14 Fractional Integral and Hardy-Littlewood-Sobolev Inequality

Theorem

If $T:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a Lipschitz map, that is, there exists $c$ such that $|Tx-Ty|⩽c|x-y|$, $\forall x,y\in {\mathbb{R}}^n$, then $T$ maps measurable sets into measurable sets.

\begin{proof} We can prove that:

• $T$ maps $F_{\sigma }$-set to $F_{\sigma }$-set, because continuous map preserves compact sets and $F_{\sigma }$-set can be covered by countable union of compact sets;
• $T$ maps sets of measure zero into sets of measure zero by considering the cover of sets with cubes.

Recall that for any measurable set $E$, there exists $F_{\sigma }$-set $H$ and $Z$ of measure zero such that $E=H\cup Z$. Then we finish the proof. \end{proof}

Lemma

If $f(x)$ is measurable in ${\mathbb{R}}^n$, then the function $F(x,t)=f(x-t)$ is measurable in ${\mathbb{R}}^n\times {\mathbb{R}}^n={\mathbb{R}}^{2n}$.

\begin{proof} Define $g(x,t)=f(x)$, then $g$ is measurable because $g^{-1}((a,\infty ))=\{x\in {\mathbb{R}}^n:f(x)>a\}\times {\mathbb{R}}^n$. Define $T(x,t)=(x-t,x+t)$, then $T$ is Lipschitz. Note that $F(x,t)=g\circ T(x,t)$ and $F^{-1}((a,\infty ))=T^{-1}\circ g^{-1}((a,\infty ))$ is measurable by ^179d20. Thus $F$ is measurable. \end{proof}

Remark. If we take $T(x,t)=(x-t,x+t)$, the proof also works. ChatGPT tells me the choice of $T(x,t)=(x-t,x+t)$ might simply reflect a convention or aesthetic preference.

Lemma

Assume that $f,g\in L^1({\mathbb{R}}^n)$.

• $f\ast g\in L^1\left({\mathbb{R}}^n\right)$, $||f\ast g|{|}_1⩽||f|{|}_1||g|{|}_1$ and

$${\int }_{{\mathbb{R}}^n}f\ast gdx=\left({\int }_{{\mathbb{R}}^n}fdx\right)\left({\int }_{{\mathbb{R}}^n}gdx\right).$$

• If $1<p⩽\infty ,f\in L^1$, and $g\in L^p$, then $\Vert f\ast g{\Vert }_p⩽\Vert f{\Vert }_1\Vert g{\Vert }_p$.

\begin{proof} i) Note that if $f,g⩾0$, then

$$||f\ast g|{|}_1={\int }_{{\mathbb{R}}^n}f\ast gdx=||f|{|}_1||g|{|}_1.$$

Then for any $f,g\in L^1({\mathbb{R}}^n)$, we have $||f\ast g|{|}_1⩽||f|{|}_1||g|{|}_1$ and so $f\ast g\in L^1({\mathbb{R}}^n)$. Then by Fubini’s theorem, there is ${\int }_{{\mathbb{R}}^n}f\ast gdx=\left({\int }_{{\mathbb{R}}^n}fdx\right)\left({\int }_{{\mathbb{R}}^n}gdx\right)$.

ii) Let $1⩽p<\infty$, and let $q$ be the conjugate exponent to $p$. By Holder’s inequality,

$$\begin{array}{rl}& \left|{\int }_{{\mathbb{R}}^n}f(y)g(x-y)dy\right|\\ ⩽& {\int }_{{\mathbb{R}}^n}|f(y){|}^{\frac{1}{q}}|f(y){|}^{1-\frac{1}{q}}|g(x-y)|dy\\ ⩽& {\left({\int }_{{\mathbb{R}}^n}|f(y)|dy\right)}^{1/q}{\left({\int }_{{\mathbb{R}}^n}|f(y){|}^{p\left(1-\frac{1}{4}\right)}|g(x-y){|}^{p}dy\right)}^{1/p}.\end{array}$$

Then using the Fubini theorem, there is

$$\begin{array}{rl}{\int }_{{\mathbb{R}}^n}|f\ast g(x){|}^{p}dx& ⩽{\int }_{{\mathbb{R}}^n}{\left({\int }_{{\mathbb{R}}^n}|f(y)|dy\right)}^{p/q}{\int }_{{\mathbb{R}}^n}|f(y)\Vert g(x-y){|}^{p}dydx\\ & =\Vert f{\Vert }_1^{p/q}\Vert g{\Vert }_p^p{\int }_{{\mathbb{R}}^n}|f(y)|dy\\ & =\Vert f{\Vert }_1^{1+p/q}\Vert g{\Vert }_p^p\\ & =\Vert f{\Vert }_1^p\Vert g{\Vert }_p^p\end{array}$$

and then taking $p^{th}$ roots gives our desire result.

