## Category Archives: Real Analysis

### BV function and its property involves translation

Theorem 1 Suppose ${u\in L^1(\mathbb{R})}$, then ${u\in \text{BV}}$ if and only if ${\exists\, C}$ such that

$\displaystyle ||\tau_hu-u||_{L^1(\mathbb{R})}\leq C|h|,\quad \forall\, h$

Moreover, one can take ${C=|u|_{BV}}$. Here ${\tau_hu(\cdot)=u(\cdot+h)}$ is the tanslation operator.

Proof: Firstly suppose ${u\in \text{BV}}$. Let us prove

$\displaystyle \left|\int_{\mathbb{R}}(\tau_hu(x)-u(x))\phi(x)dx\right|\leq |u|_{BV}|\phi|_{L^\infty}|h|,\quad \forall \phi\in C^\infty_c(\mathbb{R}) \ \ \ \ \ (1)$

To show that

$\displaystyle LHS=\left|\int_{\mathbb{R}}u(x)(\phi(x-h)-\phi(x))dx\right|$

$\displaystyle =\left|\int_{\mathbb{R}}u(x)\psi(x)hdx\right|$

$\displaystyle \leq |u|_{BV}|\psi|_{L^\infty}|h|$

where

$\displaystyle \psi(x)=\int^x_{-\infty}\frac{\phi(s-h)-\phi(s)}{h}ds\in C_c^\infty(\mathbb{R})$

it is easy to verify ${|\psi|_\infty=|\phi|_\infty}$ therefore (1) is proved. Next one can choose such ${\phi_n\rightarrow sign(\tau_hu-u)\in L^1}$ with ${|\phi_n|\leq 1}$(it is easy to show by mollification). By dominating theorem, one get

$\displaystyle \int_{\mathbb{R}}|\tau_hu(x)-u(x)|dx\leq |u|_{BV}|h|.$

The other direction need more analysis. $\Box$

### Sobolev functions on puncture ball

Let us first see the prove of sobolev embedding of ${W^{1,1}\rightarrow L^2}$ on the plane.

Lemma: Suppose ${f\in W^{1,1}(\mathbb{R}^2)}$ with compact support. Then

$\displaystyle ||f||_{L^2}\leq ||\nabla f||_{L^1}$

Proof: Let us suppose ${f\in C^\infty_c(\mathbb{R}^2)}$, the general case can be proved by approximation. Since ${f}$ has compact support, then

$\displaystyle |f(x,y)|=\left|\int_{-\infty}^x\frac{\partial f}{\partial x}(t,y)dt\right|\leq \int_{-\infty}^{\infty}|\nabla f|(t,y)dt=F(y)$

$\displaystyle |f(x,y)|=\left|\int_{-\infty}^y\frac{\partial f}{\partial y}(x,s)ds\right|\leq \int_{-\infty}^{\infty}|\nabla f|(x,s)ds=G(x)$

Then

$\displaystyle \int_{\mathbb{R}^2}f^2(x,y)dxdy\leq \int_{-\infty}^\infty\int_{-\infty}^\infty F(y)G(x)dxdy$

$\displaystyle =\int_{-\infty}^\infty F(y)dy\int_{-\infty}^\infty G(x)dx\\ =\left(\int_{-\infty}^\infty\int_{-\infty}^\infty|\nabla f|dxdy\right)^2$

$\Box$

Suppose we have a function ${u\in W^{1,1}_{loc}(D\backslash\{0\})}$, where ${D}$ is the unit disc in ${\mathbb{R}^2}$, ${u}$ can blow up wildly near the origin. However if we know ${\nabla u\in L^1(D)}$, then actually ${u\in L^2(D)}$ and ${u\in W^{1,1}(D)}$.

Proof: Because the bad thing happened only at origin, we can suppose ${u}$ has spt inside ${\frac{1}{4}D}$ or ${D_{1/4}}$. Put a substantially large square box ${B_\epsilon}$ with length ${\frac 12}$ inside the left half of the disc ${D^-}$whose distance to the origin is ${\epsilon}$ see the picture.

puncture disc

Then on the three sides, ${l_1}$, ${l_2}$, ${l_3}$, ${u=0}$. Using the proof of the above, one can prove

$\displaystyle ||u||_{L^2(B_\epsilon)}\leq ||\nabla u||_{L^1(B_\epsilon)}\leq ||\nabla u||_{L^1(D)}$

Letting ${\epsilon\rightarrow 0}$, we get ${u\in L^2(D^-)}$. The same proof works for the right part ${D^+}$. Finally ${u\in L^2(D)}$. Choose a cut off function ${\zeta_\epsilon=\zeta(x/\epsilon)}$. Then

$\displaystyle u\zeta_\epsilon\rightarrow u\text{ in }L^1$

$\displaystyle \nabla(u\zeta_\epsilon)\rightarrow \nabla u\text{ in }L^1$

So ${u\in W^{1,1}(D)}$. $\Box$

Remark: This is called the removable singularity. There is a more general theorem related to this. Assume ${n\geq 2}$, ${K\subset\subset\Omega}$ such that ${\mathcal{H}^{n-2}(K)=0}$. Suppose ${u\in W^{1,1}_{loc}(\Omega\backslash K)}$ and ${\int_{\Omega\backslash K}|\nabla u|dx<\infty}$. Then ${u\in W^{1,1}(\Omega)}$.

