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I've been stuck trying to hunt down a proof/reference for a certain regularity result which I need for my thesis (all the authors I've consulted refer to it as "well-known"). Consider the Dirichlet problem

$$ \begin{cases} - \Delta v + v = g & \mathrm{in} \, B_r(x)\\ - v = u^* & \mathrm{on} \, \partial B_r(x) \,, \end{cases}\ $$

with $g \in L^\infty$, $u^* \in W^{1,2}(B_r(x))$ and $B_r(x) \subseteq \mathbb{R}^2$. Apparantly one should be able to deduce through "standard" regularity arguments that this implies $v \in C^1(B_r(x))$ for any weak solution $v$. Any idea what this could be and where I could find a reference for it?

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    $\begingroup$ Errrr, do you mean $\Delta$ instead of $\nabla$ and $v\in C^1$ instead of $u^*\in C^1$? If so this is indeed a standard elliptic regularity result, you can check for example in the "elliptic bible" Gilbarg-Trudinger. One more comment: you should be careful about the meaning of $u^*\in W^{1,2}$: do you mean $u^*\in W^{1,2}(\partial B_r)$, or that $u^*$ is the boundary trace of some function $U^*\in W^{1,2}(B_r)$? $\endgroup$
    – leo monsaingeon
    Commented May 9, 2022 at 20:13
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    $\begingroup$ Huuuum, if you really mean the latter ($u^*\in W^{1,2}(B_r)$) then I gues this is not true, at least not "up to the boundary". For, obviously there exist functions $u^*\in W^{1,2}(B_r)$ whose trace is not $C^1(\partial\Omega)$, which should hold if the solution $v$ were really to belong to $C^1(\overline B_r)$. And actually the same works if you meant $u^*\in W^{1,2}(\partial B_r)$. Just one possible hint, tough: one possible way to get all the way to $C^1$ (and actually $C^{1,\alpha}$) is to get to $W^{2,p}$ for large $p\gg 1$. These are "strong solutions", chapter 9 in Gilbarg-Trufinger $\endgroup$
    – leo monsaingeon
    Commented May 9, 2022 at 20:32
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    $\begingroup$ well, for the local regularity theory you can simply ignore the boundary conditions... $\endgroup$
    – leo monsaingeon
    Commented May 9, 2022 at 21:58
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    $\begingroup$ Also, you should specify in which sense your equation is understood. Are you talking about strong solutions, weak solutions, distributional solutions [...] ? I can prove that strong solutions are $C^{1,1}_{loc}$, for example $\endgroup$
    – leo monsaingeon
    Commented May 9, 2022 at 22:20
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    $\begingroup$ By $v\in C^1(B_r)$ you seem to mean $v\in C^1_{\rm loc} (B_r)$. The notation $C^1(B_r)$ is somewhat ambiguous (what norm is on that space?). By superposition, you can consider the solution for $u^*$ without $g$, say, $v_1$ and then the solution with $g$ without $u^*$, say $v_2$. The first one is analytic in the interior, so $C^{1,1}$. For the second one, you can use the Green function of the problem, I am pretty sure it is known (if the $+v$ is a problem, you can forget it since by the maximum principle $\|v\|_{\infty} \leq \|g\|_{\infty}$, so replace $g$ by $g-v$, but I think not). $\endgroup$
    – username
    Commented May 10, 2022 at 18:39

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