The variation of the maximal function of a radial function

TL;DR

This paper investigates how the variation of the Hardy-Littlewood maximal function behaves for radial functions in higher dimensions, establishing that their variations are comparable.

Contribution

It proves that the variation of the non-centered Hardy-Littlewood maximal function of a radial function is comparable to the variation of the original function, extending understanding in higher dimensions.

Findings

Variation of the maximal function is comparable to the original function's variation.

Results apply to non-centered Hardy-Littlewood maximal functions.

Focus on radial functions in higher dimensions.

Abstract

We study the problem concerning the variation of the Hardy-Littlewood maximal function in higher dimensions. As the main result, we prove that the variation of the non-centered Hardy-Littlewood maximal function of a radial function is comparable to the variation of the function itself.

Equations

$$Mf(x)=\underset{B(z,r)\ni x}{\sup }\frac{1}{|B(z,r)|}{\int }_{B(z,r)}|f(y)|dy=:\underset{B(z,r)\ni x}{\sup }{\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B(z,r)}|f(y)|dy$$

$$|D{M}_{c}f(x)|\le M_c(|Df|)(x)$$

$${\left|\left|DMf\right|\right|}_1\le C_n{\left|\left|Df\right|\right|}_1?$$

$$DMf(x)={\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B}Df(y)dy.$$

$${\int }_{B}Df(y)\cdot (y-x)dy=0.$$

$$|DMf(x)|\le \frac{C_n}{|x|}{\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B(0,|x|)}|Df(y)||y|dy.$$

$$M^{I}f(x)=\underset{x\in B(z,r),r\le |x|/4}{\sup }{\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B(z,r)}|f(y)|dy,$$

$${\left|\left|D{M}^{I}f\right|\right|}_1\le C_n{\left|\left|Df\right|\right|}_1.$$

$$f_{/4}(x):=\underset{x\in B(z,|x|/4)}{\sup }{\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B(z,|x|/4)}|f(y)|dy,$$

$${\int }_{\{2^k\le |y|\le 2^{k+1}\}}D{M}^{I}f(y)dy\le C_n{\int }_{\{2^{k-1}\le |y|\le 2^{k+2}\}}|D|f|(y)|dy.$$

$$\widetilde{M}F(x)=\sup\bigg{\{}\,\bigg{|}\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB(z,r)}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B(z,r)}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B(z,r)}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B(z,r)}}F\bigg{|}\,:\,x\in B(z,r)\,,\,\,\int_{B(z,r)}F(y)\cdot(y-x)dy\,=0\,\bigg{\}}\,.$$

$$D_{v}f(x):=\underset{h\to 0}{\lim }\frac{1}{h}(f(x+hv)-f(x)),$$

$${\mathcal{B}}_x:=$$

$$DMf(x)={\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B}D|f|(y)dy\text{ and }{D}_{v}Mf(x)={\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_B{D}_v|f|(y)dy.$$

$$\frac{DMf(x)}{|DMf(x)|}=\frac{z-x}{|z-x|}.$$

$${\int }_{B}D|f|(y)\cdot (y-x)dy=0.$$

$$|DMf(x)|=\frac{1}{r}{\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B}D|f|(y)\cdot (z-y)dy.$$

$$DMf(x)\cdot \frac{x}{|x|}=\frac{1}{|x|}{\text{vrule}width=5.40005pt,height=2.7pt,depth=-2.24997pt\int }_{B}D|f|(y)\cdot ydy.$$

$$\underset{i\to \infty }{\lim }\frac{L_i(y)-y}{h_i}=g(y).$$

$$\lim_{i\to\infty}\frac{1}{h_{i}}\bigg{(}\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB_{i}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{i}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{i}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{i}}}f(y)\,dy\,-\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}f(y)\,dy\,\bigg{)}\,=\,\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}Df(y)\cdot g(y)\,dy\,.$$

$$\displaystyle\frac{1}{h_{i}}\bigg{(}\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB_{i}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{i}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{i}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{i}}}f(y)\,dy\,-\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}f(y)\,dy\,\bigg{)}=\frac{1}{h_{i}}\bigg{(}\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptL_{i}(B)}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{L_{i}(B)}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{L_{i}(B)}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{L_{i}(B)}}f(y)\,dy\,-\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}f(y)\,dy\,\bigg{)}$$