I learned this from Prof. Brezis’s class. Also see his book: Sobolev maps with values into the circle.

### Sharpness of Morrey’s inequality

Thm: Suppose ${n, then there exists a constant ${C}$ such that

$\displaystyle ||u||_{C^{0,\gamma}(\mathbb{R}^n)}\leq C(n,p)||u||_{W^{1,p}(\mathbb{R}^n)}$

for all ${u\in C^1(\mathbb{R}^n)}$, where ${\gamma=1-n/p}$. For functions in ${W^{1,p}(\Omega)}$, we have

Thm: Suppose ${\Omega}$ is bounded domain in ${\mathbb{R}^n}$, and ${\partial\Omega\in C^1}$. Assume ${n and ${u\in W^{1,p}(\Omega)}$, there exists one version ${u^*\in C^{0,\gamma}(\bar{\Omega})}$, for ${\gamma=1-n/p}$, such that

$\displaystyle ||u^*||_{C^{0,\gamma}(\mathbb{R}^n)}\leq C(n,p,\Omega)||u||_{W^{1,p}(\mathbb{R}^n)}$

Remark: This ${\gamma}$ is critical number. For ${\beta\in (\gamma,1]}$, we can choose ${\alpha\in (\gamma,\beta)}$. Consider the function ${u(x)=|x|^\alpha}$ on the unit ball ${B_1}$. Then

$\displaystyle D_iu=\alpha x_i|x|^{\alpha-2}$

${u\in W^{1,p}(B_1)}$ if and only if ${(1-\alpha)p, which is ${\alpha>\gamma}$.

However consider the ${[u]_\beta}$ which is

$\displaystyle [u]_{\beta;B_1}=\sup_{x,y\in B_1}\frac{\big||x|^\alpha-|y|^\beta\big|}{|x-y|^\beta}\geq \sup_{x\in B_1}\frac{|x|^\alpha}{|x|^\beta}=+\infty$

So ${u\not\in C^{0,\beta}(B_1)}$.

### Sharpness of Sobolev embedding

Consider the Sobolev embedding

Thm 1: Suppose ${u\in W^{1,p}(\Omega)}$, ${\Omega}$ is a bounded domain in ${\mathbb{R}^n}$ with ${C^1}$ boundary, ${p, then

$\displaystyle ||u||_{L^q(\Omega)}\leq C(n,p,q,\Omega)||u||_{W^{1,p}(\Omega)}$

for any ${1\leq q\leq \frac{np}{n-p}}$.

Consider the sharpness of Sobolev embedding, which means ${q}$ can not be bigger than ${\frac{np}{n-p}}$.

WLOG assume ${\Omega=B_1(0)}$. Choose a particular function ${u(x)=\frac{1}{|x|^\alpha}}$, then ${u\in W^{1,p}(B_1)}$ when ${\alpha<\frac{n-p}{p}}$. The reason is the following

${u}$ is smooth away from ${0}$ with

$\displaystyle D_iu=\frac{-\alpha x_i}{|x|^{\alpha+2}}$

For any ${\phi\in C_c^\infty(B_1)}$

$\displaystyle \int_{B_1-B_\epsilon}uD_i\phi=-\int_{B_1-B_\epsilon}D_iu\phi+\int_{\partial B_\epsilon}u\phi\nu_ids$

Let ${\epsilon \rightarrow 0}$, we have

$\displaystyle \left|\int_{\partial B_\epsilon}u\phi\nu_ids\right|\leq C\int_{\partial B_\epsilon}\epsilon^{-\alpha}ds\leq \epsilon^{-\alpha+n-1}\rightarrow 0$

Then

$\displaystyle \int_B uD_i\phi=-\int_BD_iu\phi$

for any ${\phi\in C^\infty_c(B)}$ when ${\alpha. ${u}$ has weak derivative ${D_iu}$, ${Du\in L^p(B)}$ only when ${\alpha<\frac{n-p}{p}}$.

Let us calculate

$\displaystyle ||u||_{L^q}=C\left(\int_0^1r^{-\alpha q+n-1}dr\right)^{1/q}$

${u\in L^q(B_1)}$ if and only if ${\alpha<\frac{n}{q}}$.

So if ${q>\frac{np}{n-p}}$, we can find one ${\alpha}$ such that ${\alpha\in (\frac{n}{q}, \frac{n-p}{p})}$. By the above analysis, such ${u\in W^{1,p}(B_1)}$ but ${u\not\in L^q(B_1)}$.

Thm 2: Suppose ${u\in W^{1,p}(\Omega)}$, ${\Omega}$ is a bounded domain in ${\mathbb{R}^n}$ with ${C^1}$ boundary, ${p, then the embedding ${W^{1,p}(\Omega)\rightarrow L^q(\Omega)}$ is compact when ${q\in [1,\frac{n-p}{np})}$.

Consider the sharpness of compact embedding, ${q}$ must be strictly less than ${\frac{n-p}{np}}$. Actually we can find a sequence of ${u_n\in W^{1,p}(\Omega)}$ but ${u_n}$ does not have convergent subsequence in ${L^{p*}(\Omega)}$.

