## 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.