Does weakly differentiable and $L^{\infty}$ imply continuity

Question

Suppose $\Omega \subset \mathbb{R}^d$ is open, connected and bounded. Is $$W^{1,1}(\Omega)\cap L^{\infty}(\Omega) \subset C(\bar{\Omega})?$$ Here $W^{1,1}$ denotes the space of all weakly differentiable functions with weak derivative in $L^1$ and $C$ the space of continuous functions. For $d=1$ the answer is yes as already $W^{1,1}(\Omega) \subset C(\bar{\Omega})$. For $d\geq 2$ I know that there exist functions in $W^{1,1}$ which are not continuous (e.g. $\log(\log(1/|x|))$). However this function is not in $L^\infty$.

Answer 1

No. Take the function $v(x) = \log(\log(1/\|x\|))$. Then define $w(x):=\sin( v(x))$. Then $w\in W^{1,1}(\Omega)\cap L^\infty(\Omega)$. But $w$ is not continuous at the origin.