MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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? – timur Feb 26 '13 at 4:16
Severe mistake: I changed the question: it MUST have been $C^{k,\alpha}$. Thanks! – Analysis Now Feb 26 '13 at 4:47
up vote 1 down vote accepted

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.

share|cite|improve this answer
Thank you very much! – Analysis Now Mar 2 '13 at 21:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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