# An Odd Rearrangement ofL1(Rn)

Authors: Zheng Wang; Jiecheng Chen
Journal: Journal of Function Spaces
 (2014)

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## Abstract

We introduce an odd rearrangementf* defined by π(f)(x)=f*(x)=sgn(x1)f*(νn|x|n), x∈Rn, where f* is a decreasing rearrangement of the measurable function f. With the help of this odd rearrangement, we show that for each f∈L1(Rn), there exists a g∈H1(Rn) such that df=dg, where df is an distribution function of f. Moreover, we study the boundedness of singular integral operators when they are restricted to odd rearrangement of L1(Rn), and we give some results on Hilbert transform.


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## Body

## 1. Introduction

From pages 39 to 40 in [1], we know the following relations between the Hardy space H1(Rn) and the Lebesgue space L1(Rn):
(1)H1(Rn)⊂L1(Rn),∥f∥L1≤∥f∥H1,
where H1(Rn) does not coincide with L1(Rn) since the elements of H1(Rn) have integral zero. This means that if f∈L1(Rn) and ∫Rn‍f(x)dx≠0, then f∉H1(Rn). On the other hand, it is well known that both Lebesgue spaces and Lorentz spaces can be characterized by distribution functions or decreasing rearrangements. This naturally raises the following question.Question 1.
For eachf∈L1, does there exist a function g∈H1(Rn) which is an appropriate modification of f satisfying df=dg? Here df is a distribution function of measurable function f (see Definition 1).In this paper, we will present a positive answer to Question1 by introducing an odd rearrangement. We will first address this problem in the case of n=1. Then in a similar way, motivated by the decreasing rearrangement defined on [0,∞) in [2, pp. 44–46] that keeps the distribution function, we introduce an odd rearrangement to solve the problem on Rn. To obtain applications of our result, the singular integral characterization of Hardy space attracts our attention. It is known that we can characterizeH1(R)by the Hilbert transform and characterize H1(Rn), when n≥2, by the Riesz transforms (see [1]). As it is pointed out in [2], although it is important to study the boundedness of the singular integral operators on L1(Rn), it may be impossible for us to give some conditions on singular integral operators to ensure they map L1(Rn) to L1(Rn). In this paper, we will show that some singular integral operators are of strong type (1,1) when some singular integral operators is restricted to odd rearrangements of L1(Rn).The distribution function and decreasing rearrangement are two important tools for us to study Lebesgue space and Lorentz space. Below, we briefly review their definition on the Euclidean space.Definition 1.
For a measurable functionf on Rn, the distribution function of f is the function df defined on [0,∞) by
(2)df(α)=μ({x∈Rn:|f(x)|>α}),
where μ is the Lebesgue measure.Definition 2.
Letf be a complex-valued function defined on Rn. The decreasing rearrangement of f is the function defined on [0,∞) by
(3)f*(t)=inf⁡{s>0:df(s)≤t}.
Here, we adopt the convention inf⁡∅=∞; thus f*=∞ whenever df(α)>t for all α≥0.We easily observe that when two measurable functions have the same distribution function or decreasing rearrangement, then they have the sameLp-norm.The following definition of odd rearrangement is the main issue in our study.Definition 3.
Letf be a complex-valued function defined on Rn. The odd rearrangement of f is the function f* defined on Rn by
(4)π(f)(x)=f*(x)=sgn⁡(x1)f*(νn|x|n),x∈Rn,
where f* is a decreasing rearrangement of the measurable function f and π is a projection: f(x)→f*(x).In this paper, we will show thatf and f* have the same distribution function and decreasing rearrangement. This indicates that the Lp-norm of f remains unchanged under the action of projection π. For convenience, we note that 𝒜(Rn) by the subset of L1(Rn) consists of all odd rearrangements of L1-functions.As a first step, we study the properties of the odd rearrangement in Proposition4. Next, we give criteria to determine the boundedness of the singular integral operators when it is restricted to odd rearrangements of L1(Rn) via Theorem 6. Thirdly, we consider three kinds of singular operators: singular integral operators with rough kernels, general singular integrals, and singular integrals of nonconvolutional type. Using the boundedness of Riesz transforms and Hilbert transform, we establish the relationships between the Hardy space and the Lebesgue space. Finally, we use the rearrangement which is an extension of the odd rearrangement to obtain some results on Hilbert transform.This paper is organized as follows. In Section2, we list some properties of odd rearrangement and seek sufficient conditions that ensure singular integral operators be of strong type (1,1) when restricted to odd rearrangements of L1(Rn). In Section 3, with the help of the rearrangement, we construct an odd function in ⋂0<p≤∞‍Lp(R) but not in H1(R) and show that for each f∈L1(R) and c∈(0,∞) there exists a function g∈H1(R) such that df=dg and ∥H(g)∥L1/∥g∥L1=c.Throughout, the letterC always denotes (possibly different) constants which are independent of all essential variables.


## 2. The Odd Rearrangement and Singular Integrals

From pages 46 to 48 in [2], we know many properties of the decreasing rearrangement. We find that the odd rearrangement has some similar properties with the decreasing rearrangement. The following are some properties of the odd rearrangement f*.Proposition 4.
