## Signing Off

Dear All,

You may have noticed that this blog has not been updated in several months. I thought I should announce that I will no longer be updating and/or maintaining this blog (this includes responding to comments). For the time being the content on this site will remain available, but that may change. I’m mulling the idea of creating a new blog more relevant to my graduate studies and interests (harmonic analysis, pde, etc.), which would be at a higher technical level. But nothing is in the works at this time. One final time, I thank you all for your readership.

Sincerely,
Matt

## Open Question on Sufficiency To Be a Density Basis

This post is adapted from a question I posted on Math.StackExchange approximately two weeks ago, which has not received a satisfactory answer. If you have a solution or a counterexample, please add it to Math.SE instead of here.

Let ${\mathcal{B}}$ be a differentiation basis. Given a measurable set ${E}$, we say that ${\mathcal{V}\subset\mathcal{B}}$ is a Vitali covering of ${A}$ if for every ${x\in E}$ there exists a sequence ${\left\{R_{k}(x)\right\}\subset\mathcal{V}}$ containing and with diameter tending to zero.

Question 1 Suppose that for every measurable set ${A}$ of finite measure, every Vitali covering ${\mathcal{V}}$ of ${A}$, and every ${\varepsilon>0}$, there exists a countable subcollection ${\left\{R_{k}\right\}\subset\mathcal{V}}$ with the following properties:

1. (Underflow condition) ${\left|A\setminus\bigcup R_{k}\right|=0}$
2. (Overflow condition) ${\left|\bigcup R_{k}\setminus A\right|\leq\varepsilon}$
3. (Overlap condition) ${\left\|\sum\chi_{R_{k}}\right\|_{L^{1}}\leq C\left|A\right|}$

where ${C}$ is a positive constant independent of ${A}$, ${\mathcal{V}}$, and ${\varepsilon}$. Does it follow that ${\mathcal{B}}$ satisfies the density property

$\displaystyle \lim_{k\rightarrow\infty}\dfrac{\left|A\cap R_{k}\right|}{\left|R_{k}\right|}=\chi_{A}(x) \ \text{ a.e.} (*)$

where ${x\in R_{k}\in\mathcal{B}}$ for all ${k}$ and the diameter of ${R_{k}}$ tends to zero.

My motivation for posing this question is that when ${q>1}$ and ${1/p+1/q=1}$, a variant of this theorem holds when replace the overlap condition above with ${\left\|\sum\chi_{R_{k}}\right\|_{L^{q}}\leq C\left|A\right|^{1/q}}$ and (*) by

$\displaystyle \lim_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f=f(x) \ \text{ a.e.},$

for every sequence ${\left\{R_{k}\right\}}$ as above and ${f\in L_{loc}^{p}}$. (There is no issue in considering only simple functions above, as opposed to the entire space ${L_{loc}^{\infty}}$, since the differentiaton of the integrals of each space are equivalent.) In fact, the above three conditions are a special case of what is called the ${V_{q}}$ (${1\leq q\leq\infty}$) covering property:

Definition 1 A differentiation basis ${\mathcal{B}}$ has the ${V_{q}}$ covering property if for every measurable set ${A}$ with ${\left|A\right|<\infty}$, every Vitali covering ${\mathcal{V}}$ of ${A}$, and every ${\varepsilon>0}$, there exists a countable subcollection ${\left\{R_{k}\right\}\subset\mathcal{V}}$ satisfying

1. ${\left|A\setminus\bigcup R_{k}\right|=0}$
2. ${\left|\bigcup R_{k}\setminus A\right|\leq\varepsilon}$
3. ${\left\|\sum\chi_{R_{k}}\right\|_{L^{q}}\leq C\left|A\right|^{1/q}}$

where ${C}$ is a constant independent of ${A}$, ${\mathcal{V}}$, and ${\varepsilon}$.

Theorem 2 If ${\mathcal{B}}$ satisfies the ${V_{q}}$ property, ${1, then ${\mathcal{B}}$ differentiates the integrals of functions in ${L_{loc}^{p}({{\mathbb R}}^{n})}$.

Briefly we introduce the notation

$\displaystyle \overline{D}(\int f,x):=\sup_{\left\{R_{k}\right\}}\limsup_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f \ \underline{D}(\int f,x):=\inf_{\left\{R_{k}\right\}}\liminf_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}f$

where the supremum and infimum are taken over all sequences in ${\left\{R_{k}\right\}}$ containing ${x}$ and with diameter tending to zero. When both ${\overline{D}(\int f,x)}$ and ${\underline{D}(\int f,x)}$ are equal, we simply write ${D(\int f,x)}$.

Proof: It suffices to show that the set

$\displaystyle E_{\alpha}:=\left\{x\in{{\mathbb R}}^{n}:\left|\overline{D}(\int f, x)-f(x)\right|>\alpha\right\}$

has measue zero for all ${\alpha>0}$. Without loss of generality we may assume that ${f}$ has compact support. Let ${g}$ be a continuous function of compact support, such that ${\left\|f-g\right\|_{L^{p}}<\varepsilon}$. Observe that

$\displaystyle \limsup_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}\left|f(y)-f(x)\right|dy\leq\limsup_{k\rightarrow\infty}\dfrac{1}{\left|R_{k}\right|}\int_{R_{k}}\left|f-g\right|+\left|f(x)-g(x)\right|$

So if ${x\in E_{\alpha}}$, then ${x\in F_{\alpha/2}\cup G_{\alpha/2}}$, where

$\displaystyle F_{\alpha/2}:=\left\{x\in{{\mathbb R}}^{n}:\overline{D}(\int\left|f-g\right|,x)>\alpha/2\right\},\indent G_{\alpha/2}:=\left\{x\in{{\mathbb R}}^{n}:\left|f(x)-g(x)\right|>\alpha/2\right\}$

Intersecting ${F_{\alpha/2}}$ with a compact set if necessary, we may assume that ${\left|F_{\alpha/2}\right|<\infty}$. By definition, for each ${x\in F_{\alpha/2}}$ there exists a sequence ${\left\{R_{k}^{x}\right\}\subset\mathcal{B}}$ containing ${x}$ and with diameter tending to zero. The collection of these sets forms a Vitali cover of ${F_{\alpha/2}}$, whence there exists a subcollection ${\left\{R_{k}\right\}}$ satisfying the ${V_{q}}$ conditions. By Hölder’s inequality,

$\displaystyle \begin{array}{lcl}\displaystyle\left|F_{\alpha/2}\right|\leq\sum_{k}\left|R_{k}\right|\leq\dfrac{2}{\alpha}\int_{\bigcup R_{k}}\left|f-g\right|\sum\chi_{R_{k}}&\leq&\displaystyle\dfrac{2}{\alpha}\left\|f-g\right\|_{L^{p}}\left\|\sum\chi_{R_{k}}\right\|_{L^{q}}\\[2 em] &\leq&\displaystyle\dfrac{2C}{\alpha}\varepsilon\left|F_{\alpha/2}\right|^{1/q}\end{array}\$

Dividing both sides of the inequality by ${\left|F_{\alpha/2}\right|^{1/q}}$ gives the desired estimate. By Chebyshev’s inequality,

$\displaystyle \left|G_{\alpha/2}\right|\leq\left(\dfrac{2}{\alpha}\left\|f-g\right\|_{L^{p}}\right)^{p}\leq\left(\dfrac{2}{\alpha}\varepsilon\right)^{p}$

Putting these two estimates together and noting that ${\varepsilon}$ was arbitrary completes the proof. The proof that ${\underline{D}(\int f,x)=f(x)}$ a.e. is completely analogous. $\Box$

It is easy to see that the above proof breaks down in the endpoint case ${q=1}$. Continuous functions are not dense in ${L^{\infty}}$, so an ${L^{1}}$ estimate for ${\sum\chi_{R_{k}}}$ and Hölder’s inequality is not good enough. If we had some “better” control over the overlap of the ${R_{k}}$, then we would be able to show that the density property holds.

Suppose that instead of the overlap condition for the ${V_{q}}$ covering property stated above, we have condition 3′

$\displaystyle \left\|\sum\chi_{R_{k}}-\chi_{\bigcup R_{k}}\right\|_{L^{q}}<\varepsilon$

Given a measurable set ${A}$ with ${\left|A\right|<\infty}$ and ${0<\alpha<1}$, we will show that

$\displaystyle E_{\alpha,N}:=\left\{x\in A^{c} : \left|x\right|\leq N, \exists\left\{R_{k}(x)\right\}\text{ s.t. }\dfrac{\left|R_{k}(x)\cap A\right|}{\left|R_{k}(x)\right|}>\alpha \ \forall k\right\}$

has measure zero. It is clear that the union of all ${\left\{R_{k}(x)\right\}}$ forms a Vitali cover of ${E_{\alpha,N}}$, whence there exists a countable subcollection satisfying conditions 1,2, and 3′. Observe that

$\displaystyle \begin{array}{lcl} \displaystyle\left|E_{\alpha,N}\right|\leq\sum_{k}\left|R_{k}\right|\leq\alpha^{-1}\int_{A\cap\bigcup R_{k}}\sum\chi_{R_{k}}&=&\displaystyle\alpha^{-1}\int_{A}\left[\sum\chi_{R_{k}}-\chi_{\bigcup R_{k}}\right]+\alpha^{-1}\left|A\cap\bigcup R_{k}\right|\\ [2 em]&\leq&\displaystyle\alpha^{-1}\varepsilon+\alpha^{-1}\left|A\cap\bigcup R_{k}\right|\\ [2 em]&\leq&\displaystyle\alpha^{-1}\varepsilon+\alpha^{-1}\underbrace{\left|\bigcup R_{k}\setminus E_{\alpha,N}\right|}_{\leq\varepsilon}, \end{array}$

since ${A\subset E_{\alpha,N}^{c}}$. Since ${\varepsilon>0}$ was arbitrary, we conclude that ${\left|E_{\alpha,N}\right|=0}$, whence ${E_{\alpha}:=\bigcup_{N}E_{\alpha,N}}$ has measure zero. We conclude that ${D(\chi_{A},x)=0}$ for a.e. ${x\in A^{c}}$. We apply the same argument to the set ${A'=A^{c}}$ and conclude that ${D(\chi_{A^{c}},x)=0}$ for a.e. ${x\in A}$, which implies ${D(\chi_{A},x)=1}$ for a.e. ${x\in A}$.

