen later. If u is the first nonconstant eigenfunction on M , satisfying ∆M u = −λ1 (M )u, then let f be the solution of ∆f = 0,

on D,

with boundary condition f = u,

on M .

Applying Theorem 8.1, we have Z

Z

0 ≥ −2λ1 (M )

Z

ufν + M

hαβ uα uβ + mR M

|∇f |2 .

D

On the other hand Z 2

Z ufν = 2 f fν M M Z = ∆(f 2 ) D Z =2 |∇f |2 . D

Hence, we have Z (8.10)

(2λ1 (M ) − mR) D

|∇f |2 ≥

Z hαβ uα uβ . M

Let us observe that the right hand side is independent of the extended function f . If we choose a different component of N \ M to perform this computation, the second fundamental form will differ by a sign, hence we may choose a component, say N1 , so that Z hαβ uα uβ ≥ 0. M

Hence together with (8.10), we conclude that either λ1 (M ) ≥ mR 2 , or ∇f = 0 on N1 . However, the latter is impossible because f has boundary value u which is nonconstant. This proves the estimate. 

62

PETER LI

§9 Isoperimetric Inequalities and Sobolev Inequalities In this section, we will show that a class of isoperimetric inequalities which occur in geometry are in fact equivalent to a class of Sobolev type inequalities. The relationship between these inequalities were exploited in the study of eigenvalues of the Laplacian as early as the 1920’s by Faber [F] and Krahn [K]. The equivalence was first formally established by Federer-Fleming [FF] (also see [Bm]) in 1960. In 1970, Cheeger [C] observed that the same argument can apply to estimating the first eigenvalue of the Laplacian. We will first define the isoperimetric and Sobolev constants on a manifold. Let us assume that M is a compact Riemannian manifold with or without boundary ∂M . Definition 9.1. If ∂M 6= φ, we define the Dirichlet α-isoperimetric constant of M by IDα (M ) =

A(∂Ω)

inf

Ω⊂M ∂Ω∩∂M =∅

1

V (Ω) α

where the infimum is taken over all subdomains Ω ⊂ M with the properties that ∂Ω is a hypersurface not intersecting ∂M . Similarly, we define the Neumann α-isoperimetric constant of M . Definition 9.2. The Neumann α-isoperimetric constant of M is defined by INα (M ) =

inf

A(S)

∂Ω1 =S=∂Ω2 M =Ω1 ∪S∪Ω2

1

min{V (Ω1 ), V (Ω2 )} α

where infimum is taken over all hypersurfaces S dividing M into 2 parts denoted by Ω1 and Ω2 . Note that in this case, there is no assumption on whether M has boundary or not. Definition 9.3. If ∂M 6= ∅, we define the Dirichlet α-Sobolev constant of M by R |∇f | SDα (M ) = inf RM 1 f ∈H1,1 (M ) ( M |f |α ) α f| =0 ∂M

where infimum is taken over all functions f in the first Sobolev space with Dirichlet boundary condition. We also define the Neumann α-Sobolev constant of M. Definition 9.4. The Neumann α-Sobolev constant of M is defined by R |∇f | SNα (M ) = inf RM 1 f ∈H1,1 (M ) (inf k∈R |f − k|α ) α M where the first infimum is taken over all functions f in the first Sobolev space, and the second infimum is taken over all real numbers k. Again, there is no assumption on whether M has boundary or not. Theorem 9.1. For any α > 0, we have IDα (M ) = SDα (M ). Proof. To see that IDα (M ) ≤ SDα (M ), it suffices to show that for any Lipschitz function f defined on M with boundary condition f |∂M ≡ 0 must satisfy Z  α1 Z α . |∇f | ≥ IDα (M ) |f | M

M

Without loss of generality, we may assume that f ≥ 0. Let us define Mt = {x ∈ M |f (x) > t} to be the sublevel set of f . By the co-area formula, Z Z ∞ (9.1) |∇f | = A(∂Mt )dt M 0 Z ∞ 1 ≥ IDα (M ) V (Mt ) α dt. 0

GEOMETRIC ANALYSIS

63

We now claim that for any s ≥ 0, we have the inequality α Z s Z s 1 tα−1 V (Mt )dt. V (Mt ) α dt ≥α 0

0

This is obvious for the case s = 0. Differentiating both sides as functions of s, we obtain Z s Z s α α−1 1 1 1 d V (Mt ) α dt (9.2) V (Mt ) α dt =α V (Ms ) α ds 0 0 and  Z s  d α−1 (9.3) α t V (Mt )dt = αsα−1 V (Ms ). ds 0 Rs 1 1 Observing that 0 V (Mt ) α dt ≥ sV (Ms ) α , because Ms ⊂ Mt for t ≤ s, we conclude that (9.2) is greater than or equal to (9.3). Integrating from 0 to s yields the inequality as claimed. Applying this inequality to (9.1) yields  Z ∞  α1 Z α−1 |∇f | ≥ IDα (M ) α V (Mt )dt . t M

0

However, the co-area formula implies that Z Z Z ∞ Z ∞ dAs d(tα ) ∞ dsdt α tα−1 V (Mt )dt = dt |∇f | t ∂Ms 0 0 Z ∞ Z dAt = tα dt |∇f | 0 ∂Mt Z = f α. M

This proves IDα (M ) ≤ SDα (M ). We will now prove that IDα (M ) ≥ SDα (M ). Let Ω be a subdomain of M with smooth boundary ∂Ω such that ∂Ω ∩ ∂M = φ. We denote the -neighborhood of ∂Ω in Ω by N = {x ∈ Ω|d(x, ∂Ω) < }. Note that for  > 0 sufficiently small, the distance function d(·, ∂Ω) to ∂Ω is a smooth function on N . Define  0, on M \ Ω    1 f (x) = d(x, ∂Ω), on N    1, on Ω \ N , Clearly f is a Lipschitz function defined on M with Dirichlet boundary condition. Moreover, Z Z  1 |∇f | = A(∂Nt \ ∂Ω)dt. M 0  On the other hand, we have Z  α1 Z α |∇f | ≥ SDα (M ) |f | M

M 1

≥ SDα (M )V (Ω \ N ) α . Hence

1 

Z



1

A(∂Nt \ ∂Ω)dt ≥ SDα (M )V (Ω \ N ) α , 0

and letting  → 0 yields 1

A(∂Ω) ≥ SDα (M )V (Ω) α . Since Ω is arbitrary, this proves IDα (M ) ≥ SDα (M ). 

64

PETER LI

Theorem 9.2. For any α > 0, we have 1

min{1, 21− α } INα (M ) ≤ SNα (M ) and

1

SNα (M ) ≤ max{1, 21− α } INα (M ). Proof. Let f be a Lipschitz function defined on M and k ∈ R be chosen such that M+ = {x ∈ M |f (x) − k > 0} and M− = {x ∈ M |f (x) − k < 0} 1 2V

satisfy the conditions that V (M+ ) ≤

(M ) and V (M− ) ≤ 12 V (M ). To show that 1

SNα (M ) ≥ min{1, 21− α } INα (M ), it suffices to show that Z

1

|∇u| ≥ min{1, 21− α } INα (M )

Z

M

|u|α

 α1

M

for u = f − k. Note that if Mt = {x ∈ M |u(x) > t} then for t > 0, we have V (Mt ) ≤ V (M+ ) ≤

1 V (M ). 2

This implies that min{V (Mt ), V (M \ Mt ))} = V (Mt ), 1 α

and A(∂Mt ) ≥ INα (M )V (Mt ) . Therefore by the same argument as in the proof of Theorem 9.1, we have Z

! α1

Z

α

|∇u| ≥ INα (M )

|u|

M+

.

M+

Similar, we also obtain Z

! α1

Z

|u|α

|∇u| ≥ INα (M ) M−

.

M−

Hence ! α1

 Z

Z |∇u| ≥ INα (M ) 

M

|u|α

≥ min{1, 2

|u|α

+

M+ 1 1− α

! α1 

Z



M−

Z } INα (M )

α

|u|

 α1 .

M 1

This proves SNα (M ) ≥ min{1, 21− α } INα (M ). 1 To prove that max{1, 21− α }INα (M ) ≥ SNα (M ), we consider any hypersurface S dividing M into two components denoted by Ω1 and Ω2 . Let us assume that V (Ω2 ) ≤ V (Ω1 ). For  > 0 sufficiently small, let us define N = {x ∈ Ω2 | d(x, S) < }

GEOMETRIC ANALYSIS

and the function

65

1, on Ω1 1 f (x) = 1 − d(x, S), on N     0, on Ω2 − N ,    

Let 0 ≤ k ≤ 1 be chosen such that Z

|f − k |α = inf

k∈R

M

Z

|f − k|α .

M

By using a similar argument as in the proof of Theorem 9.1, we have Z Z (9.4) |∇f | = |∇f | M

N

Z |f − k |

≥ SNα (M )

α

 α1

M

Z ≥ SNα (M )

|f − k |α +

! α1

Z

|f − k |α

Ω2 \N

Ω1


1 ≥ SNα (M ) (1 − k )α V (Ω1 ) + kα V (Ω2 \ N ) α
1 1 ≥ SNα (M ) (1 − k )α + kα α V (Ω2 \ N ) α . We now observe that (1 − k)α + k α ≥ 21−α for all 0 ≤ k ≤ 1 and α > 1, also (1 − k)α + k α ≥ 1 for all 0 ≤ k ≤ 1 and α ≤ 1. Hence by taking  → 0, the left hand side of (9.4) tends to A(S) while the right 1−α 1 hand side of (9.4) is bounded from below by SNα (M ) min{1, 2 α }V (Ω2 ) α . This establishes the inequality 1 max{1, 21− α }INα (M ) ≥ SNα (M ).  Let us point out that when the dimension of M is m and α > geodesic balls of radius r behaves like αm−1 m V (r) ∼ r m and the area of the their boundary is asymptotic to

m m−1 ,

then by the fact that the volume of

A(r) ∼ αm−1 rm−1 , it is clear that both IDα (M ) = 0 = INα (M ). Hence it is only interesting to consider those α ≤

m m−1 .

Corollary 9.1. (Cheeger [C]) Let M be a compact Riemannian manifold. If ∂M 6= ∅, let us denote µ1 (M ) to be the first Dirichlet eigenvalue on M and λ1 (M ) to be its first nonzero Neumann eigenvalue for the Laplacian. When ∂M = ∅, we will denote the first nonzero eigenvalue of M by λ1 (M ) also. Then µ1 (M ) ≥

ID1 (M )2 4

λ1 (M ) ≥

IN1 (M )2 . 4

µ1 (M ) ≥

ID1 (M )2 , 4

and

Proof. By Theorem 9.1, to see that

66

PETER LI

it suffices to show that for any Lipschitz function f with Dirichlet boundary condition, it must satisfy Z Z SD1 (M )2 |∇f |2 ≥ f 2. 4 M M Applying the definition of SD1 (M ) to the function f 2 , we have Z Z (9.5) |∇f 2 | ≥ SD1 (M ) f 2. M

M

On the other hand, Z

|∇f 2 | = 2

M

Z |f ||∇f | M

Z ≤2

f

2

 12 Z |∇f |

2

 21 .

M

M

Hence, the desired inequality follows from this and (9.5). For the Neumann eigenvalue, we simply observe that if u is the first eigenfunction satisfying ∆u = −λ1 (M )u, then u must change sign. If we denote M+ = {x ∈ M |u(x) > 0} and M− = {x ∈ M |u(x) < 0}, then µ1 (M+ ) = λ1 (M ) = µ1 (M− ). Let us assume that V (M+ ) ≤ V (M− ). In particular, this implies that ID1 (M+ ) ≥ IN1 (M ). Hence by our previous argument, λ1 (M ) = µ1 (M+ ) ID1 (M+ )2 4 IN1 (M )2 ≥ , 4 ≥

proving the corollary.



Corollary 9.2. Let M be a compact Riemannian manifold with boundary. For any function f ∈ H 1,2 (M ) and f |∂M ≡ 0, we have  2 Z  2−α Z α 2α 2−α 2 2−α |∇f | ≥ IDα (M ) |f | . 2 M M 2

Proof. By applying Theorem 9.1 and the definition of SDα (M ) to the function |f | 2−α , we obtain Z |∇|f | M

2 2−α

Z | ≥ IDα (M )

|f |

2α 2−α

 α1 .

M

On the other hand, Schwarz’s inequality implies that Z Z 2 α 2 |∇|f | 2−α | = |f | 2−α |∇f | 2 − α M M Z  12 Z  12 2α 2 2 |f | 2−α ≤ |∇f | . 2−α M M This proves the corollary.



GEOMETRIC ANALYSIS

67

Corollary 9.3. Let M be a complete Riemannian manifold with or without boundary. There exists constants C1 , C2 > 0 depending only on α, such that, Z

Z

|∇f |2 ≥ C1 SNα (M )2

|f |

M

2α 2−α

 2−α α

− 2−2α α

!

Z

|f |2

− C2 V (M )

M

M

for all f ∈ H 1,2 (M ). Proof. Let us first observe that a function g satisfies Z sgn(g) |g|α−1 = 0 M

if and only if Z

|g|α = inf

Z

k∈R

M

|g − k|α .

M

In particular, Z (9.6)

|g|

SNα (M )

α

 α1

Z ≤

|∇g|. M

M

For any given f ∈ H 1,2 (M ), let us choose k ∈ R such that Z 2α−2 (9.7) sgn(f − k) |f − k| 2−α = 0. M 2

Using g = sgn(f − k) |f − k| 2−α , (9.6) implies that 2 2−α

Z |f − k|

α 2−α

Z |∇f | ≥ SNα (M )

|f − k|

2α 2−α

 α1 .

M

M

Applying Schwarz’s inequality as in the proof of Corollary 9.3 yields Z



2

|∇f | ≥

(9.8) M

 ≥  ≥

2−α SNα (M ) 2

2 Z

2−α SNα (M ) 2

2 

2−α SNα (M ) 2

2

|f − k|

2α 2−α

 2−α α

M 1−α

Z |f |

2

2α 2−α

− V (M ) |k|

2α 2−α

 2−α α

M

2

(2−α)(1−α) α

Z |f |

2α 2−α

 2−α α

! − V (M )

2−α α

|k|2

.

M

By changing the sign of f if necessary, we may assume k ≥ 0. We can estimate k from above as follows. Let us define M+ = {x ∈ M | f (x) − k ≥ 0} and M− = {x ∈ M | f (x) − k < 0}. The condition (9.7) implies that Z (f − k) M+

2α−2 2−α

Z

2α−2

(k − f ) 2−α .

= M−

68

PETER LI

However, since Z (f − k)

2α−2 2−α

Z ≤ C3

M+

and

Z

2α−2

|f | 2−α − V (M+ ) k

2α−2 2−α

M+

2α−2

(k − f ) 2−α ≥ C4 V (M− ) k

2α−2 2−α

Z −

M−

2α−2

|f | 2−α , M−

for some constants C3 , C4 > 0 depending on α only, we conclude that Z Z 2α−2 2α−2 2α−2 C3 |f | 2−α − V (M+ ) k 2−α ≥ C4 V (M− ) k 2−α − M+

2α−2

|f | 2−α .

M−

This implies that Z

2α−2

|f | 2−α ≥ V (M ) k

C5

2α−2 2−α

M

for some constant C5 depending only on α. Applying H¨older inequality to the left hand side, we obtain Z C6 |f |2 ≥ V (M ) k 2 . M

Substituting into (9.8) yields the corollary.



GEOMETRIC ANALYSIS

69

§10 The Heat Equation In this section, we will discuss the existence and basic properties of the fundamental solution of the heat equation (heat kernel). The heat equation on M × (0, T ), for some constant 0 < T ≤ ∞, is given by 



∂ ∆− ∂t

f (x, t) = 0.

Typically, we consider solving it by prescribing an initial data f0 on M, i.e., lim f (x, t) = f0 (x).

t→0

The fundamental solution for the heat equation is a kernel function H(x, y, t) defined on M × M × (0, ∞) with the property that the function f (x, t) defined by Z f (x, t) =

H(x, y, t) f0 (y) dy M

solves the heat equation  (10.1)

∂ ∆− ∂t

 f (x, t) = 0

on

M × (0, ∞)

with initial condition (10.2)

lim f (x, t) = f0 (x).

t→0

When M is a compact manifold with boundary, the two natural boundary conditions are the Dirichlet and the Neumann boundary conditions. For the first case, the solution is required to satisfy (10.3)

f (x, t) = 0

∂M × (0, ∞)

on

in addition to (10.1) and (10.2). In the case of Neumann boundary condition, the solution should satisfy (10.4)

∂f (x, t) = 0 ∂ν

on

∂M × (0, ∞),

where ν is the outward normal to ∂M. No boundary condition is necessary when M is a compact manifold without boundary. Let us first consider Dirichlet boundary condition. Elliptic theory asserts that there is a set of eigenvalues {µ1 < µ2 ≤ · · · ≤ µi ≤ . . . } with corresponding eigenfunctions {φi } satisfying ∆φi = −µi φi , such that they form an orthonormal basis with respect to the L2 -norm. In particular, any function f0 ∈ L2 (M ) can be written in the form ∞ X f0 (x) = ai φi (x) i=1

with

Z ai =

f0 φi dx. M

70

PETER LI

Formally, the function given by f (x, t) =

∞ X

e−µi t ai φi (x)

i=1

will satisfy (10.1) with boundary condition (10.3) and initial condition (10.2). This is equivalent to saying that Z f (x, t) = H(x, y, t) f0 (y) dy, M

if we define the kernel by

∞ X

H(x, y, t) =

e−µi t φi (x) φi (y).

i=1

To justify the above discussion, we first prove a theorem on H(x, y, t). The key estimates follow from that in [L2] where they were derived for the more general setting of differential forms. It is also important to point out that H(x, y, t) is obviously symmetric in the x and y variables. Theorem 10.1. Let M be a compact manifold with boundary. The kernel function H(x, y, t) =

∞ X

e−µi t φi (x) φi (y)

i=1

is well defined on M × M × (0, ∞). It is the unique kernel, such that for any f0 ∈ L2 (M ), the function given by Z f (x, t) = H(x, y, t) f0 (y) dy M

solves the heat equation 



∂ ∆− ∂t

f (x, t) = 0

M × (0, ∞)

on

and f (x, t) = 0

∂M × (0, ∞),

on

with lim f (x, t) = f0 (x).

t→0

Moreover, H(x, y, t) is positive on (M \ ∂M ) × (M \ ∂M ) × (0, ∞) and it satisfies Z H(x, y, t) ≤ 1. M

Proof. We begin by estimating the supremum norm of an eigenfunction. Let us recall that Sobolev inequality asserts that, for m ≥ 3, there exists a constant CSD > 0 depending only on M , such that, Z

|∇f |2 ≥ CSD

(10.5)

Z

M

2m

 m−2 m

|f | m−2 M

for all f ∈ Hc1,2 (M ). In fact, Corollary 9.2 implied that CSD ≥ Sobolev inequality can be taken to be Z

2

Z

|∇f | ≥ CSD M

p

|f | M



m−2 2m−2

 p2

2 m (M ) ID m−1 . When m = 2, the

GEOMETRIC ANALYSIS

71

for all f ∈ Hc1,2 (M ) and for any fixed 2 ≤ p < ∞. For our purpose, we do not need the precise value of the Sobolev constant CSD , beyond the fact that it only depends on the manifold M. Let us assume that m ≥ 3 since the case when m = 2 can be done similarly. For any nonnegative function u ∈ Hc1,2 (M ) and for any constant k ≥ 2, integrating by parts and applying the Sobolev inequality gives Z Z (10.6) uk−1 ∆u = −(k − 1) uk−2 |∇u|2 M M Z k 4(k − 1) =− |∇(u 2 )|2 2 k M Z  m−2 m km 4(k − 1)CSD m−2 ≤− |u| k2 M Z  m−2 m km 2CSD m−2 ≤− |u| . k M In particular, if u = |φi |, then u satisfies ∆u ≥ −µi u. Hence (10.6) asserts that Z

|φi |k ≥

M

2CSD k µi

which can be rewritten as 

for all k ≥ 2, with β = have

m m−2

2CSD k µi

Z

 m−2 m

km

|φi | m−2

,

M

 k1 kφi kkβ ≤ kφi kk

and kφi kk denoting the Lk -norm of φi . Setting k = 2β j for j = 0, 1, 2, . . . , we  kφi k2β j+1 ≤

β j µi CSD



1 2β j

kφi k2β j .

Iterating this estimate and using kφi k2 = 1, we conclude that 1

kφi k2β j+1 ≤

 j  ` Y β µi 2β` `=0

CSD

.

Letting j → ∞ and applying the fact that lim –kφi kp = kφi k∞ ,

p→∞

we obtain m

(10.7)

kφi k∞ ≤ C1 µi4 ,

where C1 > 0 is a constant depending on CSD . We will now estimate µk using k. Let E be the k-dimensional vector space spanned by the first k th eigenfunctions {φ1 , φ2 , . . . , φk }. We will show that for any u ∈ E, the estimate m−1

(10.8)

kuk∞ ≤ C2 µk 2 V (M )

m−1 m

–kuk2

must hold for some constant C2 > 0 depending only on m and CSD . Combining with Lemma 7.2, this estimate will imply (10.9)

1

2

µk ≥ C3 k m−1 V (M )− m

72

PETER LI

for some constant C3 depending only on m and CSD . To prove (10.8), we observe that for a C ∞ function u, since |∇u|2 = |∇|u||2 , the identities ∆(u2 ) = 2u ∆u + 2|∇u|2 and ∆|u|2 = 2|u| ∆|u| + 2|∇|u||2 imply that u ∆u = |u| ∆|u|. Hence applying (10.6) to |u|, it can be written as Z |u|

(10.10)

k−2

M

2CSD u ∆u ≤ − k

Z |u|

km m−2

 m−2 m .

M

We claim that for any u ∈ E and j ≥ 0, we have the estimate j Y

–kuk2β j+1 ≤

(10.11)

2

µk V (M ) m β ` C

`=0

1 2β ` −1

!

–kuk2 .

To see this, for each j ≥ 0, let us choose h ∈ E, such that, khk2β j+1 kuk2β j+1 = max . u∈E khk2 kuk2

(10.12)

Setting u = h and k = 2β j in (10.10), we have (10.13)

CSD βj

Z

|h|2β

j+1

 β1

Z

|h|2β

≤−

M

j

−2

h∆h

M

Z

2β j+1

≤



|∆h|

1 2β j+1

Z

M

2β j

 2βj −1 j

|h| M

To estimate the first term on the right hand side for h ∈ E, we write h=

k X

a` φ` ,

`=1

hence ∆h = −

k X

µ` a` φ` .

`=1

Observe that, for p ≥ 2 and 1 ≤ i ≤ k, the function Z F (µi ) =

|∆h|p

M

is convex as a function of µi because p k X µ` a` φ` M `=1 p−2 Z X k = p(p − 1) µ` a` φ` a2i φ2i M

∂2F = ∂µ2i

Z

∂2 ∂µ2i

`=1

≥ 0.

2β

β−1

V (M ) 2βj+1 .

GEOMETRIC ANALYSIS

73

This implies that for a fixed 1 ≤ i ≤ k, F (µi ) is bounded above by either F (0) or F (µk ). In particular, we conclude that there exists a subset {α} of {1, 2, . . . , k} such that Z

2β j+1 X . µk aα φα ≤ M α Z

2β j+1

|∆h|

(10.14) M

Since the function

P

α

aα φα ∈ E, the property of h asserts that 2β j+1 X ≤ aα φα M α

Z

X

a2α

!β j+1 Z

|h|2β

≤

2

|h| j+1

−β j+1

h

M

α

Z

Z

2β j+1

M

.

M

Combining with (10.13) and (10.14), we have C βj

Z

|h|2β

j+1

 β1

M

β−1

j

2β −1 V (M ) 2βj+1 , ≤ µk ||h||2β j+1 ||h||2β j

hence 2

µk V (M ) m β j C

–khk2β j+1 ≤

(10.15)

!

1 2β j −1

–khk2β j .

For j = 0, the extremal property of h implies (10.11) for that case. We now assume that (10.11) is valid for j − 1. To show that (10.11) is valid for j, we consider h satisfying (10.12). The induction hypothesis together with (10.15) yield !

2

–khk2β j+1 ≤

µk V (M ) m β j C

≤

µk V (M ) m β j C

–khk2β j !

2

=

j Y `=0

1 2β j −1

1 2β j −1

j−1 Y `=0

2

µk V (M ) m β ` C

!

2

µk V (M ) m β ` C

!

1 2β ` −1

–khk2

1 2β ` −1

–khk2 .

Again, the extremal property of h implies (10.11) for any j, asserting the claim. Letting j → ∞, we conclude the validity of inequality (10.8), hence also of (10.9). The convergence of the infinite series ∞ X

e−µi t φi (x) φi (y)

i=1

can now be seen by using (10.7) to get e−µi t φi (x) φi (y) ≤ e−µi t kφi k2∞ m

≤ C1 e−µi t µi2 ≤ C1 C(t) e−

µi t 2

,

74

PETER LI

where we have used the fact that m

e−x t x 2 ≤ C(t) e−

xt 2

for some constant C(t) > 0 depending only on t and for all 0 ≤ x < ∞. Applying (10.8), we conclude that ∞ X

e−µi t φi (x) φi (y) ≤ C1 C(t)

i=1

∞ X

e−C4 i

1 m−1

t

,

i=1

which clearly converges uniformly on M × M × [a, ∞) for any a > 0. Hence the kernel H(x, y, t) =

∞ X

e−µi t φi (x) φi (y)

i=1

is well defined and satisfies Dirichlet boundary condition in both x and y for t > 0. Moreover, if f0 (x) = P∞ 2 i=1 ai φi (x) in L (M ), then Z f (x, t) = =

H(x, y, t) f0 (y) dy M ∞ X

e−µi t ai φi (x),

i=1

which obviously satisfies the initial condition lim f (x, t) = f0 (x).

t→0

We also note that Z h∇φi , ∇φj i = δij µi M

implies that the finite sum 2 k Z X k X −µi t e−(µi +µj ) t φi (x) φj (x) h∇φi (y), ∇φj (y)i dy e φi (x) ∇φi (y) dy = M M

Z

i,j=1

i=1

=

k X

e2µi t µi φi (x) φi (x)

i=1

Pk converges as k → ∞ according to our previous argument. Hence the truncated sum i=1 e−µi t φi (x) φi (y) converges to H(x, y, t) weak in H 1,2 (M ). In particular, H(x, y, t) is a weakly solution to the heat equation since each truncated sum solves the heat equation. Regularity theory implies that H(x, y, t) is a smooth solution and also f (x, t) solves the Dirichlet heat equation R (10.1) and (10.3), with initial condition f0 . The strong maximum principle asserts that f (x, t) = M H(x, y, t) f0 (y) dy is positive on (M \∂M )×(0, ∞) whenever f0 ≥ 0 on M. This implies that H(x, y, t) is positive on (M \ ∂M ) × (M \ ∂M ) × (0, ∞). If f¯(x, t) is another solution to the heat equation with initial condition f0 , then the maximum principle implies that solution of the heat equation f (x, t) − f¯(x, t) must be identically R 0 since it vanishes on (M × {0}) ∪ (∂M × (0, ∞)). In particular, H(x, y, t) is unique. The property that M H(x, y, t) dy ≤ 1 follows from the fact that it solves the heat equation with initial condition f0 = 1 and the maximum principle. The case of m = 2 can be proved in a similar manner using the 2-dimensional Sobolev inequality. 

GEOMETRIC ANALYSIS

75

Note that the same argument can be used to prove the existence of the heat kernel for compact manifolds without boundary, or for compact manifolds with Neumann boundary condition. In either of these cases, the heat kernel is given by ∞ X K(x, y, t) = e−λi t ψi (x) ψi (y) i=0

where ψi is the (Neumann) eigenfunction with eigenvalue λi satisfying ∆ψi = −λi ψi . 1

Since λ0 = 0 and ψ0 = V (M )− 2 , we only need to derive estimates on ψi and λi for i ≥ 1. To see this, we m replace the Sobolev inequality by that from Corollary 9.3 setting α = m−1 when m ≥ 3, and α = 43 when m = 2. It takes the form  m−2 Z Z Z m 2m 2 m−2 − C2 |∇f | ≥ C1 |f | |f |2 . M

M

M

where C1 and C2 are constants depending only on M. Inequality (10.6) will now take the form Z −λi

2C1 u ≤− k M k

Z u

km m−2

 m−2 m

M

2C2 + k

Z

uk .

M

with u = |ψ|, hence  kψi k2β j+1 ≤ and

β j λi + C2 C1



1 2β j

kψi k2β j

 m  kψi k∞ ≤ C3 λi4 + C4 .

We can also estimate λi using a similar method and the existence of K(x, y, t) is established. The positivity of K(x, y, t) can be seen by applying the Hopf boundary point lemma asserting that if (x, t) ∈ ∂M × (0, ∞) is a maximum Rpoint for a solution to the heat equation, f (x, t), then ∂f ∂ν (x, t) > 0. In particular, if f0 ≥ 0, then f (x, t) = M K(x, y, t) f0 (y) dy must have it’s maximum occur on M × {0} since ∂f ∂ν = 0 on ∂M × (0, ∞), hence f (x, t) ≥ 0. Again R this implies that K(x, y, t) is positive on M × M × (0, ∞). Uniqueness of K also follows. In particular, M K(x, y, t) dy = 1 because the constant 1 also solves the Neumann heat equation with 1 as initial condition. We now have the following version of Theorem 10.1 for the Neumann heat kernel. Theorem 10.2. The kernel function K(x, y, t) =

∞ X

e−λi t ψi (x) ψi (y)

i=0

is well defined on M × M × (0, ∞). It is the unique kernel such that, for any f0 ∈ L2 (M ), the function given by Z f (x, t) = K(x, y, t) f0 (y) dy M

solves the heat equation  and

∂ ∆− ∂t

 f (x, t) = 0

∂f (x, t) = 0 ∂ν

on

on

M × (0, ∞)

∂M × (0, ∞),

76

PETER LI

with lim f (x, t) = f0 (x).

t→0

Moreover, K(x, y, t) is positive on M × M × (0, ∞) and it has the property that Z K(x, y, t) = 1. M

We will now establish the fact that the Dirichlet heat kernel has a minimal property among all heat kernels. Definition 10.1. Let M be a manifold with or without boundary. We say that H(x, y, t) is a heat kernel if it is positive, symmetric in the x and y variables, and satisfies the heat equation   ∂ H(x, y, t) = 0 ∆− ∂t with initial condition lim H(x, y, t) = δx (y),

t→0

where δx (y) denotes the point mass delta function at x. Note that if f (x, t) and g(x, t) are two solutions to the heat equation, then the maximum principle allows us to compare f (x, t) with g(x, t) by considering the difference f (x, t) − g(x, t). However, if K(x, y, t) and H(x, y, t) are heat kernels, it is difficult to compare their values at t = 0. The following proposition gives an effective way of comparing and it is referred to as the Duhamel principle. Proposition 10.1. Let M be a Riemannian manifold with boundary. If H(x, y, t) and K(x, y, t) are two heat kernels, then       Z tZ ∂ ∂ K(x, z, t) − H(x, z, t) = K(y, z, s) − H(x, y, t − s) K(x, y, s) dy ds H(x, y, t − s) ∂νy ∂νy 0 ∂M on M × M × (0, ∞), where

∂ ∂νy

denotes the outward normal derivative with respect to the y-variable.

Proof. Observe that since the two kernels H(x, y, t) and K(x, y, t) are delta functions at t = 0, it follows that Z t Z ∂ H(x, y, t − s) K(y, z, s) dy ds = K(x, z, t) − H(x, z, t). 0 ∂s M On the other hand, since they both satisfy the heat equation, we have, Z t Z ∂ H(x, y, t − s) K(y, z, s) dy ds 0 ∂s M  Z tZ  ∂ =− H(x, y, t − s) K(y, z, s) dy ds ∂(t − s) 0 M   Z tZ ∂ + H(x, y, t − s) K(y, z, s) dy ds ∂s 0 M Z tZ Z tZ =− ∆y H(x, y, t − s) K(y, z, s) dy ds + H(x, y, t − s) ∆y K(y, z, s) dy ds 0 M 0 M   Z tZ ∂ =− H(x, y, t − s) K(y, z, s) dy ds ∂νy 0 ∂M   Z tZ ∂ + H(x, y, t − s) K(y, z, s) dy ds. ∂νy 0 ∂M

GEOMETRIC ANALYSIS

77

Hence we conclude that Z tZ



 ∂ K(x, z, t) − H(x, z, t) = H(x, y, t − s) K(y, z, s) dy ds ∂νy 0 ∂M   Z tZ ∂ H(x, y, t − s) K(y, z, s) dy ds. − ∂νy 0 ∂M



Corollary 10.1. Let M be a manifold with boundary. The Dirichlet heat kernel H(x, y, t) is minimal among all heat kernels. In particular, the Neumann heat kernel K(x, y, t) dominates H(x, y, t) with K(x, y, t) > H(x, y, t) on M × M × (0, ∞). Proof. According to the previous proposition, the boundary condition of H(x, y, t) asserts that Z tZ K(x, z, t) − H(x, z, t) = − 0

∂M



 ∂ H(x, y, t − s) K(x, y, s) dy ds. ∂νy

Using the fact that K(x, y, s) > 0, H(x, y, t − s) > 0 with y ∈ M \ ∂M , and H(x, y, t − s) = 0 with y ∈ ∂M , we conclude that the right hand side is positive by the Hopf boundary lemma, this proves the inequality.  We also have the following monotonicity property for the Dirichlet heat kernel. Corollary 10.2. Let Ω1 and Ω2 be two compact subdomains of M with the property that Ω1 ⊂ Ω2 . Suppose H1 (x, y, t) and H2 (x, y, t) be their corresponding Dirichlet heat kernel. Then H1 (x, y, t) ≤ H2 (x, y, t) on Ω1 × Ω1 × (0, ∞).

78

PETER LI

§11 Properties and Estimates of the Heat Kernel Recall that on Euclidean space Rm , the heat kernel is given by the formula H(x, y, t) = (4πt)

−m 2

 2  r (x, y) exp − . 4t

Since all manifolds are locally Euclidean, this formula is in fact an approximation of any heat kernel as t → 0. Theorem 11.1. Let M be a complete Riemannian manifold (with or without boundary). Suppose H(x, y, t) is a heat kernel. Then for any x ∈ M \ ∂M , −m 2

H(x, y, t) ∼ (4πt)

r2 (x, y) exp − 4t 



as t → 0 and r(x, y) → 0. Also, for fixed x, y ∈ M \ ∂M with x 6= y, we have H(x, y, t) → 0 as t → 0. Proof. For any integer k > 0, let us define the function −m 2

G(x, y, t) = (4πt)

r2 (x, y) exp − 4t 

X k

ui (x, y) ti .

i=0

Taking the Laplacian with respect to the y-variable, we have  2 m r ∆G = (4πt)− 2 exp − 4t !  k  k k X rX 1 r2 X r ∆r i i i ui t − − + h∇r, ∇ui i t + (∆ui ) t , × − 2t 2t 4t2 i=0 t i=0 i=0 also  2 m ∂G r = (4πt)− 2 exp − ∂t 4t



m r2 − + 2 2t 4t

X k

i

ui t +

i=0

k X

! i−1

i ui t

.

i=1

Hence,    2 ∂ r −m 2 G = (4πt) exp − ∆− ∂t 4t !   k−1 k−1 k k−1 X X X r∆r (m − 1) X i i i i × − + ui+1 t − r h∇r, ∇ui+1 i t + (∆ui ) t − (i + 1) ui+1 t . 2 2 i=−1 i=−1 i=0 i=0 For a fixed x, let y be within the injectivity radius of x, we now choose ui (x, y) as functions of y to satisfy  (11.1)

r∆r m − 1 − 2 2

 u0 + r h∇r, ∇u0 i = 0,

and then inductively  (11.2) for 0 ≤ i ≤ k − 1.

r∆r m − 1 − 2 2

 ui+1 + r h∇r, ∇ui+1 i + (i + 1) ui+1 = ∆ui

GEOMETRIC ANALYSIS

79

To see that this can be done, we observe that ∆r(x, y) = J −1 (y) ∂J ∂r (y), where J(y) is the area element of the sphere of radius r(x, y) at the point y. Equation (11.1) can be written as ∂u0 1 ∂J (m − 1) + u0 − u0 = 0, ∂r 2J ∂r 2r and u0 = C r

m−1 2

1

J − 2 with the constant C chosen to satisfy

(11.3)

u0 (x, y) = 1,

is a solution of (11.1). Equation (11.2) can be written as ∂ui+1 1 ∂J m − 3 − 2i + ui+1 − ui+1 = r−1 ∆ui . ∂r 2J ∂r 2r which can be solved by setting ui+1 = r

m−3−2i 2

1

J−2

Z

r

s

2i+1−m 2

1

J 2 ∆ui ds.

0

In particular, we obtain  2   r ∂ −m 2 exp − ∆− G = (4πt) ∆uk tk . ∂t 4t Note that these computations are valid when y ∈ / ∂M is within injectivity radius of x. Let us choose φ(y) to be a cutoff function satisfying ( φ(y) =

1

on

Bx (ρ)

0

on

M \ Bx (2ρ),

where 2ρ is chosen to be less the injectivity radius at x and the distance from x to ∂M. For a fixed x ∈ / ∂M , we now define F (x, y, t) = φ(y) G(x, y, t), which is supported on Bx (2ρ) and it is equal to G(x, y, t) on Bx (ρ). In particular, on Bx (2ρ) \ Bx (ρ), (11.4)

   2 k ∆ − ∂ F (x, y, t) = φ (4πt)− m2 exp − r ∆uk t + 2h∇φ, ∇Gi + G ∆ φ ∂t 4t  2 ρ −m 2 ≤ C2 t exp − . 4t

Note that since all manifolds are locally Euclidean and because of (11.3), limt→0 G(x, y, t) = δx (y). In particular, F (x, y, t) − H(x, y, t) Z t Z ∂ H(z, y, t − s) F (x, z, s) dz ds = 0 ∂s M   Z tZ Z tZ ∂ =− ∆z H(z, y, t − s) F (x, z, s) dz ds + H(z, y, t − s) F (x, z, s) dz ds ∂s 0 M 0 M   Z tZ ∂ H(z, y, t − s) ∆ − F (x, z, s) dz ds. =− ∂s 0 M

80

PETER LI

Hence using the fact that

R M

H(x, y, t) dy ≤ 1 and (11.4), we have t

Z

m

sk− 2

|H(x, y, t) − F (x, y, t)| ≤ C1

Z H(z, y, s) dz ds

0

Bx (ρ)

Z

t

+ C2 0

 2 Z m ρ H(z, y, s) dz ds s− 2 exp − 4s Bx (2ρ)\Bx (ρ)

m

≤ C3 tk+1− 2 . If y ∈ / Bx (ρ), since F (x, y, t) = φ(y)G(x, y, t) and G(x, y, t) → 0

as

t → 0,

H(x, y, t) → 0

as

t → 0.

we conclude that On the other hand, if y ∈ Bx (ρ) then  k  2 m r (x, y) X −m i ui (x, y) t ≤ C3 tk+1− 2 H(x, y, t) − (4πt) 2 exp − 4t i=0

and the theorem follows because of (11.3).  Another important property P∞ of the heat kernel is the semi-group property. In fact, this follows from the definition of H(x, y, t) = i=1 e−µi t φi (x) φi (y). In particular, using the fact that φi are orthonormal with respect to L2 , Z (11.5)

H(x, z, t) H(z, y, s) dz = M

=

X i,j ∞ X

e

−µi t −µj s

e

Z φi (x) φj (y)

φi (z) φj (z) dz M

e−µi (t+s) φi (x) φi (y)

i=1

= H(x, y, t + s). In particular, Z (11.6)

H(x, x, 2t) =

H 2 (x, y, t) dy

M

and (11.7)

1

1

H(x, y, t + s) ≤ H 2 (x, x, 2t) H 2 (y, y, 2s).

This can be used to estimate the heat kernel from above. Theorem 11.2. Let M be a Riemannian manifold with boundary. The Dirichlet heat kernel is bounded from above by  − m2 m 2CSD H(x, y, t) ≤ t− 2 , m

GEOMETRIC ANALYSIS

81

for all x, y ∈ M and t > 0, where CSD > 0 is the Sobolev constant from (10.5). Moreover, the k-th Dirichlet eigenvalue must satisfy the estimate 

2e CSD µk ≥ m

 m2

k V (M )

,

and the k-th Dirichlet eigenfunction must satisfy 

kφk k2∞

≤e

2CSD m

− m2

m

µk2 ||φk ||22 .

Proof. Differentiating (11.6), we have ∂ H(x, x, 2t) = 2 ∂t

(11.8)

Z H(x, y, t) ZM

=2

∂ H(x, y, t) dy ∂t

H(x, y, t) ∆H(x, y, t) dy M

Z

|∇H(x, y, t)|2 dy.

= −2 M

The Sobolev inequality (10.5) asserts that Z

Z

2

|∇H(x, y, t)| dy ≥ CSD

(11.9) M

|H(x, y, t)|

2m m−2

 m−2 m dy

M

where CSD > 0 is the Sobolev constant. On the other hand, H¨older inequality implies that Z

H 2 (x, y, t) dy ≤

M

Z

2m

 m−2 Z m+2

2m

 m−2 m+2

H m−2 (x, y, t) dy M

Z ≤

4  m+2

H(x, y, t) dy M

H m−2 (x, y, t) dy M

Combining with (11.8) and (11.9), we have m+2 ∂ H(x, x, 2t) ≤ −2CSD H m (x, x, 2t) ∂t

implying 2CSD ∂ −2 H m (x, x, s) ≥ . ∂s m Integrating from over  ≤ s ≤ t yields 2

2

H − m (x, x, s) − H − m (x, x, ) ≥ Since Theorem 11.1 asserts that

2CSD (s − ). m

m

H(x, x, ) ∼ (4π)− 2 , we conclude that

2

lim H m (x, x, ) = 0

→0

82

PETER LI

hence  H(x, x, s) ≤

− m2 2CSD . s m

Inequality (11.7) then completes the first part of the theorem. The second part follows from the fact that ∞ X

e−µi t =

Z H(x, x, t) dx, M

i=1

hence combining with the estimate from the first part gives ∞ X

e

−µi t

 ≤

i=1

2CSD m

− m2

m

t− 2 V (M ).

Setting t = µ−1 k , we have ke−1 ≤

k X

µ

e

− µi

k

i=1

 ≤

2CSD m

− m2

m

µk2 V (M ).

We also have e−µk t φ2k (x) ≤ H(x, x, t)  − m2 m 2CSD ≤ t− 2 , m where φk is the k-th eigenfunction with ||φk ||2 = 1. Again setting t = µ−1 k yields the estimate kφk k2∞ ≤ e



2CSD m

− m2

m

µk2 . 

Note that the estimate on µk is an improvement from that obtained in the proof of Theorem 10.1. We also have the corresponding theorem for the Neumann heat kernel. In this case, the Sobolev inequality required is of the form Z

Z

2

|∇f | ≥ CSN

(11.10)

|f |

M

2m m−2

 m−2 m

M

R for all f ∈ H 1,2 (M ) satisfying M f = 0. Observe that the Sobolev inequality given in Corollary 9.3 implies the above one since the L2 -norm of f can be estimated by Z

2

Z

f ≤

λ1 (M ) M

M

|∇f |2 .

GEOMETRIC ANALYSIS

83

Theorem 11.3. Let M be a Riemannian manifold with boundary. The Neumann heat kernel is bounded from above and below by  − m2 m 1 2CSN t− 2 ≤ K(x, y, t) − V (M ) m  − m2 m 1 2CSN ≤ t− 2 , + V (M ) m for all x, y ∈ M and t > 0, where CSN > 0 is the Sobolev constant from (11.10). Moreover, the k-th Neumann eigenvalue must satisfy the estimate λk ≥

2

m−4 m



eCSN m

k V (M )

 m2 ,

and the k-th Neumann eigenfunction must satisfy 2

kψk k2∞ ≤ e

m−4 m

CSN m

!− m2

m

λk2 ||ψk ||22 .

Proof. We will estimate the function K1 (x, y, t) = K(x, y, t) − =

∞ X

1 V (M )

e−λi t ψi (x) ψi (y)

i=1

instead, since ψ0 = V

− 21

(M ). One checks readily that K1 also satisfies the semi-group property. Moreover, Z K1 (x, y, t) dy = 0 M

and Z

Z |K1 (x, y, t)| dy ≤

M

K(x, y, t) dy + 1 M

≤ 2. The same argument as in the proof of Theorem 11.2 now works on K1 and produces !− m2 m−4 2 m CSN K1 (x, x, s) ≤ s . m Hence together with the inequality 1

1

|K1 (x, y, t)| ≤ K12 (x, x, t) K12 (y, y, t) yields the desired estimate on K1 . The rest of the proof is identical to that of Theorem 11.2.  To summarize, we have proved that Sobolev inequalites of the form (10.6) and (11.10) imply upper bounds of the Dirichlet and the Neumann heat kernels, respectively. The converse is also true as we will demonstrate in this section. The following theorem was proved by Varopoulos in [V]. Let us first state the Marcinkiewicz interpolation theorem as Lemma 11.1 without proof. The interested reader shall refer to Appendix B of [St] for the detail.

84

PETER LI

Definition 11.1. Let T be a subadditive operator on Lp (M ), i.e., for all f, g ∈ Lp (M ), |T (f + g)(x)| ≤ |T (f )(x)| + |T (g)(x)| for all x ∈ M. T is of weak (p, q)-type if there exists a constant A, such that for f ∈ Lp (M ),  m{x | |T (f )(x)| > β} ≤

A kf kp β

q

for all β > 0. If q = ∞, this means kT (f )k∞ ≤ A kf kp for all f ∈ Lp (M ). Lemma 11.1. Suppose that 1 ≤ pi ≤ qi ≤ ∞, for i = 1, 2 be such that p1 < p2 and q1 6= q2 . Assume that T is an operator of weak (p1 , q1 )-type and weak (p2 , q2 )-type. Then for any 0 < θ < 1 with 1−θ θ 1 = + p p1 p2

1 1−θ θ = + , q q1 q2

and

T is a bounded operator from Lp (M ) to Lq (M ). Theorem 11.4. Let M be a Riemannian manifold with boundary. Suppose there is a constant A1 > 0, such that, the Dirichlet heat kernel satisfies the estimate m

H(x, y, t) ≤ A1 t− 2

for all x, y ∈ M and for all t ∈ (0, ∞), then there exists a constant C1 > 0 depending only on m such that Z

2

|∇f | ≥

Z

−2 C1 A1 m

|f |

M

2m m−2

 m−2 m

M

for all f ∈ H 1,2 (M ) with f |∂M = 0. If there is a constant A2 > 0, such that, the Neumann heat kernel satisfies the estimate K(x, y, t) ≤

m 1 + A2 t− 2 V (M )

for all x, y ∈ M and for all t ∈ (0, ∞), then there exists a constant C2 > 0 depending only on m, such that, Z

2 −m

|∇f |2 ≥ C2 A2

Z

M

for all f ∈ H 1,2 (M ) with

R M

2m

|f | m−2

 m−2 m .

M

f = 0.

Proof. We will first prove the case for the Dirichlet heat kernel. Let {φi } be the set of orthonormal eigenfunctions with eigenvalues {µi }. For all f ∈ H 1,2 (M ) with f |∂M = 0, we write f=

∞ X

ai φi

i=1

and hence ∆f = −

∞ X i=1

µi ai φi .

GEOMETRIC ANALYSIS

85

For any α ∈ R, we define the psuedo-differential operator (−∆)α by ∞ X

(−∆)α f =

µα i ai φi .

i=1

We can rewrite Z

|∇f |2 = −

Z f ∆f

M

M

=

∞ X

µi a2i

i=1

Z

1

|(−∆) 2 f |2 ,

= M

hence it suffices to show that Z

1 2

2

|(−∆) f | ≥

−2 C 1 A1 m

Z |f |

M

2m m−2

 m−2 m .

M 1

1

Let us denote g = (−∆) 2 f , hence f = (−∆)− 2 g. We need to show that the operator 2m

1

(−∆)− 2 : L2 (M ) → L m−2 (M ) is bounded and satisfied the bound Z |(−∆)

− 21

g|

2m m−2

 m−2 m

2

M

P∞

1

(−∆)

|g|2 .

M

Expressing (−∆) 2 g in terms of the heat kernel, if g = − 12

Z

≤ C A1m

g(x) =

∞ X

− 21

µi

i=1 bi

φi , then we have

bi φi (x)

i=1 ∞ X

Z =

− 12

µi

φi (x) φi (y) g(y) dy

M i=1 Z ∞Z

1

t− 2 e−µi t φi (x) φi (y) g(y) dy 0 M Z ∞ Z 1 =B t− 2 H(x, y, t) g(y) dy,

=B

0

where B −1 =

R∞ 0

M

1

s− 2 e−s ds. For 0 < T < ∞, let us define the functionals T1 and T2 by T

Z T1 (g) = B

t

− 21

Z H(x, y, t) g(y) dy dt

0

and

Z

∞

T2 (g) = B T

M

1

t− 2

Z H(x, y, t) g(y) dy dt, M

86

PETER LI 1

1

hence (−∆)− 2 = T1 + T2 . For β > 0, obviously the measure of the sublevel set {(−∆)− 2 g ≥ β} can be estimated by     β β − 21 m{x | |(−∆) g(x)| > β} ≤ m x | |T1 g(x)| > + m x | |T2 g(x)| > . 2 2 On the other hand, for

1 p

+

1 q

= 1, using the upper bound of H, we have ∞

Z

Z

1

t− 2

|T2 g(x)| ≤ B

H(x, y, t) |g(y)| dy dt

T

M ∞

Z ≤ B kgkq

t ∞

≤ B kgkq

t

− 21

sup H

q−m 2q

≤ C A1 T

p−1 p

 p1 H(x, y, t) dt dt

Z (x, y, t)

y∈M

T 1 q

p

M

T

Z

 p1 H (x, y, t) dt dt

Z

− 21

M

kgkq .

For 1 < q < m, by choosing T such that 1

C A1q T

(11.11)

q−m 2q

β , 2

kgkq =

we have   1 β m{x | |(−∆)− 2 g(x)| > β} ≤ m x | |T1 g(x)| > 2  −q β kT1 (g)kqq . ≤ 2

(11.12)

To estimate kT1 gkq , we consider q

|T1 g(x)| ≤ B

T

Z

q

− 21

H(x, y, t) |g(y)| dy dt

t 0

≤B

!q

Z

T

Z

q

M − 21

t

! pq Z dt

0

≤CT

q 2p

R M

Z

− 21

Z

T

− 12

Z

q

Z H(x, y, t) |g(y)| dy

t

M

0

H(x, y, t) |g(y)|q dy dt

t

M

0

where we have used the fact that

T

H(x, y, t) dy ≤ 1. This implies that q

|T1 g(x)|q dx ≤ C T 2p

M

Z

T

1

t− 2 kgkqq dt

0 q 2

= C T kgkqq . Combining with (11.11) and (11.12), we have − 12

m{x | |(−∆)

 −q q β g(x)| > β} ≤ C T 2 kgkqq 2 q

mq

mq

≤ C A1m−q β − m−q kgkqm−q .

dt

GEOMETRIC ANALYSIS 1

This established that the operator (−∆)− 2 is of weak 1 < q1 < 2 < q2 < m and 0 < θ < 1, such that,



87

 mq -type, for 1 < q < m. We now choose q, m−q

1 1−θ θ = + 2 q1 q2 implying m−2 (1 − θ)(m − q1 ) θ(m − q2 ) = + , 2m mq1 mq2 and the theorem follows from Lemma 11.1. − 12 on the subspace The same argument R also works for Neumann heat kernel by defining the operator (−∆) 2 of S = {g ∈ L (M ) | M g = 0} by 1

(−∆)− 2 g(x) =

∞ X

− 21

λ1

Z ψi (y) g(y) dy.

ψi (x) M

i=1

where {ψi } are an orthonormal basis of eigenfunctions corresponding to the non-zero eigenvalues {λi }. In this case, we also have Z ∞Z 1 1 (−∆)− 2 g(x) = B t− 2 K1 (x, y, t) g(y) dy dt 0

M

Using the estimate on K1 , the theorem follows for the Neumann boundary condition.



88

PETER LI

§12 Gradient Estimate and Harnack Inequality for the Heat Equation In this section, we will derive estimates for positive solutions of the heat equation. The gradient estimate given by Theorem 12.1, and the Harnack inequality that follows (Corollary 12.1), have fundamental importance in the study of parabolic equations. There are many parallel developments of similar type estimates for other nonlinear partial differential equations. The estimates presented herein were proved by Li-Yau in [LY2]. Lemma 12.1. Let M m be a manifold whose Ricci curvature is bounded by Rij ≥ −(m − 1)R for some constant R. Suppose g(x, t) is a smooth function defined on M × [0, ∞) satisfying the differential equation   ∂ g(x, t) = −|∇g|2 (x, t). ∆− ∂t Then for any given α ≥ 1, the function G(x, t) = t(|∇g|2 (x, t) − αgt (x, t)) satisfies the inequality   ∂ G ≥ −2h∇g, ∇Gi − t−1 G + 2α−2 m−1 t−1 G2 + 4(α − 1)m−1 α−2 |∇g|2 G ∆− ∂t − 2−1 m(m − 1)2 α2 (α − 1)−2 tR2 . Proof. Let {e1 , . . . em } be a set of local orthonormal frame on M . Differentiating in the direction of ei , we have Gi = t(2gj gji − α gti ). Covariant differentiating again, we obtain (12.1)

2 ∆G = t(2gji + 2gj gjii − α gtii )   2 2 2 ≥t (∆g) + 2h∇g, ∇∆gi − 2(m − 1)R|∇g| − α(∆g)t , m

where we have used the inequality X i, j

2 gji

P ( i gii )2 ≥ m

and the commutation formula gj gjii = gj giij + Rij gi gj ≥ h∇g, ∇∆gi − (m − 1)R |∇g|2 . However, since ∆g = −|∇g|2 + gt = −α−1 t−1 G − α−1 (α − 1) |∇g|2

GEOMETRIC ANALYSIS

89

(12.1) becomes ∆G ≥

=

2tα−2 −1 (t G + (α − 1) |∇g|2 )2 − 2α−1 h∇g, ∇Gi − 2α−1 (α − 1)th∇g, ∇|∇g|2 i m − 2(m − 1)R t|∇g|2 + Gt − t−1 G + t(α − 1) (|∇g|2 )t 2tα−2 −2 2 (t G + 2(α − 1)t−1 |∇g|2 G + (α − 1)2 |∇g|4 ) − 2α−1 h∇g, ∇Gi m − 2α−1 (α − 1)h∇g, ∇Gi − 2(m − 1)R t|∇g|2 + Gt − t−1 G.

The lemma follows from applying the inequality (α − 1)2 α−2 |∇g|4 − m(m − 1)R |∇g|2 ≥ −4−1 m2 (m − 1)2 α2 (α − 1)−2 R2 .  Theorem 12.1. Let M m be a complete manifold with boundary. Assume that p ∈ M and ρ > 0 so that the geodesic ball Bp (4ρ) does not intersect the boundary of M. Suppose the Ricci curvature of M on Bp (4ρ) is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. If f (x, t) is a positive solution of the equation   ∂ f (x, t) = 0 ∆− ∂t on M × [0, T ], then for any 1 < α, the function g(x, t) = log f (x, t) must satisfy the estimate |∇g|2 − αgt ≤

m 2 −1 α t + C1 α2 (α − 1)−1 (ρ−2 + R) 2

on Bp (2ρ) × (0, T ], where C1 is a constant depending only on m. Proof. Let φ(r(x)) be a cutoff function of the distance to p defined on Bp (4ρ) with the properties ( φ(r(x)) =

1

on Bp (2ρ)

0

on M \ Bp (4ρ), 1

−C ρ−1 ≤ φ− 2 φ0 ≤ 0, and φ00 ≥ −C ρ−2 where φ0 denotes differentiation with respect to r. For G = t(|∇g|2 − αgt ) we consider the function φ G with support on Bp (4ρ) × (0, T ]. Let (x0 , t0 ) ∈ Bp (4ρ) × (0, T ] be the maximum point of φ G. We may assume that φ G is positive at (x0 , t0 ) otherwise the theorem is trivial. Note that at (x0 , t0 ) we have the properties that (12.2)

∇(φ G) = 0,

(12.3)

∂ (φ G) ≥ 0, ∂t

90

PETER LI

and ∆(φ G) ≤ 0.

(12.4)

Also using the comparison theorem and the property of φ we have √ ∆φ ≥ −C2 ρ−2 (ρ R + 1). Applying Lemma 12.1 and this estimate to the identity ∆(φ G) = (∆φ) G + 2h∇φ, ∇Gi + φ (∆G) we obtain √ ∆(φ G) ≥ −C2 ρ−2 (ρ R + 1) G + 2φ−1 h∇φ, ∇(φ G)i − 2φ−1 |∇φ|2 G + φ Gt − 2h∇g, ∇Gi − t−1 G + 2α−2 m−1 t−1 G2 + 4(α − 1)m−1 α−2 |∇g|2 G − 2−1 m(m − 1)2 α2 (α − 1)−2 tφ R2 . Evaluating at (x0 , t0 ) and using (12.2), (12.3), and (12.4) yield √ 0 ≥ −C2 ρ−2 (ρ R + 1) G − 2φ−1 |∇φ|2 G + 2Gh∇φ, ∇gi −2 −1 −1 − φt−1 m t0 φ G2 + 4(α − 1)m−1 α−2 |∇g|2 φ G 0 G + 2α

− 2−1 m(m − 1)2 α2 (α − 1)−2 t0 φ R2 . Multiplying through by φ t0 and using the estimate for |∇φ|, we conclude that √ 3 0 ≥ −C3 ρ−2 (ρ R + 1)t0 φ G − φ2 G − C4 t0 φ 2 ρ−1 G |∇g|

(12.5)

+ 2α−2 m−1 φ2 G2 + 4(α − 1)m−1 α−2 t0 |∇g|2 φ2 G − 2−1 m(m − 1)2 α2 (α − 1)−2 t20 φ2 R2 . Observe that there exists a constant C5 > 0 such that (12.6)

1

4m−1 (α − 1)α−2 φ |∇g|2 − C4 ρ−1 φ 2 |∇g| ≥ −C5 α2 (α − 1)−1 ρ−2 .

Hence combining with (12.5), we conclude that   √ 0 ≥ 2m−1 α−2 (φ G)2 − C6 ρ−2 t0 (ρ R + 1 + α2 (α − 1)−1 ) + 1 φ G − 2−1 m(m − 1)2 α2 (α − 1)−2 t20 R2 for some constants C6 > 0. This implies that at the maximum point (x0 , t0 ) φG ≤

  √ mα2 + C7 t0 (ρ−2 α2 (ρ R + 1 + α2 (α − 1)−1 ) + α2 (α − 1)−1 R . 2

In particular, since t0 ≤ T, when restricted on Bp (2ρ) × {T } we have |∇g|2 − α gt ≤

  √ mα2 + C7 (ρ−2 α2 (ρ R + 1 + α2 (α − 1)−1 ) + α2 (α − 1)−1 R . 2T

The theorem follows by letting t = T . 




GEOMETRIC ANALYSIS

91

Corollary 12.1. Under the same hypotheses of Theorem 12.1,  f (x, t1 ) ≤ f (y, t2 )

t2 t1

 mα 2

 exp

 α r2 (x, y) + C1 α (α − 1)−1 (ρ−2 + R) (t2 − t1 ) 4(t2 − t1 )

for any x, y ∈ Bp (ρ) and 0 < t1 < t2 ≤ T, where r(x, y) is the geodesic distance between x and y. Proof. Let γ : [0, 1] → M be a minimizing geodesic joining y to x. Since x, y ∈ Bp (ρ), the triangle inequality asserts that `(γ) = r(x, y) ≤ r(x) + r(y) ≤ 2ρ, where `(γ) denotes the length of γ and r(x) is the distance from p to x. We now claim that γ ⊂ Bp (2ρ). Indeed if this were not the case and if γ(s) ∈ / Bp (2ρ) for some s ∈ [0, 1], then the minimizing property of γ would imply that `(γ|[0,s] ) > ρ and `(γ|[s,1] ) > ρ violating the fact that `(γ) = r(x, y) ≤ 2ρ. We define η : [0, 1] → Bp (ρ) × [t1 , t2 ] by η(s) = (γ(s), (1 − s)t2 + st1 ). Clearly, η(0) = (y, t2 ) and η(1) = (x, t1 ). Integrating along η, we obtain Z (12.7)

1



d log f ds

log f (x, t1 ) − log f (y, t2 ) = 0

Z =

 ds

1

(hγ 0 , ∇(log f )i − (t2 − t1 )(log f )t ) ds.

