Qα Analysis on Euclidean Spaces 9783110600285, 9783110601121

Table of contents :
Preface
Contents
1. Groundwork
2 Estimation of John–Nirenberg type
3. Functional Hausdorff content
4. Gauss or Poisson lifting
5. Preduality and decomposition
6. Incompressible Navier–Stokes system
7. Left or right composition
8. Extension or restriction principle
Bibliography
Index

Citation preview

Jie Xiao Qα Analysis on Euclidean Spaces

Advances in Analysis and Geometry

|

Editor in Chief Jie Xiao, Memorial University, Canada Editorial Board Der-Chen Chang, Georgetown University, USA Goong Chen, Texas A&M University, USA Andrea Colesanti, University of Florence, Italy Robert McCann, University of Toronto, Canada De-Qi Zhang, National University of Singapore, Singapore Kehe Zhu, University at Albany, USA

Volume 1

Jie Xiao

Qα Analysis on Euclidean Spaces |

Mathematics Subject Classification 2010 Primary: 42B37; Secondary: 35Q30 Author Prof. Jie Xiao Memorial University Department of Mathematics & Statistics St. John’s, NL A1C 5S7 Canada [email protected]

ISBN 978-3-11-060112-1 e-ISBN (PDF) 978-3-11-060028-5 e-ISBN (EPUB) 978-3-11-060010-0 ISSN 2511-0438 Library of Congress Control Number: 2018966447 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2019 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Preface In their 2000 paper [22], M. Essén, S. Janson, L. Peng and J. Xiao introduced such a new function space in several real variables that if α ∈ ℝ = (−∞, ∞) then Qα (ℝn ) comprises all Lebesgue measurable functions f : ℝn = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ℝ × ⋅⋅⋅ × ℝ → ℝ 2≤n copies

with the bounded double-integral-mean of a fractional difference quotient sup

(r,x)∈ℝ+ ×ℝn

1 ˆ 2 2 |f (y) − f (z)| − ( ) dydz) < ∞, n −1 y − r −1 z| 2 +α |r B(x,r) B(x,r)

ˆ ( −

where: ℝ+ = (0, ∞); { { { { n n { { {B(x, r) = {y ∈ ℝ : |y − x| < r} & B(o, r) = {y ∈ ℝ : |y| < r} = B(0, r); { dz, dy, dx = the Lebesgue measures associated with z, y, x ∈ ℝn ; { { ˆ { { { { − = the integral mean under Lebesgue measure on ℝn . { Needless to say, Qα (ℝn ) module of real constants, denoted by K(ℝn ), is a Banach space, in particular, ▷ if α ∈ (−∞, 0) = ℝ− then Qα (ℝn ) coincides with John–Nirenberg’s BMO(ℝn ) (cf. [38]) of all Lebesgue measurable functions f : ℝn → ℝ enjoying ˆ − |f (y) − fB(x,r) | dy < ∞, sup sup |f − fB(x,r) |B(x,r) = (r,x)∈ℝ+ ×ℝn

▷ ▷

(r,x)∈ℝ+ ×ℝn B(x,r)

where fB(x,r) stands for the integral mean of f on B(x, r) with respect to the Lebesgue measure on ℝn . If α ∈ [1, ∞), then Qα (ℝn ) is identified with K(ℝn ). [0, 1) ∋ α 󳨃→ Qα (ℝn ) is a decreasing map, that is, 0 ≤ α1 < α2 < 1 ⇒ BMO(ℝn ) ⊋ Qα1 (ℝn ) ⊋ Qα2 (ℝn ) ⊋ K(ℝn ).

The above facts actually induced a list of five open problems in the final section of [22], the initially-BMO-oriented study of Qα (ℝn ). Since 2000, those problems have received a lot of attention. Not only have the problems been essentially resolved, but also they have been applied to an investigation of the evolutionary incompressible Navier–Stokes system and its stationary formulation. This has driven me to write a relatively self-contained book documenting almost all major results obtained in the past years. Below is an outline of the book. https://doi.org/10.1515/9783110600285-201

VI | Preface

▷

▷

▷ ▷

▷

▷

▷

▷

Chapter 1 introduces the reader to the basic building blocks of Qα (ℝn )—definition and invariance, monotonicity and relationship with Besov spaces, mean oscillation via successive dyadic partitions and wavelet decomposition. Chapter 2 uses the dyadic divisions of a cube to show a John–Nirenberg inequality with precise constants for BMO(ℝn ) and a John–Nirenberg inequality for Qα (ℝn ) with coarse constants. Chapter 3 gives a short account of the functional Hausdorff content and its Choquet integration, Carleson measure and tent space. Chapter 4 focuses on lifting Qα (ℝn ) to ℝn+1 + via the Gauss integral or the Poisson integral, characterizing the divergence of (Qα (ℝn ))n and linking Qα (ℝn ) up with the square Campanato space C2,2α (ℝn ). n Chapter 5 looks at the predual of either Qα (ℝn ) or Q−1 α (ℝ ) via the Hardy–Hausdorff space or the Hardy–Hausdorff–Sobolev space, and decomposes Qα (ℝn ) through the Fefferman–Stein type based on L∞ (ℝn ) ∩ Qα (ℝn ). n Chapter 6 applies Qα (ℝn ) and Q−1 α (ℝ ), as well as their relatives: conformal Morrey space M2,2 (ℝn ), BMO(ℝn ) and conformal Besov space Ḃ 1∞,∞ (ℝn ), to studying the mild solution of the evolutionary incompressible Navier–Stokes system in ℝn+1 + and the zero solution of the stationary incompressible Navier–Stokes system in ℝn . Chapter 7 deals with the left or right composition problem for Qα (ℝn ), thereby finding a relation between Qα (ℝn )∗ and Ap (ℝn ) and resolving [22, problem 8.4] via the local/global self-similar Minkowski dimension and the A1 -class living on a degeneracy set, and leading to an investigation of the well-posedness of the transport equation in Qα (ℝn ). Chapter 8 settles the extension or restriction problem for Qα (ℝn ) posed in [22, problem 8.5] through the Hardy–Littlewood maximal operator which is actually bounded on Qα (ℝn ).

Below are two fundamental conventions which will be frequently used in this book: ▷ X ≈ Y represents X ≲ Y ≲ X; namely, there is a constant κ > 0 such that κ−1 X ≤ Y ≤ κX; ▷ 1S stands for the characteristic function of a subset S ⊆ ℝn . It is my hope that this book will be useful for researchers, graduate students and even senior undergraduate students working in functional-harmonic-potential analysis and partial differential equations as well as the closely related fields. Last but not least, it is a great pleasure to thank my coauthors who communicated directly or indirectly with me in this endeavor. This work was supported in part by the Natural Sciences and Engineering Research Council of Canada (#20171864). Fall 2017–Summer 2018, St. John’s, Canada

Jie Xiao

Contents Preface | V 1 1.1 1.2 1.3 1.4 1.5

Groundwork | 1 Definition and invariance | 1 Monotonicity and relationship with Besov space | 3 Mean oscillation via successive dyadic partition | 10 Wavelet decomposition | 16 Notes | 22

2 2.1 2.2 2.3

Estimation of John–Nirenberg type | 25 J-N-BMO inequality with special constant | 25 J-N-Qα inequality with general constant | 31 Notes | 38

3 3.1 3.2 3.3

Functional Hausdorff content | 39 Hausdorff capacity and Choquet integration | 39 Carleson measure and tent space | 47 Notes | 59

4 4.1 4.2 4.3 4.4

Gauss or Poisson lifting | 61 Extension via either Gauss or Poisson integral | 61 Divergence of Qn | 66 Connection with square Campanato space | 70 Notes | 82

5 5.1 5.2 5.3 5.4

Preduality and decomposition | 87 Hardy–Hausdorff space as a predual | 87 Hardy–Hausdorff–Sobolev space as a predual | 96 Fefferman–Stein type decomposition | 99 Notes | 102

6 6.1 6.2

Incompressible Navier–Stokes system | 105 Evolutionary N-S system in (Q−1 )n | 105 n Evolutionary N-S system in (M2,2 )n or (Ḃ −1 ∞,∞ ) | 115

6.3 6.4 7 7.1

2n

Stationary N-S system in (BMO−1 ∩ C2 ‖f ‖Ḃαp,q (ℝn ) = { 1 ´ p −α p n as q = ∞. {supy∈ℝ |y| ( ℝn |f (x + y) − f (x)| dx)

Clearly, if

𝒯r0 ,x0 (x) = r0 x + x0 ,

then α− pn

‖f ∘ 𝒯r0 ,x0 ‖Ḃαp,q (ℝn ) = r0

‖f ‖Ḃαp,q (ℝn ) .

Moreover, if p = q, then there are two constants cp,n,0 and cp,n,1 such that (cf. [62, 8]) 1

lim α p ‖f ‖Ḃαp,p (ℝn ) = cp,n,0 ‖f ‖Lp (ℝn )

α→0

&

1

lim(1 − α) p ‖f ‖Ḃαp,p (ℝn ) = cp,n,1 ‖∇f ‖Lp (ℝn )

α→1

n hold for any f in the class C∞ 0 (ℝ ) of all infinitely differentiable functions

g : ℝn → ℝ with compact support supp(g) ⊆ ℝn . Of course, ‖ ⋅ ‖Lp (ℝn ) stands for the Lebesgue p-norm on ℝn ; ∇ is the standard gradient in spacial variable x ∈ ℝn .

8 | 1 Groundwork Theorem 1.4. Let (α, n − 1) ∈ [0, 1) × ℕ. Then: (i) n ̇β { {B nβ ,2 (ℝ ) as 0 < α = β < 1 ≤ q ≤ 2 ̇Bβn (ℝn ) ⊊ ⊊ Qα (ℝn ); { ,q {Ḃ βn (ℝn ) as 0 < α < β < 1 ≤ q < ∞ β { β ,∞ (ii)

Ẇ 1,n (ℝn ) ⊆ CIS(ℝn ) ⊊ Qα (ℝn ) n q

→ 1 in (i), where a weakly differentiable function f : ℝn → ℝ lies in the conformal Sobolev energy space Ẇ 1,n (ℝn ) or the conformal invariant Sobolev space CIS(ℝn ) if and only if exists as the limit β =

1 ˆ n { n { { ‖f ‖Ẇ1,n (ℝn ) = ( |∇f (x)| dx) < ∞ or { { { { ℝn 1 ˆ { { 2 { { 2 2−n { n < ∞. |∇f (x)| dx) ‖f ‖ = sup (ℓ(I) { CIS(ℝ ) { I∈C (ℝn ) { I

Proof. (i) Note that (cf. [69, Chapter 3. Theorems 3–4]) β

n ̇ { {B nβ ,2 (ℝ ) as 0 < β = α < 1 ≤ q ≤ 2 B n ,q (ℝ ) ⊊ { β {Ḃ n (ℝn ) as 0 < α < β < 1 ≤ q < ∞. β { β ,∞

̇β

n

So, it is enough to verify β

n ̇ { {B nβ ,2 (ℝ ) as 0 < β = α < 1 Qα (ℝ ) ⊋ { β {Ḃ n (ℝn ) as 0 < α < β < 1. { β ,∞ n

To get the inclusion, we consider the following two cases. ▷ If 0 < β = α < 1, then Theorem 1.2 and Hölder’s inequality with exponents n give n−2α ˆ

‖f ‖2Qα (ℝn ) ≈ sup ℓ(I)2α−n I∈C (ℝn )

≲ sup ℓ(I)n−2α I∈C (ℝn )

≲

‖f ‖2Ḃαn (ℝn ) . α ,2

|y| −2−1 n, then −1 x, y ∈ J ∈ Dk (I) ⇒ 2k ≤ ℓ(I)|x − y|−1 ∞ ≲ ℓ(I)|x − y|

where |x − y|∞ = |(x1 − y1 , . . . , xn − yn )|∞ = max |xj − yj |, 1≤j≤n

and hence KI,α (x, y) ≤

∑

2k ≤ℓ(I)|x−y|−1 ∞

2(2α+n)k ℓ(I)−2n ≲ ℓ(I)2α−n |x − y|−2α−n .