(
1
)  df*=df*,
(
2
)
(
f
*
)
*
=
f
*,
(
3
)  (f*)*=f*,
(
4
)  (f*+g*)*=f*+g*,
(
5
)  ∥f∥Lp,q=∥f*∥Lp,q,
(
6
)  ∥f*+g*∥L1=∥f*∥L1+∥g*∥L1,
(
7
)  (kf)*=|k|f*,
(
8
)  |fn|↗|f|a.e.implies|(fn)*|↗|f*|,
(
9
)  |f|≤liminf⁡n→∞|fn|a.e.implies|f*| ≤ liminf⁡n→∞|(fn)*|a.e.
(
10
)  (|f|p)*=sgn⁡(x1)|f*|p,
(
11
)  ∥f∥L∞=lim⁡sgn⁡x→0,x1≠0(x1)f*(x).Proof.
We only need to prove thatdf*=df*; since other properties can be showed similar with the proof of Proposition  1.4.5 in [2], we know that f*(x)=sgn⁡(x1)f*(νn|x|n), ∀α>0.
(i) Ifdf*(α)=0, then f*(t)≤α,∀t>0. So we have
(5)|f*(x)|≤f*(νn|x|n)≤α,∀x≠0.
This means that df*(α)=0.
(ii) Ifdf*(α)>0, ∀ϵ>0(ϵ<df*(α)), for convenience, we choose
(6)t1=df*(α)-ϵ,t2=df*(α)+ϵ.
Since f* is a nonnegative decreasing function defined on [0,∞), it follows that
(7)f*(t1)>α,f*(t2)≤α;
then we have
(8)f*(νn(νn-1/nt11/n)n)>α,f*(νn(νn-1/nt21/n)n)≤α.
So if x∈B(0,νn-1/nt11/n)⋂‍{x∈Rn;x1≠0}, then we obtain that
(9)|f*(x)|=f*(νn|x|n)≥f*(t1)>α;
this means that
(10)df*(α)≥|B(0,νn-1/nt11/n)|=t1=df*(α)-ϵ.
And if x∉B(0,νn-1/nt21/n), then we obtain that
(11)|f*(x)|≤f*(νn|x|n)≤f*(t2)≤α;
this means that
(12)df*(α)≤|B(0,νn-1/nt21/n)|=t2=df*(α)+ϵ.
Let ϵ→0; we conclude that df*=df*.Firstly, we show that for eachf∈𝒜(Rn), we can find one sequence of simple functions that converge to f.Lemma 5.
For eachf∈𝒜(Rn), there exists one sequence {fk}k⊂𝒜(Rn) satisfying
(13)fk(x)=12k∑j=122ksgn⁡(x1)χB(0,Rk,j)(x),|fk(x)|↗|f(x)|(k⟶∞),
where B(0,Rk,j+1)⊂B(0,Rk,j).Proof.
We take(14)fk=sgn⁡(x1)12k∑j=122k‍χ{x∈Rn:f*(νn|x|n)>j/2k}.
Since f*(t) is a nonnegative decreasing right continuous function defined on [0,∞), it follows that {f*(t)>α} is an open set of [0,∞), which means that {x∈Rn:f*(νn|x|n)>j/2k} is an open ball of Rn. We take B(0,Rk,j)={x∈Rn:f*(νn|x|n)>j/2k}; then the theorem can be showed easily.Note thatℬ(Rn)={sgn⁡(x1)f(x):f(x) is a radical function in L1(Rn)}. It is easy for us to show that ℬ(Rn) is a subspace of L1(Rn). Now we give criteria to determine the boundedness of the singular integral operators when they are restricted to odd rearrangements of L1(Rn) via Theorem 6, Corollary 7, and Corollary 8.Theorem 6.
Suppose thatT is a singular integral operator satisfying the following.(1)
There exists a constantC independent of R such that
(15)∥T(sgn⁡(x1)χB(0,R))∥L1≤C∥sgn⁡(x1)χB(0,R)∥L1,∀R>0.(2)
T is weak type of (1,1) when restricted to the functions of ℬ(Rn).Then T∘π is strong type of (1,1) with norm at most C.Proof.
For eachf∈𝒜(Rn), according to Lemma 5, there exists one sequence {fk}k satisfying
(16){fk}k⊂𝒜(Rn),fk(x)=12k∑j=122k‍sgn⁡(x1)χB(0,Rk,j)(x);|fk(x)|↗|f(x)|(k⟶∞),
where B(0,Rk,j+1)⊂B(0,Rk,j). We can easily obtain that
(17)∥fk-f∥L1⟶0(k⟶∞).
Due to the fact that T is weak type of (1,1), we obtain that
(18)∥T(fk)-T(f)∥L1,∞⟶0(k⟶∞).
Then T(fk) converges to T(f) in measure, so there exists some subsequence {T(fki)}i>0converges to T(f)a.e. Via Fatou’s lemma, we conclude that
(19)∥T(f)∥L1=∥lim⁡i→∞T(fki)∥L1≤liminf⁡i→∞∥T(fki)∥L1≤liminf⁡i→∞C∥fki∥L1≤C∥f∥L1.Theorem6 is a criterion to determine the boundedness of the singular integral operators when they are restricted to odd rearrangements of L1(Rn). By Theorem 6, we can find two simpler criteria. Corollary 7 is an immediate consequence of Theorem 6, while Corollary 8 can be showed later.Corollary 7.