As before assume ${\Omega=B_1}$. Choose ${u\in C^1_0(B_1)}$, define ${u_{\lambda}(x)=\lambda^\alpha u(\lambda x)}$, for ${\lambda\geq 1}$, then ${u_\lambda\in C^1_0(B_{1/\lambda})}$

$\displaystyle ||u_\lambda||_{L^{p*}(B_1)}=||u_\lambda||_{L^{p*}(\mathbb{R}^n)}=\lambda^{\alpha-n/p*}||u||_{L^{p*}(B_1)}$

$\displaystyle ||Du_\lambda||_{L^p(B_1)}=\lambda^{\alpha+1-n/p}||Du||_{L^p(B_1)}$

Since ${u_\lambda}$ has compact support, then

$\displaystyle ||u||_{W^{1,p}}\leq C||Du||_{L^p}\leq C\lambda^{\alpha+1-n/p}||Du||_{L^p(B_1)}$

Choose ${\alpha=n/p*=n/p-1}$, then ${||u||_{W^{1,p}}}$ is bounded. However, ${u_\lambda}$ has no convergent subsequence. Otherwise as ${\lambda\rightarrow \infty}$, ${u_\lambda(x)\rightarrow 0}$ when ${x\neq 0}$, such subsequence must converge to 0 in ${L^{p*}(\Omega)}$. Since we have ${||u_\lambda||_{L^{p*}}=||u||_{L^{p*}}}$, apparently this can not be true.

### Small things about Hausdorff measure

Suppose ${X}$ is a metric space, ${\mathscr{F}}$ is a families of subsets of ${X}$, and ${\phi}$ is a function on ${\mathscr{F}}$ such that

$\displaystyle 0\leq \phi(A)\leq \infty\quad\text{ whenever }A\in \mathscr{F}$

we can obtain a measure through a prodecure called Caratheodory’s construction. Define

$\displaystyle \tilde{\phi}_\delta(E)=\inf\left\{\sum_i\phi(A_i)\big| E\subset\cup_i A_i\text{ and diam}A_i<\delta\right\}$

for any set ${E\subset X}$. Then

$\displaystyle \tilde{\phi}_\delta\geq\tilde{\phi}_\sigma \text{ if }0<\delta<\sigma<\infty$

$\displaystyle \tilde{\phi}_\delta(E\cup F)\geq \tilde{\phi}_\delta(E)+\tilde{\phi}_\delta(F)\text{ if dist}(E,F)>\delta>0$

This means ${\tilde{\phi}_\delta(E)}$ is monotone, then it is reasonable to define

$\displaystyle \lambda(E)=\lim\limits_{\delta\rightarrow 0+}\tilde{\phi}_\delta(E)=\sup_{\delta>0}\tilde{\phi}_\delta(E)$

${\tilde{\phi}_\delta}$ and ${\lambda}$ are (outer) measures. Moreover

$\displaystyle \lambda(E\cup F)\geq \lambda(E)+\lambda(F)\text{ if dist}(E,F)>0$

This means all open set are measurable in ${\lambda}$. If ${\mathscr{F}}$ contains Borel sets of ${X}$, then ${\lambda}$ is Borel regular measure.

Suppose ${X=\mathbb{R}^n}$, ${\mathscr{F}}$ is the Borel set of ${\mathbb{R}^n}$, ${m\geq0}$

$\displaystyle \phi(B)=w_m\left(\frac{\text{diam }B}{2}\right)^m, \quad B\in \mathscr{F}$

Then the Caratheodory’s construction will yield the ${m-}$dimensional Hausdorff measure.

Lemma: ${X}$ is a (separable?) metric space, ${\mu}$ is a measure on ${Y}$, ${f:X\rightarrow Y}$, ${f(A)}$ is measurable whenever ${A}$ is a Borel subset of ${X}$, define

$\displaystyle \phi(A)=\mu(f(A))$

${\lambda}$ is obtained from the Caratheodory’s construction from ${\phi}$ on all Borel subsets of ${X}$. Then

$\displaystyle \lambda(A)=\int_YN(f|A,y)d\mu(y)\text{ whenever }A\text{ is a Borel set }$

where ${N(f|A,y)=card\{x\in A|f(x)=y\}}$. Proof: Find Borel partitions ${\mathscr{P}_i}$ of ${A}$ such that ${\mathscr{P}_{i+1}}$ is a refinement of ${\mathscr{P}_i}$, or all subsets in ${\mathscr{P}_i}$ are unions of that of ${\mathscr{P}_{i+1}}$. And also assume

$\displaystyle \lim_{i\rightarrow \infty} \sup_{E\subset \mathscr{P}_i} \text{diam}(E)=0$

Let ${\chi^i_{f(E)}}$ be the characteristic function of ${f(E)}$, ${E}$ is from partition ${\mathscr{P}_i}$.

$\displaystyle \sum_{E\in \mathscr{P}_i}\chi^i_{f(E)}(y)\nearrow N(f|A,y)$

From Levy’s monotone convergence thm,

$\displaystyle \int_Y N(f|A,y)d\mu(y)=\lim_{i\rightarrow \infty}\int_Y \sum_{E\in \mathscr{P}_i}\chi^i_{f(E)}(y) d\mu(y)=\lim_{i\rightarrow \infty}\sum_{E\in \mathscr{P}_i}\mu(f(E))$

$\displaystyle =\lim_{i\rightarrow \infty}\sum_{E\in \mathscr{P}_i}\phi(E)\geq \lambda(A)$

by the defintion of ${\lambda(A)}$.