If we could somehow pass from condition 3 to a countable subcollection satisfying conditions 1,2 and 3′, then the question would be resolved as the argument above shows. But it is not clear to this author how to do that, nor do any potential counterexamples come to mind.

## A Differentiation Basis without the Vitali Covering Property

— 1. Introduction —

A differentiation basis ${\mathcal{B}}$ is a fine collection of bounded measurable sets ${R}$; i.e. for every ${x\in{{\mathbb R}}^{n}}$ there exists a sequence ${\left\{R_{k}\right\}\subset\mathcal{B}(x)}$ with diameter tending to zero. Classical examples of differentiation bases are the collections of balls and cubes. Since measures are ${\sigma}$-additive for countable collections of disjoint measurable sets, we are often interested in extracting disjoint subcollections which cover or “almost cover” some set. As is well-known, balls (cubes) satisfy a strong covering property proven by Vitali: given a measurable set ${A}$ with finite (Lebesgue) measure, there exists a countable disjoint collection ${\mathcal{G}}$ of balls (cubes) satisfying

$\displaystyle \left|A\triangle\bigcup_{B\in\mathcal{G}}B\right|=0$

We can abstract this covering property, hereafterto referred to as the Vitali covering property, and ask which differentiation bases possess it. It turns out that this question is intimately linked to the differentiation of integrals of functions (in the sense of Lebesgue’s theorem), but we save the exploration of that topic for another day.

In today’s post, we are going to construct an example of a differentiation basis ${\mathcal{B}}$ which does not possess the Vitali covering property: there exists a set of finite measure which cannot be almost covered (i.e. modulo a null set) by any countable disjoint subcollection of ${\mathcal{B}}$. The proof is not difficult, but perhaps notationally cumbersome. The construction rests on a lemma which tells us that we can cover a set ${A}$ of finite measure by disjoint homothetic copies of any compact set ${K}$, whose diameters can be taken to be arbitrarily uniformly “small”.

Section 2 reviews the classical Vitali covering theorem for nondegenerate closed balls in ${{{\mathbb R}}^{n}}$. The reader is welcome to skip over it to Section 3, in which we prove the aforementioned lemma and construct the counterexample. The material for this post is mostly adapted from [L. Evans and F. Gariepy Measure Theory and Fine Properties of Functions] and [M.D. Guzman, Differentiation of Integrals in ${{{\mathbb R}}^{n}}$]. The latter reference is highly recommended to the reader further interested in the covering properties of differentiation bases.

— 2. Vitali Covering Lemma —

For a fixed constant ${C>3}$, we let ${\hat{B}}$ to denote the concentric closed ball with radius ${C}$ times the radius of ${B}$. Explanation of the condition on ${C}$ is provided below.

Lemma 1 Let ${\mathcal{F}}$ be a collection of nondegenerate closed balls in ${{{\mathbb R}}^{n}}$ with ${\delta:=\sup\left\{\text{ diam} B : B\in\mathcal{F}\right\}<\infty}$. Then there exists a countable family ${\mathcal{G}}$ of disjoint balls in ${\mathcal{F}}$ such that for every ${B\in\mathcal{F}}$ there exists a ball ${B'\in\mathcal{G}}$ with ${B\cap B'\neq\emptyset}$ and ${B\subset\widehat{B'}}$. In particular,

$\displaystyle \bigcup_{B\in\mathcal{F}}B\subset\bigcup_{B\in\mathcal{G}}\hat{B}$

For the proof of the lemma, we remind the reader that a collection of nondegenerate disjoint balls in ${{{\mathbb R}}^{n}}$ is necessarily countable by the countability of the rationals.

Proof: We partition ${\mathcal{F}}$ by setting ${\mathcal{F}_{j}:=\left\{B\in\mathcal{F} : \delta/c^{j}<\text{ diam} B\leq \delta/c^{j-1}\right\}}$ for all ${j\in\mathbb{N}}$, where ${C=2c+1>3}$. Let ${\mathcal{G}_{1}}$ be a maximal (with respect to inclusion) disjoint collection of balls in ${\mathcal{F}_{1}}$. Assuming we have defined ${\mathcal{G}_{1},\ldots,\mathcal{G}_{k-1}}$, we define ${\mathcal{G}_{k}}$ to be any maximal disjoint subcollection of

$\displaystyle \left\{B\in\mathcal{F}_{k} : B\cap B'=\emptyset \ \forall B'\in\bigcup_{j=1}^{k-1}\mathcal{G}_{j}\right\}$

I.e. the largest collection of disjoint balls in ${\mathcal{F}_{k}}$ which do not intersect any of the balls in the previous ${G_{j}}$. Set ${\mathcal{G}:=\bigcup_{j=1}^{\infty}\mathcal{G}_{j}}$, which is by construction a countable collection of disjoint balls in ${\mathcal{F}}$.

We claim that for every ${B\in\mathcal{F}}$, there exists a ball ${B'\in\mathcal{G}}$ such that ${B\cap B'\neq\emptyset}$ and ${B\subset\widehat{B'}}$. Indeed, ${B\in\mathcal{F}_{k}}$, for some ${k}$, whence by the maximality of ${\mathcal{G}_{k}}$, ${B\cap B'\neq\emptyset}$ for some ${B'\in\mathcal{G}_{k}}$. It follows from the triangle inequality that ${B}$ is contained in the ball concentric to ${B'}$ of radius

$\displaystyle r=\text{ diam}(B')/2+\text{ diam}(B)\leq \text{ diam}(B')/2+c(\delta/c^{j})<\text{ diam}(B')(1/2+c)$

$\Box$

Lemma 2 Let ${\mathcal{F}}$ be a fine cover of a measurable set ${A}$ by closed balls with uniformly bounded diameter. Then there exists a countable subcollection ${\mathcal{G}}$ of disjoint balls in ${\mathcal{F}}$ such that for every finite subset ${\left\{B_{1},\ldots,B_{m}\right\}\subset\mathcal{F}}$,

$\displaystyle A\setminus\bigcup_{j=1}^{m}B_{j}\subset\bigcup_{B\in\mathcal{G}\setminus\left\{B_{1},\ldots,B_{m}\right\}}\hat{B}$

Proof: Let ${\mathcal{G}}$ be as in Lemma 1, and let ${\left\{B_{1},\ldots,B_{m}\right\}\subset\mathcal{F}}$. If ${A\subset\bigcup_{i=1}^{m}B_{i}}$, then there is nothing to prove. Otherwise, let ${x\in A\setminus\bigcup_{i=1}^{m}B_{i}}$. Since ${\bigcup_{i=1}^{m}B_{i}}$ is closed and ${\mathcal{F}}$ is fine, there exists a ball ${B\in\mathcal{F}}$ containing ${x}$ and ${B\cap B_{i}=\emptyset}$ for ${1\leq i\leq m}$. But there exists ${B'\in\mathcal{G}}$ with ${B\cap B'\neq\emptyset}$, whence ${B\subset\widehat{B'}}$. $\Box$

Theorem 3 Let ${U\subset{{\mathbb R}}^{n}}$ be open and ${\delta>0}$. With ${\mathcal{F}}$ as above, there exists a countable subcollection ${\mathcal{G}}$ of disjoint closed balls in ${U}$ such that ${\text{ diam} B\leq\delta}$ for all ${B\in\mathcal{G}}$ and

$\displaystyle \left|U\setminus\bigcup_{B\in\mathcal{G}}B\right|=0$

Proof: We first assume that ${\left|U\right|<\infty}$. Fix ${1-C^{-n}<\theta<1}$. We claim that there is a finite collection ${\left\{B_{i}\right\}_{i=1}^{M_{1}}}$ of disjoint closed balls in ${U}$ such that ${\text{ diam}(B_{i})<\delta}$ for all ${1\leq i\leq M_{1}}$ and

$\displaystyle \left|U\setminus\bigcup_{i=1}^{M_{1}}B_{i}\right|\leq\theta\left|U\right|$

Indeed, let ${\mathcal{F}_{1}:=\left\{B : B \text{ is an open ball}, B\subset U, \text{ diam} B<\delta\right\}}$. By Lemma 1, there exists a countable disjoint subcollection ${\mathcal{G}_{1}\subset\mathcal{F}_{1}}$ satisfying

$\displaystyle U\subset\bigcup_{B\in\mathcal{G}_{1}}\hat{B}$

Whence by the dilation property of Lebesgue measure and ${\sigma}$-additivity,

$\displaystyle \left|U\right|\leq\sum_{B\in\mathcal{G}_{1}}\left|\hat{B}\right|=C^{n}\sum_{B\in\mathcal{G}_{1}}\left|B\right|=C^{n}\left|\bigcup_{B\in\mathcal{G}_{1}}B\right|$

Therefore

$\displaystyle \left|U\setminus\bigcup_{B\in\mathcal{G}_{1}}B\right|\leq\left(1-\dfrac{1}{C^{n}}\right)\left|U\right|<\theta\left|U\right|$

By the continuity of measure, there exists an ${M_{1}}$ which satisfies the claim above.

