# Using Morera's Theorem to Extend a Holomorphic Function by a Point

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??

• 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 $G'(z)$. – reuns Jun 18 at 21:48

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 $$0$$.
• 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? – thesundayscientist Jun 18 at 21:43
• 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. – Robert Israel Jun 19 at 1:52
• Really sorry, but from where did we get the idea that $f$ is uniformly continuous on compacts? – thesundayscientist Jun 19 at 9:45