0

On the other hand, Theorem 12.1 implies that −(log f )t ≤

m −1 αt + A − α−1 |∇(log f )|2 2

where A = C1 α (α − 1)−1 (ρ−2 + R). Hence (12.7) becomes  log

f (x, t1 ) f (y, t2 )



1

Z


|γ 0 | |∇(log f )| − (t2 − t1 ) α−1 |∇(log f )|2 ds

≤ 0

Z + (t2 − t1 ) 0

1



A+

m −1  αt ds 2

where t = (1 − s)t2 + st1 . Using the fact that −(t2 − t1 ) α−1 z 2 + |γ 0 | z ≤

α |γ 0 |2 4(t2 − t1 )

for any z, we conclude that  log

f (x, t1 ) f (y, t2 )



α r2 (x, y) mα ≤ + log 4(t2 − t1 ) 2



t2 t1

 + A(t2 − t1 )

where we have used r(x, y) = `(γ) = |γ 0 |. The corollary follows by exponentiating both sides .  We are now ready to prove the existence of a minimal heat kernel on any complete (not necessarily compact) manifold.

92

PETER LI

Theorem 12.2. Let M m be a complete, noncompact manifold. There exists a positive, symmetric heat kernel H(x, y, t) on M with the property that, for any f0 ∈ L2 (M ), the function given by Z f (x, t) = H(x, y, t) f0 (y) dy M

solves the heat equation  ∆−

∂ ∂t

 f (x, t) = 0

M × (0, ∞)

on

with initial condition lim f (x, t) = f0 (x).

t→0

Moreover, H(x, y, t) satisfies Z H(x, y, t) dy ≤ 1, M

¯ and it is the unique minimal heat kernel on M × M × (0, ∞), namely, if H(x, y, t) is another positive heat kernel defined on M then ¯ H(x, y, t) ≤ H(x, y, t) for all x, y ∈ M and t ∈ (0, ∞). Proof. Let {Ωi } be a compact exhaustion of M satisfying Ωi ⊂ Ωj for i ≤ j and ∪i Ωi = M. On each Ωi , let Hi (x, y, t) be the Dirichlet heat kernel given by Theorem 10.1. The maximum principle implies that Hi (x, y, t) ≤ Hj (x, y, t) for i ≤ j. For a fixed point p ∈ M and a fixed radius ρ > 0, there exists i0 sufficiently large such that Bp (4ρ) ⊂ Ωi for i ≥ i0 . Corollary 12.1 asserts that, there exists a constant C depending only on ρ, t, and the lower bound of the Ricci curvature on Bp (4ρ), such that, Hi (z, x, t) ≤ C Hi (z, y, 2t) R for all z ∈ Ωi and x, y ∈ Bp (ρ). Combining with the fact that Ωi Hi (z, y, 2t) dy ≤ 1, this yields Hi (z, x, t) ≤ C Vp−1 (ρ). Hence the monotonically increasing sequence {Hi (z, x, t)} converges uniformly on compact subsets of M × M × (0, ∞) to a kernel function H(z, x, t). Obviously, H(z, x, t) is positive and symmetric in the space variables, and since Z Z H(x, y, t) dy ≤ lim Hi (x, y, t) dy i→∞

Bp (ρ)

Bp (ρ)

≤ 1, we also obtain

Z H(x, y, t) dy ≤ 1. M

Moreover, the local gradient estimate of Theorem 12.1 asserts that |∇f |2 − α f ft ≤ C f 2

on

Bp (ρ)

for f (x, t) = Hi (z, x, t). In particular, if φ is a nonnegative cutoff function supported on Bp (ρ) then Z Z Z 2 2 2 φ |∇f | ≤ α φ f ft + C φ2 f 2 M M M Z Z ≤α φ2 f ∆f + C φ2 f 2 M M Z Z Z 2 2 ≤ −α φ |∇f | − 2α φ f h∇φ, ∇f i + C φ2 f 2 . M

M

M

GEOMETRIC ANALYSIS

Applying the Schwarz inequality Z −2

Z

2

φ f h∇φ, ∇f i ≤

93

Z

2

|∇φ|2 f 2 ,

φ |∇f | +

M

M

M

we obtain the estimate Z

2

Z

2

|∇f | ≤ C

2

Z

φ f +α

Bp ( ρ 2)

M

|∇φ|2 f 2

M

≤ C1 . by choosing φ = 1 on Bp ( ρ2 ). Hence the sequence Hi (z, x, t) converges in H 1,2 -norm over any compact subset of M. In particular, H(z, x, t) solves the heat equation. The semi-group property of Hi (z, x, t) asserts that Z Z Hi (z, y, t) Hi (x, y, s) dy ≤ Hi (z, y, t) Hi (x, y, s) dy Ωj

Ωi

= Hi (z, x, t + s), for i ≥ j. Hence letting i → ∞, we conclude that Z H(z, y, t) H(x, y, s) dy ≤ H(z, x, t + s). Ωj

In particular, by setting x = z and t = s, this implies that H(z, y, t) is in L2 with respect to the y-variable, hence also the z-variable because of symmetry. Letting j → ∞, we also conclude that Z H(z, y, t) H(x, y, s) dy ≤ H(z, x, t + s). M

On the other hand, since {Hi } is monotonically increasing, we have Z Z H(z, y, t) H(x, y, s) dy ≥ Hi (z, y, t) Hi (x, y, s) dy Ωi

Ωi

= Hi (z, x, t + s), hence

Z H(z, y, t) H(x, y, s) dy ≥ H(z, x, t + s). M

Therefore H also satisfies the semi-group property Z H(z, y, t) H(x, y, s) dy = H(z, x, t + s). M

Obviously, if f0 is a continuous function with compact support, then Z f (x, t) = H(x, y, t) f0 (y) dy M

will satisfy the heat equation with initial data f0 , since this is the case for the function Z fi (x, t) = Hi (x, y, t) f0 (y) dy M

94

PETER LI

when i is sufficiently large. For f0 just being continuous and in L2 (M ), we see that this is also the case by a approximation argument since H ∈ L2 (M ).  In the above proof, the estimates on Hi (x, y, t) are interior estimates and it does not depend on the boundary condition. In particular, the same estimates are valid for Neumann heat kernels Ki (x, y, t) on Ωi . Though the sequence {Ki (x, y, t)} are not necessarily monotonic, but the kernels are uniformly bounded on compact subsets, and one can extract a convergence subsequence replacing the originally sequence Ki (x, y, t) converging to a kernel K(x, y, t). However, Corollary 10.1 asserts that Hi (x, y, t) < Ki (x, y, t) and by taking the limit, we conclude that H(x, y, t) ≤ K(x, y, t).

(12.8)

An interesting R question is to ask if H(x, y, t) = K(x, y, t). Recall that show that M K(x, y, t) dy ≤ 1. Hence if

R M

H(x, y, t) dy ≤ 1, and one can also

Z H(x, y, t) dy = 1

(12.9) M

then combining with (12.8) imply that H(x, y, t) = K(x, y, t). We will address the validity of (12.9) in a later section. The estimates in Theorem 12.1 and Corollary 12.1 take much simpler forms when the function is defined globally on a complete manifold with nonnegative Ricci curvature. In fact, the estimates are sharp on Rm . Corollary 12.2. Let M m be a complete manifold with nonnegative Ricci curvature. If f (x, t) is a positive solution of the heat equation   ∂ f (x, t) = 0 ∆− ∂t on M × [0, ∞), then |∇f |2 ft m − ≤ f2 f 2t on M × [0, ∞). Moreover,  f (x, t1 ) ≤ f (y, t2 )

t2 t1

 m2

 exp

r2 (x, y) 4(t2 − t1 )



for all x, y ∈ M and 0 < t1 < t2 < ∞. Proof. Apply Theorem 12.1 by first taking ρ → ∞, then α → 1.  The gradient estimate is also valid for manifolds with boundary when the boundary is assumed to satisfy some convexity hypothesis. In the case of the Dirichlet boundary condition, one needs to assume that the boundary has nonnegative mean curvature with respect to the outward normal. For the Neumann boundary condition, one needs to assume that the boundary is convex in the sense that it has nonnegative second fundamental form with respect to the outward normal. We will state and prove the Neumann boundary condition case since this will be used in §14. Theorem 12.3. Let M be a complete manifold with convex boundary ∂M. Suppose the Ricci curvature of M is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. If f (x, t) is a positive solution of the equation   ∂ f (x, t) = 0 ∆− ∂t

GEOMETRIC ANALYSIS

95

on M × [0, T ] with Neumann boundary condition fν (x, t) = 0

on

∂M × (0, T ],

then for any 1 < α, the function g(x, t) = log f (x, t) satisfies the estimate |∇g|2 − αgt ≤

m 2 −1 α t + C1 α2 (α − 1)−1 R 2

on M × (0, T ], where C1 is a constant depending only on m. Moreover,     mα α `2 (γ) t2 2 exp + C1 α (α − 1)−1 R (t2 − t1 ) f (x, t1 ) ≤ f (y, t2 ) t1 4(t2 − t1 ) for any x, y ∈ M and 0 < t1 < t2 ≤ T, where `(γ) is the length of the shortest curve γ ⊂ M joining x to y. Proof. The argument for the gradient estimate is again by applying the maximum principle to the function G(x, t) = t(|∇g|2 (x, t) − α gt (x, t)). In this case, since M is compact, we will not need to use a cutoff function. In particular, if (x0 , t0 ) is the maximum point for G on M × [0, T0 ] with T0 ≤ T and if t0 = 0, then G(x, t) ≤ 0 for all (x, t) ∈ M × [0, T0 ]. Hence we may assumed that t0 > 0. If x0 ∈ ∂M , then the Hopf boundary point lemma asserts that (12.10)

Gν (x0 , t0 ) > 0

where ν is the outward normal to ∂M. However, since t(2giν gi − αgtν ) = t(2f −2 fiν fi − 2f −3 |∇f |2 fν − α f −1 ftν + α f −2 ft fν ), the boundary condition on f implies that Gν (x0 , t0 ) = 2t0 f −2 (x0 , t0 ) fiν (x0 , t0 ) fi (x0 , t0 ). On the other hand, the computation in the proof of Corollary 5.1 implies that 2fiν fi = −2hβγ fβ fγ ≤0 where hβγ is the second fundamental form of ∂M. This gives a contradiction to (12.10), hence x0 ∈ / ∂M. We can now apply the maximum principle to G at the point (x0 , t0 ). At this point, Lemma 12.1 implies that 0 ≥ −G + 2α−2 m−1 G2 − 2−1 m(m − 1)2 α2 (α − 1)−2 t20 R2 , hence we conclude that G(x, t) ≤ G(x0 , t0 ) m ≤ α2 + C1 α2 (α − 1)−1 t0 R. 2 In particular |∇g|2 (x, T0 ) − αgt (x, T0 ) ≤

m 2 −1 α T0 + C1 α2 (α − 1)−1 R. 2

and the first part of the theorem follows. The Harnack inequality follows from the same argument as in the proof of Corollary 12.1. 

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PETER LI

§13 Upper and Lower Bounds for the Heat Kernel In this section, we will derive estimates for positive solutions of the heat equation. In particular, upper and lower bounds for the fundamental solution of the heat equation will be established for manifolds with Ricci curvature bounded from below. Much of the following argument was developed by Li and Yau in [LY2]. However, part of the argument has since been simplified and generalized. In particular, the integral estimate of the heat kernel given in Lemma 13.2 is due to Davies [D]. Lemma 13.1. Let M be a compact domain with boundary. Suppose f (x, t) be a nonnegative subsolution for the heat equation on M × (0, T ) with Dirichlet boundary condition f (x, t) = 0

on

∂M × (0, T ).

Assume that g(x, t) is a function defined on M × (0, T ) satisfying the differential inequality |∇g|2 + gt ≤ 0, and µ1 (M ) > 0 denotes the first eigenvalue of the Dirichlet Laplacian on M . Then the function Z F (t) = exp(2µ1 (M ) t) exp(2g(x, t)) f 2 (x, t) dx M

is a non-increasing function of t ∈ (0, T ). Proof. Using the variational characterization of µ1 (M ), we have that Z Z µ1 (M ) exp(2g) f 2 ≤ |∇(exp(g) f )|2 M M Z = |∇ exp(g)|2 f 2 M Z Z +2 exp(g) f h∇ exp(g), ∇f i + exp(2g) |∇f |2 M M Z 2 2 = |∇ exp(g)| f M Z Z 1 2 h∇ exp(2g), ∇(f )i + exp(2g) |∇f |2 . + 2 M M Integration by parts yields Z Z 2 h∇ exp(2g), ∇(f )i = − exp(2g) ∆(f 2 ) M M Z Z = −2 exp(2g) f ∆f − 2 M

exp(2g) |∇f |2 ,

M

hence Z (13.1)

µ1 (M )

Z

exp(2g) f 2 ≤

M

|∇ exp(g)|2 f 2 −

ZM ≤

Z exp(2g) f ∆f M

exp(2g) f 2 |∇g|2 −

M

Z exp(2g) f ft . M

On the other hand, ∂ ∂t

Z M

exp(2g) f 2



Z

exp(2g) f 2 gt + 2

=2 M

Z

exp(2g) f ft Z exp(2g) f 2 |∇g|2 + 2 exp(2g) f ft . M

Z ≤ −2 M

M

GEOMETRIC ANALYSIS

97

Hence, combining with (13.1), we have Z

∂ 2µ1 (M ) exp(2g) f ≤ − ∂t M 2

Z exp(2g) f

2

 .

M

This implies that ∂ ∂t and the lemma follows.



Z

exp(2g) f 2

exp(2µ1 (M )t)

 ≤0

M



R Corollary 13.1. Let M be a complete Riemannian manifold and f (x, t) = M H(x, y, t) u(y) dy be a solution of the heat equation defined on M × [0, ∞). Suppose µ1 (M ) ≥ 0 denotes the greatest lower bound of the L2 spectrum of the Laplacian on M. Assume that g(x, t) is a function defined on M × (0, ∞) satisfying |∇g|2 + gt ≤ 0, then the function Z

exp(2g(x, t)) f 2 (x, t) dx

F (x, t) = exp(2µ1 (M ) t) M

is non-increasing in t ∈ (0, ∞). Proof. Let {Ωi } be a compact exhaustion of M such that Ωi ⊂ Ωi+1 and ∪i Ωi = M. R

Define fi (x, t) = Ωi Hi (x, y, t) u(y) dy where Hi (x, y, t) is the Dirichlet heat kernel defined on Ωi . According to Lemma 13.1, the function Z Fi (x, t) = exp(2µ1 (Ωi ) t) exp(2g(x, t)) fi2 (x, t) dx Ωi

is a non-increasing function in t. The corollary follows by letting i → ∞ and observing that µ1 (Ωi ) → µ1 (M ) and fi → f.  Lemma 13.2. (Davies) Let M be a complete manifold. Suppose B1 and B2 are bounded subsets of M . Then  2  Z Z 1 1 −d (B1 , B2 ) , H(x, y, t) dy dx ≤ exp(−µ1 (M ) t)V 2 (B1 ) V 2 (B2 ) exp 4t B1 B2 where V (Bi ) denotes the volume of the set Bi for i = 1, 2 and d(B1 , B2 ) denotes the distance between the sets B1 and B2 . If M is a manifold with boundary, and H(x, y, t) is the Dirichlet heat kernel on M , then the same estimate still holds providing that d(B1 , B2 ) is now interpreted as the distance between the two sets by curves in M. In particular, d(B1 , B2 ) is given by the infimum of the length of all curves contained in M joining B1 to B2 . Proof. For i = 1, 2, let us denote Z fi (x, t) =

H(x, y, t) dy Bi

and gi (x, t) =

d2 (x, Bi ) , 4(t + )

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PETER LI

where d(x, Bi ) is the distance between x and the set Bi . A direct computation shows that d∇d 2 2 |∇gi | = 2(t + ) 2 d (x, Bi ) ≤ 4(t + )2 and

∂gi −d2 (x, Bi ) . = ∂t 4(t + )2 Hence applying Corollary 13.1 to fi and gi , we conclude that Z Z exp(2gi (x, t)) fi2 (x, t) dx ≤ exp(−2µ1 (M ) t) M

exp(2gi (x, 0)) fi2 (x, 0) dx

M

However, since Z fi (x, 0) =

H(x, y, 0) dy Bi

= χ(Bi ) is the characteristic function of Bi , we conclude that Z exp(2gi (x, 0)) fi2 (x, 0) = V (Bi ), M

and Z (13.2)

exp(2gi (x, t)) fi2 (x, t) dx ≤ V (Bi ) exp(−2µ1 (M ) t)

M

for i = 1, 2. On the other hand, triangle inequality asserts that d2 (B1 , B2 ) ≤ (d(x, B1 ) + d(x, B2 ))2 ≤ 2d2 (x, B1 ) + 2d2 (x, B2 ), hence this implies that  exp

d2 (B1 , B2 ) 8(t + )

 ≤ exp(g1 + g2 ).

Combining with (13.2), we have  2  Z Z d (B1 , B2 ) (13.3) exp f1 f2 ≤ exp(g1 ) exp(g2 ) f1 f2 8(t + ) M M Z  21 Z  21 ≤ exp(2g1 ) f12 exp(2g2 ) f22 M 1

M 1

≤ V 2 (B1 ) V 2 (B2 ) exp(−2µ1 (M ) t). Using the semigroup property, the left hand side can be written as  2  Z d (B1 , B2 ) exp f1 f2 8(t + ) M  2 Z Z Z d (B1 , B2 ) = exp H(x, z, t) H(x, y, t) dz dy dx 8(t + ) M B1 B2  2 Z Z d (B1 , B2 ) = exp H(y, z, 2t) dz dy. 8(t + ) B1 B2 Lemma 13.2 follows by combing this with (13.3) and by letting  → 0.  We are now ready to give an upper bound on the fundamental solution of the heat equation.

GEOMETRIC ANALYSIS

99

Theorem 13.1. (Li-Yau) Let M m be a complete manifold and H(x, y, t) denotes the minimal symmetric heat kernel defined on M × M × (0, ∞) with the properties that   ∂ ∆y − H(x, y, t) = 0 ∂t and lim H(x, y, t) = δx (y).

t→0

For any p ∈ M ,  > 0, and ρ > 0, if the Ricci curvature of M on Bp (4ρ) is bounded from below by Rij ≥ −(m − 1)R then there exists constants C1 , C2 > 0 depending only on m and , such that, −1 √ −1 √ H(x, y, t) ≤ C1 exp(−µ1 (M ) t) Vx 2 ( t) Vy 2 ( t)   p r2 (x, y) −2 + C2 (ρ + R) t × exp − 4(1 + 2)t

for any x, y ∈ Bp (ρ) and t ≤

ρ2 4 .

In particular, if the Ricci curvature of M is bounded by Rij ≥ −(m − 1)R,

then −1 √ −1 √ H(x, y, t) ≤ C1 exp(−µ1 (M ) t) Vx 2 ( t) Vy 2 ( t)   √ r2 (x, y) × exp − + C2 R t 4(1 + 2)t

for all x, y ∈ M and t ∈ (0, ∞). Proof. For a fixed y ∈ Bp (ρ) and δ > 0, applying Corollary 12.1 to the positive solution f (x, t) = H(x, y, t) by taking t1 = t and t2 = (1 + δ)t, we have  2  mα α r (x, x0 ) 0 2 H(x, y, t) ≤ H(x , y, (1 + δ)t)(1 + δ) exp + Aδt , 4δt √ where A = C1 α(α − 1)−1 (ρ−2 + R). Integrating over x0 ∈ Bx ( t), this gives α  √ Z mα −1 0 0 2 (13.4) H(x, y, t) ≤ (1 + δ) exp + Aδt Vx ( t) √ H(x , y, (1 + δ)t) dx . 4δ Bx ( t) Applying Corollary 12.1 and the same argument to the positive solution Z 0 0 f (y, s) = √ H(x , y, s) dx Bx ( t)

by taking t1 = (1 + δ)t and t2 = (1 + 2δ)t, we obtain Z √ Bx ( t)

H(x0 , y, (1 + δ)t) dx0 ≤



 mα α  √ 1 + 2δ 2 exp + Aδt Vy−1 ( t) 1+δ 4δ Z Z 0 0 0 0 × √ √ H(x , y , (1 + 2δ)t) dx dy . By ( t)

Bx ( t)

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PETER LI

Hence combining with (13.4), we have α

 + 2Aδt 2δ Z √ Z √ −1 −1 × Vx ( t) Vy ( t) √

H(x, y, t) ≤ (1 + 2δ)

(13.5)

mα 2

exp

By ( t)

√ Bx ( t)

H(x0 , y 0 , (1 + 2δ)t) dx0 dy 0 .

On the other hand, Lemma 13.2 implies that Z Z 0 0 0 0 (13.6) √ √ H(x , y , (1 + 2δ)t) dx dy By ( t)

Bx ( t)

√ 1 √ ≤ exp(−µ1 (1 + 2δ)t) Vx ( t) Vy2 ( t) exp 1 2

Observing that √ √ d(Bx ( t), By ( t)) =

(

0

√ r(x, y) − 2 t



√ √  −d2 (Bx ( t), By ( t)) . 4(1 + 2δ)t

√ if r(x, y) ≤ 2 t √ if r(x, y) > 2 t,

hence √ √ −d2 (Bx ( t), By ( t)) =0 4(1 + 2δ)t ≤1−

r2 (x, y) 4(1 + 4δ)t

√ when r(x, y) ≤ 2 t, and √ √ √ −d2 (Bx ( t), By ( t)) −(r(x, y) − 2 t)2 ≤ 4(1 + 2δ)t 4(1 + 2δ)t 2 −r (x, y) 1 ≤ + 4(1 + 4δ)t 2δ otherwise. In any case, there exists a constant C3 > 0, such that, when combining with (13.5) and (13.6) we obtain   mα (α + 1) 2 (13.7) H(x, y, t) ≤ C3 (1 + 2δ) exp + (2δAt − µ1 (1 + 2δ)t 2δ   −r2 (x, y) − 12 √ − 12 √ . × Vx ( t) Vy ( t) exp 4(1 + 4δ)t) When At > 1, we choose 2δ =

√ AT

, and (13.7) can be further estimated by

  mα    2 √  (α + 1) 1+ √ exp + At exp(−µ1 t)  AT ! −r2 (x, y) − 12 √ − 21 √ × Vx ( t) Vy ( t) exp )t) 4(1 + √2 AT    √ mα (α + 1) ≤ C3 (1 + ) 2 exp + At exp(−µ1 t)    −r2 (x, y) − 12 √ − 21 √ . × Vx ( t) Vy ( t) exp 4(1 + 2)t)

H(x, y, t) ≤ C3

GEOMETRIC ANALYSIS

101

√ When At ≤ 1, we choose 2δ =  and conclude from (13.7) and At ≤ At that   √ mα (α + 1) 2 H(x, y, t) ≤ C3 (1 + ) exp +  At exp(−µ1 t)    −r2 (x, y) − 12 √ − 12 √ × Vx ( t) Vy ( t) exp . 4(1 + 2)t) In either case, the theorem follows by choosing α = 2.  Due to the fact that both Corollary 12.1 and Lemma 13.2 are valid on manifolds with boundary after suitable localization, a corresponding upper bound for the Dirichlet heat kernel can also be derived following the above proof. Corollary 13.2. Let M m be a manifold with boundary and H(x, y, t) be its Dirichlet heat kernel. Let us assume that Bp (4ρ) ∩ ∂M = ∅ and the Ricci curvature of M on Bp (4ρ) is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. For any  > 0, there exists constants C1 > 0 depending only on m and , and C2 depending only on m, such that, −1 √ −1 √ H(x, y, t) ≤ C1 exp(−µ1 (M ) t) Vx 2 ( t) Vy 2 ( t)   p r2 (x, y) × exp − + C2 (ρ−2 + R) t 4(1 + 2)t

for any x, y ∈ Bp (ρ) and t ≤

ρ2 4 .

We will now derive a lower bound for the heat kernel. The following comparison type theorems were first proved by Cheeger-Yau [CY] and adapted to the present form in [LY2]. Lemma 13.3. Let M m be a complete manifold with boundary. Suppose Bp (ρ) is the geodesic ball centered at some fixed point p ∈ M with radius 0 < ρ ≤ d(p, ∂M ), at most the distance from p to ∂M . Assume that the Ricci curvature of M is bounded from below by Rij ≥ (m − 1)K

on

Bp (ρ),

¯ is a simply connected space form with constant sectional curvature K and for some constant K. Suppose M ¯ of radius ρ. Let H(x, y, t) and H(¯ ¯ x, y¯, t) be the Bp¯(ρ) is a geodesic ball centered at some fixed point p¯ ∈ M Dirichlet heat kernels on M and Bp¯(ρ), respectively. Then for any nonnegative function f0 (r) defined on [0, ρ] satisfying the properties f00 (0) = 0, f0 (ρ) = 0, and f00 (r) ≤ 0, the inequality Z Z ¯ z , y¯, t) f0 (¯ H(z, y, t) f0 (r(z)) dz ≥ H(¯ r(¯ z ))d¯ z M

Bp¯(ρ)

is valid for all y ∈ Bp (ρ) with y¯ ∈ Bp¯(ρ) such that r(y) = r¯(¯ y ), where r(y) and r¯(¯ y ) denote the distance to ¯ , respectively. In particular, the inequalities p ∈ M and the distance to p¯ ∈ M Z Z ¯ z , y¯, t) d¯ H(z, y, t) dz ≥ H(¯ z Bp (ρ1 )

for all ρ1 ≤ ρ, and

Bp¯(ρ1 )

¯ p, y¯, t) H(p, y, t) ≥ H(¯

are valid for r(y) = r¯(¯ y ) and for t ∈ (0, ∞).

102

PETER LI

Proof. We will prove the inequality for ρ < d(p, ∂M ) and the general case follows by taking the limit as ρ → d(p, ∂M ). For ρ < d(p, ∂M ), let f0 (x) = f0 (r(x)) be a nonnegative function of the distance r to the center point p satisfying the stated properties. Let Z f (x, t) =

H(z, x, t) f0 (z) dz M

be the solution of the Dirichlet heat equation with f0 as initial data on M. We also consider the solution of the heat equation on Bp¯(ρ) given by f¯(¯ x, t) =

Z

¯ z, x H(¯ ¯, t) f0 (¯ r(¯ z )) d¯ z,

Bp¯(ρ)

¯ to the point p¯. By the uniqueness of the heat equation and where r¯(¯ z ) is the distance function on M the fact that f0 is rotationally symmetric, f¯ must also be rotationally symmetric. Hence, we can write f¯(¯ y , t) = f¯(¯ r(¯ y ), t). In particular, ∂ f¯ (0, t) = 0, f¯0 (0, t) = ∂ r¯ and also by the Dirichlet boundary condition, ∂ f¯ f¯0 (ρ, t) = (ρ, t) < 0. ∂ r¯

(13.8) We claim that

f¯0 (¯ r, t) ≤ 0

(13.9)

¯ can for all 0 ≤ r¯ ≤ ρ and for all t ≥ 0. Indeed, since f¯ is a function of r¯ and t alone, the heat equation on M be written as   ∂ ¯ ¯ (13.10) f (¯ r, t) 0= ∆− ∂t ¯ r + f¯00 − f¯t , = f¯0 ∆¯ with ¯r = ∆¯

   

√ √ (m − 1) K cot( Kr), −1

(m − 1)r ,    (m − 1)√−K coth(√−Kr),

for K > 0 for K = 0 for K < 0.

Differentiating this with respect to r¯ again yields ¯ r + u (∆¯ ¯ r)0 + u00 − ut 0 = u0 ∆¯ with u = f¯0 and

 √  −(m − 1)K csc2 ( Kr),   ¯ r)0 = (∆¯ −(m − 1)r−2 ,    (m − 1)K csch2 (√−Kr),

for K > 0 for K = 0 for K < 0.

¯ r)0 ≤ 0 for any choices of K. Hence, the boundary conditions u(0, t) = 0 and u(ρ, t) < 0, the Note that (∆¯ initial condition u(¯ r, 0) ≤ 0, together with the maximum principle imply that u(¯ r, t) ≤ 0.

GEOMETRIC ANALYSIS

103

We now consider the transplanted function define by f¯(y, t) = f¯(r(y), t). The Laplacian comparison theorem, (13.9), and (13.10) assert that   ∂ ¯ ∆− f (r(y), t) = f¯0 ∆r + f¯00 − f¯t ∂t ¯ r + f¯00 − f¯t ≥ f¯0 ∆¯ =0 in the sense of distribution on M, with initial condition f¯(y, 0) = f0 (y). The difference F = f − f¯ is therefore a supersolution of the heat equation of Bp (ρ) × (0, ∞) which vanishes on Bp (ρ) × {0}. Moreover, the Dirichlet boundary condition yields F (y, t) = f (y, t) ≥ 0

on

∂Bp (ρ) × (0, ∞).

The parabolic maximum principle implies that F ≥ 0, hence f (y, t) ≥ f¯(r(y), t). The second and the third inequalities follow by approximating the characteristic function of Bp (ρ1 ) and the point mass function at p by a sequence of smooth functions satisfying the properties required for f0 .  Lemma 13.4. Let M m be a complete manifold with boundary. Suppose there exists a point p ∈ M such that M is geodesically star-shape with respect to p, and ρ0 is the maximum distance from p to any point in ∂M. Assume that the Ricci curvature of M is bounded from below by Rij ≥ (m − 1)K ¯ is a simply connected space form with constant sectional curvature K and for some constant K. Suppose M ¯ of radius ρ0 . Let K(x, y, t) and K(¯ ¯ x, y¯, t) be Bp¯(ρ0 ) is a geodesic ball centered at some fixed point p¯ ∈ M the Neumann heat kernels on M and Bp¯(ρ0 ), respectively. Suppose ρ1 ≤ d(p, ∂M ), then for any nonnegative function f0 (r) defined on [0, ρ1 ] satisfying properties that f00 (0) = 0, f0 (ρ1 ) = 0, and f00 (r) ≤ 0, the inequality Z Z ¯ z , y¯, t) f0 (¯ K(¯ r(¯ z ))d¯ z K(z, y, t) f0 (r(z)) dz ≥ M

Bp¯(ρ0 )

is valid for all y ∈ M with y¯ ∈ Bp¯(ρ0 ) such that r(y) = r¯(¯ y ), where r(y) and r¯(¯ y ) denote the distance to ¯ , respectively. In particular, the inequalities p ∈ M and the distance to p¯ ∈ M Z Z ¯ z , y¯, t) d¯ K(z, y, t) dz ≥ K(¯ z Bp (ρ1 )

for all ρ1 ≤ d(p, ∂M ), and

Bp¯(ρ1 )

¯ p, y¯, t) K(p, y, t) ≥ K(¯

are valid for all r(y) = r¯(¯ y ) and for all t ∈ (0, ∞). Proof. The proof for the Neumann boundary condition follows along the same argument as that of the previous lemma. By solving the heat equation with Neumann boundary condition on Bp¯(ρ0 ), we consider the rotationally symmetric solution Z f¯(¯ y , t) = K(¯ z , y¯, t) f0 (¯ z ) d¯ z. Bp¯(ρ0 )

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PETER LI

The Neumann boundary condition asserts that f¯0 (ρ0 , t) = 0 instead of (13.8). As in the previous proof, the maximum principle implies that u ¯(¯ r, t) = f¯0 (¯ r, t) ≤ 0. Again we consider the difference F (y, t) = f (y, t) − f¯(r(y), t) with

on

M × (0, ∞),

Z f (y, t) =

K(z, y, t) f0 (z) dz. M

We now claim that the normal derivative ∂F ∂ν with respect to the outward normal must be nonnegative on ∂M. Clearly, by the boundary condition of F, it suffices to check that ∂ f¯ ≤ 0. ∂ν Indeed, using the fact that f¯ is rotationally symmetric, for any y ∈ ∂M, we have ∂ ∂ f¯ (y, t) = f¯0 (r(y), t) h , νi(y). ∂ν ∂r ∂ The assertion follows from the fact that f¯0 ≤ 0, and the fact that h ∂r , νi ≥ 0 on ∂M because M is geodesically star-shape with respect to p. Hence

∂F (y, t) ≥ 0 ∂ν

on

∂M × (0, ∞),

implying that the minimum of F cannot occur on ∂M ×(0, ∞), because of the Hopf boundary lemma. Hence, the minimum of F must achieved on M × {0}, which has value 0. This implies that f ≥ f¯ and the rest of the lemma follows.  Theorem 13.2. Let M m be a complete manifold with boundary. Let p ∈ M and ρ > 0 be such that Bp (4ρ) ∩ M = ∅. Assume that the Ricci curvature is bounded from below by Rij ≥ −(m − 1)R

on

Bp (4ρ),

¯ x, y¯, t) is the Dirichlet for some constant R ≥ 0. Suppose H(x, y, t) is Dirichlet heat kernel on Bp (4ρ) and H(¯ heat kernel on Bx¯ (3ρ) in the simply connected, constant −R curvature, space form. Then for  > 0, there exists constant C4 > 0 depending on m, such that, √ 1 √ 1 √
¯ x, y¯, (1 − )t) V¯x¯ ( t) Vx− 2 ( t) Vy− 2 ( t) exp −2−1 −  C4 (1 + Rt) H(x, y, t) ≥ (1 − )m H(¯ and

√ √
¯ x, y¯, (1 − )t) V¯x¯ ( t) Vx−1 ( t) exp −2−1 −  C4 (1 + Rt) , H(x, y, t) ≥ (1 − )m H(¯

for x, y ∈ Bp (ρ) and 0 < t ≤ ρ2 . ¯ z , y¯, t) is the Dirichlet Proof. Let x, y ∈ Bp (ρ), the comparison theorem of Lemma 13.3 asserts that if H(¯ ¯ heat kernel on Bx¯ (3ρ) ⊂ M , then Z Z ¯ (13.10) H(¯ z , y¯, (1 − δ)t) d¯ z≤ H(z, y, (1 − δ)t) dz Bx ¯ (ρ1 )

Bx (ρ1 )

GEOMETRIC ANALYSIS

105

for ρ1 ≤ 3ρ. However, Corollary 12.1 implies that H(z, y, (1 − δ)t) ≤ H(x, y, t) (1 − δ)−



αm 2

exp

 αρ21 + Aδt 4δt

for z ∈ Bx (ρ1 ), with A = C1 α(α − 1)−1 (ρ−2 + R). Integrating over Bx (ρ1 ), we conclude that  2  Z αm αρ1 (13.11) + Aδt . H(z, y, (1 − δ)t) dz ≤ H(x, y, t) Vx (ρ1 ) (1 − δ)− 2 exp 4δt Bx (ρ1 ) ¯ x, y¯, t), we have Applying the same argument to H(¯ ¯ x, y¯, (1 − 2δ)t) Vx¯ (ρ1 ) ≤ H(¯

Z

¯ z , y¯, (1 − δ)t) d¯ H(¯ z



Bx ¯ (ρ1 )

Combining with (13.10), (13.11), and setting ρ1 =

√

1−δ 1 − 2δ

 αm 2

 exp

 αρ21 + Aδt . 4δt

t, we obtain

  α √ √ ¯ x, y¯, (1 − 2δ)t) V¯x¯ ( t) Vx−1 ( t) (1 − 2δ) αm 2 H(x, y, t) ≥ H(¯ exp − − 2Aδt . 2δ Setting α = 2,  = 2δ, and using t ≤ ρ2 , we conclude that   √ √ ¯ x, y¯, (1 − )t) V¯x¯ ( t) V −1 ( t) exp − 2 −  C4 (1 + Rt) H(x, y, t) ≥ (1 − )m H(¯ x  for some constant C4 > 0 depending only on m. If y ∈ Bp (ρ) also, then we can symmetrize the estimate and yield   √ 1 √ 1 √ ¯ x, y¯, (1 − )t) V¯x¯ ( t) Vx− 2 ( t) Vy− 2 ( t) exp − 2 −  C4 (1 + Rt) .  H(x, y, t) ≥ (1 − )m H(¯  Corollary 13.3. Let M m be a complete, noncompact manifold without boundary. Suppose M has nonnegative Ricci curvature. For any  > 0, there exists a constant C5 > 0 depending on m and , such that, the minimal heat kernel H(x, y, t) satisfies the lower bounds   r2 (x, y) −1 √ −1 √ H(x, y, t) ≥ C5 Vx 2 ( t) Vy 2 ( t) exp − 4(1 − )t and H(x, y, t) ≥

C5 Vx−1 (

√

r2 (x, y) t) exp − 4(1 − )t 



for all x, y ∈ M and t ∈ (0, ∞). ¯ x, y¯, t) on M ¯ Proof. According to Theorem 13.2, by setting ρ = ∞, it suffices to estimate the heat kernel H(¯ from below. Note that if R = 0, then   √ r¯2 (¯ x, y¯) ¯ ¯ H(¯ x, y¯, (1 − )t) Vx¯ ( t) = C exp − 4(1 − )t and our estimate becomes H(x, y, t) ≥ since r¯(¯ x, y¯) = r(x, y). 

−1 √ −1 √ C4 Vx 2 ( t) Vy 2 ( t)

r2 (x, y) exp − 4(1 − )t 



106

PETER LI

´ Inequality and Parabolic Mean Value Inequality §14 Sobolev Inequality, Poincare In this section, we will use the heat kernel to estimate the Sobolev constant and the first non-zero Neumann eigenvalue for geodesic balls. These estimates are useful in the studies of various partial differential equations. In particular, these estimates will provide control over the constants involved in the DiGiorgi-Nash-Moser theory in terms of the lower bound of the Ricci curvature and the diameter of the underlying geodesic ball. The Sobolev inequality was proved by Saloff-Coste [SC] adopting Varopoulos’ argument (Theorem 11.4) [V] to this situation. We will also use the heat kernel estimate to prove a parabolic mean value inequality established by Li-Tam in [LT5]. A corollary to this parabolic version will be a generalization to the mean value inequality given in §7. We will first establish a lemma on the decay rate of the heat kernel. Lemma 14.1. Let M be a manifold with boundary. If µ1 (M ) is the first Dirichlet eigenvalue of M, then the Dirichlet heat kernel must satisfy H(x, x, t) ≤ H(x, x, t0 ) exp(−µ1 (M ) (t − t0 )) for all x ∈ M and t ≥ t0 > 0. If λ1 (M ) is the first non-zero Neumann eigenvalue of M , then the Neumann heat kernel must satisfy K(x, x, t) − V −1 (M ) ≤ (K(x, x, t0 ) − V −1 (M )) exp(−λ1 (M ) (t − t0 )) for all x ∈ M and t ≥ t0 > 0. Proof. The semi-group property of the heat kernel implies that ∂ ∂ H(x, x, 2t) = ∂t ∂t Z =2

Z

H 2 (x, y, t) dy

M

H(x, y, t) ∆H(x, y, t) dy

M

Z

|∇H|2 (x, y, t) dy Z ≤ −2µ1 (M ) H 2 (x, y, t) dy

= −2

M

M

= −2µ1 (M )H(x, x, 2t). Rewriting the inequality into the form ∂ (log H(x, x, t)) ≤ −µ1 (M ) ∂t and integrating from t0 to t yields that desired estimate. The estimate on the Neumann heat kernel follows exactly the same manner by considering K1 (x, x, t) = K(x, x, t) − V −1 (M ) that has the property Z K1 (x, y, t) dy = 0, M

and using the variational principle for λ1 (M ). 

GEOMETRIC ANALYSIS

107

Theorem 14.1. Let M m be a complete manifold and Bp (ρ) be a geodesic ball centered at p with radius ρ. Assume that M has Ricci curvature bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. Let H(x, y, t) be the Dirichlet heat kernel defined on Bp (ρ). Then there exists constants C15 , C16 > 0 depending only on m such that for all x ∈ Bp (ρ) and t > 0, we have √ m H(x, x, t) ≤ C15 t− 2 Vp−1 (ρ) ρm exp(C16 ρ R), where Vp (ρ) is the volume of Bp (ρ). Proof. Let HM (x, y, t) be the heat kernel on the complete manifold M. Recall that an estimate of Li-Yau implies that √ √ HM (x, x, t) ≤ C1 Vx−1 ( t) exp(C2 Rt) for some constant C1 , C2 > 0. By the monotonicity property of the heat kernel, we conclude that for x ∈ Bp (ρ) and for t ≤ ρ2 , √ √ √ √ H(x, x, t) ≤ C1 V¯ −1 ( t)(V¯ ( t)Vx−1 ( t)) exp(C2 ρ R) √ √ where V¯ ( t) is the volume of the ball of radius t in the simply connected space form of constant curvature −R. By the volume comparison theorem, since t ≤ ρ2 , we have √ √ V¯ ( t)Vx−1 ( t) ≤ V¯ (ρ)Vx−1 (ρ), Hence (14.1)

√ √ H(x, x, t) ≤ C1 V¯ −1 ( t)(V¯ (ρ)Vx−1 (ρ)) exp(C2 ρ R) √ m ≤ C1 t− 2 (V¯ (ρ)Vx−1 (ρ)) exp(C2 ρ R).

On the other hand, for t ≥ ρ2 we can use the estimate given by Lemma 14.1 and conclude that (14.2)

H(x, x, t) ≤ H(x, x, ρ2 ) exp(−µ1 (t − ρ2 )) √ ≤ C1 Vx−1 (ρ) exp(C2 ρ R − µ1 (t − ρ2 )).

Let us now consider the function

m

f (t) = t 2 exp(−µ1 (t − ρ2 )). It has a local maximum at the point t =  f

m 2µ1

m 2µ1



with maximum value 

=

Moreover, f (t) is a decreasing function for t >

(14.4)

 m2

m 2µ1 .

 m  exp − + µ1 ρ2 . 2

Hence if

m 2µ1

≤ ρ2 then

f (t) ≤ f (ρ2 ) = ρm

(14.3) for t ≥ ρ2 . On the other hand, if

m 2µ1

m 2µ1

≥ ρ2 then   m f (t) ≤ f 2µ1  m  m  m 2 exp − + µ1 ρ2 = 2µ1 2   m2 m ≤ . 2µ1

108

PETER LI

√ However, by the estimate of Li-Schoen (Theorem 5.4), µ1 ≥ C3 ρ−2 exp(−C4 ρ R). Applying this to (14.4) and together with (14.3) yields √ f (t) ≤ C5 ρm exp(C6 ρ R) for all t ≥ ρ2 . Therefore, substituting this into (14.2), we have √ m H(x, x, t) ≤ C7 t− 2 Vx−1 (ρ) ρm exp(C8 ρ R) for t ≥ ρ2 . √ On the other hand, using V¯ (ρ) ≤ C9 ρm exp(C10 ρ R) for some constants C9 , C10 > 0 depending only on m, the estimate (14.1) becomes √ m (14.5) H(x, x, t) ≤ C11 t− 2 Vx−1 (ρ) ρm exp(C12 ρ R) for all t ≤ ρ2 , hence is valid for all t > 0. The Bishop volume comparison theorem and the fact that x ∈ Bp (ρ) implies that V¯ (ρ) Vx (ρ) ≥ Vx (2ρ) ¯ V (2ρ) √ ≥ C13 Vp (ρ) exp(−C14 ρ R). The theorem follows by combining this with (14.5).  Combining with Theorem 11.4, we obtain the following Sobolev inequality on a geodesic ball. Theorem 14.2. Let M m be a complete manifold, of dimension m ≥ 3, with Ricci curvature bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. There exists constants C17 , C18 > 0 depending only on m such that for any function f ∈ Hc1,2 (Bp (ρ)) with compact support in some geodesic ball Bp (ρ), it must satisfy the Sobolev inequality ! m−2 Z Z m √ 2m 2 |∇f |2 ≥ C17 ρ−2 exp(−C18 ρ R) (Vp (ρ)) m |f | m−2 . Bp (ρ)

Bp (ρ)

Remark 1: The Sobolev inequality can be written in the form Z —

|∇f |2 ≥ C17 ρ−2

! m−2 m

√

Z exp(−C18 ρ R) —

Bp (ρ)

|f |

2m m−2

,

Bp (ρ)

√ R R where —Bp (ρ) g denotes Vp−1 (ρ) Bp (ρ) g. This has the advantage that the constant C17 ρ−2 exp(−C18 ρ R) is now independent on the volume of the ball. Remark 2: When M has nonnegative Ricci curvature, the Sobolev inequality now takes the form ! m−2 Z Z m 2m

2

|∇f |2 ≥ C17 ρ−2 (Vp (ρ)) m

|f | m−2

.

Bp (ρ)

Bp (ρ)

In particular, when M has maximal volume growth, i.e., Vp (ρ) ≥ C ρm , the Sobolev constant is uniformly bounded independent of ρ, as in the case of Rm . Conversely, one can show in §16 that if a complete manifold has uniformly bounded Sobolev constant of the form ! m−2 Z Z m 2m

|∇f |2 ≥ C

Bp (ρ)

|f | m−2 Bp (ρ)

then the manifold must have volume growth bounded by Vp (ρ) ≥ C ρm .

,

GEOMETRIC ANALYSIS

109

Theorem 14.3. Let M m be a compact manifold without boundary with dimension m ≥ 3. Suppose the Ricci curvature of M is bounded below by Rij ≥ −(m − 1)R for some constant R ≥ 0, and let us denote d(M ) to be diameter of M. Then these exists constants C1 , C2 > 0 depending only on m, such that, Z

−2

2

|∇f | + d M

Z (M )

√ f ≥ C1 exp(−C2 d(M ) R)V 2

2 m

Z |f |

(M )

2m m−2

 m−2 m

M

M

for all f ∈ H 1,2 (M ). Moreover, there are also constants C3 , C4 > 0 depending only on m, such that, the inequality Z  m−2 Z m √ 2m 2 2 m−2 m |∇f | ≥ C1 exp(−C2 d(M ) R)V (M ) |f | M

is also valid for f ∈ H

1,2

M

(M ) satisfying the extra condition

R M

f = 0.

Proof. Let {xi }ki=1 be a maximal set of points in M with the properties that r(xi , xj ) ≥ where d = d(M ). Clearly this implies that     d d (14.6) Bxi ∩ Bxj =∅ 8 8

d 4

for all i 6= j,

for i 6= j, hence k X

(14.7)

V xi

i=1

On the other hand, if x ∈ M \ ∪ki=1 Bxi set {xi }, hence we have (14.8)

d 4




  d ≤ V (M ). 8

, then r(x, xi ) ≥

M = ∪ki=1 Bxi

d 4,

contradicting the maximal property of the

  d 4

and (14.9)

k X

  d V (M ) ≤ V xi . 4 i=1

However, the volume comparison theorem asserts that         d d ¯ d ¯ −1 d V xi ≤ V xi V V , 4 8 4 8 where V¯ (r) denotes the volume of the geodesic ball of radius r in the simply connected space form with constant −R curvature. Combining with (14.7), we conclude that k X

  d 4 i=1     k   d ¯ −1 d X d ≤ V¯ V Vxi 4 8 i=1 8     d ¯ −1 d V V (M ) ≤ V¯ 4 8

kV (M ) ≤ k

V xi

110

PETER LI

implying the bound     d ¯ −1 d V 4 8 √ ≤ C exp(Cd R).

k ≤ V¯

To prove the Sobolev inequality, we may consider only nonnegative function f ∈ H 1,2 (M ). Let φ be a nonnegative cutoff function with the properties that    d  1 on Bxi  4   φ=  d  0 on M \ Bxi 2 and |∇φ|2 ≤ Cd−2 . Applying Theorem 14.2 to the function φ f yields √

Z

! m−2   Z m 2m d m−2 . |φ f | 2 Bxi ( d 2)

2 m xi

|∇(φ f )|2 ≥ C17 d−2 exp(−C18 d R)V

(14.10) Bx i ( d 2)

Using Z

−2

2

Z

|∇f | + 2Cd

2 M

Z

2

2

f ≥2

Z

2

|∇φ|2 f 2

φ |∇f | + 2 Bxi ( d 2)

M

Z

Bxi ( d 2)

|∇(φ f )|2

≥ Bxi ( d 2)

and

Z

Z

2m

2m

|φ f | m−2 ≥ Bxi (

d 2

)

|f | m−2 Bxi (

d 4

)

inequality (14.10) becomes Z

|∇f |2 + Cd−2

M

Z

f 2 ≥ C19 d−2

√ 2 exp(−C18 d R)Vxm i

M

! m−2   Z m 2m d |f | m−2 . 2 Bxi ( d 4)

Note that the volume comparison theorem allows us to estimate   √ d Vxi ≥ C exp(−Cd R) Vxi (d) 2 √ = C exp(−Cd R) V (M ). Summing over i = 1, . . . , k and using the inequality Z k X i=1

! m−2 m |f |

2m m−2

≥

Bxi ( d 4)

! m−2 m

k Z X i=1

Bxi ( d 4)

Z |f |

= M

|f |

2m m−2

2m m−2

 m−2 m

GEOMETRIC ANALYSIS

111

because of (14.8), we conclude that Z k

|∇f |2 + kCd−2

M

Z

√ f 2 ≥ C19 d−2 exp(−C20 d R) V

2 m

Z

2m

|f | m−2

(M )

 m−2 m .

M

M

The first part of the theorem follows by applying the upper bound for k. The second part follows from the Poincar´e inequality Z

Z

2

|∇f | ≥ λ1 (M ) M

for f ∈ H 1,2 (M ) satisfying

R M

f2

M

f = 0, and lower bound of λ1 (M ) provided by Theorem 5.3. 

We will next present an estimate of the first nonzero Neumann eigenvalue for star-shape domains. As a consequence, it gives a lower bound of the Neumann eigenvalue for geodesic balls. This was first proved by Buser [Bu] using geometric argument via the isoperimetric inequality. The analytic argument which we present below was due to Chen-Li [CnL]. Theorem 14.4. Let M be an m-dimensional manifold with boundary ∂M . Assume that M is geodesically star-shape with respect to a point p ∈ M. Suppose that the Ricci curvature of M is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. Let ρ be the radius of the largest geodesic ball centered at p contained in M , and ρ0 be the radius of the smallest geodesic ball centered at p containing M. Then there exists a constant C1 > 0 depending only on m, such that the first nonzero Neumann eigenvalue λ1 (M ) has a lower bound given by λ1 (M ) ≥

√ ρm m+2 exp(−C1 (1 + ρ0 R)). ρ0

Proof. By the variational principle, it suffices to show that there exists a constant C1 > 0 such that Z M

|∇f |2 ≥

√ ρm inf m+2 exp(−C1 (1 + ρ0 R)) a∈R ρ0

Z

(f − a)2

M

for any smooth function f0 defined on M. Let K(x, y, t) be the Neumann heat kernel defined on M. The function Z f (x, t) = K(x, y, t) f0 (y) dy M

solves the heat equation   ∂ ∆− f (x, t) = 0 ∂t on M with Neumann boundary condition ∂f =0 ∂ν on ∂M and initial condition f (x, 0) = f0 (x). Let us now consider the function Z g(x, t) = M

K(x, y, t) (f0 (y) − f (x, t))2 dy.

112

PETER LI

Clearly, Z

Z

Z

K(x, y, t) f02 (y) dy dx − ZM M Z = f02 (y) dy − f 2 (x, t) dx M M Z Z t ∂ f 2 (x, s) dx =− ∂s M 0 Z tZ = −2 f (x, s) ∆f (x, s) dx 0 M Z tZ |∇f |2 (x, s) dx. =2

Z

g(x, t) dx = M

0

f 2 (x, t) dx

M

M

However, if we consider ∂ ∂t

Z

|∇f |2 (x, t) dx = 2

Z h∇f, ∇ft i(x, t) dx

M

M

Z = −2

h∆f, ft i(x, t) dx ZM

= −2

ft2 (x, t) dx,

M

we conclude that

Z

|∇f |2 (x, t) dx ≤

Z

M

|∇f0 |2 (x) dx.

M

Hence, we have the estimate Z

Z

|∇f0 |2 (x) dx.

g(x, t) dx ≤ 2t

(14.11) M

M

On the other hand, since g is also nonnegative, we have Z Z (14.12) g(x, t) dx ≥ g(x, t) dx Bp ( ρ 2)

M

Z

Z = Bp ( ρ 2)

K(x, y, t) f0 (y) − f (x, t))2 dy dx

M

Z ≥

Z

(f0 (y) − f (x, t))2 dy dx

inf K(x, y, t)

y∈M Bp ( ρ 2)

M

Z ≥ inf

a∈R

2

Z

(f0 (y) − a) dy M

inf K(x, y, t) dx.

y∈M Bp ( ρ 2)

Using (14.11) and (14.12), we conclude that λ1 (M ) ≥ (2t)−1

Z inf K(x, y, t) dx

y∈M Bp ( ρ 2)

for all t > 0. Note that the Harnack inequality (Corollary 12.1) of Li-Yau asserts that, for any x, z ∈ Bp ( ρ2 ), we have     2  t ρ t K(x, y, t) ≥ K z, y, C4 exp −C5 + Rt + 2 2 t ρ

GEOMETRIC ANALYSIS

113

for some constants C4 , C5 > 0 depending only on m. Hence Z inf K(x, y, t) dx y∈M Bp ( ρ 2)

ρ ≥ Vp ( ) K(x, y, t) inf 2 x∈Bp ( ρ2 ),y∈M    2   Z ρ t t inf ≥ C4 exp −C5 + Rt + 2 dz, K z, y, y∈M B ( ρ ) t ρ 2 p 2 and (14.13)

   2   Z C4 ρ t t λ1 (M ) ≥ inf exp −C5 + Rt + 2 K z, y, dz. y∈M B ( ρ ) t t ρ 2 p 2

To estimate the right hand side, we apply Lemma 13.4 and obtain     Z t t K z, y, dz ≥ φ¯ r¯(¯ y ), . (14.14) 2 2 Bp ( ρ 2) where

   Z  t t ¯ ¯ K z¯, y¯, φ r¯(¯ y ), = d¯ z 2 2 Bp¯( ρ 2)

¯ being the Neumann heat kernel on Bp¯(ρ0 ) with r¯(¯ with K y ) = r(y). The proof of Lemma 13.4 asserts that 0 ¯ φ ≤ 0, hence     t t ¯ ¯ ≥ φ ρ0 , . (14.15) φ r¯(¯ y ), 2 2 On the other hand, since Bp¯(ρ0 ) is convex, we can apply Theorem 12.3 to the function φ¯ and obtain (14.16)

       2 t t ρ0 φ¯ ρ0 , ≥ C6 φ¯ 0, exp −C7 + Rt 2 4 t

for some constants C6 , C7 > 0 depending only on m. Using φ¯0 ≤ 0 again, we conclude that p¯ is a maximum point of φ¯ for any t ∈ (0, ∞). Therefore, the heat equation implies that ∂ ¯ ¯¯ φ = ∆φ(0, t) ∂t ≤ 0, hence   t ¯ ¯ t) φ 0, ≥ lim φ(0, t→∞ 4 Z = V¯p¯−1 (ρ0 ) d¯ z Bp¯( ρ 2)

where V¯p¯(ρ0 ) is the volume of Bp¯(ρ0 ). Combining with (14.13), (14.14), (14.15), and (14.16), we have −1

λ1 (M ) ≥ C21 (t)

 exp −C22



 ρ2 t ρ20 + 2+ + Rt t ρ t

V¯p¯( ρ2 ) , V¯p¯(ρ0 )

114

PETER LI

for some constants C21 , C22 > 0. Setting t=

ρ20 √ 1 + ρ0 R

and using the estimates t−1 ≥ ρ−2 0 and

√ ρ20 R √ ≤ 1 + ρ0 R, 1 + ρ0 R

we conclude that

√ V¯p¯( ρ2 ) λ1 (M ) ≥ ρ−2 0 exp(−C23 (1 + ρ0 R)) ¯ Vp¯(ρ0 )

for some constant C23 > 0 depending only on m. The theorem now follows by observing that there exists constants C24 > 0 and C25 > 0 depending only on m such that  √  r) r¯m exp C24 (1 + r¯ R) ≤ V¯p¯(¯  √  ≤ r¯m exp C25 (1 + r¯ R) . This completes the proof of Theorem 14.4.  Corollary 14.1. Let Bp (ρ) be a geodesic ball centered at a point p ∈ M with radius ρ > 0 such that Bp (ρ) ∩ ∂M = ∅. Suppose that the Ricci curvature of Bp (ρ) is bounded from below by Rij ≥ −(m − 1)R for some constant R ≥ 0. Then there exists a constant C2 > 0 depending only on m, such that the first nonzero Neumann eigenvalue λ1 (Bp (ρ)) has a lower bound given by √ λ1 (Bp (ρ)) ≥ ρ−2 exp(−C2 (1 + ρ R)). The following theorem gives a parabolic mean value inequality for subsolutions of the heat equation. It was first proved by Li-Tam in [LT5]. Theorem 14.5. Let M m be a complete noncompact Riemannian manifold with boundary. Let p ∈ M and ρ > 0 be such that Bp (4ρ) ∩ ∂M = ∅. Suppose g(x, t) is a nonnegative function defined on Bp (4ρ) × [0, T ] for 2 some 0 < T ≤ ρ4 satisfying the differential inequality ∆g −

∂g ≥ 0. ∂t

If the Ricci curvature of Bp (4ρ) is bounded by Rij ≥ −(m − 1)R for some constant R ≥ 0, then for any q > 0, there exists positive constants C1 and C2 depending only on m and q such that for any 0 < τ < T , 0 < δ < 21 , and 0 < η < 12 , sup Bp ((1−δ)ρ)×[τ,T ]

 √ V¯ (2ρ)  √ ρ R + 1 exp(C2 R T ) Vp (ρ)  m+2 Z T Z 1 1 × +√ ds g q (y, s) dy, δρ ητ (1−η)τ Bp (ρ)

g q ≤ C1

GEOMETRIC ANALYSIS

115

where V¯ (r) is the volume of the geodesic ball of radius r in m-dimensional, simply connected space form with constant sectional curvature −R. ˜ ψ˜ ≤ 1, Proof. For any 0 < δ, η < 41 , let φ˜ and ψ˜ be two smooth functions on [0, ∞) so that 0 ≤ φ, 1

for

0≤s≤1−δ

0

for

s ≥ 1,

0

for

0 ≤ s ≤ 1 − 2η

1

for

s ≥ 1 − η.