Now set ${U_{2}:=U\setminus\bigcup_{i=1}^{M_{1}}B_{i}}$ and

$\displaystyle \mathcal{F}_{2}:=\left\{B : B\text{ is open}, B\subset U_{2}, \text{ diam} B<\delta\right\}$

Repeating the argument, we can find finitely many disjoint balls ${B_{M_{1}+1},\ldots,B_{M_{2}}\in\mathcal{F}_{2}}$ satisfying

$\displaystyle \left|U\setminus\bigcup_{i=1}^{M_{2}}\right|=\left|U_{2}\setminus\bigcup_{i=M_{1}+1}^{M_{2}}B_{i}\right|\leq\theta\left|U_{2}\right|\leq\theta^{2}\left|U\right|$

We continue in this fashion to obtain a countable collection of disjoint balls and an increasing sequence ${(M_{k})}$ such that

$\displaystyle \left|U\setminus\bigcup_{i=1}^{M_{k}}B_{i}\right|\leq\theta^{k}\left|U\right|,\indent\forall k\in\mathbb{N}$

Since ${\theta^{k}\rightarrow 0}$ as ${k\rightarrow\infty}$, the proof is complete.

If ${\left|U\right|=\infty}$, then we can apply the preceding argument to the disjoint sets ${U_{m}:=\left\{x\in U: m-1<\left|x\right|\leq m\right\}}$, for ${m\geq 1}$. $\Box$

As an aside, we remark that ${3}$ cannot be taken as the constant in Vitali’s lemma, as the following counterexample in one dimension due to O. Schramm shows that it cannot. Let ${\mathcal{F}:=\left\{B(x,r):\left|x\right|< 1/2,\left|x\right|. Observe that every ball ${B\in\mathcal{F}}$ contains the origin, hence any disjoint subcollection of ${\mathcal{F}}$ consists of one ball. It is clear that ${\bigcup_{\mathcal{F}}B=(-1,1)}$, but ${3B}$ is a proper subset of ${(-1,1)}$.

— 3. Counterexample —

We present a version of a counterexample due to H. Bohr which has been simplified by M.D. Guzman. At the end of the section, we will make some remarks on how to extend the counterexample to ${n}$ dimensions, when ${n\geq 2}$.

Lemma 4  Let ${U\subset{{\mathbb R}}^{n}}$ be any bounded open set, and let ${K}$ be a compact set with positive measure. For ${r>0}$ there exists a disjoint sequence ${\left\{K_{k}\right\}}$ of sets homothetic to ${K}$ and contained in ${U}$ such that ${\left|U\setminus\bigcup K_{k}\right|=0}$ and ${\text{ diam}{K_{k}}.

Proof: Let ${Q}$ be a half-open cube such that ${K}$ is contained in the interior of ${Q}$, and let ${\alpha\left|Q\right|=\left|K\right|}$ with ${0<\alpha<1}$. Using the hypothesis that ${U}$ is open together with a stopping-time argument, we can partition ${U}$ into a sequence of disjoint half-open cubes ${\left\{Q_{j}\right\}}$ of diameter less than ${r}$. For each cube ${Q_{j}}$, let ${T_{j}}$ be the homothecy mapping ${Q}$ onto ${Q_{j}}$, and let ${K_{j}'=T_{j}(K)}$. Note that by construction ${K_{j}'\subset Q_{j}}$. For an integer ${N_{1}\geq 1}$, define an open set ${U_{1}:=U\setminus\bigcup_{i=1}^{N_{1}}K_{j}'}$ and observe that

$\displaystyle \begin{array}{lcl}\displaystyle\left|U_{1}\right|=\left|\bigcup_{j=1}^{\infty}Q_{j}\setminus\bigcup_{j=1}^{N_{1}}K_{j}'\right|&=&\displaystyle\left|\bigcup_{j=1}^{N_{1}}Q_{j}\setminus K_{j}'\right|+\left|\bigcup_{j=N_{1}+1}^{\infty}Q_{j}\right|\\[2 em]&=&\displaystyle(1-\alpha)\left|\bigcup_{j=1}^{N_{1}}Q_{j}\right|+\left|\bigcup_{j=N_{1}+1}^{\infty}Q_{j}\right|\end{array}$

Now if we take ${N_{1}}$ sufficiently large, then the last quantity is ${<(1-\alpha/2)\left|U\right|}$. Set ${K_{j}:=K_{j}'}$ for ${j=1,\ldots,N_{1}}$. Suppose we constructed sets ${U_{1},\ldots,U_{k}}$ and ${\left\{K_{j}\right\}_{j=1}^{N_{k}}}$ such that

$\displaystyle U_{i}:=U_{i-1}\setminus\bigcup_{i=N_{j-1}+1}^{N_{j}}K_{j}, \left|U_{i}\right|\leq\left(1-\dfrac{\alpha}{2}\right)^{i}\left|U\right|$

for ${2\leq i\leq k}$. Repeating the argument above, we see that we can find sets ${\left\{K_{j}\right\}_{j=N_{k}+1}^{N_{k+1}}}$ so that

$\displaystyle \left|U_{k+1}\right|=\left|U_{k}\setminus\bigcup_{j=N_{k}+1}^{N_{k+1}}K_{j}\right|\leq\left(1-\dfrac{\alpha}{2}\right)^{k+1}\left|U\right|$

The conclusion of the lemma follows by induction. $\Box$

Theorem 5 Let ${Q}$ be the unit square in ${{{\mathbb R}}^{2}}$. There exists a subset ${F\subset Q}$ with ${\left|F\right|=1}$ with the following property: there exists a sequence ${\left\{I_{k}(x)\right\}}$ of open rectangles containing ${x}$ and with diameter tending to zero, such that for each disjoint sequence ${\left\{R_{k}\right\}}$ from ${\left\{I_{k}(x)\right\}_{x\in F,k\geq 1}}$,

$\displaystyle \left|F\setminus\bigcup_{k}R_{k}\right|>\dfrac{1}{2}$

Before diving into the proof of Theorem 5, we give an auxillary geometric construction which we will need. Let ${N}$ be an integer greater than ${1}$, and consider in ${{{\mathbb R}}^{2}}$ the rectangles ${I_{1},\ldots,I_{N}}$ where the vertices of ${I_{j}}$ are ${(0,0)}$, ${(j,0)}$, ${(0,N/j)}$, and ${(j,N/j)}$. It is evident that ${\left|I_{j}\right|=N}$ for all ${j}$ and ${\left|E:=\bigcap_{j=1}^{N}I_{j}\right|=1}$. If we set ${J_{N}:=\bigcup_{j=1}^{N}I_{j}}$, then it is easy to verify that

$\displaystyle \left|J_{N}\right|=N\underbrace{\sum_{j=1}^{N}\dfrac{1}{j}}_{\alpha(N)}$

Proof: Let ${\left\{N_{k}\right\}}$ be an increasing sequence of natural numbers. For ${N_{i}}$ fixed, we can use Lemma 4 to almost cover the open unit cube ${Q}$ with a disjoint sequence ${\left\{S_{k}^{i}\right\}_{k\geq 1}}$ of sets homothetic to ${J_{H_{i}}}$ (as defined above), with each ${S_{k}^{i}}$ contained in ${Q}$ and with diameter ${<2^{-i}}$.