Note that the defition of ${\phi(A)}$ implies ${\tilde{\phi}_\delta\geq \phi(A)}$ for any ${\delta>0}$, hence ${\lambda(A)\geq \phi(A)}$ for very Borel set ${A}$. Since all Borel set are ${\lambda}$ measurable, for any partition ${\mathscr{P}}$

$\displaystyle \lambda(A)=\sum_{E\in \mathscr{P}}\lambda(E)\geq \sum_{E\in \mathscr{P}}\phi(E)=\sum_{E\in \mathscr{P}}\mu(f(E))$

combining with the result before,

$\displaystyle \lambda(A)=\int_YN(f|A,y)d\mu(y)$

$\Box$

Theorem: Suppose ${f:\mathbb{R}^l\rightarrow \mathbb{R}^n}$ is a Lipschitzian map. ${m}$ is nonegative number, then

$\displaystyle \int_{\mathbb{R}^n}N(f|A,y)d\mathscr{H}^m(y)\leq (Lip f)^m\mathscr{H}^m(A)$

whenever ${A\subset\mathbb{R}^l}$ is a Borel subset. Proof: Let us apply the above lemma with ${\mu=\mathscr{H}^m}$ on ${\mathbb{R}^n}$, then

$\displaystyle \lambda(A)=\int_{\mathbb{R}^n}N(f|A,y)d\mathscr{H}^m(y)$

For any Borel set ${B}$, any covering of ${B}$ by sets diameter ${\leq \delta}$ yields a covering of ${f(B)}$ of diameter ${\leq (Lip f)\delta}$, then

$\displaystyle \phi(B)=\mathscr{H}^m(f(B))\leq (Lip f)^m\mathscr{H}^m(B)$

$\displaystyle \tilde{\phi}_\delta(A)\leq \sum_{\text{diam}B_i<\delta, B_i\in \mathscr{P}}\phi(B_i)\leq (Lip f)^m\mathscr{H}^m(A)$

By taking the limit, ${\lambda(A)\leq (Lip f)^m\mathscr{H}^m(A)}$. $\Box$

Corollary: For any connected set ${E\subset \mathbb{R}^n}$

$\displaystyle \mathscr{H}^1(E)\geq \text{diam}\,E$

Proof: Since ${\mathscr{H}^1}$ is a regular measure on ${\mathbb{R}^n}$, we can find Borel set ${B}$ containing ${E}$ with the same ${\mathscr{H}^1}$ measure. Choose ${a\in B}$, define ${f:\mathbb{R}^n\rightarrow \mathbb{R}^1}$, ${f(x)=|x-a|}$. Then ${f}$ is Lipschitzian map and ${Lip\, f=1}$. By the previous theorem

$\displaystyle \mathscr{H}^1(B)\geq\int_{\mathbb{R}^1}N(f|B,y)d\mathscr{H}^1(y)\geq \mathscr{H}^1(f(E))=\mathscr{L}^1(f(E))\geq\text{diam}\,E$

$\Box$

### Product rule or Product formula for weak derivative

Problem: Derive the product formula for weak derivative

$\displaystyle D(uv)=uDv+vDu$

holds for all ${u,v\in W^1(\Omega)}$ such that ${uv}$ and ${uDv+vDu\in L^1_{loc}(\Omega)}$.

Proof: Step 1: prove the case ${u\in W^1(\Omega)}$, ${v\in C^1(\Omega)}$

Step 2: prove the case ${u,v\in W^{1}(\Omega)\cap L^\infty_{loc}(\Omega)}$ by step 1. Readers can also see the proof on page269 of Functional Analysis, Sobolev Spaces and Partial Differential Equations (Haim Brezis)

Step 3: Define ${f\in C^0(\mathbb{R}^n)}$

$\displaystyle f_n(t)=\begin{cases}n,\quad t>n\\ t,\quad |t|\leq n\\ -n,\quad t<-n\end{cases}$

then ${f_n}$ is piecewise smooth in ${\mathbb{R}}$ and ${f_n'\in L^\infty (\mathbb{R})}$. So ${f_n(u)\in W^1(\Omega)}$ by lemma 7.8 in Gilbarg and Trudinger’s bk and

$\displaystyle D(f_n(u))(x)=\begin{cases}Du(x),\quad |u(x)|\leq n,\\ 0,\quad\quad |u(x)|>n.\end{cases}$

Denote ${u_n=f_n(u)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)}$ and ${v_n=f_n(v)\in W^1(\Omega)\cap L^\infty_{loc}(\Omega)}$. By step 2 we have ${u_nv_n\in W^1(\Omega)}$ and

$\displaystyle \int_{\Omega} u_nv_n D\phi dx=-\int_{\Omega}(u_nDv_n+v_n Du_n)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)$

Note that by assumption

$\displaystyle |u_nv_n|\leq |uv|\in L^1_{loc}(\Omega)$

$\displaystyle |u_nDv_n|\leq |uDv|\in L^1_{loc}(\Omega)$

$\displaystyle |v_nDu_n|\leq |vDu|\in L^1_{loc}(\Omega)$

By dominating convergence theorem, letting ${n\rightarrow \infty}$

$\displaystyle \int_{\Omega} uv D\phi dx=-\int_{\Omega}(uDv+v Du)\phi dx\quad \forall\, \phi\in C^\infty_0(\Omega)$

Step 4:Consider the problem with additional assumption ${u\geq 0}$ and ${v\geq 0}$.

