Is the complex harmonic extension of a $\mathcal{C^r}$ map from $S^1$ to $\mathbb{C}$ is smooth upto the boundary ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:10:26Z http://mathoverflow.net/feeds/question/61072 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61072/is-the-complex-harmonic-extension-of-a-mathcalcr-map-from-s1-to-mathbb Is the complex harmonic extension of a $\mathcal{C^r}$ map from $S^1$ to $\mathbb{C}$ is smooth upto the boundary ? Analysis Now 2011-04-08T15:00:23Z 2011-04-08T15:00:23Z <p>Suppose we have a map $ h : S^1\to \mathbb{C} $ that we know is a $\mathcal{C^r} $ map ( in the sense of a map between 1-manifold ( or in the sense of a $2\pi$ periodic map from $\mathbb{R}\to \mathbb{R} $ ). Consider the complex harmonic extension $H$ of that map onto the whole closed unit disk $\mathbb{D} $. Is $H, $ as a map from the closed unit disk, a $\mathcal{C^k}(\mathbb{\bar{D}}) $map ? From PDE or complex analysis books, $H$ is definitely a $ \mathcal{C^\infty}(\mathbb{D})\cap \mathcal{C}(\mathbb{\bar{D}}) $ map , but my question is : is there an open set $U$ containing the closed unit disk $\bar{D}$ such that $H \in \mathcal{C^\infty}(U)$ ?</p> <p>Just suggest a reference if the answer is pretty long . Thanks.</p>