Let ${\left\{I_{k,j}^{i}\right\}_{j=1}^{N_{i}}}$ be the ${N_{i}}$ open rectangles constituting ${S_{k}^{i}}$, homothetic to the intervals ${I_{j}}$ of ${J_{N_{i}}}$. Define a family of rectangles ${A_{i}:=\left\{I_{k,j}^{i}\right\}_{k=1,j=1}^{\infty, N_{i}}}$. If we denote the union of the elements of ${A_{i}}$ by ${F_{i}}$, then it is clear that ${\left|F^{i}\right|=\left|Q\right|=1}$. Observe that if ${\left\{R_{l}^{i}\right\}=\left\{I_{k_{l},j_{l}}^{i}\right\}}$ is any disjoint subcollection of ${A _{i}}$, then ${\left\{R_{l}^{i}\right\}}$ contains at most one rectangle ${I_{k,j}^{i}}$ of those constituting ${S_{k}^{i}}$, since ${I_{k,j}^{i}\cap I_{k,m}^{i}\neq\emptyset}$ for ${1\leq j\leq m\leq N_{i}}$. Also observe that since

$\displaystyle \dfrac{\left|I_{k,j}^{i}\right|}{\left|S_{k}^{i}\right|}=\dfrac{1}{\alpha(N_{i})}\forall k\geq 1,1\leq j\leq N_{i} \text{ and }\sum_{k}\left|S_{k}^{i}\right|=1$

we have that

$\displaystyle \sum_{l=1}^{\infty}\left|R_{l}^{i}\right|=\sum_{l=1}^{\infty}\dfrac{\left|I_{k_{l},j_{l}}^{i}\right|}{\left|S_{k_{l}}^{i}\right|}\cdot\left|S_{k_{l}}^{i}\right|\leq\dfrac{1}{\alpha(N_{i})}$

Now set ${F:=\bigcap_{i=1}^{\infty}F_{i}}$. Since ${F_{i}\subset Q}$ and ${\left|Q\setminus F_{i}\right|=0}$, it follows that ${\left|F\right|=1}$. Consider the family of rectangles ${A:=\bigcup_{i=1}^{\infty}A_{i}}$. It is evident that for each ${x\in F}$ there is a sequence of rectangles ${\left\{I_{k}(x)\right\}}$ of ${A}$ containing ${x}$ and having diameter tending to zero. For any disjoint sequence ${\left\{R_{k}\right\}\subset A}$, we have

$\displaystyle \sum_{k=1}^{\infty}\left|R_{k}\right|\leq\sum_{k=1}^{\infty}\dfrac{1}{\alpha(N_{k})}$

Since ${\alpha(N_{k})>\log(N_{k})/2}$, we can chose our original sequence ${\left\{N_{i}\right\}}$ to satisfy

$\displaystyle \sum_{i=1}^{\infty}\dfrac{1}{\alpha(N_{i})}<\dfrac{1}{2},$

whence ${\left|F\setminus\bigcup R_{k}\right|>1/2}$. $\Box$

To generalize the counterexample to higher dimensions, we simply modify the rectangles ${I_{j}}$ in the auxilary construction by ${\tilde{I}_{j}:=I_{j}\times[0,1]^{n-2}}$, so ${\bigcap_{j=1}^{N}\tilde{I}_{j}}$ has volume ${1}$ and ${\tilde{J}_{N}:=\bigcup_{j=1}^{N}\tilde{I}_{j}}$ has volume ${N\alpha(N)}$. We can then repeat the argument in the proof above with ${Q}$ now the unit cube in ${{{\mathbb R}}^{n}}$.

Suppose we replace the Lebesgue measure $\left|\cdot\right|$ on ${\mathbb{R}}^{n}$ by a Radon measure $\mu$. Does there exist a differentiation basis for $\mu$ which does not possess the Vitali covering property? The above construction relied on the fact that if two measurable sets $A$ and $B$ have measures have ratio $\alpha:=\left|A\right|/\left|B\right|$, then $\alpha=\left|T(A)\right|/\left|T(B)\right|$ for any homothety $T$. This is a consequence of the translation and dilation properties of the Lebesgue measure. It is not clear to us how to modify the above construction to work without these properties. We are curious to hear anyone else’s thoughts on the question.

## Kuran’s Theorem

It is well-known that a harmonic function ${u:\Omega\rightarrow\mathbb{R}}$ satisfies the mean value property

$\displaystyle u(x)=\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}u(y)dy=\dfrac{1}{\left|\partial B(x,r)\right|}\int_{\partial B(x,r)}u(y)dS(y),$

for all balls ${B(x,r)}$ and spheres ${\partial B(x,r)}$ contained in ${\Omega}$. (We abuse notation and use ${\left|\cdot\right|}$ to denote both the ${n}$-dimensional Lebesgue measure and the ${(n-1)}$-dimensional surface measure.) This mean value property actually characterizes harmonic functions. A locally integrable function ${u}$ which satisfies the mean value property is harmonic; this is a form of Weyl’s lemma. Given this result, we could say that the mean value property over balls/spheres completely characterizes harmonic functions.

What if, instead, we know that a connected, open set (a domain) ${\Omega}$ has the property that every function ${u}$ which is harmonic in ${\Omega}$ satisfies the volume mean value property

$\displaystyle u(x)=\dfrac{1}{\left|\Omega\right|}\int_{\Omega}u(y)dy,\indent\forall x\in\Omega$

Is ${\Omega}$ necessarily an open ball of some radius ${r}$? After limited results were proved by B. Epstein [2] and M.M. Schiffer [3], \”{U}. Kuran provided a very elegant answer in the affirmative to this question for bounded domains [5]. As shown in [4], the hypothesis that ${\Omega}$ is connected is unnecessary, so we will trivially deviate from Kuran’s original result by dispensing with it.

Theorem 1 (Kuran) Let ${\Omega\subset\mathbb{R}^{n}}$ (${n\geq 2}$) be a bounded, open set. Suppose there is a point ${x_{0}\in\Omega}$ such that for every function harmonic in and integrable over ${\Omega}$,

$\displaystyle u(x_{0})=\dfrac{1}{\left|\Omega\right|}\int_{\Omega}u(x)dx$

Then ${\Omega}$ is an open ball (disk) centered at ${x_{0}}$.

Proof: The idea of the proof is pretty simple. Since ${\Omega}$ has compact closure, there exists a point ${x_{1}\in\partial\Omega}$ satisfying the equality

$\displaystyle r:=\left|x_{1}-x_{0}\right|=\inf_{x\in\Omega^{c}}\left|x-x_{0}\right|$

So ${B:=B(x_{0},r)\subset\Omega}$. If we can show that ${\Omega\subset\overline{B}}$, then since ${\text{int}(\overline{B})=B}$, it follows that ${\Omega=B}$. Assume the contrary: ${\Omega\setminus\overline{B}\neq\emptyset}$ and therefore has positive measure, being open.

Now suppose there is a function ${u}$ which vanishes at ${x_{0}}$ in ${\Omega}$ and is harmonic on the domain. Then it follows from the hypotheses and the ordinary mean value property applied to the ball ${B}$ that

$\displaystyle 0=u(x_{0})=\int_{\Omega}u(x)dx=\int_{\Omega\setminus B}u(x)dx+\underbrace{\int_{B}u(x)dx}_{=0}=\int_{\Omega\setminus B}u(x)dx$

From the last expression, we see that a nonnegative function ${u}$ cannot be strictly positive on a subset of ${\Omega\setminus B}$ of positive measure. This last observation suggests that by a sensible choice for ${u}$, we can arrive at a contradiction.

The Poisson kernel suggests that we consider the expression

$\displaystyle K(x):=\dfrac{\left|x-x_{0}\right|^{2}-r^{2}}{r^{2-n}\left|x-x_{1}\right|^{n}},$

which equals ${-1}$ at ${x_{0}}$. Since we can add a constant term without changing harmonicity, we are led to consider the function

$\displaystyle u(x):=\dfrac{\left|x-x_{0}\right|^{2}}{r^{2-n}\left|x-x_{1}\right|^{n}}+1$

Lemma 2 The function ${u}$ defined above vanishes at ${x_{0}}$, is harmonic on ${\mathbb{R}^{n}\setminus\left\{x_{1}\right\}}$, is integrable on ${\Omega}$, and is minorized by ${1}$ on ${\Omega\setminus B}$.

Proof: Computing partial derivatives (it’s a good exercise in calculus to work it out yourself), we see that

$\displaystyle \begin{array}{lcl} \displaystyle u_{x_{i}x_{i}}(x)&=&\displaystyle\dfrac{2}{r^{2-n}\left|x-x_{1}\right|^{n}}-\dfrac{4n(x_{i}-x_{0,i})(x_{i}-x_{1,i})}{r^{2-n}\left|x-x_{1}\right|^{n+2}}-\dfrac{n\left(\left|x-x_{0}\right|^{2}-r^{2}\right)}{r^{2-n}\left|x-x_{1}\right|^{n+2}}+\dfrac{n(n+2)\left(\left|x-x_{0}\right|^{2}-r^{2}\right)\left(x_{i}-x_{1,i}\right)^{2}}{r^{2-n}\left|x-x_{1}\right|^{n+4}} \end{array}$

whence

$\displaystyle \begin{array}{lcl}\displaystyle\Delta u(x)&=&\displaystyle\dfrac{2n}{r^{2-n}\left|x-x_{1}\right|^{n}}-\dfrac{4n(x-x_{0})\cdot(x-x_{1})}{r^{2-n}\left|x-x_{1}\right|^{n+2}}+\dfrac{2n\left(\left|x-x_{0}\right|^{2}-r^{2}\right)}{r^{2-n}\left|x-x_{1}\right|^{n+2}}\\[2 em] \displaystyle&=&\displaystyle\dfrac{2n\left|x-x_{1}\right|^{2}-4n(x-x_{0})\cdot(x-x_{1})+2n(\left|x-x_{0}\right|^{2}-r^{2})}{r^{2-n}\left|x-x_{1}\right|^{n+2}}\\ [2 em]\displaystyle&=&\displaystyle\dfrac{2n\left|(x-x_{1})-(x-x_{0})\right|^{2}-2nr^{2}}{r^{2-n}\left|x-x_{1}\right|^{n+2}}\\ [2 em]\displaystyle&=&\displaystyle0\end{array}$

It is evident that ${u\geq 1}$ on ${\Omega\setminus B}$. To see that ${u\in L^{1}(\Omega)}$, observe that ${\left|K(x)\right|\lesssim\left|x-x_{1}\right|^{1-n}}$ near the singularity ${x_{1}}$. $\Box$

We conclude that

$\displaystyle 0=\int_{\Omega}u(x)dx\geq\int_{\Omega\setminus B}1dx>0,$

which is a contradiction. This completes the proof. $\Box$

We conclude by mentioning some related results for the benefit of the interested reader. In a cutely titled paper [1], D. Aharonov, M.M. Schiffer, and L. Zalcman prove a related result for a class of spaces which the authors call potatoes. In [6], N. Suzuki and N.A. Watson prove the analogous inverse mean value theorem for solutions of the heat equation. Lastly, we mention that Kuran’s theorem has application to approximation theory. M. Goldstein, W. Haussman, and L. Rogge use Kuran’s theorem to prove existence and uniqueness of the best harmonic approximant in ${L^{1}}$ norm to a subharmonic function [4].

