Let $\Omega$ be a bounded domain, $f\in L^{\infty}(\Omega),$ and $0\leq u\in H\_0^1(\Omega)$ is a non-negative solution of $\Delta u=f$. My question is as follows:


* Can we conclude that $u\in C^0(\bar{\Omega})$ without any boundary smoothness assumption on $\Omega$? If not, could you please provide a counterexample?


By employing De Giorgi iteration and the Harnack inequality, we can establish that $u$ belongs to $L^{\infty}(\Omega)$ and exhibits Hölder continuity in $\Omega$. Given that $u\in H\_0^1(\Omega)$, my intuition is that $u$ remains continuous throughout the entire closure $\bar{\Omega}$ and equals zero on the boundary. However, since no assumptions have been made regarding the boundary $\partial \Omega$, I have been unable to prove this conjecture.



Continuous up to the boundary without boundary smoothness
Here is an example in the plane which is genuinely discontinuous up to the boundary. I try to keep it as elementary as possible. There is an if and only if criterion for continuity up to the boundary of harmonic functions, due to Wiener: this example is essentially how one shows the only if part.


Take $B\_1$, and remove from it a collection of balls $\bar{B}\_{r\_k}(2^{-k}),$ with $k \geq 1$, centered at points on the real axis, of radii to be chosen but with $r\_k \ll 2^{-k}$. Set $U\_k = B\_1 \setminus \cup\_{j = 1}^k\bar{B}\_{r\_j}(2^{-j})$, and $U = (\cap\_k U\_k) \setminus \{0\}$ ($0$ is also removed to make $U$ open). Solve for $u\_k$ satisfying $\Delta u\_k = - 1$ on $U\_k$ and $u\_k \in H^1\_0(U\_k)$, and similarly $u$ with $\Delta u = -1$ on $U$ with $u \in H^1\_0(U)$ (there isn't any issue with solving this using standard energy methods, even on $U$).


From the maximum principle, $u\_{k+1}\leq u\_k$ on $U\_{k+1}$, and $u \leq u\_k$ on $U$. Let us note here that $u\_0$ (with no balls removed) has $u\_0 \leq \frac{1}{4} < 1$ (just by directly computing $u\_0 = (1 - |x|^2)/4$).


We also have that $u\_k \rightarrow u$ pointwise on $U$: indeed, the $u\_k$ are uniformly bounded in $H^1(U)$ by integrating by parts,
$$
\int\_{U\_k} |\nabla u\_k|^2 = \int\_{U\_k} u\_k \leq C \|\nabla u\_k\|\_{L^2(U\_k)}.
$$
Extract a weakly convergent subsequence $u\_k \rightharpoonup v \in H^1\_0(B\_1)$. It is straightforward to check that $v$ solves $\Delta v = -1$ on $U$, and that $v \in H^1\_0(U\_k)$ for every $k$. I claim (see below) that $v \in H^1\_0(U)$; then from uniqueness/weak maximum principle it follows that $v = u$, that $u\_k \rightarrow u$ in $L^2(U)$, and from regularity the convergence is pointwise, or even locally in $C^j$, etc.


Consider, then, $w = u\_k - u\_{k + 1}$. This function is harmonic on $U\_{k+1}$, and vanishes on the boundary except along $\partial B\_{r\_{k+1}}(2^{-k})$--along which it has $w < 1$. The function $h = 1 - \frac{\log (|x - 2^{-k}|/r\_k)}{\log (2/ r\_k)}$ is also harmonic on $U\_k$, is nonnegative on $B\_1$, and is $1$ along $\partial B\_{r\_{k+1}}(2^{-k})$: by the maximum principle, $w \leq h$. For $x$ in the interval $[-1, 0]$, we have that
$$
w \leq 1 - \frac{\log (2^{-k}/r\_k)}{\log (2/r\_k)} = \frac{k\log 2}{\log (2/r\_k)}.
$$
Now, we get to choose $r\_k$ as small as we like: for example, we can choose it to be $2e^{-t 2^k}$ for a small $t$, so that $w \leq 2k t 2^{-k}$ along this entire line segment [the main point here is to choose it so that $u\_k - u\_{k + 1}$ is summable with small sum].


Then summing this, we see that for any $x \in [-1, 0)$,
$$
|u - u\_0| \leq \sum\_{k = 0}^\infty |u\_k - u\_{k + 1}| \leq C t.
$$
Choosing $t$ small enough and using that $u\_0 > 1/8$ along $[-1/2, 0]$, this gives $ u > 1/16$ on this segment.


On the other hand, on each of the boundaries of balls composing $\partial U \setminus \{0\}$, we have $u = 0$. We conclude that $u$ had to be discontinuous at $0$.


**To verify the claim**: let $h\_r = \frac{[\log|x| - \log r^2)]\_+}{|\log r|}$. This function vanishes for $|x| \leq r^2$, and for $|x| \geq r$ it has $h\_r \geq 1 > v$. In between, we can integrate
$$
\int\_{r^2 < |x| < r} |\nabla h\_r|^2 = \int\_{r^2}^r \frac{1}{s |\log r|^2}ds \leq \frac{C}{|\log r|}.
$$
The point of this is, set $v\_r = \min\{v, h\_r\}$. Then $v\_r \in H^1\_0(U)$ (it vanishes on $B\_{r^2}$, no issues near $0$ here), and $v\_r = v$ outside of $B\_r$. We then have
$$
\int |\nabla (v\_r - v)|^2 = C \int\_{|x|< r} |\nabla v|^2 + |\nabla h\_r|^2,
$$
with the first term going to $0$ because $|\nabla v|^2$ is integrable, while the second term is controlled by $1/|\log r|\rightarrow 0$ from above. So $v\_r \rightarrow v$ in $H^1(B\_1)$, and $v\_r \in H^1\_0(U)$: it follows that $v \in H^1\_0(U)$.