˜ = φ(s)



˜ ψ(s) =



and

We can choose φ˜ and ψ˜ such that, 1 −C δ −1 ≤ φ˜0 φ˜− 2 ≤ 0,

φ˜00 ≥ −C δ −2 , and 0 ≤ ψ˜0 ≤ C η −1 , for some constant C > 0 which is independent of δ and η. ˜ ˜ τ −1 ). Since Let r(x) denote the distance function from p, we define φ(x) = φ(r(x) R−1 ), and ψ(t) = ψ(t g is a nonnegative subsolution of the heat equation, for q > 1, we have 

 ∂ ∆− (φ(x) ψ(t) g q (x, t)) ∂t ≥ (∆φ) ψ g q + 2qψ g q−1 h∇φ, ∇gi + q(q − 1)φ ψ g q−2 |∇g|2 − φ ψ 0 g q ≥ (∆φ) ψ g q −

|∇φ|2 q ψ gq − φ ψ0 gq . q−1 φ

On the other hand, the Laplacian comparison theorem asserts that  ∆φ −

(14.17)

q |∇φ|2 q−1 φ



√ ≥ −C3 (ρ R + 1)δ −2 ρ−2 ,

for some constant C3 depending only on m and q. If H(x, y, t) is the Dirichlet heat kernel of M , by using (14.17) and the definition of φ and ψ, there exists a constant C4 depending only on m and q so that, (14.18) g q (x, t) = g q (x, t) φ(x) ψ(t)   Z t Z ∂ (g q (y, s) φ(y) ψ(s)) dy =− ds H(x, y, t − s) ∆y − ∂s 0 M   Z t Z ∂ =− ds H(x, y, t − s) ∆y − (g q (y, s) φ(y) ψ(s)) dy ∂s (1−2η)τ Bp (ρ) ( Z T Z √ −2 −2 ds H(x, y, t − s) g q (y, s) dy ≤ C4 (ρ R + 1) δ ρ (1−2η)τ

1 + ητ

Z

(1−η)τ

Z

) q

H(x, y, t − s) g (y, s) dy

ds (1−2η)τ

Bp (ρ)\Bp ((1−δ)ρ)

Bp (ρ)

,

116

PETER LI

for any x ∈ Bp ((1 − 2δ)ρ) and τ < t < T . Using the upper bound of heat kernel given by Corollary 13.2, if 2 x, y ∈ Bp (R) and t − s ≤ R4 then we have √ −1 √ ( t − s) Vy 2 ( t − s)   p r2 (x, y) −2 × exp −C6 + C7 (ρ + R)(t − s) , t−s − 21

H(x, y, t − s) ≤ C5 Vx

where C5 , C6 and C7 are constants depending only on m, On the other hand, the volume comparison theorem 2 implies that, for x ∈ Bp (ρ) and 0 < s < t ≤ T ≤ ρ4 , m Vp (ρ) Vx (2ρ) V¯ (2ρ) √ √ ≤ ≤ ¯ √ ≤ C8 (t − s)− 2 V¯ (2ρ), Vx ( t − s)) Vx ( t − s) V ( t − s)

for some constant C8 depending only on m. Hence (14.19)   p m V¯ (2ρ) r2 (x, y) (t − s)− 2 exp −C6 + C7 (ρ−2 + R)(t − s) Vp (ρ) t−s   ¯ p V (2ρ) ≤ C10 (δρ)−m exp C7 (ρ−2 + R)(t − s) , Vp (ρ)

H(x, y, t − s) ≤ C9

2

for x ∈ Bp ((1 − 2δ)ρ), y ∈ Bp (ρ) \ Bp ((1 − δ)ρ) and 0 < s ≤ t ≤ T ≤ ρ4 , where C9 and C10 are constants depending only on m. Also for t > τ , and s ≤ (1 − η)τ , we can find a constant C11 depending only on m so that m

H(x, y, t − s) ≤ C11 (ητ )− 2

(14.20)

 p  V¯ (2ρ) exp C7 (ρ−2 + R)(t − s) . Vp (ρ)

Combining (14.18), (14.19), and (14.20), we conclude that (14.21)   p V¯ (2ρ) √ (ρ R + 1) exp C13 (ρ−2 + R)T Vp (ρ) Z  Z T m+2 × (δρ)−(m+2) + (ητ )− 2 ds

g q ≤ C12

sup B((1−2δ)ρ)×[τ,T ]

(1−2η)τ

g q (y, s) dy,

Bp (ρ)

2

for 0 < τ < T ≤ ρ4 , 0 < δ, η ≤ 41 and q > 1, where C12 and C13 depend only on m and q. This completes the proof of the theorem for the case when q > 1 if we replace 2δ by δ and 2η by η. In order to prove the case when q ≤ 1, we can proceed as in [LS]. Take q = 2 in (14.21), we have (14.22)

  p V¯ (2ρ) √ (ρ R + 1) exp C13 (ρ−2 + R)T Vp (ρ) Z  Z T m+2 × (δρ)−(m+2) + (ητ )− 2 ds g 2 (y, s) dy,

g 2 ≤ C12

sup Bp ((1−δ)ρ)×[τ,T ]

(1−η)τ

for 0 < δ, η < 12 , and T ≤

ρ2 4 .

For k ≥ 0 and l ≥ 1, let M (k, l) =

sup Bp ((1−δk )ρ)×[τl ,T ]

g2 ,

Bp (ρ)

GEOMETRIC ANALYSIS

where δk = 2−k δ, and τl = (1 − η

Pl

i=1

117

2−i )τ . Inequality (14.22) implies that

  p V¯ (2ρ) √ (ρ R + 1) exp C13 (ρ−2 + R)T Vp (ρ)   (l+1)(m+2) m+2 2 (ητ )− 2 × 2(k+1)(m+2) (δρ)−(m+2) + 2 Z T Z × M λ (k + 1, l + 1) ds g q (y, s) dy,

M (k, l) ≤ C12

(1−η)τ

Bp (ρ)

˜ (k) = M (k, 2k + 1), then where λ = 1 − 2q . Let M ˜ (k) ≤ A(2m+2 )k+1 M ˜ λ (k + 1), M where    p m+2 V¯ (2ρ) √ (ρ R + 1) exp C13 (ρ−2 + R)T (δρ)−(m+2) + (ητ )− 2 Vp (ρ) Z T Z × ds g q (y, s) dy.

A = C12

(1−η)τ

Bp (ρ)

Iterating this inequality, we get 1 ˜ (0) ≤ C14 A 1−λ , M

where C14 is a constant depending only on m and q. From this, it is easy to conclude that the theorem is also true for 0 < q ≤ 1.  Corollary 14.2. Let M m be a complete manifold with boundary and p be a fixed point in M. Suppose ρ > 0 be such that Bp (4ρ) ∩ ∂M = ∅ and assume that the Ricci curvature of M on Bp (4ρ) satisfies the bound Rij ≥ −(m − 1)R for some constant R ≥ 0. Let 0 < δ < 21 , q > 0, and λ ≥ 0 be fixed constants. Then there exists a constant √ C > 0 depending only on δ, q, λ ρ2 , m, and ρ R, such that, for any nonnegative function f defined on Bp (2ρ) satisfying the differential inequality ∆f ≥ −λ f, we have sup

f q ≤ C Vp−1 (ρ)

Z

Bp ((1−δ)ρ)

f q (y) dy.

Bp (ρ)

Proof. Let us consider the function g(x, t) = e−λt f (x) which satisfies the differential inequality 

∂ ∆− ∂t

 g(x, t) ≥ 0

on Bp (2ρ) × [0, ∞). Applying Theorem 14.5 to g(x, t) and by setting T = sup 2 Bp ((1−δ)ρ)×[ ρ8

e 2 , ρ4

]

f (x) ≤ C1 V¯ (2ρ) ρ

−pλ t q

−(m+2)

Z

ρ2 4 3ρ2 32

ρ2 4 ,

τ=

ρ2 8 ,

and η = 14 , we obtain

e−qλ s ds

Z  √ √  × (ρ R + 1) exp C2 ρ R Vp−1 (ρ) Bp (ρ)

f q (y) dy,

118

PETER LI

where C1 > 0 depends on m, q, and δ. This implies that   √ f q (x) ≤ C4 exp C2 ρ R + C3 λ ρ2 V¯ (2ρ) ρ−m

sup

(14.23)

Bp ((1−δ)ρ)

×

Vp−1 (ρ)

Z

f q (y) dy.

Bp (ρ)

Obviously the corollary follows from this estimate.



Corollary 14.3. Let M m be a complete manifold with boundary and p be a fixed point in M. Suppose ρ > 0 be such that Bp (4ρ) ∩ ∂M = ∅ and assume that the Ricci curvature of M is nonnegative. Let 0 < δ < 12 and q > 0 be fixed constants. Then there exists a constant C > 0 depending only on δ, q, and m, such that, for any nonnegative function f defined on Bp (2ρ) satisfying the differential inequality ∆f ≥ 0, we have sup

f q ≤ C Vp−1 (ρ)

Bp ((1−δ)ρ)

Z

f q (y) dy.

Bp (ρ)

Proof. Applying the assumptions to (14.23) by setting αm−1 m V¯ (2ρ) = ρ , m and λ = 0, we obtain sup Bp ((1−δ)ρ)

This proves the corollary.



q

f (x) ≤ C1 ρ

−2

Z

ρ2 4 3ρ2 32

ds Vp−1 (ρ)

Z Bp (ρ)

f q (y) dy.

GEOMETRIC ANALYSIS

119

§15 Uniqueness and Maximum Principle for the Heat Equation As eluded in our previous discussion, if H(x, y, t) is the minimal heat kernel on a complete manifold, constructed by Theorem 12.2, then the question if Z (15.1) H(x, y, t) dy = 1 M

is related to the uniqueness of the heat kernel. The validity (15.1) implies that H(x, y, t) = K(x, y, t) where K(x, y, t) is a limiting heat kernel obtained from a sequence of Neumann heat kernels. Note that the inequality Z (15.2) H(x, y, t) ≤ 1 M

implies that the (minimal) heat semi-group is contractive on L∞ , because Z H(x, y, t) f0 (y) dy |f (x, t)| = M Z ≤ kf0 k∞ H(x, y, t) dy M

≤ kf0 k∞ . implying kf (·, t)k∞ ≤ kf0 k∞ . Observe that the constant function 1 and heat equation with the same initial data, hence if Z H(x, y, t) dy < 1

R M

H(x, y, t) dy are solutions to the

M

for some x ∈ M then there yields two distinct L∞ solutions to the heat equation with initial data 1. In general, let f (x, t) be any L∞ solution to the heat equation with f (x, 0) = f0 (x). If |f (x, t)| < A on M × [0, T ] then the functions A + f and A − f are both positive solutions to the heat equation with initial data given by A + f0 and A − f0 , respectively. The minimality of the heat kernel H implies that Z H(x, y, t) (A + f0 (y)) dy ≤ A + f (x, t) M

and

Z H(x, y, t) (A − f0 (y)) dy ≤ A − f (x, t). M

Therefore, we conclude that  Z Z H(x, y, t) f0 (y) dy − A 1 − M

 H(x, y, t) dy

M

≤ f (x, t)  Z Z ≤ H(x, y, t) f0 (y) dy + A 1 − M

 H(x, y, t) dy .

M

R

In particular, the validity of (15.1) implies that f (x, t) = M H(x, y, t) f0 (y) dy, asserting the uniqueness of all bounded solutions is equivalent to (15.1). Inequality (15.2) also implies that the semi-group is contractive on L1 , since Z Z Z |f (x, t)| dy ≤ H(x, y, t) |f0 |(y) dy dx M M M Z ≤ |f0 |(y) dy. M

120

PETER LI

When the equality (15.1) is valid, then for nonnegative initial data f0 , we have Z

Z

Z

f (x, t) dy = M

H(x, y, t) f0 (y) dy dx ZM

=

M

f0 (y) dy, M

asserting that the heat semi-group preserves L1 -norm on nonnegative functions. Also, note that the heat equation is contractive on Lp for 1 < p < ∞. To see this, let f0 ∈ Lp and Z f (x, t) =

H(x, y, t) f0 (y) dy M

be the solution of the heat equation with initial data f0 . We will first assume that f0 ≥ 0. In this case, since H(x, y, t) is the limit of aR sequence of Dirichlet heat kernels Hi (x, y, t) on a compact exhaustion {Ωi }, f (x, t) is the limit of fi (x, t) = M Hi (x, y, t) f0 dy where we extend Hi to be 0 on M \ Ωi . However, since ∂ ∂t

Z M

Z

fip = p

fip−1 ∆fi M Z = −p(p − 1) fip−2 |∇fi |2 M

≤ 0, we have kfi (·, t)kp ≤ kfi (·, 0)kp ≤ kf0 kp for all t. Taking the limiting as i → ∞, we conclude that kf (·, t)kp ≤ kf0 kp , as claimed. In general, we observe that Z |f (x, t)| = H(x, y, t) f0 (y) dy ZM ≤ H(x, y, t) |f0 (y)| dy. M

hence the contractivity of the minimal heat equation on Lp is established. Proposition 15.1. Let M m be a complete manifold. Let f (x, t) be a nonnegative subsolution solution of the heat equation. If f (·, t) ∈ Lp (M ) for all [0, T ], for 1 < p < ∞, then kf (·, t)kp ≤ kf (·, 0)kp for all 0 ≤ t ≤ T. Moreover, the heat equation (not necessarily given by the minimal heat kernel) is contractive in Lp and it is uniquely determined by it initial condition, hence has to be given by Z f (x, t) =

H(x, y, t) f0 (y) dy. M

GEOMETRIC ANALYSIS

121

Proof. Using the fact that f (x, t) is a nonnegative subsolution, for a nonnegative cutoff function φ defined on M , we have Z Z ∂ 2 p φ (x) f (x, t) dx = p φ2 (x) f p−1 (x, t) ft (x, t) dx (15.3) ∂t M M Z ≤p φ2 (x) f p−1 (x, t) ∆f (x, t) dx M Z = −p(p − 1) φ2 (x) f p−2 (x, t) |∇f |2 (x, t) dx M Z − 2p φ(x) f p−1 (x, t) h∇φ(x), ∇f (x, t)i dx. M

Schwarz inequality asserts that Z Z −2p φ f p−1 h∇φ, ∇f i dx ≤ p(p − 1) M

φ2 f p−2 |∇f |2 dx + (p − 1)−1 p

M

Z

|∇φ|2 f p .

M

Combining with (15.3), we conclude that Z Z ∂ 2 p −1 φ (x) f (x, t) dx ≤ (p − 1) p |∇φ|2 f p . (15.4) ∂t M M Choosing the cutoff function φ to satisfy ( φ=

1

on

Bq (ρ)

0

on

M \ Bq (2ρ)

and |∇φ|2 ≤ 2 ρ−2 , inequality (15.4) yields Z Z f p (x, t) dx − Bq (ρ)

f0p (x) dx ≤

Bq (2ρ)

Z

φ2 (x) f p (x, t) dx −

M

Z

φ2 (x) f0p (x) dx

M −1

Z tZ

|∇φ|2 f p (x) dx dt Z tZ −1 −2 ≤ 2p(p − 1) ρ f p (x) dx dt.

≤ (p − 1)

p

0

M

0

M

Using the assumption that f (·, t) ∈ Lp (M ) and letting ρ → ∞, we obtain Z Z p f (x, t) dx ≤ f p (x, 0) dx. M

M

Observing that by taking absolute value of a solution yields a nonnegative subsolution, we conclude that the heat equation (not necessarily minimal) is contractive on nonnegative Lp functions. In particular, by taking the difference of two Lp solutions, this implies that Lp solutions are uniquely determined by their initial conditions.  We are now ready to address the issue of uniqueness of L∞ solutions. The theorem presented below was proved by Karp-Li [KL] in a unpublished paper. It was also independently proved by Grigor’yan [G1] with an slight weaker assumption.

122

PETER LI

Theorem 15.1. (Karp-Li) Let M m be a complete manifold. Suppose f (x, t) is a subsolution to the heat equation satisfying   ∂ ∆− f (x, t) ≥ 0 on M × (0, ∞) ∂t with f (x, 0) ≤ 0

on

M. 2

If there exists p ∈ M and a constant A > 0 such that the L -norm of f over the geodesic ball centered at p of radius ρ satisfies Z f 2 (x, t) ≤ exp(A ρ2 ) for all t ∈ (0, T ), Bp (ρ)

then f (x, t) ≤ 0. Proof. Let us define f+ (x, t) = max{f (x, t), 0}, which is a subsolution satisfying  ∆−

∂ ∂t

 f+ (x, t) ≥ 0

on

M × (0, ∞)

with f+ (x, 0) = 0. It suffices to show that f+ = 0 on M × [0, ∞]. For simplicity of notation, we will replace f+ by f . Let us consider the function (15.5)

g(x, t) =

−r2 (x) 4(2T − t)

for x ∈ M and 0 ≤ t < T, where r(x) is the distance function to the fixed point p ∈ M. One verifies that |∇g|2 + gt = 0.

(15.6)

For  > 0 and ρ > 0, let φ be a nonnegative cutoff function with the properties that ( 1 on Bp (ρ) (15.7) φ= 0 on M \ Bp (ρ + 1), with 0 ≤ φ ≤ 1 and |∇φ|2 ≤ 2. Using the fact f is a nonnegative subsolution to the heat equation and its initial condition, we have (15.8) Z

τ

Z

0

M τ Z

φ2 exp(g) f ∆f −

Z

τ

Z

φ2 exp(g) f ft Z Z Z τZ 2 2 =− φ exp(g) |∇f | − 2 φ exp(g) f h∇φ, ∇f i − φ2 exp(g) f h∇g, ∇f i 0 M 0 M 0 M Z Z Z 1 τ 1 2 2 φ exp(g) f t=τ + φ2 exp(g) f 2 gt . − 2 M 2 0 M

0≤

0

ZMτ

However, Schwarz inequality implies that Z τZ Z Z Z Z 1 τ 1 τ − φ2 exp(g) f h∇g, ∇f i ≤ φ2 exp(g) f 2 |∇g|2 + φ2 exp(g) |∇f |2 2 0 M 2 0 M 0 M

GEOMETRIC ANALYSIS

and

Z

τ

Z

−

φ exp(g) f h∇φ, ∇f i ≤ 0

M

1 4

τ

Z

Z

0

123

φ2 exp(g) |∇f |2 +

M

Z

τ

Z

0

|∇φ|2 exp(g) f 2 ,

M

hence (15.8) becomes τ

Z Z 1 τ 0≤2 |∇φ| exp(g) f + φ2 exp(g) f 2 |∇g|2 2 0 M 0 M Z Z Z 1 1 τ 2 2 − φ2 exp(g) f 2 gt . φ exp(g) f t=τ + 2 M 2 0 M Z

Z

2

2

Combining with (15.6) and (15.7), we obtain Z Z (15.9) exp(g) f 2 t=τ ≤

M

Bp (ρ)

φ2 exp(g) f 2 t=τ

Z

τ

Z

exp(g) f 2 .

≤4 0

Bp (ρ+1)\Bp (ρ)

However, (15.5) implies that ρ2 exp(g) ≤ exp − 8T 

 on

(Bp (ρ + 1) \ Bp (ρ)) × [0, T ],

and (15.9) yields ρ2 exp(g) f (x, τ ) dx ≤ 4 exp − 8T Bp (ρ)

Z



2

Z

τ

0

Z

f 2 (x, t) dx dt

Bp (ρ+1)\Bp (ρ)

for all 0 ≤ τ ≤ T. Applying the growth assumption on f , we conclude that   Z ρ2 2 2 + A(ρ + 1) . exp(g) f (x, τ ) dx ≤ 4τ exp − 8T Bp (ρ) Choosing T
0, such that, the volume growth of M satisfy the estimate Vp (ρ) ≤ exp(Aρ2 ) for all ρ > 0. Then any L∞ solution of the heat equation is uniquely determined by its initial data. In particular, Z H(x, y, t) = 1. M

Proof. By considering the difference of two L∞ solutions, it suffices to prove that if kf (·, t)k∞ ≤ A and solves the heat equation on M with f (x, 0) = 0, then f must be identically 0. This follows by applying Theorem 15.1 to the function g = |f |.  We have yet address the uniqueness of L1 solutions. It turns out that with some mild curvature assumption, L1 solutions to the heat equation is also unique. We will delay our discussion in a latter section (§31). Note that Theorem 15.1 can be viewed as a form of maximum principle on complete manifolds. Various generalizations based on the argument given in the proof of Theorem 15.1 was developed. The interested readers shall refer to [LoT] and [NT] for further reference.

124

PETER LI

§16 Large Time Behavior of the Heat Kernel In this section, we will discuss the behavior of the heat equation when t → ∞. When M is a compact manifold with (or without) boundary, due to the eigenfunction expansion, the Dirichlet and Neumann heat kernels are asymptotically given by the first eigenfunction and the first eigenvalue. In particular, H(x, y, t) ∼ e−µ1 t ψ1 (x) ψ1 (y) and K(x, y, t) ∼ V −1 (M ) as t → ∞. However, if M is complete and noncompact, then the behavior of H as t → ∞ is not so clear. If the bottom of the L2 spectrum µ1 (M ) is positive, it is still rather easy to see that t−1 log H(x, y, t) ∼ −µ1 (M ). The following discussion gives a sharp asymptotic behavior of H for manifolds with nonnegative Ricci curvature with maximal volume growth. As a corollary (Corollary 16.1), one concludes that such manifolds must have finite fundamental group. This was proved by the author in [L5]. Lemma 16.1. Let M m be a complete manifold with nonnegative Ricci curvature. If there exists a constant θ > 0 such that Vp (ρ) lim inf m = θ ρ→∞ ρ for some point p ∈ M, then Ap (ρ) =θ ρ→∞ m ρm−1 lim

where Ap (ρ) denotes the area of the geodesic sphere ∂Bp (ρ). ¯ Proof. The volume comparison theorem asserts that if A(ρ) is the area of the sphere of radius ρ in Rm , then Ap (ρ) ¯ A(ρ) is a decreasing function of ρ. In particular, the limit Ap (ρ) ρ→∞ m ρm−1 lim

exists. Note that since

Vp (ρ) V¯ (ρ)

is also a decreasing function, we have the limit lim

ρ→∞

Vp (ρ) = θ. ρm

L’Hˆ opital’s rule implies that lim

ρ→∞

which proves the lemma. 

Ap (ρ) Vp (ρ) = lim , ρ→∞ m ρm−1 ρm

GEOMETRIC ANALYSIS

125

Theorem 16.1. (Li) Let M m be a complete manifold with nonnegative Ricci curvature. If there exists a constant θ > 0 such that Vp (ρ) lim inf m = θ ρ→∞ ρ for some point p ∈ M, then

√ m lim Vp ( t) H(x, y, t) = m−1 αm−1 (4π)− 2 ,

t→∞

where αm−1 denotes the area of the unit sphere Sm−1 ⊂ Rm . Proof. The estimate of Corollary 12.2 asserts that  H(p, p, t1 ) ≤ H(p, p, t2 ) for t1 ≤ t2 . This implies that

t2 t1

 m2

m

t 2 H(p, p, t) is a nondecreasing function of t. However, Theorem 13.1 asserts that √ H(p, p, t) ≤ C Vp−1 ( t). Together with the assumption on the volume growth, we conclude that m

t 2 H(p, p, t) is bounded from above. Hence the limit m

lim t 2 H(p, p, t) = α

t→∞

exists and

m

t 2 H(p, p, t) % α. For x, y ∈ M , Corollary 12.2 asserts that m

 r2 (x) 4δt  2  m r (x) + r2 (y) ≤ ((1 + 2δ)t) 2 H(p, p, (1 + 2δ)t) exp 4δt  2  r (x) + r2 (y) ≤ α exp 4δt 

m

t 2 H(x, y, t) ≤ ((1 + δ)t) 2 H(p, y, (1 + δ)t) exp

for δ > 0, where r(·) denotes the to the point p. This implies that m

lim sup t 2 H(x, y, t) ≤ α. t→∞

The same argument also show that  2  m m r (x) + r2 (y) t 2 H(x, y, t) ≥ ((1 − 2δ)t) 2 H(p, p, (1 − 2δ)t) exp − 4δt implying that

m

lim inf t 2 H(x, y, t) ≥ α, t→∞

126

PETER LI

hence m

(16.1)

lim t 2 H(x, y, t) = α.

t→∞

To compute the value of α, we use the heat kernel comparison in Lemma 13.3 and obtain Z

Z

¯ p, y¯, t) d¯ H(¯ y

H(p, y, t) dy ≥ ¯ p¯(ρ) B

Bp (ρ)

¯ is the heat kernel of Rm with r(y) = r¯(¯ where H y ). In particular m

Z

m

Z

¯ p, y¯, t) d¯ H(¯ y

H(p, y, t) dy ≥ lim t 2

lim t 2

t→∞

t→∞

Bp (ρ)

¯ p¯(ρ) B m

= lim t 2 t→∞

 2  m r¯ (¯ y) (4πt)− 2 exp − d¯ y 4t ¯ Bp¯(ρ)

Z

m = (4π)− 2 V¯ (ρ),

implying

m αVp (ρ) ≥ (4π)− 2 V¯ (ρ).

Letting ρ → ∞, we conclude that m

α ≥ (4π)− 2 m−1 αm−1 θ−1 .

(16.2)

To obtain the upper bound for α, we apply Corollary 12.2 again and get r2 (y) H (p, p, t) exp − 2δt 

2



≤ (1 + δ)m H 2 (p, y, (1 + δ)t).

Integrating over ∂Bp (ρ) yields ρ2 Ap (ρ) exp − 2δt 



2

H (p, p, t) ≤ (1 + δ)

m

Z

H 2 (p, y, (1 + δ)t) dy.

∂Bp (ρ)

Integrating over 0 ≤ ρ ≤ ∞ and using the semi-group property, we obtain Z (16.3) 0

∞

  Z ρ2 Ap (ρ) exp − H 2 (p, p, t) dρ ≤ (1 + δ)m H 2 (p, y, (1 + δ)t) dy 2δt M = (1 + δ)m H(p, p, 2(1 + δ)t).

However, Lemma 16.1 asserts that Ap (ρ) & θ, m ρm−1 hence Ap (ρ) ≥ θ m ρm−1 . ¯ Observe that since A(ρ) = αm−1 ρm−1 , after combining with (16.3), we conclude that −1 θ m αm−1

Z 0

∞

  ρ2 ¯ A(ρ) exp − H 2 (p, p, t) dρ ≤ (1 + δ)m H(p, p, 2(1 + δ)t). 2δt

GEOMETRIC ANALYSIS

127

m

Multiplying both side by (2(1 + δ)t) 2 and using (16.1), we get (16.4)

m

−1 θ m αm−1 (2(1 + δ)t) 2

∞

Z 0

  ρ2 ¯ A(ρ) exp − H 2 (p, p, t) dρ ≤ (1 + δ)m α. 2δt

Noting that m ¯ p, p¯, 2δt) (8πδt)− 2 = H(¯ Z ¯ 2 (¯ = H p, y¯, δt) d¯ y Rm   Z ∞ ρ2 = A¯p¯(ρ)(4πδt)−m exp − dρ 2δt 0

we can rewrite (16.4) as m

m

−1 θ m αm−1 (2(1 + δ)) 2 (4πδt)m H 2 (p, p, t) (8πδ)− 2 ≤ (1 + δ)m α.

Letting t → ∞ and using (16.1), we obtain m

m

−1 θ m αm−1 (1 + δ)− 2 (4πδ) 2 α2 ≤ α

hence

m

m

α ≤ θ−1 m−1 αm−1 (1 + δ) 2 (4πδ)− 2 . Letting δ → ∞ and combining with (16.2), we obtain m

α = θ−1 m−1 αm−1 (4π)− 2 . The theorem follows by substituting the definition of θ.  Corollary 16.1. Let M be a complete manifold with nonnegative Ricci curvature. Suppose there exists a constant θ > 0 such that Vp (ρ) lim inf m = θ ρ→∞ ρ for some point p ∈ M. Then |π1 (M )| < ∞. Moreover, |π1 (M )| = lim

ρ→∞

Vp˜(ρ) Vp (ρ)

˜ is a preimage point in the universal cover M ˜ of M and Vp˜(ρ) is the volume of geodesic ball of where p˜ ∈ M ˜ radius ρ centered at p˜ in M . ˜ → M be the projection map from the universal cover M ˜ to M. We equip M ˜ with the Proof. Let π : M pull-back metric from M so that π is a local isometry and the elements of the fundamental group π1 (M ) ˜ . If we denote H and H ˜ to be minimal heat kernels on M and M ˜ , respectively, and acts as isometries on M then we claim that X ˜ x, g(˜ H(π(˜ x), π(˜ y ), t) = H(˜ y ), t) g∈π1 (M )

˜ . In fact, we first claim that for any x ˜, y˜ ∈ M (16.5)

H1 (x, y, t) =

X g∈π1 (M )

˜ x, g(˜ H(˜ y ), t)

128

PETER LI

for x, y ∈ M with π(˜ x) = x and π(˜ y ) = y is well-defined. To see this, for any x y ∈ M , let By (ρ) be a sufficiently small geodesic ball around y such that their preimages {Bg(˜y) (ρ) | g ∈ π1 (M )} are all disjoint. In particular, Z X Z ˜ x, z˜, t) d˜ ˜ x, z˜, t) d˜ (16.6) H(˜ z≤ H(˜ z ˜ M

Bg(y) ˜ (ρ)

g∈π1 (M )

≤ 1. On the other hand, Corollary 12.2 asserts that ˜ x, y˜, t) ≤ C Vg(˜y) (ρ) H(˜

Z

˜ x, z˜, 2t) d˜ H(˜ z exp

Bg(y) ˜ (ρ)



ρ2 4t

 .

Combining with (16.6), we conclude that (16.5) is summable and hence well-defined. Obviously   ∂ H1 (x, y, t) = 0 on M × (0, ∞) ∆y − ∂t with H1 (x, y, 0) = δx (y). Duhamel principle now implies that for y ∈ Bx (ρ), H(x, y, t) − H1 (x, y, t) Z Z t ∂ φ2 (z) H1 (x, z, t − s) H(z, y, s) dz ds = 0 ∂s M Z tZ =− φ2 (z) ∆H1 (x, z, t − s) H(z, y, s) dz ds 0 M Z tZ + φ2 (z) H1 (x, z, t − s) ∆H(z, y, s) dz ds 0 M Z tZ =2 φ(z) H(z, y, s) h∇H1 (x, z, t − s), ∇φ(z)i dz ds 0 M Z tZ −2 φ(z) H1 (x, z, t − s) h∇H(z, y, s), ∇φ(z)i dz ds

(16.7)

0

M

for any nonnegative cutoff function φ satisfying ( 1 φ= 0

on

Bx (ρ)

on

M \ Bx (2ρ)

and |∇φ|2 ≤ 2ρ−2 . However, Schwarz inequality asserts that (16.8) Z t Z 0

M

≤

φ(z) H(z, y, s) h∇H1 (x, z, t − s), ∇φ(z)i dz ds

√

2ρ−1

Z tZ H(z, y, s) |∇H1 (x, z, t − s)| dz ds 0

≤

√

2ρ

−1

Z

Bx (2ρ)\Bx (ρ) t

Z

! 21 2

Z

|∇H1 (x, z, t − s)| dz

H (z, y, s) dz 0

Bx (2ρ)\Bx (ρ)

! 12 2

Bx (2ρ)\Bx (ρ)

ds.

GEOMETRIC ANALYSIS

129

Similarly, we have (16.9) Z t Z 0

M

≤

φ(z) H1 (x, z, t − s) h∇H(z, y, s), ∇φ(z)i dz ds

√

2ρ

−1

Z tZ H1 (x, z, t − s) |∇H(z, y, s)| dz ds 0

≤

√

2ρ−1

Z 0

Bx (2ρ)\Bx (ρ) t

! 12

Z

H12 (x, z, t − s) dz

! 12

Z

|∇H(z, y, s)|2 dz

ds.

Bx (2ρ)\Bx (ρ)

Bx (2ρ)\Bx (ρ)

On the other hand, the semi-group property asserts that Z H 2 (z, y, s) dz = H(y, y, 2s), M

hence H ∈ L2 (M ). Also, for the cutoff function φ defined above, we have Z

2

2

Z

φ (z) |∇H| (z, y, s) dz = − M

Z

2

φ H(z, y, s) ∆H(z, y, s) dz − 2 φ(z) H(z, y, s) h∇φ(z), ∇H(z, y, s)i dz Z  ZM 1 1 ∂ φ2 (z) H 2 (z, y, s) dz + φ2 (z) |∇H(z, y, s)|2 dz ≤− 2 ∂s 2 M Z M 2 2 +2 |∇φ| (z) H (z, y, s) dz, M

M

implying that Z  Z ∂ 2 2 φ (z) |∇H(z, y, s)| dz ≤ − φ (z) H (z, y, s) dz + 4 |∇φ|2 (z) H 2 (z, y, s) dz ∂s M M M Z  Z ∂ 2 2 −2 ≤− φ (z) H (z, y, s) dz + 2ρ H 2 (z, y, s) dz. ∂s M M

Z

2

2

Letting ρ → ∞ and using the fact that H ∈ L2 , we conclude that Z  ∂ 2 |∇H| (z, y, s) dz ≤ − H (z, y, s) dz ∂s M M ∂ = H(y, y, 2s), ∂s

Z

2

˜ 1 is a fixed fundamental domain of M in hence |∇H| ∈ L2 (M ). In the case of H1 , we also observe that if M ˜ ˜ ˜ M and use the fact that H(g(˜ x), g(˜ y ), t) = H(˜ x, y˜, t) for any g ∈ π1 (M ), then Z

H12 (x, z, t − s) dz =

M

X g, h∈π1 (M )

=

X g, h∈π1 (M )

=

X g, h∈π1 (M )

Z

˜ x, g(˜ ˜ x, h(˜ H(˜ z ), t − s) H(˜ z ) t − s) d˜ z

˜1 M

Z

˜ x, g(˜ ˜ ◦ h−1 (˜ H(˜ z ), t − s) H(g x), g(˜ z ) t − s) d˜ z

˜1 M

Z ˜ 1) g(M

˜ x, z˜, t − s) H(g ˜ ◦ h−1 (˜ H(˜ x), z˜, t − s) d˜ z.

130

PETER LI

However, since X

X

˜ ◦ h−1 (˜ H(g x), z˜, t − s) =

h∈π1 (M )

˜ x), z˜, t − s), H(k(˜

k∈π1 (M )

we conclude that Z

H12 (x, z, t − s) dz =

M

Z

X

=

Z

X k∈π1 (M )

=

X

˜ x, z˜, t − s) H(k(˜ ˜ x), z˜, t − s) d˜ H(˜ z

˜ 1) g(M

g, k∈π1 (M )

˜ x, z˜, t − s) H(k(˜ ˜ x), z˜, t − s) d˜ H(˜ z

˜ M

˜ x, k(˜ H(˜ x), 2(t − s))

k∈π1 (M )

= H1 (˜ x, x ˜, 2(t − s)), hence H1 ∈ L2 (M ). The same argument we used before also imply that |∇H1 | ∈ L2 (M ). Therefore the right hand sides of (16.8) and (16.9) tends to 0 as ρ → ∞, and(16.7) yields H(x, y, t) = H1 (x, y, t). Applying Theorem 16.1, we have √ m m−1 αm−1 (4π)− 2 = lim Vp ( t) H(p, p, t) t→∞ X √ ˜ p, g(˜ = lim Vp ( t) H(˜ p), t)

(16.10)

t→∞

g∈π1 (M )

√ √ Vp ( t) ˜ p, g(˜ √ Vp˜( t) H(˜ p), t). V ( t) (M ) p˜

X

= lim

t→∞ g∈π1

√ √ ˜ also has nonnegative curvature and Vp˜( t) ≥ Vp ( t), hence we can On the other hand, we observe that M ˜ and obtain apply Theorem 16.1 to H √ √ X Vp ( t) √ Vp ( t) −1 −m ˜ √ Vp˜( t) H(˜ √ . p, g(˜ p), t) = |π1 (M )| m αm−1 (4π) 2 lim lim t→∞ Vp˜( t) t→∞ V ( t) g∈π (M ) p˜ 1

Combining with (16.10), we obtain

√ Vp ( t) √ = 1, |π1 (M )| lim t→∞ Vp˜( t)

as claimed. In particular, since lim inf ρ→0

Vp (ρ) =θ ρm

and Vp˜(ρ) ≤ m−1 αm−1 ρm , hence we conclude that |π1 (M )| ≤ m−1 αm−1 θ−1 is finite.



The following lemma proved in [L5] asserts a mean value type property for subharmonic functions at infinite of a complete manifold with nonnegative Ricci curvature. This will be useful in a later section when we study linear growth harmonic functions.

GEOMETRIC ANALYSIS

131

Lemma 16.2. (Li) Let M m be a complete manifold with nonnegative Ricci curvature. Suppose f is a bounded subharmonic function defined on M, then Z

lim Vp−1 (ρ)

f (x) dx = sup f.

ρ→∞

M

Bp (ρ)

Proof. Let us define g0 (x) = supM f − f (x), then g0 satisfies ∆g0 ≤ 0 and g0 ≥ 0. It suffices to show that lim

ρ→∞

Vp−1 (ρ)

Z g0 (x) dx = 0. Bp (ρ)

Let us consider the solution to heat equation Z g(x, t) =

H(x, y, t) g0 (y) dy. M

Observe that if Hi (x, y, t) is a Dirichlet heat kernel on a compact subdomain Ωi ⊂ M , then ∂ ∂t

Z Ωi

Hi (x, y, t) g0 (y) dy Z ∂Hi = (x, y, t) g0 (y) dy Ωi ∂t Z ∆y Hi (x, y, t) g0 (y) dy = Ωi Z Z = Hi (x, y, t) ∆y g0 (y) dy + Ωi

∂Ωi

∂Hi (x, y, t) g0 (y) dy ∂ν

≤ 0, R implying the function gi (x, t) = Ωi Hi (x, y, t) g0 (y) dy is nonincreasing in t. Since H(x, y, t) is the limit of a sequence of Dirichlet heat kernels, we conclude that g(x, t) is also nonincreasing in t. However, the maximum principle Theorem 15.1 asserts that g(x, t) ≥ 0, hence g(x, t) converges to a nonnegative, bounded function g∞ (x) as t → ∞. Moreover, g∞ is harmonic because ∆g(x, t) = which tends to 0 as t → ∞ by the fact that Z 0

∞

∂g ∂t

∂g(x, t) ∂t

≤ 0 and

∂g (x, t) dt = g∞ (x) − g0 (x). ∂t

In particular, we conclude that ∆g∞ = 0. On the hand, since M has nonnegative Ricci curvature, the only bounded harmonic functions are constants. The facts that g∞ ≤ g0 and inf M g0 = 0 imply that g∞ = 0, hence g(x, t) → 0.

132

PETER LI

Note that for p ∈ M , we have Z g(p, t) =

(16.11)

H(p, y, t) g0 (y) dy ZM

≥

√ Bp ( t)

H(p, y, t) g0 (y) dy.

Applying the lower bound (Corollary 13.3), we obtain √ H(p, y, t) ≥ C Vp−1 ( t) √ for all y ∈ Bp ( t). Hence combining with (16.11) yields √ Z g(p, t) ≥ C Vp−1 ( t)

√ g0 (y) dy, Bp ( t)

and the lemma follows.



GEOMETRIC ANALYSIS

133

§17 Green’s Function m

Let M be a compact manifold of dimension m with boundary ∂M. Let ∆ be the Laplacian defined on functions with Dirichlet boundary condition on M. Then standard elliptic theory asserts that there exists a Green’s function G(x, y) defined on M × M \ D, where D = {(x, x) | x ∈ M }, so that Z G(x, y) ∆f (y) dy = −f (x)

(17.1) M

for all functions f satisfying the Dirichlet boundary condition f |∂M = 0. Moreover, G(x, y) = 0 for y ∈ ∂M and x ∈ M \∂M. Since both G and f satisfy Dirichlet boundary condition, after integration by parts, (17.1) becomes Z ∆y G(x, y) f (y) dy = −f (x)

(17.2) M

which is equivalent to saying that ∆y G(x, y) = −δx (y),

(17.3)

where δx (y) is the delta function at x. If we let f (x) = G(z, x), then (17.2) yields Z G(z, x) = −

∆y G(x, y) G(z, y) dy. M

However, applying (17.1) to the right hand side, we obtain (17.4)

G(z, x) = G(x, z).

This shows that G(x, y) must be symmetric in the variables x and y. We also observe that by letting f (x) = ∆x G(z, x) in (17.2) Z ∆x G(z, x) = − ∆y G(x, y) ∆y G(z, y) dy. M

Since the right hand side is symmetric in x and z, we conclude that ∆x G(z, x) = ∆z G(x, z). On the other hand, ∆z G(x, z) = ∆z G(z, x) by the symmetry of G, we conclude that ∆x G(z, x) = ∆z G(z, x). Therefore (17.2) can be written as Z (17.5)

∆x

Z G(x, y) f (y) dy =

M

∆x G(x, y) f (y) dy M

= −f (x)

134

PETER LI

Hence if we define the operator ∆−1 by ∆−1 f (x) = −

Z G(x, y) f (y) dy, M

then (17.1) and (17.5) gives ∆ ∆−1 = I and ∆−1 ∆ = I, respectively. When M m is a complete, non-compact manifold without boundary, one would like to obtain a Green’s function G(x, y) also satisfying the properties (17.1), (17.5) and (17.4) for any compactly supported function f ∈ Cc∞ (M ). For the case M = Rm , it is well known that the function ( ((m − 2)αm−1 )−1 r(x, y)2−m for m≥3 G(x, y) = −1 −(α1 ) log(r(x, y)) for m = 2, with αm−1 being the area of the unit (m − 1)-sphere in Rm , will satisfy these properties. In fact, these formulas, when interpreted appropriately, reflect the ideal situations for manifolds also. Let us now assume that M m has a point p around which the metric is rotationally symmetric. This is equivalent to saying that if we take polar coordinates (r, θ2 , . . . , θm ) around p, then ds2M = dr2 + gαβ dθα dθβ with gαβ (r) being functions of r alone. The Laplacian with respect to this coordinate system will take the form √ 1 ∂ g ∂ ∂2 + ∆∂Bp (r) ∆= 2 +√ ∂r g ∂r ∂r where g = det(gαβ ) and ∆∂Bp (r) denotes the Laplacian defined on the sphere, ∂Bp (r), of radius r centered at p. If we let Ap (r) to be the area of ∂Bp (r), then Z √ g dθ = Ap (r). Since

√

g is independent of the θα ’s,

√ A0p (r) 1 ∂ g = , √ g ∂r Ap (r)

and we have ∆=

A0p (r) ∂ ∂2 + + ∆∂Bp (r) . 2 ∂r Ap (r) ∂r

Let us now consider the function Z G(y) = − 1

r(y)

dt Ap (t)

where r(y) is the distance from p to y. Direct computation gives ∂ 2 G A0p (r) ∂G + ∂r2 Ap (r) ∂r =0

∆G =

GEOMETRIC ANALYSIS

135

for y 6= p. Moreover, if f ∈ Cc∞ (M ) is a compactly supported smooth function, then Z Z (17.6) G(y) ∆f (y) dy = ∆G(y) f (y) dy M \Bp ()

M \Bp ()

Z − ∂Bp () 

Z = 1

Z

dt Ap (t)

∂Bp ()

Z

∂G f dθ ∂Bp () ∂r Z ∂f 1 dθ − f dθ. ∂r Ap () ∂Bp ()

∂f dθ + ∂r

G(y)

Note that the continuity of f implies that Z

1 Ap ()

f dθ → f (p) ∂Bp ()

as  → 0. Also the continuity of ∆f and the divergence theorem imply that Z Z ∂f 1 1 dθ = ∆f dy Vp () ∂Bp () ∂r Vp () Bp () → ∆f (p) as  → 0, where Vp (r) is the volume of Bp (r). This implies that Z ∂Bp ()

∂f dθ ∼ ∆f (p) Vp (). ∂r

On the other hand, since Ap (t) ∼ αm−1 tm−1 where αm−1 is the area of the unit m − 1-sphere in Rm , we have  1   Z  − 2−m for  dt (m − 2) αm−1 ∼ 1  1 Ap (t)   log() for α1

m≥3 m = 2.

Therefore, by letting  → 0, the right hand side of (17.6) becomes Z lim

→0

1



dt Ap (t)

Z

hence

∂Bp ()

1 ∂f dθ − ∂r Ap ()

f dθ ∂Bp ()

Z G(y) ∆f (y) dy = −f (p). M

In particular, this implies that one can take Z (17.7)

G(p, y) = − 1

to be the Green’s function. Note that if

Z 1

∞

r(y)

!

Z

dt Ap (t)

dt 0 be any positive constant, the function ¯ y) (1 + α) gi (y) − ωi−1 G(p, ¯ and (17.10). It is also nonnegative on ∂Bp (ρ2 ). tends to ∞ as y → p because of the asymptotic behavior of G Hence the maximum principle implies that it must be nonnegative on Bp (ρ2 ). Letting α → 0, we obtain the lower bound for gi . The upper bound follows similarly by considering the function ¯ y) − 1 (1 − α) gi (y) − ωi−1 G(p, and using (17.9) and (17.11). The assumption that ωi → ∞, (17.12), and Lemma 17.1 imply that there is a subsequence of gi (y) that converges uniformly on compact subsets of Bp (ρ2 ) \ {p}. We also denote this subsequence by gi (y). Moreover the limiting function g is harmonic and bounded between 0 and 1. However, the removable singularity theorem for harmonic functions asserts that g can be extended to a harmonic function defined on Bp (ρ2 ). On the other hand, the property that si (r) is monotonically decreasing implies that sup gi (y) y∈∂Bp (r)

is decreasing in r. Passing to the limit, we conclude that sup

g(y)

y∈∂Bp (r)

is non-increasing in r. In particular, g(p) ≥ supy∈∂Bp (r) g(y) for all r > 0, and p must be a local maximum for g. However, this violates the maximum principle unless g is identically constant. On the other hand, the fact that si (r) is monotonically decreasing implies that the supremum of gi on M \ Bp (ρ1 ) is achieved on ∂Bp (ρ1 ). Using (17.9) and (17.11), we conclude that 1 − gi (y) is nonnegative on M \ Bp (ρ1 ). Using Lemma 17.1, we conclude that 1 − gi must have a uniformly convergent subsequence since gi → g on Bp (ρ2 ) \ {p}. Hence gi → g which is a constant function on M. However, by passing to the limit (17.9) implies that oscy∈Bp (ρ2 )\Bp (ρ1 ) g(y) = 1, which is a contradiction. Hence the oscillation of G(p, y) must be bounded on Bp (ρ2 ) \ Bp (ρ1 ).  Theorem 17.1. Let M m be a complete manifold without boundary. There exists a Green’s function G(x, y) which is smooth on (M × M ) \ D satisfying properties (17.1), (17.4), and (17.5). Moreover, G(x, y) can be taken to be positive if and only if there exists a positive superharmonic function f on M \ Bp (ρ) with the property that lim inf f (x) < inf f (x). x→∞

x∈∂Bp (ρ)

GEOMETRIC ANALYSIS

139

Proof. Let us first observe that if G(x, y) is a positive symmetric Green’s function, then the maximum principle asserts that the infimum of G(p, y) must achieved at infinity of M. The function f (y) given by the restriction of G(p, y) on M \ Bp (ρ) is a positive harmonic function with the desired property. We will first construct G(p, y) assuming that there is a nonconstant, positive, superharmonic function f defined on M \ Bp (ρ) whose infimum is achieved at infinity of M. Let {Ωi }∞ i=1 be a compact exhaustion of M with the property that Bp (ρ) ⊂ Ωi for all i. Let Gi (p, y) be the Dirichlet Green’s function defined on Ωi . Note that Gi (p, y) are monotonically increasing. In fact, the maximum principle and the asymptotic behavior of Gi (p, y) as y → p asserts that (1 + α)Gj (p, y) ≥ Gi (p, y) for α > 0 and i < j, and the monotonicity follows by taking α → 0. As before, let us denote si (ρ) =

sup

Gi (p, y),

y∈∂Bp (ρ)

then si (ρ) is a monotonically increasing sequence in i. We now claim that si (ρ) is bounded from above. To see this, let us consider the sequence of functions s−1 i (ρ) Gi (p, y). By possible addition and multiplication by constants, we may assume that inf

f (x) = 1

x∈∂Bp (ρ)

and (17.13)

lim inf f (x) = 0. x→∞

The maximum principle asserts that s−1 i (ρ) Gi (p, y) ≤ f (y)

on

M \ Bp (ρ).

By passing to a subsequence, the sequence of functions {s−1 i (ρ) Gi (p, y)} converges uniformly on compact subsets of M \ Bp (ρ) to a harmonic function g satisfying the properties that g(y) ≤ f (y)

on M \ Bp (ρ)

and (17.14)

sup

g(x) = 1.

x∈∂Bp (ρ)

Lemma 17.1 asserts that the subsequence, in fact, converges on compact subset of M \{p} and g is a harmonic function on M \ {p}. However, the maximum principle implies that −1 −1 ¯ ¯ s−1 i (ρ) G(p, y) ≤ si (ρ) Gi (p, y) ≤ si (ρ) G(p, y) + 1

¯ is the Dirichlet Green’s function on Bp (ρ). If si (ρ) → ∞, then similar to the argument of Lemma where G 17.2 we conclude that g must be identically 0 due to (17.13), contradicting (17.14). Hence si (ρ) must be bounded.

140

PETER LI

Let us denote s(ρ) = limi→∞ si (ρ). The maximum principle again implies that Gi (p, y) ≤ s(ρ) f (y)

M \ Bp (ρ).

on

Arguing as before, we conclude that the monotonic sequence Gi (p, y) converges uniformly on compact subsets of M \ {p} to a harmonic function G(p, y). In particular, given any  > 0, for sufficiently large i, we have 0 ≤ G(p, y) − Gi (p, y) ≤ 

on

∂Bp (1).

The maximum principle then asserts that Gi (p, y) ≤ G(p, y) ≤ Gi (p, y) + 

Bp (1) \ {p}.

on

In particular, if f ∈ Cc∞ (M ), then Z

Z G(p, y) ∆f (y) dy −

M

ZM ≤ M

Z ≤

Gi (p, y) ∆f (y) |G(p, y) − Gi (p, y)| |∆f |(y) dy Z |∆f |(y) dy + |G(p, y) − Gi (p, y)| |∆f |(y) dy. M \Bp (1)

Bp (1)

Note that the second term on the right hand side tends to 0 as i → ∞, and using Z Gi (p, y) ∆f (y) dy = −f (p), M

we conclude that

Z

M

Z G(p, y) ∆f (y) dy + f (p) ≤ 

|∆f |(y) dy.

Bp (1)

Letting  → 0, this yields property (17.1) for G(p, y). Observe that the above argument shows that Gi (q, y) → G(q, y) as long as q ∈ Bp ( ρ2 ). For arbitrary q ∈ M, we just need to take ρ sufficiently large such that q ∈ Bp ( ρ2 ). Hence Gi (q, y) → G(q, y) for any q ∈ M. Since each Gi satisfies (17.4) and (17.5), G will have the same properties. Note that the function G obtained in this manner will have the minimal property, namely, ˜ y) ≥ G(x, y), G(x, ˜ Indeed, since the maximum principle implies that for any positive Green’s function G. ˜ y) ≥ Gi (x, y) G(x,

on

Ωi ,

the minimal property of G follows by taking the limit as i → ∞. We now assume that f does not exist on M \ Bp (ρ). This is equivalent to saying that si (ρ) → ∞

GEOMETRIC ANALYSIS

141

as i → ∞ for all fixed ρ. Let us denote ai =

inf y∈∂Bp (1)

Gi (p, y).

The sequence of functions {hi (p, y) = Gi (p, y) − ai } will have the property that inf y∈∂Bp (1)

hi (p, y) = 0.

Also, Lemma 17.2 asserts that there is a constant ω such that the oscillations of Gi (p, y) over the set Bp (2) \ Bp ( 21 ) is bounded by ωi (1, 2) ≤ ω. In particular, hi (p, y) ≤ ω

sup y∈∂Bp (1)

for all i. Hence the sequence of positive harmonic functions {hi + ω} when restricted to Bp (2) \ {p} has a convergent subsequence, also denoted by {hi + ω}, such that {hi (p, y) + ω} converges uniformly on compact subsets of Bp (2) \ {p} to a harmonic function G(p, y) + ω. Similarly, the sequence of positive harmonic functions {ω − hi } when restricted to M \ Bp ( 21 ) has a subsequence converging to ω − G1 (p, y) on compact subsets of M \ Bp ( 21 ). Since G1 (p, y) = G(p, y) on Bp (2) \ Bp ( 12 ), we can denote G1 by G also, and G(p, y) is the uniform limit of hi (p, y) on compact subsets of M \ {p}. Moreover, since Z

Z Gi (p, y) ∆f (y) dy − ai

hi (p, y) ∆f (y) dy = M

Z

M

∆f (y) dy M

= −f (p) for f ∈ Cc∞ (M ), the previous argument implies that G(p, y) also satisfies property (17.1). We also observe ¯ y) is the Dirichlet Green’s function on Bp (1), then the maximum principle implies that that if G(p, ¯ y) ≤ hi (p, y) ≤ G(p, ¯ y) + ω. G(p, Hence by passing through the limit, we conclude that ¯ y) ≤ G(p, y) ≤ G(p, ¯ y) + ω. G(p,

(17.15)

For another point q 6= p, the same argument implies that there is a subsequence ij of i and a sequence of constants bij such that Gij (q, y) − bij converges uniformly on compact subset of M \ {q}. Setting y = p, we observe that since (Gij (q, p) − bij ) = (Gij (p, q) − aij ) + (aij − bij ), by taking the limit as j → ∞, we conclude that lim Gij (q, p) = G(p, q) + lim (aij − bij ).

j→∞

j→∞

Hence the limit lim (aij − bij ) = c

j→∞

142

PETER LI

must exist and the sequence of functions Gij (q, y) − aij = (Gij (q, y) − bij ) − (aij − bij ) also converges. Let us denote lim Gij (q, y) − aij = J(q, y).

j→∞

Again, setting y = p, we conclude that G(p, q) = J(q, p). We now claim that the original sequence Gi (q, y) − ai must converge to J(q, y) also. To see this, it suffices to show that if Gik (q, y) − aik is any other convergent subsequence of i, then it must converge to J(q, y). Let lim (Gik (q, y) − aik ) = K(q, y)

k→∞

denote its limit. The same argument as above shows that K(q, p) = G(p, q), hence (17.16)

K(q, p) = J(q, p).

On the other hand, if we let ρ be sufficiently large so that Bq (1) ⊂ Bp (ρ), then G(p, y) − K(q, y) = lim (Gik (p, y) − aik ) − lim (Gik (q, y) − aik ) k→∞

k→∞

= lim (Gik (p, y) − (Gik (q, y)). k→∞

Since Gik satisfies Dirichlet boundary condition on Ωik , for sufficiently large k such that Bp (ρ) ⊂ Ωik , the maximum principle asserts that the harmonic function Gik (p, y) − Gik (q, y) when restricted on Ωik \ Bp (ρ) must have its maximum achieved on ∂Bp (ρ). Passing to the limit, we conclude that sup (G(p, y) − K(q, y)) = sup (G(p, y) − K(q, y)) y∈M \Bp (ρ)

y∈∂Bp (ρ)

and G(p, y) − K(q, y) is bounded on M \ Bp (ρ). A similar argument implies that G(p, y) − J(q, y) is also bounded on M \ Bp (ρ) and therefore K(q, y) − J(q, y) is bounded on M \ Bp (ρ). This together with estimate (17.15), when applied to both K and J at the point q, implies that K(q, y) − J(q, y) is bounded on M \ {q}. Again the removable singularity theorem asserts that K(q, y) − J(q, y) is a bounded harmonic function in y. However, this will imply that M admits a harmonic function on M \ Bp (ρ) with its infimum achieved at infinity unless K(q, y) − J(q, y) is identically constant. The identity (17.16) implies that K(q, y) = J(q, y). In particular, we write this as G(q, y). Properties (17.4) and (17.5) obviously follows since G is the limit of functions satisfying these properties. 

GEOMETRIC ANALYSIS

143

§18 Parabolicity In view of the construction in §17, the existence and nonexistence of a positive Green’s function divides the class of complete manifolds into two categories. In general the methods in dealing with function theory on these manifolds are different, hence it is important to understand the difference between the two categories. Definition 18.1. A complete manifold is said to be parabolic if it does not admit a positive Green’s function. Otherwise it is said to be nonparabolic. As pointed out in Theorem 17.1, a manifold is nonparabolic if and only if there exists a positive superharmonic function whose infimum is achieved at infinity. This property can be localized at any unbounded component at infinity. Definition 18.2. An end, E, with respect to a compact subset Ω ⊂ M is an unbounded connected component of M \ Ω. The number of ends with respect of Ω, denoted by NΩ (M ), is the number of unbounded connected components of M \ Ω. It is obvious that if Ω1 ⊂ Ω2 , then NΩ1 (M ) ≤ NΩ2 (M ). Hence if {Ωi } is a compact exhaustion of M , then NΩi (M ) is a monotonically nondecreasing sequence. If this sequence is bounded, then we say that M has finitely many ends. In this case, we denote the number of ends of M by N (M ) = max NΩi (M ). i→∞

One readily checks that this is independent of the compact exhaustion {Ωi }. In fact, it is also easy to see that there must be an i such that N (M ) = NΩi (M ). Moreover, for any compact set Ω containing Ωi , NΩ (M ) = N (M ). Hence for all practical purposes, we may assume that M \ Bp (ρ0 ) has N (M ) number of unbounded connect components, for some ρ0 . In general, when we say that E is an end we mean that it is an end with respect to some compact subset ¯ Ω. In particular, its boundary ∂E is given by ∂Ω ∩ E. Definition 18.3. An end E is said to be parabolic if it does not admit a positive harmonic function f satisfying f =1 on ∂E and lim inf f (y) < 1, y→E(∞)

where E(∞) denotes the infinity of E. Otherwise, E is said to be nonparabolic and the function f is said to be a barrier function of E. Observe that by possibly subtraction and multiplication of constants, we may assume that the barrier function satisfies lim inf f (y) = 0. y→E(∞)

With this notion, we can also count the number of nonparabolic (parabolic) ends as in the definition of N (M ). Definition 18.4. Let {Ωi } be a compact exhaustion of M . We denote Ni0 (M ) to be the number of nonparabolic ends with respect to the compact set Ωi . We say that M has finitely many nonparabolic ends if the sequence {Ni0 (M )} is bounded. In this case, we say that N 0 (M ) = limi→∞ Ni0 (M ) is the number of nonparabolic ends of M .

144

PETER LI

Definition 18.5. Let {Ωi } be a compact exhaustion of M . We denote Ni0 (M ) to be the number of parabolic ends with respect to the compact set Ωi . We say that M has finitely many parabolic ends if the sequence {Ni0 (M )} is bounded. In this case, we say that N 0 (M ) = limi→∞ Ni0 (M ) is the number of parabolic ends of M. Note that if E is a nonparabolic end of M , then by extending f to be identically 1 on (M \ Ω) \ E, it can be used to construct a positive Green’s function on M. Hence, M is nonparabolic if and only if M has a nonparabolic end. Of course, it is possible for a nonparabolic manifold to have many parabolic ends. Let us also point out that E being nonparabolic is equivalent to saying that E has a positive Green’s function with Neumann boundary condition. This can be seen by repeating the contruction of Green’s function in the previous section on a manifold with boundary by taking Neumann boundary condition. One needs to check that the maximum principle arguments are still valid in this case because of the presence of ∂E. However, the Hopf boundary lemma asserts that at a boundary maximum point of a harmonic function f , the outward pointing normal derivative must be positive violating the Neumann boundary condition. For the purpose of application, it is often important to obtain appropriate estimates on the barrier function. We will give a canonical method of constructing barrier functions which will help in obtaining estimates later on. Lemma 18.1. An end E with respect to the compact set Bp (ρ0 ) is nonparabolic if and only if the sequence of positive harmonic functions {fi }, defined on Ep (ρi ) = E ∩ Bp (ρi ) for ρ0 < ρ1 < ρ2 < · · · → ∞, satisfying fi = 1

on

∂E

and fi = 0

∂Bp (ρi ) ∩ E,

on

converges uniformly on compact subsets of E ∪ ∂E to a barrier function f . The barrier constructed in this manner is minimal among all barrier functions, and f has finite Dirichlet integral on E. Proof. If E is nonparabolic, then there exists a barrier function g satisfying g=1

on

∂E

and lim inf g(y) = 0. y→E(∞)

For each i, the maximum principle asserts that fi ≤ g

Bp (ρi ) ∩ E

on

and the sequence of functions {fi } is monotonically increasing in i. Lemma 17.1 implies that fi must converge uniformly on compact subsets of E ∪ ∂E to a nonnegative function f satisfying (18.1)

f ≤g

on

E

with boundary condition f =1

on

∂E.

and lim inf f (y) = 0. y→E(∞)

Moreover, f is harmonic on E and we claim that the convergence is Lipschitz up to the boundary ∂E. To see this, it suffices to show that |∇ log fi |2 are also uniformly bounded on compact subsets of E ∪ ∂E. Following the argument for the gradient estimate in §6, we can consider the function φ |∇ log fi |2

GEOMETRIC ANALYSIS

where φ(r) is the cutoff function satisfying the properties  1 for φ(r) = 0 for

145

r ≤ R0 r ≥ 2R0 ,

−C R0−1 ≤ φ0 ≤ 0, and |φ00 | ≤ C R0−2 . If the maximum point of this function occurs in the interior of E, then the same argument as in Theorem 6.1 will give the estimate of |∇ log fi |2 up to ∂E. Therefore, we only need to show that we can still estimate |∇ log fi |2 if its maximum point occurs on ∂E. If x0 ∈ ∂E is the maximum point of φ |∇ log fi |2 , hence a maximum point of |∇ log fi |2 , then by setting h = log fi , the strong maximum principle asserts that ∂|∇h|2 (x0 ) < 0. ∂r On the other hand, since h = 0 on ∂E, we can choosing orthonormal frame {e1 , . . . , em } at x0 such that ∇h ∂ e1 = |∇h| = − ∂r . Hence we have ∂|∇h|2 ∂r = −2h1 h11

(18.2)

0>

= −2|∇h| h11 . However, the Laplacian of h is given by −|∇h|2 = ∆h = h11 + H h1 , where H is the mean curvature of ∂E with respect to the outward normal vector e1 . Therefore combining with (18.2), we have 0 > −|∇h| (−|∇h|2 − H h1 ) = |∇h|3 + H |∇h|2 . This implies that the maximum point cannot occur on ∂E if min∂E H ≥ 0. If −H0 = min∂E H is negative, then |∇h|2 (x0 ) ≤ H02 . Since x0 is the maximum point for φ |∇h|2 , we obtain an estimate of |∇h|2 on ∂E. Note that in this argument we assume that ∂E is smooth so that H0 < ∞. If this is not the case, we can always smooth out ∂E and the statement of the lemma is still valid for the smooth boundary. In particular, this proved that f is a barrier function and the minimal property of f follows from (18.1) because it applies to any barrier function g. The boundary conditions on fi and the fact that it is harmonic imply that Z Z ∂fi 2 |∇fi | = − Ep (ρi ) ∂E ∂r for each i. Hence for any fixed ρ < ρi , Z Ep (ρ)

|∇fi |2 ≤ −

Z ∂E

∂fi . ∂r

146

PETER LI

Since the right hand side is uniformly bounded according to the above gradient estimate, by letting i → ∞, we conclude that Z |∇f |2 ≤ C, Ep (ρ)

for some constant 0 < C < ∞. In fact, since f is harmonic, we see that Z ∂E

∂f = ∂r

Z ∂Bp (ρ)∩E

∂f ∂r

Z = lim

i→∞

∂Bp (ρ)∩E

Z = lim

i→∞

∂E

∂fi ∂r

∂fi , ∂r

R hence C can be taken to be ∂E ∂f ∂r . Conversely, if the sequence fi converges uniformly on compact subsets of E to a barrier function f then E is nonparabolic by definition.  If E is parabolic, the sequence fi will still converge to a harmonic function f because of Lemma 17.1. However, f will be identically 1. In this case, by renormilizing the sequence {fi }, we can still construct a harmonic function on E that reflects the parabolicity property of E. Lemma 18.2. Let E be a parabolic end with respect to Bp (ρ0 ). Let ρi be an increasing sequence such that ρ0 < ρ1 < ρ2 < · · · < ρi → ∞. There exists a sequence of constants Ci → ∞ such the sequence of positive harmonic functions gi , defined on Ep (ρi ) = Bp (ρi ) ∩ E, satisfying gi = 0

on

∂E

and gi = Ci

∂Bp (ρi ) ∩ E

on

has a convergent subsequence that converges uniformly on compact subsets of E ∪ ∂E to a positive harmonic function g. Moreover g will have the properties that g=0

on

∂E

and sup g = ∞. y∈E

Proof. The fact that E is parabolic implies that the sequence fi in Lemma 17.1 converges to the constant function 1. Let us define gi = Ci (1 − fi ). Obviously gi will have the appropriate boundary conditions. If we set Ci = (1 −

inf ∂Bp (ρ1 )∩E

fi )−1 ,

then sup∂Bp (ρ1 )∩E gi = 1. Since fi → 1, Ci → ∞. Also Lemma 17.1 implies that, by passing to a subsequence, gi → g on compact subset of E to a harmonic function. We will also call g as a barrier function of E.  Using this characterization of parabolicity, a theorem of Royden [Ro] follows as a corollary.

