# proof of Morera’s theorem

We provide a proof of Morera’s theorem under the hypothesis that $\int_{\Gamma}f(z)dz=0$ for any circuit $\Gamma$ contained in $G$. This is apparently more restrictive, but actually equivalent, to supposing $\int_{\partial\Delta}f(z)dz=0$ for any triangle $\Delta\subseteq G$, 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 $z\in G$, $n\in\mathbb{N}$, $n\geq 1$.

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 $F:G\to\mathbb{C}$ 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 $\Gamma$ 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

 $\int_{\Gamma}f(w)dw=\int_{\gamma_{1}}f(w)dw+\int_{-\gamma_{2}}f(w)dw=\int_{% \gamma_{1}}f(w)dw-\int_{\gamma_{2}}f(w)dw\;;$ (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 $0<|\Delta{z}|: then we can choose as a path from $z$ to $z+\Delta{z}$ the segment $\gamma:[0,1]\to G$ parameterized by $t\mapsto z+t\Delta{z}$. Write $f=u+iv$ with $u,v:G\to\mathbb{R}$: by additivity of integral and the mean value theorem,

 $\displaystyle\frac{F(z+\Delta{z})-F(z)}{\Delta{z}}$ $\displaystyle=$ $\displaystyle\frac{1}{\Delta{z}}\int_{\gamma}f(w)dw$ $\displaystyle=$ $\displaystyle\frac{1}{\Delta{z}}\int_{0}^{1}f(z+t\Delta{z})\Delta{z}dt$ $\displaystyle=$ $\displaystyle u(z+\theta_{u}\Delta{z})+iv(z+\theta_{v}\Delta{z})$

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

 $\lim_{\Delta{z}\to 0}\frac{F(z+\Delta{z})-F(z)}{\Delta{z}}=u(z)+iv(z)=f(z)\,.$

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

Title proof of Morera’s theorem ProofOfMorerasTheorem 2013-03-22 18:53:34 2013-03-22 18:53:34 Ziosilvio (18733) Ziosilvio (18733) 10 Ziosilvio (18733) Proof msc 30D20