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This is a long shot but...

In Gilbarg and Trudinger, they prove a specific case for Kellogg's theorem for a ball (Theorem 4.13):

let $B$ be a ball in $\mathbb{R}^n$ and $u$ and $f$ functions on $\bar{B}$ satisfying $u\in C^2(B)\cap C^0 (\bar{B})$, $f\in C^\alpha (\bar{B})$, $\Delta u =f$ in $B$, $u=0$ on $\partial B$. Then $u\in C^{2,\alpha}(\bar B)$.

As opposed to the general theory for linear elliptic operators, this special case is proven using the Kelvin transform.

Could anyone explain to me the logic of the proof. I get the feeling that details have been left out and I can't seem to bridge the gap. How exactly is Theorem 4.11 used?

(I'm guessing only people with the text can really answer this question, sorry)

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By earlier results (Theorem 4.6 should do) we already know that $u \in C^{2,\alpha}(B\setminus \partial B)$, so the question is whether this regularity continues up to the boundary. Thus it suffices to show that each point on the boundary has a neighbourhood in which $u$ is $C^{2,\alpha}$.

Theorem 4.11 gives the same result but with the ball $B$ replaced by a half-ball $B^+$ - the idea is to transfer this result to $B$ via the Kelvin transform. If we translate our ball so that $p \in \partial B$ is the origin, then the Kelvin transform sends $B$ to a half-space, and more pertinently sends some neighbourhood of the boundary point $q$ opposite $p$ to a half-ball. Here's an illustration with $B$ having radius $1$ (small half-ball in red):

Kelvin transform domains

The transformed versions of $u$ and $f$ satisfy the hypotheses of Theorem 4.11 on a small enough half-ball; so the transform $v$ of $u$ is $C^{2,\alpha}$ there. You should be able to verify that the transform preserves regularity, so $u \in C^{2,\alpha}$ on the region highlighted in blue, which is a neighbourhood of $q$.

Since any $q \in \partial B$ has a corresponding opposite point $p$, we can get regularity near $q$ by performing this inversion about $p$; and thus we have regularity on the entire boundary.

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