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Use Morera's Theorem to prove the following: Let $\Omega\subset_{\text{open}} \mathbb{C}$,$z_{0}\in\Omega$. Suppose $f$ is continuous in $\Omega$ and analytic in $\Omega \backslash \{z_{0}\}$. Then $f$ is analytic in $\Omega$

I am not really sure how to proceed. Morera's theorem states:

Theorem: Morera If $g$ is a continuous function in the open disc $D$ such that for any triangle $T$ contained in $D$: $$ \oint_{T} g(z) \, dz = 0 $$ Then $g$ is holomorphic in $D$.

  • $\Omega$ is an open subset of $\mathbb{C}$. So at $z_{0}$ there is some open disk of radius $r$ contained in $\Omega$, $D(z_{0},r)\subset \Omega$. If I can show that Morera's thoerem holds on this disk, then I am done, since holomorphic functions are always analytic - so is $f$ is holomorphic on $D(z_{0},r) \supset \{z_{0}\}$ then it is analytic at $z_{0}$.

  • I know need only concern myself with triangles $T$ such that one of the vertices is $z_{0}$, since all other cases follow automatically from analyticity (holomorphicity) of $f$ and Cauchy's/Goursat's theorem(s).

  • However, how can I evaluate these integrals??

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  • $\begingroup$ Also note $(z-z_0) g(z)$ is holomorphic (if $g$ is only bounded near $z_0$ look at $(z-z_0)^2 g(z)$) thus it is analytic thus $g(z)$ is meromorphic with a pole at $z_0$ and its boundedness implies the pole is of order $0$ ie. $g(z)$ is analytic. For Morera it is the same idea except you'll look at $G(z)=\int_a^z g(s)ds$ which is holomorphic thus analytic and so is $G'(z)$. $\endgroup$ – reuns Jun 18 at 21:48
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It's not true that you need only consider triangles with one vertex $z_0$. Indeed those are not a problem: the integral over a triangle with one vertex $z_0$ is the limit of integrals over triangles that $z_0$ is outside of, which are $0$ by Cauchy.

But you can't use Cauchy/Goursat when $z_0$ is inside your triangle. To handle those, break up such a triangle into three triangles, each of which has $z_0$ at one vertex. The integral over your triangle is the sum of the integrals over the three sub-triangles (as the integrals over the interior edges cancel). And by the previous paragraph, the integrals over those triangles are $0$.

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  • $\begingroup$ What is the formal way in which I would define that limit? - I can see intuitively that it should follow imeadiately from the continuity of $f$ there shouldn't be a discontinuity on the boundary, but how would I show this rigrously? $\endgroup$ – thesundayscientist Jun 18 at 21:43
  • $\begingroup$ If $\Gamma_p$ is a positively-oriented triangle with corners $a, b, p$, and $f$ is continuous on a region containing $\Gamma_p$ for $p$ in a neighbourhood of $z_0$, then $\lim_{p \to z_0} \oint_{\Gamma_p} f(z)\; dz = \oint_{\Gamma_{z_0}} f(z)\; dz$. This follows from uniform continuity of $f$ on a compact set. $\endgroup$ – Robert Israel Jun 19 at 1:52
  • $\begingroup$ Really sorry, but from where did we get the idea that $f$ is uniformly continuous on compacts? $\endgroup$ – thesundayscientist Jun 19 at 9:45
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    $\begingroup$ Heine-Cantor theorem: a continuous function from a compact metric space to a metric space is uniformly continuous. $\endgroup$ – Robert Israel Jun 19 at 12:09
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    $\begingroup$ He was talking about the triangle. $\endgroup$ – cmk Jun 20 at 11:40

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