Suppose thatT is a singular integral operator satisfying(1)
for anya>0, T(δaf)=δaT(f));(2)
∥T(sgn⁡(x1)χB(0,1))∥L1<∞;(3)
T is weak type of (1,1) when restricted to the functions of ℬ(Rn).Then T∘π is strong type of (1,1).Corollary 8.
Suppose thatT is a singular integral operator satisfying the following:(1)
T is a bounded operator from H1(Rn) to L1(Rn);(2)
T is weak type of (1,1).Then T∘π is strong type of (1,1).Now let us study the boundedness of singular integral operators with the help of Theorem6 and Corollary 7. Firstly, we consider the boundedness of singular integral operators with rough kernels. The method of rotations is used here; thus we need to consider the boundedness of Hilbert transform firstly.Lemma 9.
Consider the characteristic functionχ(-1,1) of an interval (-1,1),
(20)H(sgn⁡(·)χ(-1,1))(x)=1πlog⁡x2|x2-1|,∥H(sgn⁡(·)χ(-1,1))(x)∥L1=8πlog⁡(1+2).Proof.
As in [2, pp. 251-252], for the characteristic function χ[a,b] of an interval [a,b],
(21)H(χ[a,b])(x)=1πlog⁡|x-a||x-b|.
So it is not difficult for us to get the above results.Proposition 10.
For eachf∈L1(R), we haveH(f*)∈L1(R),
(22)∥H∘π(f)∥L1≤4πlog⁡(1+2)∥f∥L1.
Equality can occur if f=sgn⁡(x)χ(-1,1).Proof.
Since Hilbert transformH is bounded linear operator mapping L1(R) to L1,∞(R) and satisfying
(23)T(δaf)=δaT(f),∀a>0.
Using Lemma 9, it is an immediate consequence of Corollary 7.With the help of boundedness of Hilbert transform, we establish the relationship betweenL1(R) andH1(R)in the following corollary.Corollary 11.
For allg∈𝒜(R),
(24)∥g∥H1~∥g∥L1.Proof.
We use singular integral characterization ofH1(R) in [2]; following Proposition 10, we get that
(25)∥g∥H1=∥g∥L1+∥H(g)∥L1~∥g∥L1,
for all g∈𝒜(R).Since the boundedness of Hilbert transform has been known, we can use the method of rotations to study singular integral operatorTΩ where Ω is odd and integrable over Sn-1.Lemma 12.
IfΩ is odd and integrable over Sn-1, then
(26)∥TΩ(sgn⁡(x1)χB(0,1))∥L1≤4log⁡⁡(1+2)∥Ω∥L1(Sn-1)×∥sgn⁡(x1)χB(0,1)∥L1.Proof.
Via the method of rotation we can obtain(27)TΩ(sgn(x1)χB(0,1))(y)=π2∫Sn-1‍Ω(θ)Hθ(sgn⁡(x1)χB(0,1))(y)dθ,
where Hθ are the directional Hilbert transforms defined by
(28)Hθ(f)(x)=1πp.v.∫-∞+∞‍f(x-tθ)dtt.
Since the following identity is valid for all matrices, A∈O(n):
(29)HA(e1)(f)(x)=He1(f∘A)(A-1x),
we take g=sgn⁡(x1)χB(0,1); then
(30)∫Sn-1‍Ω(θ)Hθ(g)(x)dθ=∫Sn-1⋂‍{θ:θ1>0}‍Ω(θ)Hθ(g)(x)dθ+∫Sn-1⋂‍{θ:θ1<0}‍Ω(θ)Hθ(g)(x)dθ=∫Sn-1⋂‍{θ:θ1>0}‍Ω(θ)Hθ(g)(x)dθ-∫Sn-1⋂‍{θ:θ1>0}‍Ω(θ)H-θ(g)(x)dθ=∫Sn-1⋂‍{θ:θ1>0}‍Ω(θ)(Hθ-H-θ)(g)(x)dθ.
Since for any a,b>0 the following inequality holds:
(31)∥H(χ(0,a)-χ(-b,0))(x)+H(χ(0,a)-χ(-b,0))(-x)∥L1(R)≤2∥H(χ(0,(a+b)/2)-χ(-(a+b)/2,0))(x)∥L1(R),
we obtain
(32)∥(Hθ-H-θ)(g)(x)∥L1=∥(HA(e1)-H-A(e1))(g)(x)∥L1=∥He1(g∘A)(A-1x)-He1(g∘(-A))(-A-1x)∥L1=∥He1(g∘A)(x)+He1(g∘A)(-x)∥L1≤2∥He1(g)(x)∥L1=8πlog⁡(1+2)∥g(x)∥L1.
Then, via Minkowski’s integral inequality, we obtain
(33)∥TΩ(sgn⁡(x1)χB(0,1))∥L1=π2∥∫Sn-1⋂‍{θ:θ1>0}‍Ω(θ)(Hθ-H-θ)(g)(x)dθ∥L1≤π2∫Sn-1⋂‍{θ:θ1>0}‍|Ω(θ)|·∥(Hθ-H-θ)(g)(x)∥L1dθ=4log⁡(1+2)∥Ω∥L1(Sn-1)∥sgn(x1)χB(0,1)∥L1.According to [3], if Ω∈Llog⁡+L(Sn-1) with ∫Sn-1‍Ω(θ)dθ=0, then TΩ is weak type of (1,1). With the help of Corollary 8 and Lemma 12, we easily get the following proposition.Proposition 13.