Consequently, Ψf ,α (I) ≲ Θf ,α (I). ▷

If α ≤ −2−1 n, then

and hence

Ψ (I) ≈ Φ (I) { { { f ,α ˆ ˆ f −2n { |f (x) − f (y)|2 dxdy ≲ Θf ,α (I), {ℓ(I) { I I { Ψf ,α (I) ≲ Θf ,α (I).

Accordingly, sup Ψf ,α (I) ≲ sup Θf ,α (I) = ‖f ‖2Qα (ℝn ) .

I∈C (ℝn )

I∈C (ℝn )

(iii)⇒(ii) This is trivial. (ii)⇒(i) This follows from the following two-fold treatment. On the one hand, it is evident to get the inequality ˆ ∞ Ψf ,α (I) ≲ ∑ 2(2α−n)k ∑ ℓ(J)−n |f (x) − cJ |2 dx ∀ I ∈ C (ℝn ). k=0

J∈Dk (I)

J

On the other hand, we are about to prove the following inequality ˆ Ψf ,α (I + z) dy ∀ I ∈ C (ℝn ), Θf ,α (I) ≲ Ψf ,α (I) + ℓ(I)−n |z|∞ k, there exist at most c2(j−k) (for an absolute constant c > 0) cubes J ∈ Dj (I) with (j, k, l) ∈ Λ2 (J), each contributing 2−j(n−2α+2−ϵ)−k(ϵ−n−2) ℓ(I)−n |a(j, k, l)|2 to the second double sum mentioned above. Putting them together yields at most 22αk |a(j, k, l)|2 ℓ(I)−n . Consequently, ∞

Ψf ,α (I) ≲ ∑ ∑

∑

J∈ℰ(I) k=0 I(j,k,l)∈Dk (J)

22αk ℓ(I)−n |a(j, k, l)|2 + 1.

We have proved that Ψf ,α (I) ≲ 1 holds for any cube I of dyadic edge-length. Since the same estimate applies to any translate I + z, it follows that ℓ(I)2α−n

ˆ ˆ I

I

|f (x) − f (y)|2 dxdy ≲ sup Ψf ,α (I + z) |x − y|n+2α |z|∞ t}) ∀ I ∈ C (ℝn ), and define τ(t) = sup

I∈C (ℝn )

mf ,I (t) m(I)

.

For t≥e

& I ∈ C (ℝn ),

let F (I, t) be the family of all elements J ∈ D (I) with ˆ − |f (x) − fI | dx > t. J

Then I ∉ F (I, t) due to

ˆ − |f (x) − fI | dx ≤ ‖f ‖BMO(ℝn ) = 1 < e ≤ t. I

Next, we introduce the dyadic maximal function ˆ ℳI,d (|f − fI |)(x) = sup{ − |f (y) − fI | dy : x ∈ J ∈ D (I)}, J

thereby finding {x ∈ I : ℳI,d (|f − fI |)(x) > t} =

⋃

J∈F (I,t)

Since t < |f (x) − fI | ≤ ℳI,d (|f − fI |)(x), it follows that {x ∈ I : |f (x) − fI | > t} ⊆

⋃

J∈F (I,t)

J.

J.

2.1 J-N-BMO inequality with special constant | 27

Note that if J ∈ F (I, t), then it is contained in a cube J ∗ of twice the side-length, which is not in F (I, t). So ˆ − |f (x) − fI | dx ≤ t. J∗

This in turn implies ˆ ˆ m(J ∗ ) t < − |f (y) − fI | dy| ≤ − |f (y) − fI | dy| ≤ 2n t m(J) J

and

∀ J ∈ F (I, t),

J∗

ˆ |f (x) − fI | ≤ |f (x) − fJ | + − |f (y) − fI | dy ≤ |f (x) − fJ | + 2n t

∀ J ∈ F (I, t).

J

Consequently, if |f (x) − fI | > s > 2n t, then |f (x) − fJ | > s − 2n t

∀ J ∈ F (I, t),

and hence mf ,I (s) = m({x ∈ I : |f (x) − fI | > s} ∩ {x ∈ I : |f (x) − fI | > t}) ≤ ≤

∑

J∈F (I,t)

m({x ∈ I : |f (x) − fI | > s} ∩ J)

∑ (

J∈F (I,t)

m({x ∈ J : |f (x) − fJ | > s − 2n t}) )m(J) m(J)

≤ τ(s − 2n t) n

∑

J∈F (I,t)

≤ τ(s − 2 t)t

−1

m(J) ˆ

∑

J∈F (I,t)

≤ τ(s − 2n t)t −1 m(I).

|f (y) − fI | dy

J

In short, we have τ(s) ≤ t −1 τ(s − 2n t) as s > 2n t. In what follows, let us choose t = e. Note that τ(s) ≤ 1

∀ s > 0.

28 | 2 Estimation of John–Nirenberg type So the last inequality on τ gives τ(s) ≤ exp(1 −

s ) e2n

∀ s ∈ (0, e2n ].

Consequently, if s ∈ (e2n , 2e2n ], then τ(s − 2n e) ≤ exp(1 −

s − e2n ) e2n

as s > e2n ,

and hence τ(s) ≤ e−1 τ(s − 2n e) ≤ exp(1 −

s ) e2n

as s ∈ (e2n , 2e2n ].

Since ∞

(0, ∞) = ⋃ (ke2n , (k + 1)e2n ], k=0

if τ(s) ≤ exp(1 −

s ) e2n

as s ∈ (ke2n , (k + 1)e2n ],

then τ(s) ≤ e−1 τ(s − e2n ) ≤ e−1 exp(1 − = exp(1 −

s − e2n ) e2n

s ) e2n

as s ∈ ((k + 1)e2n , (k + 2)e2n ],

and hence mf ,I (s) m(I)

≤ τ(s) ≤ exp(1 −

s ) e2n

as s ∈ (0, ∞).

For the second part, we utilize 0 < c < (e2n )−1 and ˆ I

c|f (x) − fI | c ) dx = ( ) exp( ‖f ‖BMO(ℝn ) ‖f ‖BMO(ℝn )

ˆ∞ 0

exp(

ct )mf ,I (t) dt ‖f ‖BMO(ℝn )

2.1 J-N-BMO inequality with special constant | 29

≤(

ˆ∞

ec ‖f ‖BMO(ℝn )

)m(I)

exp(−

0

t((e2n )−1 − c) ) dt ‖f ‖BMO(ℝn )

ec = ( n −1 )m(I) (e2 ) − c to get the desired inequality. (ii) It follows from et ≥ t

∀ t ≥ 0.

Proposition 2.2. For a Lebesgue measurable function f on ℝn , let 1 ´ ( n |f (x)|p dx) p as 1 ≤ p < ∞ ‖f ‖Lp (ℝn ) = { ℝ n ess supx∈ℝ |f (x)| as p = ∞,

and ´ ℐα f (x) = {

Then: (i)

f (y) ℝn |y−x|n−α

dy

f (x)

as

‖ℐ n f ‖BMO(ℝn ) ≲ ‖f ‖Lp (ℝn ) p

(ii)

p

1− p

as

q q ‖f ‖Lq (ℝn ) ≲ ‖f ‖BMO(ℝ n ) ‖f ‖Lp (ℝn )

for

0 t}) j=1

∞

≤ ∑ m({x ∈ Ij : |f (x) − fIj | > t − |fIj |}) j=1

n

c2 (t − 2 p ‖f ‖BMO(ℝn ) ) ≤ ∑ c1 exp(− )m(Ij ) ‖f ‖BMO(ℝn ) j=1 ∞

2.2 J-N-Qα inequality with general constant | 31 n

c2 (t − 2 p ‖f ‖BMO(ℝn ) ) ‖f ‖Lp (ℝn ) p ≤ c1 exp(− )( ) . ‖f ‖BMO(ℝn ) ‖f ‖BMO(ℝn ) This, along with the standard layer-cake formula, yields n

2 p ‖f ‖BMO(ℝn )

ˆ

‖f ‖qLq (ℝn ) =

0

m({x ∈ ℝn : |f (x)| > t}) dt q ˆ∞

m({x ∈ ℝn : |f (x)| > t}) dt q

+ n

2 p ‖f ‖BMO(ℝn ) n

2 p ‖f ‖BMO(ℝn )

ˆ

t −p ‖f ‖pLp (ℝn ) dt q

≲ 0

ˆ∞

n

+ n

2 p ‖f ‖BMO(ℝn )

‖f ‖Lp (ℝn ) p q c2 (t − 2 p ‖f ‖BMO(ℝn ) ) )( ) dt c1 exp(− ‖f ‖BMO(ℝn ) ‖f ‖BMO(ℝn )

‖f ‖pLp (ℝn ) . ≲ ‖f ‖q−p BMO(ℝn )

2.2 J-N-Qα inequality with general constant Although Theorem 2.1(i) is globally applicable for Qα (ℝn ) ⊆ BMO(ℝn ), it is not the exact form for Qα (ℝn ). In [22, Problem 8.1] it is conjectured that if f ∈ Qα (ℝn ) { { { {α ∈ (0, 1) { { {t ∈ ℝ+ , then there is a constant c > 0 such that mf ,J (t)

∞

sup ∑ 2(2α−n)k

I∈C (ℝn )

k=0

∑

J∈Dk (I)

m(J)

≲ t −1 exp(−

ct ). ‖f ‖Qα (ℝn )

But, according to Theorem 1.5, if there is a constant c > 0 such that ∞

sup ∑ 2(2α−n)k

I∈C (ℝn )

then f ∈ Qα (ℝn ).

k=0

∑

J∈Dk (I)

mf ,J (t) m(J)

≲ t −1 exp(−ct),

32 | 2 Estimation of John–Nirenberg type Theorem 2.3. Let (α, n − 1) ∈ [0, 1) × ℕ. (i) If f ∈ Qα (ℝn ), then there is a constant c > 0 such that mf ,J (t)

∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

∑

J∈Dk (I)

m(J)

≲

1 + (t −1 ‖f ‖Qα (ℝn ) )2 exp(ct‖f ‖−1 ) Q (ℝn )

∀ t > 0;

α

(ii) Conversely, if for p ∈ [0, 2) there is a constants c > 0 such that mf ,J (t)

∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

∑

J∈Dk (I)

≲

m(J)

1 + t −p exp(ct)

∀ t > 0,

then f ∈ Qα (ℝn ). Proof. (i) If 0 = ‖f ‖Qα (ℝn ) , then the result is trivial, and hence in the sequel it is always assumed that 0 < ‖f ‖Qα (ℝn ) < ∞. Two cases are considered for 1 2

⦀f ⦀Qα (ℝn ) = sup (Ψf ,α (I)) ≈ ‖f ‖Qα (ℝn ) . I∈C (ℝn )

Case t ≤ ⦀f ⦀Qα (ℝn ) . This ensures 1 ≲ exp(−ct‖f ‖−1 Qα (ℝn ) )

∀ c > 0.