Firstly we assume ${v\geq 1}$. Define ${\tilde{u}_n=\min\{u,\frac{n}{v}\}=u+(\frac{n}{v}-u)^+}$. Since ${u,\frac{n}{v} \in W^1(\Omega)}$, then ${\tilde{u}_n\in W^1(\Omega)}$, satisfies ${0\leq \tilde{u}_n v\leq n}$, moreover

$\displaystyle D\tilde{u}_n=\begin{cases}Du,\quad uv

$\displaystyle vD\tilde{u}_n+\tilde{u}_n Dv=\begin{cases}vDu+uDv,\quad uv

Suppose ${\displaystyle f_\epsilon(t)=\frac{t}{1+\epsilon t}}$ defined on ${\mathbb{R}_+^1}$, where ${\epsilon>0}$. Then ${f_\epsilon}$ is a piecewise smooth function in ${\mathbb{R}_+^1}$ and ${f'_\epsilon\in L^\infty(\mathbb{R}_+^1)}$. By lemma 7.8 on Gilbarg and trudinger’s book, ${f_\epsilon(\tilde{u}_n)\in W^1(\Omega)}$. Note that ${f_\epsilon(u)\in L^\infty(\Omega)}$ by the conclusion of step 3,

$\displaystyle \int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi=-\int_{\Omega}[Df_\epsilon(\tilde{u}_n)v+f_\epsilon(\tilde{u}_n)Dv]\phi$

that is

$\displaystyle \int_{\Omega}\frac{\tilde{u}_n}{1+\epsilon \tilde{u}_n}vD\phi=-\int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon \tilde{u}_n}\phi+\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon \tilde{u}_n)^2}\phi\quad (1)$

By dominating convergence theorem, as ${\epsilon\rightarrow 0}$

$\displaystyle \int_{\Omega}f_\epsilon(\tilde{u}_n) vD\phi\rightarrow \int_{\Omega}\tilde{u}_nvD\phi$

$\displaystyle \int_{\Omega}\frac{\tilde{u}_nDv+vD\tilde{u}_n}{1+\epsilon\tilde{u}_n}\phi\rightarrow \int_{\Omega}(\tilde{u}_nDv+vD\tilde{u}_n)\phi$

And

$\displaystyle \left|\int_{\Omega}\frac{\epsilon \tilde{u}_nvD\tilde{u}_n}{(1+\epsilon\tilde{u}_n)^2}\phi\right|\leq \epsilon n\int_{\Omega}|D\tilde{u}_n||\phi|\rightarrow 0\text{ as }\epsilon\rightarrow 0$

Letting ${\epsilon\rightarrow 0}$, ${(1)}$ implies

$\displaystyle \int_{\Omega}\tilde{u}_nvD\phi=-\int_{\Omega}[vD\tilde{u}_n+\tilde{u}_nDv]\phi$

Since for any ${n>0}$

$\displaystyle |\tilde{u}_nv|\leq |uv|$

$\displaystyle |vD\tilde{u}_n+\tilde{u}_nDv|\leq |vDu+uDv|$

by dominating convergence theorem, letting ${n\rightarrow \infty}$

$\displaystyle \int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi$

If we only know ${v\geq 0}$, consider ${u}$ and ${v+1}$, repeat the above proof

$\displaystyle \int_{\Omega}u(v+1)D\phi=-\int_{\Omega}[(v+1)Du+uD(v+1)]\phi$

note that ${u\in W^1(\Omega)}$, this is equivalent to

$\displaystyle \int_{\Omega}uvD\phi=-\int_{\Omega}[vDu+uDv]\phi$

Step 5: Consider the most general case with no extra assumption. Since

$\displaystyle u^+v^+, u^+v^-, u^-v^+,u^-v^-\in L^\infty_{loc}(\Omega)$

$\displaystyle v^{\pm}Du^\pm+u^\pm Dv^\pm\in L^\infty_{loc}(\Omega)\text{ respectively }$

step 4 will imply

$\displaystyle u^\pm v^\pm\in W^1(\Omega),\quad D(u^\pm v^\pm)=v^{\pm}Du^\pm+u^\pm Dv^\pm$

So

$\displaystyle uv= u^+v^+-u^+v^-- u^-v^++u^-v^-\in W^1(\Omega)$

and

$\displaystyle D(uv)=uDv+vDu$

$\Box$

Remark: Who can simplify this proof? It is ugly.

### Regularity of Newtonian Potential

Suppose ${\Gamma(\cdot)}$ is the fundamental solution of laplace equation.

