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Let $D=\{z:|z|\leq 1\}$ be the unit disc in $\mathbb{C}$. Say $f$ is analytic on $D$ and $g$ is analytic on $\overline{D^c}$, and that $f|_{\partial D}=g|_{\partial D}$.

Is there necessarily an analytic function $h$ such that $h|_D=f$ and $h|_{D^c}=g$?

S.B.
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  • When you say analytic on a closed set, do you mean analytic on a neighbourhhod of the set? – mrf Dec 16 '14 at 20:08
  • It is analytic on the interior of the set, and attains some value on the boundary. For the unit disc, Cauchy's integral formula will hold for any interior point, using the boundary as the path. – S.B. Dec 16 '14 at 20:11
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    Continuous up to the boundary? – mrf Dec 16 '14 at 20:11
  • Yes. To simplify, If they can be extended, then the extension is obviously unique. So there is an extension of them both, $h$, and I'm asking whether $h$ must be entire, i.e. analytic on all the complex field, or not. – S.B. Dec 16 '14 at 20:14
  • Did you think about a conformal map from the unit disk to the upper half-plane with the boundary being mapped to the real axis? You could try the Schwarz reflection principle, but the real values on $\mathbb{R}$ is a problem... – Igor Deruga Dec 17 '14 at 17:58
  • So, what you suggest is something like: Take $f$, use conformal map to map it to the upper half plane, use Schwarz reflection principle to extend it, and then use the reverse conformal map to map it to the function $h$ on the entire plain? – S.B. Dec 17 '14 at 18:04

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Define $$ h(z) = \begin{cases} f(z), & |z|\le 1 \\ g(z), & |z| > 1. \end{cases} $$ By assumption, $h$ is continous on $\mathbb{C}$, so by Morera's theorem, it suffices to check whether $$ \int_\gamma h(z)\,dz = 0 $$ for all closed curves $\gamma$. In fact, it is enough to check that the integral vanishes for all triangles.

Let $\gamma$ be an arbitrary triangle. There are a few cases to consider. If $\gamma$ encloses the unit disc, it follows by deformation of contours that $$ \int_\gamma h(z)\,dz = \int_{|z| = 1} h(z)\,dz = \int_{|z| = 1} f(z)\,dz = 0 $$ by Cauchy's integral theorem. Similarly, if $\gamma$ is in the interior of the unit disc, the integral vanishes again by Cauchy's integral theorem.

Finally, if the triangle intersects the unit circle, we can write the triangle as the union of a number of curves lying completely inside or completely outside the unit disc. These curves consist of line segments that together form the original triangle together with arcs along the circle traversed twice; once in each direction. All in all, the integral along each such closed curve vanishes, yet again by Cauchy's integral theorem, so our $h$ is indeed an entire function.

Note: we need a version of Cauchy's integral theorem valid for functions holomorphic on (or outside) the unit disc, continous up to the boundary. Not all textbooks prove this version, but it's fairly straight-forward.

mrf
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