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Let $f:S^n\to C$ be a continuous function, $n\geq 1$. When $n=1$, this is a well-known theorem, called Kellog's theorem (or sometimes Kellog-Warschawski's theorem) which states the following

Theorem: Fix $k \geq 0, 0<\alpha<1$. Let $f\in C^{k,\alpha}(S^1)$. Then its harmonic extension $H(f)$, which is the solution to the Dirichlet problem on the unit disk $D$ with boundary value $f$, is in $C^{k, \alpha}(D)$.

My main question is: is the above true for $n\geq 2$ as well? Any refernces/ suggestions?

While I don't know exactly a complete reference for the proof, but I have read the following theorem mentioned in the book "Boundary Behaviour of Conformal maps" by Christian Pommerenke which states:

Let $F:D\to\Omega $ be a conformal homeomorphism of $D$ onto a Jordan domain $\Omega$ whose boundary curve $\partial\Omega$ has a $C^{k,\alpha}$ -parametrization. Then $f\in C^{k, \alpha}(D)$. Note that any conformal homeomorphism $F$ of $D$ onto a Jordan domain extends to the boundary of $D$, by Caratheodory's extension theorem.

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    $\begingroup$ Supposing that by $C^{k+1,\alpha}(D)$ you mean a result up to the boundary (otherwise harmonic functions are analytic in the interior), how is it possible for $H(f)$ to be more regular than its boundary values? $\endgroup$
    – timur
    Feb 26, 2013 at 4:16
  • $\begingroup$ Severe mistake: I changed the question: it MUST have been $C^{k,\alpha}$. Thanks! $\endgroup$ Feb 26, 2013 at 4:47

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It follows from the Schauder theory. You can also establish Kellog's theorem directly. One approach is given in DiBenedetto's PDE book, where he uses Kellog's theorem in the proof of Schauder estimates.

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In respect of your main question, the answer is yes. Please refer to the Algebraic Lemma on pp378 of the article C^(1/,1/3)- regularity in the Dirichlet problem, available at: https://www.sciencedirect.com/science/article/pii/S0898122107001927

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