$$=$$

$$\approx$$

$$\underset{h\to 0}{\lim }\frac{L_h(y)-y}{h}=g(y).$$

$${\int }_{B}D|f|(y)\cdot g(y)dy=0.$$

$$\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}D|f|(y)\cdot g(y)\,dy\,=\,\lim_{h\to 0}\frac{1}{h}\bigg{(}\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB_{h}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{h}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{h}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{h}}}|f|(y)-\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}|f|(y)\bigg{)}\,.$$

$$D_{v}Mf(x)=\underset{h\to 0}{\lim }\frac{1}{h}(Mf(x+hv)-Mf(x))$$

$$\ge$$

$$=$$

$$D_{v}Mf(x)=\underset{h\to 0}{\lim }\frac{1}{h}(Mf(x)-Mf(x-hv))$$

$$\le$$

$$=$$

$$L_{i}(y)=y+\bigg{(}\frac{|z-(x+h_{i}v)|-|z-x|}{|z-x|}\bigg{)}(y-z)\,.$$

Full text

The author was supported by the Academy of Finland, project no. 292797

Abstract. It is shown for the non-centered Hardy-Littlewood maximal operator $M$ that $\left|\left|DMf\right|\right|_{1}\leq C_{n}\left|\left|Df\right|\right|_{1}$ for all radial functions in $W^{1,1}(\mathbb{R}^{n})\,$.

1. Introduction

The non-centered Hardy-Littlewood maximal operator $M$ is defined by setting for $f\in L^{1}_{loc}(\mathbb{R}^{n})\,$ that

[TABLE]

for every $x\in\mathbb{R}^{n}\,$. The centered version of $M$, denoted by $M_{c}$, is defined by taking the supremum over all balls centered at $x$. The classical theorem of Hardy, Littlewood and Wiener asserts that $M$ (and $M_{c}$) is bounded on $L^{p}(\mathbb{R}^{n})\,$ for $1<p\leq\infty\,$. This result is one of the cornerstones of the harmonic analysis. While the absolute size of a maximal function is usually the principal interest, the applications in Sobolev-spaces and in the potential theory have motivated the active research of the regularity properties of maximal functions. The first observation was made by Kinnunen who verified [Ki] that $M_{c}$ is bounded in Sobolev-space $W^{1,p}(\mathbb{R}^{n})\,$ if $1<p\leq\infty\,$, and inequality

[TABLE]

holds for all $x\in\mathbb{R}^{n}$. The proof is relatively simple and inequality (1.2) (and the boundedness) holds also for $M$ and many other variants.

The most challenging open problem in this field is so called ’$W^{1,1}$-problem’: Does it hold for all $f\in W^{1,1}(\mathbb{R}^{n})\,$, that $Mf\in W^{1,1}(\mathbb{R}^{n})$ and

[TABLE]

This problem has been discussed (and studied) for example in [AlPe], [CaHu], [CaMa], [HO], [HM], [Ku] and [Ta]. The fundamental obstacle is that $M$ is not bounded in $L^{1}$ and therefore inequality (1.2) is not enough to solve the problem. In the case $n=1$ the answer is known to be positive, as was proved by Tanaka [Ta]. For $M_{c}$ the problem turns out to be very complicated also when $n=1$. However, Kurka [Ku] managed to show that the answer is positive also in this case.

The goal of this paper is to develop technology for $W^{1,1}$-problem in higher dimensions, where the problem is still completely open. The known proofs in the one-dimensional case are strongly based on the simplicity of the topology: the crucial trick (in the non-centered case) is that $Mf$ does not have a strict local maximum outside the set $\{Mf(x)=f(x)\}$. This fact is a strong tool when $n=1$ but is far from sufficient for higher dimensions.

The formula for the derivative of the maximal function (see Lemma 2.2 or [L]) has an important role in the paper. It says that if $Mf(x)=\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}|f|$, $|f(x)|<Mf(x)<\infty$, and $Mf$ is differentiable at $x$, then

[TABLE]

From this formula one can see immediately the validity of the estimate (1.2) for $M$. However, since $B$ is exactly the ball which gives the maximal average (for $|f|$), it is expected that one can derive from (1.3) much more sophisticated estimates than (1.2). In Section 2 (Lemma 2.2), we perform basic analysis related to this issue. The key observation we make is that if $B$ is as above, then

[TABLE]

In the backround of this equality stands a more general princinple, concerning other maximal operators as well: if the value of the maximal function is attained to ball (or other permissible object) $B$, then the weighted integral of $|Df|$ over $B$ is zero for a set of weights depending on the maximal operator. We believe that the utilization of this principle is a key for a possible solution of $W^{1,1}$-problem.

