A "special" inclusion of $W^{1,p}_0(\Omega)$ in $L^{\infty}(\Omega)$

Question

I'm in trouble with this stuff.

If $\Omega$ is a bounded open set in $\mathbb{R}^n$ of class $C^1$ and $f\in W^{1,p}_0(\Omega)$, is it true or not that $f\in L^{\infty}(\Omega)$? I think so, but I cannot sketch a proof.

Can anyone help me? Thanks.

Answer 1

No. Consider $n=3$, $p=1$ and $\Omega$ the unit ball. Let $g(x)=\frac1{|x|^{1/2}}$. Then $g_{x_i}(x)=-\frac{x_i}{2|x|^{5/2}}$, so using polar coordinates it is straightforward go show that $g,g_{x_i}\in L^1(\Omega)$ and so $g\in W^{1,1}(\Omega)$. Hence if $\varphi\in C^\infty_c(\Omega)$ is such that $\varphi(x)=1$ for $|x|\le\frac12$, we define $f=g\varphi$ and so $f\in W^{1,1}(\Omega)\setminus L^\infty(\Omega)$.

Answer 2

It is false for $n \ge 2$: if $x_0 \in \Omega$ then $|x - x_0|^{-\alpha}$ is unbounded and belongs to $W^{1,1}(\Omega)$ provided $0 < \alpha < n - 1$. Tinker with this function away from the singularity to make it compactly supported in $\Omega$.

If $n = 1$, then any Sobolev function has a continuous representative which will in turn be bounded on the compact set $\overline{\Omega}$.