IfΩ is an odd function defined on Sn-1 and Ω∈Llog⁡+L(Sn-1), then
(34)∥TΩ∘π(f)∥L1≤4log⁡(1+2)∥Ω∥L1(Sn-1)∥f∥L1.We know that Riesz transforms satisfy the conditions in Proposition13, so Rj∘π maps L1(Rn) to L1(Rn). Then we establish one relationship between L1(Rn) and H1(Rn) in the following corollary.Corollary 14.
(i) For allf∈L1(Rn), then
(35)∥Rj∘π(f)∥L1≤4log⁡(1+2)ωn∥f∥L1.
Equality can occur if and only if f=0 a.e.
(ii) For allg∈𝒜(Rn),
(36)∥g∥H1~∥g∥L1.Remark 15.
Corollary14 is the high dimensional extension of Corollary 11.Here we can prove Corollary8.Proof of Corollary8.
SinceT is a bounded operator from H1(Rn) to L1(Rn) and sgn⁡(x1)χB(0,R)∈𝒜(Rn), then via Corollary 14, we obtain that
(37)∥T(sgn⁡(x1)χB(0,R))∥L1≲∥sgn⁡(x1)χB(0,R)∥H1≲∥sgn⁡(x1)χB(0,R)∥L1.
So Corollary 8 is an immediate consequence of Theorem 6 and Corollary 14.Remark 16.
WhenΩ is odd and integrable over Sn-1, whether TΩ is weak type of (1,1) is an open problem. So we suppose Ω∈Llog⁡+L(Sn-1) in Proposition 13. As showed in [4], TΩ is weak type (1,1) when restricted to radial functions in L1(Rn). But we can try to show that TΩ can be weak type (1,1) when restricted to functions in ℬ(Rn), and then the same conclusion of Proposition 13 can be obtained. On the other hand, we try to omit the “odd” condition in the following proposition.Proposition 17.
LetΩ∈Llog⁡+L(Sn-1) with ∫Sn-1‍Ω(θ)dθ=0 and suppose for some r>0 that
(38)∫B(0,2r)c‍inf⁡lx∈C∫B(x,r)‍|Ω(y)|y|n-lx|dydx<∞.
Then
(39)∥TΩ(sgn⁡(x1)χB(0,1))∥L1<∞,
and there exists a constant C, such that for all f∈L1(Rn),
(40)∥TΩ∘π(f)∥L1≤C∥f∥L1.Proof.
SinceΩ∈Llog⁡+L(Sn-1), according to [2, pp. 274-275], the singular integral operator TΩ is bounded on L2(Rn). Then we obtain(41)∥TΩ(sgn⁡(x1)χB(0,1))∥L1(Rn)=∥δr-1TΩ(sgn⁡(x1)χB(0,r))∥L1(Rn)=r-n∥TΩ(sgn⁡(x1)χB(0,r))∥L1(Rn)=r-n∥TΩ(sgn⁡(x1)χB(0,r))∥L1(B(0,r))+r-n∥TΩ(sgn⁡(x1)χB(0,r))∥L1(B(0,2r)c)=r-nI1+r-nI2,I1=∥TΩ(sgn⁡(x1)χB(0,r))∥L1(B(0,r))=∥TΩ(sgn⁡(x1)χB(0,r))∥L2×∥χB(0,2r)∥L2<∞.
For each lx∈C,
(42)|TΩ(sgn⁡(x1)χB(0,r))|=|∫Rn‍Ω(y/|y|)|y|nsgn⁡(x1-y1)χB(0,r)(x-y)dy|=|∫B(x,r)‍Ω(y/|y|)|y|nsgn⁡(x1-y1)dy|=|∫B(x,r)‍(Ω(y/|y|)|y|n-lx)sgn⁡(x1-y1)dy|≤∫B(x,r)‍|Ω(y/|y|)|y|n-lx|dy;
this means that
(43)|TΩ(sgn⁡(x1)χB(0,r))|≤inf⁡lx∈C∫B(x,r)‍|Ω(y/|y|)|y|n-lx|dy,I2=∫B(0,2r)c‍|TΩ(sgn⁡(x1)χB(0,r))|dx≤∫B(0,2r)c‍inf⁡lx∈C∫B(x,r)‍|Ω(y/|y|)|y|n-lx|dydx<∞,
According to [3], TΩ is weak type of (1,1), so Proposition 17 is an immediate consequence of Corollary 7.Now let us apply Corollary8 to the general singular integral operators and show their boundedness on L1(Rn) when restricted to odd rearrangement.The following is the definition of the general singular operators.The functionK defined Rn∖{0} that satisfies size estimates,
(44)sup⁡0<R<∞1R∫|x|≤R‍|K(x)||x|dx≤A1,
the smooth estimate, expressed in terms of Hömander’s condition,
(45)sup⁡|y|≠0∫|x|>2|y|‍|K(x-y)-K(x)|dx≤A2,
and cancelation condition,
(46)sup⁡0<R1<R2|∫R1<|x|<R2‍K(x)dx|≤A3.