As the main result of this paper, we employ equality (1.4) to show that in the case of radial functions the answer to $W^{1,1}$-problem is positive (Theorem 3.11). Even in this case the problem is evidently non-trivial and truly differs from the one-dimensional case. To become convinced about this, consider the important special case where $f$ is radially decreasing ($f(x)=g(|x|)$, where $g:[0,\infty)\to\mathbb{R}$ is decreasing). In this case $Mf$ is radially decreasing as well and $Mf(0)=f(0)$. If $n=1$, these facts immediately imply that $\left|\left|DMf\right|\right|_{1}=\left|\left|Df\right|\right|_{1}$, but if $n\geq 2$ this is definitely not the case: the additional estimates are necessary. This type of estimate for radially decreasing functions can be derived from (1.3) and (1.4), saying that

[TABLE]

By using this inequality, the positive answer to $W^{1,1}$-problem for radially decreasing functions follows straightforwardly by Fubini Theorem (Corollary 3.1).

For general radial functions, inequality (1.5) turns out to hold only if the maximal average is achieved in a ball with radius comparable to $|x|$. To overcome this problem, we study the auxiliary maximal function $M^{I}$, defined for $f\in L^{1}_{loc}(\mathbb{R}^{n})$ by

[TABLE]

and prove (Lemma 3.2) that for all radial $f\in W^{1,1}(\mathbb{R}^{n})$ it holds that

[TABLE]

The proof of this auxiliary result resembles the proof of $W^{1,1}$-problem (for $M$) in the case $n=1$. As the first step, we prove by straightforward calculation that for the ’endpoint operator’ of $M^{I}$, defined by

[TABLE]

it holds that $\left|\left|Df_{/4}\right|\right|_{1}\leq C\left|\left|Df\right|\right|_{1}$ for all $f\in W^{1,1}(\mathbb{R}^{n})$. Recall again the fact that $Mf$ does not have a local maximum in $\{Mf(x)>|f(x)|\}$, leading to the estimate $\left|\left|DMf\right|\right|_{1}\leq\left|\left|Df\right|\right|_{1}$ in the case $n=1$. As a multidimensional counterpart for radial functions, we show that $M^{I}f$ does not have a local maximum in $\{M^{I}f(x)>\max\{|f(x)|,f_{/4}(x)\}\}$ and for every $k\in\mathbb{Z}$ it holds that

[TABLE]

Estimate (1.6) can be easily derived from this fact. The main result follows by combining (1.6) and exploiting the estimate (1.5) in $\{Mf(x)>M^{I}f(x)\}$.

Question

The analysis presented in this paper raises the interest towards the study of the integrability properties of some conditional maximal operators. As an example, (1.3) and (1.4) yield that $|DMf(x)|\leq\widetilde{M}(D|f|)(x)$, where $\widetilde{M}$ is defined for all locally integrable gradient fields $F:\mathbb{R}^{n}\to\mathbb{R}^{n}$ by

[TABLE]

It is clear that $\widetilde{M}F$ is bounded by $M(|F|)$, but does it hold that $\widetilde{M}$ has even better integrability properties than $M$? What about the boundedness in the Hardy-space $H^{1}$ or even in $L^{1}$? Notice that the boundedness of $\widetilde{M}$ in $L^{1}$ would imply the solution to $W^{1,1}$-problem. This problem is almost completely open, even in the case $n=1$. Counterexamples would be highly interesting as well.

Acknowledgements. The author would like to thank Antti Vähäkangas for useful comments on the manuscript and inspiring discussions.

2. Preliminaries and general results

Let us introduce some notation. The boundary of the $n$-dimensional unit ball is denoted by $S^{n-1}$. The $s$-dimensional Hausdorff measure is denoted by $\mathcal{H}^{s}$. The volume of the $n$-dimensional unit ball is denoted by $\omega_{n}$ and the $\mathcal{H}^{n-1}$-measure of $S^{n-1}$ by $\sigma_{n}$. The weak derivative of $f$ (if exists) is denoted by $Df$. If $v\in S^{n-1}$, then

[TABLE]

in the case the limit exists.