$\displaystyle \Gamma(x-y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}, n>2\\\frac{1}{2\pi}\log|x-y|, n=2.\end{cases}$

We know it has property

$\displaystyle |D_{y_i}\Gamma(x-y)|\leq C\frac{1}{|x-y|^{n-1}}$

We can define the Newtonianial potential of ${f}$

$\displaystyle Nf(x)=\int_{\Omega}\Gamma(x-y)f(y)dy\quad x\in\mathbb{R}^n$

${Nf}$ is well defined. We have the following propery of ${Nf}$

Thm: Suppose ${\Omega\subset\mathbb{R}^n}$ is a bounded domain, ${n\geq 2}$. Assume ${f}$ is bounded and local integrable in ${\Omega}$, then ${Nf}$ is ${C^1(\mathbb{R}^n)}$ and

$\displaystyle D_{x_i}Nf=\int_{\Omega} D_{x_i}\Gamma(x-y)f(y)dy$

Proof: Fix ${0<\epsilon<1}$, let ${B_\epsilon(x)=\{y\in\Omega||x-y|<\epsilon\}}$ and ${B_\epsilon^c(x)=\{y\in\Omega||x-y|\geq\epsilon\}}$.

$\displaystyle Nf=\int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy+\int_{B^c_\epsilon(x)}\Gamma(x-y)f(y)dy=I+II$

Since ${D_{x_i}\Gamma(x-y)}$ is uniformly bounded in ${B^c_\epsilon(x)}$, by the Lebesgue dominating theorem, ${II}$ is differentiable and

$\displaystyle D_{x_i}II=\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy$

Considering ${I}$, we will prove if ${|x-z|<\epsilon/2}$

$\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}$

where ${C=C(||f||_{L^\infty},n,\Omega)}$.

$\displaystyle \int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy-\int_{B_\epsilon(z)}\Gamma(z-y)f(y)dy=\int_{B_\epsilon(x)\cap B^c_\epsilon(z)}\Gamma(x-y)f(y)dy$

$\displaystyle +\int_{B_\epsilon(x)\cap B_\epsilon(z)}[\Gamma(x-y)-\Gamma(z-y)]f(y)dy-\int_{B^c_\epsilon(x)\cap B_\epsilon(z)}\Gamma(z-y)f(y)dy$

${=\mathcal{A}+\mathcal{B}+\mathcal{C}}$

For any ${y\in B_\epsilon(x)\cap B^c_\epsilon(z)}$, ${|y-x|\geq |y-z|-|z-x|\geq \epsilon/2}$. So

$\displaystyle |\mathcal{A}|\leq C\frac{1}{\epsilon^{n-2}} |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\frac{\epsilon^n-(\epsilon-|x-z|/2)^n}{\epsilon^{n-2}}\leq C\epsilon |x-z|\text{ if } n>2$

$\displaystyle |\mathcal{A}|\leq C|\log\frac{\epsilon}{2}| |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2$

Similarly for ${C}$, we also have

$\displaystyle |\mathcal{C}|\leq\begin{cases}C\epsilon|x-z|\quad \quad \text{ if } n>2\\ C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}$

Considering ${\mathcal{B}}$, we have

$\displaystyle |\mathcal{B}|\leq C\int_{B_\epsilon(x)\cap B_\epsilon(z)}\int_0^1|D_{x_i}\Gamma(tx+(1-t)z-y)||x-z|dtdy$

$\displaystyle =C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)}|D_{x_i}\Gamma(tx+(1-t)z-y)|dydt$

$\displaystyle \leq C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)} \frac{1}{|tx+(1-t)z-y|^{n-1}}dydt$

$\displaystyle \leq C|x-z|\int^1_0 C\epsilon dt\leq C\epsilon |x-z|$

Combing the fact about ${\mathcal{A},\mathcal{B},\mathcal{C}}$, we get

$\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}$

So

$\displaystyle \left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|$

$\displaystyle \leq \left|\frac{II(x)-II(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\frac{I(x)-I(z)}{x-z}\right|$

Applying the fact that ${II}$ is differentiable and the fact about ${I}$

$\displaystyle \overline{\lim\limits_{z\rightarrow x}}\left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|$

$\displaystyle \leq \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}$

$\displaystyle \leq C\epsilon+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}$

Let ${\epsilon\rightarrow 0}$, we get ${D_{x_i}Nf}$ exists and

$\displaystyle D_{x_i}Nf=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy$

Next we shall prove ${D_{x_i}Nf}$ is continuous for ${i=1,\cdots,n}$, then ${Nf\in C^1(\mathbb{R}^n)}$

$\displaystyle \left|\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy-\int_{\Omega}D_{x_i}\Gamma(z-y)f(y)dy\right|$

$\displaystyle \leq\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|$

$\displaystyle +\left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|$

By the dominating theorem

$\displaystyle \lim\limits_{x\rightarrow z}\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|=0$

$\displaystyle \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq C\epsilon$

$\displaystyle \left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|\leq C\epsilon$

where ${C=C(||f||_{L^\infty}, n)}$. So ${\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy=D_{x_i}Nf}$ is continuous everywhere.

Remark: This revised version also have errors. It is a shame. What do you mean ${II}$ is differentiable? Don’t you notice that the integral domain of ${II}$ also has relation with ${x}$?

To prove this theorem rigorously and neatly, we need the following lemma from calculus

Lemma: Suppose ${u_n{x}}$ is a sequence of differntiable function on ${[a,b]\rightarrow \mathbb{R}}$. If ${u_n}$ converges to another function ${u}$ uniformly(${u}$ is finite somewhere) and ${u'_n}$ converges uniformly to ${v}$ in ${\Omega}$ then ${u}$ is differentiable and ${u'=v}$.

