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.


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