GEOMETRIC ANALYSIS

147

Corollary 18.1. (Royden) Parabolicity is a quasi-isometric invariant. Proof. It suffices to show that if E is a nonparabolic end with respect to the metric ds2 , then for any other metric ds21 which is uniformly equivalent to ds2 , the end E remains nonparabolic with respect to ds21 . Let us denote ∇1 , ∆1 and ν1 to be the gradient, the Laplacian, and the unit outward normal to ∂E with respect to the metric ds21 . According to Lemma 18.1, there exists a sequence of harmonic functions, {fi }, with respect to ds2 , converging to a non-constant harmonic function f . Moreover, they satisfy the property stated in Lemma 18.1. In particular, Z Z ∂fi dA = |∇fi |2 dV. (18.3) Ep (ρi ) ∂E ∂ν On the other hand, let hi be the harmonic function with respect to ds21 satisfying hi = 1

on ∂E

and hi = 0

∂Bp (ρi ) ∩ E,

on

then we also have Z (18.4) ∂E

∂hi dA1 = ∂ν1

Z

|∇1 hi |2 dV1 .

Ep (ρi )

However, the fact that there exists a constant C > 1 such that C −1 ds2 ≤ ds21 ≤ C ds2 implies that Z

|∇1 hi |2 dV1 ≥ C1

(18.5) Ep (ρi )

Z

|∇hi |2 dV

Ep (ρi )

for some constant C1 > 0. On the other hand, the Dirichlet integral minimizing property of harmonic function implies that Z Z |∇hi |2 dV ≥

Ep (ρi )

|∇fi |2 dV.

Ep (ρi )

Hence combining with (18.3), (18.4), and (18.5) we conclude that Z Z ∂fi ∂hi dA1 ≥ C1 dA. ∂ν 1 ∂E ∂ν ∂E Since we know that fi → f by passing to a subsequence, and similarly hi → h, we have Z Z ∂h ∂f dA1 ≥ C1 dA. ∂ν 1 ∂E ∂E ∂ν The fact that f is harmonic and nonconstant implies that the right hand side must be positive. On the other hand, we also know that h is harmonic and the fact that Z ∂h dA1 > 0 ∂E ∂ν1 implies that h is non-constant, hence its is a barrier function with respect to the metric ds21 .  The following theorem from [LT4] gives a necessary condition for a manifold to be nonparabolic.

148

PETER LI

Theorem 18.1. (Li-Tam) Let M be a complete manifold. If M is nonparabolic then for any point p ∈ M Z ∞ dt < ∞, A p (t) 1 where Ap (r) denotes the area of ∂Bp (r). Moreover, if G(p, y) is the minimal Green’s function, then Z r dt ≤ sup G(p, y) − inf G(p, y) A y∈∂Bp (r) p (t) y∈∂Bp (1) 1 for all r > 1. Proof. For p ∈ M , let G(p, y) be the minimal positive Green’s function with a pole at p. In view of the construction in §17, we may assume that G(p, y) is the limit of the sequence {Gi (p, y)}, where Gi is the Dirichlet Green’s function defined on Bp (ρi ). For any 1 < ρ < ρi , let us denote si (1) = sup Gi (p, y) y∈∂Bp (1)

and ii (ρ) =

inf y∈∂Bp (ρ)

Gi (p, y).

Let f be the harmonic function defined on Bp (ρ) \ Bp (1) satisfying the boundary conditions f (y) = si (1)

on

∂Bp (1)

and f (y) = Gi (p, y)

on

∂Bp (ρ).

The maximum principle implies that f (y) ≥ Gi (p, y)

Bp (ρ) \ Bp (1).

on

In particular, we have (18.6)

∂Gi ∂f ≤ ∂r ∂r

on

∂Bp (ρ).

On the other hand, since f (y) is harmonic, Stokes theorem implies that Z 0= ∆f Bp (ρ)\Bp (1)

Z = ∂Bp (ρ)

∂f − ∂r

Z ∂Bp (1)

∂f . ∂r

Also, we observe that Z ∂Bp (ρ)

∂Gi = ∂r

Z ∆Gi Bp (ρ)

= −1 and (18.6) implies that Z (18.7) ∂Bp (1)

∂f ≤ −1. ∂r

GEOMETRIC ANALYSIS

149

Let us now consider a harmonic function h defined on Bp (ρ) \ Bp (1) satisfying the boundary conditions h(y) = si (1)

on

∂Bp (1)

h(y) = ii (ρ)

on

∂Bp (ρ).

and Again the maximum principle implies that h(y) ≤ f (y)

Bp (ρ) \ Bp (1),

on

and

∂f ∂h ≤ on ∂Bp (1). ∂r ∂r Hence combining with (18.7), we obtain Z Z ∂h ∂h (18.8) = ∂Bp (1) ∂r ∂Bp (ρ) ∂r ≤ −1. If we define the function ρ

Z g(r) = (si (1) − ii (ρ)) 1

dt Ap (t)

−1 Z

ρ

r

dt + ii (ρ) Ap (t)

then g(r(y)) will have the same boundary conditions as h(y). The Dirichlet integral minimizing property for harmonic functions implies that Z Z 2 |∇h| ≤ |∇g|2 (18.9) Bp (ρ)\Bp (1)

Bp (ρ)\Bp (1)

Z

ρ

Z (si (1) − ii (ρ))

= 1

1

= (si (1) − ii (ρ))

2

Z

ρ

1

dt Ap (t) −1

dt Ap (t)

On the other hand, integration by parts and (18.8) yield Z Z |∇h|2 = ii (ρ) Bp (ρ)\Bp (1)

ρ

∂Bp (ρ)

−1

1 Ap (r)

!2 Ap (r) dr

.

∂h − si (1) ∂r

Z ∂Bp (1)

∂h ∂r

≥ si (1) − ii (ρ) because si (1) ≥ si (ρ) ≥ ii (ρ), since si (r) =

sup

Gi (p, y)

y∈∂Bp (r)

is a decreasing function of r. Combining this with (18.9) gives the estimate Z ρ dt ≤ si (1) − ii (ρ). A p (t) 1 Letting i → ∞ and using the fact that G is nonconstant, we arrive with the estimate Z ρ dt ≤ s(1) − i(ρ) 1 Ap (t)

150

PETER LI

where s(1) =

sup

G(p, y)

y∈∂Bp (1)

and i(ρ) =

inf

G(p, y).

y∈∂Bp (ρ)

Letting ρ → ∞, the finiteness of Z

∞

1

follows.

dt Ap (t)



The following lemma gives a criterion for an end to be nonparabolic. It was first proved by Cao, Shen and Zhu in [CSZ] for minimal submanifolds. However, their arugment can be generalized to the following context. Lemma 18.3. Let E be an end of a complete Riemannian manifold. Suppose for some ν ≥ 1 and C > 0, E satisfies a Sobolev type inequality of the form Z  ν1 Z 2ν |u| ≤C |∇u|2 E

for all compactly supported function u ∈ nonparabolic.

E

Hc1,2 (E)

defined on E, then E must either have finite volume or be

Proof. Let E be an end of M . For ρ sufficiently large, let us consider the set E(ρ) = E ∩ Bp (ρ), where Bp (ρ) is the geodesic ball of radius ρ in M centered at some point p ∈ M. Let us denote by r the distance function of M to the point p. Suppose the function fρ is the solution to the equation ∆fρ = 0

on

E(ρ),

fρ = 1

on

∂E

with boundary conditions and fρ = 0

E ∩ ∂Bp (ρ).

on

According to Lemma 18.1, fρ converges to a nonconstant harmonic function f if and only if E is nonparabolic. For a fixed 0 < ρ0 < ρ such that E(ρ0 ) 6= ∅, let φ be a nonnegative cutoff function satisfying the properties that φ=1 on E(ρ) \ E(ρ0 ), φ=0

on

∂E,

and |∇φ| ≤ C1 . Applying the inequality in the assumption and using the fact that fρ is harmonic, we obtain ! ν1 Z Z 2ν (φ fρ ) ≤C |∇(φ fρ )|2 E(ρ)

E(ρ)

Z

|∇φ|2 fρ2 + 2

=C E(ρ)

Z

|∇φ|

fρ2

E(ρ)

Z =C E(ρ)

|∇φ|2 fρ2 .

1 + 2

!

Z

φ2 |∇fρ |2

φ fρ h∇φ, ∇fρ i + E(ρ)

2

=C

Z

E(ρ)

Z h∇(φ E(ρ)

2

), ∇(fρ2 )i

Z

! 2

2

φ |∇fρ |

+ E(ρ)

GEOMETRIC ANALYSIS

151

In particular, for a fixed ρ1 satisfying ρ0 < ρ1 < ρ, we have ! ν1

Z

fρ2ν

Z

fρ2 .

≤ C2

E(ρ1 )\E(ρ0 )

E(ρ0 )

If E is parabolic, then the limiting function f is identically 1. Letting ρ → ∞, we obtain 1

(VE (ρ1 ) − VE (ρ0 )) ν ≤ C VE (ρ0 ), where VE (ρ) denotes the volume of the set E(ρ). Since ρ1 > ρ0 is arbitrary, this implies that E must have finite volume and the theorem is proved.  Lemma 18.4. Let M be a complete Riemannian manifold. Given a geodesic ball Bp (ρ0 ), suppose there exists constants ν > 1 and C > 0, such that a Sobolev type inequality of the form ! ν1

Z

2ν

Z

|∇u|2

≤C

|u|

Bp (ρ0 )

Bp (ρ0 )

is valid for all compactly supported functions u ∈ Hc1,2 (Bp (ρ0 )). Then there exists a constant C1 > 0 depending only on C and ν, such that, 2ν

Vp (ρ0 ) ≥ C1 ρ0ν−1 . In particular, if an end E of M with respect to the compact set Bp (ρ0 ) satisfies the Sobolev inequality Z

2ν

 ν1

|u| E

Z ≤C

|∇u|2

E

for all compactly supported functions u ∈ Hc1,2 (E), then the end must have volume growth given by 2ν

VE (ρ0 + ρ) ≥ C1 ρ ν−1 for all ρ > 0, where VE (ρ0 + ρ) denotes the volume of the set Ep (ρ0 + ρ). Proof. For a fixed point p ∈ M and 0 < ρ < ρ0 < ρ0 , let us consider the function

u(x) =

 1     ρ0 − r(x)  ρ0 − ρ    0

on Bp (ρ) on Bp (ρ0 ) \ Bp (ρ) on M \ Bp (ρ0 ).

where r(x) is the distance function to the point p. Plugging this into the Sobolev inequality, we conclude that 1

Vpν (ρ) ≤ C(ρ0 − ρ)−2 Vp (ρ0 ).

(18.10) Let us define ρk = (1 −

Pk

i=1

2−(i+1) )ρ0 for k = 1, 2, . . . . Using ρ0 = ρk−1 and ρ = ρk , (18.10) becomes 1

Vpν (ρk ) ≤ C4k+1 ρ−2 0 Vp (ρk−1 ).

152

PETER LI

This can be rewritten as  ν1

 (18.11)

2ν

 Vp (ρk ) 

k+1

≤ C4

2ν

ρkν−1

= C4k+1

ρ−2 0

ν−1 ρk−1 Vp (ρk−1 ) 2

2ν

ν−1 ρkν−1 ρk−1 Pk−1 2ν (1 − i=1 2−(i+1) ) ν−1 Vp (ρk−1 ) . Pk 2ν 2 ν−1 (1 − i=1 2−(i+1) ) ν−1 ρk−1

Using the estimate k X 1 ≤1− 2−(i+1) ≤ 1 2 i=1

for all 1 ≤ k (18.11) yields  ν1



Vp (ρk−1 ) k)   Vp (ρ ≤ C1 4k+1 . 2ν 2ν ν−1 ν−1 ρk ρk−1 where C1 > 0 depends on C and ν. Iterating this inequality k-times, we obtain  (18.12)

ν −k

k)   Vp (ρ 2ν ν−1 ρk

Pk ≤ C1

i=1

ν −i

Pk −i V (ρ ) p 0 . 4 i=1 (i+1)ν 2ν ρ0ν−1

Observing that ρk → ρ20 and ν −k → 0 as k → ∞, we conclude that the left hand side of (18.12) tends to 1. Hence (18.12) becomes P∞ −i P∞ 2ν −i − ν C1 i=1 4− i=1 (i+1)ν ρ0ν−1 ≤ Vp (ρ0 ). The lower bound of the volume follows by observing that ∞ X

ν −i
1. If the Sobolev inequality is valid on an end E with respect to Bp (ρ0 ), then we apply the above volume estimate to the ball Bx (ρ) with x ∈ ∂Bp (ρ0 + ρ) ∩ E. Hence we obtain VE (ρ0 + 2ρ) ≥ Vx (ρ) 2ν

≥ C1 ρ ν−1 , and the volume growth of E follows.



Combining Lemma 18.3 and Lemma 18.4, we obtain the following corollary for the Sobolev type inequality with ν > 1. The case when ν = 1 is just the Dirichlet Poincar´e inequality and will be addressed separately in a later section. In that case, it is possible to have a finite volume end given by a cusp. Corollary 18.2. Let E be an end of a complete Riemannian manifold. Suppose that for some ν > 1 and C > 0, E satisfies a Sobolev type inequality of the form Z

|u|2ν

 ν1

E

for all compactly supported functions u ∈

Hc1,2 (E)

Z ≤C

|∇u|2 ,

E

defined on E, then E must be nonparabolic.

GEOMETRIC ANALYSIS

153

§19 Harmonic Functions and Ends In this section, we will construct harmonic functions defined on M by extending the barrier functions defined at all of the ends of M. This construction was first proved by Tam and Li in [LT1] for manifolds with nonnegative sectional curvature near infinity. They later gave a construction for arbitrary complete manifolds in [LT6]. In [STW], Sung, Tam, and Wang presented the construction in a more systematic manner. It is their version that is given here. Theorem 19.1. (Li-Tam and Sung-Tam-Wang) Let M be a complete manifold and let Ω be a smooth compact subdomain in M. Suppose g is a harmonic function defined on M \ Ω which is smooth up to the boundary ∂Ω of Ω. If M is nonparabolic, then there exists a harmonic function h defined on M and a constant C > 0, such that, |g(x) − h(x)| ≤ C G(p, x), where G(p, x) is the minimal positive Green’s function and p ∈ Ω. Moreover, the function g − h has finite Dirichlet integral on M \ Ω. Proof. Let p ∈ Ω be a fixed point and let ρ0 > 0 be such that Ω ⊂ Bp (ρ0 ). Let φ be a smooth nonnegative function satisfying ( 0 on Bp (ρ0 ) φ= 1 on M \ Bp (2ρ0 ). If M is nonparabolic and G(x, y) is the minimal positive Green’s function on M, then we define Z h(x) = φ(x) g(x) + G(x, y) ∆(φ g)(y) dy. M

Since ∆(φ g) = ∆g = 0

M \ Bp (2ρ0 ),

on

the function ∆(φ g) has compact support and h is harmonic on M. Observing that for x ∈ / Bp (4ρ0 ), we have |h(x) − g(x)| = |h(x) − φ(x) g(x)| Z G(x, y) ∆(φ g)(y) dy = Bp (2ρ0 ) Z ≤ sup G(x, y) |∆(φ g)|(y) dy. y∈Bp (2ρ0 )

Bp (2ρ0 )

Since x ∈ / Bp (4ρ0 ), G(x, y) as a function of y is harmonic on Bp (4ρ0 ). Hence the local Harnack inequality asserts that G(x, y) ≤ C G(x, p), where C is a constant depending only on the lower bound of the Ricci curvature of Bp (4ρ0 ), m, and ρ0 . This proves that |h(x) − g(x)| ≤ C G(p, x) for all x ∈ / Bp (4ρ0 ). Since G is positive, compactness implies that the same bound is valid for all x ∈ M by possibly adjusting the constant C. Note that from the construction of G in §17, the bound on the function |h − g| can be written as (19.1)

|h(x) − g(x)| ≤ C f (x)

on

M \ Bp (4ρ0 )

where f is the minimal barrier function of §18 satisfying the properties that f =1

on

∂Bp (4ρ0 ),

154

PETER LI

inf f (y) = 0,

y→∞

and

Z

|∇f |2 < ∞.

M \Bp (4ρ0 )

For any ρ > 4ρ0 , let uρ be the harmonic function defined on Bp (ρ) \ Bp (4ρ0 ) satisfying the boundary conditions uρ = h − g

(19.2)

on

∂Bp (4ρ0 ),

and (19.3)

uρ = 0

on

∂Bp (ρ).

Since (19.1) implies that |uρ − (h − g)| ≤ C f

∂Bp (4ρ0 ) ∪ ∂Bp (ρ),

on

the maximum principle asserts that |uρ − (h − g)| ≤ C f

Bp (ρ) \ Bp (4ρ0 ).

on

Hence there exists a sequence of {ρi } such that ui − (h − g), with ui = uρi , converges uniformly on compact subset of M \ Bp (4ρ0 ) to a harmonic function u − (h − g). Moreover, u − (h − g) = 0

on

∂Bp (4ρ0 )

and |u − (h − g)| ≤ C f. In particular, the harmonic function f − C −1 (u − (h − g)) is nonnegative and has boundary value 1 on ∂Bp (4ρ0 ). Hence by the minimality property of f , f ≤ f − C −1 (u − (h − g)) and u − (h − g) ≤ 0. Running through the same argument using the function (h − g) − u, we conclude that (h − g) − u ≤ 0, hence h − g = u. In particular, the sequence of functions {ui } converges to h − g. However, since Z Z ∂ui |∇ui |2 = − ui ∂r Bp (ρi )\Bp (4ρ0 ) ∂Bp (4ρ0 ) Z ∂ui =− (h − g) , ∂r ∂Bp (4ρ0 ) we conclude that for any ρ > 4ρ0 Z

|∇ui |2 ≤ −

Bp (ρ)\Bp (4ρ0 )

Z (h − g)

∂ui . ∂r

(h − g)

∂(h − g) . ∂r

∂Bp (4ρ0 )

Letting i → ∞ yields Z Bp (ρ)\Bp (4ρ0 )

|∇(h − g)|2 ≤ −

Z ∂Bp (4ρ0 )

Since ρ is arbitrary, this completes the proof of the theorem. 

GEOMETRIC ANALYSIS

155

Theorem 19.2. (Li-Tam and Sung-Tam-Wang) Let M be a complete manifold and Ω be a smooth compact subdomain in M. Suppose g is a harmonic function defined on M \ Ω which is smooth up to the boundary ∂Ω of Ω. If M is parabolic, then there exists a harmonic function h defined on M such that |g − h| is bounded on M \ Ω if Z ∂g = 0, ∂Ω ∂ν where ν is the outward unit normal to ∂Ω. Moreover, |g − h| must have finite Dirichlet integral on M \ Ω. Proof. Following a similar argument as in the proof of Theorem 19.1, we define the harmonic function Z h(x) = φ(x) g(x) + G(x, y) ∆(φ g)(y) dy M

where G(x, y) is a symmetric Green’s function constructed in §17, and φ given by ( 0 on Bp (ρ0 ) φ= 1 on M \ Bp (2ρ0 ). Again when for x ∈ / Bp (4ρ0 ), we have (19.4)

|h(x) − g(x)| = |h(x) − φ(x) g(x)| Z G(x, y) ∆(φ g)(y) dy = Bp (2ρ0 ) Z ≤ (G(x, y) − G(x, p)) ∆(φ g)(y) dy Bp (2ρ0 ) Z + G(x, p) ∆(φ g)(y) dy . Bp (2ρ0 )

However, since Z

Z ∆(φ g)(y) dy =

Bp (2ρ0 )

∂Bp (2ρ0 )

Z = ∂Ω

∂g ∂r

∂g ∂ν

=0 and Z (G(x, y) − G(x, p)) ∆(φ g)(y) dy Bp (2ρ0 ) Z ≤ sup |G(x, y) − G(x, p)| y∈Bp (2ρ0 )

|∆(φ g)(y)| dy,

Bp (2ρ0 )

(19.4) becomes |h(x) − g(x)| ≤ C

sup

|G(x, y) − G(x, p)|.

y∈Bp (2ρ0 )

On the other hand, from the construction of G, we know that G(y, x) − G(p, x) is a bounded harmonic function on M \ Bp (4ρ0 ) for y ∈ Bp (2ρ0 ). This implies that |h − g| is bounded on M \ Ω.

156

PETER LI

To see that h − g has finite Dirichlet integral, we follow a similar argument as in the proof of Theorem 19.1. By solving the Dirichlet boundary problem (19.2) and (19.3), we obtain a harmonic function u with the properties that u − (h − g) = 0 on ∂Bp (4ρ0 ) and |u − (h − g)| ≤ C. Unless u − (g − h) is identically 0, then either u − (g − h) = α

sup M \Bp (8ρ0 )

or inf

M \Bp (8ρ0 )

u − (g − h) = −α

for some α > 0. In the first case, the function f = α − u + (h − g) will be a nonconstant positive harmonic function whose infimum is achieved at infinity. In the second case, the function f = u − (h − g) − α will have the same property. This contradicts the assumption that M is parabolic, and we must have u = h − g. The rest of the argument is exactly the same as in the proof of Theorem 19.1.  There is one more property resulting from the above constructions that is important to point out. Recall that when M is nonparabolic, then the minimal Green’s function G is given by the limit of a sequence of Dirichlet Green’s function defined on a compact exhaustion. Let us assume that the compact exhaustion are given by {Bp (ρi )}. Similarly, for a parabolic manifold, the constructed Green’s function can be obtained by taking the limit of Gi − ai where ai is a sequence of constants. In either case, if we define the sequence of harmonic functions Z hi (x) = φ(x) g(x) + Gi (x, y) ∆(φ g)(y) dy Bp (ρi )

for ρi ≥ 4ρ0 , then hi will solve the Dirichlet problem (19.5)

∆hi = 0

on

Bp (ρi )

and (19.6)

hi = g

on

∂Bp (ρi ).

In the nonparabolic case, obviously hi → h since Gi → G. In the parabolic case, this is also true because Z ai ∆(φ g)(y) dy = 0 Bp (ρi )

hence

Z (Gi (x, y) − ai ) ∆(φ g)(y) dy

hi (x) = φ(x) g(x) + Bp (ρi )

which converges to h. This implies that the harmonic function constructed in Theorem 19.1 and Theorem 19.2 can be obtained by taking limits of harmonic functions hi solving the Dirichlet problem (19.5) and (19.6). By the maximum principle, we conclude that inf ∂Bp (ρi )

g ≤ hi (x) ≤ sup g ∂Bp (ρi )

for all x ∈ Bp (ρi ). Taking the limit as i → ∞, we obtain lim inf g ≤ h(x) ≤ lim sup g. y→∞

y→∞

We are now ready to construct harmonic functions that reflects the geometry and topology of a complete manifold.

GEOMETRIC ANALYSIS

157

Theorem 19.3. (Li-Tam) Let M be a complete manifold that is nonparabolic. There exists spaces of harmonic functions K0 (M ) and K0 (M ), with (possibly infinite) dimensions given by k 0 (M ) and k 0 (M ), respectively, such that k 0 (M ) = N 0 (M ) and k 0 (M ) = N 0 (M ). In particular, k 0 (M ) + k 0 (M ) = N (M ). Moreover, K0 (M ) is a subspace of the space of bounded harmonic functions with finite Dirichlet integral on M, and K0 (M ) is spanned by a set of positive harmonic functions. Proof. The assumption that M is nonparabolic implies that N 0 (M ) ≥ 1. If N 0 (M ) = 1, then the space K0 (M ) is given by the constant functions. Let us now assume that N0 (M ) ≥ 2 and E1 , . . . , ENρ0 (M ) be the set of all nonparabolic ends with respect to the compact set Bp (ρ). It suffices to show that we can construct Nρ0 (M ) many linearly independent bounded harmonic functions with finite Dirichlet integral. Since ρ is arbitrary, this will prove that K0 (M ) exists. For each 1 ≤ i ≤ Nρ0 (M ), let us define the harmonic function ψi on M \ Bp (ρ) given by  ψi =

1

on

Ei

0

on

M \ (Bp (ρ) ∪ Ei ).

By Theorem 19.1, there exists a harmonic function hi defined on M such that hi − ψi is bounded and has finite Dirichlet integral on M \ Bp (ρ). By the remark following Theorem 19.2, hi must also be bounded between 0 and 1 and has finite Dirichlet integral on M. Moreover, on any nonparabolic end Ek , Theorem 19.1 also asserts that |ψi − hi | ≤ C fk on Ek . where fk is the minimal barrier function of Ek . Applying to Ei , this implies that if {xij } is a sequence of points in Ei with xij → Ei (∞) as j → ∞, such that, fi (xij ) → 0, then hi (xij ) → 1. Similarly, this also show α α that for any sequence {xα j } ⊂ Eα with xj → Eα (∞) for α 6= i as j → ∞, such that, fα (xj ) → 0 then α hi (xj ) → 0. Obviously, this construction yields Nρ0 (M ) many bounded harmonic functions {hi } with finite Dirichlet integrals, and satisfying lim hi (xij ) = 1 j→∞

and lim hi (xα j)=0

j→∞

for

α 6= i.

Linearly independency of this set of functions follows and K0 is constructed. 0 Let us now denote E10 , . . . EN to be the set of parabolic ends with respect to Bp (ρ). We will construct 0 ρ (M ) 0 a set of Nρ (M ) linearly independent positive harmonic functions. Again, since ρ is arbitrary, this will prove that a space K0 exists with the properties that dim K0 = k 0 (M ) = N 0 (M ) and K0 is spanned by positive harmonic functions. Therefore K(M ) = K0 (M ) + K0 (M ) and the Theorem follows. Modifying the above argument, for each Ei0 , let gi be a barrier function for Ei0 given by Lemma 18.2. Let us define the function ψi by ( gi on Ei0 ψi = 0 on M \ (Bp (ρ) ∪ Ei0 ). Using Theorem 19.1 again, there exists a harmonic function hi with the property that |ψi − hi | ≤ C

on

Eα0

for

α 6= i

158

PETER LI

and |ψi − hi | ≤ C fk In particular, since gi (y) → ∞ as y →

Ei0 (∞),

on

Ek .

we conclude that

lim sup hi (y) = ∞.

y→Ei0 (∞)

Also hi is bounded on M \ Ei0 . Note that according to the remark proceeding to Theorem 19.2, hi must be positive since it can be viewed as a limit of a sequence of positive harmonic functions satisfying (19.5) and (19.6). Moreover, since there is a nonparabolic end, inf M hi = 0. Clearly, the set of functions {hi } are linearly independent and they span the space Kρ0 (M ). This completes the theorem.  A similar construction also gives a corresponding theorem for parabolic manifolds. Theorem 19.4. (Li-Tam) Let M be a complete manifold that is parabolic. There exists a space of harmonic functions K(M ), with (possibly infinite) dimension given by k(M ), such that k(M ) = N (M ). Moreover, K(M ) is spanned by a set of harmonic functions which are bounded either from above or below when restricted on each end of M. Proof. Let Ei0 for 1 ≤ i ≤ Nρ (M ) be the set of parabolic ends with respect to Bp (ρ). Following the notation as before, we denote gi be the barrier functions for the ends. Note that for each i, from the construction gi achieves its minimum value on ∂Ei0 . Therefore, by the maximum principle, we conclude that Z ∂gi = αi ∂Ei0 ∂r for some αi > 0. We define the function g1 α1 ψi = − gi αi    0    

on on on

E10 Ei0 M \ (E10 ∪ Ei0 )

Clearly, Z ∂Bp (ρ)

∂ψi =0 ∂r

and we can apply Theorem 19.2 to obtain a harmonic function hi on M with the property that |ψi − hi | is bounded and has finite Dirichlet integral. In particular, the function hi has the properties that lim sup hi (y) = ∞,

y→E10 (∞)

lim inf hi (y) = −∞,

y→Ei0 (∞)

and |hi (y)| ≤ C

on

M − (Bp (ρ) ∪ E10 ∪ Ei0 ).

Moreover, since the barrier functions gj are positive, hi must be bounded either from above or from below on each end. Clearly, the set of functions {hi } together with a constant function form a linear independent set and span the space Kρ (M ). This completes the proof of the theorem.  Combining with Theorem 7.2, we obtain the following finiteness theorem.

GEOMETRIC ANALYSIS

159

Theorem 19.5. (Li-Tam) Let M be an m-dimensional complete noncompact Riemannian manifold without boundary. Suppose that the Ricci curvature of M is nonnegative on M \ Bp (1) for some unit geodesic ball centered at p ∈ M . Let us assume that the lower bound of the Ricci curvature on Bp (1) is given by Rij ≥ −(m − 1)R for some constant R ≥ 0. Then M must have finitely many ends. Moreover, there exists a constant C(m, R) > 0 depending only on m and R, such that, N (M ) ≤ C(m, R). L Proof. If M is nonparabolic, we define the space of harmonic functions H0 (M ) = K0 (M ) K0 (M ) as constructed in Theorem 19.3. Since functions in K0 (M ) are all bounded harmonic functions and any finite dimensional subspace of K0 (M ) is spanned by harmonic functions that are positive and each tends to infinity at only end, their linear combination must be either bounded above or bounded below on each end. Hence we can apply Theorem 7.2 to estimate dim H0 (M ), therefore obtaining the bound N (M ) ≤ C(m, R) by combining with Theorem 19.3. If M is parabolic, we simply combine Theorem 19.4 and Theorem 7.2 to obtain the estimate on N (M ).  We would like to point out that when M has nonnegative Ricci curvature everywhere, this argument also recovers the splitting theorem (Theorem 4.2) of Cheeger and Gromoll [CG2] . The interested reader may note that the estimate on the number of ends actually holds on manifolds whose Ricci curvature need not be nonnegative at infinity. In fact, it was proved by Li and Tam [LT1] that if we assume that the Ricci curvature of M is bounded from below by Rij (x) ≥ −(m − 1)k(r(x)) for some nonincreasing function k(r) ≥ 0 satisfying the property Z

∞

k(r) rm−1 dr < ∞,

1

then N (M ) is finite and can be estimated. However, the proof is more involved and might not substantially add to our overall educational purpose.

160

PETER LI

§20 Manifolds with Positive Spectrum In this section, we will consider manifolds whose spectrum of the Laplacian, ∆, acting on L2 functions are bounded from below by a positive number. If we denote µ1 (M ) = inf Spec(∆) to be the infimum of the spectrum of ∆, it can be characterized by R µ1 (M ) =

|∇f |2 f2 M

M R

inf 1,2

f ∈Hc (M )

where the infimum is taken over all compactly supported functions in the Sobolev space H 1,2 (M ). In particular, µ1 (M ) = inf µ1 (Ωi ) i→∞

for any compact exhaustion {Ωi }, where µ1 (Ωi ) is the Dirichlet first eigenvalue of ∆ on Ωi . In 1975, Cheng and Yau [CgY] gave a necessary condition for a complete manifold to have µ1 (M ) > 0, by showing that if M has polynomial volume growth then µ1 (M ) = 0. Another important quantity of the spectrum is the infimum of the essential spectrum µe (M ) of ∆. It has the property that µe (M ) ≥ µ1 (M ) and given any  > 0, there exists a compact set Ω ⊂ M , such that µe (M ) ≤ µ1 (Ω0 ) +  for any compact set Ω0 ⊂ M \ Ω. In view of this remark, most of the statements proved in this section with the assumption of µ1 (E) > 0 on an end can be stated as µe (M ) > 0 Throughout this section, we will assume that E is an end of M with respect to the compact set Bp (ρ0 ). We also assume that the infimum of the Dirichlet spectrum of ∆ on E is positive. In particular, R 0 < µ1 (E) ≤

|∇f |2 f2 E

E R

for any compactly supported function f ∈ Hc1,2 (E) defined on E. Our goal is to give decay L2 -estimates on the harmonic functions f ∈ K0 (M ) constructed in Theorem 19.1. This is also equivalent to estimating the barrier functions on a nonparabolic end, hence the Green’s function on a nonparabolic manifold. Note that by taking ρ0 sufficiently large, the condition µ1 (E) > 0 is guaranteed if µe (M ) > 0. In 1981, Brooks [Br] improved Cheng-Yau’s theorem and showed that if M has infinite volume and if we deonte the volume entropy, τ (M ), by τ (M ) = lim sup ρ−1 log Vp (ρ), ρ→∞

then µe (M ) ≤

τ 2 (M ) . 4

Recently, in a paper of Li and Wang [LW5], the authors proved a sharp estimate of the barrier functions. As one of the consequences, this implies an estimate on the volume growth (Corollary 20.3) and can be viewed as an improvement of Brooks’ result.

GEOMETRIC ANALYSIS

161

Theorem 20.1. (Li-Wang) Let M be a complete Riemannian manifold. Suppose E is an end of M with respect to Bp (ρ0 ), such that µ1 (E) > µ, for some constant µ ≥ 0. Let f be a nonnegative function defined on E satisfying the differential inequality ∆f ≥ −µ f. If f satisfies the growth condition Z

f 2 e−2ar = o(ρ)

E(ρ)

p

as ρ → ∞, with a =

µ1 (E) − µ, then it must satisfy the decay estimate

Z

f 2 ≤ C(a)(1 + (ρ − ρ0 )−1 ) e−2aρ

E(ρ+1)\E(ρ)

Z

e2ar f 2

E(ρ0 +1)\E(ρ0 )

for some constant C(a) > 0 depending on a and for all ρ ≥ 2(ρ0 + 1), where E(ρ) = Bp (ρ) ∩ E. Proof. We will first prove that for any 0 < δ < 1, there exist a constant 0 < C < ∞ such that, Z

e2δar f 2 ≤ C.

E

Indeed, let φ(r(x)) be a nonnegative cutoff function with support on E with r(x) being the geodesic distance to the fixed point p. Then for any function, h(r(x)), integration by parts yields Z (20.3) E

|∇(φ eh f )|2 Z Z Z h 2 2 h 2 2 = |∇(φ e )| f + (φ e ) |∇f | + 2 φ eh f h∇(φ eh ), ∇f i E E E Z Z Z 1 = |∇(φ eh )|2 f 2 + φ2 e2h |∇f |2 + h∇(φ2 e2h ), ∇(f 2 )i 2 ZE ZE ZE 1 h 2 2 2 2h 2 φ2 e2h ∆(f 2 ) = |∇(φ e )| f + φ e |∇f | − 2 E E E Z Z h 2 2 ≤ |∇(φ e )| f + µ φ2 e2h f 2 E E Z Z = |∇φ|2 e2h f 2 + 2 φ e2h h∇φ, ∇hif 2 E E Z Z 2 2 2h 2 + φ |∇h| e f + µ φ2 e2h f 2 . E

E

On the other hand, using the variational principle for µ1 (E), we have Z µ1 (E)

φ2 e2h f 2 ≤

E

Z

|∇(φ eh f )|2 ,

E

hence (20.3) becomes (20.4)

a2

Z

φ2 e2h f 2 E Z Z Z ≤ |∇φ|2 e2h f 2 + 2 φ e2h h∇φ, ∇hif 2 + φ2 |∇h|2 e2h f 2 . E

E

E

162

PETER LI

Let us now choose

φ(r(x)) =

 r(x) − ρ0      1

on

E(ρ0 + 1) \ E(ρ0 )

on

E(ρ) \ E(ρ0 + 1),

−1

 ρ (2ρ−r(x))     0 on

E(2ρ) \ E(ρ)

on

E \ E(2ρ),

and     δar

h(r) =

   A − ar

for

for some fixed constant A > (ρ0 + 1)(1 + δ)a. When ρ ≥

|∇h|2 =

A (1 + δ)a A r≥ , (1 + δ)a

r≤

for

 2 2   δ a

A (1+δ)a ,

A (1 + δ)a A r≥ (1 + δ)a r≤

for

   a2

we see that

for

and h∇φ, ∇hi =

  

δa ρ

−1

 

a

0

on

E(ρ0 + 1) \ E(ρ0 )

on

E(2ρ) \ E(ρ)

otherwise.

Substituting into (20.4), we obtain a2

Z E

φ2 e2h f 2 Z ≤

e

2h

2

f +ρ

−2

Z

E(ρ0 +1)\E(ρ0 )

e2h f 2 Z

E(2ρ)\E(ρ)

Z

e2h f 2 + 2ρ−1 a

+ 2δa E(ρ0 +1)\E(ρ0 ) 2 2

e2h f 2

E(2ρ)\E(ρ)

Z

2 2h

+δ a

φ e

2

2

Z

φ2 e2h f 2 .

f +a

E(A((1+δ)a)−1 )\E(ρ0 ))

E(2ρ)\E(A((1+δ)a)−1 )

This can be rewritten as a2

Z

e2h f 2

E(A((1+δ)a)−1 )\E(ρ0 +1)

≤ a2 Z ≤

Z

φ2 e2h f 2

E(A((1+δ)a)−1 )

e2h f 2 + ρ−2

E(ρ0 +1)\E(ρ0 )

Z + 2δa

Z

e2h f 2 E(2ρ)\E(ρ) Z 2h 2 e f + 2ρ−1 a

E(ρ0 +1)\E(ρ0 ) 2 2

Z

+δ a

E(2ρ)\E(ρ) 2 2h

φ e E(A((1+δ)a)−1 )\E(ρ0 ))

2

f ,

e2h f 2

GEOMETRIC ANALYSIS

163

hence 2

2

Z

e2h f 2

(1 − δ )a

E(A((1+δ)a)−1 )\E(ρ0 +1))

Z

2 2

≤ (δ a + 2δa + 1) Z −2 +ρ

e2h f 2 Z 2h 2 −1 e f + 2ρ a

E(ρ0 +1)\E(ρ0 ))

E(2ρ)\E(ρ)

e2h f 2 .

E(2ρ)\E(ρ)

The definition of h and the assumption on the growth estimate on f imply that the last two terms on the right hand side tends to 0 as ρ → ∞. Hence we obtain the estimate Z Z (1 − δ 2 )a2 e2δar f 2 ≤ (δ 2 a2 + 2δa + 1) e2δar f 2 . E(A((1+δ)a)−1 )\E(ρ0 +1))

E(ρ0 +1)\E(ρ0 ))

Since the right hand side is independent of A, by letting A → ∞ we conclude that Z (20.5) e2δar f 2 ≤ C1 , E\E(ρ0 +1)

with C1 =

δ 2 a2 + 2δa + 1 (1 − δ 2 )a2

Z

e2δar f 2 .

E(ρ0 +1)\E(ρ0 )

Our next step is to improve this estimate by setting h = ar in the preceding argument. Note that (20.4) asserts that Z −2a φ e2ar h∇φ, ∇ri f 2 E Z ≤ |∇φ|2 e2ar f 2 . E

For ρ0 < ρ1 < ρ, let us now choose φ to be  r(x) − ρ0    ρ1 − ρ0 φ(x) =  ρ − r(x)   ρ − ρ1

on

E(ρ1 ) \ E(ρ0 )

on

E(ρ) \ E(ρ1 ).

We conclude that 2a (ρ − ρ1 )2

Z E(ρ)\E(ρ1 )

≤

(ρ − r) e2ar f 2 Z

1 1 e2ar f 2 + 2 (ρ1 − ρ0 ) E(ρ1 )\E(ρ0 ) (ρ − ρ1 )2 Z 2a + (r − ρ0 ) e2ar f 2 . (ρ1 − ρ0 )2 E(ρ1 )\E(ρ0 )

Z E(ρ)\E(ρ1 )

On the other hand, for any 0 < t < ρ − ρ1 , since Z 2at e2ar f 2 (ρ − ρ1 )2 E(ρ−t)\E(ρ1 ) Z 2a ≤ (ρ − r) e2ar f 2 , (ρ − ρ1 )2 E(ρ)\E(ρ1 )

e2ar f 2

164

PETER LI

we deduce that (20.6)

2at (ρ − ρ1 )2  ≤

Z

e2ar f 2

E(ρ−t)\E(ρ1 )

Z 2a 1 + e2ar f 2 ρ1 − ρ0 (ρ1 − ρ0 )2 E(ρ1 )\E(ρ0 ) Z 1 + e2ar f 2 . (ρ − ρ1 )2 E(ρ)\E(ρ1 )

Observe that if we take ρ1 = ρ0 + 1, t = a−1 , and set Z g(ρ) =

e2ar f 2 ,

E(ρ)\E(ρ0 +1)

then the inequality (20.6) can be written as g(ρ − a−1 ) ≤ C2 ρ2 + where

2a + 1 C2 = 2

1 g(ρ), 2

Z

e2ar f 2

E(ρ0 +1)\E(ρ0 )

is independent of ρ. Iterating this inequality, we obtain for any positive integer k and ρ ≥ 1 k X (ρ + ia−1 )2

g(ρ) ≤ C2

i=1 ∞ X 2

≤ C2 ρ

i=1

2i−1

+ 2−k g(ρ + ka−1 )

(1 + ia−1 )2 + 2−k g(ρ + ka−1 ) 2i−1

≤ C3 ρ + 2−k g(ρ + ka−1 ), 2

where C3 = C2

∞ X (1 + ia−1 )2

2i−1

i=1

.

However, our previous estimate (20.5) implies that g(ρ + ka−1 ) =

Z

e2ar f 2

E(ρ+ka−1 )\E(ρ0 +1)

≤ e2a(ρ+ka

−1

)(1−δ)

Z

e2δar f 2

E(ρ+ka−1 )\E(ρ

≤Ce

2a(ρ+ka−1 )(1−δ)

.

Hence, 2−k g(ρ + ka−1 ) → 0 as k → ∞ by choosing 2(1 − δ) < ln 2. This proves the estimate Z (20.7) E(ρ)\E(ρ0 +1)

e2ar f 2 ≤ C3 ρ2

0 +1)

GEOMETRIC ANALYSIS

165

for all ρ ≥ ρ0 + 1. Using inequality (20.6) again and by choosing ρ1 = ρ0 + 1 and t = Z aρ e2ar f 2

ρ 2

this time, we conclude that

E( ρ 2 )\E(ρ0 +1)

≤ (2a + 1)(ρ − ρ0 − 1)2

Z

Z

e2ar f 2 +

E(ρ0 +1)\E(ρ0 )

e2ar f 2 .

E(ρ)\E(ρ0 +1)

Applying the estimate (20.7) to the second term on the right hand side, we have Z e2ar f 2 ≤ C5 ρ, E( ρ 2 )\E(ρ0 +1)

where

∞

−1

C5 = (2a + 1)a

(ρ − ρ0 − 1)2 X (1 + ia−1 )2 + ρ2 2i i=1

!Z

Therefore, for ρ ≥ 2(ρ0 + 1), we have Z Z (20.8) e2ar f 2 ≤ C(a) ρ E(ρ)

e2ar f 2 .

E(ρ0 +1)\E(ρ0 )

e2ar f 2 ,

E(ρ0 +1)\E(ρ0 )

where C(a) denotes a constant depending only on a. We are now ready to prove the lemma by using (20.8). Setting t = 2a−1 and ρ1 = ρ − 4a−1 in (20.6), we obtain Z

e2ar f 2

E(ρ−2a−1 )\E(ρ−4a−1 )



8 4 ≤ + a(ρ − ρ0 − 4a−1 ) a2 (ρ − ρ0 − 4a−1 )2 Z 1 e2ar f 2 . + 4 E(ρ)\E(ρ−4a−1 )

Z

e2ar f 2

E(ρ−4a−1 )\E(ρ0 )

According to (20.8), the first term of the right hand side is bounded by Z −1 −1 C(a)(1 + (ρ − ρ0 − 4a ) ) e2ar f 2 E(ρ0 +1)\E(ρ0 )

for ρ ≥ 2(ρ0 + 1). Hence, by renaming ρ, the above inequality can be rewritten as Z Z Z 1 2ar 2 −1 2ar 2 e2ar f 2 . e f ≤ C(a)(1 + (ρ − ρ0 ) ) e f + 3 E(ρ+4a−1 )\E(ρ+2a−1 ) E(ρ+2a−1 )\E(ρ) E(ρ0 +1)\E(ρ0 ) Iterating this inequality k times, we arrive at Z Z k−1 X 2ar 2 −1 −i −k e f ≤ C(a)(1 + (ρ − ρ0 ) ) 3 +3 E(ρ+2a−1 )\E(ρ)

i=0

e2ar f 2 .

E(ρ+2a−1 (k+1))\E(ρ+2a−1 k)

However, using (20.8) again, we conclude that the second term is bounded by Z Z 3−k e2ar f 2 ≤ C(a) 3−k (ρ + 2(k + 1)) E(ρ+2(k+1))\E(ρ+2k)

which tends to 0 as k → ∞. Hence Z

e2ar f 2

E(ρ0 +1)\E(ρ0 )

e2ar f 2 ≤ C(a)(1 + (ρ − ρ0 )−1 ).

E(ρ+2a−1 )\E(ρ)

for ρ + 4a−1 ≥ 2(ρ0 + 1), and the theorem follows when a ≤ 2. If a ≥ 2, we simply sum up the above estimate [ a2 ] times by dividing the the interval [ρ, ρ + 1] into [ a2 ] components.  The following proposition is helpful in finding a lower bound for µ1 (M ).

166

PETER LI

Proposition 20.1. Let M be a compact Riemannian manifold with smooth boundary ∂M . If there exists a positive function f defined on M satisfying ∆f ≤ −µf, then the first Dirichlet eigenvalue, µ1 (M ), of M must satisfy µ1 (M ) ≥ µ. In particular, if M is complete, noncompact, without boundary, and the positive function f described above is defined on M, then µ1 (M ) ≥ µ. Proof. Let us first assume that M is compact with smooth boundary ∂M. Let u be the first eigenfunction satisfying ∆u = −µ1 (M ) u on M and u=0

on

∂M.

We may assume that u ≥ 0 on M , and regularity of u asserts that u > 0 in the interior of M. Integration by parts yields Z Z Z (µ1 (M ) − µ) uf ≥ u ∆f − f ∆u M ZM ZM ∂u ∂f − f = u ∂ν ∂ν ∂M ∂M ≥ 0, where ν is the outward unit normal of M . Hence µ1 (M ) ≥ µ and the first part of the proposition is proved. When M is complete, noncompact, without boundary, we apply the previous argument to any compact smooth subdomain D ⊂ M and obtain µ1 (D) ≥ µ. The second part of the proposition follows from the fact that inf µ1 (D) = µ1 (M ). 

D⊂M

Note that the hypothesis of Theorem 20.1 is best possible. Indeed, if we consider the hyperbolic space form of constant −1 sectional curvature, then the metric in terms of polar coordinates is given by ds2Hm = dr2 + sinh2 t ds2Sm−1 . The volume growth is given by r

Z

sinhm−1 t dt ∼ C e(m−1)r

Vp (r) = αm−1 0

and (20.9)

µ1 (Hm ) ≤

(m − 1)2 4

by Cheng’s theorem (Corollary 6.2). In fact, if we consider β(x) = lim (t − r(γ(t), x)) t→∞

GEOMETRIC ANALYSIS

167

to be the Buseman function with respect to some geodesic ray γ, then following the computation in §4, we conclude that ∆β(x) = − lim ∆r(γ(t), x) t→∞

= −(m − 1). Defining the positive function  h(x) = exp

 m−1 β(x) , 2

a direct computation implies that ∆h = −

(m − 1)2 h. 4

2

, hence combining with (20.9) yields µ1 (Hm ) = Proposition 20.1 implies that µ1 (Hm ) ≥ (m−1) 4 Let us now consider any nonconstant bounded harmonic function f , then Z √ f 2 e−2 µ1 r = O(ρ).

(m−1)2 . 4

Bp (ρ)

Of course, if the conclusion of Theorem 20.1 is valid, then f will be in L2 (Hm ) which will imply that f is identically constant by Yau’s theorem (Lemma 7.1). However, it is known that Hm has an infinite dimensional space of bounded harmonic function. This implies that the hypothesis of Theorem 20.1 cannot be relaxed. Corollary 20.1. Let M be a complete Riemannian manifold. Suppose E is an end of M such that µ1 (E) > 0. Then for any harmonic function f ∈ K0 (M ), there exists a constant a such that f − a must be in L2 (E). Moreover, the function f − a must satisfy the decay estimate Z p (f − a)2 ≤ C exp(−2ρ µ1 (E)) E(ρ+1)\E(ρ)

for some constant C > 0 depending on f , µ1 (E) and m, where E(ρ) = Bp (ρ) ∩ E. Proof. It suffices to prove the corollary for those functions f constructed in Theorem 19.1 because the decay property is preserved under linear combinations. Following the remark proceeding Theorem 19.2, for a nonparabolic end E1 , let fρ be a sequence of harmonic functions by solving the Dirichlet boundary problem ∆fρ = 0

on

Bp (ρ),

fρ = 1

on

∂Bp (ρ) ∩ E1

fρ = 0

on

∂Bp (ρ) \ E1 .

and 0

A subsequence of this sequence converges to f ∈ K (M ) uniformly on compact subsets of M. For any fixed end E, since fρ has boundary value either 0 or 1 on ∂E(ρ), by considering either the function fρ or 1 − fρ , we may assume that fρ has boundary value 0 on ∂Bp (ρ) ∩ E. Let us define the function gρ by ( fρ (x) on E(ρ) gρ (x) = 0 on E \ E(ρ). Clearly, gρ is a nonnegative subharmonic function defined on E. Moreover, since gρ has compact support, the hypothesis of Theorem 20.1 is satisfied and the decay estimate holds for gρ . The corollary follows by taking ρ → ∞.  We point out that Corollary 20.1 also holds for any function f with a = 0 provided that f is the limit of a sequence of harmonic functions fρ on E(ρ) satisfying fρ = 0 on ∂E(ρ) regardless of their boundary values on ∂E. In particular, when applying Corollary 20.1 to the Green’s function, we obtain the following sharp decay estimate.

168

PETER LI

Corollary 20.2. Let M be a complete manifold with µ1 (M ) > 0. Then the minimal positive Green’s function G(p, ·) with pole at p ∈ M must satisfy the decay estimate Z p G2 (p, x) dx ≤ C exp(−2ρ µ1 (M )) Bp (ρ+1)\Bp (ρ)

for ρ ≥ 1. In the case when M = Hm , equation (17.8) asserts that the Green’s function is given by Z ∞ dt G(p, x) = r(x) Ap (t) where Ap (t) = αm−1 sinh(m−1) t is the area of the boundary of the geodesic ball of radius t centered at p ∈ Hm . One computes readily that the integral Z G2 (p, x) dx ∼ C exp(−(m − 1) ρ). Bp (ρ+1)\Bp (ρ) 2

p Since µ1 (Hm ) = (m−1) , the quantity 2 µ1 (Hm ) is exactly (m − 1), indicating the sharpness of Corollary 4 20.2. Applying Corollary 20.1, we will obtain volume estimates for those ends with positive spectrum. As pointed out in the introduction, these estimates are sharp. The sharp growth estimate is realized by the hyperbolic space Hm , while the sharp decay estimate is realized by a hyperbolic cusp (see Example 21.1). To state our estimate, let us denote the volume of the set E(ρ) by VE (ρ) and the volume of the end E by V (E). Theorem 20.2. (Li-Wang) Let E be an end of complete manifold M with µ1 (E) > 0. (1) If E is a parabolic end, then E must have exponential volume decay given by   p VE (ρ + 1) − VE (ρ) ≤ C (1 + (ρ − ρ0 )−1 ) (VE (ρ0 + 1) − VE (ρ0 )) exp −2(ρ − ρ0 ) µ1 (E) for some constant C > 0 depending on the µ1 (E). In particular, we have   p V (E) − VE (ρ) ≤ C (VE (ρ0 + 1) − VE (ρ0 )) exp −2(ρ − ρ0 ) µ1 (E) . (2) If E is a non-parabolic end, then E must have exponential volume growth given by VE (ρ) ≥ C exp(2ρ

p

µ1 (E))

for all ρ ≥ ρ0 + 1 and some constant C depending on the end E. Proof. Let fρ be the harmonic function on E(ρ) with fρ = 1 on ∂E and fρ = 0 on ∂E(ρ). The assumption that E is parabolic implies that fρ converges to f = 1 as ρ → ∞. Hence the estimate from Theorem 20.1 yields the first estimate of the theorem. Letting ρ = ρ + i for i = 0, 1, . . . and summing over i, we conclude that V (E) − VE (ρ) ≤ C (VE (ρ0 + 1) − VE (ρ0 ))

∞   X p (1 + (ρ + i − ρ0 )−1 ) exp −2(ρ + i − ρ0 ) µ1 (E) i=0

  p ≤ C (VE (ρ0 + 1) − VE (ρ0 )) exp −2(ρ − ρ0 ) µ1 (E) . This proves the volume decay estimate for the case of parabolic ends.

GEOMETRIC ANALYSIS

169

If E is non-parabolic, then fρ converges to a nonconstant harmonic function f on E. Thus, we conclude that there exists a positive constant C such that for r ≥ ρ0 Z ∂f C= ∂ν Z∂E ∂f = ∂B (r)∩E ∂ν Z p ≤ |∇f | ∂Bp (r)∩E 1 2

≤ AE (r)

!1/2

Z |∇f |

2

,

∂Bp (r)∩E

or equivalently, C ≤ AE (r)

Z

|∇f |2 .

∂Bp (r)∩E

Integrating the preceding inequality with respect to r from ρ to ρ + 1 and using Corollary 20.1 and Lemma 7.1, we obtain Z Z ρ+1 1 dr ≤ C |∇f |2 AE (r) E(ρ+1)\E(ρ) ρ   p ≤ C exp −2ρ µ1 (E) . Therefore, ρ+1 1 AE (r) dr dr A E (r) ρ ρ   p ≤ C exp −2ρ µ1 (E) (VE (ρ + 1) − VE (ρ))   p ≤ C exp −2ρ µ1 (E) VE (ρ + 1).

Z

ρ+1

Z

1≤

Since ρ is arbitrary, we conclude that VE (ρ) ≥ C exp(2ρ

p

µ1 (E))

by adjusting the constant C, and the theorem is proved.  Corollary 20.3. Let M be a complete manifold with infinite volume. Suppose the essential spectrum of M has a positive lower bound, i.e., µe (M ) > 0, then for any  > 0, there exists C > 0 depending on , such that, p Vp (ρ) ≥ C exp(2ρ µe (M ) − ). Proof. For  > 0, let ρ0 be sufficient large such that µ1 (M \ Bp (ρ0 )) +  ≥ µe (M ). Theorem 20.2 and the infinite volume assumption on M asserts that M must have at least one infinite volume end E with respect to Bp (ρ0 ). In fact, the volume growth of this end must satisfy p VE (ρ) ≥ C exp(2ρ µe (M ) − ), hence Vp (ρ) ≥ C exp(2ρ

p

µe (M ) − ). 

Our last corollary concerns Lq harmonic functions on an end with positive bottom spectrum.

170

PETER LI

Corollary 20.4. Let E be an end of M with µ1 (E) > 0. Let Lq (E) be the space of Lq harmonic functions on E. When q ≥ 2, if u ∈ Lq (E), then u must be bounded and it must satisfy the estimate Z

u2 ≤ C exp(−2ρ

p µ1 (E)).

E(ρ+1)\E(ρ)

When 1 < q < 2, the same conclusion is true provided that the volume growth of E is bounded by  VE (ρ) ≤ C exp

 p 2q ρ µ1 (E) . 2−q

Proof. Let u ∈ Lq (E) be an Lq -harmonic function. Define fρ to be the harmonic function on E(ρ) satisfying fρ = 0

∂Bp (ρ) ∩ E

on

and fρ = u

on

∂E.

Clearly, the maximum principle asserts that a subsequence of fρ as ρ → ∞ will converge to a function f ∈ L∞ (E) with u = f on ∂E. Moreover, by Corollary 20.1, f satisfies the estimate Z

p f 2 ≤ C exp(−2ρ µ1 (E)).

(20.10) E(ρ+1)\E(ρ)

If q ≥ 2, the boundedness of f and (20.10) imply that f ∈ Lq (E). In particular, the function u−f is in Lq (E) with 0 boundary condition on ∂E. Applying the uniqueness theorem of Yau [Y] for Lq harmonic functions, we conclude that u = f. For 1 < q < 2, the Schwarz inequality, (20.10) and the volume growth bound give Z

q

! q2

Z

f ≤ E(ρ+1)\E(ρ)

f

2

(VE (ρ + 1) − VE (ρ))

2−q 2

E(ρ+1)\E(ρ)

p p ≤ C exp(−q ρ µ1 (E)) exp(q ρ µ1 (E)) ≤ C, implying that the Lq -norm of f is at most of linear growth. Again, by slightly modifying Yau’s uniqueness theorem we conclude that f = u. Indeed for the completeness sake, we will outline a modification of Yau’s argument for this case. Observe that the function g = |f − u| is a subharmonic function defined on E with boundary condition g=0

on

∂E.

Let φ be a cutoff function satisfying ( φ=

1

on

E(ρ)

0

on

E \ E(2ρ)

and |∇φ| ≤ C ρ−1

on

E(2ρ) \ E(ρ).

GEOMETRIC ANALYSIS

171

Integration by parts yields Z

φ2 g q−1 ∆g Z Z = −2 φ g q−1 h∇φ, ∇gi − (q − 1) φ2 g q−2 |∇g|2 .

0≤

E

E

E

On the other hand, applying the Schwarz inequality Z Z Z q−1 2 q−1 2 q−2 2 −2 φg h∇φ, ∇gi ≤ φ g |∇g| + |∇φ|2 g q , 2 q−1 E E E we conclude that

Z

2

φ g E

q−2

4 |∇g| ≤ (q − 1)2 2

Z

|∇φ|2 g q .