1. D. Aharonov, M.M. Schiffer, and L. Zalcman, Potato Kugel, Israel Jour. of Math. 40 (1981), 331-339.
2. B. Epstein, On the Mean-Value Property of Harmonic Functions, Proc. Amer. Math. Soc. 13 (1962), 830.
3. B. Epstein and M.M. Schiffer, On the Mean-Value Property of Harmonic Functions, J. Analyse Math. 14 (1965), 109-111.
4. M. Goldstein, W. Haussman, and L. Rogge, On the Mean Value Property of Harmonic Functions and Best Harmonic $L^{1}$-Approximation, Trans. Amer. Math. Soc. 305 (1988), 505-515.
5. U. Kuran, On the Mean-Value Property of Harmonic Functions, Bull. Lon. Math. Soc. (1972), 311-312.
6. N. Suzuki and N.A. Watson, A Characterization of Heat Balls by a Mean Value Property for Temperatures, Proc. Amer. Math. Soc. 129 (2001), 2709-2713.

## A Banach algebra proof of the prime number theorem

The prime number theorem can be expressed as the assertion

$latex displaystyle sum_{n leq x} Lambda(n) = x + o(x) (1)&fg=000000$

as $latex {x rightarrow infty}&fg=000000$, where

$latex displaystyle Lambda(n) := sum_{d|n} mu(d) log frac{n}{d}&fg=000000$

is the von Mangoldt function. It is a basic result in analytic number theory, but requires a bit of effort to prove. One “elementary” proof of this theorem proceeds through the Selberg symmetry formula

$latex displaystyle sum_{n leq x} Lambda_2(n) = 2 x log x + O(x) (2)&fg=000000$

where the second von Mangoldt function $latex {Lambda_2}&fg=000000$ is defined by the formula

$latex displaystyle Lambda_2(n) := sum_{d|n} mu(d) log^2 frac{n}{d} (3)&fg=000000$

or equivalently

$latex displaystyle Lambda_2(n) = Lambda(n) log n + sum_{d|n} Lambda(d) Lambda(frac{n}{d}). (4)&fg=000000$

(We are avoiding the use of the $latex {*}&fg=000000$ symbol here to denote Dirichlet convolution, as we will need this symbol to denote ordinary convolution shortly.) For the convenience of…

View original post 3,616 more words

## Heat Ball and Heat Sphere Mean Value Property

Like solutions to the Laplace equation, (classical) solutions to the heat equation ${\partial_{t}u-\Delta u=0}$ satisfy a mean value property. But instead of integrals over balls or spheres, the heat mean value property involves integrals over a heat ball. The heat ball of radius ${r>0}$ centered at a point of ${(x,t)}$ in ${\mathbb{R}^{n}\times\mathbb{R}}$, denoted ${E(x,t;r)}$ is the set

$\displaystyle E(x,t;r):=\left\{(y,s)\in\mathbb{R}^{n}\times\mathbb{R}:s\leq t \text{ and }\Phi(x-y,t-s)\geq r^{-n}\right\},$

where ${\Phi}$ denotes the heat kernel

$\displaystyle \Phi(x-y,t-s)=\dfrac{1}{(4\pi(t-s))^{n/2}}e^{-\frac{\left|x-y\right|^{2}}{4(t-s)}}$

When ${(x,t)=(0,0)}$, we sometimes write ${E(r)}$ instead of ${E(0,0;r)}$.

After playing around with the inequality ${\Phi(x-y,t-s)\geq r^{-n}}$, one can verify that

$\displaystyle E(x,y;r)=\left\{(y,s)\in\mathbb{R}^{n+1}:t-\frac{r^{2}}{4\pi}\leq s\leq t, \left|x-y\right|^{2}\leq 2n(t-s)\log\left(\frac{r^{2}}{4\pi (t-s)}\right)\right\}$

and

$\displaystyle \partial E(x,t;r)=\left\{(y,s)\in\mathbb{R}^{n+1}:t-\frac{r^{2}}{4\pi}\leq s\leq t,\left|x-y\right|^{2}=2n(t-s)\log\left(\dfrac{r^{2}}{4\pi (t-s)}\right)\right\}$

From the above expressions, it is easy to see that ${E(x,y;r)}$ is bounded and closed. Furthermore, the “center” ${(x,t)}$ actually lies on the boundary, so it is somewhat of a misnomer. I leave it to the reader to verify that ${E(x,t;r)=(x,t)+E(r)}$ and ${E(r)}$ is obtained from ${E(1)}$ by the parabolic scaling ${(y,s)\mapsto(ry,r^{2}s)}$.

The following theorem of N.A. Watson establishes the heat ball mean value property.

Theorem 1 Let ${U\subset\mathbb{R}^{n}}$ be a bounded open set, and let ${T>0}$. Set ${U_{T}:=U\times (0,T)}$. If ${u}$ is a ${C_{1}^{2}(U_{T})}$ subsolution of the heat equation, then

$\displaystyle u(x,t)=\dfrac{1}{4r^{n}}\iint_{E(x,t;r)}u(y,s)\dfrac{\left|x-y\right|^{2}}{(t-s)^{2}}dyds$

for all ${E(x,t;r)\subset U_{T}}$, where equality holds if ${u}$ is a solution of the heat equation.

After seeing the proof of this surprisingly recent theorem, I wondered if there was a corresponding result for the boundary of ${\partial E(x,t;r)}$, a heat sphere mean value property. There was no proof of reference for a proof of such a result in the book I was reading, and Watson’s paper makes no mention of such a result either. I asked a professor at Georgia Tech if he knew of mean value property for the heat sphere. He did not, but he gave me a very helpful suggestion which led to the following result; many thanks to Professor F. Bonetto.

Theorem 2 Let ${U\subset\mathbb{R}^{n}}$ be a bounded open set, and let ${T>0}$. If ${u}$ is a ${C_{1}^{2}(U_{T})}$ solution of the heat equation on ${U_{T}}$, then

$\displaystyle u(x,t)=-\dfrac{1}{2r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(x+R\omega,t+s)\dfrac{R^{n}}{s}d\omega ds$

for all ${E(x,t;r)\subset U_{T}}$.

It appears Theorem 2 is at least folklore, as I found the result contained in some very nice lecture notes of Professor G. Menon.

Interestingly, proving the mean value property for the heat equation is done in the reverse order of with the Laplace equation. With harmonic functions, one shows the mean value property holds over spheres and then uses integration in polar coordinates to show the property holds over balls. In contrast, we will show that the mean value property holds over heat balls and then use this to show that it holds over heat spheres.

To prove Theorem 1, I will follows [Evans].