Upon applying ˆ mf ,J (t) ≤ t −2 − |f (x) − fJ |2 dx

∀ t > 0,

J

we get ∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

∑

J∈Dk (I)

mf ,J (t) m(J)

≤ sup Ψf ,α (I) I∈C (ℝn )

≲ t −2 ‖f ‖2Qα (ℝn ) ≲

1 + (t −1 ‖f ‖Qα (ℝn ) )2 exp(ct‖f ‖−1 ) Q (ℝn ) α

.

2.2 J-N-Qα inequality with general constant | 33

Case t > ⦀f ⦀Qα (ℝn ) . Given I ∈ C (ℝn ) { { { k−1∈ℕ { { { {r ∈ (0, ∞), we apply the classical Calderón–Zygmund decomposition theorem to |f (x) − fJ | on each J ∈ Dk (I), thereby getting J = Ωf ,J (r) ∪ (J \ Ωf ,J (r)) such that: ▷

∞

Ωf ,J (r) = ⋃ Jl , l=1

namely, the union of cubes Jl ∈ D (I) whose interiors are disjoint; ▷

|f (x) − fJ | ≤ r

for a. e. x ∈ J \ Ωf ,J (r);

ˆ r < − |f (x) − fJ | dx ≤ 2n r

▷

∀ Jl ;

Jl

▷

rm(U) ≤

ˆ

|f (x) − fJ | dx ≤ 2n rm(U) for any union U of some cubes in {Jl }.

U

Note that ‖f ‖BMO(ℝn )

ˆ = sup − |f (x) − fI | dy ≤ ⦀f ⦀Qα (ℝn ) . I∈C (ℝn )

I

So for each cube K in the decomposition of Ωf ,J (r) there is K ∗ ⊆ J such that ∗

ℓ(K ) = 2ℓ(K) &

|f

K∗

ˆ − fJ | ≤ − |f (x) − fJ | dx ≤ r, K∗

and thus

ˆ r ≤ − |f (x) − fJ | dx K

34 | 2 Estimation of John–Nirenberg type ˆ ≤ − |f (x) − fK ∗ | dx + |fK ∗ − fJ | K

ˆ ≤ 2n − |f (x) − fK ∗ | dx + r n

K∗

≤ 2 ⦀f ⦀Qα (ℝn ) + r. Meanwhile, an application of the Cauchy–Schwarz inequality gives that if U is the union of some cubes in {Jl } then ˆ r 2 m(U) ≤ |f (x) − fJ |2 dx U

and hence r 2 m(Ωf ,J (r)) ≤

ˆ

|f (x) − fJ |2 dx.

Ωf ,J (r)

Although r1 < r2 ⇒ Ωf ,J (r2 ) ⊆ Ωf ,J (r1 ), we want to establish the following inequality m(Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) )) ≤ 2−n m(Ωf ,J (r)). To do so, taking K ∈ {Jl } we have ˆ − |f (x) − fJ | dx ≤ 2n ⦀f ⦀Qα (ℝn ) + r < 2n+1 ⦀f ⦀Qα (ℝn ) + r, K

thereby finding that K does not belong to the family of cubes making Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ). If K ∩ Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ) ≠ 0, then this set must be the union of some cubes from the decomposition of Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ), and hence n+1

r+2

⦀f ⦀Qα

(ℝn )

≤

ˆ − K∩Ωf ,J (r+2n+1 ⦀f ⦀Qα (ℝn ) )

|f (x) − fJ | dx

2.2 J-N-Qα inequality with general constant | 35

ˆ −

≤ K∩Ωf ,J

≤(

(r+2n+1

|f (x) − fK | dx + |fK − fJ | ⦀f ⦀Qα (ℝn ) )

m(K) )⦀f ⦀Qα (ℝn ) + 2n ⦀f ⦀Qα (ℝn ) + r. m(K ∩ Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ))

Consequently, we read off m(K ∩ Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) )) ≤ 2−n m(K). This, along with a summation of those K and the fact that Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ) = Ωf ,J (r) ∩ Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ), derives the desired inequality. Consequently, the following inequality ∞

∑2

(2α−n)k

∑

m(Ωf ,J (r + 2n+1 ⦀f ⦀Qα (ℝn ) ) m(J)

J∈Dk (I)

k=0

∞

≤ 2−n ∑ 2(2α−n)k k=0

∑

m(Ωf ,J (r)) m(J)

J∈Dk (I)

is valid. Note that mf ,J (t) ≤ m(Ωf ,J (t)). So upon letting j be the integer part of t − ⦀f ⦀Qα (ℝn )

2n+1 ⦀f ⦀Qα (ℝn ) and setting s = (1 + j2n+1 )⦀f ⦀Qα (ℝn ) , we find ⦀f ⦀Qα (ℝn ) ≤ s ≤ t, thereby estimating ∞

∑ 2(2α−n)k

k=0

∑

J∈Dk (I)

mf ,J (t) m(J)

∞

≤ ∑ 2(2α−n)k k=0

∑

J∈Dk (I)

∞

≤ ∑ 2(2α−n)k k=0

∑

J∈Dk (I)

mf ,J (s) m(J)

m(Ωf ,J (s)) m(J)

36 | 2 Estimation of John–Nirenberg type m(Ωf ,J (1 + (j − 1)2n+1 ⦀f ⦀Qα (ℝn ) ))

∞

≤ 2−n ∑ 2(2α−n)k k=0

∑

m(J)

J∈Dk (I)

.

An iteration with j times derives ∞

∑ 2(2α−n)k

k=0

∑

J∈Dk (I)

mf ,J (t) m(J)

m(Ωf ,J (⦀f ⦀Qα (ℝn ) ))

∞

≤ 2−nj ∑ 2(2α−n)k k=0

≤2 ≤2

∑

m(J)

J∈Dk (I)

−jn −n( −

≤ (2

t−⦀f ⦀ Qα (ℝn ) 2n+1 ⦀f ⦀ Qα (ℝn )

nt 2n+1 ⦀f ⦀ Qα (ℝn )

−1) n

)2 2n +1 +n .

Putting c = 2−n−1 n ln 2 produces mf ,J (t)

∞

∑ 2(2α−n)k

k=0

∑

m(J)

J∈Dk (I)

≲ exp(−ct⦀f ⦀−1 Qα (ℝn ) ),

and a positive constant c∗ depending on c such that mf ,J (t)

∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

∑

m(J)

J∈Dk (I)

≲

1 + (t −1 ‖f ‖Qα (ℝn ) )2

) exp(c∗ t‖f ‖−1 Q (ℝn )

.

α

(ii) Suppose that there is a constant c > 0 obeying ∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

∑

mf ,J (t) m(J)

J∈Dk (I)

≲

1 + t −p exp(ct)

for p ∈ [0, 2).

Then ∞

(2α−n)k

sup ∑ 2

I∈C (ℝn ) k=0

−1

∑ m(J)

J∈Dk (I)

ˆ∞

tmf ,J (t)dt ≲

0

ˆ∞ 0

t(1 + t −p ) dt < ∞, exp(ct)

and hence f ∈ Qα (ℝn ) follows from Theorem 1.5. Not only Proposition 2.2(i) under p = n can be slightly improved, but also Proposition 2.2(ii) under q → p = 2 can be understood under Qα (ℝn )-structure. Proposition 2.4. Let (α, β, γ, n − 1) ∈ (−∞, 1) × (−∞, 1) × (−∞, 1) × ℕ. Then: (i) ‖ℐ1 f ‖Qα (ℝn ) ≲ ‖f ‖Ln (ℝn ) ;

2.2 J-N-Qα inequality with general constant | 37

(ii)

for

‖f ‖Qα (ℝn ) ≲ ‖f ‖Qβ (ℝn ) ‖f ‖Qγ (ℝn )

β+γ . 2

α=

Proof. (i) Note that if ℛj (f )(x) =

Γ( n+1 ) 2 π

n+1 2

p. v.

ˆ

ℝn

(xj − yj )f (y) |x − y|n+1

dy

is the j-th Riesz transform of f with Γ(⋅) being the gamma function then (Rj (f ))nj=1 exists as the gradient of the 1-order Riesz singular integral ˆ ℐ1 f (x) = f (y)|y − x|1−n dy, ℝn

up to a dimensional constant, namely (ℛ1 (f ), . . . , ℛn (f )) =

Γ( n+1 ) 2

(1 − n)π

n+1 2

∇ℐ1 f .

According to Theorem 1.4(ii), we have ‖f ‖Qα (ℝn ) ≲ ‖∇f ‖Ln (ℝn ) thereby utilizing the boundedness of the Riesz transformation (ℛj , . . . , ℛn ) on Ln (ℝn ) to obtain

n

‖ℐ1 f ‖Qα (ℝn ) ≲ ‖∇ℐ1 f ‖Ln (ℝn ) ≲ ∑ ‖ℛj (f )‖Ln (ℝn ) ≲ ‖f ‖Ln (ℝn ) , j=1

namely, n

n

n

ℐ1 L (ℝ ) ⊆ Qα (ℝ ).

(ii) Since α = 2−1 (β + γ), an application of the Cauchy–Schwarz inequality gives that ˆ ˆ I

I

1 1 ˆ ˆ ˆ ˆ 2 2 |f (y) − f (z)|2 |f (y) − f (z)|2 |f (y) − f (z)|2 dydz) ( dydz ≤ ( dydz) |y − z|n+2α |y − z|n+2γ |y − z|n+2β

I

I

holds for any I ∈ C (ℝn ) so the desired inequality follows.