E

Using the property of φ, this implies that Z Z C q−2 2 g |∇g| ≤ 2 gq . ρ E(2ρ)\E(ρ) E(ρ) The growth estimate on the Lq -norm of f and the fact that u ∈ Lq implies that the right hand side tends to 0 as ρ → ∞. Hence g must be identically constant. The boundary condition of g asserts that it must be identically 0, and f = u. In both cases, since f = u, the function u must satisfy (20.9). This concludes the corollary. 

172

PETER LI

§21 Manifolds with Ricci Curvature Bounded from Below In this section, we will assume that M m has Ricci curvature bounded from below by −(m − 1)R for some constant R > 0. After normalizing, we may assume that R = 1. Note that the Bishop volume comparison theorem asserts that for any x ∈ M , the ratio of the volumes of geodesic balls Bx (ρ1 ) and Bx (ρ2 ) for ρ1 < ρ2 must satisfy (21.1)

V¯ (ρ2 ) Vx (ρ2 ) ≤ ¯ , Vx (ρ1 ) V (ρ1 )

where V¯ (ρ) is the volume of a geodesic ball of radius ρ in the m- dimensional hyperbolic space form Hm of constant −1 curvature. In particular, by taking x = p, ρ1 = 0 and ρ2 = ρ this implies that (21.2)

Vp (ρ) ≤ C3 exp((m − 1)ρ)

for sufficiently large ρ. On the other hand, if we let x ∈ ∂Bp (ρ), ρ1 = 1 and ρ2 = ρ + 1, (21.1) implies (21.3)

Vx (1) ≥ C4 Vx (ρ + 1) exp(−(m − 1)ρ) ≥ C4 Vp (1) exp(−(m − 1)ρ).

Hence we have the following volume estimate. Proposition 21.1. Let M m be a complete manifold with Ricci curvature bounded from below by Rij ≥ −(m − 1). The volume growth of M must satisfy the upper bound Vp (ρ) ≤ V¯ (ρ) ≤ C3 exp((m − 1)ρ) for some constant C3 > 0 depending only on m. Also the volume cannot decay faster than exponentially and must satisfy the estimate Vx (1) ≥ C4 Vp (1) exp(−(m − 1)r(x)) where r(x) denotes the distance from p to x. In particular, Vp (ρ + 1) − Vp (ρ − 1) ≥ C4 Vp (1) exp(−(m − 1)ρ). Note that in viewed of Theorem 20.2, if we also assume that µ1 (M ) > 0, then one has both upper and lower control of the volume growth for a parabolic end and also for a nonparabolic end. Recall that by Cheng’s theorem (Corollary 6.2), under the Ricci curvature assumption µ1 (M ) ≤

(m − 1)2 . 4

While Hm achieves equality, it is not the only manifold with this property. In fact, let us consider the following example. Example 21.1. Let M m = R × N m−1 be the complete manifold with the warped product metric ds2M = dt2 + exp(2t) ds2N . According to the computation in Appendix A (A.3) and (A.4), by setting f (t) = exp t, the Ricci curvature on M is given by R1j = −(m − 1)δ1j

GEOMETRIC ANALYSIS

173

and ˜ αβ − (m − 1)δαβ , Rαβ = exp(−2t) R ˜ αβ is the Ricci tensor on N and e1 = ∂ . In particular, if the Ricci curvature of N is nonnegative, where R ∂t then Rij ≥ −(m − 1). Moreover, N is Ricci flat if and only if M is Einstein with Rij = −(m − 1). Let us consider f = exp(−αt) for a constant

m−1 2

≤ α ≤ m − 1. One computes that

d2 f df + (m − 1) 2 dt dt = α2 f − (m − 1)α f

∆f =

= −α(m − 1 − α)f and |∇f |2 = α2 f 2 . If we let α =

m−1 2 ,

then the function 

(m − 1)t f (t) = exp − 2



satisfies the equation ∆f = −

(m − 1)2 f. 4

Since f is positive, Proposition 20.1 implies that µ1 (M ) ≥ (m−1)2 . 4

(m−1)2 . 4

Together with the upper bound of Cheng,

we conclude that µ1 (M ) = It turns out that Example 21.1 gives the only class of manifolds with dimension m ≥ 3 that have more than one end on which Cheng’s inequality is achieved. There is another important example that is relevant to the issues being considered in this section. Example 21.2. For m ≥ 3, let M m = R × N m−1 be the product manifold endowed with the warped product metric ds2M = dt2 + cosh2 t ds2N , where N is a compact manifold with Ricci curvature bounded from below by ˜ αβ ≥ −(m − 2). R Following the notation and the computation in Appendix A, by setting f (t) = cosh t, the Ricci curvature on M is given by R1j = −(m − 1)δ1j and
˜ αβ − 1 + (m − 2) tanh2 (t) δαβ . Rαβ = cosh−2 (t) R ˜ αβ ≥ −(m − 2). Moreover M is Einstein with Hence Rij ≥ −(m − 1) because of the assumption that R ˜ αβ = −(m − 2). Rij = −(m − 1) if and only if N is Einstein with R

174

PETER LI

Note that if we define f (t) = cosh−(m−2) t then  2  ∂ sinh t ∂ ∆f = + (m − 1) cosh−(m−2) t ∂t2 cosh t ∂t = −(m − 2) cosh−(m−2) t = −(m − 2)f (t). Proposition 20.1 implies that µ1 (M ) ≥ (m − 2). If we compute the L2 -norm of f on the set [−t, t] × N , then Z Z t 2 f = V (N ) coshm−1 t cosh−2(m−2) t dt −t

[−t,t]×N

Z

t

= V (N )

cosh3−m t dt.

−t

This implies that f ∈ L2 (M ) for m > 3, and f is an eigenfunction, hence µ1 (M ) ≤ m − 2. When m = 3, the L2 -norm of f over [−t, t] × N grows linearly in t. Although f is not an eigenfunction, but it is still sufficient to show that µ1 (M ) ≤ m − 2, as will be shown in the next theorem. In any event, we have µ1 (M ) = m − 2. The following improved version of the Bochner formula for harmonic functions was first proved by Yau [Y1]. It is by now standard in the literature. Lemma 21.1. (Yau) Let M m be a complete manifold whose Ricci curvature is bounded by Rij (x) ≥ −(m − 1)k(x) for some function k(x). Suppse f is a harmonic function defined on M , then   m 1 q q + q − 2 |∇f |−q |∇(|∇f |q )|2 ∆|∇f | ≥ −q(m − 1)k(x) |∇f | + q m−1 for all q > 0. In particular, if q ≥

m−2 m−1

then

∆|∇f |q ≥ −q(m − 1)k(x) |∇f |q . Proof. Let {e1 , . . . , em } be an orthonormal frame near a point x ∈ M so that |∇f | e1 = ∇f at x. We will follow a similar computation as in the proof of Lemma 5.4. Since f is harmonic, a direct computation and the commutation formula yield 2 ∆|∇f |2 ≥ 2fij − 2(m − 1)k(x) |∇f |2 .

Also, just like (5.13) and (5.14), we have |∇|∇f |2 |2 = 4|∇f |2

m X

2 f1j

j=1

and 2 2 fij ≥ f11 +2

m X α=2 m

2 f1α +

(∆f − f11 )2 m−1

m X 2 ≥ f . m − 1 j=1 1j Combining all these, we obtain m |∇f |−2 |∇|∇f |2 |2 − 2(m − 1)k(x) |∇f |2 2(m − 1) 2m |∇|∇f ||2 − 2(m − 1)k(x) |∇f |2 . = (m − 1) The lemma follows directly from this inequality.  ∆|∇f |2 ≥

GEOMETRIC ANALYSIS

175

Theorem 21.1. (Li-Wang) Let M m be a complete Riemannian manifold of dimension m ≥ 3. Suppose Rij ≥ −(m − 1) and µ1 (M ) ≥ m − 2, then either (1) M has only one end with infinite volume; or (2) M is the warped product given by Example 21.2 Proof. Assume that M has more than one infinite volume end. According to Theorem 19.3, there exists a nonconstant function f ∈ K0 (M ) given by the construction. Let g be the function defined by m−2

g = |∇f | m−1 . Lemma 21.1 asserts that ∆g ≥ −(m − 2)g. We now claim that the function g must satisfy the integral condition Z g 2 ≤ C ρ. Bp (2ρ)\Bp (ρ)

To see this, let us apply H¨ older’s inequality and get Z g2 (21.4) Bp (2ρ)\Bp (ρ)

Z



≤

exp 2r

p

 µ1 (M ) |∇f |2



p

! m−2 m−1

Bp (2ρ)\Bp (ρ)

Z ×

exp −2(m − 2)r

 µ1 (M )

1 ! m−1

.

Bp (2ρ)\Bp (ρ)

Using the lower bound on the Ricci curvature, the volume comparison theorem asserts that ¯ Ap (r) ≤ A(r), ¯ where A(r) is the area of the geodesic sphere of radius r in the constant curvature hyperbolic space form Hm . A direct computation yields that Z   p (21.5) exp −2(m − 2)r µ1 (M ) Bp (2ρ)\Bp (ρ)

Z

2ρ

≤C

  p exp −2(m − 2)r µ1 (M ) exp((m − 1)r) dr

ρ

Z =C

2ρ

  p exp (m − 1)r − 2(m − 2)r µ1 (M ) dr.

ρ

Using the lower bound of µ1 (M ), we conclude that the right hand side of (21.5) is at most linear in ρ when m = 3, and exponentially decays to 0 when m ≥ 4.

176

PETER LI

On the other hand, recall that the decay estimate in Lemma 20.1 yields Z   p (f − a)2 ≤ C exp −2ρ µ1 (M ) E(ρ+1)\E(ρ)

where a is the asymptotic value of f at infinity of an end E given by either 1 or 0. Combining with Lemma 7.1 by choosing α = 1, ρ1 = ρ, ρ2 = ρ + 1, ρ3 = ρ + 2 and ρ4 = ρ + 3, we have Z   p |∇f |2 ≤ 8C exp −2ρ µ1 (M ) . E(ρ+2)\E(ρ+1)

In particular, there exists a constant C1 > 0 independent of ρ such that Z  p  exp 2r µ1 (M ) |∇f |2 ≤ C1 , E(ρ+2)\E(ρ+1)

and

Z

  p exp 2r µ1 (M ) |∇f |2 ≤ C1 ρ.

Bp (2ρ)\Bp (ρ)

Applying this and (21.5) to (21.4), we conclude that Z

g 2 ≤ C2 ρ.

Bp (2ρ)\Bp (ρ)

for the case when n = 3, and Z

g2 → 0

Bp (2ρ)\Bp (ρ)

when n ≥ 4. This proves our claim on the L2 estimate of g. To complete our proof of the theorem, we consider φ to be a nonnegative compactly supported function on M . Then Z Z Z Z (21.6) |∇(φ g)|2 = |∇φ|2 g 2 + 2 φ g h∇φ, ∇gi + φ2 |∇g|2 . M

M

M

M

The second term on the right hand side can be written as Z Z 1 (21.7) 2 φ g h∇φ, ∇gi = h∇(φ2 ), ∇(g 2 )i 2 M M Z Z 2 =− φ g ∆g − φ2 |∇g|2 M M Z Z Z = (m − 2) φ2 g 2 − φ2 |∇g|2 − M

M

φ2 g (∆g + (m − 2) g).

M

Combining with (21.6) and the variational charaterization of µ1 (M ) ≥ m − 2, this implies that Z Z 2 2 (m − 2) φ g ≤ |∇(φ g)|2 M M Z Z = (m − 2) φ2 g 2 + |∇φ|2 g 2 M M Z 2 − φ g (∆g + (m − 2) g) . M

GEOMETRIC ANALYSIS

177

Hence, we have Z

φ2 g (∆g + (m − 2) g) ≤

(21.8) M

Z

|∇φ|2 g 2 .

M

For ρ > 0, let us choose φ to satisfy the properties that ( φ=

1

on

Bp (ρ)

0

on

M \ Bp (2ρ)

and |∇φ| ≤ C ρ−1

Bp (2ρ) \ Bp (ρ)

on

for some constant C > 0. Then the right hand side of (21.8) can be estimated by Z

|∇φ|2 g 2 ≤ C ρ−2

M

Z

g2 .

Bp (2ρ)\Bp (ρ)

By the L2 estimate of g, this tends to 0 as ρ → ∞. Hence together with (21.8), we conclude that g either must be identically 0 or it must satisfy ∆g = −(m − 2) g. If M has more than one infinite volume end, then by Theorem 19.3 there must exist a non-constant f , hence g 6= 0. So all the inequalities used in deriving Lemma 21.1 become equalities. We can now argue to conclude that M = R × N with the warped product metric ds2 = dt2 + cosh2 t ds2N ˜ ≥ −(m − 2). for some compact manifold N with Ricci curvature satisfying R Indeed, since ∆f = 0, the Hessian of f must be of the form −(m − 1)µ 0 0 · 0 µ 0 ·   0 0 µ ·  (fij ) =  · · · ·   · · · 0 0 0 · 

· · ·

 0 0  0 .  

· · µ

The fact that f1α = 0 for all α 6= 1 implies that |∇f | is identically constant along the level set of f. In particular, the level sets of |∇f | and f coincide. Moreover, µ δαβ = fαβ = hαβ f1 with (hαβ ) being the second fundamental form of the level set of f with respect to e1 . Hence (21.9)

f11 = −H f1

where H is the mean curvature of the level set of f with respect to e1 . Applying the same computation on the function g, we obtain (21.10)

−(m − 2) g = ∆g = g11 + H g1 .

178

PETER LI m−2

On the other hand, since g = |∇f | m−1 , we have   m−2 g1 = |∇f | m−1

1

m m−2 |∇f |− m−1 fi fi1 = m−1 1 m − 2 − m−1 f11 . = f1 m−1

Hence, combining with (21.9), we conclude that H = −f1−1 f11 m−1 =− g1 g −1 . m−2

(21.11)

Substituting into (21.10), this yields g11 − 1

m−1 (g1 )2 g −1 + (m − 2) g = 0. m−2

1

Setting u = g − m−2 = |∇f |− m−1 , this differential equation becomes u11 − u = 0. Viewing this as an ordinary differential equation along the integral curve generated by the vector field e1 , one concludes that u(t) = A exp(t) + B exp(−t). Since u must be nonnegative, A and B must be nonnegative. Moreover, ∇f 6= 0. Note that M is assumed to have at least two infinite volume ends. We claim that any fixed level set N of |∇f | must be compact. Indeed, by the fact that f has no critical points and that the level set of f coincides with the level set of |∇f |, M must be topologically the product R × N. If N is non-compact then M will have only one end, hence N must be compact. If both A and B are nonzero, then the function u must have its minimum along N , hence by reparameterizing, we may assume N is given by t = 0. Moreover, by scaling f , we may also assume that N = {|∇f | = 1}. Therefore, 0 = u0 (0) = A − B. and 1 = u(0) = A + B. This implies that u(t) = cosh t and g(t) = cosh−(m−2) t. Using (21.11), we conclude that H(t) = (m − 1) tanh t and

− → II αβ (t) = δαβ tanh t.

This implies that the metric on M = R × N must be of the form dsM = dt2 + cosh2 t ds2N

GEOMETRIC ANALYSIS

179

as claimed. If either A or B is 0, say B = 0, then u(t) = A exp t. In this case, after normalizing A = 1, g(t) = exp−(m−2)t . Using (21.11), we conclude that H(t) = (m − 1) and hαβ (t) = δαβ . This implies that the metric on M = R × N must be of the form ds2M = dt2 + exp(2t) ds2N . However, this implies that M has only one infinite volume end which contradicts our assumption. This completes the proof of the theorem. The curvature restriction on N follows by direct computation and from the curvature assumption on M.  Theorem 21.2. (Li-Wang) Let M m be a complete m-dimensional manifold with m ≥ 3. Suppose that Rij ≥ −(m − 1) and µ1 (M ) ≥

(m − 1)2 . 4

Then either (1) M has only one end; (2) M = R × N with the warped product metric ds2M = dt2 + exp(2t) ds2N , where N is a compact manifold with nonnegative Ricci curvature; or (3) M is of dimension 3 and M 3 = R × N 2 with the warped product metric ds2M = dt2 + cosh2 t ds2N , where N is a compact surface with Gaussian curvature bounded from below by −1. Proof. Assuming that M has more than one end, then Theorem 21.1 implies that there must only be one 2 nonparabolic end unless m = 3, in which case m − 2 = (m−1) . This accounts for case (3) in our theorem. 4 We may now assume that M has only one nonparabolic end E1 and at least one parabolic end E2 . Theorem 19.3 implies that there exists a nonconstant positive harmonic function f ∈ K0 (M ) defined on M with the property that lim supx→E2 (∞) f (x) = ∞. The gradient estimate of Theorem 6.1 asserts that |∇f |2 ≤ (m − 1)2 f 2 . Combining this estimate with the fact that f is harmonic we obtain (21.12)

1 1 3 ∆f 2 = − f − 2 |∇f |2 4 (m − 1)2 1 ≤− f2 4

180

PETER LI 1

If we denote h = f 2 , then for any nonnegative cutoff function φ we have Z Z Z 2 2 2 |∇(φ h)| = |∇φ| h − φ2 h ∆h M M M Z Z (m − 1)2 2 2 φ2 h2 ≤ |∇φ| h + 4 M M   Z (m − 1)2 2 − φ h h + ∆h . 4 M On the other hand, the characterization of µ1 (M ) ≥ (m − 1)2 4

Z

(m−1)2 4

φ2 h2 ≤

M

implies that

Z

|∇(φ h)|2 .

M

Hence, we conclude that Z (21.13) M

φ2 h



 Z (m − 1)2 h + ∆h ≤ |∇φ|2 h2 . 4 M

Integrating the gradient estimate of Theorem 6.1 along geodesics, we deduce that f must satisfy the growth estimate f (x) ≤ C exp((m − 1)r(x)), where r(x) is the geodesic distance from x to a fixed point p ∈ M. In particular, when restricted on the parabolic end E2 , together with the volume estimate of Theorem 20.2, we conclude that Z (21.14) f ≤ C ρ. E2 (ρ)

On the other hand, Corollary 20.1 asserts that on E1 , the function f must satisfy the decay estimate Z f 2 ≤ C exp(−(m − 1)ρ) E1 (ρ+1)\E1 (ρ)

for ρ sufficiently large. In particular, Schwarz inequality implies that   Z 1 (m − 1) f ≤ C exp − ρ VE21 (ρ + 1), 2 E1 (ρ+1)\E1 (ρ) where VE1 (r) denotes the volume of E1 (r). Combining this with the volume estimate of Proposition 21.1, we conclude that Z f ≤C E1 (ρ+1)\E1 (ρ)

for some constant C independent of ρ. In particular, Z f ≤ C ρ, E1 (ρ)

and together with (21.14) we obtain the growth estimate Z (21.15) f ≤ C ρ. Bp (ρ)

GEOMETRIC ANALYSIS

181

Choosing φ to be φ=

 1     2ρ − r    

ρ 0

on

Bp (ρ)

on

Bp (2ρ) \ Bp (ρ)

on

M \ Bp (2ρ),

in (21.13), we conclude that the right hand side is given by Z Z |∇φ|2 h2 = ρ−2 M

h2 .

Bp (2ρ)\Bp (ρ)

Inequality (21.15) implies that Z

|∇φ|2 h2 → 0

M

as ρ → ∞. Hence we obtain ∆h = −

(m − 1)2 h 4

∆h ≥ −

(m − 1)2 h 4

and the inequalities in deriving the estimate

are all equalities. In particular, |∇f | = (m − 1)f and |∇(log f )|2 = (m − 1)2 .

(21.16)

Hence the inequalities used to prove Theorem 6.1 must all be equalities. More specifically, (log f )1α = 0

for all

2 ≤ α ≤ m,

and δαβ |∇(log f )|2 m−1 = −(m − 1)δαβ for all

(log f )αβ = −

2 ≤ α, β ≤ m.

On the other hand, since e1 is the unit normal to the level set of log f, the second fundamental form (hαβ ) of the level set is given by (log f )αβ = (log f )1 hαβ = (m − 1)hαβ , hence implying that hαβ = −δαβ . log f Moreover, (21.16) also implies that if we set t = m−1 , then t must be the distance function between the level sets of f, hence also for the level set of log f. The fact that hαβ = −δαβ implies that the metric on M can be written as ds2M = dt2 + exp(−2t) ds2N . Since M has two ends, N must be compact. A direct computation shows that the condition Rij ≥ −(m − 1) ˜ ij ≥ 0. This proves the theorem.  implies that R

182

PETER LI

§22 Manifolds with Finite Volume In this section, we assume that M m is a complete manifold with finite volume. Suppose that the Ricci curvature is bounded from below by Rij ≥ −(m − 1). Since the constant functions are L2 harmonic functions, this implies that 0 is an eigenvalue for the L2 spectrum of the Laplacian. We define the quantity λ1 (M ) by the Rayleigh quotient R |∇φ|2 M R , λ1 (M ) = inf R φ2 φ∈H 1,2 (M ), φ=0 M M R where the infimum is taken over all functions φ in the Sobolev space H 1,2 (M ) satisfying M φ = 0. This plays the role of a generalized first non-zero Neumann eigenvalue, although λ1 (M ) might not necessarily be an eigenvalue. Note that λ1 (M ) ≤ max{µ1 (Ω1 ), µ1 (Ω2 )}, for any two disjoint domains Ω1 and Ω2 of M, where µ1 (Ω1 ) and µ1 (Ω2 ) are their first Dirichlet eigenvalues respectively. In particular, we have λ1 (M ) ≤ µe (M ), the smallest essential spectrum of M. Therefore, according to Cheng’s Theorem [Cg] (Corollary 6.2), one always has (m − 1)2 . 4 The main purpose of this section is to prove the following theorem. λ1 (M ) ≤

Theorem 22.1. Let M m be a complete Riemannian manifold with Ricci curvature bounded from below by Rij ≥ −(m − 1). Assume that M has finite volume given by V (M ), and λ1 (M ) ≥

(m − 1)2 . 4

Then there exists a constant C(m) > 0 depending only on m, such that,  N (M ) ≤ C(m)

V (M ) Vp (1)

2

 ln

 V (M ) +1 , Vp (1)

where Vp (1) denotes the volume of the unit ball centered at any point p ∈ M. (m−1)2 implies 4 (m−1)2 ∪ [ 4 , ∞).

The assumption λ1 (M ) ≥

that 0 is the only eigenvalue below

(m−1)2 . 4

So the spectrum of

the M satisfies σ(M ) ⊂ {0} In the special case when m = 2, our estimate is less effective than using the Cohn-Vossen-Hartman formula (see [LT4]). Indeed, the assumption that K ≥ −1 implies that the negative part of the Gaussian curvature defined by  0 if K > 0 K− = −K if K ≤ 0, is at most 1. In particular, Z K− ≤ V (M ) M

GEOMETRIC ANALYSIS

183

and M has finite total curvature. Hartman’s theorem then implies that M must be conformally equivalent to a compact Riemannian surface of genus g with N (M ) punctures. Moreover, since M has finite volume, the Cohn-Vossen-Hartman formula (see [LT4]) asserts that Z −V (M ) ≤

K M

= 2πχ(M ). In particular, using the Euler characteristic is given by χ(M ) = 2 − 2g − N (M ), we conclude that N (M ) ≤ 2 − 2g + (2π)−1 V (M ). This indicates that the dependency in V (M ) is better than the one provided by Theorem 22.1, and the value of the theorem lies in the case when m ≥ 3. Note that when M = Hm /Γ is a hyperbolic manifold, with its universal covering given by the hyperbolic 2 , ∞). In this special case, the m-space Hm , the L2 -spectrum of the Laplacian on Hm is the interval [ (m−1) 4 2

can be expressed as λ1 (M ) = µ1 (Hm ), where µ1 (Hm ) is the greatest lower assumption that λ1 (M ) ≥ (m−1) 4 m bound of the spectrum of ∆ on H . With this point of view, it is also possible to prove a theorem analogous to Theorem 22.1 for locally symmetric spaces of finite volume. However, stronger results are available by using Margulis’ thick-thin decomposition. For any finite volume locally symmetric space M = X/Γ, where X is a symmetric space and Γ a discrete subgroup of the isometry group of X, the number of ends of M is always bounded by N (M ) ≤ C(X) V (M ), where the constant C depends on X and V (M ) is the volume of M. Theorem 22.2. Let M m be a complete Riemannian manifold with Ricci curvature bounded from below by Rij ≥ −(m − 1). Assume that M has finite volume given by V (M ), and µ1 (M \ Bp (ρ0 )) ≥

(m − 1)2 . 4

Then there exists a constant C(m) > 0 depending only on m, such that, the number of ends of M satisfies N (M ) ≤ C(m) V (M ) Vp−1 (1) exp((m − 1)ρ0 ), where Vp (1) denotes the volume of the unit ball centered at point p ∈ M. Proof. According to Theorem 20.2, if we denote Vp (ρ) to be the volume of the geodesic ball Bp (ρ), then for all ρ ≥ 2(ρ0 + 1) we have (22.1)

Vp (ρ + 2) − Vp (ρ) ≤ C (1 + (ρ − ρ0 )−1 ) exp(−(m − 1)(ρ − ρ0 )) (Vp (ρ0 + 1) − Vp (ρ0 ))

On the other hand, Proposition 21.1 asserts that if y ∈ ∂Bp (ρ + 1) then (22.2)

Vy (1) ≥ C1−1 exp(−(m − 1)ρ) Vp (1).

184

PETER LI

Obviously, if Nρ (M ) denotes the number of ends with respect to Bp (ρ), then there exists Nρ (M ) number of points {yi ∈ ∂Bp (ρ)} such that Byi (1) ∩ Byj (1) = ∅ for i 6= j. In particular applying (22.2) to each of the yi and combining with (22.1), we conclude that Nρ (M )

Nρ (M ) C1−1

exp(−(m − 1)ρ) Vp (1) ≤

X

Vyi (1)

i=1

≤ Vp (ρ + 2) − Vp (ρ) ≤ C (1 + (ρ − ρ0 )−1 ) exp(−(m − 1)(ρ − ρ0 )) (Vp (ρ0 + 1) − Vp (ρ0 )). This implies that Nρ (M ) ≤ C C1 (1 + (ρ − ρ0 )−1 ) exp((m − 1)ρ0 ) (Vp (ρ0 + 1) − Vp (ρ0 )) Vp−1 (1). Letting ρ → ∞, we conclude that the number of ends N (M ) of M is bounded by (22.3)

N (M ) ≤ C C1 exp((m − 1)ρ0 ) (Vp (ρ0 + 1) − Vp (ρ0 )) Vp−1 (1)

and the result follows.



We remark that if M has finitely many eigenvalues 0 < µ1 ≤ µ2 ≤ · · · ≤ µk below to see there exists ρ0 > 0 such that µ1 (M \ Bp (ρ0 )) ≥

(m−1)2 , 4

then it is easy

(m − 1)2 . 4

In particular, Theorem 22.1 implies that M must have finitely many ends. However, the estimate of the number of ends is not effective as it is unclear to us at this moment how to control the size of ρ0 in terms 2 of the eigenvalues below (m−1) . On the other hand, we will demonstrate below that ρ0 can be effectively 4 controlled if k = 1. We also remark that the proof of Theorem 22.1 can be applied to any finite volume locally symmetric space M = X/Γ. Indeed, it is known in this case that there exists ρ0 > 0 such that µ1 (M \ Bp (ρ0 )) = µ1 (X) and an appropriate version of the volume comparison theorem (in terms of the balls defined by the Buseman functions) holds on M, hence such M must have finitely many ends. Of course, as pointed out in the introduction, stronger result holds true. We would also like to point out that the question if λ1 (M ) = µ1 (X) is an important one, as indicated by the famous Selberg conjecture asserting that this is the case if M is an arithmetic surface. The following lemma allows us to estimate µ1 (Bp (ρ)) of a geodesic ball centered at p with radius ρ in terms of the volume of the ball. Note that we do not need to impose any curvature assumptions on M. Lemma 22.1. Let M be a complete Riemannian manifold. Then for any 0 < δ < 1, ρ > 2 and p ∈ M , we have     2 1 Vp (ρ) 12 µ1 (Bp (ρ)) ≤ 2 ln + ln . 4δ (ρ − 1)2 Vp (1) 1−δ Proof. For the ease of notation, we will use µ1 to denote µ1 (Bp (ρ)). We may assume µ1 ≥ 4ρ−2 as otherwise the conclusion automatically holds true. Let r(x) denote the distance function to a fixed point p ∈ M . The variational characterization of µ1 (Bp (ρ)) implies that

GEOMETRIC ANALYSIS

Z µ1

√ φ2 exp(−2δ µ1 r) ≤

M

185

√ |∇(φ exp(−δ µ1 r)|2

Z ZM

=

√ √ |∇φ|2 exp(−2δ µ1 r) − 2δ µ1 M Z √ 2 + δ µ1 φ2 exp(−2δ µ1 r)

√ φ exp(−2δ µ1 r)h∇φ, ∇ri

Z M

M

for any nonnegative Lipschitz function φ with support in Bp (ρ). In particular, for ρ > 2, if we choose

φ=

   

1

√ µ1 (ρ − r)    0

−1

on

Bp (ρ − µ1 2 )

on

Bp (ρ) \ Bp (ρ − µ1 2 )

on

M \ Bp (ρ)

−1

then we have √ (1 − δ ) µ1 exp(−2δ µ1 ) Vp (1) ≤ (1 − δ ) µ1 φ2 exp(−2δ µ1 r) M Z Z √ √ √ 2 = |∇φ| exp(−2δ µ1 r) − 2δ µ1 φ exp(−2δ µ1 r)h∇φ, ∇ri M M √ ≤ (1 + 2δ) µ1 exp(−2δ (ρ µ1 − 1))) Vp (ρ). 2

√

2

Z

Therefore, √ exp(2δ ( µ1 (ρ − 1))) ≤ and 2δ

√

 µ1 (ρ − 1) ≤ ln

The lemma follows by rewriting this inequality.

12 1−δ

3e2 Vp (ρ) 2(1 − δ) Vp (1) 

 + ln

Vp (ρ) Vp (1)

 .



As a corollary of this lemma, the bottom spectrum of any complete manifold M can be estimated in terms of its volume entropy. This result was proved in Corollary 20.3 with a different argument. Corollary 22.1. Let M be a complete manifold and µ1 (M ) its bottom spectrum. Then 1 µ1 (M ) ≤ 4

 2 ln Vp (ρ) lim inf . ρ→∞ ρ

Proof. Note that µ1 (M ) = limρ→∞ µ1 (Bp (ρ)). Now the result follows by first letting ρ go to infinity and then δ go to 1 in the estimate of Lemma 22.1.  We are now ready to prove Theorem 22.1. Proof of Theorem 22.1. Let p ∈ M be a fixed point. For any 0 < δ < 1, let      1 12 V (M ) ρ0 = ln + ln + 1. (m − 1)δ 1−δ Vp (1) Then according to Lemma 22.1,

186

PETER LI

µ1 (Bp (ρ0 )) ≤

(22.4)

(m − 1)2 . 4

We now observe that µ1 (M \ Bp (ρ0 )) ≥ Indeed, if (22.4) is valid and also µ1 (M \ Bp (ρ0 )) < (m−1)2 , 4

(m − 1)2 . 4

(m−1)2 , 4

then the variational principle will imply that

λ1 (M ) < contradicting our assumption. By Theorem 22.2, we have N (M ) ≤ C(m) V (M ) Vp−1 (1) exp((m − 1)ρ0 ). If V (M ) > eVp (1) then the claimed estimate follows by plugging in the value of ρ0 and setting δ =1−

1 . ln(V (M ) Vp−1 (1))

On the other hand, if V (M ) ≤ eVp (1), then ρ0 is bounded from above and hence N (M ) ≤ C(m) implying the estimate again.



As pointed out earlier, the key ingredients in the proof of Theorem 22.1 relied on the decay estimate of the volume (22.2) given by the upper bound of the greatest lower bound of the spectrum of the model manifold hyperbolic space Hm . Similar type theorems for K¨ahler manifolds and quaternionic K¨ahler manifolds also follow similarly by using the corresponding comparison results from [LW8] and [KLZ], respectively.

GEOMETRIC ANALYSIS

187

§23 Stability of Minimal Hypersurfaces in a 3-Manifold In the next two sections, we will apply the theory of harmonic functions to the study of complete minimal hypersurfaces in a complete manifold with nonnegative curvature. Let N m+1 be a complete manifold with nonnegative Ricci curvature. Suppose M m is a complete minimal − → hypersurface in N . If | II |2 denotes the square of the length of the second fundamental form of M and RN νν is the Ricci curvature of N in the direction of the unit normal ν to M , then M being stable in N is characterized by the stability inequality (1.4) Z (23.1) M

− → ψ 2 | II |2 +

Z M

ψ 2 RN νν ≤

Z

|∇ψ|2

M

for any compactly supported function ψ ∈ Hc1,2 (M ). Geometrically the stability inequality is derived from the second variation formula for the volume functional under normal variations. Hence a stable minimal hypersurface is not only a critical point of the volume functional but its second derivative is nonnegative with respect to any normal variations. The elliptic operator associated to the stability inequality is given by − → L = ∆ + | II |2 + RN νν . The stability of M is equivalent to the fact that the operator −L is nonnegative. We say that M has finite index when the operator −L has only finitely many negative eigenvalues. This has the geometric interpretation that there are only a finite dimensional space of normal variations violating the stability inequality. The study of stable minimal hypersurfaces can be viewed as an effort of proving a generalized Bernstein’s theorem. Bernstein first established that an entire minimal graph in R3 must be a plane. Recall that a minimal graph is a minimal hypersurface which is given by a graph of a function defined on R2 . The validity of Bernstein theorem in higher dimension was established for entire minimal graph in Rm+1 for m ≤ 7 by Simons [S], and many other authors, such as Fleming [Fl], Almgren [A], and DeGiorgi [De], for the lower dimensional cases. Counter-examples for m ≥ 8 was found by Bombieri, DeGiorgi, and Guisti [BDG]. Since entire minimal graphs are area minimizing, a natural question is to ask if a Bernstein type theorem is valid for stable minimal hypersurfaces in Rm+1 . In 1979, do Carmo and Peng [dCP] proved that a complete, stable, minimally immersed hypersurface M in R3 must be planar. At the same time, Fischer-Colbrie and Schoen [FCS] independently showed that a complete, stable, minimally immersed hypersurface M in a complete 3-dimensional manifold N with nonnegative scalar curvature must be either conformally a plane R2 or conformally a cylinder R × S1 . For the special case when N is R3 , they also proved that M must be planar. In 1984, Gulliver [Gu1] studied a yet larger class of submanifolds in R3 . He proved that a complete, oriented, minimally immersed hypersurface with finite index in R3 must have finite total curvature. In particular, applying Huber’s theorem, one concludes that the hypersurface must be conformally equivalent to a compact Riemann surface with finitely many punctures. The same result was also independently proved by Fischer-Colbrie in [FC]. In addition, she also proved that a complete, oriented, minimally immersed hypersurface with finite index in a complete 3-dimensional manifold with nonnegative scalar curvature must be conformally equivalent to a compact Riemann surface with finite punctures. Shortly after, Gulliver [Gu2] improved the result of Fischer-Colbrie and proved that if the ambient manifold has nonnegative scalar curvature then a minimal hypersurface with finite index must have quadratic area growth, finite topological type, and the square of the length of the second fundamental form must be integrable. Indeed, a complete surface with quadratic area growth and finite topological type must be conformally equivalent to a compact Riemann surface with finitely many punctures. In 1997, Cao, Shen, and Zhu [CSZ] considered the high dimensional cases of the theorem of do CarmoPeng and Fischer-Colbrie-Schoen. They proved that a complete, oriented, stable, minimally immersed hypersurface M n in Rn+1 must have only one end. This theorem was generalized by Li and Wang [LW4],

188

PETER LI

where they showed that a complete, oriented, minimally immersed hypersurface M n in Rn+1 with finite index must have finitely many ends. In a later paper [LW7], Li and Wang also generalized their theorem to minimal hypersurfaces with finite index in a complete manifold with nonnegative sectional curvature. We will first present the relationship between harmonic functions and stability of minimal hypersurfaces in this section. Some two dimensional results will be proved using this technique, while the higher dimensional results will be presented in the next section. Note that when the ambient manifold is Euclidean space and n ≥ 3, then M is non-parabolic by applying Corollary 18.2. The key issue is the validity of Sobolev inequality proved by Michael and Simon [MS] in the form  m−2 Z Z m 2m m−2 ≤C |∇u|2 |u| M

M n

on any minimal submanifolds of R . However, when N is only assumed to have nonnegative Ricci (or sectional) curvature, then M can be parabolic as in the case of the cylinder M = R × P in N = R2 × P. The next theorem states that the case M being parabolic is a very special situation. Theorem 23.1. Let M m be a complete, minimally immersed, stable, hypersurface in a manifold, N m+1 , with nonnegative Ricci curvature. If M is parabolic, then M must be totally geodesic in N . Moreover, the Ricci curvature Rνν of N in the normal direction also vanishes, and M must have nonnegative scalar curvature. Proof. According to Lemma 18.1 and the discussion thereafter, the assumption that M is parabolic can be characterized by the following construction. For a fixed p ∈ M , a given ρ > 0, and a sequence ρi > ρ with ρi → ∞, let fi be a sequence of harmonic functions satisfying ∆fi = 0,

Bp (ρi ) \ Bp (ρ),

on

with boundary conditions fi = 1,

on

∂Bp (ρ)

and fi = 0,

on

∂Bp (ρi ).

Then the manifold M is parabolic if and only if fi converges uniformly on compact subsets of M \ Bp (ρ) to the constant function 1. If this is the case, since Z

|∇fi |2 =

Z fi

Bp (ρi )\Bp (ρ)

∂Bp (ρi )

Z

∂fi − ∂r

Z fi ∂Bp (ρ)

∂fi ∂r

∂fi , ∂r

=− ∂Bp (ρ)

the fact that fi → 1 implies that the right hand side must tend to 0 as ρi → ∞. To prove the proposition, we consider the compactly supported function    1 ψ = fi   0

on

Bp (ρ)

on

Bp (ρi ) \ Bp (ρ)

on

M \ Bp (ρi ).

Using this as test function in the stability inequality (23.1), we conclude that Z Bp (ρ)

− → | II |2 +

Z Bp (ρ)

RN νν ≤

Z Bp (ρi )\Bp (ρ)

|∇fi |2 .

GEOMETRIC ANALYSIS

189

Since the right hand side vanishes as i → ∞, we have − → | II |2 +

Z

Z

Bp (ρ)

RN νν = 0,

Bp (ρ)

− → hence | II |2 and RN νν vanish identically on Bp (ρ). Due to the fact that ρ is arbitrary, this implies the vanishing − → N of II and Rνν on M. The nonnegativity of the scalar curvature of M follows by applying the Gauss curvature equation and using the assumption that N has nonnegative Ricci curvature.  Theorem 23.1 reduces our study of stable minimal hypersurfaces to the nonparabolic case. In this case, we will recall the lemma of Schoen and Yau [SY1]. Lemma 23.1. (Schoen-Yau) Let M m be a complete, minimally immersed, stable, hypersurface in N m+1 . Let us denote RN and K N to be the Ricci curvature and the sectional curvature of N , respectively. Suppose u is a harmonic function defined on M . Then the inequality Z

1 |∇φ| |∇u| ≥ m M 2

2

− → φ | II |2 |∇u|2 +

Z

+

2

M m X α=1

Z

2 φ2 R N νν |∇u|

M

Z

φ2 K N (e1 , eα ) |∇u|2 +

M

1 m−1

Z

φ2 |∇|∇u||2

M

holds for any compactly supported, nonnegative function φ ∈ Hc1,2 (M ). Proof. The Bochner formula of Lemma 21.1 asserts that (23.2)

∆|∇u|2 ≥ 2Rij ui uj +

m |∇u|−2 |∇|∇u|2 |2 . 2(m − 1)

If {e1 , e2 , . . . , em } is an orthonormal frame of M and (hij ) is the second fundamental form of M , then the Gauss curvature equation asserts that R11 = =

m X α=2 m X

K N (e1 , eα ) +

m X

h11 hαα −

m X

h21α

α=2

α=2

K N (e1 , eα ) − h211 −

α=2

m X

h21α ,

α=2

where we have used the assumption that M is minimal and K N (e1 , eα ) is the sectional curvature for the 2-plane section spanned by e1 and eα . On the other hand, using the inequality m X − → | II |2 = h2ij i,j=1

≥ h211 +

m X

h2αα + 2

α=2

≥

h211

≥

m m−1

+

(

m X

h21α

α=2

m X hαα )2 +2 h21α m−1 α=2 ! m X 2 2 h11 + h1α ,

Pm

α=2

α=2

190

PETER LI

we conclude that

m X

R11 ≥

m−1 − → | II |2 . m

K N (e1 , eα ) −

α=2

Choosing the orthonormal frame so that ∇u = |∇u| e1 , this implies that Rij ui uj ≥

m X

K N (e1 , eα ) |∇u|2 −

α=2

m−1 − → | II |2 |∇u|2 . m

Substituting this into (23.2) yields m X

∆|∇u| ≥

(23.3)

m−1 − |∇|∇u||2 → | II |2 |∇u| + . m (m − 1)|∇u|

K N (e1 , eα ) |∇u| −

α=2

By setting ψ = φ|∇u| with φ being a nonnegative compactly supported function on M , the stability inequality (23.1) implies that Z

− → φ2 | II |2 |∇u|2 +

M

Z

Z |∇φ|2 |∇u|2 + 2 φ |∇u| h∇φ, ∇|∇u|i M M Z + φ2 |∇|∇u||2 M Z Z 2 2 = |∇φ| |∇u| − φ2 |∇u| ∆|∇u|.

2 φ2 RN νν |∇u| ≤

M

Z

M

M

Combining with (23.3), we conclude that

1 m

Z

− → φ2 | II |2 |∇u|2 +

M

Z

2 φ2 RN νν |∇u| +

M

Z ≤

m Z X α=1

φ2 K N (e1 , eα ) |∇u|2 +

M

1 m−1

Z

φ2 |∇|∇u||2

M

|∇φ|2 |∇u|2 . 

M

As a corollary of Theorem 23.1 and Lemma 23.1, we readily recover the theorems of Fischer-Colbrie and Schoen [FCS]. This theorem was also independently proved by do Carmo and Peng [dCP] for the special case when N = R3 . In the case of minimal surface in a 3-dimensional manifold, if we denote S N to be the scalar curvature of N and K to be the Gaussian curvature of M , then using the Gauss curvature equation the stability inequality can be written as (23.4)

1 2

1 − → ψ 2 | II |2 + 2 M

Z

Z M

ψ2 S N −

Z

ψ2 K ≤

M

Z

|∇ψ|2

M

for any nonnegative, compactly supported function ψ ∈ Hc1,2 (M ). Theorem 23.2. (Fischer-Colbrie-Schoen) Let M 2 be an oriented, complete, stable, minimal hypersurface in a complete manifold N 3 with nonnegative scalar curvature. Then M must be conformally equivalent to either the complex plane C or the cylinder R × S1 . If M is conformally equivalent to the cylinder and has finite total curvature, then it must be totally geodesic and the scalar curvature of N along M must be identically 0.

GEOMETRIC ANALYSIS

191

˜ be the universal covering of M. By the uniformization theorem, M ˜ must be conformally Proof. Let M 2 ˜ equivalent to either the unit disk D or the complex plane C. We claim that M cannot be conformally equivalent to D2 . → ˜ by lifting the functions |− To see this, we first observe that the stability inequality still holds on M II |2 ˜ . In fact, if ψ is a nonnegative, compactly supported function on M ˜ and π : M ˜ → M is the and S N to M covering map, then the function ψ¯ defined by X ψ¯2 (x) = ψ 2 (˜ x) x ˜∈π −1 (x)

is a nonnegative, compactly supported function on M. The stability inequality on M asserts that Z Z Z 1 1 − → ψ 2 | II |2 + ψ2 S N − ψ2 K 2 M˜ 2 M˜ ˜ M Z Z Z 1 →2 1 2 − ¯ = ψ | II | + ψ¯2 S N − ψ¯2 K 2 M 2 M M Z ¯ 2. ≤ |∇ψ| M

On the other hand, Schwarz inequality implies that 2 X 2 ¯ ¯ |ψ∇ψ| = ψ∇ψ x˜∈π−1 (x) X X ≤ ψ2 x ˜∈π −1 (x)

x ˜∈π −1 (x)

X

= ψ¯2

|∇ψ|2

|∇ψ|2 .

x ˜∈π −1 (x)

Hence

Z

¯2≤ |∇ψ|

Z

|∇ψ|2

˜ M

M

˜ . If M ˜ is conformally equivalent to D2 , the invariance of the and the stability inequality is valid on M Laplace operator in dimension 2 asserts that there exist non-constant, bounded harmonic functions with ˜ . Moreover, if u is such a harmonic function then Lemma 7.2 asserts that finite Dirichlet integral on M ∆|∇u| ≥ K |∇u| +

|∇|∇u|| |∇u|

˜ imply that and the proof of Lemma 23.1 applying to the stability inequality (23.4) on M Z Z Z Z 1 1 →2 2 − 2 2 N 2 2 2 (23.5) φ | II | |∇u| + φ S |∇u| + φ |∇|∇u|| ≤ |∇φ|2 |∇u|2 2 M˜ 2 M˜ ˜ ˜ M M for any nonnegative, compactly supported function  1     2ρ − r φ= ρ     0

˜ ). However, choosing φ ∈ Hc1,2 (M on

Bp (ρ)

on

Bp (2ρ)

on

˜ \ Bp (2ρ), M

192

PETER LI

the right hand side becomes Z

Z

1 |∇φ| |∇u| = 2 ρ ˜ M 2

2

|∇u|2 .

Bp (2ρ)\Bp (ρ)

The fact that u has finite Dirichlet integral implies that this tends to 0 as ρ → ∞. Therefore, inequality (23.5) − → ˜ has finite asserts the vanishing of II and |∇|∇u||. In particular, |∇u| must be identically constant and M ˜ volume because u has finite Dirichlet integral. This contradicts the assumption and M must be conformally equivalent to C. Using the uniformization theorem again, we conclude that M must be conformally equivalent to either the complex plane C or the cylinder R × S1 . This proves the first part of the theorem. If M is conformally equivalent to R×S1 and has finite total curvature, then applying the proof of Theorem 23.1 to the stability inequality (23.4) we conclude that (23.6)

1 2

1 − → | II |2 + 2 M

Z

Z

SN −

M

Z K ≤ 0. M

In particular, the assumption that N has nonnegative scalar curvature implies that Z K ≥ 0. M

Cohn-Vossen inequality [CV] then asserts that Z 2π χ(M ) ≥

K ≥ 0. M

Since M is the cylinder, we conclude that 0 on M as claimed. 

R M

− → K = 0 and (23.6) asserts that | II |2 and S N are both identically

Combining this theorem with Theorem 23.1, we obtained the following corollary, which was also proved in [FCS]. Corollary 23.1. (Fischer-Colbrie-Schoen) Let M 2 be an oriented, complete, stable, minimal hypersurface in a complete manifold N 3 with nonnegative Ricci curvature. Then M must be totally geodesic in N and the Ricci curvature of N in the normal direction to M must be identically zero along M. Moreover, M is either (1) conformally equivalent to the complex plane C; or (2) isometrically the cylinder R × S1 . Moreover, if N = R3 , then M must be planar in R3 .

GEOMETRIC ANALYSIS

193

§24 Stability of Minimal Hypersurfaces in a Higher Dimensional Manifold Let us now assume that N m+1 is a complete manifold with nonnegative sectional curvature. For a fixed point p ∈ N , let γ : [0, ∞) → N be a normal geodesic ray emanating from p. Recall that the Buseman function, βγ , with respect to γ is defined by βγ (x) = lim (t − d(x, γ(t)). t→∞

We define that the Buseman function, β, with respect to the point p by β(x) = sup βγ (x). γ

It was proved in [CG1] that β(x) is a convex exhaustion function of N . Suppose M n is a minimally immersed hypersurface in N m+1 , then a direct computation implies that the restriction of β on M is subharmonic with respect to the induced metric. The following lemma is useful for the construction of harmonic functions on a manifold possessing subharmonic functions with certain properties. Since the manifold is only assumed to be complete, it is likely that there are other applications of this lemma. Lemma 24.1. (Li-Wang) Let M m be a complete manifold, and E is an end of M with respect to Bp (1). Suppose g is a subharmonic function defined on E with the property that its maximum is not achieved on ∂E. Let us define E(ρ) = Bp (ρ) ∩ E and s(ρ) = sup∂Bp (ρ)∩E g. For any sequence {ρi } with ρi → ∞, there exists a subsequence, which we also denote by {ρi }, and a sequence of positive constants {Ai }, such that, the solutions {ui } to the boundary value problem ∆ui = 0

on

E(ρi ),

ui = 0

on

∂E,

and ui = Ai

∂E(ρi ) \ ∂E,

on

converges on compact subsets of E to a positive harmonic function u with boundary value u=0

on

∂E.

Moreover, the sequence {Ai } satisfies the bound 0 < Ai ≤ C s(ρi ) for some constant 0 < C < ∞, and Z

|∇ui |2 = Ai .

E(ρi )

Proof. Lemma 18.2 asserts that there exists a sequence ρi → ∞, a sequence of constants Ci → ∞ and a sequence of positive functions fi satisfying ∆fi = 0

on

E(ρi ),

fi = 0

on

∂E,

and fi = Ci

on

∂Bp (ρi ) ∩ E,

194

PETER LI

which converges to a positive harmonic function f. Moreover, Z

|∇fi |2 =

Z

∂fi − ∂r

fi

E(ρi )

∂Bp (ρi )∩E

Z = Ci ∂Bp (ρi )∩E

Z = Ci ∂E

ui = ∂E

then Z ui = Ci ∂E

∂fi ∂r

fi ∂E

∂fi ∂r

∂fi ∂r

∂fi . ∂r

Let us define Z

Z

∂fi ∂r

−1 fi ,

−1 on ∂Bp (ρi ) ∩ E.

The sequence ui will also converge to the harmonic function Z u= ∂E

∂f ∂r

−1 f.

We now claim that there exists a constant 0 < C < ∞ such that the sequence {Ai = Ci satisfies lim sup i→∞

R

∂fi ∂E ∂r

−1

}

Ai = C. s(ρi )

Indeed, the maximum principle asserts that Ai (g(x) − s(1)) ≤ ui (x) s(ρi )

(24.1)

on E(ρi ) since this inequality is satisfied on ∂E(ρi ). The assumption that g does not achieve its maximum on ∂E and the maximum principle imply that s(ρ) > s(1) for some ρ > 1. For this particular ρ, let x ∈ ∂E(ρ)\∂E such that g(x) = s(ρ). Applying (24.1) and letting i → ∞, we conclude that

lim sup i→∞

Ai u(x) ≤ s(ρi ) s(ρ) − s(1) < ∞,

confirming the existence of C. Integrating by parts yields Z E(ρi )

Z

∂ui ∂r ∂E(ρi )\∂E Z ∂ui = Ai . ∂E(ρi )\∂E ∂r

|∇ui |2 =

ui

GEOMETRIC ANALYSIS

195

On the other hand, since Z 0=

∆ui E(ρi )

Z = ∂E(ρi )\∂E

Z

∂ui − ∂r

∂E

∂ui , ∂r

we conclude that Z

|∇ui |2 = Ai

E(ρi )

Z ∂E

∂ui ∂r

= Ai .  Theorem 24.1. (Li-Wang) Let M m be a complete, stable, minimally immersed hypersurface in N m+1 . Suppose N is a complete manifold with nonnegative sectional curvature. If M is parabolic, then it must be totally geodesic and has nonnegative sectional curvature. In particular, M either has only one end, or M = R × P with the product metric, where P is compact with nonnegative sectional curvature. If M is non-parabolic, then it must only have one non-parabolic end. In this case, any parabolic end of M must be contained in a bounded subset of N. Proof. If M is parabolic, Theorem 23.1 implies that M must be totally geodesic and the Ricci curvature of N in the normal direction must vanish on M . In particular, the Gauss curvature equation asserts that M has nonnegative sectional curvature. Moreover, if M has more than one end, then Theorem 4.2 implies that M must be isometrically R × P for some compact manifold P with nonnegative sectional curvature. From this point on, we may assume that M is non-parabolic and has at least two ends. Suppose E and F are two ends of M with respect to the compact set Bp (ρ0 ). The non-parabolicity of M implies that at least one of the ends is non-parabolic. For the sake of argument, let E be the non-parabolic end. Let us assume that F is either non-parabolic, or that the Buseman function β when restricted to F is unbounded, hence satisfying the hypothesis of Lemma 24.1. Note that if M is properly immersed in N then β will be unbounded on each end of M. Using the Buseman function β in the role of g in Lemma 24.1, we observe that because |∇β| ≤ 1, s(ρ) is at most linear growth. In particular, there exists a sequence of harmonic functions {ui } defined on F (ρi ) that converges to a positive harmonic function u defined on F. Moreover, they satisfy ui = 0 and

on

Z ∂F

∂F

∂ui = 1. ∂r

Similarly, because E is assumed to be non-parabolic, by passing to a subsequence if necessary, there exists a sequence of harmonic functions {vi } defined on E(ρi ) that converges to a bounded harmonic function v defined on E. The sequence also satisfy vi = 0 and

Z ∂E

∂vi = ∂r

on Z E(ρi )

∂E

|∇vi |2 = 1.

196

PETER LI

Let us define a sequence of functions ki on Bp (ρi ) by

ki (x) =

  

vi (x)

on

E(ρi )

−ui (x)

on

F (ρi )

0

on

Bp (ρi ) \ (E(ρi ) ∪ F (ρi )).

 

Obviously, ki is harmonic on E(ρi ) ∪ F (ρi ) and Z Bp (1)

∂ki = ∂r

Z ∂E

∂vi − ∂r

Z ∂F

∂ui ∂r

= 0. Let wi be the solution to the boundary value problem ∆wi = 0

on

Bp (ρi )

and w i = ki

on

∂Bp (ρi ).

The fact that wi minimizes Dirichlet integral implies that Z

Z

2

|∇ki |2

|∇wi | ≤ Bp (ρi )

Bp (ρi )

Z

2

Z

|∇vi | +

= Ei

|∇ui |2

Fi

≤ 2C ρi . On the other hand, according to Theorem 19.2, since ki converges to the function  on E   v(x) on F k(x) = −u(x)   0 on M \ (E ∪ F ), there exists a constant C1 > 0 independent of i such that a subsequence of {wi } converges to a harmonic function w on M satisfying |w − k| ≤ C1 . Applying Lemma 23.1, we have Z (24.2)

(m − 1) M

− → φ2 |∇wi |2 | II |2 + m

Z M

φ2 |∇|∇wi ||2 ≤ m(m − 1)

Z

|∇φ|2 |∇wi |2

M

for any nonnegative function φ ∈ Hc1,2 (M ) supported on Bp (ρi ). In particular, for any fixed ρ0 < ρ < ρi , if we set  1 on Bp (ρ)  φ(x) = ρi − r(x)  on Bp (ρi ) \ Bp (ρ), ρi − ρ

GEOMETRIC ANALYSIS

197

then we have Z (24.3)

(m − 1)

− → |∇wi | | II |2 + m 2

Bp (ρ)

Z

|∇|∇wi ||2

Bp (ρ)

− → φ |∇wi | | II |2 + m

Z

2

≤ (m − 1)

2

M

φ2 |∇|∇wi ||2

M

Z

|∇φ|2 |∇wi |2

≤ m(m − 1) m(m − 1) ≤ (ρi − ρ)2

Z

ZM

|∇wi |2

Bp (ρi )

C ρi . (ρi − ρ)2

≤ Letting i → ∞, this implies that

− → |∇w|2 | II |2 = 0

Z Bp (ρ)

and

Z

|∇|∇w||2 = 0.

Bp (ρ)

Since ρ is arbitrary, we conclude that |∇w| is constant. Because w is a non-constant harmonic function, this − → means that |∇w| = 6 0, hence | II |2 must be identically 0 and M is a totally geodesic submanifold. The Gauss curvature equation then asserts that M must have nonnegative sectional curvature also. The assumption that M has at least two ends and Theorem 4.2 implies that M must split isometrically into R × P, where P must be a compact manifold with nonnegative sectional curvature. However, this contradicts the assumption that M is non-parabolic, hence M must have only one end.  Corollary 24.1. Let M m be a complete, properly immersed, stable, minimal hypersurface in N m+1 . Suppose N is a complete manifold with nonnegative sectional curvature. Then either (1) M has only one end; or (2) M = R × P with the product metric, where P is compact with nonnegative sectional curvature, and M is totally geodesic in N . We will now consider the case when the minimal hypersurface has finite index. The following lemma proved by Fischer-Colbrie [FC] give an effective way to use the finite index assumption. Lemma 24.2. (Fischer-Colbrie) Let M be a complete manifold and V be a locally bounded function defined on M. Suppose the spectrum, Spec(−L), of the elliptic operator L = ∆ + V (x) on the negative axis is given by only finitely many negative eigenvalues, then there exists a compact set D such that −L is nonnegative on M \ D. Proof. Let Ω ⊂ M be any compact subset of M. The variational principle asserts that the lowest eigenvalue of L is given by R R |∇φ|2 − Ω V φ2 Ω R µ1 (L, Ω) = inf . φ2 Ω where the infimum is taken over all compactly supported φ ∈ Hc1,2 (Ω). In particular, since Z Z Z Z 2 2 2 |∇φ| − V φ ≥ µ1 (Ω) φ − sup V φ2 Ω

Ω

Ω

Ω

Ω

198

PETER LI

where µ1 (Ω) is the first Dirichlet eigenvalue of the Laplacian on Ω, we conclude that (24.4)

µ1 (L, Ω) ≥ µ1 (Ω) − sup V. Ω

For a fixed p ∈ M, let us consider the geodesic ball Bp (ρ). Inequality (24.4) asserts that µ1 (L, Bp (ρ)) ≥ 0 for ρ sufficiently small because µ1 (Bp (ρ)) → ∞ as ρ → ∞. Let us define ρ1 = 2 sup{ρ | µ1 (L, Bp (ρ)) ≥ 0}. If ρ1 = ∞, then −L is nonnegative on M and the lemma is proved. Hence we may assume that ρ1 < ∞ and define ρ2 = 2 sup{ρ | µ1 (L, Bp (ρ) \ Bp (ρ1 )) ≥ 0}. Again, if ρ2 = ∞, then −L is nonnegative on M \ Bp (ρ1 ). Inductively we can define ρi = 2 sup{ρ | µ1 (L, Bp (ρ) \ Bp (ρi−1 )) ≥ 0}. We claim that some ρi must be infinite, hence −L is nonnegative on M \ Bp (ρi−1 ). If not, we get a infinite sequence of ρi → ∞. Moreover the definition of ρi and the monotonicity of µ1 implies that −L is negative on all the annuli Bp (ρi ) \ Bp (ρi−1 ). Let fi be the eigenfunction on Bp (ρi ) \ Bp (ρi−1 ) such that Lfi = −µ1 (L, Bp (ρi ) \ Bp (ρi−1 )) fi , where µ1 (L, Bp (ρi )\Bp (ρi−1 )) < 0. Since the set of {fi } has disjoint support, they spanned an ∞-dimensional subspace of Hc1,2 (M ) such that −L is negative. This contradicts the assumption that −L has only finitely many negative eigenvalues and the lemma is proved.  Theorem 24.2. (Li-Wang) Let M m be a complete, minimally immersed hypersurface with finite index in N m+1 . Suppose N is a complete manifold with nonnegative sectional curvature. Let N (M ) be the number of ends of M that is either non-parabolic, or parabolic but not contained in any bounded subset of N , then there exists a constant C > 0 depending on a compact set of the M, such that, N (M ) ≤ C. Proof. Lemma 24.2 asserts that the assumption that M has finite index implies that there exists a compact subset D such that M \ D is stable. In particular, the stability inequality (23.1) holds for any compactly supported function ψ defined on M \ D. Following the argument as in the proof of Theorem 24.1, for each pair of such ends E and F , the sequence of harmonic functions {wi } converges to a harmonic function w on M satisfying (24.2) with φ being a compactly support function on M \ D. Moreover, Theorem 19.4 asserts that the space of harmonic functions, K = {w}, constructed above will have dimension equal to the number of ends. Hence, it suffices to estimate the dimension of K. Note that for any w ∈ K, we may assume that it can be approximated by a sequence of harmonic functions wi whose Dirichlet integral is at most linear growth, as indicated by Lemma 24.1. Let ρ0 be sufficiently large such that D ⊂ Bp (ρ0 ) and set φ to be  0 on Bp (ρ0 )      on Bp (ρ0 + 1) \ Bp (ρ0 )  r(x) − ρ0 φ(x) = 1 on Bp (ρ0 + 2) \ Bp (ρ0 + 1)       ρi − r(x) on Bp (ρi ) \ Bp (ρ0 + 2) ρi − ρ0 − 2 for ρ0 + 2 < ρi . Following the argument for (24.3), we conclude that Z Z →2 2 2 − (m − 1) φ |∇wi | | II | + m φ2 |∇|∇wi ||2 M M Z ≤ m(m − 1) |∇φ|2 |∇wi |2 ZM Z m(m − 1) ≤ m(m − 1) |∇wi |2 + |∇wi |2 2 (ρ − ρ − 2) i 0 B (ρ +1)\Bp (ρ0 ) Bp (ρi )\Bp (ρ) Z p 0 C ρi 2 . ≤ m(m − 1) |∇wi | + (ρi − ρ0 − 2)2 Bp (ρ0 +1)\Bp (ρ0 )

GEOMETRIC ANALYSIS

Letting i → ∞, we obtain the estimate Z (m − 1) (24.5)

− → φ2 |∇w|2 | II |2 + m M Z ≤ m(m − 1)

Z

199

φ2 |∇|∇w||2

M

|∇w|2 .

