# Higher regularity for solutions of elliptic equations

Let $$\Omega$$ be a bounded domain in $$\mathbb{R}^d$$. Let $$f\in L^\infty(\Omega)$$. For the problem $$-\Delta u=f\mbox{ in }\Omega\\ ~~~~~~~~~u=0\mbox{ on }\partial\Omega,$$ one could seek solutions in two ways. One is through Lax-Milgram Lemma, since $$L^2(\Omega)\subset L^\infty(\Omega)$$, $$f\in L^2(\Omega)$$ and hence a solution $$u$$ exists to the problem in $$H^1_0(\Omega)$$. However, $$f\in L^p(\Omega)$$ for all $$p\in(1,\infty)$$, and hence one could mimic the monotone methods which are used to prove existence of solutions to monotone nonlinear problems to prove that the solution also exists in $$W^{1,p}_0(\Omega)$$ for all $$p\in(1,\infty)$$. My question is:

By uniqueness, all these solutions are the same functions measure theoretically, hence does this not imply a higher regularity result for the solution of this equation? In fact, since the solution is in all the $$W^{1,p}_0$$, would this not imply that the solution is in fact Holder continuous?

It feels to me that there is something wrong with my argument, but I am not able to figure it out.

Your argument is correct, assuming you can show existence in $$W^{1,p}_0(\Omega).$$ They key point is that you have uniqueness of solutions in $$H^1_0(\Omega),$$ so higher regularity solutions automatically coincide with the $$H^1_0(\Omega)$$ solution you obtain from Lax-Milgram. The idea is the same as when you prove elliptic regularity (the case $$f \in C^{\infty}$$); you start with a weak solution but end up proving it's actually smooth.
The fact that $$u$$ is Hölder continuous doesn't contradict anything; the point is that it's second derivative is not Hölder continuous in general, so this is consistent with the fact that $$\Delta u \in L^{\infty}.$$
In fact, using more sophisticated techniques one can show that $$u \in W^{2,p}(\Omega)$$ for all $$p < \infty.$$ In particular, its first derivatives are $$\alpha$$-Hölder continous for all $$\alpha \in (0,1)$$ by Sobolev embedding.