Proof of Theorem 1: As the translate of ${u}$ is again a solution of the heat equation (on a translated domain), we may assume without loss of generality that ${(x,t)=(0,0)}$. For ${r>0}$, define

$\displaystyle \phi(r)=\dfrac{1}{r^{n}}\iint_{E(r)}u(y,s)\dfrac{\left|y\right|^{2}}{s^{2}}dyds=\iint_{E(1)}u(ry,r^{2}s)\dfrac{\left|y\right|^{2}}{s^{2}}dyds,$

where we make the change of variable ${(y,s)\mapsto (ry,r^{2}s)}$. Since ${u}$ is ${C_{1}^{2}(U)}$ and ${U}$ is a bounded domain, we can differentiate inside the integral to obtain

$\displaystyle \begin{array}{lcl} \displaystyle\phi'(r)&=&\displaystyle\iint_{E(1)}\sum_{i=1}^{n}u_{y_{i}}y_{i}\dfrac{\left|y\right|^{2}}{s^{2}}+2ru_{s}\dfrac{\left|y\right|^{2}}{s}dyds\\ [2 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}\sum_{i=1}^{n}u_{y_{i}}y_{i}\dfrac{\left|y\right|^{2}}{s^{2}}+2u_{s}\dfrac{\left|y\right|^{2}}{s}dyds\\ [2 em]&=:&\displaystyle A+B \end{array}$

We introduce the function

$\displaystyle \psi(y,s):=-\dfrac{n}{2}\log(-4\pi s)+\dfrac{\left|y\right|^{2}}{4s}+n\log r,$

which can be obtained by taking the logarithm of ${r^{n}\Phi(-y,-s)}$. Observe that ${\Phi(y,-s)=r^{-n}}$ on ${\partial E(r)}$ and therefore ${\psi=0}$ on ${\partial E(r)}$. We can write the expression ${B}$ as

$\displaystyle \begin{array}{lcl} \displaystyle B&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}4u_{s}\sum_{i=1}^{n}y_{i}\psi_{y_{i}}dyds\\ [1.5 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}4nu_{s}\psi+4\sum_{i=1}^{n}u_{sy_{i}}y_{i}\psi dyds; \end{array}$

where we integrate by parts with respect to ${y}$ and use the fact that ${\psi}$ vanishes on ${\partial E(r)}$ for the boundary term. Integrating by parts now with respect to ${s}$, we obtain

$\displaystyle \begin{array}{lcl} \displaystyle B&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi+4\sum_{i=1}^{n}u_{y_{i}}y_{i}\psi_{s}dyds\\ [2 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi+4\sum_{i=1}^{n}u_{y_{i}}y_{i}\left(-\dfrac{n}{2s}-\dfrac{\left|y\right|^{2}}{4s^{2}}\right)dyds\\ [2 em]&=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi-\dfrac{2n}{s}\sum_{i=1}^{n}u_{y_{i}}y_{i}dyds-A \end{array}$

Since ${u}$ is a subsolution of the heat equation, we have the inequality

$\displaystyle \begin{array}{lcl} \displaystyle\phi'(r)&=&\displaystyle A+B\\ [2 em] &=&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4nu_{s}\psi-\dfrac{2n}{s}\sum_{i=1}^{n}u_{y_{i}}y_{i}dyds\\ [2 em]&\geq&\displaystyle\dfrac{1}{r^{n+1}}\iint_{E(r)}-4n\Delta u\psi-\dfrac{2n}{s}\sum_{i=1}^{n}u_{y_{i}}y_{i}dyds\\ [2 em]&=&\displaystyle\sum_{i=1}^{n}\dfrac{1}{r^{n+1}}\iint_{E(r)}4nu_{y_{i}}\psi_{y_{i}}-\dfrac{2n}{s}u_{y_{i}}y_{i}dyds\\ [2 em]&=&\displaystyle\sum_{i=1}^{n}\dfrac{1}{r^{n+1}}\iint_{E(r)}\dfrac{2n}{s}u_{y_{i}}y_{i}-\dfrac{2n}{s}u_{y_{i}}y_{i}dyds=0 \end{array}$

where we integrate by parts the first term (w.r.t. ${y}$) to obtain the penultimate expression. Therefore ${\phi}$ is nondecreasing. Using the dominated convergence theorem and the continuity of ${u}$, we obtain

$\displaystyle \begin{array}{lcl} \displaystyle\phi(r)\geq\lim_{t\rightarrow 0^{+}}\phi(t)&=&\displaystyle\iint_{E(1)}\left(\lim_{t\rightarrow 0^{+}}u(ty,t^{2}s)\right)\dfrac{\left|y\right|^{2}}{s^{2}}dyds\\ [2 em]&=&\displaystyle u(0,0)\iint_{E(1)}\dfrac{\left|y\right|^{2}}{s^{2}}dyds\\ [2 em]&=&\displaystyle 4u(0,0) \end{array}$

The result ${4=\iint_{E(1)}\left|y\right|^{2}s^{-2}dyds}$, as it is an exercise in integration, but I have uploaded a PDF of the computation. Dividing both sides by ${r}$ yields the desired inequality. In particular, if ${u}$ is a solution then equality holds. $\Box$

Using the preceding result, we now prove Theorem 2. For convenience, we introduce the notation

$\displaystyle R=R(r,s)=\left(-2ns\log\left(\dfrac{r^{2}}{-4\pi s}\right)\right)^{1/2}$

Proof of Theorem 2: Since the translate of ${u}$ is again a solution of the heat equation, we may assume without loss of generality that ${(x,t)=(0,0)}$. Using polar coordinates, we can write

$\displaystyle \dfrac{1}{4r^{n}}\iint_{E(r)}u(y,s)\dfrac{\left|y\right|^{2}}{\left|s\right|^{2}}dyds=\dfrac{1}{4r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{0}^{R(r,s)}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho ds$

Differentiating ${\phi}$ with respect to ${r}$, we obtain

$\displaystyle \begin{array}{lcl} \displaystyle\phi'(r)&=&\displaystyle-\dfrac{n}{4r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{0}^{R(r,s)}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho ds\\ [2 em]&+&\displaystyle\dfrac{1}{4r^{n}}\int_{0}^{R(r,-\frac{r^{2}}{4\pi})}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho\vert_{s=-\frac{r^{2}}{4\pi}}\\ [2 em]&+&\displaystyle\dfrac{1}{4r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}\cdot\dfrac{dR}{dr}d\omega\vert_{\rho=R(s,r)}ds \end{array}$

Note that the second term in the expression above vanishes since ${R(r,-\frac{r^{2}}{4\pi})=0}$. We compute

$\displaystyle \dfrac{dR}{dr}=\dfrac{(-2ns)^{1/2}}{r\log\left(\frac{r^{2}}{-4\pi s}\right)^{1/2}}$

and note that

$\displaystyle R\cdot\dfrac{dR}{dr}=\left(-2ns\log\left(\frac{r^{2}}{-4\pi s}\right)\right)^{1/2}\dfrac{(-2ns)^{1/2}}{r\left(\log\frac{r^{2}}{-4\pi s}\right)^{1/2}}=\dfrac{-2ns}{r}$

If ${u}$ is a (classical) solution to the heat equation, then our proof of the mean value property over the heat ball shows that ${\phi'(r)=0}$. Rearranging terms, we obtain

$\displaystyle \begin{array}{lcl} \displaystyle\dfrac{n}{r}\iint_{E(r)}u(y,s)\dfrac{\left|y\right|^{2}}{\left|s\right|^{2}}dyds &=&\displaystyle\dfrac{n}{4r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{0}^{R(r,s)}\int_{S^{n-1}}u(\rho\omega,s)\dfrac{\rho^{n+1}}{\left|s\right|^{2}}d\omega d\rho ds\\ [2 em]&=&\displaystyle\dfrac{1}{4r^{n}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u\left(R\omega,s\right)\dfrac{R^{n+1}}{\left|s\right|^{2}}\cdot\dfrac{dR}{dr}d\omega ds\\ [2 em]&=&\displaystyle\dfrac{-2n}{4r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(R\omega,s)\dfrac{R^{n}}{s}d\omega ds\\ [2 em] &=&\displaystyle\dfrac{-n}{2r^{n+1}}\int_{-\frac{r^{2}}{4\pi}}^{0}\int_{S^{n-1}}u(R\omega,s)\dfrac{R^{n}}{s}d\omega ds \end{array}$

Multiplying both sides of the equality by ${r/n}$ completes the proof. $\Box$

1. L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.
2. G. Menov, “Lectures on Partial Differential Equations”, Accessed: 10/11/14.
3. N.A. Watson, “A Theory of Subtemperatures in Several Variables”, Proc. Lon. Math. Soc. s3-26 (1973), 385-417.
4. X. Yu, “Heat Equation – Maximum Principles”, Accessed: 10/1/14.

## An Extension of a Theorem of Marczewski

In [3], Marczewski proved that convergence almost surely (a.s.) is equivalent to convergence in probability precisely when ${\Omega}$ is the (at most) countable union of disjoint atoms (i.e. purely atomic). In particular, convergence a.s. is defined by a topology when ${\Omega}$ is purely atomic. It is well known that convergence a.s. need not be defined by a topology when ${\Omega}$ is not purely atomic. Try to construct a counter-example on the unit interval with Borel ${\sigma}$-algebra and Lebesgue measure.

Today, I want to give an extension of Marczewski’s result which characterizes when convergence a.s. is defined by a topology. Using arguments similar to those in my last post, we will prove the following theorem.

Theorem 1 Let ${(X,\mathcal{A},\mu)}$ be a finite measure space. There exists a topology ${\tau}$ on the space of measurable functions ${L^{0}(\mu)}$ coinciding with convergence a.s. if and only if ${X}$ is purely atomic.

