Solution to the Dirichlet problem is smooth up to the boundary if the boundary data is smooth?

where can I get a quick exposition to the boundary regularity problem for the Laplacian operator ?

In other words, suppose $h:S^1 \to \mathbb C$ and let $H: \bar{D}\to \mathbb C$ be its complex harmonic extension, i.e. $H(z) = \int h(t)p(z,t)\mathrm dt \quad \forall z\in D$ be the complex harmonic (but NOT holomorphic) extension of $h$ , where the integral is taken over $S^1$ and $p(z,t)$= Poisson kernel.

I want to quickly study the proof of the theorem : if the boundary data $h$ is $C^k$ , then the extension $H$ is $C^r(\bar{D})$ for some r .

I was told that the PDE book by Gilbarg-Trudinger is a source, but is there a quicker source where I can read everything in a short time ? Again, I just need in two dimensions.

Thanks !

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Do you have any kind of reference at all for your claimed result? It is certainly true that Holder continuous boundary data leads to a Holder solution, same Holder exponent, after that I am not so sure. At the other extreme, if the boundary data has a finite Fourier series, the solution is the real part of a polynomial and extends to the whole plane. In between is less certain. – Will Jagy May 4 '11 at 5:13
Actually I talked to people about that subject based on my guess, since I do not know in detail the regularity theory for elliptic PDEs,and they refered me to Gilberg-Trudinger. So I do not have a trusted source where I have seen the exact result being used. – Mathmath May 4 '11 at 17:09
It's not true for $r=k$ unless you also have Holder continuity (and is true for $r < k$). Someone asked this on mathoverflow not so long ago. Don't think it got a proper answer, but it is not too hard to find a counterexample for the non-Holder case. – George Lowther May 4 '11 at 20:55

You may want to read Kellogg's book on Potential Theory...

From Gilbarg and Trudinger, page 66,

Corollary 4.14. Let $\varphi \in C^{2,\alpha}(\bar{B}), \; \; f \in C^{\alpha}(\bar{B}) .$ Then the Dirichlet problem, $\Delta u = f$ in $B, \; \; u = \varphi$ on $\partial B,$ is uniquely solvable for a function $u\in C^{2,\alpha}(\bar{B}).$

In your case $f = 0.$ The Hölder spaces are defined on page 51.

A stronger version of this is indeed called Kellogg's Theorem, and the reference is Foundations of Potential Theory by O. D. Kellogg, a Dover reprint.

A more recent book, though no more elementary, is Axler et al

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I usually type umlauts using their html-tags, &auml; gives ä, &ouml; gives ö, etc. Doesn't seem to work in comments, though... I just entered them directly via my keyboard. – t.b. May 4 '11 at 20:06
Thank you, Theo. I guess there will always be tricks needed unless a site takes pure Latex. There was some very good reason to make MathJax or the like the dominant math symbol processor, I will need to just deal with it. – Will Jagy May 4 '11 at 20:19
@ will Jagy : For the statement of Kellog's theorem that you stated above ( or the same statement in Gilberg-Trudinger's book ), I have a question : It says the extension $\phi$ of the boundary data satisfies $\phi \in C^{2,\alpha} (\bar{B})$, but it does not say that the derivative of the boundary data $\phi: S^1\to S^1$ alone satisfies : $|\phi"(t) - \phi"(s) | \leq M |t-s|^\alpha \forall t,s \in S^1$, that is, it uses the information about the values of $\phi$ on the interior $B$, not JUST the boundary $S^1$. Are these two kinds of Holder continuity equivalent ? – Mathmath May 19 '11 at 3:08