proof of Morera’s theorem

We provide a proof of Morera’s theoremunder the hypothesis that${\int }_{\mathrm{\Gamma }}f(z)𝑑z=0$for any circuit $\mathrm{\Gamma }$ contained in $G$.This is apparently more restrictive, but actually equivalent,to supposing${\int }_{\partial \mathrm{\Delta }}f(z)𝑑z=0$for any triangle $\mathrm{\Delta }\subseteq G$,provided that $f$ is continuous in $G$.

The idea is to prove that $f$ has an antiderivative $F$ in $G$.Then $F$, being holomorphic in $G$,will have derivatives of any order in $G$;but $F^{(n)}(z)=f^{(n-1)}(z)$ for all $z\in G$, $n\in \mathbb{N}$, $n\ge 1$.

First, suppose $G$ is connected.Then $G$, being open, is also pathwise connected.

Fix $z_0\in G$.For any $z\in G$ define $F(z)$ as

$$F(z)={\int }_{\gamma (z_0,z)}f(w)𝑑w,$$(1)

where $\gamma (z_0,z)$ is a path entirely contained in $G$with initial point $z_0$ and final point $z$.

The function $F:G\to ℂ$ is well defined.In fact, let ${\gamma }_1$ and ${\gamma }_2$be any two paths entirely contained in $G$with initial point $z_0$ and final point $z$;define a circuit $\mathrm{\Gamma }$ by joining ${\gamma }_1$ and $-{\gamma }_2$,the path obtained from ${\gamma }_2$ by“reversing the parameter direction”.Then by linearity and additivity of integral

$${\int }_{\mathrm{\Gamma }}f(w)𝑑w={\int }_{{\gamma }_1}f(w)𝑑w+{\int }_{-{\gamma }_2}f(w)𝑑w={\int }_{{\gamma }_1}f(w)𝑑w-{\int }_{{\gamma }_2}f(w)𝑑w;$$(2)

but the left-hand side is 0 by hypothesis,thus the two integrals on the right-hand side are equal.

We must now prove that $F^{\prime }=f$ in $G$.Given $z\in G$, there exists $r>0$such that the ball $B_r(z)$ of radius $r$ centered in $z$is contained in $G$.Suppose :then we can choose as a path from $z$ to $z+\mathrm{\Delta }z$the segment $\gamma :[0,1]\to G$ parameterized by $t\mapsto z+t\mathrm{\Delta }z$.Write $f=u+iv$ with $u,v:G\to ℝ$:by additivity of integral and the mean value theorem,

${\displaystyle \frac{F(z+\mathrm{\Delta }z)-F(z)}{\mathrm{\Delta }z}}$$=$${\displaystyle \frac{1}{\mathrm{\Delta }z}}{\displaystyle {\int }_{\gamma }}f(w)𝑑w$

$=$${\displaystyle \frac{1}{\mathrm{\Delta }z}}{\displaystyle {\int }_0^1}f(z+t\mathrm{\Delta }z)\mathrm{\Delta }z𝑑t$

$=$$u(z+{\theta }_u\mathrm{\Delta }z)+iv(z+{\theta }_v\mathrm{\Delta }z)$

for some ${\theta }_u,{\theta }_v\in (0,1)$.Since $f$ is continuous, so are $u$ and $v$, and

$$\underset{\mathrm{\Delta }z\to 0}{lim}\frac{F(z+\mathrm{\Delta }z)-F(z)}{\mathrm{\Delta }z}=u(z)+iv(z)=f(z).$$

In the general case, we just repeat the procedureonce for each connected component of $G$.