I

I

38 | 2 Estimation of John–Nirenberg type

2.3 Notes Below are some historical remarks on Chapter 2. Section 2.1. Theorem 2.1 comes from [53] and Proposition 2.2 comes from a combination of two notes [77, 20]. Theorem 2.1 indeed induces the so-called BMO(γ, ℝn ) (cf. [82]) consisting of all locally integrable functions f on ℝn with 1 ˆ p p sup ( − |f (x) − fI | dx) < ∞,

− γ1

sup p

p∈[1,∞)

I∈C (ℝn )

I

equivalently, ∃ a constant c > 0 such that ˆ sup − (exp |c|f (x) − fI |γ ) dx < ∞. I∈C (ℝn )

I

Since γ ∈ (1, ∞) ⇒ BMO(γ, ℝn ) ⊆ BMO(ℝn ), an interesting problem is to investigate the relationship between Qα (ℝn ) and BMO(γ, ℝn ) under (α, γ) ∈ [0, 1) × (1, ∞). Section 2.2. Theorem 2.3 comes from a modification of the first and third sections of [103] and Proposition 2.4 comes from another look at [91, Theorem 4.1]. Moreover, [103, Theorem 3] shows that under α ∈ (0, 2−1 ): ▷ if p ∈ [0, 2), then there exists no constant c > 0 such that mf ,J (t)

∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

▷

∑

m(J)

J∈Dk (I)

≲

1 + (t −1 ‖f ‖Qα (ℝn ) )p exp(ct‖f ‖−1 ) Q (ℝn )

∀t > 0

α

holds for any function f ∈ Qα (ℝn ); there is a function f ∉ Qα (ℝn ) such that ∞

sup ∑ 2(2α−n)k

I∈C (ℝn ) k=0

∑

J∈Dk (I)

holds for some positive constant c.

mf ,J (t) m(J)

≲

1 + t −2 exp(ct)

∀t > 0

3 Functional Hausdorff content In order to solve [22, problems 8.2–8.3] via an appropriate lift of f ∈ Qα (ℝn ) to ℝn+1 + , we are led to consider a brief theory of functional Hausdorff content.

3.1 Hausdorff capacity and Choquet integration Due to the fractional phenomenon driven by Theorem 1.4, it is very natural to bring the Hausdorff capacity and the associated Choquet integration into play. Definition 3.1. Let d ∈ (0, n] and E ⊆ ℝn . (i) The d-dimensional Hausdorff capacity (or content) of E is defined by ∞

Λ(∞) (E) = inf{∑ rjd : E ⊆ ⋃ B(xj , rj )}, d j=1

j

where the infimum is taken over all coverings of E by countable families of balls with radii rj . (ii) The d-dimensional Hausdorff measure of E is given by Λd (E) = lim Λ(ϵ) (E), d ϵ→0

where Λ(ϵ) (E) is defined by (i) with the additional requirement that the radii of the d covering balls satisfy 0 < rj ≤ ϵ. (iii) The d-dimensional dyadic Hausdorff capacity of E is determined by ∞

∘

(E) = inf{∑ ℓ(Ij )d : E ⊆ (⋃ Ij ) }, Λ̃ (∞) d j

j=1

where the infimum ranges only over covers of E by {Ij } ⊂ D (ℝn ) with the interior ∞

∘

(⋃ I j ) j=1

∞

of ⋃ Ij . j=1

(iv) The Choquet integral of f : ℝn → [0, ∞] with respect to Λ(∞) is decided by d ˆ ℝn

f

dΛ(∞) d

ˆ∞ = 0

({x ∈ ℝn : f (x) > t}) dt. Λ(∞) d

} and the Cho, Λ̃ (∞) Proposition 3.2. Below is a list of basic properties on the pair {Λ(∞) d d n quet integral of an arbitrary nonnegative function f on ℝ under d ∈ (0, n]. https://doi.org/10.1515/9783110600285-003

40 | 3 Functional Hausdorff content (i) (ii)

Both capacities are nonnegative, monotone, countably subadditive set functions on all subsets of ℝn . Λ(∞) is finite on any bounded subset of ℝn , while having the same null sets as Λd , d that is, Λ(∞) (E) = 0 ⇔ Λd (E) = 0. d

(iii) There is 6−n Λ̃ (∞) (E) ≤ Λ(∞) (E) ≤ nn Λ̃ (∞) (E) ∀ E ⊆ ℝn . d d d (iv) Λ̃ (∞) (⋅) satisfies: d (iv)-a the outer regularity—if E ⊆ ℝn then Λ̃ (∞) (E) = inf{Λ̃ (∞) (O) : open O ⊇ E}; d d (iv)-b the strong subadditivity—if E1 , E2 are two subsets of ℝn then Λ̃ (∞) (E1 ∪ E2 ) + Λ̃ (∞) (E1 ∩ E2 ) ≤ Λ̃ (∞) (E1 ) + Λ̃ (∞) (E2 ); d d d d (iv)-c the upward continuity—if {Ei } is an increasing sequence of subsets of ℝn then Λ̃ (∞) (⋃ Ei ) = lim Λ̃ (∞) (Ei ); d d i→∞

i

(iv)-d the downward continuity—if {Ei } is a decreasing sequence of compact subsets of ℝn then Λ̃ (∞) (⋂ Ei ) = lim Λ̃ (∞) (Ei ). d d i→∞

i

(v)

ˆ ℝn

(vi)

holds for any E ⊆ ℝn .

ˆ ℝn

(vii)

(∞)

1E dΛd

= Λ(∞) (E) d

cf dΛ(∞) =c d

ˆ ℝn

f dΛ(∞) d

holds for any constant c. ˆ ℝn

f dΛ(∞) =0⇔f =0 d

(E) = 0. -a. e. on ℝn , that is, f = 0 on ℝn holds except on E with Λ(∞) holds Λ(∞) d d

3.1 Hausdorff capacity and Choquet integration

(viii)

ˆ

f

d n

ℝn

(ix)

dΛ(∞) n

| 41

n ˆ d (∞) ≤ ( f dΛd )

ℝn

holds for any constant d ∈ (0, n]. ˆ ∞ ∞ˆ (∞) ̃ ∑ fj dΛd ≤ ∑ fj dΛ̃ (∞) d ℝn

j=1

j=1

ℝn

holds for any fj ≥ 0 on ℝn . Proof. The first seven properties are well known; see, for example, [1]. To prove (viii), we utilize the inequality n

(∞) d Λ(∞) n (E) ≤ (Λd (E))

∀ E ⊆ ℝn

to produce ˆ ℝn

f

n d

dΛ(∞) n

ˆ∞ n−d n d dt (Λ(∞) n ({x ∈ ℝ : f (x) > t}))t

n = d

0

ˆ∞ ≤ 0

n

ˆt d d d n n ds) dt ( (Λ(∞) ({x ∈ ℝ : f (x) > s})) n dt 0

n

ˆ∞ d n . ≤ ( Λ(∞) ({x ∈ ℝ : f (x) > t}) dt) d 0

Of course, (ix) follows from (iv)-b. Theorem 3.3. For a locally integrable function f on ℝn , denoted by f ∈ L1loc (ℝn ), and x ∈ ℝn , let ˆ ˆ ℳf (x) = sup − |f (y)| dy & ℳd f (x) = sup − |f (y)| dy x∈I∈C (ℝn )

x∈I∈D(ℝn )

I

I

be the Hardy–Littlewood maximal function of f and its dyadic version. If d ∈ (0, n], then: (i) Λ(∞) ({x ∈ ℝn : ℳd f (x) > t}) ≤ Λ(∞) ({x ∈ ℝn : ℳf (x) > t}) d d ≤ 3d Λ(∞) ({x ∈ ℝn : ℳd f (x) > 4−n t}) ∀ t > 0; d

(ii)

ˆ ℝn

´

≳{ |f |p dΛ(∞) d

(∞) p ℝn (ℳf ) dΛd d ({y supt∈(0,∞) t n Λ(∞) d

as p ∈ ( dn , ∞)

∈ ℝn : ℳf (y) > t}) as p = dn .

42 | 3 Functional Hausdorff content Proof. (i) This follows from [27, p. 136] and ∀ x ∈ ℝn .

ℳd f (x) ≤ ℳf (x)

(ii) Two cases are handled below. Case p ∈ ( dn , ∞). If xI is the center of I ∈ C (ℝn ), then 1ℐ (x) ≲ min{1, (

and hence an application of ˆ

d n

n

ℓ(I) ) } ∀ x ∈ ℝn , |x − xI |

< p gives p

(ℳ1I ) ℝn

dΛ(∞) d

ˆ1

d

t

≲ ℓ(I) (1 +

d − pn

dt).

0

Next, for each integer k let {Ij,k } ⊆ D (ℝn ) be a nonoverlapping family such that {x ∈ ℝn : 2k < f (x) ≤ 2k+1 } ⊆ ⋃ Ij,k { { { j { d (∞) n k k+1 { {∑ ℓ(Ij,k ) ≤ 2Λd ({x ∈ ℝ : 2 < f (x) ≤ 2 }). { j Then Jk = ⋃ Ij,k j

▷

& g = ∑ 2(k+1)p 1Jk ⇒ f p ≤ g. k

In case of p ≥ 1, we have (ℳf )p ≤ ℳg ≤ ∑ 2(k+1)p ∑ ℳ1Ij,k , j

k

thereby getting via the above maximal estimate and Propositions 3.2(iii) & (ix), ˆ ˆ ≲ (ℳf )p dΛ(∞) ∑ 2(k+1)p ∑ ℳ1Ij,k dΛ(∞) d d ℝn

ℝn

j

k

≲ ∑ 2(k+1)p ∑ j

k

(k+1)p

≲ ∑2 k

≲

ˆ

ℝn

∑ ℓ(Ij,k )d j

({x ∑ 2(k+1)p Λ(∞) d k

ˆ ≲

ℝn

(∞)

ℳ1Ij,k dΛd

. f p dΛ(∞) d

∈ ℝn : 2k < f (x) ≤ 2k+1 })

3.1 Hausdorff capacity and Choquet integration

▷

In case of

d n

| 43

< p < 1, we use f ≤ ∑ 2k+1 1Jk k

to get (ℳf )p ≤ ∑ 2(k+1)p ∑(ℳ1Ij ,k )p , j

k

whence

ˆ ℝn

(ℳf )p dΛ(∞) ≲ ∑ 2(k+1)p ∑ ℓ(Ij,k )d ≲ d j

k

ˆ ℝn

f p dΛ(∞) . d

Case p = dn . We make the following consideration. ▷ Observe first that if {Ij } ⊆ C (ℝn ) is a sequence of nonoverlapping cubes then there is a subsequence {Ijk } obeying ∑ ℓ(Ijk )d ≤ 2ℓ(I)d ∀ I ∈ D (ℝn ) { { { {Ijk ⊆I { { (∞) d { {Λd (⋃ Ij ) ≤ 2 ∑ ℓ(Ijk ) . j k { In fact, if j1 = 1

&

j1 , . . . , jk

are selected for the first inequality, then jk+1 is defined as the first index such that {Ij1 , . . . , Ijk+1 } obeys the first inequality. Upon taking j ∈ (jl , jl+1 ) for some l ∈ ℕ, we can find Ij∗ ∈ D (ℝn ) such that Ij∗ ⊇ Ij

&

∑

d

Ijk ⊆Il∗ & k≤l

ℓ(Ijk )d + ℓ(Ij )d > 2ℓ(Ij∗ ) ,

whence ∑

Ijk ⊆Il∗ & k≤l

d

ℓ(Ijk )d ≥ ℓ(Ij∗ ) .

