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)


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.