Definition 2.1**.**

For $f\in L^{1}_{loc}(\mathbb{R}^{n})$ let

[TABLE]

It is easy to see that if $f\in L^{1}(\mathbb{R}^{n})$ and $|f(x)|<Mf(x)<\infty$, then $\mathcal{B}_{x}\not=\emptyset\,$.

The following lemma is the main result of this section. We point out that below $(6)$ is especially useful in the case of radial functions.

Lemma 2.2**.**

Suppose that $f\in W^{1,1}(\mathbb{R}^{n})$, $Mf(x)>f(x)$ and $Mf$ is differentiable at $x$. Then

1. (1)

For all $v\in S^{n-1}$ and $B\in\mathcal{B}_{x}\,$, it holds that

[TABLE] 2. (2)

If $x\in B$ for some $B\in\mathcal{B}_{x}$, then $DMf(x)=0\,.$ 3. (3)

If $x\in\partial B$, $B=B(z,r)\in\mathcal{B}_{x}$ and $DMf(x)\not=0$, then

[TABLE] 4. (4)

If $B\in\mathcal{B}_{x}$, then

[TABLE] 5. (5)

If $x\in\partial B$, $B=B(z,r)\in\mathcal{B}_{x}$, then

[TABLE] 6. (6)

If $B\in\mathcal{B}_{x}$, then

[TABLE]

The proof of Lemma 2.2 is essentially based on the following auxiliary propositions.

Proposition 2.3**.**

Suppose that $f\in W^{1,1}(\mathbb{R}^{n})$, $B$ is a ball, $h_{i}\in\mathbb{R}$ such that $h_{i}\to 0$ as $i\to\infty$, and $B_{i}=L_{i}(B)$, where $L_{i}$ are affine mappings and

[TABLE]

Then

[TABLE]

Proof.

The proof is a simple calculation:

[TABLE]

if $i\to\infty\,$. ∎

Lemma 2.4**.**

Let $f\in W^{1,1}(\mathbb{R}^{n})$, $x\in\mathbb{R}^{n}$, $B\in\mathcal{B}_{x}$, $\delta>0$, and let $L_{h}$, $h\in[-\delta,\delta]$, be affine mappings such that $x\in L_{h}(\bar{B})$ and

[TABLE]

Then

[TABLE]

Proof.

Let us denote $B_{h}:=L_{h}(B)$. By Proposition 2.3 it holds that

[TABLE]

Since $B\in\mathcal{B}_{x}$ and $x\in\bar{B}_{h}$, the sign of the quantity inside the large parentheses is non-positive for all $h\in[-\delta,\delta]$. However, the sign of $1/h$ depends on the sign of $h$. The conclusion is that the above equality is possible only if (2.12) is valid. ∎

Proof of Lemma 2.2

1. (1)

The claim is counterpart for the formula for $DM_{c}f$, which was first time proved in [L]. Suppose that $B=B(z,r)\in\mathcal{B}_{x}$ and let $B_{h}:=B(z+hv,r)$. Then it holds that

[TABLE]

On the other hand, if $B_{h}:=B(z-hv,r)$, then

[TABLE]

These inequalities imply the claim. 2. (2)

If $B\in\mathcal{B}_{x}$ and $x\in B$, then $y\in B$ if $|y-x|$ is small enough, and thus $Mf(y)\geq Mf(x)$. 3. (3)

Let $B=B(z,r)\in\mathcal{B}_{x}$, $v\in S^{n-1}$ such that $v\cdot(z-x)=0$, and $h_{i}\in(0,\infty)$, $h_{i}\to 0$ as $i\to\infty\,$. Moreover, let us denote $B_{i}:=B(z,|z-(x+h_{i}v)|)$. Then it clearly holds that $x+h_{i}v\in\bar{B}_{i}$ and it is also easy to see that $B_{i}=L_{i}(B)$ for an affine mapping $L_{i}$ given by

[TABLE]

By the assumption $v\cdot(z-x)=0$ it follows that

[TABLE]

Therefore, Proposition 2.3 implies that

[TABLE]