Condition (3) implies that there exists a sequence δj↘0 as j→0 such that the following limit exists:
(47)lim⁡j→∞∫δj≤|x|≤1‍K(x)dx.
The tempered distribution W has the form
(48)W(φ)=lim⁡j→∞∫|x|≥δj‍K(x)φ(x)dx,φ∈S(Rn).Proposition 18.
Let the function K be defined onRn∖{0} that satisfies (45), (46), and (47). Suppose that the operator T is given by convolution with W. Then ∥T∘π(f)∥L1≤Cn(A1+A2+A3)∥f∥L1.Proof.
By the results of [1, pp. 93-94], T admits an extension that is Lp bounded for 1<p<∞ and is of weak type (1,1). Also, there exists a constant Cn such that for all f∈H1(Rn),
(49)∥T(f)∥L1≤Cn(A1+A2+A3)∥f∥H1.
Thus Proposition 18 is an immediate consequence of Corollary 8.Finally, we give results ofC-Z operators. Corollary 8 is used again.Proposition 19.
Assume thatK(x,y) is in SK(δ,A), and T is an element of CZO(δ,A,B) that is associated with the kernel K. Then for all f∈L1(Rn),
(50)∥T∘π(f)∥L1≤Cn,δ(A+B)∥f∥L1,
where Cn is a constant that depends on the dimension.Proof.
By the results of [1, pp. 171–190], T admits an extension that is Lp bounded for 1<p<∞ and is of weak type (1,1). Also, there exists a constant Cn,δ such that for all f∈H1(Rn),
(51)∥T(f)∥L1≤Cn,δ(A+B)∥f∥H1.
Thus Proposition19 is an immediate consequence of Corollary 8 too.


## 3. Some Results on Hilbert Transform

In Section2, we find that if f=χE∈L1(R), by odd rearrangement, we obtain df=df*and ∥H(f*)∥L1/∥f*∥=C1=(4/π)log⁡(1+2). In this section, we use the rearrangement, which is like the odd rearrangement, to show that for given f∈L1(R) and c∈(0,∞), there exists a function g∈H1(R) satisfying df=dg and ∥H(g)∥L1/∥g∥L1=c.Proposition 20.
For givenf=χE∈L1(R) and c∈(0,∞), there exists a function g∈H1(R) satisfying df=dg and ∥H(g)∥L1/∥g∥L1=c.Proof.
Since∥H(f*)∥L1/∥f*∥L1=C1=(4/π)log⁡(1+2), we discuss it in two cases.
Case
1. Given that f=χE∈L1(R) and c∈(C1,∞), then there exists a function g∈H1(R) satisfying df=dg and ∥H(g)∥L1/∥g∥L1=c.
Case
2. Given that f=χE∈L1(R) and c∈(0,C1), then there exists a function g∈H1(R) satisfying df=dg and ∥H(g)∥L1/∥g∥L1=c.
Proof of Case
1. We take F(t)=tlog⁡t-t, t>0, and F(0)=0. Then F(t) is a continuous function defined on [0,∞). And we consider the function
(52)h(t,x)=χ(1,t)(x)-χ(-t,-1)(x),t>1.
Then we obtain ∥H(h(t,·))(x)∥L1(R)=(4/π)[F(t+(t1+1)/2)-F(t-(t1+1)/2)-F((t1+1)/2-1)−F((t1+1)/2+1)]. We take
(53)G(t)=F(t+t1+12)-F(t-t1+12)-F(t1+12-1)-F(t1+12+1);
then
(54)∥H(h(t,·))(x)∥L1(R)∥h(t,x)∥L1(R)=2πG(t)t-1.
Here, we use three steps to prove {(2/π)(G(t)/(t-1));t>1}=(C1,∞).
Step
1. Consider
(55)lim⁡t→1+G(t)t-1=+∞.
Proof. We take (1+t2)/2=h; then t=2h2-1. Consider
(56)lim⁡t→1+G(t)t-1=lim⁡h→1+(F(2h2-1+h)-F(2h2-1-h)-F(h-1)-F(h+1)F(2h2-1+h))×(2h2-1-1)-1=lim⁡h→1+12h/2h2-1×[(2h2h2-1+1)log⁡⁡(2h2-1+h)-(2h2h2-1-1)log⁡⁡(2h2-1-h)-log⁡⁡(h-1)-log⁡⁡(h+1)(2h2h2-1+1)]=lim⁡h→1+12h/2h2-1×[2h2h2-1log⁡2h2-1+h2h2-1-h+log⁡⁡(2h2-1+h)+log⁡⁡(2h2-1-h)-log⁡⁡(h-1)-log⁡⁡(h+1)2h2h2-1log⁡2h2-1+h2h2-1-h]=lim⁡h→1+⁡(2h/2h2-1)log⁡(2h2-1+h/2h2-1-h)2h/2h2-1=lim⁡h→1+⁡log⁡2h2-1+h2h2-1-h=+∞.