Proof: Marczewski’s result proves sufficiency, so we prove necessity. If ${X}$ is not purely atomic, then there exists a measurable subset ${A}$ which is not an atom and on which ${\mu}$ satisfies the intermediate value property. In particular, for each ${n\in\mathbb{N}}$, there exists a partition of ${A}$ into ${m_{n}}$ measurable sets ${A_{1}^{n},\ldots,A_{m_{n}}^{n}}$ with ${0<\mu(A_{i}^{n})\leq n^{-1}}$. Consider the sequence ${f_{k}}$ formed by indicator functions of the ${A_{i}^{n}}$:

$\displaystyle 1_{A_{1}^{1}},\ldots,1_{A_{m_{1}}^{1}},1_{A_{1}^{2}},\ldots,1_{A_{m_{2}}^{2}},1_{A_{1}^{3}},\ldots,$

I claim that ${f_{k}}$ converges to zero in measure. Indeed, ${f_{k}}$ is supported on ${A_{i}^{n}}$, for some ${n}$ and ${1\leq i\leq m_{n}}$. Note tht ${\mu(A_{i}^{n})\rightarrow 0}$ as ${n\rightarrow\infty}$ and ${k\rightarrow\infty}$ implies ${n\rightarrow\infty}$. For any ${x\in X}$, ${f_{k}(x)=1}$ for infinitely many ${k}$, whence ${f_{k}}$ does not converge to zero a.s. Recall that if every subsequence of ${f_{k}}$ has a further subsequence which converges to some ${f}$ in a topology ${\tau}$, then ${f_{k}\rightarrow f}$ in ${\tau}$. But convergence in measure implies convergence of a subsequence a.s. to the same limit, which is a contradiction. $\Box$

1. S.K. Berberian, Lectures in Functional Analysis and Operator Theory, Spring Verlag 1974.
1. V.I. Bogachev, Measure Theory Volume I, Springer 2007.
2. E. Marczewski, “Remarks on the Convergence of Measurable Sets and Measurable Functions”, Colloquium Math. (1955), 118-124.

## On the Normability of the Space of Measurable Functions

Suppose ${(X,\mathcal{A},\mu)}$ is a finite measure space, and let ${L^{0}(X)}$ denote the space of real- or complex-valued measurable functions on ${X}$. We can topologize ${L^{0}(X)}$ by saying a sequence ${f_{n}\rightarrow f}$ in ${L^{0}(X)}$ if and only if

$\displaystyle \lim_{n\rightarrow\infty}\mu\left\{\left|f_{n}-f\right|\geq\varepsilon\right\}=0,\indent\forall\varepsilon>0$

We say that ${f_{n}}$ converges in measure to ${f}$. It is an easy exercise to verify that convergence in measure can be realized by the metric topologies

$\displaystyle \rho_{1}(f,g):=\int_{X}\dfrac{\left|f-g\right|}{1+\left|f-g\right|}d\mu,\indent\rho_{2}(f,g):=\int_{X}\min\left(\left|f-g\right|,1\right)d\mu$

and in fact, these metrics are complete.

For ${p\geq 1}$, the ${L^{p}}$ spaces have a metric topology which is actually given by the ${L^{p}}$ norm. The integrands in the expressions for ${\rho_{1}}$ and ${\rho_{2}}$ are not homogeneous in their arguments, so we should not expect these metrics to be norms. However, there could still be a norm lurking out there on ${L^{0}(X)}$ with an equivalent norm topology. This past summer, I became interested in the following question after reading a comment on Math.SE: What are necessary and sufficient conditions for ${L^{0}(X)}$ to be normable (i.e., there exists a norm ${\left\|\cdot\right\|}$ which induces the same topology on ${L^{0}(X)}$ as convergence in measure)?

For “typical” spaces such as the measurable functions on ${[0,1]}$, it is easy to disprove normability with simple counterexamples. Suppose there is a norm ${\left\|\cdot\right\|}$ on ${L^{0}[0,1]}$. Consider the measurable functions defined by

$\displaystyle f_{n}:=\left\|1_{A_{n}}\right\|^{-1}1_{A_{n}},\indent A_{n}:=\left\{\frac{n-1}{n}\leq x<\frac{n}{n+1}\right\}$

It is evident that ${f_{n}}$ converges to zero in measure since ${\left|A_{n}\right|\rightarrow 0}$, as ${n\rightarrow\infty}$. However, ${\left\|f_{n}\right\|=1}$ for all ${n}$. Normability fails because we able to find measurable subsets of ${[0,1]}$ of arbitrarily small measure. This is was possible because the unit interval with Lebesgue measure has no atoms.

It turns out that my question was answered by A.J. Thomasian in 1957, who showed that the space of random variables on some probability space ${(\Omega,\mathcal{F},P)}$ is normable if and only if ${\Omega}$ is the finite union of disjoint atoms. It follows from normalizing a finite measure that Thomasian’s result holds for finite measure spaces.

The other day, I was thinking about Thomasian’s result in terms of the Kolmogorov normability criterion, which says that a separated topological vector space (tvs) is normable if and only if it has a bounded, convex neighborhood of zero. And it seemed to be that this was the right context for his result, rather than the specific language of probability theory. So today I’m going to relate the two.

First, we prove a technically convenient characterization of boundedness for tvs.

Lemma 1 A subset ${A}$ of a tvs ${(X,\tau)}$ is bounded if and only if for any sequence of elements ${x_{n}\in A}$ and scalars ${\lambda_{n}\rightarrow 0}$, the sequence ${\lambda_{n}x_{n}\rightarrow 0}$.

Proof: Suppose ${A}$ is bounded. Let ${U}$ be an open neighborhood of zero, and let ${\lambda_{n}}$ a sequence of scalars tending to ${0}$. Without loss of generaliy, we may assume that ${U}$ is balanced, so that ${A\subset\lambda_{n}^{-1}U}$ for almost all ${n}$. Given a sequence ${x_{n}\in A}$, it follows that ${\lambda_{n}x_{n}\in U}$ for almost all ${n}$.

If ${A}$ is unbounded, then there exists an open neighborhood ${U}$ of zero such that for all ${n\in\mathbb{N}}$, there is an element ${x_{n}\in A\setminus (n\cdot U)\Leftrightarrow n^{-1}x_{n}\notin U}$. $\Box$

Suppose ${X}$ is not the finite union of disjoint atoms. Recall that we can write the space ${X}$ as ${X=A\cup\bigcup_{n=1}^{\infty}A_{n}}$, where ${A}$ is either of positive measure and not an atom or of measure zero and ${\left\{A_{n}\right\}}$ are pairwise disjoint atoms. There are two cases to consider.

1. ${\mu(A)>0}$: By the intermediate value property for measures, for any ${\varepsilon>0}$ there exists a measurable set ${B\subset A}$ with ${\mu(B)=\varepsilon}$. Define a sequence of measurable functions by ${f_{n}:=n^{2}\cdot1_{B}}$. Given any open neighborhood ${U}$ of zero in ${L^{0}(X)}$, we can take ${\varepsilon}$ sufficiently small so that ${f_{n}\in U}$. But ${n^{-1}\cdot f_{n}}$ does not converge to zero in measure, so by Lemma 1, ${U}$ is not bounded.
2. ${\mu(A_{n})>0}$ for all ${n\in\mathbb{N}}$: It follows from ${\sigma}$-additivity that ${\mu(A_{n})\rightarrow 0}$ as ${n\rightarrow\infty}$. Given ${\varepsilon>0}$, we can find ${B:=A_{n}}$ with ${\mu(A_{n})<\varepsilon}$. The argument in the first case shows any open neighborhood ${U}$ of zero is not bounded.

Now suppose ${X}$ is the finite union of disjoint atoms ${A_{1},\ldots,A_{n}}$. Any measurable function ${f}$ is a.s. constant on each ${A_{j}}$, so we can write ${f=\sum_{j=1}^{n}a_{j}1_{A_{j}}}$, where ${a_{1},\ldots,a_{n}}$ are constants. Set ${\varepsilon:=\min_{j}\mu(A_{j})/2}$. With ${\rho_{1}}$ as above, we have

$\displaystyle \varepsilon>\rho_{1}(f,0)=\sum_{j=1}^{n}\frac{a_{j}}{1+a_{j}}\mu(A_{j})\geq\sum_{j\atop{a_{j}\neq0}}\mu(A_{j})\geq\sum_{j\atop{a_{j}\neq 0}}\varepsilon,$

which implies that ${\left\{j:a_{j}\neq 0\right\}=\emptyset}$. So ${\left\{0\right\}=\left\{f\in L^{0}(X):\rho_{1}(f,0)<\varepsilon\right\}}$ is an open neighborhood of zero which is trivially bounded, whence by Kolmogorov’s criterion, ${L^{0}(X)}$ is normable.

For completeness, it’s worth mentioning that Thomasian extended an earlier result of Marczewski which showed that convergence in probability implies convergence a.s. if and only if ${\Omega}$ is the (at most) countable union of disjoint atoms. So in particular, if ${L^{0}(X)}$ is normable, then convergence in measure implies converge a.s.

1. A.J. Thomasian, “Metrics and Norms on Spaces of Random Variables”, The Annals of Mathematical Statistics 28 (1957), 512-514.

2. E. Marczewski, “Remarks on the Convergence of Measurable Sets and Measurable Functions”, Colloquium Math. (1955), 118-124.

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## Harmonic Functions: Weyl Lemma

In this post, we prove a generalization of the result that any (clasically) harmonic function ${u}$ defined on a bounded, open set ${\Omega\subset\mathbb{R}^{n}}$ is analytic. If ${u\in L_{loc}^{1}(\Omega)}$ is weakly harmonic, i.e. ${\int(\Delta u)\varphi=\int u\Delta\varphi=0}$ for any compactly supported ${C^{\infty}}$ function ${\varphi}$ (the space of such functions is denoted ${C_{0}^{\infty}(\mathbb{R}^{n})}$, then ${u}$ is strongly or classically harmonic and therefore equal a.e. to an analytic function. This result is due to Hermann Weyl and appears to be known in the literature as the Weyl Lemma.