This shows that $D_{v}Mf(x)=0$ for all $v$ orthogonal to $(z-x)$. In particular, it follows that $DMf(x)$ is parallel to $z-x$ or $x-z$. The final claim follows easily by the fact that $Mf(x+h(z-x))\geq Mf(x)$ if $0<h\leq 2$. 4. (4)

Let $B\in\mathcal{B}_{x}$ and $L_{h}(y):=y+h(y-x)\,$, $h\in\mathbb{R}$. Then it holds that $L_{h}$ is affine mapping, $L_{h}(x)=x$, and so $x\in L_{h}(B)=:B_{h}$, and $(L_{h}(y)-y)/h=y-x\,$ for all $h\in\mathbb{R}\,$. Therefore, Lemma 2.4 implies that

[TABLE] 5. (5)

By combining $(1)$, $(3)$ and $(4)$ the claim follows by

[TABLE] 6. (6)

The claim follows from $(1)$ and $(4)$.

$\Box$

3. $W^{1,1}$-problem for radial functions

Radial functions and notation

In what follows, we will interpret a radial function on $\mathbb{R}^{n}$ as a function on $(0,\infty)$ in a natural way. To be more precise, if $f\in W^{1,1}_{loc}(\mathbb{R}^{n})$ is radial, it is well known fact that there exists continuous function $\tilde{f}:(0,\infty)\to\mathbb{R}$ such that $\tilde{f}$ is weakly differentiable,

[TABLE]

and (by a possible redefinition of $f$ in a set of measure zero) for all $t\in(0,\infty)$ it holds that $f(x)=\tilde{f}(t)$ and $D_{x/|x|}f(x)=\tilde{f}^{\prime}(t)$ if $|x|=t$. In what follows, we will simplify the notation and use $f$ to denote $\tilde{f}$ as well. To avoid the possibility of misuderstanding, we usually use variable $t$ and notation $f^{\prime}$ (instead of $Df$) when we are actually working with $\tilde{f}$. We also say that $f$ is radially decreasing if $f$ is radial and $f(t_{1})\leq f(t_{2})$ if $t_{1}>t_{2}$. Notice also that if $f$ is radial then $Mf$ is also radial.

The following result is an easy consequence of Lemma 2.2.

Corollary 3.1**.**

If $f\in W^{1,1}(\mathbb{R}^{n})$ is radially decreasing, then $DMf\in W^{1,1}(\mathbb{R}^{n})$ and $\left|\left|DMf\right|\right|_{1}\leq C_{n}\left|\left|Df\right|\right|_{1}\,$.

Proof.

Since $f$ is radially decreasing, it is easy to show (the rigorous proof is left to the reader) that if $Mf(x)\not=0$ and $B\in\mathcal{B}_{x}$, then $0\in\bar{B}$ and $\bar{B}\subset\bar{B}(0,|x|)$. Especially, we get by Lemma 2.2, $(6)$, that

[TABLE]

Then the claim follows by Fubini theorem:

[TABLE]

∎

In the case of general radial functions, (1.5) is in general valid (and useful) only for those $x$ for which the radius of $B\in\mathcal{B}_{x}$ is comparable to $|x|$. As it was explained in the introduction, the main auxiliary tool in the case of general radial functions is the following result (recall the definition of $M^{I}$ in the introduction):

Lemma 3.2**.**

*If $f\in W^{1,1}(\mathbb{R}^{n})$ is radial, then $M^{I}f\in W^{1,1}(\mathbb{R}^{n})$ and

$\left|\left|DM^{I}f\right|\right|_{1}\leq C_{n}\left|\left|Df\right|\right|_{1}\,$.*

Before the actual proof of this result, we prove several auxiliary results. The first of them is well known.

Proposition 3.3**.**

Suppose that $E\subset\mathbb{R}$ is open. Then there exist disjoint intervals $(a_{i},b_{i})$ such that $E=\cup_{i=1}^{\infty}(a_{i},b_{i})\,$ and $a_{i},b_{i}\in\partial E\cup\{-\infty,\infty\}$ for all $i\in{\mathbb{N}}\,$.

The following auxiliary result is repeatedly utilized in the proof. The result is well known but we express the proof for readers convenience.

Lemma 3.4**.**

Suppose that $\Omega\subset\mathbb{R}^{n}$, $f\in W^{1,1}(\Omega)$ is continuous, $g:\Omega\to\mathbb{R}$ is continuous and weakly differentiable in $E:=\{x\in\Omega:g(x)>f(x)\}\,$, and $\int_{E}|Dg|<\infty\,$. Then $\max\{f,g\}$ is weakly differentiable in $\Omega\,$ and

[TABLE]

Proof.