When $p=\infty$, for any given $x$, we have

$$f\ast g(x)={\int }_{{\mathbb{R}}^n}f(y)g(x-y)dy⩽\Vert g{\Vert }_{\infty }{\int }_{{\mathbb{R}}^n}|f(y)|dy=\Vert f{\Vert }_1\Vert g{\Vert }_{\infty }.$$

By the arbitrary of $x$, we finish the proof. \end{proof}

The definition of $I_{\alpha }f$ is written in this remark.

Hardy-Littlewood-Sobolev

Let $0<\alpha <n,1⩽p<\frac{n}{\alpha },\frac{1}{q}=\frac{1}{p}-\frac{\alpha }{n}$. Then for every $f\in L^p\left({\mathbb{R}}^n\right),I_{\alpha }f$ is measurable in ${\mathbb{R}}^n$. Moreover,

• if $1<p<\frac{n}{\alpha }$, then ${\Vert I_{\alpha }f\Vert }_q⩽c\Vert f{\Vert }_p$ for a constant $c$ that depends only on $\alpha ,n$, and $p$;
• if $p=1$, then

$$\Vert I_{\alpha }f{\Vert }_{q,w}=\underset{\lambda >0}{\sup }\lambda {\left|\left\{x\in {\mathbb{R}}^n:\left|I_{\alpha }f(x)\right|>\lambda \right\}\right|}^{1/q}⩽c\Vert f{\Vert }_1.$$

\begin{proof} See here.

读了一遍，大概拿中文写个大纲：

• 先考虑nonnegative f的情况，估计 $I_{\alpha }f$ 的大小。
  • 把 $I_{\alpha }f$ 分成 $|x-y|<\delta$ 和 $|x-y|>\delta$ 两部分来算积分，后者用Holder估计，前者用 the HardyLittlewood maximal function 在一系列圆环来做（因为这样可以避开奇点）
  • 取一个 $\delta$ 把前面两部分得到的丑陋的两个常数操作一下，得到 $I_{\alpha }f(x)\le c\Vert f{\Vert }_{p^{\frac{\alpha P}{N^n}}}^{\ast }{f}^{\ast }(x{)}^{1-\frac{\alpha p}{n}}$.
• 用^4tmmbv算 $||I_{\alpha }f|{|}_q$. 现在我们完成了 i) 的对于 nonnegative 的证明。
• 当 $p=1$ 时，还是用 $I_{\alpha }f$ 的估计，用一下 Hardy-Littlewood. 现在我们完成了 ii) 的对于 nonnegative 的证明。
• 现在考虑一般的 $f$，注意到 ${\Vert I_{\alpha }f\Vert }_q\le {\Vert I_{\alpha }(|f|)\Vert }_q$ 就可以了。 \end{proof}

Remark. Here is a motivation of ^z20dwo. Define

$$I_{\alpha }f(x)={\int }_{{\mathbb{R}}^n}f(y)\frac{1}{|x-y{|}^{n-\alpha }}dy,x\in {\mathbb{R}}^n,$$

and note that it is a convolution of $f$ and $|x{|}^{\alpha -n}$. By Fubini’s theorem, we know ${\int }_{{\mathbb{R}}^n}f\ast gdx=\left({\int }_{{\mathbb{R}}^n}fdx\right)\left({\int }_{{\mathbb{R}}^n}gdx\right)$ and so $I_{\alpha }f$ exists but it is not necessary finite.

We aim to find $p$ and $q$ such that $I_{\alpha }$ is a bounded operator from $L^p$ to $L^q$. It will turn out that the values of $p$ and $q$ for which the norm inequality

$${\Vert I_{\alpha }f\Vert }_q⩽c\Vert f{\Vert }_p\text{ for all }f\in L^p\left({\mathbb{R}}^n\right)\text{\hspace{1em}(*)}$$

is true, for some $c$ independent of $f$, are limited to

$$1<p<\frac{n}{\alpha }\text{ and }\frac{1}{q}=\frac{1}{p}-\frac{\alpha }{n}.$$

To explain it, we need to verify the following observation:

• If $p⩾\frac{n}{\alpha }$, there exists $f\in L^p\left({\mathbb{R}}^n\right)$ such that $I_{\alpha }f=\infty$ everywhere in ${\mathbb{R}}^n$. That is, $(\ast )$ cannot hold if $p⩾\frac{n}{\alpha }$.
• If $p=1$ and $\frac{1}{q}=1-\frac{\alpha }{n}$, there exists $f\in L^1\left({\mathbb{R}}^n\right)$ such that ${\Vert I_{\alpha }f\Vert }_q=\infty$. Thus, $(\ast )$ fails when $p=1$ and $q=n/(n-\alpha )$.
• If $1⩽p<\frac{n}{\alpha }$, the only value of $q$ for which $(\ast )$ can possibly hold for all $f\in L^p\left({\mathbb{R}}^n\right)$ satisfies $\frac{1}{q}=\frac{1}{p}-\frac{\alpha }{n}$.

The computation is omitted.

Remark. 这个证明不考。