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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 $H \in \mathcal{C^\infty}(U)$ ?

Just suggest a reference if the answer is pretty long . Thanks.

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  • $\begingroup$ When $r=0$, it is well-known that the harmonic extension exists, is unique and is continuous over $\overline D$. If $h$ is more regular, you get more regularity of $H$ over $\overline D$ by the method of translations (here, method of rotations). Therefore $H\in{\mathcal C}^r(\overline D)$ as well. $\endgroup$
    – Denis Serre
    Commented Apr 8, 2011 at 15:11
  • $\begingroup$ @ Dennis Serre : Thanks ! But could you cite a reference for the method of rotations/translation or explain i in more detail ? I looked at Evan's PDE book, apparently I could not find it. $\endgroup$
    – Analysis Now
    Commented Apr 8, 2011 at 15:14
  • $\begingroup$ You can work this out by calculating the derivative of $H$ normal to the circle. After translating the circle to the real line by a Möbius transformation, you get something like the integral of $h$ against the principal value of $1/x^2$. This is not well defined for general $h$ in $C^1$, so the answer is no. If $h$ is $C^2$ then you get $H$ in $C^1$. Actually, the derivative of $h$ being Holder continuous should be enough to have $H$ in $C^1$. $\endgroup$
    – George Lowther
    Commented Apr 8, 2011 at 21:26
  • $\begingroup$ Your second question is clearly not true (if I understood correctly, there's two questions here, right?). If $h$ is only $C^r$ for $r < \infty$ then you can't extend $H$ as a $C^\infty$ function. If $r=\infty$ then you can. Unless you want to extend $H$ as a harmonic function in which case, no, you can't. $\endgroup$
    – George Lowther
    Commented Apr 8, 2011 at 22:07
  • $\begingroup$ My first comment was too optimistic. You certainly get a ${\mathcal C}^r$ regularity in the tangential direction my the method of translations (I'll come back later on that). But this is not true in the normal direction when $r$ is an integer. I remember vaguely that otherwise (that is if $r$ is not an integer), the answer is yes. In particular, if $h$ is ${\mathcal C}^\infty$, you obtain $H\in{\mathcal C}^(\overline D)$. But this does not mean that $H$ be ${\mathcal C}^\infty(U)$ for an open set $U$ containing $\overline D$. Only $H$ is ${\mathcal C}^r(U_r)$ for every $r$. $\endgroup$
    – Denis Serre
    Commented Apr 9, 2011 at 7:03

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