Suppose that $\phi$ is a smooth test function, compactly supported in $\Omega$, $1\leq i\leq n$, $L(t)=p+te_{i}$, $p\in\mathbb{R}^{n}\,$, and let $L$ denote the line $L(\mathbb{R})$. By Proposition 3.3, $E\cap L$ can be written as a union of disjoint and open (in $\Omega\cap L$) line segments $E_{j}=L((a_{j},b_{j}))$, $j\in{\mathbb{N}}$, such that $L(a_{j}),L(b_{j})\in\partial E$ (with respect to $\Omega\cap L$) or $a_{j}=-\infty$ or $b_{j}=\infty$. In particular, $f(L(a_{j}))=g(L(a_{j}))$ if $a_{j}\not=-\infty$ and $f(L(b_{j}))=g(L(b_{j}))$ if $b_{j}\not=\infty$. Since $\phi$ is compactly supported, it follows that

[TABLE]

Therefore, by using the assumptions for $g$, it holds that

[TABLE]

for all $j\in{\mathbb{N}}\,$. Then

[TABLE]

This implies the claim. ∎

Definition 3.5**.**

Let $f:\Omega\to\mathbb{R}$, where $\Omega\subset\mathbb{R}$ is open. We say that $x$ is a local strict maximum of $f$ in $(a,b)\subset\Omega$, $-\infty\leq a<b\leq\infty$, if there exist $a^{\prime},b^{\prime}\in(a,b)$ such that $a^{\prime}<x<b^{\prime}$, $f(t)\leq f(x)$ if $t\in(a^{\prime},b^{\prime})$, and $\max\{f(a^{\prime}),f(b^{\prime})\}<f(x)$.

Proposition 3.6**.**

Suppose that $f:[a,b]\to\mathbb{R}$ is continuous and $c\in(a,b)$ such that $f(c)>\max\{f(a),f(b)\}$. Then $f$ has a local strict maximum on $(a,c)$.

Proof.

It is easy to see that now any maximum point $c$ ($f(c)=\max f$), which is known to exist, is also a local strict maximum of $f$. ∎

Proposition 3.7**.**

Suppose that $f:[a,b]\to\mathbb{R}$ is continuous and does not have a local strict maximum on $(a,b)$. Then there exists $c\in[a,b]$ such that $f$ is non-increasing on $[a,c]$ and non-decreasing on $[c,b]$.

Proof.

Since $f$ is continuous, we can choose $c\in[a,b]$ such that $f(c)=\min f$. To show that $f$ is non-decreasing on $[c,b]$, let $c<y_{1}<y_{2}<b$ and assume, on the contrary, that $f(y_{2})<f(y_{1})$. This implies that $f(y_{1})>\max\{f(c),f(y_{2})\}$, and thus $f$ has a strict local maximum on $(c,y_{2})$ by Proposition 3.6. This is the desired contradiction. To show that $f$ is non-increasing on $[a,c]$, let $a<y_{1}<y_{2}<c$ and assume, on the contrary, that $f(y_{1})<f(y_{2})$. This implies that $f(y_{2})>\max\{f(y_{1}),f(c)\}$, and thus $f$ has a strict local maximum on $(y_{1},c)$ by Proposition 3.6. This is the desired contradiction. ∎

Let us define for $0<a\leq b<\infty$ the annular domains

[TABLE]

Lemma 3.8**.**

If $f\in W^{1,1}(\mathbb{R}^{n})$ is radial, then $Mf$ does not have a local strict maximum in $\{t\in(0,\infty):\,Mf(t)>f(t)\}\,$.

Proof.