44 | 3 Functional Hausdorff content Without loss of generality, we may assume ∑ ℓ(Ijk )d < ∞, k

thereby seeing that is bounded. Upon considering {Ik̃ } as a sequence of maximal cubes taken from {Ij∗ }, we get {ℓ(Ij∗ )}

⋃ Ij ⊆ (⋃ Ijk ) ∪ (⋃ Ik̃ ), j

k

k

thereby reaching the second inequality Λ(∞) (⋃ Ij ) ≤ 2 ∑ ℓ(Ijk )d . d j

▷

k

Next, for t > 0 let {Ij } ⊆ D (ℝn ) be the sequence of maximal cubes with ˆ − |f (x)| dx > t. Ij

Then {x ∈ ℝn : ℳd f (x) > t} = ⋃ Ij . j

Now, an application of Proposition 3.2(viii) derives d

ℓ(Ij ) ≤ (t

−1

ˆ

d n

f (x) dx) ≲ t

− dn

Ij

ˆ Ij

d

. f n dΛ(∞) d

By (i) and the first observation above, we obtain a subsequence {Ijk } of {Ij } such that ({x ∈ ℝn : ℳd f (x) > t}) ({x ∈ ℝn : ℳf (x) > 4n t}) ≤ 3d Λ(∞) Λ(∞) d d ≤ 31+d ∑ ℓ(Ijk )d k

≲t ≲t

− dn

k

− dn

ˆ

∑ ˆ ℝn

Ij k

d

|f | n dΛ(∞) d d

|f | n dΛ(∞) d

holds, thereby reaching the desired conclusion in (ii).

3.1 Hausdorff capacity and Choquet integration |

45

Below is the Adams preduality. Theorem 3.4. For d ∈ (0, n], let L1,d (ℝn ) be Morrey space of all Radon measures μ on ℝn (denoted by μ ∈ RM(ℝn )) with ⦀μ⦀L1,d (ℝn ) = sup

I∈C (ℝn )

|μ|(I) < ∞, ℓ(I)d

where |μ| = μ+ + μ−

& μ± = max{0, ±μ}.

(i) If L1 (Λ(∞) ) is the completion of C0 (ℝn ) of all continuous functions with compact supd n port in ℝ with respect to ˆ , |f | dΛ(∞) d ℝn

´ then L1,d (ℝn ) is isomorphic to the dual of L1 (Λ(∞) ) under the pairing ℝn f dμ. d (ii) If f is nonnegative and lower semicontinuous on ℝn , then ˆ ˆ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 f dμ󵄨󵄨󵄨 : 0 ≤ μ ∈ L1,d (ℝn ) & ‖μ‖ 1,d n ≤ 1}. f dΛ(∞) ≈ sup{ L (ℝ ) 󵄨󵄨 󵄨󵄨 d 󵄨n 󵄨 n ℝ

ℝ

Proof. (i) Clearly, μ ∈ L1,d (ℝn ) induces a bounded linear functional on L1 (Λ(∞) ) via d ˆ ℒ(f ) = f dμ ℝn

thanks to |μ|(E) ≤ ⦀μ⦀L1,d (ℝn ) Λ(∞) (E) ∀ E ⊆ ℝn d and

ˆ |ℒ(f )| ≤

|f | d|μ| ≤ ⦀μ⦀L1,d (ℝn )

ℝn

ˆ ℝn

|f | dΛ(∞) d

∀ f ∈ L1 (Λ(∞) ). d

Conversely, suppose that ℒ is a linear functional on L1 (Λ(∞) ) with its norm ‖ℒ‖ < d ∞, then there exists a μ ∈ RM(ℝn ) such that ℒ is determined by ˆ ℒ(f ) = ). f dμ ∀ f ∈ C0 (ℝn ) ⊆ L1 (Λ(∞) d ℝn

46 | 3 Functional Hausdorff content However, for any g ∈ C0 (ℝn ) we have 󵄨󵄨ˆ 󵄨󵄨 ˆ 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 g d|μ|󵄨󵄨󵄨 ≤ |g| d|μ| 󵄨n 󵄨 ℝ ℝn ˆ ≤ sup{ f dμ : f ∈ C0 (ℝn ) & |f | ≤ |g|} ℝn

ˆ ≤ ‖ℒ‖ sup{ |f | dΛ(∞) : f ∈ C0 (ℝn ) & |f | ≤ |g|} d ˆ ≤ ‖ℒ‖ ℝn

ℝn

|g| dΛ(∞) . d

For (I, ϵ) ∈ C (ℝn ) × ℝ+ let Iϵ ∈ C (ℝn ) have the same center as I’s and the edge-length ϵ + ℓ(I). Upon choosing a function g ∈ C0 (ℝn )

1 with g = { 0

on I

on ℝn \ Iϵ ,

we achieve |μ|(I) ≤ ‖ℒ‖Λ(∞) (Iϵ ), d whence ⦀μ⦀L1,d (ℝn ) ≤ ‖ℒ‖. ) to its second dual has the fol(ii) Note that the canonical mapping from L1 (Λ(∞) d 1 (∞) lowing norm of g ∈ L (Λd ) ˆ

ℝn

󵄨󵄨ˆ 󵄨󵄨 󵄨 󵄨 (∞) ̃ |g| dΛd = sup{󵄨󵄨󵄨 g dμ󵄨󵄨󵄨 : 0 ≤ μ ∈ L1,d (ℝn ) & ⦀μ⦀L1,d (ℝn ) ≤ 1}. 󵄨󵄨 󵄨󵄨 n ℝ

So, if f is nonnegative and lower semicontinuous on ℝn , then it can be approximated from below by a nonnegative and increasing sequence {fj } ⊆ C0 (ℝn ), and hence

ˆ ℝn

󵄨󵄨ˆ 󵄨󵄨 󵄨󵄨 󵄨 1,d n ≤ sup{ fj dΛ̃ (∞) 󵄨󵄨 f dμ󵄨󵄨󵄨 : 0 ≤ μ ∈ L (ℝ ) & ‖μ‖L1,d (ℝn ) ≤ 1}. d 󵄨󵄨 󵄨󵄨 n ℝ

3.2 Carleson measure and tent space

| 47

Proposition 3.2(iv) is used to derive ({x ∈ ℝn : fj (x) ≥ t}) = Λ̃ (∞) ({x ∈ ℝn : f (x) ≥ t}) ∀ t > 0, lim Λ̃ (∞) d d

j→∞

whence ˆ ℝn

󵄨󵄨 󵄨 f dΛ̃ (∞) ≈ sup{󵄨󵄨󵄨 d 󵄨󵄨

ˆ

ℝn

󵄨󵄨 󵄨 f dμ󵄨󵄨󵄨 : 0 ≤ μ ∈ L1,d (ℝn ) & ‖μ‖L1,d (ℝn ) ≤ 1}. 󵄨󵄨

3.2 Carleson measure and tent space A focus on the upper half-space ℝn+1 + leads to the concepts of a Carleson measure and a nontangential maximal function. Definition 3.5. Let p > 0. n+1 (i) A Radon measure μ on ℝn+1 + (denoted by μ ∈ RM(ℝ+ )) is a p-Carleson measure provided ⦀μ⦀CMp = sup

|μ|(S(I)) < ∞, ℓ(I)pn

where |μ| = μ+ + μ−

& μ± = max{0, ±μ}

and the supremum is taken over all Carleson boxes S(I) = {(x, t) : x ∈ I, t ∈ (0, ℓ(I))} based on I ∈ C (ℝn ). (ii) The nontangential maximal function 𝒩 (f ) of a measurable function f on ℝn+1 + is defined by 𝒩 (f )(x) =

sup |f (y, t)|,

(y,t)∈Λ(x)

where Λ(x) = {(y, t) ∈ ℝn+1 + : |y − x| < t} is the cone at x ∈ ℝn .

48 | 3 Functional Hausdorff content Theorem 3.6. Let d ∈ (0, n] { { { { { {d < p < ∞ { { { n { { { |y|2 { − n2 { { exp(− G (y) = (4πt) ) { t 4t { n+1 { n + 1 − n+1 2 2 − 2 { 2 t(t + |y| ) { P (y) = Γ( )π { t { 2 { { { ˆ∞ { { { s−1 −t { { { {Γ(s) = t e ds 0 {

∀ (y, t) ∈ ℝn+1 + ∀ (y, t) ∈ ℝn+1 + ∀s > 0

and μ be a Radon measure on ℝn+1 + . (i) The following five statements are equivalent: (i)-a ⦀μ⦀CM d < ∞; n

(i)-b

¨

sup

(x,s)∈ℝn+1 + ℝn+1 +

(i)-c

¨

(t(|x − y|2 + (t − s)2 )

|f (y, t)| d|μ|(y, t) ≲

− n+1 2

ˆ ℝn

ℝn+1 +

holds for all measurable functions f on ℝn+1 + ; (i)-d ¨ ˆ |f (y, t)|p d|μ|(y, t) ≲

ℝn

ℝn+1 +

d

) n d|μ|(y, t) < ∞;

(∞)

𝒩 (f ) dΛd

|f0 |p dΛ(∞) d

holds for all measurable functions f (y, t) = f0 ∗ Gt 2 (y) or (i)-e

p

sup s |μ|({x ∈ s>0

f (y, t) = f0 ∗ Pt (y) with

ℝn+1 +

: |f (y, t)| > s}) ≲

ˆ ℝn

f0 ∈ L1loc (ℝn );

|f0 |p dΛ(∞) d

holds for all measurable functions f (y, t) = f0 ∗ Gt 2 (y) or

f (y, t) = f0 ∗ Pt (y) with

f0 ∈ L1loc (ℝn ).

| 49

3.2 Carleson measure and tent space

(ii) If CM d (ℝn+1 + ) is the class of all n

μ ∈ RM(ℝn+1 + )

with ⦀μ⦀CM d < ∞ n

and LN1 (Λ(∞) ) is the completion of all continuous functions f on ℝn+1 + with compact d support (denoted by f ∈ C0 (ℝn+1 )) with respect to + ˆ ‖f ‖LN1 (Λ(∞) ) = d

ℝn

(∞)

𝒩 (f ) dΛd ,

1 (∞) then CM d (ℝn+1 + ) is isomorphic to the dual of LN (Λd ) under the pairing n

˜

ℝn+1 +

f dμ.

