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lectures14-Earl.pdf page78
Theorem 232 (Morera's Theorem) Let $f: U \rightarrow \mathbb{C}$ be a continuous function on a domain such that
\[
\int_\gamma f(z) \mathrm{d} z=0
\]
for any closed path $\gamma$. Then $f$ is holomorphic.
Proof Let $z_0 \in U$. As $U$ is open and connected then it is path-connected and so for any $z \in U$ there is a path $\gamma(z)$ connecting $z_0$ to $z$. We will then define
\[
F(z)=\int_{\gamma(z)} f(w) \mathrm{d} w .
\]
Note that if $\gamma_1$ and $\gamma_2$ are two such paths then $\gamma_1 \cup\left(-\gamma_2\right)$ is a closed path and hence by hypothesis
\[
0=\int_{\gamma_1 \gamma_2^{-1}} f(w) \mathrm{d} w=\int_{\gamma_1} f(w) \mathrm{d} w-\int_{\gamma_2} f(w) \mathrm{d} w
\]
and we see that $F(z)$ is well-defined.
Now take $z_0 \in U$ and $r>0$ such that $D\left(z_0, r\right) \subseteq U$ and $h \in D\left(z_0, r\right)$. Then
\[
F\left(z_0+h\right)=F\left(z_0\right)+\int_{\left[z_0, z_0+h\right]} f(w) \mathrm{d} w .
\]
Hence
\[
\begin{aligned}
\left|\frac{F\left(z_0+h\right)-F\left(z_0\right)}{h}-f\left(z_0\right)\right| &=\left|\left(\frac{1}{h} \int_{\left[z_0, z_0+h\right]} f(w) \mathrm{d} w\right)-f\left(z_0\right)\right| \\
&=\left|\frac{1}{h} \int_{\left[z_0, z_0+h\right]}\left(f(w)-f\left(z_0\right)\right) \mathrm{d} w\right| \\
& \leqslant \frac{1}{|h|}|h| \sup _{\left[z_0, z_0+h\right]}\left|f(w)-f\left(z_0\right)\right| \quad \text { [by the Estimation Theorem] } \\
&=\sup _{\left[z_0, z_0+h\right]}\left|f(w)-f\left(z_0\right)\right| \rightarrow 0 \text { as } h \rightarrow 0 \text { by the continuity of } f \text { at } z_0
\end{aligned}
\]
Hence $F$ is holomorphic and $F^{\prime}=f$. By Corollary 225, $f$ is holomorphic. |
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