## Tag Archives: sobolev space

### 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.

### 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.

### 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