Bp (ρ0 +1)\Bp (ρ0 )

On the other hand, the Poincar´e inequality for mixed boundary condition asserts that there is a constant C > 0 depending on the set Bp (ρ0 + 2) \ Bp (ρ0 ) such that Z Z C f2 ≤ |∇f |2 Bp (ρ0 +2)\Bp (ρ0 )

Bp (ρ0 +2)\Bp (ρ0 )

for any f ∈ H 1,2 (Bp (ρ0 + 2) \ Bp (ρ0 )) with boundary condition f =0

on

∂Bp (ρ0 ).

Applying this to the function φ |∇w|, we conclude that Z C φ2 |∇w|2 Bp (ρ0 +2)\Bp (ρ0 )

Z

|∇(φ |∇w|)|2

≤ Bp (ρ0 +2)\Bp (ρ0 )

Z

2

≤2

Z

2

φ2 |∇|∇w||2

|∇φ| |∇w| + 2 Bp (ρ0 +1)\Bp (ρ0 )

Z

Bp (ρ0 +2)\Bp (ρ0 )

|∇w|2 + 2

≤2

Z

Bp (ρ0 +1)\Bp (ρ0 )

φ2 |∇|∇w||2 .

Bp (ρ0 +2)\Bp (ρ0 )

Combining with (24.5), we have Z (24.6)

|∇w|2

C Bp (ρ0 +2)\Bp (ρ0 +1)

Z

φ2 |∇w|2

≤ C1 Bp (ρ0 +2)\Bp (ρ0 )

Z

|∇w|2 .

≤ Bp (ρ0 +1)

Since the function |∇w| satisfies the differential inequality − → ∆|∇w| ≥ −| II |2 |∇w|, the local mean value inequality (Corollary 14.2) asserts that there exists a constant C > 0 depending on the set Bp (ρ0 + 2), such that, if x ∈ Bp (ρ0 + 1) then Z |∇w|2 (x) ≤ C |∇w|2 . Bx (1)

Together with (24.5), this implies sup Bp (ρ0 +1)

|∇w|2 ≤ C

Z Bp (ρ0 +1)

|∇w|2 .

200

PETER LI

The dimension estimate on K then follows from Lemma 7.2 and the same argument as in the proof of Theorem 7.2.  We would like to remark that the properness assumption of M in N in Corollary 24.1 is only used to ensure that the Buseman functions satisfy the hypothesis of Lemma 24.1 on a parabolic end. In the event when the end is not properly immersed, it is still possible that one can find a Buseman function satisfying the property that its maximum is not achieved on ∂E. For example, if the end E is not contained in a compact set of N then the Buseman function must not be bounded on E and the hypothesis of Lemma 24.1 is met. If this is the case, the proof of Corollary 24.1 is still valid for that end. When N = Rm+1 , we can also recover the theorems in [CSZ] and [LW4]. Corollary 24.2. (Cao-Shen-Zhu) Let M m be a minimal hypersurface in Rm+1 with m ≥ 3. If M is stable, then M must have only one end. Proof. According to Corollary 24.1 and the above remark, we only need to rule out the case when M has a parabolic end. Indeed, by a Sobolev inequality of Michael and Simon [MS] and Corollary 18.2, any end E must be nonparablic, and the corollary follows.  The above argument also proves the finite index case. Corollary 24.3. (Li-Wang) Let M m be a minimally immersed hypersurface in Rm+1 with m ≥ 3. If M has finite index then there exist a constant C > 0 depending on M such that N (M ) ≤ C.

GEOMETRIC ANALYSIS

201

§25 Linear Growth Harmonic Functions In the forthcoming three sections, we will consider polynomial growth harmonic functions on a complete manifold. Recall that any polynomial growth harmonic function in Rm is necessarily a polynomial with respect to the variables in rectangular coordintates. Hence the space of polynomial growth harmonic functions of at most order d are given by the space of harmonic polynomials of degree at most d. In particular, these spaces are all finite dimensional (see Appendix B). Definition 25.1. Let us denote Hd (M ) to be the vector space of all polynomial growth harmonic functions defined on M of order at most d, i.e., Hd (M ) = {f | ∆f = 0, and |f (x)| ≤ C rd (x) for some constant C > 0}, where r(x) is the distance from x to a fixed point p ∈ M. We also denote the dimension of the vector space Hd (M ) by hd (M ). Using this notation, it was computed in Appendix B that     m+d−1 m+d−2 d m h (R ) = + . d d−1 On the other hand, Cheng’s theorem (Corollary 6.3) asserts that if M has nonnegative Ricci curvature, then hd (M ) = 1 for all d < 1. In view of this, Yau conjectured that hd (M ) must be finite dimensional if M has nonnegative Ricci curvature. In 1989, Li and Tam proved that h1 (M ) ≤ m + 1 if M has nonnegative Ricci curvature. Note that this estimate is sharp and it is achieved by Rm . In fact, Li [L7] showed that if M is K¨ ahler and has nonnegative holomorphic bisectional curvature then equality holds if and on if M = Cn with 2n = m. The real case of this theorem was later proved by Cheeger, Colding, and Minicozzi [CCM]. They showed that if M has nonnegative Ricci curvature and h1 (M ) = m + 1, then M = Rm . We will first present the theorem of Li and Tam [LT3], which is a finer version of the statement mentioned above. The proof relies on Lemma 16.2, which can be viewed as a sharp mean value inequality at infinity Theorem 25.1. (Li-Tam) Let M be a complete manifold with nonnegative Ricci curvature. Suppose the volume growth of M satisfies lim sup ρ−n Vp (ρ) < ∞ ρ→∞

for some n ≤ m, then h1 (M ) ≤ n + 1 ≤ m + 1. Proof. For any f ∈ H1 (M ), the curvature assumption, Lemma 3.2 and Corollary 6.3 imply that |∇f |2 is a bounded subharmonic function on M . In particular, Lemma 16.2 asserts that Z −1 (25.1) lim Vp (ρ) |∇f |2 = sup |∇f |2 . ρ→∞

M

Bp (ρ)

For a fixed point p ∈ M , let us define the subspace of H1 (M ) given by H0 = {f | f (p) = 0, f ∈ H1 (M )} and ¯ on H0 by a bilinear form D Z ¯ g) = lim Vp−1 (ρ) D(f, h∇f, ∇gi. ρ→∞

Bp (ρ)

202

PETER LI

Observe that the fact that f − f (p) ∈ H0 for any f ∈ H1 (M ) implies that the codimension of H0 = H1 (M ) ¯ is an inner product on H0 because of (25.1). Given any finite dimensional subspace H00 is 1. Note that D 0 of H with dim H00 = k, let {f1 , . . . , fk } be an orthonormal basis of H00 with respect to this inner product. Note that if we define the function F (x) by F 2 (x) =

k X

fi2 (x),

i=1

then it is independent of orthogonal transformation on the set {fi }. Similar to the argument in Lemma 7.2, for each x ∈ M , we may assume that fi (x) = 0 for all i 6= 1. At this point, we see that F 2 (x) = f12 (x) and F (x) ∇F (x) = f1 (x) ∇f1 (x). Hence |∇F |(x) ≤ |∇f1 |(x) ≤ 1

(25.2)

because supM |∇f1 |2 = 1 due to (25.1). Also, integrating along a geodesic and using the fact that F (p) = 0, we have F (x) ≤ r(x).

(25.3) Applying (25.2) to the integral (25.4)

2

k Z X i=1

|∇fi |2 =

Z

Bp (ρ)

∆(F 2 )

Bp (ρ)

Z =2

F ∂Bp (ρ)

∂F ∂r

Z ≤2

F ∂Bp (ρ)

and combining with (25.3), this implies that (25.5)

Vp−1 (ρ)

k Z X i=1

|∇fi |2 ≤

Bp (ρ)

ρ Ap (ρ) . Vp (ρ)

¯ implies that for any  > 0, there The fact that {fi } are orthonormal with respect to the inner product D exists ρ sufficiently large such that k Z X k −  ≤ Vp−1 (ρ) |∇fi |2 . i=1

Bp (ρ)

Hence together with (25.5) yield k− Ap (ρ) ≤ ρ Vp (ρ) for ρ ≥ ρ0 with ρ0 sufficiently large. Integrating this inequality from ρ0 to ρ gives  k− ρ Vp (ρ) ≤ . ρ0 Vp (ρ0 ) This implies that k −  ≤ n. Since  is arbitrary, the theorem follows. 1



We are now ready to consider the equality case when h (M ) = m + 1 for the above theorem. First we need the following lemmas.

GEOMETRIC ANALYSIS

203

Lemma 25.1. Let M m be a complete manifold with nonnegative Ricci curvature. If h1 (M ) = m + 1, then the function m X

F (x) =

! 21 fi2 (x)

,

i=1

defined in the proof of Theorem 25.1, must satisfy lim

x→∞

F (x) = 1. r(x)

Proof. Inequality (25.7) asserts that m X

Z

Vp−1 (ρ)

(25.6)

|∇fi |2 ≤

Bp (ρ) i=1

ρ Vp (ρ)

Z ∂Bp (ρ)

F . ρ

On the other hand, since Rij ≥ 0, the Laplacian comparison theorem asserts that ∆r2 ≤ 2m, hence, Z 2mVp (ρ) ≥ ∆r2 Bp (ρ)

= 2ρ Ap (ρ). Combining this estimate with (25.6), we obtain Vp−1 (ρ)

Z

m X

2

|∇fi | ≤

mA−1 p (ρ)

Z ∂Bp (ρ)

Bp (ρ) i=1

F . ρ

Letting ρ → ∞ and using (25.3), we conclude that lim A−1 p (ρ)

(25.7)

ρ→∞

Z ∂Bp (ρ)

F = 1. ρ

For any given δ > 0, let mδ (ρ) be the (m − 1)-dimensional measure of the set   F (x) x ∈ ∂Bp (ρ) 0 and a sequence of {xi } with xi → ∞ such that F (xi ) r−1 (xi ) ≤ 1 − α. i) For any y ∈ Bxi ( α r(x ), inequality (25.2) asserts that 4

F (y) r−1 (y) ≤ F (xi ) r−1 (y) + r(xi , y) r−1 (y) α ≤ (1 − α)r(xi ) r−1 (y) + r(xi ) r−1 (y), 4

204

PETER LI

where r(xi , y) is the distance between xi and y. However, since r(y) ≥ (1 − α4 )r(xi ), this implies that F (y) r−1 (y) ≤ 1 − Setting δ =

2α 4−α

2α . 4−α

and using (25.8), we see that for any  > 0, by taking i sufficiently large, we have, 

(25.9)

V xi

α r(xi ) 4



Z

r(xi )+

≤ r(xi )−

Z

α r(xi ) 4

α r(xi ) 4

r(xi )+

≤ r(xi )−

mδ (r) dr

α r(xi ) 4

α r(xi ) 4

Ap (r) dr

  α r(xi ) . ≤  Vp r(xi ) + 4 On the other hand, the volume comparison theorem asserts that     α n α r(xi ) Vxi (3r(xi )) ≥ Vxi 4 12  α n ≥ Vp (2r(xi )). 12 Combining with (25.9), we conclude that

 α n

≤ . 12 Since  is arbitrary, this give a contradiction, and the lemme is proved.  Using the orthonormal basis {f1 , . . . , fm } of H0 = {f ∈ H1 (M ) | f (p) = 0}, we define the map Φ : M → Rm by Φ(x) = (f1 , . . . , fm ). m

Lemma 25.2. Suppose X : Bp (ρ) → R is a function defined on Bp (ρ) with |X(x)| = 1 for all x ∈ Bp (ρ). For any subset A ⊂ Bp (ρ) and any θ > 0, let us define the set Cθ,A = {v ∈ S(A) | |dΦ(v) − X(π(v))| < θ} where S(A) = {v ∈ Tq (M ) | q ∈ A, |v| = 1} is the unit sphere bundle over A. Then for any 0 <  ≤ θ4 , we can take ρ sufficiently large, such that   V (Cθ,A ) V¯ (θ − 2) V (A) ≥ − , V (S(Bp (ρ)) V (Sm−1 ) Vp (ρ) where V¯ (θ) denotes the (m − 1)-dimensional measure of the set given by {v ∈ Sm−1 | |v − v0 | ≤ θ}, for some fixed vector v0 ∈ Sm−1 . Proof. For any  > 0, let us define the set D = {y ∈ Bp (ρ) | |hv, wi − hdΦ(v), dΦ(w)i| < , for all v, w ∈ Sy (M )}.

GEOMETRIC ANALYSIS

205

¯ ρ implies that The fact that {fi } form an orthonormal basis with respect to the inner product D lim Vp−1 (ρ)

ρ→∞

Z |h∇fi , ∇fj i − δij | = 0 Bp (ρ)

for all 1 ≤ i, j ≤ m, hence V (D) ≥1− Vp (ρ) for ρ sufficiently large. This implies that V (A ∩ D) ≥ V (A) −  Vp (ρ). For any fixed y ∈ A∩D, and an orthonormal basis {ei } of Ty (M ), the set {dΦ(ei )} form for dΦ(Ty (M )) Pambasis 2 if  is sufficiently small. This means that there exists a set of numbers {ai }m with a = 1, such that, i=1 i=1 i m X ai dΦ(ei ) < . X(y) − i=1

Clearly, the set defined by ( P =

v ∈ Sy (M )

) m X ai ei < θ − 2, with y ∈ D ∩ A v − i=1

satisfies P ⊂ Cθ,A . Therefore, V (Cθ,A ) ≥ V (P ) = V¯ (θ − 2) V (D ∩ A) ≥ V¯ (θ − 2)(V (A) − Vp (ρ)), and this implies the lemma.



We are now ready to consider the equality case in Theorem 25.1 Theorem 25.2. (Cheeger-Colding-Minicozzi) Let M m be a complete manifold with nonnegative Ricci curvature. If h1 (M ) = m + 1 then M = Rm . Proof. According to the Bishop volume comparison theorem, to show that M = Rm , it suffices to show that for any  > 0, the volume of geodesic balls satisfy Vp (ρ) ≥ (1 − ) ωm ρm , for all ρ sufficiently large. Indeed, this will imply that lim ρ−m Vp (ρ) = ωm

ρ→∞

and the equality case of the volume comparison will imply that M = Rm .

206

PETER LI

For any  > 0, let us define A to be a subset of Rm by A = B0 ((1 − )ρ) \ Φ(Bp (ρ)). Let us choose a set of open balls B1 = {Bxi (ρi )} using the following procedure. For any x ∈ A, let ρx be the supremum of those values r such that Bx (r) ⊂ A. In particular, ∂Bx (ρx ) ∩ ∂A 6= ∅. If ∂Bx (ρx ) ∩ Φ(Bp (ρ)) 6= ∅, then we say that Bx (ρx ) ∈ B1 . If ∂Bx (ρx ) ∩ Φ(Bp (ρ)) = ∅, then there must be a point y ∈ ∂Bx (ρx ) ∩ B0 ((1 − )ρ) and x must contained in the line segment γ(t) joining y to 0. Suppose x = γ(t0 ) for some 0 < t0 < (1 − )ρ. Let us consider the balls Bγ(t) (t) centered at γ(t) of radius t for t ≥ t0 . It is clear that y ∈ ∂Bγ(t) (t) and x ∈ Bγ(t) (t). Let t1 be the first t > t0 such that Bγ(t1 ) (t1 ) ∩ Φ(Bp (ρ)) 6= ∅. Note that t1 < (1−)ρ must exists since 0 = Φ(p) and Bγ(t1 ) (t1 ) ⊂ B0 ((1 − )ρ). In particular, Bγ(t1 ) (t1 ) ⊂ A, and we 2 include Bγ(t1 ) (t1 ) in our collection B1 . Clearly, B1 covers A and ρx ≤ (1−)ρ for all Bx (ρx ) ∈ B1 . 2 We will choose the set {Bxi (ρi )} among the collection B1 inductively. Let Bx1 (ρ1 ) = Bx1 (ρx1 ) be chosen, such that, ρx1 ≥ 12 supx∈A ρx . For a fixed α > 1, we define B2 = {Bx (ρx ) ∈ B1 | Bx (α ρx ) ∩ Bx1 (α ρ1 ) = ∅}. Let us choose Bx2 (ρ2 ) = Bx2 (ρx2 ) such that ρx2 ≥

1 2

supBx (ρx )∈B2 ρx . Inductively, we define Bi+1

 

   i  X = Bx (ρx ) ∈ B1 Bx (α ρx ) ∩  Bxj (α ρj ) = ∅ .   j=1

If Bi+1 = ∅, then {Bxj (ρj )}ij=1 is our collection. If Bi+1 6= ∅, then we choose Bxi+1 (ρi+1 ) = Bxi+1 (ρxi+1 ) such that ρxi+1 ≥ 21 supBx (ρx )∈Bi+1 ρx . In any case, we will get a collection B = {Bxj (ρj )} whether it is finite or not. This collection will have the properties that (25.10)

Bxi (Ri ) ∩ Φ(Bp (ρ)) 6= ∅,

and (25.11)

Bxi (α ρi ) ∩ Bxj (α ρj ) = ∅,

for i 6= j.

We now claim that (25.12)

A ⊂ ∪B Bx (5α ρx ).

To see this, suppose y ∈ A \ ∪B Bx (5α ρx ) then there exists Bx (ρx ) ∈ B1 , such that, y ∈ Bx (ρx ) and Bx (α ρx ) ∩ (∪B Bx (α ρx )) 6= ∅, otherwise the inductive procedure will continue. Let i be the smallest such number so that Bx (α ρx ) ∩ Bxi (α ρi ) 6= ∅. This implies that Bx (ρx ) ∈ Bi

GEOMETRIC ANALYSIS

and hence ρi ≥

207

1 ρx . 2

Therefore, we conclude that y ∈ Bxi (5α ρi ), which is a contradiction, hence (25.12) is valid. Let D be any relatively compact subdomain D ⊂ A such that 2V (D) ≥ V (A). Since D is compact, there exists a finite subcover {Bxi (ρi )} of the cover B such that D ⊂ ∪i Bxi (5α ρi ). In particular, this implies that V (D) ≤

X

Vxi (5α ρi )

i

= (5α)m

X

Vxi (ρi ).

i

Hence the finite collection {Bxi (ρi )} will not only satisfy (25.10) and (25.11), it also satisfies V (A) ≤ 2(5α)m

(25.13)

X

Vxi (ρi ).

i

Let yi ∈ ∂Bxi (ρi ) ∩ Φ(Bp (ρ)) and pi ∈ Bp (ρ) such that Φ(pi ) = yi . Since |∇fi |2 ≤ 1, we have |dΦ(v)| ≤

√

m|v|

for any tangent vector v. Hence, we have √ Φ(Bpi ((2 m)−1 ρi )) ⊂ Byi (2−1 ρi )

(25.14)

⊂ Bxi ((2−1 + 1)ρi ). For any θ > 0, let us define  Cθ,i =

 v ∈ S Bpi



ρ √i 2 m

  |dΦ(v) + ∇rxi (Φ(π(v)))| < θ ,

where rxi is the distance function to xi . We also define and Cθ = ∪i Cθ,i . Note that by taking α = 32 , (25.14) and (25.11) imply that √ √ Φ(Bpi ((2 m)−1 ρi ) ∩ Φ(Bpj ((2 m)−1 ρj ) = ∅ and Cθ,i ∩ Cθ,j = ∅

208

for i 6= j. Also Bpi that (25.15)

PETER LI



ρi √ 2 m



√ ⊂ Bp ((1 + (4 m)−1 ) ρ) since ρi ≤

ρ 2,

after applying Lemma 25.2, we conclude

V (Cθ ) √ V (S(Bp ((1 + (4 m)−1 ) ρ))) !
  X √ −1 V¯ θ2 ρi −1 √ ≥ Vpi Vp ((1 + (4 m) ) ρ) −  . V (Sm−1 ) 2 m i

On the other hand, volume comparison theorem asserts that     √ √ ρ ρ √i √i Vpi Vp−1 ((1 + (4 m)−1 ) ρ) ≥ Vpi Vp−1 ((2 + (4 m)−1 ) ρ) i 2 m 2 m  m ρi √ . ≥ (4 m + 2−1 ) ρ Therefore combining with (25.15), we have V¯ ( θ2 ) V (Cθ ) √ −1 ≥ V (Sm−1 ) V (S(Bp ((1 + (4 m) ) ρ))



P m  ρ √ i i−1 m −  . ((4 m + 2 )ρ)

However, property (25.13) asserts that  2

15 2

m ωm

X

ρm i ≥ V (A)

i

hence (25.16)


  V¯ θ2 V (A) V (Cθ ) √ −1 ≥ C −  , 1 V (Sm−1 ) ρm V (S(Bp ((1 + (4 m) ) ρ)))

for some constant 0 depending only on m.  C1 >  ρ i Let v ∈ S Bpi 2√m be a unit tangent vector. We denote γv to be the normal geodesic enamating in the direction of v, i.e., γv0 (0) = v. If γv (ρi ) ∈ Bp (ρ), then Φ(γv (ρi )) ∈ / A. If γv (ρi ) ∈ / Bp (ρ), then because of Lemma 25.1, for ρ sufficiently large, ρ(γv (ρi )) > (1 − )r(γv (ρi )), hence Φ(γv (ρi )) ∈ / B0 ((1 − )ρ) and Φ(γv (ρi )) ∈ /A also. Note that since γv (0) ∈ Bpi that



ρi √ 2 m



and (25.14) imply that Φ(γv (0)) ∈ Bxi

Φ(γv (0)) − ρi ∇rxi (Φ(γv (0)) ∈ Bxi

ρ  i

2

.

Together with the fact that Φ(γv (ρi )) ∈ / A, we have 1 < −∇rxi (Φ(π(v))) − ρ−1 i (Φ(γv (ρi ) − Φ(γv (0))) 2

3ρi 2




\Bxi

ρi 2




, we conclude

GEOMETRIC ANALYSIS



for all v ∈ S Bpi



ρi √ 2 m



. Thus for θ ≤

1 4



and v ∈ Cθ ∩ S Bpi

209



ρi √ 2 m



, we obtain

1 < |dΦ(v) − ρ−1 i (Φ(γv (ρi )) − Φ(γv (0))|. 4 Hence we have V (Cθ ) ≤ 4

XZ ρ √i 2 m

Cθ ∩S Bpi

i

−1

|dΦ(v) − ρi (Φ(γv (ρi )) − Φ(γv (0))|.

Combining this with (25.16), we obtain the estimate (25.17)


  V¯ θ2 V (A) −  C 1 V (Sm−1 ) ρm 1

≤ 4V −1 (S(Bp ((1 + 2m− 2 ) ρ))) XZ −1 ×

|dΦ(v) − ρi (Φ(γv (ρi )) − Φ(γv (0))|. i

ρ √i 2 m

Cθ ∩S Bpi

We now claim that for any  > 0, we can choose ρ sufficiently large such that the right hand side of (25.17) is at most . To see this, we consider Z ρi d −1 dΦ(v) − ρ (Φ(γ (t))) dt |dΦ(v) − ρ−1 (Φ(γ (ρ )) − Φ(γ (0))| = v v i v i i dt 0 Z ρi (dΦ(v) − dΦ(γv0 (t))) dt = ρ−1 i 0 Z ρ i Z t d 0 = ρ−1 (dΦ(γ (s))) ds dt v i 0 0 ds Z ρi Z ρi ≤ ρ−1 |Hess(Φ)(γv0 (s), γv0 (s))| ds dt i 0 Z ρi 0 ≤ |Hess(Φ)(γv0 (s), γv0 (s))| ds. 0

Therefore, we have (25.18) 1

V −1 (S(Bp ((1 + 2m− 2 ) ρ)))

XZ Cθ ∩S(Bpi (ρi ))

i −1

≤2

ρV

−1

− 21

(Bp ((1 + 2m

|dΦ(v) − ρ−1 i (Φ(γv (ρi )) − Φ(γv (0))|

Z ) ρ))

1

|Hess(Φ)|.

Bp ((1+2m− 2 ) ρ)

On the hand, the Bochner formula implies that 2 ∆|∇f |2 ≥ 2fij

for any harmonic function on M. Let φ(t) be a cut-off function satisfying the properties  φ(t) =

1

for

t≤ρ

0

for

t ≥ 2ρ,

210

PETER LI

with 0 ≤ φ(t) ≤ 1, −C ρ−1 ≤ φ0 (t) ≤ 0, and |φ00 (t)| ≤ C ρ−2 . Taking the composition with the function P 2
21 F (x) = and integrate by parts, we obtain i fi Z Z 2 2 (25.19) 2 fij ≤ φ(F ) fij Bp ((1−)ρ)

M

Z ≤ ZM ≤

φ(F ) ∆(|∇f |2 − 1) ∆φ (F ) (|∇f |2 − 1)

M

Z ≤

(φ0 (F ) ∆F + φ00 (F )|∇F |2 )(|∇f |2 − 1).

M

The fact that ∆F 2 = 2

P

i

|∇fi |2 ≥ 2|∇F |2 implies that ∆F ≥ 0.

Also 2F ∆F + 2|∇F |2 ≤ ∆F 2 ≤ 2m implies that ∆F ≤ mF −1 . By taking f = fi for i = 1, . . . , m and using the fact that |∇fi | ≤ 1, (25.18) can be estimated by Z Z Z 2 −1 −1 2 −2 2 fjk ≤ C ρ F (1 − |∇f | ) + C ρ (1 − |∇f |2 ) F ≤2ρ

Bp ((1−)ρ)

≤ C ρ−2

F ≤2ρ

Z

(1 − |∇f |2 )

Bp ((2+)ρ)

≤  C ρ−2 Vp ((2 + )ρ) m  2+ . ≤  C ρ2 Vp ((1 − )ρ) 1− This implies that Z

! 12

Z |Hess(Φ)| ≤

Bp (ρ)

|Hess(Φ)|

2

1

Vp2 (ρ)

Bp (ρ)

≤  C ρ−1 Vp (ρ). Combining with (25.18) and (25.17), we conclude that V (A) ≤  C ρm . This implies that (25.20)

V (Φ(Bp (ρ))) ≥ V0 ((1 − )ρ) − V (A) ≥ (1 − )m V0 (ρ) −  Cρm .

On the other hand, since |dΦ| ≤ 1, we have Vp (ρ) ≥ V (Φ(Bp (ρ)). Hence combining with (25.19), we conclude that Vp (ρ) ≥ (1 − ) V0 (ρ) for sufficiently large ρ. This proves the theorem. 

GEOMETRIC ANALYSIS

211

§26 Polynomial Growth Harmonic Functions In the case of Yau’s conjecture, it was first proved by Colding and Minicozzi [CM1-3] and they gave the estimate hd (M ) ≤ C dm−1 on a manifold with nonnegative Ricci curvature, where C > 0 is a constant depending only on M . In [L8], Li gave a much simplified proof of this estimate which holds for a much larger class of manifolds. We will present this argument below. Definition 26.1. A manifold is said to satisfy a volume comparison condition (Vµ ) for some µ > 1, if there exists a constant CV > 0, such that, for any point x ∈ M and any real numbers 0 < ρ1 ≤ ρ2 < ∞, the volume of the geodesic balls centered at x satisfy the inequality  Vx (ρ2 ) ≤ CV Vx (ρ1 )

ρ2 ρ1

µ .

Definition 26.2. A manifold is said to satisfy a mean value inequality (M) if there exists a constant CM > 0 such that for any x ∈ M , ρ > 0, and any nonnegative subharmonic function f defined on M , it must satisfy Z f 2 (x) ≤ CM Vx−1 (ρ)

f 2 (y) dy.

Bx (ρ)

Note that if M has nonnegative Ricci curvature then M satisfies condition (Vm ) with CV = 1, by the Bishop volume comparison theorem. Moreover, condition (M) is also valid on such manifold because of Theorem 7.1. Lemma 26.1. (Li) Let K be a k-dimensional linear space of sections of a vector bundle E over M. Assume that M has polynomial volume growth at most of order µ, i.e., Vp (ρ) ≤ C ρµ for ∂ ∈ M and ρ → ∞. Suppose each section u ∈ K are polynomial growth of at most degree d, such that |u|(x) ≤ C rd (x), where r(x) is the geodesic distance to the fixed point p ∈ M. For any β > 1, δ > 0, and ρ0 > 0, there exists ρ > ρ0 such that if {ui }ki=1 is an orthonormal basis of K with respect to the inner product Aβρ (u, v) = R hu, vi, then Bp (βρ) k Z X |ui |2 ≥ k β −(2d+µ+δ) . i=1

Bp (ρ)

R Proof. For each ρ > 0, let Aρ be the nonnegative bilinear form defined on K given by Aρ (u, v) = Bp (ρ) hu, vi. Let us denote the trace of the bilinear form Aρ with respect to Aρ0 by trρ0 Aρ . Similarly, let detρ0 Aρ be the determinant of Aρ with respect to Aρ0 . Assuming that the lemma is false, then for all ρ > ρ0 , we have trβρ Aρ < k β −(2d+µ+δ) . On the other hand, the arithmetic-geometric means asserts that 1

(detβρ Aρ ) k ≤ k −1 trβρ Aρ .

212

PETER LI

This implies that detβρ Aρ ≤ β −k (2d+µ+δ) for all ρ > ρ0 . Setting ρ = β i ρ for i = 0, 1, . . . , j − 1 and iterating this inequality j times yield detβ j ρ Aρ ≤ β −jk(2d+µ+δ) .

(26.1)

However, for a fixed Aρ -orthonormal basis {ui }ki=1 of K, the assumptions on K and on the volume growth imply that there exists a constant C > 0, depending on K, such that Z

) |ui |2 ≤ C (1 + ρ2d+µ 1

Bp (ρ1 )

for all 1 ≤ i ≤ k and for all ρ1 ≥ ρ. In particular, this implies that detρ Aβ j ρ ≤ k! C β jk(2d+µ) ρk(2d+µ) . This contradicts (26.1) as j → ∞, and the lemma is proved.  Theorem 26.1. (Li) Let M m be a complete manifold satisfying conditions (Vµ ) and (M). Suppose E is a rank-n vector bundle over M. Let Sd (M, E) ⊂ Γ(E) be a linear subspace of sections of E, such that, all u ∈ Sd (M, E) satisfy (a) ∆|u| ≥ 0, and (b) |u|(x) ≤ O(rd (x)) as x → ∞. Then the dimension of Sd (M, E) is finite. Moreover, for all d ≥ 1, there exists a constant C > 0 depending only on µ such that dim Sd (M, E) ≤ n C CM dµ−1 . Proof. Let K be a finite dimensional linear subspace of Sd (M, E) with dim K = k and {ui }ki=1 be any basis of K. Then for p ∈ M, ρ > 0, and any 0 <  < 1/2, we claim that they must satisfy the estimate k Z X

(26.2)

i=1

Z

|ui |2 ≤ n C CM −(µ−1)

u∈{hA,U i}

Bp (ρ)

|u|2 ,

sup Bp ((1+)ρ)

where the supremum is taken over all u ∈ K of the form u = hA, U i for some unit vector A = (a1 , . . . , ak ) ∈ Rk with U = (u1 , . . . , uk ). PR To see this, we will estimate |ui |2 by using a argument similar to Lemma 7.2. Observe that for Bp (ρ) any x ∈ Bp (ρ), there exists a subspace Kx ⊂ K which is of at most codimension-n, such that u(x) = 0 for all u ∈ Kx . Hence, by an orthonormal change of basis, we may assume that ui ∈ Kx for n + 1 ≤ i ≤ k and Pk Pn 2 2 i=1 |ui | (x) = i=1 |ui | (x). Since ∆|ui | ≥ 0, if we denote the distance from p to x by r(x), then the mean value inequality (M) implies that

(26.3)

k X

|ui |2 (x) =

i=1

n X

|ui |2 (x)

i=1

≤

CM Vx−1 ((1

+ )ρ − r(x))

n Z X i=1

≤

CM Vx−1 ((1

|ui |2

Bx ((1+)ρ−r(x))

Z + )ρ − r(x)) n

sup u∈{hA,U i}

Bp ((1+)ρ)

|u|2 .

GEOMETRIC ANALYSIS

213

However, condition (Vµ ) and the fact that r(x) ≤ ρ imply that µ (1 + )ρ − r(x) CV Vx ((1 + )ρ − r(x)) ≥ Vx (2ρ) 2ρ  µ (1 + )ρ − r(x) ≥ Vp (ρ). 2ρ 

(26.4)

Hence, substituting into (26.3) and integrating over Bp (ρ), we have k Z X

(26.5)

|ui |2 ≤

Bp (ρ)

i=1

Z nCV CM 2µ sup u2 Vp (ρ) u∈{hA,U i} Bp ((1+)ρ) Z × ((1 + ) − ρ−1 r(x))−µ dx. Bp (ρ)

On the other hand, if we define f (r) = ((1 + ) − ρ−1 r)−µ then f 0 (r) = µ ρ−1 ((1 + ) − ρ−1 r)−(µ+1) ≥ 0 and

Z

−1

((1 + ) − ρ

−µ

r(x))

ρ

Z dx =

Ap (t) f (t) dt.

Bp (ρ)

0

To estimate this, we integrate by parts and obtain ρ

Z (26.6)

f (t) Ap (t) dt =

[f (t) Vp (t)]ρ0

Z

ρ

−

0

f 0 (t) Vp (t) dt.

0

Using f 0 (t) ≥ 0 and applying condition (Vµ ), we have Z

ρ

f 0 (t) Vp (t) dt ≥ ρ−µ Vp (ρ)

0

ρ

Z

f 0 (t) tµ dt

0

≥ρ

−µ

 Vp (ρ)

[f (t) tµ ]ρ0

Z −µ

ρ

f (t) t

µ−1

 dt .

0

Substituting this into (26.6) yields Z

ρ

f (t) Ap (t) dt = ρ−µ Vp (ρ) µ

0

≤ ρ−1 Vp (ρ) µ

Z

ρ

Z 0ρ

f (t) tµ−1 dt ((1 + ) − tρ−1 )−µ dt

0

µ Vp (ρ) (−µ+1 − (1 + )−µ+1 ) µ−1 µ ≤ Vp (ρ) −(µ−1) . µ−1 =

The claim follows by combining this with (26.5). To complete the proof of the theorem, let {ui }ki=1 be an Aβρ -orthonormal basis of any finite dimensional subspace K ⊂ Sd (M, E). Clearly, it suffices to prove the estimate for k = dim K. Note that condition (Vµ )

214

PETER LI

implies that the volume growth of M is at most of order ρµ , hence Lemma 26.1 implies that there is a ρ > 0 such that k Z X |ui |2 ≥ k β −(2d+µ+δ) . Bp (ρ)

i=1

On the other hand, by setting β = 1 + , inequality (26.2) implies that k Z X i=1

because

R Bp ((1+)ρ)

|ui |2 ≤ n C CM −(µ−1)

Bp (ρ)

|u|2 = 1 for all u ∈ {hA, U i}. For d ≥ 1, the estimate on k follows by setting  = (2d)−1

and observing that (1 + (2d)−1 )−(2d+µ+δ) is bounded from below.  Note that extra care is used to obtain the order µ − 1 in Theorem 26.1 because this is sharp on Euclidean space (B.3). In particular, we recover the theorem of Colding and Minicozzi as a corollary. Corollary 26.1. (Colding-Minicozzi) Let M m be a complete Riemannian manifold with nonnegative Ricci curvature. There exists a constant C > 0 depending only on m, such that, dimension of the Hd (M ) is bounded by hd (M ) ≤ C dm−1 for all d ≥ 1. Finite dimensionality of Hd (M ) can be obtain be relaxing both the volume comparison condition (CalVµ ) and the mean value inequality condition (M). However, the order of dependency in d will not be sharp. Definition 26.3. A complete manifold M is said to satisfy a weak mean value inequality (WM) if there exists constants CM > 0 and b > 1 such that, for any nonnegative subharmonic function f defined on M , it must satisfy f (x) ≤ CWM Vx−1 (ρ)

Z f (y) dy. Bx (bρ)

for all x ∈ M and ρ > 0. Theorem 26.2. Let M be a complete manifold satisfying the weak mean value property (WM). Suppose that the volume growth of M satisfies Vp (ρ) = O(ρµ ) as ρ → ∞ for some point p ∈ M. Then Hd (M ) is finite dimensional for all d ≥ 0 and dim Hd (M ) ≤ CWM (2b + 1)(2d+µ) . Proof. Let β = 2b + 1, where b is the constant in (WM). By Lemma 26.1, for any δ > 0, there exists ρ > 0 such that for u1 , . . . , uk , an orthonormal basis of Hd (M ) with respect to the inner product Aβρ (u, v), we have (26.7)

k Z X i=1

On the other hand, since the function exists a point q ∈ ∂Bp (ρ) such that

(26.8)

Pk

u2i ≥ k β −(2d+µ+δ) .

Bp (ρ)

i=1

u2i (x) is subharmonic, the maximum principle implies that there

k X i=1

u2i (x) ≤

k X i=1

u2i (q)

GEOMETRIC ANALYSIS

215

for all x ∈ Bp (ρ). Again, by an orthonormal change of basis, we may assume that uj (q) = 0 for 2 ≤ j ≤ k. Applying the weak mean value inequality (WM) to u21 and noting that Bp (ρ) ⊂ Bq (2ρ) ⊂ Bp ((2b + 1)ρ), we get Vp (ρ) u21 (q) ≤ Vq (2ρ) u21 (q) Z ≤ CWM

u21

Bq (2bρ)

Z

u21

≤ CWM Bp ((2b+1)ρ)

= CWM . Thus, integrating (26.8) over the ball Bp (ρ) gives k Z X

(26.9)

i=1

u2i ≤ Vp (ρ)

Bp (ρ)

k X

u2i (q)

i=1

≤ Vp (ρ) u21 (q) ≤ CWM . From (26.7) and (26.9) we conclude that k β −(2d+µ+δ) ≤ CWM , hence k ≤ CWM β (2d+µ) as δ > 0 is arbitrary. This completes the proof.



Corollary 26.2. Let M be a complete manifold. Suppose there exist constants C1 , C2 > 0 and µ > 2, such that, for all p ∈ M, ρ > 0, and for all f ∈ Hc1,2 (M ), we have Z |f |

2µ µ−2

! µ−2 µ ≤ C1 Vp (ρ)

−2 µ

ρ2

Z

|∇f |2 + C2 Vp (ρ)

Bp (ρ)

Bp (ρ)

−2 µ

Z

f 2.

Bp (ρ)

Then dim Hd (M ) < C β d for all d ≥ 0 for some constant C > 0 and β > 1 depending only on C1 , C2 , µ, and b. Proof. Applying the Moser iteration process, which will be presented in more generality in §33, the Sobolev type inequality in the hypothesis implies the weak mean value inequality (WM). Also, by choosing f to be a cut-off function satisfying ( 1 if x ∈ Bp (1) f (x) = 0 if x ∈ M \ Bp (2) and |∇f | ≤ 1, and applying to the Sobolev inequality in the hypothesis we conclude that µ−2 µ

Vp

2 −µ

(1) ≤ C1 Vp

(ρ)(C1 ρ2 + C2 )Vp (2).

This implies that Vp (ρ) = O(ρµ ) as ρ → ∞. The corollary follows from Theorem 26.2. 

216

PETER LI

In the case when M m has nonnegative sectional curvature, it was shown by Li and Wang in [LW3] that Pd if hd (M ) = dim Hd (M ), then i=1 hi (M ) has an upper bound which is asymptotically sharp as d → ∞. In particular, they showed that if 0 ≤ α0 ≤ ωm is a constant given by lim inf ρ−m Vp (ρ) = α0 , ρ→∞

then lim inf d−(m−1) hd ≤ d→∞

2α0 . (m − 1)! ωm

Moreover, lim inf d−(m−1) hd = d→∞

2 (m − 1)!

if and only if M = Rm . The proof of this will not be presented here, however this motivates the following question. Question. If M is a complete manifold with nonnegative sectional curvature, then is it true that hd (M ) ≤ hd (Rm )     m+d−1 m+d−2 = + ? d d−1

GEOMETRIC ANALYSIS

217

§27 Mean Value Constant, Liouville Property, and Minimal Submanifolds In this section, we will assume that M satisfies a slight stronger version of the mean value property (M) as in Definition 26.2. Definition 27.1. A manifold is said to satisfy an L1 mean valued inequality (M1 ) if there exists a constant CM1 > 0 such that for any x ∈ M , ρ > 0, and any nonnegative subharmonic function f defined on M , it must satisfy Z f (x) ≤ CM1 Vx−1 (ρ)

f (y) dy. Bx (ρ)

2 Note that by applying Schwarz inequality, (M1 ) implies (M), with CM < CM . It turns out that the 1 value of CM1 plays a significant role in both the regularity theory and the Liouville property of a manifold. It was proved by Li and Wang [LW2] that when CM1 is sufficiently close to 1, the mean value inequality (M1 ) becomes more powerful.

Theorem 27.1. (Li-Wang) Let M be a complete manifold satisfying the mean value inequality (M1 ). If CM1 < 2 and M has subexponential volume growth, then M has the Liouville property, i.e., M does not admit any non-constant bounded harmonic functions. Proof. Let f be a harmonic function defined on M . Let us define s(r) = sup f (x), ¯ p (r) B

i(r) = inf f (x), ¯ p (r) B

and ω(r) = s(r) − i(r). ¯p (r) such that f (x) = s(r) and f (y) = i(r). In fact, the maximum principle asserts Assume that x, y ∈ B that x, y ∈ ∂Bp (r). For any ρ > r, the mean value inequality applying to the functions f − i(ρ) and s(ρ) − f gives Z Z (27.1) CM1 (f − i(ρ)) ≥ CM1 (f − i(ρ)) Bp (ρ)

Bx (ρ−r)

≥ (s(r) − i(ρ)) Vx (ρ − r) and Z (27.2)

Z (s(ρ) − f ) ≥ CM1

CM1 Bp (ρ)

(s(ρ) − f ) By (ρ−r)

≥ (s(ρ) − i(r)) Vy (ρ − r). By taking ρ > 2r, we have Bp (ρ − 2r) ⊂ Bx (ρ − r) and Bp (ρ − 2r) ⊂ By (ρ − r), hence we have Vx (ρ − r) ≥ Vp (ρ − 2r) and Vy (ρ − r) ≥ Vp (ρ − 2r).

218

PETER LI

Combining these inequalities with (27.1) and (27.2), and adding the two estimates, we conclude that CM1 ω(ρ) Vp (ρ) ≥ (ω(r) + ω(ρ)) Vp (ρ − 2r). This implies that  (27.3)

CM1

 Vp (ρ) − 1 ω(ρ) ≥ ω(r). Vp (ρ − 2r)

Suppose now that f is a non-constant bounded harmonic function. Then by scaling we may assume that the total oscillation sup f − inf f = 1. Hence, ω(r) % 1 as r → ∞. In particular, (27.3) becomes (27.4)

C M1

For any r > 0, we claim that the sequence

Vp (ρ) − 1 ≥ ω(r). Vp (ρ − 2r)

Vp (2(i+1)r) Vp (2ir)

lim inf i→∞

must satisfy

Vp (2(i + 1)r) = 1. Vp (2ir)

Indeed, if this is not the case, then there exists i0 > 0 such that Vp (2(i + 1)r) ≥1+ Vp (2ir) for i ≥ i0 . Iterating this inequality k times beginning from i = i0 , we have Vp (2(k + i0 )r) ≥ (1 + )k Vp (2i0 r). Let k → ∞, this implies that M has exponential volume growth, contradicting the assumption. Hence there is a subsequence ij , such that Vp (2(ij + 1)r) →1 Vp (2ij r) as j → ∞. Setting ρj = 2(ij + 1)r, then Vp (ρj ) → 1. Vp (ρj − 2r) Together with (27.4), this implies that (CM1 − 1) ≥ ω(r) for all r > 0. However, this contradicts the assumptions that ω(r) % 1 and CM1 < 2, hence f must be identically constant. Theorem 27.2. (Li-Wang) Let M be a complete manifold satisfying the mean value inequality (M1 ) and the volume comparison condition Vµ . If CV CM1 < 2, then there exists a constant 0 < α < 1 depending only on ACM1 and µ such that any harmonic function f defined on M satisfying |f (x)| = O(rα (x)), where r is the distance to some fixed point p ∈ M , must be identically constant.

GEOMETRIC ANALYSIS

219

Proof. Following the proof of Theorem 27.1, inequalities (27.1) and (27.2) are valid for any harmonic functions defined on M . On the other hand, using the volume growth condition, we have  µ ρ−r CV Vx (ρ − r) ≥ Vx (ρ + r) ρ+r ≥ Vp (ρ) and similarly  CV

ρ−r ρ+r

µ Vy (ρ − r) ≥ Vp (ρ).

Hence substituting the above inequalities into (27.1) and (27.2), and taking their sum, we conclude that  µ ρ−r . CM1 ω(ρ) ≥ CV−1 (ω(r) + ω(ρ)) ρ+r This implies that  (27.5)

 CV CM1

ρ+r ρ−r

µ

 − 1 ω(ρ) ≥ ω(r).

Setting ρ = (1 + β)r, this becomes   µ  2+β CV CM1 − 1 ω((1 + β)r) ≥ ω(r). β Iterating this inequality k times yields  (27.6)

 CV CM1

2+β β

µ

k − 1 ω((1 + β)k r) ≥ ω(r).

On the other hand, the assumption on the growth rate of f implies that ω((1 + β)k r) ≤ C (1 + β)αk rα . Hence combining with (27.6) yields  (27.7)

 CV CM1

2 +1 β

µ

 k ω(r) − 1 (1 + β)α ≥ . C rα

Obviously, by taking β sufficiently large depending on µ and using the assumption that CV CM1 < 2, we may assume that µ  2 CV CM1 + 2 +1 ≤ , β 2CV CM1 which implies 

 CV CM1

2 +1 β

µ

 CV CM1 −1 ≤ . 2

Hence, there exists α sufficiently small depending on β and CV CM1 such that   µ  2 CV CM1 + 1 − 1 (1 + β)α < 1. β Under these choices, the left hand side of (27.7) will tend to 0 as k → ∞. Hence ω(r) = 0 and f is identically constant.

220

PETER LI

Theorem 27.3. (Li-Wang) Let M be a singular manifold satisfying the mean value inequality M1 and the weak volume growth condition Vµ in a neighborhood Bp (ρ) of a point p ∈ M. If CV CM1 < 2, then any bounded harmonic function defined on Bp (ρ) must be C α at p for some 0 < α < 1 depending on ACM1 and µ. Proof. As in the proof in Theorem 27.2, by setting r = βρ for some β < 1 in (27.5), we have 

 CV CM1

1−β 1+β

µ

 − 1 ω(ρ) ≥ ω(βρ).

Interating this k times, we obtain 

 CV CM1

1−β 1+β

k

µ −1

ω(ρ) ≥ ω(β k ρ).

H¨ older continuity follows if we can find α > 0 such that µ k   1−β −1 ≤ C(β k ρ)α . CV CM1 1+β This can be achieved by finding α such that  CV CM1

1−β 1+β

µ

− 1 ≤ βα.

To see this, we choose β sufficiently small so that 

1−β 1+β

µ ≤

which implies  CV CM1

1−β 1+β

2 + CV CM1 , 2CV CM1

µ −1≤

Therefore, there exists α sufficiently small so that β α ≥

CV CM1 . 2

CV CM1 2

, and the theorem follows.



Lemma 27.1. Let M m be a minimal varifold in RN . Let r¯(x, y) be the extrinsic distance function between ¯x (ρ) = {y ∈ two points x, y ∈ M obtained by the restriction of the Euclidean distance function to M and B M | r¯(x, y) < ρ} be the extrinsic ball centered at x of radius ρ. For any regular point x ∈ M and any nonnegative subharmonic function h defined on M , it must satisfy the mean value inequality ωm ρm h(x) ≤

Z h(y) dy ¯ x (ρ) B

for all x ∈ M and ρ ≥ 0. Proof. For a point x ∈ M , it is known that the extrinsic distance function r¯x (y) = r¯(x, y) satisfies |∇¯ rx | ≤ 1 and ∆¯ rx =

m−1 . r¯x

GEOMETRIC ANALYSIS

If we define

221

2−m 2−m ¯ x (y) = r¯x (y) − r G m(m − 2) ωm

with ωm = V¯ (1), then a direct computation shows that (m − 1)¯ rx−m (m − 1)¯ rx−m |∇¯ rx |2 + m ωm m ωm

¯x = − ∆G ≤ 0.

¯x (r) with pole Since x is a regular point, if we denote the Dirichlet Green’s function on the extrinsic ball B ¯ x ∼ Gx when y → x. Hence, together with the fact that Gx (y) = 0 = G ¯ x (y) when at x by Gx , then G ¯x (r), the maximum principle implies that y ∈ ∂B ¯ x (y) ≥ Gx (y). G This also implies that ¯ x (y) ∂G ∂Gx (y) ≥ ∂µ ∂ν r¯1−m (y) ∂ r¯x (y) =− x m ωm ∂ν r¯x1−m (y) ≥− m ωm ¯x (r) with ν being the outward pointing unit normal vector on ∂ B ¯x (r). on ∂ B Using the property of the Green’s function, if h is a non-negative subharmonic function, we have Z −h(x) = ¯ x (r) B

∆Gx (y) h(y) dy

Z

Z

=

Gx (y) ∆h(y) +

¯ x (r) B 1−m

r ≥− m ωm

¯ x (r) ∂B

∂Gx (y) h(y) ∂ν

Z h(y). ¯ x (r) ∂B

Multiplying both sides by rm−1 and integrating over [0, ρ], we obtain the mean value inequality Z

Z

ρ

Z

h= ¯ x (ρ) B

¯ x (r) B

0

Z

ρ

h |∇¯ rx |

Z

≥

h 0

¯ x (r) B m

≥ ωm ρ h(x).  Theorem 27.4. (Li-Wang) Let M m be a minimal varifold in RN . Let r¯(x, y) be the extrinsic distance function between two points x, y ∈ M . For a fixed point p ∈ M , suppose the density function θp (ρ) =

V¯p (ρ) ωm ρm

222

PETER LI

has an upper bound θp (ρ) ≤ θ¯ < 2, where V¯p (ρ) is the volume of the extrinsic ball of radius ρ center at p. Then there exists 0 < α < 1 depending only on θ¯ and m, such that, if f is a harmonic function satisfying the growth condition |f (x)| = O(¯ rpα (x)), then f must be identically constant. Proof. If f is a harmonic function, then applying a similar argument as in Theorem 27.2 and Lemma 27.1, we obtain Z f − V¯p (ρ) i(ρ) ≥ ω ¯ m (ρ − r)m f (x) − ωm (ρ − r)m i(ρ) ¯ p (ρ) B

and V¯p (ρ) s(ρ) −

Z ¯ p (ρ) B

f ≥ ωm (ρ − r)m s(ρ) − ωm (ρ − r)m f (y)

¯p (r). In particular, since smooth points are dense in B ¯p (r), this implies that for any smooth points x, y ∈ ∂ B 

 θp (ρ)

ρ ρ−r

m

 − 1 ωp (ρ) ≥ ωp (r).

The assumption that θp (ρ) ≤ θ¯ < 2 and the argument of Theorem 27.2 yield the desired conclusion. Combining the argument of Theorem 27.3 and Theorem 27.4, we have the following regularity theorem. However, we should point out that it was proved by Simon [Si] that under the assumption of Theorem 27.5, a Poincar´e inequality is valid. In which case, the Moser iteration argument will imply C α regularity also. Theorem 27.5. (Simon) Let M m be a stationary varifold in RN . Suppose the multiplicity at a point p ∈ M defined by θp = lim θp (ρ) ρ→0

is strictly less than 2. Then any harmonic function defined in a neighborhood of p is C α for some 0 < α < 1 depending only on θp and m. Corollary 27.1. Let M m be a minimal stationary varifold in RN . Suppose the volume growth of M is Euclidean, i.e., V¯x (ρ) ≤ θ¯ ωm ρm for some constant θ¯ ≥ ωm . Then there exists a constant C1 > 0 depending only on θ¯ and m, such that hd (M ) ≤ C dm−1 for all d ≥ 1. Proof. Lemma 27.1 and the volume assumption assert that M satisfies the mean value inequality (M) for extrinsic balls. On the other hand, since ωm ρm ≤ V¯x (ρ) ≤ θ¯ ωm ρm , ¯ The theorem follows from the argument it satisfies the weak volume comparison condition (Vm ) with CV = θ. of Theorem 26.3. 

GEOMETRIC ANALYSIS

223

§28 Massive Sets In this section, we will introduce the notion of d-massive sets. The notion of 0-massive set was first introduced by Grigor’yan [G2] where he established the relationship between massive sets and harmonic functions. While it is not clear if such a relationship exists for d > 0, but the notion of d-massive sets have important geometric and analytic implications. Definition 28.1. For any real number d ≥ 0, a d-massive set Ω is a subset of a manifold M that admits a nonnegative, subharmonic function f defined on Ω with the boundary condition f =0

on

∂Ω,

and satisfying the growth property f (x) ≤ C rd (x) for all x ∈ Ω and for some constant C > 0. The function f is called a potential function of Ω. We also denote md (M ) to be the maximum number of disjoint d-massive sets admissible on M. We will now give a proof of Grigor’yan theorem [G2]. Theorem 28.1. (Grigor’yan) Let M m be a complete Riemannian manifold. The maximum number of disjoint 0-massive sets admissible on M is given by the dimension of the space of bounded harmonic functions on M, i.e., m0 (M ) = h0 (M ). Proof. Note that m0 (M ) ≥ 1 because the constant function 1 is a nonnegative bounded harmonic function on M . Let Ωi with 1 ≤ i ≤ k be a set of disjoint 0-massive sets in M. By scaling, we may assume that there are nonnegative subharmonic functions fi on Ωi such that fi = 0

on

∂Ωi

and sup fi = 1. Ωi

For a fixed 1 ≤ i ≤ k, a point p ∈ M and ρ > 0, let us solve the Dirichlet problem ∆uρ = 0

on

Bp (ρ)

with boundary value  f on ∂Bp (ρ) ∩ Ωi   i on ∂Bp (ρ) ∩ Ωα , for α 6= i uρ = 1−fα   0 on ∂Bp (ρ) \ (∪kj=1 Ωj ). Since both the functions fi and 1 − fα are bounded between 0 and 1, the maximum principle implies that 0 ≤ uρ ≤ 1 on Bp (ρ). Applying the maximum principle on each Ωj seperately, we conclude that uρ ≥ fi

on

Bp (ρ) ∩ Ωi

and uρ ≤ 1 − fα

on

Bp (ρ) ∩ Ωα , for α 6= i.

The fact that uρ is bounded implies that there exists a sequence ρj → ∞ such that uρj → ui ,

224

PETER LI

where ui is a harmonic function defined on M satisfying 0 ≤ ui ≤ 1. Moreover, ui ≥ fi

on Ωi

and ui ≤ 1 − fα

on Ωα .

Clearly, if {xj } is a sequence of points in M such that fi (xj ) → 1 then ui (xj ) → 1. Similarly, if {yj } is a sequence of points such that fα (yj ) → 1 then ui (yj ) → 0. We claim that the set of functions {ui }ki=1 forms a linearly independent set. Indeed, let u=

k X

ai ui

i=1

be a linear combination such that u is identically 0. For each 1 ≤ i ≤ k, evaluating u on a sequence of points {xj } such that fi (xj ) → 1 we conclude that ai = 0. Hence {ui } are linearly independent and k ≤ h0 (M ). Since k is arbitrary, this implies that m0 (M ) ≤ h0 (M ). To prove the reverse inequality, we may assume that m0 (M ) is finite, otherwise the proposition is autoˆ be the Stone-Cˇech compactification of M . Then every bounded continuous function matically true. Let M 0 ˆ . Let {Ωi }m (M ) be a set of disjoint 0-massive sets in M. For each on M can be continuously extended to M i=1 i ∈ {1, . . . , m0 (M )}, let us define the set ˆ | f (ˆ Si = ∩ {ˆ x∈M x) = sup f }, where the intersection is taken over all the potential functions f of Ωi . The fact that f is subharmonic ˆ \ M. We claim that Si 6= ∅. In fact, for each together with the maximum principle implies that Si ⊂ M ˆ \ M . By the compactness of potential function f of Ωi , the set {ˆ x | f (ˆ x) = sup f } is a closed subset of M ˆ \ M , we need only to show that for any finitely many potential functions f1 , . . . , fl of Ωi , M ∩lj=1 {ˆ x | fj (ˆ x) = sup fj } = 6 ∅. We will argue by induction on l. It is trivially true for one potential function. Let us assume that it is true for l potential functions that ∩lj=1 {ˆ x | fj (ˆ x) = sup fj } = 6 ∅. After normalizing supfj = 1, if we define the function f = f1 + · · · + fl , then we have {ˆ x | f (ˆ x) = sup f } = ∩lj=1 {ˆ x | fj (ˆ x) = sup fj }. Note that both f and fl+1 are potential functions of Ωi . If {ˆ x | f (ˆ x) = sup f } ∩ {ˆ x | fl+1 (ˆ x) = sup fl+1 } = ∅, then for sufficiently small , the sets D1 = {x ∈ M | f (x) > sup f − }

GEOMETRIC ANALYSIS

225

and D2 = {x ∈ M | fl+1 (x) > sup fl+1 − } are disjoint. Clearly both D1 and D2 are subsets of Ωi with the properties that ∂D1 ∩ ∂Ωi = ∅ and ∂D2 ∩ ∂Ωi = ∅ because f = fl+1 = 0 on ∂Ωi . Also, the functions g1 = f − sup f +  and g2 = fl+1 − sup fl+1 +  are potential functions of D1 and D2 , respectively. In particular, this implies that M has m0 (M ) + 1 disjoint massive sets given by {Ω1 , . . . , Ωi−1 , D1 , D2 , Ωi+1 , . . . , Ωk0 }, which is a contradiction. Therefore, ∩l+1 x | fj (ˆ x) = sup fj } j=1 {ˆ = {ˆ x | f (ˆ x) = sup f } ∩ {ˆ x | fl+1 (ˆ x) = sup fl+1 } 6= ∅, and the claim that Si is nonempty follows. We now show that for each i there exists a potential function hi of Ωi such that {ˆ x | hi (ˆ x) = sup hi } = Si . ˆ such The function hi will be called a minimal potential function of Ωi . For an arbitrary open set U in M that Si ⊂ U , note that ˆ \U ⊂M ˆ \ Si M ˆ | f (ˆ = ∪ {ˆ x∈M x) < sup f }, ˆ \ U implies that there exist where the union is over all potential functions f of Ωi . The compactness of M finitely many potential functions f1 , . . . , fl of Ωi such that ˆ \ U ⊂ ∪lj=1 {ˆ M x | fj (ˆ x) < sup fj }. Let us define g = f1 + · · · + fl , which has the property that {ˆ x | g(ˆ x) = sup g} ⊂ U . One may assume by ˆ , n = 1, 2, . . . , scaling g that 0 ≤ g ≤ 1 on M and sup g = 1. Now choose a sequence of open sets Un ⊂ M ∞ such that Un ⊂ Un+1 and ∩n=1 Un = Si . For each Un , there exists a potential function gn of Ωi such that 0 ≤ gn ≤ 1, sup gn = 1 and {ˆ x | gn (ˆ x) = sup gn } ⊂ Un . By defining hi =

∞ X

2−n gn ,

n=1

it is clear that hi is a minimal potential function of Ωi satisfying {ˆ x | hi (ˆ x) = sup hi } = Si . From now on, we will denote hi to be a minimal potential function of Ωi .

