TITLE: Boundedness of functions in $W_0^{1,p}(\Omega)$
QUESTION [5 upvotes]: Let $\Omega\subset\mathbb{R}^N$ be a bounded regular domain and $p\in (1,\infty)$. Suppose that $u\in W_0^{1,p}(\Omega)$ and $u$ is locally essentially bounded. Does this implies that $u$ is globally essentially bounded in $\Omega$.
Thanks.

REPLY [4 votes]: I think that your conclusion is wrong for $p \le N$. Of course, the case $p > N$ is trivial, since $W^{1,p}$ embeds into $L^\infty$.
A counterexample in $W_0^{1,p}(\Omega)$, $p \le N$,  could be constructed as follows. A single point $x \in \Omega$ has capacity $0$, that is, for fixed $r > 0$, such that $B_r(x) \subset \Omega$, and fixed $\delta > 0$, $N > 0$, there is a non-negative function $f \in W_0^{1,p}(\Omega)$ of arbitrarily small norm, such that $f$ is supported in $B_r(x)$ and there exists $\varepsilon > 0$, with $f \ge N$ on $B_\varepsilon(x)$.
Now, take a sequence of points $\{x_i\}$ which converges to the boundary of $\Omega$ (that is, in a compact set $K \subset \Omega$ there is only a finite number of points). Now, take functions $f_i \ge 0$ of norm $2^{-i}$ with distinct support, which are $\ge i$ around $x_i$ (in the above sense). The series $\sum_{i=1}^\infty f_i$ is Cauchy in $W_0^{1,p}(\Omega)$, hence the limit $f$ exists. Now, it is easy to see that $f \in L^\infty_\mathrm{loc}(\Omega) \setminus L^\infty(\Omega)$.