Proof: Let ${\xi(r):C^1(\mathbb{R}_+^1)\rightarrow \mathbb{R}^1}$ and ${0\leq\eta'\leq 2}$, $\xi(r)=0\text{ if } r\leq 1$, $\xi(r)=1\text{ if }r>2$.

Define

${\displaystyle w_\epsilon=\int_{\Omega}\xi\left(\frac{|x-y|}{\epsilon}\right)\Gamma(x-y)f(y)dy}$

Let ${\xi_\epsilon(r)=\xi_{\epsilon}(r/\epsilon)}$ Since

$\displaystyle \left|D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)f(y)\right)\right|\leq \frac{C}{\epsilon^{n-1}}f(y)$

which is in ${L^1(\Omega)}$. By the Lebesgue Differentiable theorem, ${w_\epsilon}$ is differntiable and

$\displaystyle D_{x_i}w_\epsilon=\int_{\Omega}D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)\right)f(y)dy$

$\displaystyle \left|D_{x_i}w_\epsilon-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq \sup|f|\int_{\Omega}\left|D_{x_i}((1-\xi_\epsilon)\Gamma(x-y))\right|dy$

$\displaystyle \leq \sup|f|\int_{|x-y|\leq 2\epsilon}|D_{x_i}\Gamma|+\frac{2}{\epsilon}|\Gamma|dy$

$\displaystyle \leq \sup|f|\begin{cases}C\epsilon\quad \quad \text{ if }n>2\\ C\epsilon (1+|\log\epsilon|)\text{ if }n=2\end{cases}$

So ${D_{x_i}w_\epsilon}$ converges to ${\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy}$ uniformly on any compact subset of ${\mathbb{R}^n}$. And it is easy to prove ${w_\epsilon}$ converges to ${Nf}$ uniformly on any compact subset of ${\mathbb{R}^n}$. So by the lemma, we have ${Nf}$ is differentiable and has

$\displaystyle D_{x_i}Nf(x)=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy,\quad i=1,2\cdots,n$

### Measure on sphere

Considering an n-sphere, which is the unit sphere in $\mathbb{R}^n$, we usually encounter this kind of integration

$\displaystyle \int_{\mathbb{S}^n}f(x)dS_x$

here $dS_x$ is the sphere measure. For example find the value of $\displaystyle \int_{\mathbb{S}^n}x^+_ndS_x$.

To calculate this integral, we have to understand $dS_x$. What is the relation to Lebesgue measure in Euclidean space?

The key point is viewing $\mathbb{S}^n$ as a manifold, then $dS_x$ is the volume element of $\mathbb{S}^n$, which can be written as $dS_x=\sqrt{g}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}$ if $u_1,u_2,\cdots,u_{n-1}$ is the local coordinates of $\mathbb{S}^n$.

$\displaystyle \psi: B_1(0)\mapsto \mathbb{S}_+^n$

$\displaystyle (u_1,u_2,\cdots,u_{n-1})\to (u_1,u_2,\cdots,u_{n-1}, \gamma)$

where $B_1(0)\subset \mathbb{R}^{n-1}$ and $\gamma=\sqrt{1-u^2_1-u^2_2-\cdots-u^2_{n-1}}$.

Suppose the metric of $\mathbb{R}^n$ is $g_0$, then it induces a metric $\psi^*g_0$ on $\mathbb{S}^n$.

$\displaystyle \psi^*g_0\left(\frac{\partial }{\partial u_i}, \frac{\partial }{\partial u_j}\right)=\delta_{ij}+\frac{\partial \gamma}{\partial u_i}\frac{\partial \gamma}{\partial u_j}$.

Since $(u_1,u_2,\cdots,u_{n-1})$ can serve as a local coordinates of $\mathbb{S}_+^n$, then

$\displaystyle dS_x=\sqrt{\psi^*g_0}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}=\sqrt{1+|D\gamma|^2}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}.$

Or we can get from the volume form

$\displaystyle \omega = \sum_{j=1}^{n} (-1)^{j-1} x_j \,dx_1 \wedge \cdots \wedge dx_{j-1} \wedge dx_{j+1}\wedge \cdots \wedge dx_{n}$

and substitute $x_{n}$ by $\gamma$.

So

$\displaystyle \int_{\mathbb{S}^n}x^+_ndS_x=\int_{B_1(0)}\gamma\sqrt{1+|D\gamma|^2}du_1\wedge du_2\wedge \cdots \wedge du_{n-1}$

which can be calculated by the polar coordinates.

### Bound of norm on gradient implies bound on function

$\mathbf{Problem:}$ Suppose $p\in L^2_{loc}(B_1)$, $\nabla p\in L^2(B_1)$, prove that $p\in L^2(B_1)$ where $B_1\subset\mathbb{R}^n$ is a ball with radius 1.

We are going to use the following lemma.

Lemma 7.16 on GT’s book. p162

Let $\Omega$ be convex and $u\in W^{1,1}(\Omega)$. Then

$\displaystyle |u(x)-u_S|\leq \frac{d^n}{n|S|}\int_{\Omega}\frac{|Du(y)|}{|x-y|^{n-1}}dy$,   $a.e. \Omega$

where $\displaystyle u_S=\frac{1}{|S|}\int_S udx$, $d=diam\,\Omega$.

