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

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<p\leq \infty}, 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<p\leq \infty} 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<n}, 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<n}, 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<n-1}. {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<n}, 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<n\\-n\frac{Dv}{v^2},\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)

\displaystyle vD\tilde{u}_n+\tilde{u}_n Dv=\begin{cases}vDu+uDv,\quad uv<n\,\quad uv\geq n\end{cases}\in L^1_{loc}(\Omega)

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|<x_1\\1+x_1\quad if\quad x_1<0, |x_2|<-x_1\\1-x_2\quad if\quad x_2>0, |x_1|<x_2\\1+x_2\quad if\quad x_2<0, |x_1|<-x_2.\end{cases}

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