Step
2. lim⁡t→∞⁡G(t)/(t-1)=2log⁡(1+2). Consider
(57)lim⁡t→+∞G(t)t-1=lim⁡h→+∞(F(2h2-1+h)-F(2h2-1-h)-F(h-1)-F(h+1)F(2h2-1+h))×(2h2-1-1)-1=lim⁡h→+∞log⁡2h2-1+h2h2-1-h=log⁡2+12-1=log⁡(2+1)2=2log⁡(2+1).
Step
3. G(t)/(t-1) is decreasing on the interval (1,∞). Consider
(58)dhdt=t2(1+t2)/2>0,t>1,ddt(G(t)t-1)=ddh(×(2h2-1-1)-1)(F(2h2-1+h)-F(2h2-1-h)-F(h-1)-F(h+1)(F(2h2-1+h)-F(2h2-1-h))×(2h2-1-1)-1)dhdt=4h(2h2-1-1)22h2-1×log⁡h+12h2-1+h<0,(h>1).
So we obtain that
(59)ddt(G(t)t-1)<0.
Thus G(t)/(t-1) is decreasing on the interval (1,∞). Since G(t)/(t-1) is continuous on the interval (0,∞), it is easy to conclude that {(2/π)(G(t)/(t-1));t>1}=(C1,∞). Now let us discuss Case 1.
∀
c
∈
(
C
1
,
∞
), then there exists tc>1 satisfying
(60)∥H(h(tc,·))(x)∥L1(R)∥h(tc,x)∥L1(R)=c.
Taking g(x)=h(tc,(2(tc-1)/|E|)x), then we obtain that
(61)∥H(g)∥L1∥g∥L1=∥H(h(tc,·))(x)∥L1(R)∥h(tc,x)∥L1(R)=c.
At the same time, we can easily verify that df=dg.
Proof of Case
2. We know that ∥H(χ(0,1)-χ(-1,0))(x)∥L1/∥χ(0,1) − χ(-1,0)∥L1 = C1 in Section 2. If we take f1(x)=χ(0,1)-χ(-1,0), then ∥H(f1)(x)∥L1/∥f1∥L1=C1; we note that
(62)fn(x)=∑k=12n-1‍χ(k-12n-1,k2n-1)-χ(-k2n-1,-k-12n-1),∥H(fn)(x)∥L1∥fn∥L1=Cn,fn+1,1(x)=∑k=12n-1‍χ(k-12n,k2n)-χ(-k2n,-k-12n),fn+1,2(x)=∑k=2n-1+12n‍χ(k-12n,k2n)-χ(-k2n,-k-12n).
Here, we prove that Cn↘0(n→0).
We know that Hilbert transform satisfies(63)H(τh(f))=τhH(f),(h∈R),H(δa(f))=δaH(f),(a>0).
Thus we obtain
(64)∥H(fn+1,i)(x)∥L1∥fn+1,i∥L1=Cn,i=1,2.
Then
(65)Cn+1=∥H(fn+1)(x)∥L1∥fn+1∥L1=∥H(fn+1,1+fn+1,2)(x)∥L1∥fn+1∥L1=∥H(fn+1,1)(x)+H(fn+1,2)(x)∥L1∥fn+1∥L1<∥H(fn+1,1)(x)∥L1+∥H(fn+1,2)(x)∥L1∥fn+1∥L1=Cn.
Since ∥fn∥L1≡2,n≥1, so we just need an estimate of ∥H(fn)∥L1:
(66)∥H(fn)∥L1=∥∑k=12n-1‍log⁡(x-((k-1)/2n-1))2|(x-((k-1)/2n-1))2-(1/2n-1)2|∥L1=12n-1∥∑k=12n-1‍log⁡(x-k+1)2|(x-k+1)2-1|∥L1=12n-1∥log⁡∏k=12n-1‍(x-k+1)2|(x-k+1)2-1|∥L1=12n-1∥log⁡|x||x+1||x-(2n-1-1)||x-2n-1|∥L1=12n-1∥log⁡|x+2n-2-(1/2)||x+2n-2+(1/2)|×|x-(2n-2-(1/2))||x-2n-2-(1/2)|∥L1=12∥H(χ(1-(1/2n-1),1+(1/2n-1))-χ(-1-(1/2n-1),-1+(1/2n-1)))∥L1.
In order to estimate the limit of ∥H(fn)∥L1, we consider the following function:
(67)h(t)=∥H(χ(1-t,1+t)-χ(-1-t,-1+t))∥L1,t∈[0,1].
We obtain that
(68)h(t)=4[F(1+t+1+t2)-F(1+t-1+t2)-F(1+t2-1+t)-F(1+t2+1-t)];
thus
(69)lim⁡n→∞∥H(fn)∥L1=lim⁡t→012h(t)=0.
This means that Cn↘0,(n→∞). Now we construct the following functions:
(70)gn,t(x)=∑k=12n-1‍χ((k-1)/2n-1,k/2n-1)-χ(-k/2n-1,-(k-1)/2n-1)-2[χ((k-1)/2n-1,(k-1)/2n-1+t)-χ(-t-(k-1)/2n-1,-(k-1)/2n-1)].
Whent=0, we havegn,t=fn; and whent=1/2n, we have
(71)∥H(gn,0)∥Lp∥gn,0∥Lp=Cn,∥H(gn,1/2n)∥Lp∥gn,1/2n∥Lp=Cn+1.