226

PETER LI

For a bounded subharmonic function v on M , consider the set S = {ˆ x | v(ˆ x) = sup v}. We claim that S must contain some Si . Moreover, for each j, either S ∩ Sj = ∅ or Sj ⊂ S. In fact, let us first argue that S ∩ Si 6= ∅ for some i. If this is not the case, then for  > 0 sufficiently small the sets Ω = {x ∈ M | v(x) > sup v − } and ˜ i = {x ∈ M | hi (x) > sup hi − } Ω ˜ i = ∅. Clearly Ω ˜ i ⊂ Ωi , and each Ω ˜ i is a massive set with potential function hi − sup hi + . must satisfy Ω ∩ Ω Also, Ω is a massive set with potential function v − sup v + . Therefore ˜ 1, . . . , Ω ˜k } {Ω, Ω 0 are m0 (M ) + 1 disjoint massive sets of M , which is impossible. Therefore, S ∩ Si 6= ∅ for some i. To see that Si ⊂ S, let us consider the function w = hi + v. Note that {ˆ x | w(ˆ x) = sup w} = Si ∩ S ⊂ Si . Thus, for sufficiently small  > 0, the set W = {x | w(x) > sup w − } ⊂ Ωi , and f = w − sup w +  is a potential function of this massive set W . In particular, by extending f to be zero outside W , f is a potential function of Ωi with {ˆ x | f (ˆ x) = sup f } = {ˆ x | w(ˆ x) = sup w} = Si ∩ S. The minimality of Si implies that Si ⊂ Si ∩ S, hence Si ⊂ S. This argument also shows that for any j, either S ∩ Sj = ∅ or Sj ⊂ S. For each Ωi , let hi be a minimal potential function of Ωi with {ˆ x | hi (ˆ x) = sup hi } = Si . After normalization, we may assume that sup hi = 1. Using the construction discussed in the first part of this proof, there exists m0 (M ) bounded harmonic functions {fi } with the properties that 0 ≤ fi ≤ 1, fi = 1

on

Si ,

and fi = 0

on

Sα for α 6= i.

The claim that h0 (M ) ≤ m0 (M ) will follow if we show that any bounded harmonic function f can be written as linear combinations of the set {fi }. Too see this, we first observe that the constant function is spanned by {fi }. Indeed, if we let m0 (M ) X g= fi i=1

GEOMETRIC ANALYSIS

227

then g≥0 with g=1

Si for 1 ≤ i ≤ m0 (M ).

on

In fact, sup g = 1 because otherwise the set {ˆ x | g(ˆ x) = sup g} will be disjoint from ∪Si . In particular, there exists a sufficient small constant  so that the set {x | g(x) ≥ sup g − } will be a 0-massive set disjoint from ∪Si , hence contradicting the definition of m0 (M ). Let us now consider h to be any nonconstant bounded harmonic function. Let us denote S0 (+) = {ˆ x | h(ˆ x) = sup h} and S0 (−) = {ˆ x | h(ˆ x) = inf h}. The above argument showed that there is at least one Si such that Si ⊂ S¯0 (+) and at least one Si such that Si ⊂ S¯0 (−). Moreover Sj ∩ S¯0 (+) ∪ S¯0 (−) = ∅ if Sj is not containing in S¯0 (+) ∪ S¯0 (−). Let us denote I0 (+) and I0 (−) to be those i’s such that Si ⊂ S¯0 (+) and Si ⊂ S¯0 (−), respectively. If we define  g0 = a0 

 X



fi  + +b0 

i∈I0 (+)

 X

fi 

i∈I0 (−)

with a0 = sup h and b0 = inf h, then g0 = a0 on Si for all i ∈ I0 (+) and g0 = b0 on Si for all i ∈ I0 (+). In particular, the harmonic function h1 = h − g0 will satisfy the properties that h1 = 0

on

∪i∈(I0 (+)∪I0 (−)) Si

and h1 = h

Sj for j ∈ / (I0 (+) ∪ I0 (−)).

on

Applying the same process to h1 taking combinations with the remaining fj for j ∈ / (I0 (+) ∪ I0 (−)), we obtain a harmonic function h2 such that h2 = 0

on

∪i∈(I1 (+)∪I1 (−)) Si

and h2 = h1

on

Si for j ∈ / (I1 (+) ∪ I1 (−)),

where I1 (+) and I1 (−) are those i’s such that Si ⊂ {ˆ x | h1 (ˆ x) = sup h1 } and Si ⊂ {ˆ x | h1 (ˆ x) = inf h1 }, respectively. In particular, h2 = 0

on

∪i∈(I0 (+)∪I0 (−)∪I1 (+)∪I1 (−)) Si

and h2 = h

on

Si for j ∈ / (I0 (+) ∪ I0 (−) ∪ I1 (+) ∪ I1 (−)).

228

PETER LI

Since m0 (M ) < ∞, we can apply this inductive procedure finitely many times and ended up with a function hk which vanishes on Si for all 1 ≤ i ≤ m0 (M ). This function hk must be identically 0. Indeed, if this is not the case, then either sup hk or inf hk is not 0. Let us assume that sup hk > 0. then for some  > 0 the set given by {ˆ x | hk (ˆ x) = sup hk − } will be disjoint from ∪Si and will produce another 0-massive set, which is a contradiction. This proves the inequality h0 (M ) ≤ m0 (M ), hence the theorem.  As pointed out earlier, there are no direct relationship between hd (M ) and md (M ) for d > 0. However, there are parallel theories concerning the two numbers as demonstrated by the following theorem that mirrors Theorem 26.1 when applied to polynomial growth harmonic functions. Theorem 28.2. (Li-Wang) Let M m be a complete manifold satisfying conditions (Vµ ) and (M). For all d ≥ 1, there exists a constant C > 0 depending only on µ such that md (M ) ≤ C CM dµ−1 . Proof. The proof follows exactly as the proof of Theorem 26.1. Instead of using an orthonormal basis for a finite dimensional subspace of harmonic functions, we use a set of potential functions fi with normalized L2 -norm. Since they have disjoint support, they are automatically perpendicular with each other. We also do not need to change bases and consider the space Kx = {u | u(x) = 0} because for any x ∈ M there is at most one fi that does not vanish on x. The rest of the argument is exactly the same. Similarly, we also have the following finiteness theorem that mirrors Theorem 26.2. Theorem 28.3. Let M be a complete manifold satisfying the weak mean value property (WM). Suppose that the volume growth of M satisfies Vp (ρ) = O(ρµ ) as ρ → ∞ for some point p ∈ M. Then md (M ) ≤ CWM (2b + 1)(2d+µ) . The next theorem gives a sharp estimate of md on R2 . Theorem 28.4. On R2 ,

md (R2 ) ≤ 2d

for all d ≥ 0. Proof. Let {Ω1 , . . . , Ωk } be disjoint d-massive sets and {u1 , . . . , uk } be their corresponding potential functions. Note that since Z Z ∂ui 2 , |∇ui | ≤ ui ∂r Ωi ∩B(ρ) Ωi ∩∂B(ρ) Schwarz inequality implies that (28.1)

1 2

2λ1 (Ωi ∩ ∂B(ρ))

Z

|∇ui |2

Ωi ∩B(ρ)

Z

∂ui ∂r Ωi ∩∂B(ρ) Z Z 2 ≤ λ1 (Ωi ∩ ∂B(ρ)) ui + 1 2

≤ 2λ1 (Ωi ∩ ∂B(ρ))

ui

Ωi ∩∂B(ρ)

Z ≤

¯ i |2 + |∇u

Ωi ∩∂B(ρ)

Z = Ωi ∩∂B(ρ)

Z Ωi ∂B(ρ)

|∇ui |2 ,



Ωi ∩∂B(ρ) 2 ∂ui

∂r



∂ui ∂r

2

GEOMETRIC ANALYSIS

229

where λ1 (Ωi ∩ ∂B(ρ)) denotes the first Dirichlet eigenvalue on Ωi ∩ ∂Bp (ρ). Using the fact that π2 , A(Ωi ∩ ∂B(ρ))2

λ1 (Ωi ∩ ∂B(ρ)) ≥ we conclude that (28.2)

2

k X

k X

1

λ12 (Ωi ∩ ∂B(ρ)) ≥ 2π

i=1

1 . A(Ωi ∩ ∂B(ρ))

i=1

On the other hand, combining with the inequality k X

2

k ≤

!

k X

A(Ωi ∩ ∂B(ρ))

i=1

i=1 k X

≤ 2πρ

i=1

1 A(Ωi ∩ ∂B(ρ))

1 A(Ωi ∩ ∂B(ρ))

! ,

(28.1) and (28.2) imply that k X

(28.3)

i=1

|∇ui |2 k2 RΩi ∩∂B(ρ) . ≥ ρ |∇ui |2 Ωi ∩B(ρ)

R

Observing that Z

∂ |∇ui | = ∂r Ωi ∩∂B(r) 2

!

Z

2

,

ρ ρ0



|∇ui | Ωi ∩B(ρ)

(28.3) can be written as k Z Y ∂ ln |∇ui |2 ∂r i=1 Ωi ∩B(ρ)

! ≥

k2 . r

Integrating both sides from ρ0 to ρ yields ln

Ωi ∩B(ρ)

|∇ui |2

Ωi ∩B(ρ0 )

|∇ui |2

R

k Y

R i=1

!! ≥ k 2 ln



The growth assumption on the ui ’s asserts that k Z Y i=1

|∇ui |2 ≤ C ρ2d−2 V (ρ)

Ωi ∩B(ρ)

≤ C k ρ2kd , hence we have 2kd ln ρ + C1 ≥ k 2 ln ρ − k 2 ln ρ0 . Letting ρ → ∞ implies that 2d ≥ k, as to be proven.




k

.

!

230

PETER LI

§29 The Structure of Harmonic Maps into a Cartan-Hadamard Manifold In this section, we will apply the notion of massive sets to study the structure of the image of a harmonic map whose target is a hyperbolic space. In fact, Li and Wang [LW1] developed this theory that applies to the target being a strongly negatively curved Cartan-Hadamard manifold, or when it is a 2-dimensional visibility manifold. Throughout this section we shall assume that N is a Cartan-Hadamard manifold, namely, N is simply connected and has nonpositive sectional curvature. It is well known that N can be compactified by adding ¯ = N ∪ S∞ (N ) is homeomorphic to a closed a sphere at infinity S∞ (N ). The resulting compact space N Euclidean ball. Two geodesic rays γ1 and γ2 in N are called equivalent if their Hausdorff distance is finite. Then the geometric boundary S∞ (N ) is simply given by the equivalence classes of geodesic rays in N . A ¯ converges to x ∈ N ¯ if for some fixed point p ∈ N , the sequence of geodesic sequence of points {xi } in N rays {pxi } converges to a geodesic ray γ ∈ x. In this case, we say γ is the geodesic segment px joining p to x. Recall that a subset C in N is strictly convex if any geodesic segment between any two points in C is also contained in C. For a subset K in N , the convex hull of K, denoted by C(K), is defined to be the smallest strictly convex subset C in N containing K. The convex hull can also be obtained by taking the intersection of all convex sets C ⊂ N containing K. When N is a Cartan-Hadamard manifold, there is only one geodesic segment joining a pair of points in N . In this case, there is only one notion of convexity, and we will simply say a set is convex when it is a strictly convex set. For the purpose of this article, we will need a ¯ . Since a geodesic line is a geodesic segment joining the two end points in S∞ (N ), notion of convexity for N ¯ . However, it is not true in general that it still makes sense to talk about geodesics joining two points in N any two points in S∞ (N ) can always be joined by a geodesic segment given by a geodesic line, as indicated by two non-antipodal points in S∞ (Rn ). If every pair of points in S∞ (N ) can be joined by a geodesic line in N , then N is said to be a visibility manifold. This class of manifolds was extensively studied in [EO]. A typical example of a visibility manifold is a Cartan-Hadamard manifold with sectional curvature bounded from above by a negative constant −a < 0. To remedy the situation when N is not a visibility manifold, we defined a generalized notion of geodesic segment joining two points at infinity. ¯ is the limiting set of a sequence Definition 29.1. A geodesic segment γ joining a pair of points x and y in N of geodesic segments {γi } in N with end points {xi } and {yi } such that xi → x and yi → y. We will denote γ by xy. Observe that if xy∩S∞ (N ) = {x, y}, then xy must be a geodesic line in N and hence a geodesic segment in the traditional sense. For the case of two non-antipodal points in S∞ (R2 ), the shortest arc on S1 = S∞ (R2 ) joining the two points will be the geodesic segment in the sense defined above. If the two points are antipodal in S∞ (R2 ), say the northpole and the southpole, then there are infinitely many geodesic segments joining them. Each vertical line is a geodesic segment in the genuine sense. Also, both arcs on S1 joining the two poles are geodesic segments joining them. Using this definition, for a pair of points in S∞ (N ), it is possible to have more than one geodesic segments joining them. The convexity we will define will be in the sense of strictly convex. ¯ is a convex set if for every pair of points in C, any geodesic segment Definition 29.2. A subset C of N joining them is also in C. ¯ , we define its convex hull C(A) to be the smallest convex subset of Definition 29.3. For a subset A in N ¯ containing A. N ¯ unless otherwise In what follows, when we say that a subset is closed, we mean that it is closed in N ¯ ¯ noted. In general, we denote the closure for a subset A in N by A. For a given sequence of closed subsets ¯ decreases to the convex hull of {Ai } decreasing to A, it is natural to ask whether the convex hull of Ai in N A. For this purpose, we introduce the following definition. Definition 29.4. A Cartan-Hadamard manifold N is said to satisfy the separation property if for every ¯ and every point p not in A, there exists a closed convex set C properly containing closed convex subset A in N

GEOMETRIC ANALYSIS

231

A and separating p from A, i.e., A ⊂ C, A ∩ S∞ (N ) is contained in the interior of C ∩ S∞ (N ) and p is not in C. For a two-dimensional visibility manifold or a Cartan-Hadamard manifold with constant negative curvature, it is easy to check that the separation property holds. In fact, for a point p not in the closed convex set A, pick up a point q ∈ A such that r(p, q) = r(p, A). Then the convexity of A and the first variation formula imply that for z ∈ A, ∠(zq, qp) ≥ π/2. Let x be the midpoint of the geodesic segment between p and q, and C = {y ∈ N : ∠(yx, xp) ≥ π/2}. Then C is closed, convex as ∂C is evidently totally geodesic and C properly separates p from A. Lemma 29.1. A Cartan-Hadamard manifold N satisfies the separation properly if and only if for every ¯ such that ∩∞ Ai = A, then closed subset A and monotone decreasing sequence of closed subsets {Ai } in N i=1 ∩∞ i=1 C(Ai ) = C(A). Proof. Suppose that N satisfies the separation property. Let {Ai } be a decreasing sequence of closed subsets in N with ∩∞ i=1 Ai = A. Obviously, C(A) ⊂ ∩∞ i=1 C(Ai ) from the definition of convex hull. Assume the contrary that ∩∞ i=1 C(Ai ) 6= C(A). Then there exists a point p ∈ ∩∞ i=1 C(Ai ) but not in C(A). The separation property asserts that there is a closed convex subset C properly separating p from C(A). Let C = {x ∈ N : r(x, C) ≤ } be the -neighborhood of C. For sufficiently small  > 0, C¯ also properly separates p from C(A). Since A ∩ S∞ (N ) is contained in the interior of C ∩ S∞ (N ) and Ai is decreasing to A, we conclude that for i sufficiently large, Ai ⊂ C¯ . Thus, C(Ai ) ⊂ C¯ and p ∈ C¯ , which is a contradiction. Conversely, to show that N satisfies the separation property, let A be a closed convex subset and p a point ¯ with the closed unit ball of the Euclidean space endowed the canonical metric. Let not in A. We identify N Ai be the tubular neighborhood of A of size 1/i. It is then clear that Ai is a decreasing sequence of closed subsets with ∩∞ i=1 Ai = A. Hence by the assumption, ∩∞ i=1 C(Ai ) = C(A) = A. The fact that p ∈ / A implies that p ∈ / C(Ai ) for sufficiently large i. By choosing C = C(Ai ), it is clear that C properly separates p from A.  According to our definition of convex hull, it is possible that C(K) ∩ S∞ (N ) is a much bigger set than K ∩ S∞ (N ). In fact, if we consider K to be the y-axis in R2 , then K ∩ S∞ (R2 ) consists of the two poles in S1 . However, C(K) = R2 because every line given by x = constant is a geodesic joining the two poles of S1 . Hence, C(K) ∩ S∞ (R2 ) = S1 . On the other hand, if we assume in addition that N satisfies the following separation property at infinity, then C(K) ∩ S∞ (N ) = K ∩ S∞ (N ).

232

PETER LI

Definition 29.5. Let N be a Cartan-Hadamard manifold. N is said to satisfy the separation property at infinity if for any closed subset A of S∞ (N ) and any point p ∈ S∞ (N ) \ A, there exists a closed convex subset ¯ such that A is contained in the interior of C ∩ S∞ (N ) and p not in C. C in N It is easy to check that a two-dimensional visibility manifold alway satisfies separation property at infinity. On the other hand, upon improving a result of M. Anderson [An], A. Borb´ely [Bo] has shown that CartanHadamard manifold N has separation property at infinity provided that its sectional curvature satisfies −Ceαr(x) ≤ KN (x) ≤ −1 for some constant C > 0 and 0 ≤ α < 1/3, where r(x) is the distance from point x to a fixed point o ∈ N . The interested reader shall refer to [Bo] for a detail proof. The following simple lemma gives a condition equivalent to the separation property at infinity. ¯, Lemma 29.2. Let N be a Cartan-Hadamard manifold. Then for every closed set K in N C(K) ∩ S∞ (N ) = K ∩ S∞ (N ) if and only if N satisfies the separation property at infinity. Proof. Assume that N satisfies the separation property at infinity. For a given closed subset K, let A = K ∩ S∞ (N ). If A = S∞ (N ), then there is nothing to prove. So we may assume this is not the case. The closeness of K implies that A is closed. Given p ∈ S∞ (N ) \ A, there is a closed convex subset C such that A is contained in the interior of C ∩ S∞ (N ) and p is not in C. In particular, we conclude that sup r(x, C) = R < ∞. x∈K

Let us consider the R-neighborhood, CR = {x ∈ N : r(x, C) ≤ R}, of C. Then C¯R is a closed convex subset and K ⊂ C¯R . Moreover, C ∩ S∞ (N ) = C¯R ∩ S∞ (N ). / C(K)∩S∞ (N ). This shows that C(K)∩S∞ (N ) = Therefore, C(K) ⊂ C¯R and p is not in C¯R . In particular, p ∈ A. Conversely, to show that N satisfies the separation property at infinity, let A be a closed subset of S∞ (N ) and point p ∈ S∞ (N ) \ A. Then there exists a closed subset K ⊂ S∞ (N ) such that A is in the interior of K and p ∈ / K. Let C = C(K) and by the assumption, C ∩ S∞ (N ) = K. Thus, p ∈ / C and A is contained in the interior of C ∩ S∞ (N ). Thus, N satisfies the separation property at infinity and the lemma is proved.  Theorem 29.1. (Li-Wang) Let M be a complete Riemannian manifold such that the dimension of the space of bounded harmonic functions H0 (M ) is h0 (M ). Let u : M → N be a harmonic map from M into a CartanHadamard manifold, N n . Denote A = u(M ) ∩ S∞ (N ), where S∞ (N ) = Sn−1 is the geometric boundary of N . Then there exists a set of points {yi }ki=1 ⊂ u(M ) ∩ N with k ≤ h0 (M ), such that, u(M ) ⊂ ∩j C(Aj ∪ {yi }ki=1 ), ¯ properly containing A and ∩j Aj = where {Aj } is a monotonically decreasing sequence of closed subsets of N A. In addition, if we assume that N has the separation property, then u(M ) ⊂ C(A ∪ {yi }ki=1 ). Proof. Let us pick a point y0 ∈ u(M ). If u(M ) ⊂ ∩j C(Aj ∪ {y0 }),

GEOMETRIC ANALYSIS

233

then we are done. Hence we may assume that there exists an Aj properly containing A such that the set u(M ) \ C(Aj ∪ {y0 }) 6= ∅. Moreover, one can easily check that u(M ) \ C(Aj ∪ {y0 }) is bounded in N . Since u is a harmonic map and the distance function r(y, C(Aj ∪ {y0 })) to the set C(Aj ∪ {y0 }) is convex, the composition function f (x) = r(u(x), C(Aj ∪ {y0 })) is a bounded nonconstant subharmonic function on M . Thus, f attains its maximum at every point of some ˆ , a subnet of {u(xα )} supremum set S1 . In particular, for x ˆ1 ∈ S1 and a net {xα } in M converging to x ˆ1 in M converges to y1 ∈ N . Again, if u(M ) ⊂ ∩j C(Aj ∪ {y0 ∪ y1 }), then the theorem is true, otherwise by choosing a larger j if necessary, the function g(x) = r(u(x), C(Aj ∪ {y0 ∪ y1 })) is a bounded nonconstant subharmonic function on M . If g achieves its maximum on S1 , then g(ˆ x) = sup g for x ˆ ∈ S1 . In particular, sup g = g(ˆ x1 ) = r(y1 , C(Aj ∪ {y0 ∪ y1 })) = 0, which is impossible. Hence, we may assume g achieves its maximum on S2 . For a net {xα } in M converging to a point x ˆ2 in S2 , there exists a subnet of {u(xα )} that converges to y2 ∈ N . Suppose that we have chosen l points y1 , . . . , yl described in the above procedure. If u(M ) ⊂ ∩j C(Aj ∪ {yi }li=1 ), then we are done, otherwise by choosing a larger j if necessary, we define the function h(x) = r(u(x), C(Aj ∪ {yi }li=1 )), which is a bounded nonconstant subharmonic function on M. We claim that h cannot achieve its maximum on ∪li=1 Si . Indeed, if it does, then h must achieve its maximum at every point on Si for some 1 ≤ i ≤ l. Thus using a similar argument as before, h(ˆ xi ) = r(yi , C(Aj ∪ {yi }li=1 )) = 0, which is a contradiction. Hence, h achieves its maximum on some Si with i > l. We may assume that i = l + 1. Let us pick a point x ˆl+1 ∈ Sl+1 and a net {xα } converging to x ˆl+1 . Suppose yl+1 is an accumulation point of the net {u(xα )}. It is clear that this process must stop after at most m0 (M ) steps since there are only m0 (M ) massive sets. In particular, there exist k points {y1 , . . . , yk } with k ≤ m0 (M ) such that u(M ) ⊂ ∩j C(Aj ∪ {yi }ki=1 ). Moreover, yi ∈ u(M ), and the proof of the first statement is completed. The second statement follows from Lemma 29.1 

234

PETER LI

Corollary 29.1. If h0 (M ) = 1 and N is a Cartan-Hadamard manifold, then every bounded harmonic map M → N must be constant. Theorem 29.2. Let M be a complete manifold and we denote the space spanned by all positive harmonic functions on M by H+ (M ). Suppose M has the property that h0 (M ) = dim H+ (M ) < ∞. Assume that u : M → N is a harmonic map from M into a Cartan-Hadamard manifold N which either is a two-dimensional visibility manifold or has strongly negative sectional curvature, and that A = u(M )∩S∞ (N ) consists of at most one point. Then the set A is necessarily empty, and there exists a set of k points {yi }ki=1 ⊂ u(M ) ∩ N with k ≤ h0 (M ) such that u(M ) ⊂ C({yi }ki=1 ). In particular, if M has no nonconstant positive harmonic functions, then every such harmonic map must be a constant map. Proof. Theorem 29.1 implies that u(M ) ⊂ C(A ∪ {yi }ki=1 ) for some set of k points {yi }ki=1 in N with k ≤ h0 (M ). If A contains exactly one point a, let γ be a geodesic line on (−∞, +∞) such that its restriction to (0, +∞) represents a. For each yi , there exists a unique point γ(ti ) such that r(yi , γ) = r(yi , γ(ti )). Choose a point p = γ(t0 ) with t0 < ti for i = 1, 2, . . . , k. Let δ be the geodesic ray given by the restriction of γ onto (t0 , +∞), and denote the Busemann function associated to δ by β. Recall that if δ is parametrized by arclength, then β(y) = lim (t − r(y, δ(t))). t→∞

We claim that there exists a constant c such that r(y, p) ≤ β(y) + c for y ∈ C({a} ∪ {yi }ki=1 ). In fact, by the convexity of the function r(y, γ) and the choice of p, one easily checks that r(y, δ) = r(y, γ) ≤ max r(yi , γ) 1≤i≤k

=c for any y ∈ C({a} ∪ {yi }ki=1 ). Therefore, if we let y¯ ∈ δ be the point such that r(y, δ) = r(y, y¯), then r(y, p) ≤ r(y, δ) + r(¯ y , p) ≤ c + β(¯ y) ≤ 2c + β(y). This justifies the claim that r(u(x), p) ≤ β(u(x)) + c for all x ∈ M . Since u is a harmonic map and N is a Cartan-Hadamard manifold, the function r(u(x), p) is subharmonic and the function β(u(x)) + c is superharmonic. The sub-super solution method yields a harmonic function f (x) on M such that r(u(x), p) ≤ f (x) ≤ β(u(x)) + c.

GEOMETRIC ANALYSIS

235

Therefore, f is an unbounded positive harmonic function on M , contradicting to our assumption that there is no such function. In conclusion, A must be empty and C(u(M )) = C({yi }ki=1 ). This proves our first statement. The second part of the theorem follows from the first part by taking h0 (M ) = 1.  Notice that the horoball of a visibility manifold intersects the geometric boundary at exactly one point (see [BGS]). Thus, we obtain the following Liouville type theorem which partially generalizes the results in [Sh] and [T]. Corollary 29.2. Suppose M satisfies dim H+ (M ) = 1. Assume that N is either a two-dimensional visibility manifold or has strongly negative sectional curvature. Then every harmonic map from M into a horoball of N must be constant. Recall that a manifold is parabolic if it does not admit a positive Green’s function. Since a parabolic manifold has no massive subsets and every positive harmonic function must be constant, we can applying Theorem 29.1 to this case and obtain the following corollary. Corollary 29.3. Let u be a harmonic map from a parabolic manifold M into N n . Assume that N is either a two-dimensional visibility manifold or has strongly negative sectional curvature. Then u(M ) ⊂ C(A), where A = u(M ) ∩ S∞ (N ). Proof. In this case, we have h0 (M ) = 1, hence Theorem 29.1 implies that u(M ) ⊂ C(A ∪ {y}) for some y ∈ u(M ) ∩ N . Let us assume the contrary that u(M ) is not contained in C(A). In particular, the parabolicity of M implies that the function r(u(x), C(A)) is unbounded. Lemma 29.1 then asserts that u(M ) \ C(W ) is non-empty for some open set W ⊂ S∞ (N ) which properly contains A. Let us consider the function f (x) = r(u(x), C(W )), which is a nonconstant, nonnegative, bounded subharmonic function on M. However, the parabolicity assumption on M implies that such function does not exist. This completes our proof.  Theorem 29.3. Let M be a complete manifold such that the maximum number of disjoint d-massive sets of M is md (M ). Suppose u : M → N is a harmonic map from M into N, and N satisfies the separation property at infinity. Assume that there exists a point o ∈ N such that r(u(x), o) = O(rd (x)) as x → ∞. Then 0 A = u(M ) ∩ S∞ (N ) = {ai }ki=1 with k 0 ≤ md (M ) − m0 (M ). If, in addition, N is either a two-dimensional visibility manifold or has strongly negative sectional curvature, then there exist k points {yj }kj=1 ⊂ u(M ) ∩ N with k 0 + k ≤ md (M ) such that 0

u(M ) ⊂ C({ai }ki=1 ∪ {yj }kj=1 ). Proof. Since a 0-massive set is always d-massive, we have m0 (M ) ≤ md (M ). Theorem 29.1 implies that there exist k points {yj }kj=1 ⊂ u(M ) ∩ N with k ≤ m0 (M ), such that u(M ) ⊂ ∩>0 C(A ∪ {yj }kj=1 ).

236

PETER LI

0 ¯ such that Ui ∩ A 6= ∅ for If A contains at least k 0 points, then there exist disjoint open sets {Ui }ki=1 in N 0 i = 1, 2, . . . , k . Since N is assumed to satisfy the separation property at infinity, Lemma 29.2 yields that ¯ \ Ui ) ∪ {yj }k ). In particular, the function u(M ) is not a subset of C((N j=1

¯ \ Ui ) ∪ {yj }k )) fi (x) = r(u(x), C((N j=1 is not identically zero on u−1 (Ui ) and sup fi = ∞. Clearly, fi = 0 on the boundary of u−1 (Ui ) and fi (x) = O(rd (x)). This implies that each set u−1 (Ui ) is d-massive but not massive. In particular, since they are disjoint, k 0 ≤ md (M ) − m0 (M ). Thus A has at most md (M ) − m0 (M ) points, and the first conclusion follows. If, in addition, N is either a two-dimensional visibility manifold or has strongly negative sectional curvature, then Theorem 29.1 yields that 0 u(M ) ⊂ C({ai }ki=1 ∪ {yj }kj=1 ), and the estimate k 0 + k ≤ md (M ) follows from the argument. This completes our proof.  Combining with the estimates on md (M ) given by Theorem 28.2 and Theorem 28.3, we have the following corollary. Corollary 29.4. Let M be a complete manifold satisfying condition (M) and its volume growth Vp (R) = O(Rµ ) for some point p ∈ M . Suppose N is a Cartan-Hadamard manifold satisfying either one of the following conditions: (1) it has strongly negative sectional curvature; (2) it is a two-dimensional visibility manifold. Let u : M → N be a harmonic map and suppose that there exists a point o ∈ N such that r(u(x), o) = 0 O(rd (x)) as x → ∞. Then there exist sets of k 0 points {ai }ki=1 = u(M ) ∩ S∞ (N ) and k points {yj }kj=1 ⊂ u(M ) ∩ N with k 0 + k ≤ λ3(2d+µ) such that 0

u(M ) ⊂ C({ai }ki=1 ∪ {yj }kj=1 ). If M is further assumed to have property (Vµ ), then we have k 0 + k ≤ Cdµ−1 . We would like to remark that though Lemma 29.1 states that separation property is necessary and sufficient to conclude ∩∞ j=1 C(Aj ) = C(A), after careful examination of the proof of Theorem 29.1, we only need to use the fact that ∩∞ j=1 (Aj ) is bounded distance from C(A). With this in mind, using a theorem of Anderson and Borb´ely, Theorem 29.1 and hence all consequential theorems of this section is valid for N being a strongly negatively curved manifold. A complete treatment can be found in [LW1].

GEOMETRIC ANALYSIS

237

§30 Lq Harmonic Functions Recall that Yau’s theorem (Lemma 7.1) asserts that a complete manifold does not admit any nonconstant Lq harmonic functions for q ∈ (1, ∞). In this section, we will discuss the validity of this Liouville property for Lq harmonic functions for the values of q ∈ (0, 1]. The condition to ensure Liouville property for the case when q ∈ (0, 1) is quite different from the condition for the case q = 1. Both of these cases was first considered by Li and Schoen [LS], where they obtained the sharp curvature condition for q ∈ (0, 1). In the same paper, they also obtained a almost sharp condition for q = 1, while the sharp version was later proved by Li in [L4] using the heat equation method. Theorem 30.1 presented below is the argument from [LS], while Theorem 30.2 is from [L4]. Counter-examples for both theorems when the curvature conditions has been violated were given in [LS] indicating the sharpness of these conditions. Theorem 30.1. (Li-Schoen) Let M m be a complete manifold. There exists a constant δ(m) > 0 depending only on m, such that, if the Ricci curvature of M satisfies the lower bound Rij (x) ≥ −δ(m) (1 + r(x))−2 , where r(x) is the distant to a fixed point p ∈ M , then any nonnegative, Lq subharmonic function must be identically zero for q ∈ (0, 1). Proof. Let f be a nonnegative Lq subharmonic function defined on M. According to the mean value inequality (14.23) of Corollary 14.2, for nonnegative subharmonic function, we have the estimate f q (x) ≤ C1 V¯ (2ρ) ρ−m exp(C2 ρ

√

R) Vx−1 (ρ)

Z

f q,

Bx (ρ)

if Rij ≥ −(m − 1)R on Bx (2ρ), and C1 , C2 > 0 depend only on m and q. Let us now apply to any point x sufficiently far from p with 5ρ ≤ r(x). The curvature assumption asserts that V¯ (2ρ) ρ−m exp(C2 ρ

√ R)

is bounded, hence f q (x) ≤ C3 Vx−1 (ρ),

(30.1)

where C3 also depends on the Lq -norm of f. We will now estimate Vx (ρ) using volume comparison theorem. Let γ be a minimal normal geodesic joining p to x with γ(0) = p and γ(T ) = x. Define the values {ti } Pi with t0 = 0, t1 = 1 + β, and ti = 2 j=0 β j − 1 − β i , for some fixed β>

2 1

2m − 1

> 1.

Let us denote tk ≤ T to be largest value not bigger than T. We denote the points xi = γ(ti ) and they obviously satisfy the properties that r(xi , xi+1 ) = β i + β i+1 and r(xk , x) < β k + β k+1 . Moreover, the closure Pk of the set of geodesic balls {Bxi (β i )} covers γ([0, 2 j=1 β j − 1]) and they have disjoint interiors. We now claim that (30.2)

k

Vxk (β ) ≥ C4



βm (β + 2)m − β m

k Vp (1).

238

PETER LI

To see this, the volume comparison theorem implies that Vxi (β i ) ≥ Ai (Vxi (β i + 2β i−1 ) − Vxi (β i )) ≥ Ai Vxi−1 (β i−1 ), where

R β i √R(xi ,β i +2β i−1 )

sinhm−1 t dt Ai = R i 0 i−1 √ (β +2β ) R(xi ,β i +2β i−1 ) √ sinhm−1 t dt i i−1 i β

R(xi ,β +2β

)

with Rij ≥ −(m − 1)R(xi , β i + 2β i−1 ) denotes the lower bound of the Ricci curvature on Bxi (β i + 2β i−1 ). Iterating this inequality, we conclude that Vxk (β k ) ≥ Vp (1)

(30.3)

k Y

Aj .

j=1

Since r(xi ) = 2

Pi

j=0

β j − 1 − β i , the curvature assumption implies that  R(xi , β i + 2β i−1 ) ≤ δ(m) 2

i−2 X

−2 βj 

j=0

for i ≥ 2. Hence, β

i

p

p δ(m) βi R(xi , β i + 2β i−1 ) ≤ Pi−2 j 2 j=0 β p δ(m) β2 ≤ Pi−2 −j 2 j=0 β p δ(m) β(β − 1)(1 − β −i+1 )−1 ≤ 2

which can be made arbitrarily small for a fixed β by choosing δ(m) to be sufficiently small. In particular, simply approximating sinh t by t, Ai can be approximated by (β i )m (β i + 2β i−1 )m − (β i )m βm . = (β + 2)m − β m

Ai ∼

Combining this with (30.3), we conclude (30.2). To estimate Vx (β k+1 ), we consider the following cases. First, let us assume that r(x, xk ) ≤ β k (β − 1). In this case, we see that Bxk (β k ) ⊂ Bx (β k+1 ),

GEOMETRIC ANALYSIS

239

hence Vxk (β k ) ≤ Vx (β k+1 ). In particular, we conclude that Vx (β k+1 ) ≥ C4

(30.4)



βm (β + 2)m − β m

k Vp (1).

On the other hand, if r(x, xk ) > β k (β − 1) we see that Bxk (β k ) ⊂ Bx (r(x, xk ) + β k ) \ Bx (r(x, xk ) − β k ). Using the volume comparison theorem, we conclude that Vx (β k ) ≥ A (Vx (r(x, xk ) + β k ) − Vx (x(x, xk ) − β k ) ≥ A Vxk (β k ) where

R β k √R(x,r(x,xk )+β k )

sinhm−1 t dt 0 √ A= R . (r(x,xk )+β k ) R(x,r(x,xk )+β k ) m−1 √ sinh t dt k k (r(x,xk )−β )

R(x,r(x,xk )+β )

Arguing as above, since q q (r(x, xk ) + β k ) R(x, r(x, xk ) + β k ) ≤ (β k+1 + 2β k ) R(x, r(x, xk ) + β k ) p δ(m) ≤ β(β − 1) 2 can be made sufficiently small, we can approximate A by A∼

βm . (β + 2)m

Combining with (30.2), we again conclude that (30.5)

Vx (β k+1 ) ≥ C5



βm (β + 2)m − β m

k Vp (1)

for some constant C5 > 0. Note that when x → ∞, the value k → ∞. Since the choice of β ensures that βm > 1, (β + 2)m − β m hence the right hand side of (30.5) tends to infinity. Using the value R = β k+1 in (30.1), we conclude that f (x) → 0 as x → ∞. The maximum principle now asserts that f must be identically 0. 

240

PETER LI

Corollary 30.1. Let M m be a complete manifold. There exists a constant δ(m) > depending only on m, such that if the Ricci curvature of M satisfies the lower Rij (x) ≥ −δ(m) r−2 (x) then M does not admit any nontrivial Lq harmonic functions for q ∈ (0, 1). Proof. Since the absolute value of a harmonic function is subharmonic, this corollary follows directly from Theorem 30.1.  Theorem 30.2. (Li) Let M m be a complete manifold. Suppose the Ricci curvature of M satisfies the lower Rij (x) ≥ −C (1 + r(x))2 for some constant C > 0. Then any nonnegative, L1 subharmonic function must be constant. Proof. Let f be a nonnegative, L1 subharmonic function defined on M . By solving the heat equation with f as initial data, we obtain Z f (x, t) = H(x, y, t) f (y) dy M

with f (x, 0) = f (x). Recall that the heat semi-group is contractive in L1 because of Theorem 12.2, hence Z (30.6) H(x, y, t) dy ≤ 1. M

We now claim that ∂ f (x, t) ≥ 0 ∂t

(30.7)

for all x ∈ M and for all t ∈ (0, ∞). To see this, we differentiate under the integral sign and obtain Z ∂ ∂ f (x, t) = H(x, y, t) f (y) dy ∂t ∂t ZM = ∆y H(x, y, t) f (y) dy. M

If we can justify integration by parts, namely, Z Z (30.8) ∆H(x, y, t) f (y) dy = M

H(x, y, t) ∆f (y) dy,

M

then the subharmonicity of f will imply (30.7). Indeed, (30.8) follows from Green’s identity Z Z ∆H(x, y, t) f (y) dy − H(x, y, t) ∆f (y) dy Bp (ρ) Bp (ρ) Z Z ∂ ∂ = H(x, y, t) f (y) − H(x, y, t) f (y) dy ∂Bp (ρ) ∂r ∂r ∂Bp (ρ) Z Z = |∇H|(x, y, t)| f (y) dy + H(x, y, t) |∇f |(y) dy ∂Bp (ρ)

∂Bp (ρ)

GEOMETRIC ANALYSIS

241

and by showing that there exists a sequence {ρi } with ρi → ∞ such that the boundary terms on the right hand side when setting ρ = ρi tends to 0 as ρi → ∞. Using the mean value inequality, (14.23) of Corollary 14.2, we obtain the growth estimate Z √ −1 −m ¯ sup f ≤ C1 V (4ρ) ρ f exp(C2 ρ R) Vp (2ρ) Bp (ρ)

Bp (2ρ)

where −(m − 1)R is the lower bound of the Ricci curvature on Bp (4ρ). However, the assumption on the curvature of M implies that √ V¯ (4ρ) ≤ C6 exp(4(m − 1)ρ R) ≤ C6 exp(C7 ρ2 ) for some constants C6 , C7 > 0. This implies that sup f ≤ C8 Vp−1 (2ρ) exp(C9 ρ2 ),

(30.9)

Bp (ρ)

where C8 depends also on the L1 -norm of f . Let φ be a nonnegative cutoff function satisfying  0       r(x) − ρ + 1 φ(x) = 1       2ρ + 1 − r(x) ρ

on

Bp (ρ − 1) ∪ (M \ Bp (2ρ))

on

Bp (ρ) \ Bp (ρ − 1)

on

Bp (ρ) \ Bp (ρ + 1)

on

Bp (2ρ + 1) \ Bp (ρ + 1).

The subharmonicity of f implies that Z

φ2 f ∆f Z Z = −2 φ f h∇φ, ∇f i − φ2 |∇f |2 M Z M Z 1 2 2 2 ≤2 |∇φ| f − φ |∇f |2 . 2 M M

0≤

M

Hence using the definition of φ, we obtain the estimate Z Z 2 |∇f | ≤ 4 Bp (ρ+1)\Bp (ρ)

f2

Bp (2ρ+1)

for ρ ≥ 1. Combining with (30.9) and applying Schwarz inequality, we obtain Z

Z |∇f | ≤

(30.10) Bp (ρ+1)\Bp (ρ)

! 12 |∇f |2

1

Vp2 (ρ + 1)

Bp (ρ+1)\Bp (ρ)

≤ C8 exp(C10 ρ2 ). To estimate H(x, y, t) we recall that Theorem 13.1 asserting that  2  p −r (x, y) − 21 √ − 21 √ −2 + C12 (ρ + R)t , H(x, y, t) ≤ C11 Vx ( t) Vy ( t) exp (4 + )t

242

PETER LI

for all x, y ∈ Bp (ρ). For a fixed x ∈ M and for sufficiently large ρ, we may assume that x ∈ Bp ( ρ4 ). In this case, if y ∈ ∂Bp (ρ), then the curvature assumption implies that   √ −αρ2 − 21 √ − 12 √ (30.11) H(x, y, t) ≤ C11 Vx ( t) Vy ( t) exp + C12 ρ t . t Using the volume comparison theorem, we have √ √ √ Vx ( t) ≤ Vy (r(x, y) + t) − Vy (r(x, y) − t) √ √ V¯ (r + t) √ ≤ Vy ( t) V¯ ( t) √ √ m ≤ Vy ( t) C13 t 2 exp(C14 ρ(ρ + t)). Applying this to (30.11), we conclude that √ H(x, y, t) ≤ C15 Vx−1 ( t) exp

(30.12)



 √ −αρ2 + C16 ρ(ρ + t) . t

To estimate the gradient of H, we consider the integral Z φ2 (y) |∇H|2 (x, y, t) M Z Z = −2 hH(x, y, t)∇φ(y), φ(y)∇H(x, y, t)i − φ2 (y) H(x, y, t) ∆H(x, y, t) M Z M Z 1 2 2 2 2 ≤2 |∇φ| (y) H (x, y, t) + φ (y) |∇H| (x, y, t) 2 M M Z − φ2 (y) H(x, y, t) ∆H(x, y, t). M

This implies that (30.13) Z

|∇H|2

Bp (ρ+1)\Bp (ρ)

Z

φ2 |∇H|2

≤ M

Z ≤4

|∇φ|2 H 2 − 2

M

Z

φ2 H ∆H Z 2 H +2 M

Z ≤4 Bp (ρ+1)\Bp (ρ−1)

Z

2

≤4

H |∆H|

Bp (2ρ+1)\Bp (ρ−1)

Z

H +2

H

Bp (2ρ+1)\Bp (ρ−1)

Bp (2ρ+1)\Bp (ρ−1)

2

! 12 Z

(∆H) M

Using the upper bound (30.12) for H and (30.6), we can estimate Z (30.14) H 2 (x, y, t) ≤ sup H(x, y, t) y∈Bp (2ρ+1)\Bp (ρ−1)

Bp (2ρ+1)\Bp (ρ−1)

≤

√

C15 Vx−1 (

 t) exp

2

 √ −αρ2 + C16 ρ(ρ + t) . t

 21 .

GEOMETRIC ANALYSIS

243

We now claim that there exists a constant C17 > 0, such that, Z (30.15) (∆H)2 (x, y, t) ≤ C17 t−2 H(x, x, t). M

To see this, we first prove the inequality for any Dirichlet heat kernel H defined on a compact subdomain Ω of M . Using the fact that heat kernel on M can be obtained by taking limits of Dirichlet heat kernels on a compact exhaustion of M , inequality (30.15) follows. Indeed, if H(x, y, t) is a Dirichlet heat kernel defined on a compact subdomain Ω ⊂ M, we can write H using the eigenfunction expansion H(x, y, t) =

∞ X

e−λi t ψi (x, ) ψi (y),

i=1

where {ψi } are orthonormal basis of the space of L2 functions with Dirichlet boundary value satisfying the equation ∆ψi = −λi ψi . Differentiating with respect to the variable y, we have ∆H(x, y, t) = −

∞ X

λi e−λi t ψi (x) ψi (y).

i=1

Therefore, using the fact that s2 e−2s ≤ C17 e−s for all 0 ≤ s < ∞, we conclude that Z

(∆H)2 (x, y, t) dy ≤ C17 t−2

M

∞ X

e−λi t ψi2 (x)

i=1

= C17 t−2 H(x, x, t). Combining (30.13), (30.14), and (30.15), we obtain Z

2

|∇H| ≤

−1 √ C18 (Vx 2 ( t)

−1

+t



1 2

H (x, x, t)) exp

Bp (ρ+1)\Bp (ρ)

 √ −αρ2 + C19 ρ(ρ + t) . 2t

Applying Schwarz inequality and volume comparison theorem, we conclude that Z |∇H| ≤

(30.16)

−1 √ C20 (Vx 2 ( t)

+t

−1

1 2

1 2

H (x, x, t)) exp

Bp (ρ+1)\Bp (ρ)



 √ −αρ2 + C21 ρ(ρ + t) . 4t

Using (30.9), (30.10), (30.12), and (30.16), we conclude that for a fixed x ∈ M and for any sufficiently small fixed t > 0, Z Z |∇H|(x, y, t) f (y) dy + H(x, y, t) |∇f |(y) dy → 0 Bp (ρ+1)\Bp (ρ)

Bp (ρ+1)\Bp (ρ)

as ρ → ∞. Hence, the mean value theorem implies that there is a sequence of {ρi } with ρi → ∞, such that, Z

Z |∇H|(x, y, t) f (y) dy +

∂Bp (ρi )

H(x, y, t) |∇f |(y) dy → 0, ∂Bp (ρi )

244

PETER LI

and (30.8) is verified. In particular, we confirmed the monotonicity of f (x, t) for t sufficiently small. monotonicity of f (x, t) for all t follows from the semigroup property of the heat equation. Let us also observe that since Z Z ∂ H(x, y, t) dy = ∆H(x, y, t) dy, ∂t M M and because inequality (30.16) implies that Z Z ∆H(x, y, t) dy ≤ Bp (ρi )

|∇H|(x, y, t) dy

∂Bp (ρi )

→0 as ρi → ∞, we conclude that ∂ ∂t

Z H(x, y, t) dy = 0, M

hence Z (30.17)

H(x, y, t) dy = 1 M

for all x ∈ M and for all t. To finish the theorem, since (30.17) implies that Z Z Z f (x, t) dx = H(x, y, t) f (y) dy dx M M M Z = f (y) dy M

combining with the monotonicity of f (x, t), we conclude that f (x, t) = f (x), hence ∆f (x) = 0. On the other hand, for any arbitrary constant a > 0, let us we define the function g(x) = min{f (x), a}. It must be nonnegative satisfying g(x) ≤ f (x), |∇g|(x) ≤ |∇f |(x), and ∆g(x) ≤ 0. In particular, it will satisfy the same estimates, (30.9) and (30.10), as f . Hence we can show that Z Z ∂ H(x, y, t) g(y) dy = H(x, y, t) ∆g(y) dy ∂t M M ≤ 0. Since g is also L1 , the same argument as before implies that ∆g(x) = 0. Hence regularity theory asserts that g must be smooth. Since a is arbitrary, this is impossible unless f is identically constant.  Corollary 30.2. Let M m be a complete manifold. Suppose the Ricci curvature of M is bounded from below by Rij (x) ≥ −C (1 + r(x))2 for some constant C > 0. Then M does not admit any nonconstant L1 harmonic functions.

GEOMETRIC ANALYSIS

245

´ Inequality and Measured Sobolev Inequality §31 Measured Neumann Poincare In this section, we will adapt an argument by Hajtasz and Koskela [HK] that proved the validity of the Neumann Poincar´e inequality together with a volume doubling property will imply the validity of a Sobolev type inequality. In particular, we will use this argument to established a measured Sobolev inequality to be used in the next section. Definition 31.1. We say that a subdomain Ω of a manifold satisfies the volume doubling property (VD) if there exists a constant CVD (Ω) > 0, such that, Vp (2ρ) ≤ CVD (Ω) Vp (ρ) for all p ∈ Ω and for all 0 < ρ ≤ denote the smallest such constant

d(y,∂Ω) , 2

where d(y, ∂Ω) is the distance from y to ∂Ω. In particular, we

CVD (Ω) = sup

Vp (2ρ) Vp (ρ)

as the volume doubling constant on Ω. In addition to the volume doubling property, we also assume that M admits a uniform lower bound for the scaled first nonzero Neumann eigenvalue for geodesic balls. Again, one can localized the argument by introducing the following local λ1 bound. Definition 31.2. We say that CP (Ω) > 0 is a scaled local λ1 bound for the first nonzero Neumann eigenvalues over the smooth subdomain Ω if inf(ρ2 λ1 (By (ρ))) = CP (Ω), where the infimum is taken over all y ∈ Ω and all ρ ≤

d(y,∂Ω) , 2

where d(y, ∂Ω) is the distance from y to ∂Ω.

We will now state the main theorem of this section. Theorem 31.1. Let M be a complete manifold (possibly with boundary) and Bp (ρ) be a geodesic ball in M satisfying Bp (ρ) ∩ ∂M = ∅. For a any fixed 0 < δ < 1, let φ(r) be a function defined on Bp (ρ) given by 1 ρ−r φ(r) = (1 − δ)ρ    0    

if

r ≤ δρ

if

δρ ≤ r ≤ ρ

if

ρ ≤ r.

Then there exists a universal constant C1 > 0 depending on CP (Bp (ρ)) and CVD (Bp (ρ)), such that, the measured Neumann Poincar´e inequality of the form Z Z −2 2 2 ¯ C1 ρ |f (x) − f | φ (x) dV ≤ |∇f |2 (x) φ2 (x) dV Bp (ρ)

Bp (ρ)

is valid for all f ∈ C ∞ (Bp (ρ)), where f¯ =

R Bp (ρ)

f (x) φ2 (x) dV.

To prove the theorem, we will first establish a local version of the measured Neumann Poincar´e inequality. Lemma 31.1. Let Bp (ρ) ⊂ M be a geodesic ball satisfying Bp (ρ) ∩ ∂M = ∅. Assume that Bp (ρ) has a scaled local λ1 bound given by CP (Bp (ρ)) > 0. Then the measured Neumann Poincar´e inequality (31.1)

CP (Bp (ρ)) 9τ 2

Z By (τ )

|f (x) − f˜|2 φ2 (x) dV ≤

Z By (τ )

|∇f |2 (x) φ2 (x) dV

246

PETER LI

is valid for all y ∈ Bp (ρ) and for all 0 ≤ τ ≤

d(y,∂Bp (ρ)) , 2

Proof. The local λ1 bound asserts that Z Z 2 2 ˜ (31.2) |f (x) − f | φ dV = inf k∈R

By (τ )

where f˜ =

Z k∈R

By (τ )

|f (x) − k|2 dV

By (τ )

≤ CP−1 (Bp (ρ)) τ 2 sup φ2

Z

By (τ )

≤ CP−1 (Bp (ρ)) τ 2

2

sup φ = By (τ )

 

ρ − r(p, y) + τ (1 − δ)ρ

|∇f |2 (x) dV

By (τ )

supBy (τ ) φ2

Z

inf By (τ ) φ2

1 

f (x) φ2 (x) dV.

|f (x) − k|2 φ2 dV

≤ sup φ inf

  

By (τ )

By (τ ) 2

We observe that

R

|∇f |2 (x) φ2 (x) dV.

By (τ )

if

r(p, y) − τ ≤ δρ

if

r(p, y) − τ ≥ δρ

if

r(p, y) + τ ≥ δρ

if

r(p, y) + τ ≤ δρ

2

and  2   ρ − r(p, y) − τ inf φ2 = (1 − δ)ρ  By (τ )  1

(31.3)

implies supBy (τ ) φ2 inf By (τ )

φ2

        

=

1 (1 − δ)ρ ρ − r(p, y) − τ

if

r(p, y) + τ ≤ δρ

if

r(p, y) − τ ≤ δρ ≤ r(p, y) + τ

if

δρ ≤ r(p, y) − τ

2

  2     ρ − r(p, y) + τ   ρ − r(p, y) − τ

In the case when r(p, y) − τ ≤ δρ ≤ r(p, y) + τ , we have (1 − δ)ρ ρ − r(p, y) + τ ≤ ρ − r(p, y) − τ ρ − r(p, y) − τ ≤3 ρ−r(p,y)+τ since ρ−r(p,y)−τ is an increasing function of 0 ≤ τ ≤ 21 (ρ − r(p, y)). The same estimate also apply to the case when δρ ≤ r(p, y) − τ. Inequality (31.1) follows by combining these estimates with (31.2). 

We will also need the following volume doubling property for the measure φ2 dV, which follows as a consequence of the volume doubling property (VD) for the background metric. Lemma 31.2. Let Bp (ρ) ⊂ M be a geodesic ball satisfying Bp (ρ) ∩ ∂M = ∅. Suppose CVD (Bp (ρ) is the volume doubling constant for Bp (ρ). Then the inequality Z

2

Z

φ (x) dV ≤ 16CVD (Bp (ρ)) By (2τ )

is valid for all y ∈ Bp (ρ) and τ ≤

d(y,∂Bp (ρ)) . 2

By (τ )

φ2 (x) dV

GEOMETRIC ANALYSIS

Proof. If y ∈ Bp (ρ) and τ ≤ that Z (31.4)

d(y,∂Bp (ρ)) , 2

247

then the volume doubling property for the background metric asserts

φ2 (x) dV ≤ Vy (2τ ) sup φ2 By (2τ )

By (2τ )

≤ CVD (Bp (ρ)) Vy (τ ) sup φ2 By (2τ )

≤ CVD (Bp (ρ)) Since 2

sup φ = By (2τ )

    

supBy (2τ ) φ2 inf By (τ ) φ2

1 

ρ − r(p, y) + 2τ (1 − δ)ρ

Z

φ2 (x) dV.

By (τ )

if

r(p, y) − 2τ ≤ δρ

if

r(p, y) − 2τ ≥ δρ

2

combining (31.3), we have

supBy (2τ ) φ2 inf By (τ ) φ2

=

        

1 (1 − δ)ρ ρ − r(p, y) − τ

if

r(p, y) + τ ≤ δρ

if

r(p, y) − 2τ ≤ δρ ≤ r(p, y) + τ

if

δρ ≤ r(p, y) − 2τ

2

  2    ρ − r(p, y) + 2τ    ρ − r(p, y) − τ

Estimating the same way as in the proof of Lemma 31.1 by using the assumption that τ ≤ conclude that supBy (2τ ) φ2 ≤ 16 inf By (τ ) φ2

d(y,∂Bp (ρ)) , 2

we

and the lemma follows from (31.4).  Theorem 31.1 now follows by applying Lemma 31.1 and Lemma 31.2 to the more general theorem due to Hajtasz and Koskela [HK] after adaptation to include the measure φ. Theorem 31.2. Let Bp (ρ) be a geodesic ball in a complete manifold satisfying Bp (ρ) ∩ ∂M = ∅. Let φ be a nonnegative function defined on Bp (ρ). Suppose CVD,φ (Bp (ρ)) > 0 is a constant, such that, the volume doubling property with respect to the measure φ2 dV of the form Z Z φ2 dV ≤ CVD,φ (Bp (ρ)) φ2 dV By (2τ )

By (τ ) d(y,∂B (ρ))

p is valid for all y ∈ Bp (ρ) and for all τ ≤ , where d(y, ∂Bp (ρ)) = ρ − r(p, y). We also assume that 2 there exists a constant CP,φ (Bp (ρ)) > 0, such that, the measured Neumann Poincar´e inequality

CP,φ (Bp (ρ)) τ −2 inf

Z

k∈R

|f (x) − k|2 φ2 dV ≤

By (τ )

Z

|∇f |2 φ2 dV

By (τ )

d(y,∂B (ρ))

p . Then the measured Neumann Poincar´e inequality is valid is valid for all y ∈ Bp (ρ) and for all τ ≤ 2 on Bp (ρ). In particular, there exists a constant Cφ > 0 depending only on CP,φ (Bp (ρ)) and CVD,φ (Bp (ρ)), such that, Z Z Cφ ρ−2 |f − f¯|2 φ2 dV ≤ |∇f |2 φ2 dV

Bp (ρ)

Bp (ρ)

248

PETER LI

R for all f ∈ C ∞ (Bp (ρ)), where f¯ = Bp (ρ) f φ2 dV. Moreover, there exists α > 2 depending on CVD,φ (Bp (ρ)), such that, the measured Neumann Sobolev inequality of the form Cφ ρ

−2

Vφ (Bp (ρ))

is valid, where Vφ (Bp (ρ)) =

! α−2 α

Z

2 α

R Bp (ρ)

|f − f¯|

2α α−2

2

Z ≤

φ dV

Bp (ρ)

|∇f |2 φ2 dV

Bp (ρ)

φ2 dV denotes the volume of Bp (ρ) with respect to the measure φ2 dV.

Proof. In this proof, we will abbreviate our notation by using CVD,φ and CP,φ for CVD,φ (Bp (ρ)) and CP,φ (Bp (ρ)), respectively. For any y ∈ Bp (ρ), let γ(t) be a normal geodesic Pwith γ(0)  = p and γ(`) = y. Let
i ρ ρ us define the sequence of points y0 = γ(0) = p, y1 = γ 2 , and yi = γ j=1 2j . Let i0 be the largest i such that i X ρ < `. 2j j=1 Let us define Bi =

 ρ    B  yi i+1  2 

for

i < i0

  ρ     By 2i+1

for

i ≥ i0 .

For any f ∈ C ∞ (Bp (ρ)), we define fBi = Vφ−1 (Bi ) where

Z

f φ2 dV,

Bi

Z

φ2 dV

Vφ (Bi ) = Bi

denotes the volume of Bi with respect to the measure φ2 dV. We observe that fBi → f (y) as i → ∞, and |fB0 − f (y)| ≤

(31.5)

≤

∞ X i=0 ∞ X

|fBi − fBi+1 |
|fBi − fDi | + |fDi − fBi+1 |

i=0

where Di = Bzi with zi = γ

P

i ρ j=1 2j

+

3ρ 2i+3



 ρ  ⊂ Bi ∩ Bi+1 2i+3

. However, since Di ⊂ Bi , we have

|fBi − fDi | ≤

Vφ−1 (Di )

Z

Vφ−1 (Di )

Z

|f − fBi | φ2 dV

Di

|f − fBi | φ2 dV Z −1 3 ≤ CVD,φ Vφ (Bi ) |f − fBi | φ2 dV.

≤

Bi

Bi

and similarly |fBi+1 − fDi | ≤

3 CVD,φ

Vφ−1 (Bi+1 )

Z Bi+1

|f − fBi+1 | φ2 dV.

GEOMETRIC ANALYSIS

249

Hence combining with (31.5), we conclude that |fB0 − f (y)| ≤

(31.6)

3 2CVD,φ

∞ X

Vφ−1 (Bi )

Z

i=0 3 ≤ 2CVD,φ

∞  X

|f − fBi | φ2 dV

Bi

Vφ−1 (Bi )

Z

≤

3 2CVD,φ

− 12 CP,φ

i=0

ρ



2i+1

 12

Bi

i=0 ∞ X

|f − fBi |2 φ2 dV

Vφ−1 (Bi )

Z

2

2

 12

|∇f | φ dV

.

Bi

Note that for  > 0, since ∞ X

2−(i+1) < ∞,

i=0

we conclude that ∞ X

 Z ρi Vφ−1 (Bi )

|∇f |2 φ2 dV

 12

Bi

i=0

1

−3 2 ≥ C1 CVD,φ CP,φ ρ− |fB0 − f (y)|

∞ X

ρi

i=0

where ρi = 2−(i+1) ρ and C1 > 0 is a universal constant. In particular, this implies that there exists j ≥ 0, such that, Vφ−1 (Bj )

(31.7)

Z Bj

−6 |∇f |2 φ2 dV ≥ C12 CVD,φ CP,φ ρ−2 |fB0 − f (y)|2 ρ2−2 . j

Note that since r(y, yi ) ≤ ρ − r(p, yi ) ≤ρ− =

i X ρ j 2 j=1

ρ , 2i

we have Bi ⊂ By (3ρi )

for all

i.

