Boundedness of functions in $W_0^{1,p}(\Omega)$

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.

Solution 1:

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 $f \in L^\infty_\mathrm{loc}(\Omega) \setminus L^\infty(\Omega)$.

Solution 2:

Remark: This is an incomplete answer, but I felt it was worth posting what I have so far. If anybody has ways to finish the argument let me know or post or your answer.

If $p > n$, then it follows from Morrey's inequality that there exists $\tilde{u} = u$ a.e such that $u \in C^{0, \gamma} (\overline{\Omega})$ and $$ \|\tilde{u}\|_{C^{0, \gamma}(\overline{\Omega})} \le C \|u\|_{W^{1,p}(\Omega)}$$ and hence in this case it is true that $u$ is globally essentially bounded.

If $p < n$, then the result fails when $u$ doesn't have zero trace. Pick $\Omega = \mathbb{R}^n_{+} \cap B(0,1)$, the open half ball. Now, pick exponent $\alpha$ such that $p < \alpha < n$. Now choose $$ u(x) = \frac{1}{|x|^{(n - \alpha)/p}}$$ e.g. $u(r) = r^{(\alpha - n)/p}$, where $r = |x|$. Then $|u|^p = r^{\alpha - n}$, which is integrable on $\Omega$. Furthermore, the derivative $Du$ scales like $r^{-1 + (\alpha - n)/p}$, so then $|Du|^p$ scales like $r^{-p + \alpha - n}$, which is also integrable on $\Omega$. Therefore $u \in W^{1,p}(\Omega)$. $u$ only blows up at the origin, so on every compact set in the interior of $\Omega$, $u$ is bounded, but $u \notin L^{\infty}(\Omega)$.

Right now I cannot fully address your question in the case $p < n$ and $u$ with zero trace, but first note that we can take the $u(x)$ from my example above and easily make it vanish away from the semicircular portion of $\partial \Omega$. Now, multiply the resulting function by $x_n/ |x|$, where $x_n$ is the coordinate which is zero at the boundary of $\mathbb{R}^n_{+}$. Then $u \in W^{1,p}(\Omega)$, and $u = 0$ a.e. on $\partial \Omega$, so I think that this implies that $u \in W^{1,p}_0(\Omega)$, in which case $u$ would show that the result fails. However, I currently cannot rigorously argue that $u$ can be written as a limit of $C_c^{\infty}(\Omega)$ in the $W^{1,p}$ topology.