$\mathbf{Proof:}$ For any $B_r\subset B_1$ concentric with $B_1$, $p\in W^{1,1}(B_r)$, apply the lemma 7.6

$\displaystyle |p(x)-p_S|\leq \frac{r^n}{n|S|}\int_{B_r}\frac{|\nabla p(y)|}{|x-y|^{n-1}}dy\leq \frac{1}{n|S|}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy\int_{B_1}\frac{1}{|x-y|^{n-1}}dy$

Since $\displaystyle \int_{B_1}\frac{1}{|x-y|^{n-1}}dy$ is uniformly bounded for $\forall x\in\mathbb{R}^n$, then the above inequality implies that

$\displaystyle |p(x)-p_S|\leq C\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dy$

$\displaystyle \int_{B_r}|p(x)-p_S|^2dx\leq C\int_{B_1}\int_{B_1}\frac{|\nabla p(y)|^2}{|x-y|^{n-1}}dydx\leq C\int_{B_1}|\nabla p|^2dy$

Let $r\to 1^+$, we get $||p||_{L^2}\leq C(p_S+||\nabla p||_{L^2})$.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ For any bounded domain with locally lipschitz boundary, this proposition is true. Because the locally boundary enable the domain to have locally star-shaped property, we can apply this proposition locally.

### One example related to weak derivative

$\mathbf{Problem:}$ Suppose $D=\{(x_1,x_2)||x_1|<1, |x_2|<1\}$ is the open square in $\mathbb{R}^2$. $u$ is defined

$\displaystyle u(x)=\begin{cases}1-x_1\quad if\quad x_1>0, |x_2|0, |x_1|

Find the first weak derivative of $u$.

$\mathbf{Proof:}$ Suppose $v=\partial_\alpha u$ is the weak derivative, then for any $\phi\in C^\infty_0(D)$, we have

$\displaystyle \int_D v\phi dx=-\int_D u\partial_\alpha\phi dx.\quad \alpha=1,2\quad (1)$

WLOG, assume $\alpha=1$. Let us denote the four domains in the definition of $u$ as $D_1,D_2,D_3,D_4$ respectively. Note that

$\displaystyle \int_D u\partial_1\phi dx=\int_{D_1} u\partial_1 \phi dx+\int_{D_2} u\partial_1\phi dx+\int_{D_3} u\partial_1\phi dx+\int_{D_4} u\partial_1\phi dx\quad(2)$

While

$\displaystyle \int_{D_1} u\partial_1\phi dx=\iint_{D_1}(1-x_1)\partial_1\phi dx_1dx_2=\int_{0}^1\int_{-x_1}^{x_1}(1-x_1)\partial_1 \phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{-1}^1 \int_{|x_2|}^1(1-x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1-x_1)\phi\big|^1_{|x_2|} +\int_{|x_2|}^1 \phi dx_1\right]dx_2$

$\quad$                       $\displaystyle =-\int_{-1}^1(1-|x_2|)\phi(x_2,x_2)dx_2-\int_{D_1}\partial_1u\,\phi dx\quad (3)$

obtained from the integral by parts and the boundary behavior of $\phi$. Similarly for domain $D_2$

$\displaystyle \int_{D_2} u\partial_1\phi dx=\iint_{D_2}(1+x_1)\partial_1\phi dx_1dx_2=\int_{-1}^0\int_{x_1}^{-x_1}(1+x_1)\partial_1 \phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{-1}^1 \int^{-|x_2|}_{-1}(1+x_1)\partial_1 \phi dx_1dx_2=\int_{-1}^1\left[ (1+x_1)\phi\big|_{-1}^{-|x_2|} -\int_{-1}^{-|x_2|} \phi dx_1\right]dx_2$

$\quad$                      $\displaystyle =\int_{-1}^1(1-|x_2|)\phi(-x_2,x_2)dx_2-\int_{D_2}\partial_1u\,\phi dx\quad (4)$
As for the domain $D_3$ and $D_4$, we have
$\displaystyle \int_{D_3} u\partial_1\phi dx=\iint_{D_3}(1-x_2)\partial_1\phi dx_1dx_2=\int_{0}^1(1-x_2)\int_{-x_2}^{x_2}\partial_1\phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{0}^1(1-x_2)[\phi(x_2,x_2)-\phi(-x_2,x_2)] dx_1dx_2\quad (5)$

$\displaystyle \int_{D_4} u\partial_1\phi dx=\iint_{D_4}(1+x_2)\partial_1\phi dx_1dx_2=\int_{-1}^0(1+x_2)\int_{x_2}^{-x_2}\partial_1\phi dx_1dx_2$

$\quad$                      $\displaystyle =\int_{-1}^0(1+x_2)[\phi(-x_2,x_2)-\phi(x_2,x_2)] dx_1dx_2\quad (6)$
Substituting $(3-6)$ to $(2)$

$\displaystyle \int_D u\partial_1\phi dx=-\int_{D_1}\partial_1u\,\phi dx-\int_{D_2}\partial_1u\,\phi dx$.

So from $(1)$, we know

$\displaystyle v=\partial_1 u=\begin{cases}-1, x\in D_1\\1,x\in D_2\,\\ 0,x\in D_3\cup D_4. \end{cases}$

As for $\partial_2u$, one can get it in a similar way.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Evans, partial differential equation. 5.3