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Morera's Theorem as it is phrased in Wikipedia states that if $\Omega\subseteq\mathbb{C}$ is a simply connected open set in the plain and $f:\Omega\to\mathbb{C}$ is a continuous function such that for every closed piece-wise $C^{1}$ curve in $\Omega$ the following equality holds: $$\left(\star\right)\;\oint_{\gamma}f\left(z\right)dz=0$$ Then $f$ is holomorphic in $\Omega$. The proof in this case follows by constructing an anti-derivative of $f$ in $\Omega$ and deducing that $f$ is holomorphic as the derivative of a holomorphic function. My question is this: Suppose we omit the assumption that $\Omega$ is simply connected (but remains connected) and assume $f$ is a function such that for every aforementioned $\gamma$ the equality $\left(\star\right)$ holds. Pick any $z_{0}\in\Omega$ and define: $$F\left(z\right)=\int\limits _{z_{0}}^{z}f\left(\omega\right)d\omega$$ Wouldn't it follow that $F$ is an antiderivative of $f$ in $\Omega$ ? I believe the definition would be independent of the choice of path of integration from $z_{0}$ to $z$ from the assumption $\left(\star\right)$ holds.

Serpahimz
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    @AmritanshuPrasad I don't see why F would be not well-defined. An open connected set is also path connected, so there is no problem to define F. – Thomas Produit Jan 16 '14 at 09:27
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    @Serpahimz if I remember well my complex analysis: If $\Omega$ is an open connected set then "Every closed contour = 0" is equivalent to say that "Contours depens only on endpoints" thus proving that your right – Thomas Produit Jan 16 '14 at 09:31
  • Oh sorry, you are right. I was not reading this carefully. Of course, that $\Omega$ is simply connected is needed in the proof only to establish that the integral of $f$ does not depend on the path. If we already know that, we do not need this assumption. In fact, such an $f$ can be extended to a simply connected domain containing $\Omega$. – Amritanshu Prasad Jan 16 '14 at 09:36
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    You are correct. Wikipedia is deficient on this topic. http://mathworld.wolfram.com/MorerasTheorem.html – Disintegrating By Parts Jan 16 '14 at 09:37

1 Answers1

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Being holomorphic is a local property. (A property is local if the following is true: a function has it iff each point of its domain has a neighborhood in which the property holds). Other examples of local properties: continuity, differentiability, measurability, harmonicity, being real-analytic... Non-examples: integrability, being Lipschitz continuous, being bounded.

When you are to prove that a function has some local property, the global structure of its domain, and in particular its topology, is irrelevant. It suffices to establish the property on every open disk contained in the domain.

you can call me Al
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