Proof. (i) This will be verified according to the following four equivalences. n (i)-a⇔(i)-b. If (i)-a holds, then fixing (x, s) ∈ ℝn+1 + , letting I ∈ C (ℝ ) have the center x and ℓ(I) = s and noticing 1

{(|y − x|2 + |t − s|2 ) 2 ≥ s 1 { 2 2 2 m {(|y − x| + |t − s| ) ≈ 2 s

as (y, t) ∈ S(2I)

as (y, t) ∈ S(2m I),

we obtain (i)-b through ¨

− n+1 2

(t(|x − y|2 + (t − s)2 )

d

) n d|μ|(y, t) = (

ˆ

∞

m=1 m+1 S(2 I)\S(2m I)

S(2I)

ℝn+1 +

≲

ˆ

+∑

)(⋅ ⋅ ⋅)

|μ|(S(2I)) ∞ |μ|(S(2m+1 I)) +∑ md(1+n−1 ) sd sd m=1 2

≲ ⦀μ⦀CM d . n

n

Conversely, if (i)-b is valid, then for I ∈ C (ℝ ) taking (x, s) to be the center of S(I) and utilizing (y, t) ∈ S(I) ⇒ |y − x|2 + |t − s|2 ≲ ℓ(I)2 , we achieve

¨

2

n+1 2 − 2

(t(|x − y| + (t − s) )

d n

) d|μ|(y, t) ≥

ˆ (⋅ ⋅ ⋅) ≳ S(I)

ℝn+1 +

|μ|(S(I)) , ℓ(I)d

so that (i)-a follows up. (i)-a⇔(i)-c. If (i)-a is valid, then for t > 0 and a measurable function f letting Ot (f ) = {x ∈ ℝn : 𝒩 (f )(x) > t} &

Λ(∞) (Ot (f )) < ∞, d

50 | 3 Functional Hausdorff content we cover Ot (f ) by {Ij } ⊆ D (ℝn ) with ∑ ℓ(Ij )d < ∞, j

thereby being able to choose {Jk } ⊆ D (ℝn ) such that (cf. [17, Lemma 4.1]) {⋃ Ij = ⋃ Jk { { { j k { { { d { ℓ(J ) ∑ { k ≤ ∑ ℓ(Ij ) { j k { { ∗ n+1 { { {(y, t) ∈ ℝ { + : |f (y, t)| > t} ⊆ T(Ot (f )) ⊆ ⋃ T(Jk ) { { { k { { ∗ {Jk = 5√nJk , where T(O) = {(y, t) ∈ ℝn+1 + : B(y, t) ⊆ O} is the tent based on O. Accordingly, d ∗ |μ|({(y, t) ∈ ℝn+1 + : |f (y, t)| > t}) ≤ ∑ |μ|(T(Jk )) ≲ ⦀μ⦀CM d ∑ ℓ(Ij ) . n

k

j

This in turn implies (∞) |μ|({(y, t) ∈ ℝn+1 + : |f (y, t)| > t}) ≲ ⦀μ⦀CM d Λd (Ot (f )) n

and (i)-c. Conversely, if (i)-c holds, then for I ∈ C (ℝn ) letting ϕ(y, t) = 1T(I) and noticing (y, t) ∈ T(I) ∩ Λ(x) ⇔ x ∈ B(y, t) ⊆ I, we obtain 𝒩 (ϕ) = 1I ,

thereby getting (i)-a through ˆ∞ |μ|(T(I)) ≲ 0

(Ot (ϕ)) dt ≲ ℓ(I)d . Λ(∞) d

3.2 Carleson measure and tent space

| 51

(i)-a⇔(i)-d. It is enough to do this for the Gauss kernel Gt (y) since the treatment for the Poisson kernel Pt (y) is completely similar. Suppose that (i)-a holds. Note that sup |f0 ∗ Gt (y)| ≲ ℳf0 (x) ∀ x ∈ ℝn .

(y,t)∈Λ(x)

So, an application of Theorem 3.3(ii) gives (i)-d. Conversely, if (i)-d is true, then f0 = 1B of an open ball B = B(x, t) ⊆ ℝn yields that f0 ∗ Gt 2 (x) ≳

ˆ

t −n exp(−

B(x,t)

|x − y|2 ) dy ≳ 1 4t 2

and so that d

(B) ≲ m(B) n |ν|(T(B)) ≲ Λ(∞) d

holds, thereby reaching (i)-a. (i)-a⇔(i)-e. This is similar to the argument for (i)-a⇔(i)-d. (ii) The argument for (i)-a⇒(i)-b ensures that any μ ∈ CM d (ℝn+1 + ) n

induces a bounded linear functional of LN1 (Λ(∞) ) d

via

¨

f dμ.

ℝn+1 +

Conversely, if ℒ is a bounded linear functional of LN1 (Λ(∞) ) with its norm ‖ℒ‖, then an d application of 1 (∞) C0 (ℝn+1 + ) ⊆ LN (Λd )

and the Riesz representation theorem derives μ ∈ RM(ℝn+1 + ) such that ¨ { ℒ (f ) = f dμ; { { { ℝn+1 { + { { {|ℒ(f )| ≤ ‖ℒ‖‖f ‖ LN1 (Λ(∞) ) d {

∀ f ∈ C0 (ℝn+1 + ).

But then for any g ∈ C0 (ℝn+1 + ) we have ˆ 󵄨󵄨 ¨ 󵄨󵄨 󵄨󵄨 󵄨 g d|μ|󵄨󵄨󵄨 ≤ ‖ℒ‖ sup{ f dΛ(∞) : f ∈ C0 (ℝn ) & |f | ≤ |g|} ≤ ‖ℒ‖‖g‖LN1 (Λ(∞) ) . 󵄨󵄨 d 󵄨󵄨 󵄨󵄨 d n n+1 ℝ+

ℝ

52 | 3 Functional Hausdorff content Now, for (I, ϵ) ∈ C (ℝn ) × ℝ+ let Iϵ ∈ C (ℝn ) have the same center as I’s and the edge-length ϵ + ℓ(I). Upon taking a function 1 g ∈ C0 (ℝn+1 + ) with g = { 0

on S(I)

on ℝn+1 + \ S(Iϵ ),

we gain &

𝒩 (g) ≲ 1Iϵ

|μ|(S(I)) ≤ ‖ℒ‖Λ(∞) (Iϵ ) d

thereby arriving at ⦀μ⦀CM d ≲ ‖ℒ‖. n

Interestingly and importantly, Theorem 3.4 yields the so-called tent space pair n+1 1 n+1 {T∞ d (ℝ+ ), Td (ℝ+ )}

as described below. Definition 3.7. Let (d, n − 1) ∈ (0, n] × ℕ. n+1 n+1 (i) T∞ d (ℝ+ ) is the class of all measurable functions f on ℝ+ obeying ‖f ‖

T∞ d

= sup ( B⊆ℝn

¨

T(B)

d

1

dydt 2 ) 1+n ) < ∞, |f (y, t)| ( 1 t m(B) n t

2

where the supremum is taken over all open balls B ⊆ ℝn . 1 n+1 (ii) A function a on ℝn+1 + is said to be a Td (ℝ+ )-atom if there exists an open ball B = B(x, r) ⊆ ℝn such that a is supported in the tent T(B) and satisfies ¨

1

T(B)

d

dydt m(B) n ) 1−n ≤ 1. |a(y, t)| ( t t 2

n+1 (iii) T1d (ℝn+1 + ) is the class of all measurable functions f on ℝ+ satisfying

‖f ‖T1 (ℝn+1 = inf( + ) d

ω

¨

ℝn+1 +

1

ω(t, y)−1 dydt 2 ) 1−n ) < ∞, |f (y, t)| ( t td 2

3.2 Carleson measure and tent space

| 53

where the infimum is taken over all nonnegative measurable functions ω on ℝn+1 + with

ˆ ℝn

(∞)

𝒩 (ω) dΛd

≤1

and the restriction that ω is allowed to vanish only where f vanishes. Theorem 3.8. Let (d, n − 1) ∈ (0, n] × ℕ. 1 n+1 (i) f ∈ T1d (ℝn+1 + ) if and only if there is a sequence of Td (ℝ+ )-atoms {aj } and an 1 l -sequence {λj } such that f = ∑ λj aj . j

Moreover, ‖f ‖T1 (ℝn+1 ≈ inf{∑ |λj | : f = ∑ λj aj }, + ) d

j

j

where the infimum is taken over all possible atomic decompositions of f ∈ T1d (ℝn+1 + ). 1 n+1 The right-hand side thus defines a norm on Td (ℝ+ ) which makes it into a Banach space. (ii) The inequality ¨

|f (y, t)g(y, t)|

ℝn+1 +

dy dt ≲ ‖f ‖T1 (ℝn+1 ‖g‖T∞ (ℝn+1 + ) + ) d d t

∞ n+1 holds for all (f , g) ∈ T1d (ℝn+1 + ) × Td (ℝ+ ). ∞ n+1 (iii) The Banach space dual of T1d (ℝn+1 + ) can be identified with Td (ℝ+ ) under the following pairing

⟨f , g⟩ =

¨

f (y, t)g(y, t)

ℝn+1

dydt . t

Proof. (i) Suppose first that a is a T1d (ℝn+1 + )-atom. Then there is a ball B = B(xB , r) ⊆ ℝn such that supp(a) ⊆ T(B) &

¨ T(B)

|a(y, t)|2 (

1

d

dydt m(B) n ) 1−n ≤ 1. t t

54 | 3 Functional Hausdorff content For ϵ > 0 let ω(x, t) = κ r −d min{1, (

d+ϵ

r √|x − xB |2 + t 2

)

},

where κ is a constant to be determined later on. Note that in ℝn+1 + this function is identically equal to κr −d on the upper half-ball of radius r, and decays radially outside the ball. Since the Euclidean distance in ℝn+1 + −1

√ from Λ(x) to (xB , 0) ∈ ℝn+1 + is 2 |x − xB |, the nontangential maximal function of ω enjoys 𝒩 (ω)(x) ≤ κ r

−d

min{1, (

and a suitable choice of κ > 0 ensures ˆ ℝn

(∞)

𝒩 (ω) dΛd

d+ϵ

r √2 ) |x − xB |

},

≤1

via ˆ

κ −1

ℝn

ˆr

−d

(∞)

𝒩 (ω) dΛd

≤ 0

1/(d+ϵ)

Λ(∞) (B(xB , √2(r ϵ λ−1 ) d

)) dλ ≲ 1.

