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Are all function in $W_0^{1,2}(\Omega)$, $\Omega$ being a bounded domain in $\mathbb{R}^n$, $n \geq 2$, continuous and bounded w.r.t. $|.|$?. In other words, given $u\in W_0^{1,2}$ can one say that $|u(x)|\leq M$ ($M>0$) $\forall x\in \Omega$?

My line of thinking is as follows. $W_0^{1}(\Omega)\subset L^2(\Omega)\subset L^1(\Omega)\subset L^0(\Omega)$, $L^0$ being the space of measurable functions. Clearly $\forall u\in W_0^1(\Omega)$ there exists a continuous function such that $u=g$ a.e. (by virtue of being a measurable function). but $u=0$ on $\partial\Omega$, so the function $g$ also has to be zero on the boundary (This was my intuition, I might be wrong). Hence $g$ is bounded in $\overline{\Omega}$ and hence $u$ is bounded a.e.

Alexander
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Example from here works. In $n\ge 2$ dimensions, the function $$ u(x) = \begin{cases} \log \log (1+1/|x|),\quad &|x|<1/(e-1) \\ 0,\quad &|x|\ge 1/(e-1) \end{cases} $$ is unbounded, belongs to the Sobolev class $W^{1,n}$ (hence $W^{1,2}$), and vanishes in a neighborhood of the boundary of the unit ball $B$ (hence is in $W_0^{1,2}(B)$).

If you are concerned about the transition at $1/(e-1)$, note that $f$ is Lipschitz away from the origin, hence in every $W^{1,p}$ on any open set at positive distance from the origin.

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