Using the continuity of∥H(gn,t)∥L1/∥gn,t∥L1=∥H(gn,t)∥L1/2 on the interval [0,1/2n], we obtain that for each c∈(Cn+1,Cn), there exists a constant tc∈(0,1/2n) satisfying
(72)∥H(gn,tc)∥L1∥gn,tc∥L1=c.
So Case 2 is true.Here, we can get some corollaries.Corollary 21.
Suppose thatI1,I2 are two disjoint open intervals with given same length; note that d(I1,I2) is the distance between I1 and I2; then ∥H(χI1-χI2)∥L1/∥χI1-χI2∥L1 depends only on d(I1,I2). Precisely(i)
∥H(χI1-χI2)∥L1/∥χI1-χI2∥L1=C1=4/πlog⁡(1+2) when d=0;(ii)
∥H(χI1-χI2)∥L1/∥χI1-χI2∥L1↗∞,(d→∞);(iii)
{∥H(χI1-χI2)∥L1/∥χI1-χI2∥L1:I1 and I2 are disjoint open intervals}=[(4/π)log⁡(1+2),∞).Proof.
Since Hilbert transform,H, commutes with translations and dilations, we can suppose that
(73)χI1-χI2=χ(a,a+1)-χ(-a-1,-a),a≥0,d(I1,I2)=2a.
Because Hilbert transform, H, commutes with dilations, we obtain that
(74)∥H(χI1-χI2)∥L1∥χI1-χI2∥L1=∥log⁡|x2-a2|/|x2-(a+1)2|∥L12=∥log⁡|x2-1|/|x2-(1+(1/a))2|∥L12(1/a).
We take t=1+(1/a); then we obtain that
(75)∥H(χI1-χI2)∥L1∥χI1-χI2∥L1=∥H(h(t,·))(x)∥L1(R)∥h(t,x)∥L1(R)=2πG(t)t-1,
where h(t,x)=χ(1,t)(x)-χ(-t,-1)(x),t=1+(1/d)>1, and G(t) is defined in the proof of Proposition 20. According to the proof of Proposition 20, we conclude the following:(i)
lim⁡t→1+G(t)/(t-1)=+∞;(ii)
lim⁡t→∞G(t)/(t-1)=2log⁡(1+2);(iii)
G(t)/(t-1) is decreasing and continuous on the interval (1,∞).And the corollary is consequence of the three items above.As in [1], we know that the elements of H1(Rn) have integral zero. In Exercise 6.4.3 of [1], we learn that, for some p>1, if g∈Lp is compactly supported with integral zero, then g∈H1. In the following example, with the help of rearrangement, we construct an odd function in ⋂0<p≤∞‍Lp(R) while not in H1(R). Thus “compact" condition cannot be removed without modifying other conditions properly.Example 22.
According to Corollary21, we construct two sequences of intervals {I1,n} and {I2,n} satisfying the following:(i)
|I1,n|=|I2,n|=1/2n and I1,n and I2,n are disjoint intervals;(ii)
∥H(χI1,n+1-χI2,n+1)∥L1 ≥ 2∑k=1n‍∥H(χI1,k-χI2,k)∥L1, ∀n≥1;(iii)
I1,n=-I2,n and I1,n⊂(0,∞),∀n≥1;(iv)
d(I1,n+1,I2,n+1)≥d(I1,n,I2,n)+1,∀n≥1.We takefn=χI1,n-χI2,n and f=∑n≥1‍fn, since df=dχ(-1,1); we obtain that
(76)∥f∥Lp=∥χ(-1,1)∥Lp<∞,0<p≤∞,∥H(f)∥L2=∥f∥L2<∞.
Since ∑1≤k≤n‍fk→f in L2(R) as n→∞, we obtain that ∑1≤k≤n‍H(fk)→H(f) in L2(R) as n→∞. This means that there is a sequence {nk} satisfying
(77)∑1≤i≤nk‍H(fi)⟶H(f)a.e.
By noting that I1,n=(an,bn), then we obtain that
(78)0<an<bn<an+1<bn+1,an2+bn22<an+12+bn+122.
Since H(fn)=log⁡|(x2-an2)/(x2-bn2)|, we obtain that H(fn)(x)<0 if and only if |x|<(an2+bn2)/2. And we easily obtain that
(79)(∑k=1n‍H(fk))-≤(∑k=1n+1‍H(fk))-.
Since ∫|x|<(an2+bn2)/2‍H(fn)(x)dx=-∫|x|>(an2+bn2)/2‍H(fn)(x)dx, we obtain that
(80)∫R‍H(fn)(x)dx=0,∫R‍∑k=1n‍H(fk)(x)dx=0.
Then
(81)∥H(f)∥L1≥∥H(f)-∥L1=∥lim⁡k→∞[∑1≤i≤nk‍H(fi)]-∥L1=lim⁡k→∞∥[∑1≤i≤nk‍H(fi)]-∥L1=lim⁡k→∞∥[∑1≤i≤nk‍H(fi)]+∥L1≥lim⁡k→∞∥[H(fnk)]+∥L1=12lim⁡k→∞∥H(fnk)∥L1≥lim⁡k→∞∑i=1nk-1‍∥H(fi)∥L1≥lim⁡k→∞(nk-1)∥H(f1)∥L1=∞.