Hence, together with (31.7), we obtain Z By (3ρj )∩Bp (ρ)

−6 |∇f |2 φ2 dV ≥ C12 CVD,φ CP,φ Vφ (Bj ) ρ−2 |fB0 − f (y)|2 ρ2−2 j −8 ≥ C12 CVD,φ CP,φ ρ−2 |fB0 − f (y)|2 ρ2−2 Vφ (By (ρj )). j

In particular, we conclude that for all y ∈ Bp (ρ), there exists ρy > 0 such that Z (31.8) By (ρy )∩Bp (ρ) −10 where Cφ = C12 CVD,φ CP,φ .

|∇f |2 φ2 dV ≥ Cφ ρ−2 |fB0 − f (y)|2 ρ2−2 Vφ (By (ρy )), y

250

PETER LI

We now claim that the volume doubling property with respect to the measure φ2 dV implies that there exists a constant β > 0 depending only on CVD,φ , such that,  (31.9)

Vφ (Bp (ρ)) ≤ CVD,φ

ρ ρy

β Vφ (By (ρy ))

for all y ∈ Bp (ρ). Indeed, let γ be the geodesic joining y to p and we choose a sequence of points w0 , w1 , . . . , wk with w0 = y and wi ∈ γ, such that, r(wi , wi+1 ) = 2i ρy , where k is the first integer such that p ∈ Bwk (2k ρy ). Pk In particular, ρy + i=2 2i ρy ≤ r(p, y), hence   ρ k ≤ log2 −1 . ρy Note that the volume comparison asserts that
3 CVD,φ Vφ (Bwi (2i ρy )) ≥ Vφ Bwi (5(2i ρy )) ≥ Vφ (Bwi+1 (2i+1 ρy )) and CVD,φ Vφ (Bwk (2k ρy )) ≥ Vφ (Bp (2k ρy )), therefore we have (31.10)

3k+1 CVD,φ Vφ (By (ρy ))φ2 dV ≥ Vφ (Bp (2k ρy )).

On the other hand, if we set k 0 to be the smallest integer such that 0

2k (2k ρy ) ≥ ρ, then k 0 + k ≤ log2 and



ρ ρy

 +1

0

k CVD,φ Vφ (Bp (2k ρy )) ≥ Vφ (Bp (ρ)).

Combining with (31.10), we obtain 3 log2

CVD,φ

ρ ρy




+1

Vφ (Bwk (2k ρy )) ≥ Vφ (Bp (ρ)).

and the claim (31.9) is established with β = 3 log2 CVD,φ . Let us now define the sublevel set At = {y ∈ Bp (ρ) | |fB0 − f (y)| ≥ t}. Applying (31.9) to (31.8), we conclude that for all y ∈ At , there exists ρy > 0, such that, Z 2(−1)β −1 ρ2 Vφ (Bp (ρ)) |∇f |2 φ2 dV By (ρy )∩Bp (ρ)

≥ Cφ t

2



ρ ρy

2−2

1+2(−1)β −1

≥ Cφ t2 Vφ

2(−1)β −1

Vφ

(Bp (ρ)) Vφ (By (ρy ))

(By (ρy )).

GEOMETRIC ANALYSIS

251

Without loss of generality, we may assume that β ≥ 2. The Vitali covering lemma asserts that there is a countable (possibly finite), mutually disjoint, subcollection {B1 (ρ1 ), B2 (ρ2 ), . . . , Bi (ρi ), . . . } of the collection {By (ρy ) | y ∈ At } with the properties that ρ1 ≥ ρ2 ≥ · · · ≥ ρi ≥ · · · . Moreover, for any y ∈ At , there exists an i such that Bi (ρi ) ∩ By (ρy ) = ∅ and By (ρy ) ⊂ Bi (3ρi ). In particular, we have 1−2(1−)β −1

Vφ

2 (At ) ≤ CVD,φ

X

1−2(1−)β −1

Vφ

(Bi (ρi ))

i −2(1−)β −1

≤ Cφ−1 t−2 ρ2 Vφ

(Bp (ρ))

XZ Bi (ρi )∩Bp (ρ)

i −2(1−)β

≤ Cφ−1 t−2 ρ2 Vφ

|∇f |2 φ2 dV

Z

−1

|∇f |2 φ2 dV,

(Bp (ρ)) Bp (ρ)

where Cφ > 0 is a constant depending only on CVD,φ and CP,φ . Setting

B=

−2(1−)β −1 Cφ−1 ρ2 Vφ (Bp (ρ))

and if q is a constant satisfying q < Z

2 1−2(1−)β −1

|f − fB0 |q φ2 dV = q

Z

Bp (ρ)

!

Z

1 1−2(1−)β −1

|∇f |2 φ2 dV

,

Bp (ρ)

then the co-area formula implies that

∞

tq−1 Vφ (At ) dt

0

Z

T

Z

∞

Vφ (At ) dt + q tq−1 Vφ (At ) dt 0 T Z ∞ 2 q−1− 1−2(1−)β −1 dt ≤ T q Vφ (Bp (ρ)) + qB t =q

q−1

t

T

qB

q

= T Vφ (Bp (ρ)) +

2 1−2(1−)β −1

q−

T

q−

2 1−2(1−)β −1

.

Optimizing by choosing T =

−1 ρ Vφ 2 (Bp (ρ))

! 21

Z

|∇f |2 φ2 dV

,

Bp (ρ)

we obtain Z (31.11)

! q1 |f − fB0 |q φ2 dV

Bp (ρ)

Since

2 1+2(−1)β −1

≤ Cφ−1 ρ Vφ

Z (Bp (ρ))

! 12 |∇f |2 φ2 dV

Bp (ρ)

> 2, we can choose q = 2 and obtain Z

(31.12)

1 1 q−2

inf

k∈R

Bp (ρ)

|f − k|2 φ2 dV ≤

Z

|f − fB0 |2 φ2 dV

Bp (ρ)

≤ Cφ−1 ρ2 establishing the measured Neumann Poincar´e inequality.

Z Bp (ρ)

|∇f |2 φ2 dV.

.

252

PETER LI

2 In fact, since we can take 2 < q < 1+2(−1)β −1 , a measured Neumann Sobolev type inequality also follows. To see this, we apply (31.11) and obtain

(31.13) ! q1

Z

|f − f¯|q φ2 dV

! q1

Z

|f − fB0 |q φ2 dV

≤

Bp (ρ)

≤

|fB0 − f¯|q φ2 dV

+

Bp (ρ)

Cφ−1

! q1

Z Bp (ρ)

1 1 q−2

ρ Vφ

! 21

Z

2

|∇f | φ dV

(Bp (ρ))

1

+ Vφq (Bp (ρ)) |fB0 − f¯|.

2

Bp (ρ)

However, 1 q

1

Vφ (Bp (ρ)) |fB0

− f¯| ≤ Vφq (Bp (ρ)) Vφ−1 (B0 ) ≤

Z

|f − f¯| φ2 dV

B0

−1 Vφ (Bp (ρ)) Vφ 2 (B0 )

1 1 q−2

≤ CVD,φ Vφ

! 21

Z

1 q

¯2

|f − f | φ dV Bp (ρ)

Z (Bp (ρ))

! 12 |f − f¯|2 φ2 dV

with α =

2q q−2 .

.

Bp (ρ)

hence combining with (31.12) and (31.13), we conclude that ! α−2 Z Z α 2 2α −1 2 − α 2 ¯ α−2 |f − f | φ dV ≤C ρ V (Bp (ρ)) φ

Bp (ρ)

2

φ

|∇f |2 φ2 dV

Bp (ρ)



Obviously, Theorem 31.1 follows from Lemma 31.1, Lemma 31.2, and Theorem 31.2. Let us also point out that if we set φ ≡ 1, then Theorem 31.2 asserts that a local λ1 bound and volume doubling implies a Neumann Sobolev inequality. Corollary 31.1. Let Bp (ρ) be a geodesic ball in a complete manifold satisfying Bp (ρ) ∩ ∂M = ∅. Suppose CVD (Bp (ρ)) > 0 is the volume doubling constant on Bp (ρ) and CP (Bp (ρ)) > 0 is the scaled local λ1 bound as given by Definition 31.1 and Definition 31.2, respectively. Then there exists a constant C1 > 0 depending only on CP (Bp (ρ)) and CVD (Bp (ρ)), such that, the Neumann Sobolev inequality of the form ! α−2 Z Z α 2α 2 2 −α ¯ α−2 dV ≤ C1 ρ V (Bp (ρ)) |∇f |2 dV |f − f | Bp (ρ)

is valid, for all f ∈ C (Bp (ρ)), where f¯ = ∞

Bp (ρ)

R Bp (ρ)

f dV.

We would like to point out that both constants CVD (Ω) and CP (Ω) are monotonic in the sense that CVD (Ω1 ) ≤ CVD (Ω2 ) and CP (Ω1 ) ≥ CP (Ω2 ) if Ω1 ⊂ Ω2 . In particular, if M has nonnegative Ricci curvature then the volume comparison theorem asserts that CVD (Ω) = 2m for all Ω ⊂ M. Moreover, Corollary 14.1 asserts that CP (Bp (ρ)) ≥ C0 for all p ∈ M and ρ > 0, where C0 > 0 is a constant depending only on m.

GEOMETRIC ANALYSIS

253

§32 Parabolic Harnack Inequality and Regularity Theory In the section, we will present Moser’s version of the Nash-Moser’s Harnack inequality for parabolic equations. The elliptic version (also proved by DeGiorgi) can be considered as a special case when the solution is time independent. The iteration procedure of Moser was particularly useful in the theory of geometric analysis. We will attempt to cover this in as much generality as possible while keeping explicit account of the dependency of various geometric and analytic constants. In applying this type of argument in the study of geometric PDE, often the explicit geometric dependency is crucial. As a result of these estimates, one derives a mean value inequality for nonnegative subsolutions and a Harnack inequality for positive solutions of a fairly general class of parabolic operators. In particular, it gives a C α estimate for solutions of any second order parabolic (elliptic) operators of divergence form with only measurable coefficients. This regularity result was the original motivation for the development of this theory. We shall point out that the mean value inequality and the Harnack inequality derived from this argument are applicable to a more general class of equations, while the ones given in earlier sections yield stronger results but requiring more smoothness from the operator. Both approaches are important in the theory of geometric analysis, while each is more suitable for different types of situation. The following account is a slightly modified version of Moser’s argument which has been adapted to a more geometrical setting. In terms of notations, let us define the average value of a function f defined on a geodesic ball Bp (ρ) by Z Z — f dV = Vp (ρ)−1 f dV. Bp (ρ)

Bp (ρ)

q

When the point p is fixed, the average L -norm of f over Bp (ρ) is denoted by –k f kq,ρ =

! q1

Z —

q

f dV

Bp (ρ)

and the regular Lq -norm is denoted by ! q1

Z

q

kf kq,ρ =

f dV

.

Bp (ρ)

When the function is defined on Bp (ρ) × [T0 , T1 ], we denote its Lq norm over the set Bp (ρ) × [T0 , T1 ] by Z

T1

! q1

Z

f q dV dt

kf kq,ρ,[T0 ,T1 ] = T0

Bp (ρ)

and the average Lq norm by –k f kq,ρ,[T0 ,T1 ] =

−1

Z

T1

(T1 − T0 )

T0

! q1

Z —

q

f dV

.

Bp (ρ)

For the sake of further generalization to Riemannian manifolds, we will introduced a normalized Sobolev inequality on a geodesic ball centered at a point p ∈ M of radius ρ. The inequality asserts that for a fixed m constant µ ≤ m−2 , there exists a constant CSD > 0, depending on Bp (ρ), such that, (32.1)

Z —

CSD |∇φ| ≥ 2 ρ Bp (ρ) 2

Z —

! µ1 φ

2µ

,

Bp (ρ)

c for all φ ∈ H1,2 (Bp (ρ)) that has boundary condition φ = 0 on ∂Bp (ρ). Typically, the constant µ is given m by µ = m−2 when m ≥ 3, and 2 < µ < ∞ when m = 2. However, in the following context, we are not restricting the value of µ.

254

PETER LI

Lemma 32.1. Let M be a complete manifold of dimension m. Let us assume that the geodesic ball Bp (ρ) centered at p with radius ρ satisfies Bp (ρ) ∩ ∂M = ∅. Suppose that u(x, t) is a nonnegative function defined on Bp (ρ) × [T0 , T1 ] such that   ∂ u ≥ −f u ∆− ∂t in the weak sense. Assume that the function f is nonnegative on Bp (ρ) × [T0 , T1 ] and its Lq norm on Bp (ρ) µ is finite for some µ−1 < q ≤ ∞ on [T0 , T1 ], with

A = sup –k f kq,ρ = sup [T0 ,T1 ]

for

µ µ−1

! q1

Z —

[T0 ,T1 ]

q

f (x, t) dV

Bp (ρ)

< q < ∞ and A = kf k∞,ρ,[T0 ,T1 ] =

sup

f

Bp (ρ)×[T0 ,T1 ] q(µ−1) > 0 and ν = for q = ∞. If α and µ are defined by α = µ(q−1)−q constant C2 > 0 depending only on k, µ, q, and CSD , such that,

2µ−1 µ

> 1, then for any k > 1, there exists

kuk∞,θρ,[T,T1 ] ≤ C2

(kA ρ

2

−1 α CSD )



CSD ρ2

 ν−1 ν

−2

+ ((1 − θ)

ρ

−2

−1

+ (T − T0 )

 )

ρ2 CSD

ν  ν1 ! k(ν−1)

1

× (T1 − T0 ) k –kukk,ρ,[T0 ,T1 ] for any θ ∈ (0, 1) and T ∈ (T0 , T1 ). For 0 < k ≤ 1, there exists constant C3 > 0 depending only on k, µ, q and CSD , such that,  kuk∞,θρ,[T,T1 ] ≤ C3

ρ2 A CSD −1

α 

2

CSD ρ2

 ν−1 ν

+ CSDν ρ ν (T − T0 )−1

−1

+ CSDν (1 − θ)−2 ρ

ν  k(ν−1)

1 −k

Vp

ρ 2

−2(ν−1) ν

kukk,ρ,[T0 ,T1 ] .

Proof. Note that the normalized Sobolev inequality in the form described is scale invariant. Hence, without loss of generality, we may assume that Vp (ρ) = 1. For any arbitrary constant a ≥ 1, the assumption on u implies that Z Z Z φ2 f u2a −

φ2 u2a−1 ut ≥ −

φ2 u2a−1 ∆u,

for any nonnegative, compactly supported, Lipschitz function φ on Bp (ρ). The right hand side after integration by parts yields Z −

(32.2)

φ2 u2a−1 ∆u = 2

Z

φ u2a−1 h∇φ, ∇ui + (2a − 1)

Z

φ2 u2a−2 |∇u|2 .

However, using the identity Z (32.3)

|∇(φua )|2 =

Z

|∇φ|2 u2a + 2a

Z

φ u2a−1 h∇φ, ∇ui + a2

Z

φ2 u2a−2 |∇u|2 ,

GEOMETRIC ANALYSIS

255

and the assumption that a ≥ 1, we have Z

φ2 f u2a +

a

(32.4)

Z

|∇φ|2 u2a ≥

Z

1 2

|∇(φua )|2 +

CSD ≥ 2 ρ

Z

CSD ≥ 2 ρ

Z

Z

a 2µ

φ2 (u2a )t

 µ1

1 + 2

Z

 µ1

1 + 2

Z

(φ u )

φ2 (u2a )t .

When q = ∞, (32.4) implies that Z (32.5) When

aA µ µ−1

2

2a

φ u

Z

2

2a

|∇φ| u

+

a 2µ

(φ u )

φ2 (u2a )t .

< q < ∞, by H¨ older’s inequality, we have Z

(32.6)

a

φ2 f u2a ≤ aA

Z

q

(φ2 u2a ) q−1

Z

2 2a

≤ aA

 q−1 q

 µ(q−1)−q Z q(µ−1)

2 2a µ

φ u

1  q(µ−1)

.

(φ u )

However, applying the inequality 

x ≤δ by setting  =

µ(q−1)−q q(µ−1)

−1 

x + δ

1 1−



 1 −1 

and

x = (aA)

Z

q(µ−1) µ(q−1)−q

2 2a

 Z

φ u

2 2a µ

− µ1

(φ u )

,

we have (32.7) Z aA

2 2a

Z  µ(q−1)−q q(µ−1)

φ u

2 2a µ

 q−µ(q−1) qµ(µ−1)

(φ u ) ≤δ

−1 

Z

q(µ−1)

(aA) µ(q−1)−q

φ2 u2a

 Z

(φ2 u2a )µ

− µ1

1



+ δ 1−

 1 −1 . 

Multiplying through by Z

(φ2 u2a )µ

 µ1

and choosing δ so that δ

1 1−



 1 CSD −1 = ,  2ρ2

(32.6) and (32.7) yield Z a

2

2a

φ fu

 ≤ C1

CSD ρ2

−µ  µ(q−1)−q

(aA)

q(µ−1) µ(q−1)−q

Z

2 2a

φ u



CSD + 2ρ2

Z

2 2a µ

(φ u )

 µ1

256

PETER LI

for some constant C1 > 0 depending only on µ and q. Hence together with (32.4), we have  (32.8)

C1

CSD ρ2

−µ  µ(q−1)−q

CSD 2ρ2

≥

Z

q(µ−1)

φ2 u2a

(aA) µ(q−1)−q Z

(φ ua )2µ

 µ1 +

1 2

Z



Z +

|∇φ|2 u2a

φ2 (u2a )t .

In any event, (32.5) and (32.8) imply that we have the inequality  (32.9)

2C1

CSD ρ2

1−α

(aA)α

Z

CSD ≥ 2 ρ

Z

a 2µ

φ2 u2a + 2  µ1

(φ u )

Z

Z

|∇φ|2 u2a

φ2 (u2a )t .

+

q(µ−1) µ with α = µ(q−1)−q > 0 for all µ−1 < q ≤ ∞. If ψ(t) is a Lipschitz function given by

 0    t−s ψ(t) =  v   1

for

T0 ≤ t ≤ s

for

s≤t≤s+v

for

s + v ≤ t ≤ T1 ,

then (32.9) implies that (32.10)  2C1

CSD ρ2

1−α

(aA)α

t0

Z

Z

T0

Z

t0

Z

ψ 2 (t) |∇φ|2 (x) u2a (x, t) dx dt + 2

+2 Bp (ρ)

T0

CSD ≥ 2 ρ

Z

CSD = 2 ρ

Z

ψ 2 (t) φ2 (x) u2a (x, t) dx dt

Bp (ρ)

t0

t0

! µ1 a

ψ (t)

(φ(x) u (x, t))

2µ

dx

dt + ! µ1

a

ψ (t)

(φ(x) u (x, t))

T0

t0

T0

Z

2

2µ

dx

ψ(t) ψt (t) φ2 (x) u2a (x, t) dx dt

Bp (ρ)

Z

Bp (ρ)

T0 t0

Z

T0

Z

2

Z

∂ ∂t

Z

2

2

2a

ψ (t) φ (x) u (x, t) dx

dt

Bp (ρ)

ψ 2 (t0 ) φ2 (x) u2a (x, t0 ) dx,

dt +

Bp (ρ)

!

Z

Bp (ρ)

for all t0 ∈ (T0 , T1 ]. First observe that since the first term on the right hand side is nonnegative, we conclude that (32.11)  2C1

CSD ρ2

1−α (aA)

α

Z

T0

Z

T1

Z

T0

ψ 2 (t) φ2 (x) u2a (x, t) dx dt

Bp (ρ)

Bp (ρ)

Z sup s+v≤t0 ≤T1

Z

ψ 2 (t) |∇φ|2 (x) u2a (x, t) dx dt + 2

+2 ≥

T1

T1

T0 2

2a

0

φ (x) u (x, t ) dx. Bp (ρ)

Z

Z Bp (ρ)

ψ(t) ψt (t) φ2 (x) u2a (x, t) dx dt

GEOMETRIC ANALYSIS

257

By setting t0 = T1 in (32.10), we also have (32.12)  2C1

CSD ρ2

1−α

α

Z

ψ 2 (t) φ2 (x) u2a (x, t) dx dt

(aA)

T0

Z

T1

Z

T0

CSD ρ2

Z

Bp (ρ)

T1

Z

Bp (ρ)

ψ 2 (t) |∇φ|2 (x) u2a (x, t) dx dt + 2

+2

=

T1

Z

Z

T0

ψ(t) ψt (t) φ2 (x) u2a (x, t) dx dt

Bp (ρ)

! µ1

Z

s+v

T1

(φ(x) ua (x, t))

2µ

dx

dt.

Bp (ρ)

On the other hand, Schwarz inequality implies that Z

a 2

 µ−1 Z µ

(φ u )

a 2µ

 µ1

Z ≥

(φ u )

(φ ua )

2(2µ−1) µ

,

hence Z

! µ−1 Z µ

a 2

T1

! µ1

Z

a

(φ u ) dx

sup s+v≤t≤T1

(φ(x) u (x, t))

Bp (ρ)

s+v

2µ

Z

T1

Z

(φ ua )

dt ≥

dx

Bp (ρ)

s+v

2(2µ−1) µ

Bp (ρ)

Combining with (32.11) and (32.12), we obtain (32.13)  2C1

CSD ρ2

1−α (aA)

α

Z

T0

Z

T1

Z

CSD ρ2

ψ 2 (t) φ2 (x) u2a (x, t) dx dt

Bp (ρ)

Z

Bp (ρ)

T0

Z

Z

ψ 2 (t) |∇φ|2 (x) u2a (x, t) dx dt + 2

+2

≥

T1

T1

Z

s+v

T1

Z

T0

(φ(x) ua (x, t))

2(2µ−1) µ

ψ(t) ψt (t) φ2 (x) u2a (x, t) dx dt

Bp (ρ)

µ ! 2µ−1

dx dt

.

Bp (ρ)

Let us now choose φ(r(x)) to be a function of the distance to p, given by 0 τ +σ−r φ(r) =  σ   1    

on

Bp (ρ) \ Bp (τ + σ)

on

Bp (τ + σ) \ Bp (τ )

on

Bp (τ ),

and (32.13) yields µ

−1 α−1 −1 2µ−1 (2C1 (ρ2 CSD ) (aA)α + 2σ −2 + 2v −1 )(ρ2 CSD )

(32.14)

Z

T1

Z

≥

u s+v

with α =

q(µ−1) µ(q−1)−q

for all

m 2

< q ≤ ∞.

Bp (τ )

2a(2µ−1) µ

µ ! 2µ−1

Z s

T1

Z Bp (τ +σ)

u2a

.

258

PETER LI

Let us now choose the sequences of ai , τi , and σi , such that a0 = σ0 = 2−1 (1 − θ)ρ,

k , 2

a1 =

kν , 2

··· ,

σ1 = 2−2 (1 − θ)ρ,

ai =

··· ,

kν i , 2

σi = 2−(1+i) (1 − θ)ρ,

and τ0 = ρ,

τ1 = ρ − σ0 ,

··· ,

··· ,

τi = ρ −

i−1 X

σj ,

··· ,

··· .

j=0

with ν =

2µ−1 µ .

We also choose the sequences of si and vi , such that s0 = T0 ,

s1 = T0 + 2−1 (T − T0 ),

···

i X

2−i (T − T0 ),

··· ,

vi = 2−(1+i) (T − T0 ),

··· .

si = T0 +

j=1

and v0 = 2−1 (T − T0 ),

v1 = 2−2 (T − T0 ),

··· ,

Observe that limi→∞ ρi = θρ and limi→∞ si = T . Applying (32.14) to a = ai , τ = τi+1 , σ = σi , s = si , and v = vi , we have kuk2ai+1 ,τi+1 ,[si+1 ,T1 ]  1  −1 ν1 kν i −1 α−1+ ν1 iα ν + (23+2i (1 − θ)−2 ρ−2 + 22+i (T − T0 )−1 )(ρ2 CSD ) ≤ 21−α (kA)α C1 (ρ2 CSD ) × kuk2ai ,τi ,[si ,T1 ] with Z

T1

Z

kuk2ai ,τi ,[si ,T1 ] =

! 2a1 2ai

u si

i

.

Bp (τi )

Iterating this inequality, we conclude that kuk2ai+1 ,τi+1 ,[si+1 ,T1 ] ≤

i  Y

1

1

−1 α−1+ ν jα −1 ν 21−α (kA)α C1 (ρ2 CSD ) ν + (23+2j (1 − θ)−2 ρ−2 + 22+j (T − T0 )−1 )(ρ2 CSD )



1 kν j

j=0

× kuk2a0 ,τ0 ,[s0 ,T1 ] . On the other hand, we have the inequality −1

−1

lim ((T1 − T ) V (θρ)) 2ai+1 kuk2ai+1 ,τi+1 ,[si+1 ,T1 ] ≥ lim ((T1 − T ) V (θρ)) 2ai+1 kuk2ai+1 ,θρ,[T,T1 ]

i→∞

i→∞

= kuk∞,θρ,[T,T1 ] , where the right hand side is defined to be the supremum of u over the set Bp (θρ) × [T, T1 ]. Therefore, letting i → ∞, we conclude that (32.15) kuk∞,θρ,[T,T1 ] ∞   1 Y −1 α−1+ ν1 jα −1 ν1 kν j ≤ 21−α (kA)α C1 (ρ2 CSD ) ν + (23+2j (1 − θ)−2 ρ−2 + 22+j (T − T0 )−1 )(ρ2 CSD ) j=0

× kukk,ρ,[T0 ,T1 ] .

GEOMETRIC ANALYSIS

The product can be estimated by using the fact that ν = ∞ Y

Bν

−j

2µ−1 µ

259

> 1, hence the identity

ν

= B ν−1 ,

j=0

and the fact that ∞  Y

P∞

j=0

jµ−j is finite. Therefore we have 1

1

−1 ν −1 α−1+ ν jα ν + (23+2j (1 − θ)−2 ρ−2 + 22+j (T − T0 )−1 )(ρ2 CSD 21−α (kA)α C1 (ρ2 CSD ) )



1 kν j

j=0

≤

∞  Y

1

1

−1 ν −1 α−1+ ν + ((1 − θ)−2 ρ2 + (T − T0 )−1 )(ρ2 CSD 21−α (kA)α C1 (ρ2 CSD ) )



1 kν j

j

max{ν α , 4} kν j

j=0

  ν −1 α −1 ν1 k(ν−1) −1 − ν−1 ≤ C2 (kA ρ2 CSD ) (ρ2 CSD ) , ) ν + ((1 − θ)−2 ρ−2 + (T − T0 )−1 )(ρ2 CSD where C2 ≥ 0 depends only on k, µ, and q alone. Inequality (32.15) now takes the form (32.16) kuk∞,θρ,[T,T1 ] −1 α (kA ρ2 CSD )

≤ C2



CSD ρ2

 ν−1 ν

+ ((1 − θ)−2 ρ−2 + (T − T0 )−1 )

ρ2 CSD



ν  ν1 ! k(ν−1)

1

× (T1 − T0 ) k –kukk,ρ,[T0 ,T1 ] for k ≥ 2, where –kukk,ρ,[T0 ,T1 ] =

−1

(T1 − T0 )

Vp−1 (ρ)

Z

T1

! k1

Z

k

.

u Bp (ρ)

T0

We now claim that (32.16), in fact, holds for 1 < k < 2 also. To see this, we observe that the iteration process begins with a = k2 , hence for k > 1, we have 2a − 1 > 0 in (32.2). Using the inequality Z 2

2a−1

φu

−1

h∇φ, ∇ui ≥ −a

(2a − 1)

−1

Z

2

|∇φ| u

2a

Z − a(2a − 1)

φ2 u2a−2 |∇u|2

and combining with (32.2), we conclude that Z −

Z Z φ u2a−1 h∇φ, ∇ui + 2(1 − δ) φ u2a−1 h∇φ, ∇ui + (2a − 1) φ2 u2a−2 |∇u|2 Z Z −1 −1 2 2a ≥ −2δ a (2a − 1) |∇φ| u + 2(1 − δ) φ u2a−1 h∇φ, ∇ui Z + (2a − 1 − δa(2a − 1)) φ2 u2a−2 |∇u|2 .

φ2 u2a−1 ∆u = 2δ

Z

Setting δ = (2a)−1 and using (32.3), we have Z −

2

2a−1

φ u

 ∆u ≥ −

1 2a − 1 + 2 a (2a − 1) 2a2

Z

2

2a

|∇φ| u

2a − 1 + 2a2

Z

|∇(φ ua )|2 ,

260

PETER LI

hence Z

2

φ fu

2a

 +

1 2a − 1 + 2 a (2a − 1) 2a2

Z

|∇φ|2 u2a ≥

2a − 1 2a2

Z

|∇(φ ua )|2 +

(2a − 1)CSD ≥ 2a2 ρ2

Z

1 2a

a 2µ

(φ u )

Z  µ1

φ2 (u2a )t 1 + 2a

Z

φ2 (u2a )t .

Using this inequality instead of (32.4) for values of 21 < a < 1, the same argument as above yields (32.16) for k > 1. For those values of k ≤ 1, we begin with the case k = 2. In that case, (32.16) yields (32.17) kuk∞,γτ,[S,T1 ] ν ! 2(ν−1)  ν−1  2  ν1  α 
τ τ2 CSD ν −2 −2 −1 −1 ≤ C2 + 2A (1 − γ) τ + (1 − ξ) S CSD τ2 CSD 1

× (T1 − ξS) 2 –kuk2,τ,[ξS,T1 ]  α   ν−1 −2(ν−1) ρ2 CSD ν −1 ≤ C2 2A + CSDν (1 − γ)−2 τ ν CSD ρ2 ν ! 2(ν−1)  2  ν1 k 1 ρ 1− k 2 2 + (1 − ξ)−1 S −1 (T1 − ξS) 2 –kukk,τ,[ξS,T kuk∞,τ,[ξS,T , ] 1 1] CSD for any θρ ≤ τ ≤ ρ, 0 < γ < 1, T0 ≤ S ≤ T and 0 < ξ < 1. Let us choose the sequences of τi and γi to be τ−1 = θρ, τ0 = θρ + 2−1 (1 − θ)ρ, · · · , τi−1 = θρ + (1 − θ)ρ

i X

2−j , · · ·

j=1

and γi τi = τi−1 . Also, the sequence Si and ξi to be S0 = T, S1 = (1 − 2−1 )(T − T0 ) + T0 , · · · , Si = (1 −

i X

2−j )(T − T0 ) + T0 , · · ·

j=1

and ξi Si = Si+1 . Applying (32.17) by setting τ = τi , γ = γi , S = Si and ξ = ξi and iterating the inequality yields (32.18) kuk∞,θρ,[T,T1 ] ≤

j Y

C2

i=0

α   ν−1  2(ν−1) ρ2 CSD ν − ν1 −2 − ν 2A + C (1 − γ ) τ i i SD CSD ρ2 

+

ρ2 CSD

i ν ! 2(ν−1) (1− k 2)

 ν1

(1− k )j+1

2 × kuk∞,ρ,[T 0 ,T1 ]

1

−1

j Y

k

k i

2 (–kukk,τ )(1− 2 ) . i ,[Si+1 ,T1 ]

i=0

k i

(T1 − Si+1 ) 2 (1− 2 )

(Si − Si+1 )

GEOMETRIC ANALYSIS

261

We observe that since τi − τi−1 τi −(i+1) ≥2 (1 − θ),

(1 − γi ) =

we have −2(ν−1) ν

(1 − γi )−2 τi

≤ 4(i+1) (1 − θ)−2 ρ

−2(ν−1) ν

.

Also, since Si − Si+1 = 2−(i+1) (T − T0 ), (32.18) becomes

kuk∞,θρ,[T,T1 ] ≤

j Y

 C2

i=0

+

ρ2 2A CSD

−1 CSDν

α 

2 ν

CSD ρ2

 ν−1 ν

−1

+ CSDν (1 − θ)−2 ρ

j i ν  2(ν−1) Y (1− k 2)

−1

ρ (T − T0 )

2

(i+1)ν ν−1

−2(ν−1) ν

i (1− k 2)

i=0 1

j+1 (1− k 2)

× (T1 − T0 ) k kuk∞,ρ,[T0 ,T1 ]

j Y

k 2

k i

(–kukk,τi ,[Si+1 ,T1 ] )(1− 2 ) .

i=0

Letting j → ∞ and using the facts that ∞ Y

2

(i+1)ν ν−1

i (1− k 2)

< ∞,

i=0 (1− k )j+1

2 → 1, kuk∞,ρ,[T 0 ,T1 ]

as

j → ∞,

and k

1

k

2 2 –kukk,τ ≤ ((T1 − S1 ) Vp (τ0 ))− 2 kukk,ρ,[T i ,[Si+1 ,T1 ] 0 ,T1 ] 1   − ρ 2 k T1 − T0 2 ≤ Vp kukk,ρ,[T , 0 ,T1 ] 2 2

we obtain  kuk∞,θρ,[T,T1 ] ≤ C3

ρ2 2A CSD −1

2

α 

CSD ρ2

 ν−1 ν

+ CSDν ρ ν (T − T0 )−1

−1

+ CSDν (1 − θ)−2 ρ

ν  k(ν−1)

1 −k

Vp

ρ 2

−2(ν−1) ν

kukk,ρ,[T0 ,T1 ] ,

proving the desired inequality for k ≤ 1.  To complete the proof of the Harnack inequality, we will need the follow lemma.

262

PETER LI

Lemma 32.2. Let g be a positive measurable function defined on a set D ⊂ M. Suppose {Dσ | 0 < σ ≤ 1} be a family of measurable subsets of D with the properties that, Dσ0 ⊂ Dσ for 0 < σ 0 ≤ σ ≤ 1, and D1 = D. Let us assume that there are constants C4 > 0, γ > 0, 0 < δ < 1, and 0 < k0 < ∞, so that, the inequality ! k1

0

Z g

0 −γ

k0

≤ C4 (σ − σ )

V

−1


1− 1 (D) k k0

Z g

k

 k1

Dσ

Dσ0

is valid for all δ ≤ σ 0 ≤ σ ≤ 1 and 0 < k ≤ k0 . If there exists a constant C5 > 0, such that, m({x ∈ D | ln g > λ}) ≤ C5 V (D) λ−1 for all λ > 0, then there exists a constant C6 > 0 depending only on γ, δ, C4 and C5 , such that, Z g

k0

 k1

0

1 k0

≤ C6 V

(D).

Dδ

Proof. Let us normalize the volume of D to be V (D) = 1 and define the function Z 1! g k0

ψ(σ) = ln

k0

Dσ

for δ ≤ σ < 1. We may assume that ψ(σ) ≥ 2C5 , otherwise the lemma follows because of the monotonicity of ψ and by choosing C6 = exp(2C5 ). We claim that ψ(σ 0 ) ≤

(32.19)

3 ψ(σ) + 8C42 C5 (σ − σ 0 )−2γ 4

for all σ 0 < σ. To see this, we only need to verify for those σ 0 such that ψ(σ 0 ) ≥ 8C42 C5 (σ − σ 0 )−2γ .

(32.20) Let us consider Z (32.21)

g

k

 k1

! k1

Z =

g

k

+

g

Dσ ∩{ln g>ψ(σ)/2}

Dσ

! k1

Z

k

Dσ ∩{ln g≤ψ(σ)/2}

   k1 − k1   0 ψ(σ) ψ(σ) ≤ g m Dσ ∩ ln g > + exp 2 2 Dσ   1 1 ψ(σ) ≤ exp(ψ(σ))(2C5 ψ(σ)−1 ) k − k0 + exp , 2 Z

k0

 k1

0

where we have used the fact that V (Dσ ) ≤ 1. Note that we may choose k ≤ k0 to satisfy 

1 1 − k k0

−1 =

2 ln ψ(σ)



ψ(σ) 2C5

 ,

hence combining with (32.21), we obtain Z g Dσ

k

 k1

 ≤ 2 exp

ψ(σ) 2

 .

GEOMETRIC ANALYSIS

263

On the other hand, the assumption of the lemma asserts that 0

0 −γ

ψ(σ ) ≤ ln (C4 (σ − σ )

)

Z

1 1 k − k0

g

k

 k1 !

Dσ



 1 1 ψ(σ) − ln(2C4 (σ − σ 0 )−γ ) + k k0 2   ψ(σ)  ln(2C4 (σ − σ 0 )−γ )   ≤ + 1 2 ln ψ(σ) ≤

2C5

≤

3ψ(σ) 4

because of (32.20) and the monotonicity of ψ. The claim (32.19) is then validated. For some β > 1, let us now choose a sequence of {σi } with σ0 = δ, and σi+1 = σi + β −i−1 (1 − δ)(β − 1) for all i ≥ 0. Inequality (32.19) asserts that ψ(σi ) ≤

3 ψ(σi+1 ) + 8C42 C5 (1 − δ)−2γ (β − 1)−2γ β 2γ(i+1) 4

for all i ≥ 0. Iterating this inequality yields  j j  i−1 X 3 3 2 −2γ −2γ ψ(σ0 ) ≤ ψ(σj ) + 8C4 C5 (1 − δ) (β − 1) β 2γi . 4 4 i=1 Using the fact that σj = (1 − δ)(β − 1) ψ(σ0 ) ≤

Pj

i=1

8C42 C5 (1

β −i ≤ (1 − δ), and by letting j → ∞, we conclude that −2γ

− δ)

(β − 1)

−2γ

β

2γ

i ∞  X 3β 2γ i=0

By choosing 1 < β and

3β 2γ 4

4

.

< 1, we conclude that the right hand side is finite and the lemma is proved.



Theorem 32.1. Let M be a complete manifold. Suppose that the geodesic ball Bp (ρ) centered at p with radius ρ satisfies Bp (ρ) ∩ ∂M = ∅. Let u ≥ 0 be a function defined on Bp (ρ) × [0, T ], satisfying the equation   ∂ ∆− u = fu ∂t in the weak sense for some function f on Bp (ρ) whose supremum norm is given by A = kf k∞,ρ,[0,T ] . 2 Let us denote ν = m 2 for m > 2, and 1 < ν < ∞ be arbitrary when m = 2. If T ≥ ρ , then for any 0 < θ < 1, there exists a constant C16 > 0 depending only on θ, CSD , CVD , and CP , such that  2 Aρ sup u ≤ C16 exp inf 2 u. 2 2 2 Bp (θρ)×[T − ρ ,T ] B (θρ)×[T − 3ρ ,T − ρ ] p

4

4

2

Proof. Let us first prove the theorem for the case A = 0. We define w = − ln u, then ∆w =

∂w + |∇w|2 . ∂t

264

PETER LI

Let us point out that in the argument that follows, we only need the inequality ∆w ≥

∂w + |∇w|2 , ∂t

and for any compactly supported nonnegative smooth function φ ∈ Cc∞ (Bp (ρ)), we have ∂ ∂t

Z

φ2 w ≤

Z

Bp (ρ)

φ2 ∆w −

Bp (ρ)

Z

φ2 |∇w|2

Bp (ρ)

Z

Z

= −2 1 ≤− 2

φ2 |∇w|2

φ h∇φ, ∇wi − Bp (ρ)

Z

Bp (ρ) 2

Z

2

φ |∇w| + 2 Bp (ρ)

|∇φ|2 .

Bp (ρ)

In particular, choosing φ to be 1 ρ−r φ(r) = (1 − δ)ρ    0    

if

r ≤ δρ

if

δρ ≤ r ≤ ρ

if

ρ ≤ r,

we obtain ∂ ∂t

(32.22)

Z

φ2 w +

Bp (ρ)

1 2

Z

φ2 |∇w|2 ≤ (1 − δ)−2 ρ−2 Vp (ρ).

Bp (ρ)

However, Theorem 31.1 asserts that there exists a constant C7 > 0 depending on CP , CVD , and δ, such that, Z Z |∇w|2 φ2 ≥ C7 ρ−2 |w − w| ¯ 2 φ2 Bp (ρ)

Bp (ρ)

where w ¯ = Vφ−1 (Bp (ρ)) Bp (ρ) w φ2 denotes the average of w over Bp (ρ) with respect to the measure φ2 dV. Together with (32.22), we have R

(32.23) ∂ ∂w ¯ = Vφ−1 (Bp (ρ)) ∂t ∂t

Z

w φ2

Bp (ρ)

≤ −C7 ρ−2 Vφ−1 (Bp (ρ)) ≤ −C7 ρ

−2

Vp−1 (ρ)

Z Bp (ρ)

Z

|w − w| ¯ 2 φ2 + (1 − δ)−2 ρ−2 Vp (ρ) Vφ−1 (Bp (ρ))

|w − w| ¯ 2 + C8 ρ−2 ,

Bp (δρ)

because of Vp (δρ) ≤ Vφ (Bp (ρ)) ≤ Vp (ρ) and the volume doubling property. Let us define v(x, t) = w(x, t) − C8 (t − s0 ) ρ−2 and v¯(t) = w(t) ¯ − C8 (t − s0 ) ρ−2

GEOMETRIC ANALYSIS

for a fixed s0 ∈ [T −

ρ2 2 ,T

−

ρ2 4 ].

We can rewrite (32.23) as ∂¯ v + C7 ρ−2 Vp−1 (ρ) ∂t

(32.24)

265

Z

|v − v¯|2 ≤ 0.

Bp (δρ)

For any λ > 0, we define Ω+ λ (t) = {x ∈ Bp (δρ) | v(x, t) > a + λ} and Ω− λ (t) = {x ∈ Bp (δρ) | v(x, t) < a − λ} where a = v¯(s0 ) = w(s ¯ 0 ). Since

∂ ¯ ∂t v

≤ 0, we have v(x, t) − v¯(t) > a + λ − v¯(t) ≥λ

0 when x ∈ Ω+ λ (t) and t > s . Combining with (32.24), this implies that

∂ v¯(t) + C7 ρ−2 Vp−1 (ρ) (a + λ − v¯(t))2 V (Ω+ λ (t)) ≤ 0 ∂t and −C7−1 ρ2 Vp (ρ)


∂ |a + λ − v¯|−1 ≥ V (Ω+ λ (t)) ∂t

for t > s0 . Integrating over the interval (s0 , T ) yields (32.25)

C7−1 ρ2 Vp (ρ)(λ−1 − |a + λ − v¯(T )|−1 ) Z T ≥ V (Ω+ λ (t) s0

= m{(x, t) ∈ Bp (δρ) × (s0 , T ) | w(x, t) > a + λ + C8 (t − s0 ) ρ−2 }, where m{D} denotes the measure with respect to the product metric of the set D ⊂ M × R. If λ ≥ C8 (T − s0 ) ρ−2 , then (32.25) implies (32.26)

C7−1 ρ2 Vp (ρ) λ−1 ≥ m{(x, t) ∈ Bp (δρ) × (s0 , T ) | w(x, t) > a + λ + C8 (t − s0 ) ρ−2 } ≥ m{(x, t) ∈ Bp (δρ) × (s0 , T ) | w(x, t) > a + 2λ}.

On the other hand, if 0 < λ < C8 (T − s0 ) ρ−2 , then we can estimate m{(x, t) ∈ Bp (δρ) × (s0 , T ) | w(x, t) > a + 2λ} ≤ (T − s0 ) Vp (ρ) ≤ C8 (T − s0 )2 ρ−2 λ−1 Vp (ρ). However, using the fact that T − s0 ≤

ρ2 , 2

together with (32.26), we conclude that (32.27)

m{(x, t) ∈ Bp (δρ) × (s0 , T ) | w(x, t) > a + λ} ≤ C9 ρ2 Vp (ρ) λ−1 .

266

PETER LI

When t ≤ s0 , again using (32.24), we have v(x, t) − v¯(t) ≤ a − λ − v¯(t) = −λ on Ω− λ (t), hence

∂ v¯ + C7 ρ−2 Vp−1 (ρ) (a − λ − v¯(t))2 V (Ω− λ (t)) ≤ 0 ∂t

and 2 V (Ω− λ (t)) ≤ −C7 ρ Vp (ρ)

∂ (|a − λ − v¯|−1 ). ∂t

Integrating from T − ρ2 to s0 and we obtain m{(x, t) ∈ Bp (δρ) × (T − ρ2 , s0 ) | w(x, t) < a − λ + C8 (t − s0 ) ρ−2 } Z s0 = V (Ω− λ (t)) dt ≤

T −ρ2 C7 ρ2 Vp (ρ) (λ−1

− |a − λ − v¯(T − ρ2 )|−1

≤ C7 ρ2 Vp (ρ) λ−1 , hence (32.28)

m{(x, t) ∈ Bp (δρ) × (T − ρ2 , s0 ) | w(x, t) < a − λ} ≤ C7 ρ2 Vp (ρ) λ−1 .

We will now apply Lemma 32.2 to the function g = ea u. In particular, (32.28) asserts that m{(x, t) ∈ D | ln g > λ} ≤ C9 ρ2 Vp (ρ) λ−1 ≤ C10 V (D) λ−1 . with D = Bp (δρ) × (T − ρ2 , s0 ), where C10 depends on C9 , CVD , and δ. Let us define the subdomains Dσ ⊂ D by Dσ = Bp (σδρ) × (T − σρ2 ), s0 ). Obviously, Dσ0 ⊂ Dσ for 41 < σ 0 ≤ σ ≤ 1 with D1 = D. Applying Lemma 32.1 to the domain Dσ0 and Dσ we have (32.29) ν

2

1 −k

kgk∞,σ0 δρ,[T −σ0 ρ2 ,s0 ] ≤ C11 ((σ − σ 0 )δρ)−2 + (σ − σ 0 )−1 ρ−2 ) k(ν−1) (σδρ) k(ν−1) Vp 2ν 0 − k(ν−1)

≤ C11 (σ − σ ) where C11 > 0 depends on k, ν = property implies that

2µ−1 µ ,

(δρ)

2 −k

(σδρ) kgkk,σδρ,[T −σρ2 ,s0 ]

−1 Vp k (σδρ) kgkk,σδρ,[T −σρ2 ,s0 ] ,

δ, CSD , CP and CVD . Note that for σ ≥ 14 , the volume doubling V (D) = ρ2 Vp (δρ) 2 ≤ CVD ρ2 Vp (σδρ)

GEOMETRIC ANALYSIS

267

for k0 ≥ k, (32.29) asserts that ! k1

0

Z g

k0

 ≤ C12

(σ − σ 0 )

− (k

2k0 ν 0 −k)(ν−1)

V (D)−1

 k1 − k1 0 Z

gk

 k1 ,

Dσ

Dσ0

where C12 > 0 depends on k, ν, δ, CSD , CP and CVD . Lemma 32.1 now asserts that  k1

 Z —

(32.30)

0

g

k0 

≤ C13 ,

D3 4

where C13 > 0 depends on k, ν, δ, CSD , and CVD . On the other hand, applying (32.29) again, we have   2 − 1 3δρ kgkk , 3δρ ,[T − 3ρ2 ,s0 ] , kgk∞, δρ ,[T − ρ2 ,s0 ] ≤ C14 (δρ)− k0 Vp k0 0 4 2 4 4 4 hence combining with (32.30) yield kuk∞, δρ ,[T − ρ2 ,s0 ] ≤ C15 e−a .

(32.31)

2

4

where C15 > 0 depends on ν, δ, CSD , CP and CV . Similarly, using (32.27), we apply Lemma 32.2 to the function g = e−a u−1 . As above, we using Lemma 32.1 to verify ! k1

0

Z g

−k0

 ≤ C12

2k ν

0 0 − (k0 −k)(ν−1)

(σ − σ )

V (D)

−1

Dσ0

 k1 − k1 0 Z g

−k

 k1 ,

Dσ

hence the same argument implies that ku−1 k∞, δρ ,[s0 + ρ2 ,T ] ≤ C15 ea , 2

4

which can be rewritten as (32.32)

e−a ≤ C15

inf

2

u.

ρ 0 Bp ( δρ 2 )×[s + 4 ,T ]

The Harnack inequality follows by combining this with (32.31) and choosing θ = When f is not identically 0, we observe that   ∂ ∆− (e−At u) = f e−At u + Ae−At u ∂t ≥0

δ 2

and s0 = T −

and   ∂ ∆− (e−At u−1 ) = f e−At u−1 − e−At |∇u|2 u−2 + Ae−At u−1 ∂t ≤ 0.

ρ2 2 .

268

PETER LI

Hence we can apply (32.31) and (32.32) to the functions e−At u and e−At u−1 and obtain e−AT kuk∞, δρ ,[T − ρ2 ,T ] ≤ C15 e−a 2

and e−a ≤ C15 e−A(T −

2

ρ2 2

)

inf

u,

2

ρ Bp ( δρ 2 )×[T − 2 ,T ]

respectively. In particular, we conclude that sup 2

2 u ≤ C15 e

Aρ2 2

inf

2

ρ Bp ( δρ 2 )×[T − 2 ,T ]

ρ Bp ( δρ 2 )×[T − 2 ,T ]

u. 

The following corollary follows directly from Theorem 32.1. The case when M is quasi-isometric to Rm follows readily from the Nash-Moser theory. The general case when M is quasi-isometric to a complete with nonnegative Ricci curvature was first proved by Saloff-Coste [SC] and Grigor’yan [G]. Corollary 32.1. Let M be a manifold of dimension m. Suppose that ds2 is a complete metric on M such that there is a point p ∈ M and for all ρ, the quantities CP , CSD , and CVD are all bounded and positive 2 independent of ρ. Then for any metric ds (not necessarily continuous) on M which is uniformly equivalent 2 to ds2 , there does not exist any nonconstant positive harmonic functions for the Laplacian with respect to ds . In particular, any manifold which is quasi-isometric to a complete manifold with nonnegative Ricci curvature does not admit a nonconstant positive harmonic function. Proof. To see this, we first observe the uniform boundedness of CP , CSD , and CVD are a quasi-isometric invariant. Since a harmonic function is a stationary solution to the heat equation, Theorem 32.1 implies that any positive harmonic function u defined on M must satisfies Harnack’s inequality sup u ≤ C infρ u

Bp ( ρ 4)

Bp ( 4 )

for any ρ > 0. On the other hand, since u is positive, by translation, we may assume that inf M u = 0. Hence, by taking ρ → ∞, we conclude that sup u ≤ C inf u = 0. M

Therefore, u must be identically 0.

M



The next corollary yields a H¨ older regularity result for weak solutions. One important point of this is that 2 the metric ds need not be continuous as long as it is uniformly bounded with respect to some background metric ds2 . 2

Corollary 32.2. Let M be a complete manifold with metric ds2 . Suppose ds is uniformly equivalent to ds2 . Suppose u is a weak solution to the equation   ¯ − ∂ u = fu ∆ ∂t 2

defined on Bp (1) × [T − 1, T ] with respect to the metric ds . Assume that f is bounded on Bp (1) × [T − 1, T ], then u must be H¨ older continuous at the point (p, T0 ) for all T − 1 < T0 ≤ T. Proof. Let ρ be a constant such that 0 < ρ ≤ T0 − T + 1. We define s(ρ) = supBp (ρ)×[T0 −ρ2 ,T0 ] u and i(ρ) = inf Bp (ρ)×[T0 −ρ2 ,T0 ] u. Applying Theorem 32.1 to the functions s(ρ) − u and u − i(ρ), we have   ρ  s(ρ) − inf 2 u ≤ C s(ρ) − s 3ρ ρ2 2 Bp ( ρ 2 )×[T0 − 4 ,T0 − 2 ]

GEOMETRIC ANALYSIS

and u − i(ρ) ≤ C

sup 2

2

3ρ ρ Bp ( ρ 2 )×[T0 − 4 ,T0 − 2 ]

269

 ρ  i − i(ρ) . 2

Adding the two inequalities yield  osc(ρ) + osc Bp

   ρ  3ρ2 ρ2 × [T0 − , T0 − ] ≤ C osc(ρ) − osc 2 4 2 2

ρ

where osc(ρ) = s(ρ) − i(ρ) denotes the oscillation of u on Bp (ρ) × [T0 − ρ2 , T0 ]. This implies that osc for γ =

C−1 C

ρ 2

≤ γ osc(ρ)

< 1. Iterating this inequality gives osc(2−k ρ) ≤ γ k osc(ρ).

γ older continuous of This implies that u is H¨ older continuous of exponent − log log 2 in the space variable and H¨ γ exponent − log log 4 in the time variable.



270

PETER LI

Appendix A: Computation of Warped Product Metrics Let M

m

=R×N

m−1

be the product manifold endowed with the warped product metric ds2M = dt2 + f 2 (t) ds2N ,

where ds2N is a given metric on N. Our purpose is to compute the curvature on M with respect to this warped product metric. Let {˜ ω2 , . . . , ω ˜ m } be an orthonormal coframe on N with respect to ds2N . If we define ω1 = dt and 2 ωα = f (t) ω ˜ α for 2 ≤ α ≤ m, then the set {ωi }m i=1 forms an orthonormal coframe of M with respect to dsM . The first structure equations assert that dωi = ωij ∧ ωj where ωij are the connection 1-forms with the property that ωij = −ωji . On the other hand, direct exterior differentiation yields dω1 = 0 and dωα = f 0 ω1 ∧ ω ˜α + f ω ˜ αβ ∧ ω ˜β = −(log f )0 ωα ∧ ω1 + ω ˜ αβ ∧β , where ω ˜ αβ are the connection 1-forms on N and f 0 is the derivative of f with respect to t. Hence we conclude that the connection 1-forms on M are given by ω1α = −ωα1

(A.1)

= (log f )0 ωα and (A.2)

ωαβ = ω ˜ αβ .

The second structural equations also assert that dωij − ωik ∧ ωkj =

1 Rijkl ωl ∧ ωk , 2

where Rijkl is the curvature tensor on M . Exterior differentiating (A.1) yields dω1α = (log f )00 ω1 ∧ ωα + (log f )0 (−(log f )0 ωα ∧ ω1 + ω ˜ αβ ∧ ωβ ) . Hence combining with (A.1) and (A.2), we have
dω1α − ω1β ∧ ωβα = (log f )00 + ((log f )0 )2 ω1 ∧ ωα . Also, exterior differentiating (A.2) gives dωαβ = d˜ ωαβ ,

GEOMETRIC ANALYSIS

271

and dωαβ − ωα1 ∧ ω1β + ωαγ ∧ ωγβ = d˜ ωαβ − ω ˜ αγ ∧ ω ˜ γβ + ((log f )0 )2 ωα ∧ ωβ 1˜ ˜τ ∧ ω ˜ γ + ((log f )0 )2 ωα ∧ ωβ = R αβγτ ω 2 1˜ −2 = R ωτ ∧ ωγ + ((log f )0 )2 ωα ∧ ωβ , αβγτ f 2 ˜ αβγτ is the curvature tensor on N . In particular, the sectional curvature of the two plane section where R spanned by e1 and eα is given by
K(e1 , eα ) = − (log f )00 + ((log f )0 )2 , and the sectional curvature of the two plane section spanned by eα and eβ is given by ˜ α , eβ ) − ((log f )0 )2 . K(eα , eβ ) = f −2 K(e ˜ is the sectional curvature of N. Moreover the curvature tensor is given by where K

R1αjk

and

( Rαβij =

 00 0 2   (log f ) + ((log f ) ) = −(log f )00 − ((log f )0 )2   0

if

j = α, k = 1

if

j = 1, k = α otherwise.

˜ αβγτ + ((log f )0 )2 (δατ δβγ − δαγ δβτ ) f −2 R 0

if

i = γ, j = τ

otherwise.

The Ricci curvature is then given by (A.3)

R1j =

X

R1αjα

α


= −(m − 1) (log f )00 + ((log f )0 )2 δ1j and (A.4)

Rαβ =

X

Rαγβγ + Rα1β1

γ


˜ αβ − (log f )00 + (m − 1)((log f )0 )2 δαβ , = f −2 R ˜ αβ is the Ricci tensor on N. where R

272

PETER LI

Appendix B: Polynomial Growth Harmonic Functions on Euclidean Space In this section, we will determine all polynomial growth harmonic functions in Rm . We will also compute their dimensions. Recall that we denote the space of polynomial growth harmonic functions of order at most d on a complete manifold M by Hd (M ) and its dimension is denoted by hd (M ). Corollary 6.3 asserts that if M has nonnegative Ricci curvature then hd (M ) = 1 for all d < 1. Moreover, if M = Rm , with rectangular coordinates given ∂f is also harmonic in Hd−1 (Rm ) by Corollary 6.3. by {x1 , . . . , xm }, and f ∈ Hd (Rm ), then the function ∂x i ∂f Hence if d < 2 then ∂x must be constant functions for all xi . This implies that f must be a linear function i spanned by the coordinate functions {xi } and the constant functions. Therefore we conclude that hd (Rm ) = m + 1

for 1 ≤ d < 2.

We now claim that for any f ∈ Hd (Rm ), it must be a harmonic polynomial of degree at most d. To see this, we argue by induction and assume that this is true for some d ≥ 2. To prove that this is also valid for ∂f ∈ Hd (Rm ) for all 1 ≤ i ≤ m. d + 1, we consider f ∈ Hd+1 (Rm ). The above observation asserts that ∂x i ∂f Induction hypothesis implies that each ∂x = pi where pi a harmonic polynomial of degree at most d. On i the other hand, Z x1

f (x1 , 0, . . . , 0) =

p1 (t1 , 0, . . . , 0) dt1 + f (0, . . . , 0) 0

implies that f (x1 , 0, . . . , 0) is a polynomial of at most degree d + 1 in x1 . By using the formula Z xk f (x1 , . . . , xk , 0, . . . , 0) = pk (x1 , . . . , xk−1 , tk , 0 . . . , 0) dtk + f (x1 , . . . , xk−1 , 0, . . . , 0) 0

we can argue inductively that f (x1 , . . . xm ) is a polynomial of degree at most d and the claim is proved. To determine hd (Rm ) for each d ∈ Z+ , we let x = x1 and y = (x2 , . . . , xm ). For any f ∈ Hd (Rm ), we can write d X f (x, y) = ad−i (y) xi i=0

where aj are polynomials in the variables y = (x2 , . . . xm ) of degree at most j. Since f is harmonic, by separation of variables, we have 0 = ∆f ∂2 f + ∆y f ∂x2 d d X X = ad−i i(i − 1) xi−2 + ∆y ad−i xi =

i=2

=

d−2 X

i=0

(ad−2−i (i + 2)(i + 1) + ∆y ad−i ) xi + ∆y a1 xd−1 + ∆y a0 xd .

i=0

Note that since a0 and a1 are of degree 0 and 1, respectively, the last two terms vanishes. Hence we conclude that −ad−2−i (i + 2)(i + 1) = ∆y ad−i for all 0 ≤ i ≤ d − 2. This gives an inductive formula for all the coefficients once ad and ad−1 are fixed and arbitrary polynomials in y of degree at most d and d − 1, respectively. Let us denote P d (Rm−1 ) to be the space of all polynomial of degree at most d defined on Rm−1 , and pd (Rm−1 ) = dim P d (Rm−1 ), Clearly (B.1)

hd (Rm ) = pd (Rm−1 ) + pd−1 (Rm−1 ).

GEOMETRIC ANALYSIS

We now claim that d



m

m+d d

p (R ) =

273

 .

Again, to see this we write any polynomial of degree at most d as d X

f (x, y) =

ad−i (y) xi .

i=0

Obviously, each aj ∈ P j (Rm ) and hence pd (Rm ) =

d X

pi (Rm−1 ).

i=0

When m = 1, we see that pd (R1 ) = d + 1 =



d+1 d

 .

Using induction on m we have d

m

p (R ) =

 d  X m+i−1 i

i=0

.

It remains to prove that 

m+d d

 =

 d  X m+i−1 i

i=0

.

One can see this by induction on d. Obviously, the identity is true for d = 1 since 

m+1 1







and

m 0

=m+1

=1

by convention. By induction  d  X m+i−1 i

i=0

 =

m+d−1 d−1



 +

m+d−1 d

(m + d − 1)! (m + d − 1)! + m! (d − 1)! (m − 1)! d! (m + d − 1)! (d + m) = m!   d! m+d = , d =

and the identity is proved. Applying this to (B.1), we conclude that (B.2)

d

m

h (R ) =



m+d−1 d



 +

m+d−2 d−1

 .



274

PETER LI

Note that by Sterling’s formula hd (Rm ) ∼

(B.3)

2 dm−1 (m − 1)!

as d → ∞. For future reference, we also note that (B.4)

d X i=0

hi (Rm ) =

 d  X m+i−1 i=0 d

= h (Rm+1 ) 2 m d . ∼ m!

i

+

 d  X m+i−2 i=0

i−1

GEOMETRIC ANALYSIS

275

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