Suppose, on the contrary, that $t_{0}\in(0,\infty)$ is a local strict maximum of $Mf$ and $Mf(t_{0})>f(t_{0})$. Let us choose

[TABLE]

By the definition of the local strict maximum, it follows that $t_{0}\in[t^{-},t^{+}]$ and

[TABLE]

Suppose that $|x|=t_{0}$. Since $Mf(t_{0})>f(t_{0})$, it follows that there exist a ball $B$ such that $Mf(t_{0})=\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}|f|$, $x\in\bar{B}\,$. Suppose first that $B\not\subset A[t^{-},t^{+}]$. In this case there exists ${\varepsilon}>0$ such that $[t^{-}-{\varepsilon},t^{-}]\subset\{|y|:y\in\bar{B}\}$ or $[t^{+},t^{+}+{\varepsilon}]\subset\{|y|:y\in\bar{B}\}$. Especially, it follows by the definition of $M$ that $Mf(t)\geq\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B}}|f|=Mf(t_{0})$ if $t\in[t^{-}-{\varepsilon},t^{-}]$ or $t\in[t^{+},t^{+}+{\varepsilon}]$, respectively. Obviously this contradicts with the choice of $t^{-}$ and $t^{+}$. This verifies that $B\subset A[t^{-},t^{+}]$. Therefore, it holds by (3.14) that

[TABLE]

However, $f(t_{0})<Mf(t_{0})$ also implies that there exists a ball $B^{\prime}$ with positive radius such that $B^{\prime}\subset B$ and $f<Mf(t_{0})$ in $B^{\prime}$. Combining this with (3.15) yields the desired contradiction by

[TABLE]

∎

Recall the definition of $f_{/4}$ (the endpoint operator of $M^{I}$, (1.7)) from the introduction. Before showing the boundedness for $M^{I}$, we have to prove the boundedness for $f_{/4}$.

Proposition 3.9**.**

If $f\in W^{1,1}(\mathbb{R}^{n})$, then $f_{/4}\in W^{1,1}(\mathbb{R}^{n})$ and $\left|\left|Df_{/4}\right|\right|_{1}\leq C_{n}\left|\left|Df\right|\right|_{1}\,$.

Proof.

It is easy to check that $f_{/4}$ is Lipschitz outside the origin. Therefore, it suffices to verify the desired norm estimates for $Df_{/4}$. We will exploit Proposition 2.3. If $x\not=0$, we are going to show that if $h>0$ is small enough and $v\in S^{n-1}$, then

[TABLE]

To show this, we may assume that $f_{/4}(x)>f_{/4}(x+hv)$. Suppose that

[TABLE]

Especially, $x+hv\in\bar{B}_{h}$. Moreover, it is easy to compute that

[TABLE]

Then it follows by Proposition 2.3 that

[TABLE]

This proves (3.16). Then the claim follows (e.g) by using Fubini Theorem: Let us denote below $B_{x}=B(x,\frac{|x|}{2})\,$. By the above estimate,

[TABLE]

∎

The following estimate is well known.

Proposition 3.10**.**

If $f\in W^{1,1}(\mathbb{R}^{n})$ is radial and $0<a<b<\infty$, then

[TABLE]

The proof of Lemma 3.2

Let

[TABLE]

By Lemma 3.4 and Proposition 3.9 it follows that $g\in W^{1,1}(\mathbb{R}^{n})$ and $\left|\left|Dg\right|\right|_{1}\leq C_{n}\left|\left|Df\right|\right|_{1}\,$. Let

[TABLE]

It is well known that mapping $M^{I}f$ is locally Lipschitz in $E$ and, especially, $D(M^{I}f)$ exists in $E$. By Lemma 3.4, it suffices to show that $\int_{E}|DM^{I}f|\leq C_{n}\left|\left|Dg\right|\right|_{1}\,$.

First observe that since $|f|$ is radial, it follows that $M^{I}f$ and $g$ are radial as well, and continuous in $\mathbb{R}^{n}\setminus\{0\}$. In particular, if

[TABLE]

then $x\in E_{k}$ if and only if $|x|\in E^{\mathbb{R}}_{k}$. Since $E_{k}^{\mathbb{R}}$ is open, we can write

[TABLE]

such that $a_{i}<b_{i}$, $(a_{i},b_{i})$ are pairwise disjoint and $a_{i},b_{i}\in\partial E_{k}^{\mathbb{R}}$. In the other words,

[TABLE]

and (by the definition of $E_{k}$) for all $i\in\mathbb{N}$ it holds that

[TABLE]

Moreover, since $M^{I}f>f$ in $E_{k}$, Lemma 3.8 says that $M^{I}f$ does not have a strict local maximum in $E_{k}^{\mathbb{R}}$. In particular, by Proposition 3.7 there exist $c_{i}\in(a_{i},b_{i})$ such that

[TABLE]

Combining this with (3.17) implies that if $2^{-k}<a_{i}<b_{i}<2^{-k+1}$, then

[TABLE]

For the case $a_{i}=2^{-k}$ or $b_{i}=2^{-k+1}$, we employ the fact

[TABLE]

to obtain the estimates ($a_{i}=2^{-k}$ or $b_{i}=2^{-k+1}$)

[TABLE]

Combining these estimates implies that

[TABLE]

Therefore,

[TABLE]

This completes the proof. $\Box$

Then we are ready to prove our main theorem.