On the one hand, since ω−1 = r d

on T(B),

we have ¨ T(B)

|a(x, t)|2 (

ω(x, t)−1 dxdt ) 1−n = t td

¨ T(B)

d

r dxdt |a(x, t)|2 ( ) 1−n ≲ 1, t t

whence a ∈ T1d (ℝn+1 ≲ 1. + ) & ‖a‖T1 (ℝn+1 + ) d

Upon taking such T1d (ℝn+1 + )-atoms {aj } that ∑j λj aj is convergent with ‖{λj }‖1 = ∑ |λj | < ∞, j

3.2 Carleson measure and tent space

| 55

we gain ∑ λj aj = f ∈ T1d (ℝn+1 + ) & j

≲ ‖{λj }‖1 . ‖f ‖T1 (ℝn+1 + ) d

n+1 On the other hand, let f ∈ T1d (ℝn+1 + ). Choose a measurable function ω ≥ 0 on ℝ+ satisfying

ˆ ℝn

(∞)

𝒩 (ω) dΛd

≤1

&

¨

|f (x, t)|2 (

ℝn+1 +

ω(x, t)−1 dxdt . ) 1−n ≤ 2‖f ‖T1 (ℝn+1 + ) d t td

For each k ∈ ℤ, put Ek = {x ∈ ℝn : 𝒩 (ω)(x) > 2k }. Then there is a sequence {Ij,k } ⊆ D (ℝn ) with disjoint interiors such that (Ek ) & ∑ ℓ(Ij,k )d ≤ 2Λ(∞) d j

T(Ek ) ⊆ ⋃ S∗ (Ij,k ). j

Here, each tent ∗ ) over T(Ij,k

∗ = 5√nIj,k Ij,k

has been replaced by S∗ (Ij,k ) = {(y, t) ∈ ℝn+1 + : y ∈ Ij,k , t < 2 diam(Ij,k )}. Consequently, if Tj,k = S∗ (Ij,k ) \ ⋃ ⋃ S∗ (Il,m ), m>k l

then K

⋃ ⋃ Tj,k = ⋃ S∗ (Ij,−K ) \ ⋃ ⋃ S∗ (Il,m )

k=−K j

j

m>K l

⊇ T(E−K ) \ ⋃ ⋃ S∗ (Il,m ). m>K l

Note that ⋃ T(Ek ) = {(x, t) ∈ ℝn+1 + : ω(x, t) > 0} k

56 | 3 Functional Hausdorff content owing to (x, t) ∈ ̸ T(Ek ) ⇒ ω(x, t) ≤ 2k . Note also that each S∗ (Il,m ) is contained in an (n + 1)-dimensional cube of edge-length 2diam(Il,m ). Using Proposition 3.2(i) and ˆ ℝn

(∞)

𝒩 (ω) dΛd

≤ 1,

we achieve Λ(∞) ( ⋃ ⋃ S∗ (Il,m )) ≲ ∑ ∑ ℓ(Il,m )d ≲ ∑ Λ(∞) (Em ) → 0 d d m>k l

m>k l

as k → ∞,

m>k

thereby getting ⋃ ⋃ Tj,k ⊇ ⋃ T(Ek ) \ ⋂ ⋃ ⋃ S∗ (Il,m ) = {(x, t) ∈ ℝn+1 + : w(x, t) > 0} \ T∞ , k

j

k m>k l

k

where T∞ is a set whose d-dimensional Hausdorff capacity is zero, and consequently (n + 1)-dimensional Lebesgue measure is zero. Note that ω is allowed to vanish only where f vanishes. So we obtain a. e. on ℝn+1 + .

f = ∑ f 1Tj,k Upon setting

¨ − 21 { 2 dxdt ∗ d { { |f (x, t)| 1−n+d ) aj,k = f 1Tj,k (ℓ(Ij,k ) { { { t { { Tj,k 1 ¨ { { 2 { { 2 dxdt ∗ d { |f (x, t)| 1−n+d ) , { {λj,k = (ℓ(Ij,k ) { t T j,k { we have f = ∑ λj,k aj,k j,k

a. e.

&

S∗ (Ij,k ) ⊆ T(Bj,k ),

∗ where Bj,k is the ball with the same center as Ij,k and radius 2−1 ℓ(Ij,k ). Thus aj,k is supported in T(Bj,k ) and is normalized according to Definition 3.7, so each aj,k is a 1 T1d (ℝn+1 + )-atom. It remains to validate that {λj,k } is l -summable. Utilizing

ω ≤ 2k+1

c

on Tj,k ⊆ (⋃ S∗ (Il,k+1 )) ⊆ (T(Ek+1 )) l

c

3.2 Carleson measure and tent space

| 57

and the Cauchy–Schwarz inequality, we get ∗ ) ∑ |λj,k | ≤ ∑ 2(k+1)/2 ℓ(Ij,k

1 ¨ dxdt 2 ( |f (x, t)|2 ω(x, t)−1 1−n+d ) t

d/2

j,k

j,k

Tj,k

≤ (∑ 2 k,j

k+1

∗ d ℓ(Ij,k ) )

¨

1 2

1

dxdt 2 |f (x, t)| ω(x, t) 1−n+d ) t 2

(∑ j,k

Tj,k d

k

≲ ‖f ‖T1 (ℝn+1 (∑ 2 ∑ ℓ(Ij,k ) ) + ) d

≲

−1

1 2

j

k

(Ek )) ‖f ‖T1 (ℝn+1 (∑ 2k Λ(∞) d + ) d k

1 2

1 ˆ 2 (∞) ≲ ‖f ‖T1 (ℝn+1 ( 𝒩 (ω) dΛd ) + ) d

ℝn

≲ ‖f ‖T1 (ℝn+1 , + ) d

thereby reaching ∑ λj,k aj,k = f j,k

under the quasi-norm. n+1 Evidently, T∞ d (ℝ+ ) becomes a Banach space under = inf{∑ |λj | : f = ∑ λj aj }, ‖f ‖T1 (ℝn+1 ≈ ⦀f ⦀T1 (ℝn+1 + ) + ) d

d

j

j

n+1 where the infimum ranges over all possible atomic decompositions of f ∈ T∞ d (ℝ+ ). n+1 (ii) Suppose that ω is a nonnegative measurable function on ℝ+ . Let g ∈ ∞ Td (ℝn+1 + ). Then

dμg,d (x, t) = |g(x, t)|2

dxdt t 1+n−d

is a dn -Carleson measure, and hence ¨ ˆ dxdt ω(x, t)|g(x, t)|2 1+n−d ≲ ‖g‖2T∞ (ℝn+1 ) 𝒩 (ω) dΛ(∞) ≲ ‖g‖2T∞ (ℝn+1 ) . d + + d d t ℝn+1 +

ℝn

n+1 Accordingly, if f ∈ T∞ d (ℝ+ ), then the Cauchy–Schwarz inequality gives ¨ dydt |f (y, t)g(y, t)| t ℝn+1 +

58 | 3 Functional Hausdorff content ¨ ≤( ℝn+1 +

¨ ≲(

1

dydt 2 |f (y, t)| ω(y, t) 1−n+d ) ( t 2

−1

¨

1

dydt 2 |g(y, t)| ω(y, t) 1+n−d ) t 2

ℝn+1 +

|f (y, t)|2 ω(y, t)−1

ℝn+1 +

1 2

dydt ) ‖g‖T∞ (ℝn+1 . + ) d t 1−n+d

n+1 (iii) By part (ii), every g ∈ T∞ d (ℝ+ ) induces a bounded linear functional on 1 n+1 Td (ℝ+ ). So, it suffices to show the converse. Let ℒ be a bounded linear functional on n T1d (ℝn+1 + ). Fix a ball B ⊆ ℝ with radius r > 0. If f is supported in T(B) with

f ∈ L2 (T(B), t −1 dxdt), then

¨

|f (x, t)|2

T(B)

dxdt ≤ r n−d t 1−n+d

¨

|f (x, t)|2

T(B)

dxdt , t

so f is a multiple of a T1d (ℝn+1 + )-atom with ‖f ‖2T1 (ℝn+1 ) ≲ r n ‖f ‖L2 (T(B),t −1 dxdt) . d

+

Hence ℒ induces a bounded linear functional on the square space on T(B) against the weighted Lebesgue measure t −1 dxdt L2 (T(B), t −1 dxdt), and acts via the inner product with some function gB ∈ L2 (T(B), t −1 dxdt). Upon taking Bj = B(o, j) ∀ j ∈ ℕ, we have gBj = gBj+1

on T(Bj ),

so we get a single function g on ℝn+1 + which belongs locally to −1 L2 (ℝn+1 + , t dxdt),

and such that

¨ ℒ(f ) = ℝn+1 +

f (x, t)g(x, t)t −1 dxdt

3.3 Notes | 59

n+1 whenever f ∈ T∞ d (ℝ+ ) with support in some finite tent T(B). By the atomic decompon+1 sition in part (i), the subspace of such f is dense in T∞ d (ℝ+ ). Thus if we can prove n+1 g ∈ T∞ ≲ ‖ℒ‖, d (ℝ+ ) with ‖g‖T∞ (ℝn+1 + ) d

then by taking limits we can get the representation of ℒ for any n+1 f ∈ T∞ d (ℝ+ ).

To see this, fix a ball B ⊆ ℝn with radius r > 0, and for every ϵ > 0 let fϵ (x, t) = t d−n g(x, t) 1T ϵ (B) (x, t) { ϵ T (B) = T(B) ∩ {(x, t) : t > ϵ}. Then

¨ T(B)

dxdt |fϵ (x, t)| 1−n+d = t 2

¨

|g(x, t)|2

T ϵ (B)

dxdt < ∞. t 1+n−d

Accordingly, fϵ is a multiple of a T1d (ℝn+1 + )-atom with ¨ dxdt |g(x, t)|2 1+n−d ‖fϵ ‖2T1 (ℝn+1 ) ≲ r d + d t T ϵ (B)

and the constant being independent of ϵ. However, fϵ is used to derive ¨ dxdt |g(x, t)|2 1+n−d = |ℒ(fϵ )| t T ϵ (B)

≤ ‖ℒ‖‖fϵ ‖T1 (ℝn+1 + ) d

≲ ‖ℒ‖(r

d

¨ T ϵ (B)

1

dxdt 2 |g(x, t)| 1+n−d ) . t 2

Since this estimation is true for all ϵ > 0 with a constant independent of ϵ, we find the same inequality for the integral over T(B), and since that does not rely on the choice of B, we have actually demonstrated n+1 g ∈ T∞ ≲ ‖ℒ‖. d (ℝ+ ) & ‖g‖T∞ (ℝn+1 + ) d

3.3 Notes Below are some historical remarks on Chapter 3. Section 3.1. Proposition 3.2 is basically taken from [1, 17, 100]. Theorems 3.3–3.4 are formulated from a deep understanding of [1, 67]. Section 3.2. Theorems 3.6 and 3.8 come from an appropriate modification of [22, Lemma 4.1] and [17, Theorems 4.2–4.4–5.4].

