If $u \in H^1_0(\Omega)$ is such that $- \Delta u$ is $C^{0,\alpha}$, then do we have $u \in C^{2,\alpha}$ ?
The regularity theory I know tells me that $$ u \in \bigcap_{1 < p < \infty} W^{2,p}(\Omega) \subset \bigcap_{0 < \alpha < 1} C^{1,\alpha}(\bar \Omega) $$ from which I do not know how to go further. For instance $$ \bigcap_{1 < p < \infty} W^{2,p} \nsubseteq W^{2,\infty} $$ so I need better regularity theory. In case $\alpha = 0$, I know examples of $$ - \Delta u \in C^0(\bar \Omega) \nRightarrow u \in D^2(\Omega) $$ i.e. a function whose Laplacian is uniformly continuous but which is not twice differentiable everywhere. I do not know it these examples can be adapted to to Hölder regularity.
Actually this question arised when I was studying the problem of a strong solution of $$ -\Delta u = f(u)\text{ on } \Omega,\quad u = 0 \text{ on }\partial \Omega $$ being a classical solution. In this setting an obvious necessary condition is $f$ continuous. A sufficient condition is $f \in W^{1,\infty}(\mathbb R)$, in this case it is standard to show by regularity theory that $$ u \in \bigcap_{1 < p < \infty} W^{2,p}(\Omega) $$ and the superposition theory for Sobolev function tells us that $$ f \circ u \in \bigcap_{1 < p < \infty} W^{1,p}(\Omega) $$ so that using the same regularity theory $$ u \in \bigcap_{1 < p < \infty} W^{3,p}(\Omega) \subset \bigcap_{0 < \alpha < 1} C^{2,\alpha}(\bar \Omega). $$ This can be slightly weakened to $$ f \in \bigcup_{1 < p_0 < \infty } \bigcap_{p_0 < p < \infty} W^{1,p}(\mathbb R) $$ but I don't know how to relax the differentiability condition, for instance $f$ Hölder continuous. In this case the superposition theory for Sobolev functions fails but it's not clear that the regularity of $u$ also fails because as it is a solution of an elliptic PDE I hope it is more regular than "any Sobolev function".
This question is linked to Holder regularity for solutions to the Poisson equation
Yes, you can conclude that $u$ is $C^{2,\alpha}$. If $v\in C^\infty(\mathbb R^n)$ then Schauder estimates tell us that for any $\Omega’\subset \subset \Omega$, $$ \| v\|_{C^{2,\alpha}(\Omega’)}\leqslant C(\|\Delta v\|_{C^\alpha(\Omega)}+\|v\|_{L^\infty(\Omega)}). $$ The constant $C$ depends only on $n$, $\alpha$, $\Omega’$, and $\Omega$.
Since $u$ is in $H^1_0(\Omega)$ and not $C^\infty (\mathbb R^n)$, we apply Schauder to a mollification of $u$, $u_\varepsilon$, to obtain \begin{align*} \| u_\varepsilon\|_{C^{2,\alpha}(\Omega’)}&\leqslant C(\|\Delta u_\varepsilon \|_{C^\alpha(\Omega)}+\| u_\varepsilon\|_{L^\infty(\Omega)})\\ &\leqslant C(\|\Delta u\|_{C^\alpha(\Omega)}+\|u\|_{L^\infty(\Omega)}). \end{align*} Moreover, since $u\in H^1_0(\Omega)$, $\| u\|_{L^\infty(\Omega)}\leqslant C \| \Delta u\|_{L^\infty(\Omega)}$, so \begin{align*} \| u_\varepsilon\|_{C^{2,\alpha}(\Omega’)}&\leqslant C\|\Delta u\|_{C^\alpha(\Omega)}. \end{align*} Taking $\varepsilon \to 0$, gives the result.