Theorem 3.11**.**

If $f\in W^{1,1}(\mathbb{R}^{n})$ is radial, then $Mf\in W^{1,1}(\mathbb{R}^{n})$ and $\left|\left|DMf\right|\right|_{1}\leq C_{n}\left|\left|Df\right|\right|_{1}\,$.

Proof.

Let

[TABLE]

It is well known that $Mf$ is locally Lipschitz in $\{Mf(x)>f(x)\}$, implying the existence of $DMf$ in $\{Mf(x)>f(x)\}$. Since $Mf\geq M^{I}f(x)$, it holds that $Mf(x)=\max\{Mf(x),M^{I}f(x)\}$. Therefore, the theorem follows by Lemmas 3.4 and 3.2, if we can show that

[TABLE]

To show this, observe first that for all $x\in E$ there exist $r_{x}>\frac{|x|}{4}$ and $z_{x}\in\mathbb{R}^{n}$ such that $x\in B(z_{x},r_{x})\in\mathcal{B}_{x}$. Moreover, since $DMf(x)\not=0$, Lemma 2.2 ($(2)$ and $(3)$) says that $x\in\partial B(z_{x},r_{x})$ and $DMf(x)/|DMf(x)|=(z_{x}-x)/|z_{x}-x|$. On the other hand, $Mf$ is radial and so $DMf(x)/|DMf(x)|=\pm x/|x|$. We conclude that

[TABLE]

Observe that $r_{x}=|c_{x}x-x|=|c_{x}-1||x|>|x|/4$ by the assumption, and thus $|c_{x}-1|>1/4\,$. Moreover, it holds that $c_{x}\geq-1$. To see this, observe that if $c_{x}<-1$, then $-x\in B_{x}$ and, since $Mf$ is radial, $B_{x}\in\mathcal{B}_{-x}$, implying by Lemma 2.2 that $0=DMf(-x)=DMf(x)$, which contradicts with the assumption $x\in E$. Summing up, we can write $E=E_{+}\cup E_{-}$, where

[TABLE]

We are going to use different estimates for $DMf(x)$ in $E_{+}$ and $E_{-}\,$. Since $|DMf(x)|=|DMf(x)\cdot\frac{x}{|x|}|$, it follows from Lemma 2.2 (2.9) that

[TABLE]

This estimate will be used in $E_{-}$, while in $E_{+}$ we will use (easier) estimate $|DMf(x)|\leq\mathchoice{\mathop{\kern 1.79997pt\vrule width=5.40005pt,height=2.7pt,depth=-2.24997pt\kern-7.20003pt\intop}\nolimits_{\kern-2.51996ptB_{x}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{x}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{x}}}{\mathop{\kern 0.90005pt\vrule width=4.5pt,height=2.7pt,depth=-2.33997pt\kern-5.40005pt\intop}\nolimits_{B_{x}}}|D|f||\,$ (Lemma 2.2, $(1)$). We get that

[TABLE]

If $y\in B_{x}$ and $x\in E_{+}$, it follows from the definition of $E_{+}$ that $|x|\leq|y|$. Moreover, $y\in B_{x}$ and $x\in E$ imply also that $r_{x}\geq\max\{|y-x|,\frac{|x|}{4}\}\geq\frac{|y|}{6}\,$. This implies the estimate

[TABLE]

On the other hand, if $x\in E_{-}$, then $\,-1\leq c_{x}<3/4$ especially implies that $B_{x}\subset B(0,|x|)$. Therefore, if $x\in E_{-}$ and $y\in B_{x}$, then $y\in B(0,|x|)$, and thus $|x|\geq|y|\,$. Recall also that $r_{x}\geq\frac{|x|}{4}\,$. Combining these yields that

[TABLE]

for all $y\in\mathbb{R}^{n}\,$. This completes the proof. ∎