4 Gauss or Poisson lifting On the basis of Chapter 3, this chapter focuses on lifting Qα (ℝn ) to ℝn+1 + via either the n Gauss integral or the Poisson integral, characterizing the divergence of (Qα (ℝn )) and n 2,2α n on linking Qα (ℝ ) up with the square Campanato space C (ℝ ).

4.1 Extension via either Gauss or Poisson integral 2

Via either Gauss integral et Δ f (x) or Poisson integral e−t −Δ f (x), we can extend each member of Qα (ℝn ) to the upper half-space ℝn+1 + and examine its behavior. √

Theorem 4.1. For (α, n − 1) ∈ (0, 1) × ℕ { (x, y, t) ∈ ℝn × ℝn × ℝ+ , let f be a measurable function on ℝn with ˆ |x − y|2 t2 Δ 2 − n2 { e f (x) = (4πt ) )) f (y) dy (exp(− { { { 4t 2 { { n { ℝ { { { −t √−Δ n + 1 − n+1 − n+1 e f (x) = Γ( )π 2 t(t 2 + |x − y|2 ) 2 { { 2 { { { { {Δ = 𝜕x21 + ⋅ ⋅ ⋅ + 𝜕x2n { { { {∇ = (𝜕x1 , . . . , 𝜕xn ) or (𝜕t ). Then the following four statements are equivalent for F(x, t) being 2

either et Δ f (x) or e−t (i) f ∈ Qα (ℝn ); (ii) ⦀f ⦀Qα (ℝn ),1 =

sup (r

(x,r)∈ℝn+1 +

(iii) ⦀f ⦀Qα (ℝn ),2 =

sup (

(x,s)∈ℝn+1 +

(iv) ⦀f ⦀Qα (ℝn ),3 =

sup (

(x,s)∈ℝn+1 +

ℝn+1 +

ℝn+1 +

https://doi.org/10.1515/9783110600285-004

ˆr

ˆ (

0

¨

¨

2α−n

√−Δ

f (x) :

2

|∇F(y, t)| dy)

B(x,r)

|∇F(y, t)|2 (s(|x − y|2 + (t − s)2 )−

(G((x, s), (y, t)))

n−2α n

dt

t 2α−1

1 2

) < ∞;

1

n+1 2

)

2α−n n

|∇F(y, t)|

2

dydt 2 ) < ∞; t 2α−1

dydt

t 2α(1−n

−1 )

1 2

) < ∞,

62 | 4 Gauss or Poisson lifting where G((x, s), (y, t)) = |(x, s) − (y, t)|1−n − |(x, s) − (y, −t)|1−n ≥ 0 is the Green function of ℝn+1 + . Proof. Since the argument for the Poisson integral F(x, t) = e−t

√−Δ

f (x)

is completely analogous, it suffices to validate the result for 2

F(x, t) = et Δ f (x). (i)⇔(ii) Suppose that (ii) holds. Then F(x, 0) = f (x) { { { { { { F(x + y, 0) = f (x + y) { { { ˆ { { |y|2 −n { { 𝜕xj F(x, t) = 𝜕yj ((4πt 2 ) 2 exp(− 2 ))(f (x + y) − f (x)) dy { { { 4t { { n { { ˆℝ { |y|2 −n 𝜕t F(x, t) = 𝜕t ((4πt 2 ) 2 exp(− 2 ))(f (x + y) − f (x)) dy { { { 4t { { { ℝn { { 2 󵄨 󵄨 n { 󵄨󵄨 󵄨 { { 󵄨󵄨𝜕y ((4πt 2 )− 2 exp(− |y| ))󵄨󵄨󵄨 ≲ min{t −1−n , |y|−1−n } { { 󵄨󵄨 j 󵄨󵄨 2 { 4t { 󵄨 󵄨 { { { 2 󵄨 󵄨 n { 󵄨󵄨 󵄨󵄨 |y| −2 { 2 {󵄨󵄨𝜕t ((4πt ) exp(− ))󵄨󵄨󵄨 ≲ min{t −1−n , |y|−1−n }. 󵄨󵄨 2 󵄨󵄨 4t {󵄨 Note that the triangle inequality gives 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 y |y| y |y| 󵄨 󵄨 󵄨 󵄨 |f (x + y) − f (x)| ≤ 󵄨󵄨󵄨F(x + , ) − F(x, 0)󵄨󵄨󵄨 + 󵄨󵄨󵄨F(x + y − , ) − F(x + y, 0)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 2 2 = A1 + A2 . According to Theorem 1.2, it suffices to prove Bi = ℓ(I)2α−n

ˆ

|y| 2−1 r { { { −1 {Fi ∩ B(x0 , 2 r) = 0. By Lemma 8.8, we get m(Fi ) ≳ r n

∀ i = 1, 2.

Applying Lemma 8.7, we obtain 1 ≲ CQα (F1 , F2 , D) ≲ (log2 (2b)−1 )

1−n n

& b ≳ 1.

Moreover, if 1 { { { f (x) = {(log2 (2b)−1 )−1 log2 { { {0

∀ x ∈ D2 \ B(x0 , 2−1 r)

∀ x ∈ D2 ∩ (B(x0 , 2−1 r) \ B(x0 , b0 r))

|x−x0 | b0 r

otherwise,

then this function is well-defined and satisfies |∇f (x)| = (log2 (2b)−1 ) |x − x0 |−1 1D2 ∩(B(x0 ,2−1 r)\B(x0 ,b0 r)) , −1

and hence ‖f ‖Qα (ℝn ) ≤ ‖f ‖Ẇ1,n (D)

ˆ

−1 −1

≲ (log2 (2b) ) (

|z − x0 |

−n

dz)

B(x0 ,2−1 r)\B(x0 ,b0 r)

≲ (log2 (2b)−1 ) ▷

1−n n

.

Accordingly, f ∈ ℧(F1 , F2 , D). l. l. c.-b. Let x1 , x2 be an pair of points in D and x1 , x2 ∈ B(z, r) for some

(z, r) ∈ ℝn × (0, ∞).

Then 2r > |x1 − x2 |. We are about to show the existence of a rectifiable curve ζ ⊆ B(z, rb−1 )

1 n

209

210 | 8 Extension or restriction principle that connects x1 and x2 for some constant b < 1. If |x1 − x2 | ≤ max{ dist (x1 , Dc ), dist (x2 , Dc )}, then the line segment joining x1 and x2 gives the desired curve. However, if |x1 − x2 | > max{ dist (x1 , Dc ), dist (x2 , Dc )}, then upon letting Di be the connected component of D ∩ B(xi , 2−3 |x1 − x2 |) = D \ B(xi , 2−3 |x1 − x2 |)

c

containing xi for i = 1, 2, and using Lemma 8.8, we have m(Di ) ≳ |x1 − x2 |n

∀ i = 1, 2.

Obviously, D1 ∩ D2 = 0

& D1 , D2 ⊆ B(x1 , 2|x1 − x2 |).

Upon employing Lemma 8.7, we obtain CQα (D1 , D2 , D) ≳ 1. Next, we claim that there exists a positive constant N0 ≥ 2 independent of {x1 , x2 , D1 , D2 } such that D1 and D2 are in the same component of D ∩ B(x1 , N0 |x1 − x2 |). To see this, suppose that D1 , D2 are not in the same component of D ∩ B(x1 , N|x1 − x2 |) for some N > 2, say, in two different components ̃ D1 and ̃ D2 , respectively. Define 1 { { { log (N|x −x |)−log (|y−x |) 1 f (y) = { 2 1log2 (23 N) 2 2 { { {0

∀ y ∈ B(x1 , 2−3 |x1 − x2 |) ∩ ̃ D1

∀ y ∈ (B(x1 , N|x1 − x2 |) \ B(x1 , 2−3 |x1 − x2 |)) ∩ ̃ D1 otherwise.

Then ‖f ‖Qα (ℝn ) ≲ ‖f ‖Ẇ1,n (D) ≲ (log2 N)

1−n n

.

8.4 Reverse restriction principle

| 211

Upon observing 1 f ={ 0

on D1

on D2 ,

we have CQα (E, F, D) ≲ (log2 N)

1−n n

.

This inequality ensures N ≲ 1 and shows the existence of N0 as claimed. As a consequence, there exists a rectifiable curve ζ0 ⊆ D ∩ B(x1 , N0 |x1 − x2 |) joining D1 and D2 . Let x̃1 , x̃2 ∈ Di ∩ ζ0 . Since xi , x̃i ∈ Di

∀ i = 1, 2

and Di is connected, we can find a rectifiable curve ζi ⊆ Di ⊆ D ∩ B(x1 , N0 |x1 − x2 |) connecting xi and x̃i for i = 1, 2. Now D ∩ B(x1 , N1 |x1 − x2 |) ⊇ ζ = ζ0 ∪ ζ1 ∪ ζ2 . links x1 and x2 . So, upon noticing B(x1 , N0 |x1 − x2 |) ⊆ B(z, N0 |x1 − x2 | + |z − x1 |) ⊆ B(z, (N0 + 1)|x1 − x2 |)

⊆ B(z, 2(N0 + 1)r), we obtain ζ ⊆ D ∩ B(z, rb−1 ) & as required.

b = 2−1 (N0 + 1),

212 | 8 Extension or restriction principle

8.5 Notes Below are some historical remarks on Chapter 8. Section 8.1. Theorem 8.1 (partially modified by investigating the weighted Hardy operators in [64, 90, 83]) appears first in this chapter, and its the argument, along with Step 1 of the argument for Theorem 4.5, actually shows that ℳ is bounded on C−λ (ℝn ) { { { C2,2λ (ℝn ) ⊇ {BMO(ℝn ) { { 2,2λ n {M (ℝ )

as 2λ ∈ [−1, 0)

as 2λ = 0

as 2λ ∈ (0, 1],

whose “⊇” actually becomes “=” under 2λ ∈ [−1, 0] (cf. Lemma 7.1). In particular, we have the following results. ▷ ‖ℳf ‖BMO (ℝn ) ≲ ‖f ‖BMO (ℝn ). This boundedness was established initially in [5] and recently proved again in [73] via the following inequality: ˆ |ℳf (x) − (ℳf )I | dx I

ˆ ≲

|ℳ((f − f3I )13I )| dx +

I

ˆ ( I\I∩{ℳf =ℳ(f 13I )}

sup

x∈J & m(J)>m(I)

|fJ − f2J |) dx

that holds for any (f , I) ∈ BMO(ℝn ) × C (ℝn ). ▷

‖ℳf ‖Lip0