Since ∥f∥H1~∥f∥L1+∥H(f)∥L1, we get that f∉H1(R).Corollary 23.
There exist two sequences{fn}⊂H1(R) and {fn}⊂H1(R) satisfying
(82)∥fn∥L1=1,(n⟶∞),∥H(fn)∥L1⟶0,(n⟶∞),∥H(gn)∥L1⟶∞,(n⟶∞).Proof.
We takef=χ(0,1); according to Proposition 20, there exist two sequences {gn} and {fn} satisfying
(83)dfn=df,∥H(fn)∥L1∥fn∥L1=1n,∥H(gn)∥L1∥gn∥L1=n.
So we easily obtain this corollary.Similarly, via the method of rotation we used in Lemma12, we can obtain the following corollary.Corollary 24.
There exists a sequence{fm}⊂H1(Rn) satisfying
(84)∥fm∥L1=1,(m≥1),∥Rj(fm)∥L1⟶0,(m⟶∞).
where Rj(1≤j≤n) are Riesz transforms.Remark 25.
Since∥f∥L1(Rn)≲∥f∥H1(Rn), according to Corollaries 23 and 24, ∥H(f)∥L1(R) and ∑1≤j≤n‍∥Rj(f)∥L1(Rn) are not equivalent norms of Hardy spaces H1(R) and H1(Rn).Lemma 26.
LetΛ={gt,t∈I}⊂L1(R) satisfying
(85)lim⁡t→s∥gt-gs∥L1=0,∀s∈I,{∥g∥L1,g∈Λ}=(0,∞),
where I is an open interval of R. Then for all f∈L1(R),
(86)(∥f∥L1,∞)⊂{∥f+g∥L1,g∈Λ}.Proof.
Given thatf∈L1(R), ∀ϵ>0, and ∀N>0(N>ϵ), since {∥g∥L1,g∈Λ}=(0,∞), there exist t0,t1∈I satisfying
(87)∥gt0∥L1<ϵ,∥gt1∥L1>2∥f∥L1+N.
Thus, we obtain that
(88)∥gt0+f∥L1<∥f∥L1+ϵ,∥gt1+f∥L1≥∥gt0∥L1-∥f∥L1>∥f∥L1+N.
Since lim⁡t→s∥gt-gs∥L1=0,(∀s∈I), we get that ∥gt+f∥L1 is continuous on the interval I. Thus, (∥f∥L1+ϵ,∥f∥L1+N)⊂(∥gt0+f∥L1,∥gt1+f∥L1)⊂{∥f+g∥L1,g∈Λ}.
Letϵ→0 and N→∞; we obtain
(89)(∥f∥L1,∞)⊂{∥f+g∥L1,g∈Λ}.Proposition 27.
Given each simple function:f∈L1(R) and c∈(0,∞), then there exists a function g∈H1(R) satisfying
(90)df=dg,∥H(g)∥L1∥g∥=c.Proof.
Sincef is a simple function in L1(R), there are two sequences {ai}1≤i≤m and {tn}1≤i≤m satisfying
(91)f1=∑i=1m‍aiχ(2(∑k=1itk)-ti,2∑k=1itk)-∑i=1m‍aiχ(2∑k=1i-1tk,2(∑k=1i-1tk)+ti),df1=df.
We take C′′=∥H(f1)∥L1/∥f1∥L1, since
(92)f1=∑i=1m-1‍aiχ(2(∑k=1itk)-ti,2∑k=1itk)-∑i=1m-1‍aiχ(2∑k=1i-1tk,2(∑k=1i-1tk)+ti)+g1,
where g1=amχ(2(∑k=1m‍tk)-tm,2∑k=1m‍tk)-amχ(2∑k=1m-1‍tk,2(∑k=1m-1‍tk)+tm); we construct a set Γ={g1,t:t∈[0,∞)}⊂L1(R), where g1,t=am.
Alsoχ(2(∑k=1m‍tk)-tm+t,2(∑k=1m‍tk)+t)-amχ(2∑k=1m-1‍tk,2(∑k=1m-1‍tk)+tm). Using Corollary 21, g1,0=g1, and ∥g1,t∥L1≡2amtm, we obtain that
(93){∥H(h)∥L1,h∈Γ}=[∥g1∥L1,∞),
Using Lemma 26, we obtain that
(94)[∥H(f1)∥L1,∞)⊂{∥H(f1-g1+g1,t)∥L1,t∈[0,∞)}.
Since ∥f1-g1+g1,t∥L1≡2∑k=1m‍aktk, we obtain that, for each c∈[C′′,∞), there exists g=f1-g1+g1,tc satisfying
(95)dg=df,∥H(g)∥L1∥g∥L1=c.
By similar argument of Case 2 of Proposition 20, for each c∈(0,C′), there exists g satisfying
(96)dg=df,∥H(g)∥L1∥g∥L1=c.Theorem 28.
For each functionf∈L1(R) and c∈(0,∞), then there exists a function g∈H1(R) satisfying
(97)df=dg,∥H(g)∥L1∥g∥=c.Proof.
We can use the results of Proposition27 and the same method of rearrangement as in the proof of Proposition 20 to get this theorem proved; we leave the details to the reader.



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*Source: 787840-2014-02-26.xml*