The proof I will give is based on an outline in a problem set I found online.

Let ${\eta:\mathbb{R}^{n}\rightarrow\mathbb{R}}$ be the radial mollifier defined by

$\displaystyle \eta(x):=\begin{cases} C\exp\left(\dfrac{1}{\left|x\right|^{2}-1}\right) & {\left|x\right|<1}\\ 0 & {\left|x\right|\geq 1} \end{cases}$

where ${C}$ is the normalizing constant. Define an approximate identity ${\eta_{\varepsilon}=\varepsilon^{-n}\eta(-/\varepsilon)}$. I leave it to the reader to verify that ${\eta}$ is indeed ${C^{\infty}}$ (this is a standard calculus problem) and that ${\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}}$ is an approximate identity.

For ${\varepsilon>0}$, define an open subset of ${\Omega}$ by ${U_{\varepsilon}:=\left\{x\in\Omega:d(x,\partial\Omega)>\varepsilon\right\}}$ (i.e. the set of points in ${\Omega}$ which are at least ${\varepsilon}$ distance from the boundary of ${\Omega}$). As ${\eta_{\varepsilon}}$ is supported on ${B(0,\varepsilon)}$, the convolution ${u_{\varepsilon}:=u\ast\eta_{\varepsilon}}$ is well-defined. Furthermore, ${u_{\varepsilon}}$ is ${C^{\infty}}$. Iif ${x\in U_{\varepsilon}}$, then ${B(x,\varepsilon)\subset\Omega}$, so ${u_{\varepsilon}}$ is integrable on ${U_{\varepsilon}}$. By Young’s inequality, we have the estimate

$\displaystyle \left\|u_{\varepsilon}\right\|_{L^{1}(U_{\varepsilon})}\leq\left\|u\right\|_{L^{1}(\Omega)}\left\|\eta_{\varepsilon}\right\|_{L^{1}(\Omega)}=\left\|u\right\|_{L^{1}(\Omega)}$

I claim that ${u_{\varepsilon}}$ is (classicaly) harmonic on ${U_{\varepsilon}}$. Indeed, as ${\eta_{\varepsilon}\in C_{0}^{\infty}}$, we can use the Lebesgue dominated convergence theorem to differentiate inside the integral in the definition of ${u_{\varepsilon}}$ (look at the Taylor expansion of ${\eta_{\varepsilon}}$) to obtain

$\displaystyle \Delta u_{\varepsilon}(x)=\int_{\Omega}[\Delta\eta_{\varepsilon}](x-y)u(y)dy=0,$

since ${\eta_{\varepsilon}(x-\cdot)\in C_{0}^{\infty}}$ and ${u}$ is weakly harmonic. Consequently, ${u_{\varepsilon}}$ has the mean value property.

We now want to extract a sequence of the ${u_{\varepsilon}}$ which converges to some continuous function ${v}$ and then argue that this function ${v}$ must be harmonic. To accomplish the first task, we use the Arzelà-Ascoli theorem. First, we have that ${\left\{u_{\varepsilon}\right\}}$ is equicontinuous and uniformly bounded. Fix an $R>0$, and set $V:=U_{R}$.

Lemma 1 The family ${\left\{u_{\varepsilon}\right\}_{\varepsilon>0}}$ is uniformly bounded on ${\overline{V}}$ for ${0<\varepsilon.

Proof: For ${\varepsilon>0}$, ${u_{\varepsilon}}$ satisfies the mean value property. For ${B(x,r)\subset\Omega}$,

$\displaystyle \left|u_{\varepsilon}(x)\right|=\dfrac{1}{\left|B(x,r)\right|}\left|\int_{B(x,r)}u_{\varepsilon}(y)dy\right|\leq\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}\left|u_{\varepsilon}(y)\right|dy$

If ${x\in\overline{V}}$, then ${B(x,R/2)\subset\Omega}$, so we obtain the estimate

$\displaystyle \left|u_{\varepsilon}(x)\right|\leq\dfrac{1}{\left|B(x,R/2)\right|}\int_{B(x,R/2)}\left|u_{\varepsilon}(y)\right|dy=\dfrac{2^{n}}{\alpha(n)R^{n}}\leq\dfrac{2^{n}}{\alpha(n)R^{n}}\left\|u\right\|_{L^{1}(\Omega)}$

for all ${x\in\overline{V}}$. $\Box$

Lemma 2 The family ${\left\{u_{\varepsilon}\right\}_{\varepsilon>0}}$ is equicontinuous on ${\overline{V}}$ for ${0<\varepsilon.

Proof: For any ${x,y\in\overline{V}}$ and ${0<\varepsilon.

$\displaystyle \begin{array}{lcl} \displaystyle\left|u_{\varepsilon}(x)-u_{\varepsilon}(y)\right|&=&\displaystyle\left|\dfrac{1}{\left|B(x,s)\right|}\int_{B(x,s)}u_{\varepsilon}(z)dz-\dfrac{1}{\left|B(y,s)\right|}\int_{B(y,s)}u_{\varepsilon}(z)dz\right|\\ [2 em]&=&\displaystyle\dfrac{1}{\alpha(n)s^{n}}\int_{B(x,s)}u_{\varepsilon}(z)dz-\int_{B(y,s)}\left|u_{\varepsilon}(z)dz\right|\\ [2 em]&\leq&\displaystyle\dfrac{1}{\alpha(n)s^{n}}\left[\int_{B(x,s)\setminus B(y,s)}\left|u_{\varepsilon}(z)\right|dz+\int_{B(y,s)\setminus B(x,s)}\left|u_{\varepsilon}(z)\right|dz\right] \end{array}$

As ${u_{\varepsilon}}$ is uniformly bounded on ${\overline{V}}$ for all ${0<\varepsilon (with constant independent of ${\varepsilon}$) and the Lebesgue measure of the two domains depends only on ${\left|x-y\right|}$ and goes to zero as ${\left|x-y\right|\rightarrow 0}$, we obtain equicontinuity. $\Box$

By the Arzelà-Ascoli theorem, there exists a sequence ${u_{\varepsilon_{k}}}$ which converges uniformly to a continuous function ${v\in C(\overline{V})}$. Replacing ${\left\{\eta_{\varepsilon}\right\}_{\varepsilon}}$ with a countable subset with indices tending to zero, we may assume that ${\varepsilon_{k}\rightarrow 0}$.

To prove that ${v}$ is harmonic, we need the following lemma.

Lemma 3 If ${v\in L_{loc}^{1}(\Omega)}$ has the mean value property, then ${v}$ is (classically) harmonic.

Proof: Set ${\chi_{r}(x):=\left|B(0,r)\right|^{-1}\chi_{B(0,r)}(x)}$. Observe that by the Lebesgue dominated convergence theorem ${v\ast\chi_{r}}$ is continuous. By the mean value property,

$\displaystyle v\ast\chi_{r}(x)=\dfrac{1}{\left|B(0,r)\right|}\int_{\Omega}v(x-y)\chi_{B(0,r)}(y)dy=\dfrac{1}{\left|B(x,r)\right|}\int_{B(x,r)}v(z)dz=v(x),$

where we use the translation invariance of the Lebesgue measure to obtain the penultimate equality. Hence, ${v\in C(\Omega)}$.

We now show that ${v}$ is equal to a smooth function a.e. on ${\Omega}$. Set ${v_{\varepsilon}:=v\ast\eta_{\varepsilon}}$. Then

$\displaystyle \begin{array}{lcl} v_{\varepsilon}(x)&=&\displaystyle\int_{B(0,\varepsilon)}v(x-y)\eta_{\varepsilon}(y)dy\\ [2 em]&=&\displaystyle\varepsilon^{-n}\int_{B(0,\varepsilon)}v(x-y)\eta(y/\varepsilon)dy\\ [2 em]&=&\displaystyle\int_{B(0,1)}v(x-\varepsilon y)\eta(y)dy\\ [2 em]&=&\displaystyle\int_{0}^{1}\int_{\partial B(0,s)}v(x-\varepsilon z)\eta(z)dS(z)ds\\ [2 em]&=&\displaystyle\int_{0}^{1}v(x)n\alpha(n)s^{n-1}\eta(s)ds\\ [2 em]&=&\displaystyle v(x)\cdot\int_{B(0,\varepsilon)}\eta_{\varepsilon}(y)dy=v(x) \end{array}$

In particular, ${v\in C^{2}(\Omega)}$, so by the converse to the mean value property (see [Evans] Theorem 3, pp. 26), ${v}$ is harmonic. $\Box$

As ${\left\{\eta_{\varepsilon}\right\}_{\varepsilon>0}}$ is an approximate identity, ${\left\|u_{\varepsilon}-u\right\|_{L^{1}}\rightarrow 0}$ as ${\varepsilon\rightarrow 0}$. Hence, ${v=u}$ a.e. As we proved last time, $u$ is therefore equal to an analytic function a.e.

Next time, we will extend the Weyl lemma to distributions that satisfy the Laplace equation.

1. L.C. Evans, Partial Differential Equations (Second Edition), AMS, 2010.
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