Morera's Analyticity Theorem

Morera's Analyticity Theorem

We will now develop a nice theorem that gives us a criterion for when a complex function $f$ will be analytic on an open subset $A$ of $\mathbb{C}$.

 Theorem (Morera's Analyticity Theorem): Let $A \subseteq \mathbb{C}$ be open, connected and let $f : A \to \mathbb{C}$. If $f$ is continuous on $A$ and if $\displaystyle{\int_{\gamma} f(z) \: dz = 0}$ for every closed piecewise smooth curve in $A$ then $f$ is analytic on $A$.

The aim of this proof is to first define a function $F$ which is analytic on $A$ and such that $F' = f$. Then since all derivatives of $F$ exist and must also be analytic on $A$ then $F'' = f'$ exists on $A$ which shows that $f$ is analytic on $A$.

• Proof: Let $f$ be continuous on $A$ and suppose that $\displaystyle{\int_{\gamma} f(z) \: dz = 0}$ for every closed piecewise smooth curve $\gamma$ in $A$. Let $z_0 \in A$. Since $A$ is open there exists an $r > 0$ such that $D(z_0, r) \subseteq A$.
• For any curve in $D(z_0, r)$ with initial point $z_0$ and terminal point $z$, let $\displaystyle{\int_{z_0}^z f(w) \: dw}$ denote the integral of $f$ along this curve. This notation is unambiguous since if $\gamma_1$ and $\gamma_2$ are two curves in $D(z_0, r)$ with initial points $z_0$ and terminal points $z$ then $\gamma_1 + (-\gamma_2)$ is a closed piecewise smooth curve in $D(z_0, r)$ and hence in $A$ so by the hypothesis made above:
(1)
\begin{align} \quad 0 = \int_{\gamma_1 + (-\gamma_2)} f(z) \: dz = \int_{\gamma_1} f(z) \: dz - \int_{\gamma_2} f(z) \: dz \end{align}
• Hence:
(2)
\begin{align} \quad \int_{\gamma_1} f(z) \: dz = \int_{\gamma_2} f(z) \: dz \end{align}
• Define a function $F$ on $D(z_0, r)$ for all $w \in D(z_0, r)$ by:
(3)
\begin{align} \quad F(z) = \int_{z_0}^{z} f(w) \: dw \end{align}
• We will show that $F$] is analytic on [[$D(z_0, r)$. Let $z^* \in D(z_0, r)$. Then:
(4)
\begin{align} \quad F'(z^*) &= \lim_{z \to z^*} \frac{F(z) - F(z^*)}{z - z^*} \\ &= \lim_{z \to z^*} \frac{1}{z - z^*} [F(z) - F(z^*)] \\ &= \lim_{z \to z^*} \frac{1}{z - z^*} \left [ \int_{z_0}^{z} f(z) \: dz - \int_{z_0}^{z^*} f(z) \: dz \right ] \\ &= \lim_{z \to z^*} \frac{1}{z - z^*} \int_{z^*}^{z} f(z) \: dz \end{align}
• Let $\epsilon > 0$ be given. The curve from $z^*$ to $z$ is homotopic to the line segment $l$ joining $z^*$ and $z$. Note that this line segment is fully contained in $D(z_0, r)$ since $D(z_0, r)$ is a convex set. Parameterize this line segment as $l(t) = z^* + t(z - z^*)$ for $t \in [0, 1]$. Then $l'(t) = z - z^*$. So by the deformation theorem:
(5)
\begin{align} \quad \int_{z^*}^{z} f(z) \: dz & = \int_0^1 f(z^* + t(z - z^*)) \cdot (z - z^*) \: dt \end{align}
• Therefore:
(6)
\begin{align} \quad \biggr \lvert \frac{1}{z - z^*} \int_{z_1}^{z} f(w) \: dw - f(z^*) \biggr \rvert &= \biggr \lvert \frac{1}{z - z^*} \int_0^1 f(z^* + t(z - z^*)) \cdot (z - z^*) \: dt - \int_0^1 f(z^*) \: dt \biggr \rvert \\ &= \biggr \lvert \int_{0^1} [f(z^* + t(z - z^*)) - f(z^*)] \: dt \biggr \rvert \end{align}
• Since $f$ is continuous for this given $\epsilon > 0$ there exists a $\delta > 0$ such that if $\mid (z^* + t(z - z^*) - z^* \mid = \mid t(z - z^*) \mid = \mid t \mid \mid z - z^* \mid \leq \mid z - z^* \mid \leq \delta$ then $\mid f(z^* + t(z - z^*)) - f(z^*) \mid < \epsilon$. Then for this given $\delta$ we have that:
(7)
\begin{align} \quad \biggr \lvert \frac{1}{z - z^*} \int_{z_1}^{z} f(w) \: dw - f(z^*) \biggr \rvert \leq \int_0^1 \epsilon \: dt = \epsilon \end{align}
• Therefore $\displaystyle{F'(z^*) = \lim_{z \to z^*} \frac{F(z) - F(z^*)}{z - z^*} = f(z^*)}$. So $F$ is analytic at $z^*$ for each $z^* \in D(z_0, r)$, i.e., $F$ is analytic on $D(z_0, r)$ with $F' = f$. But then by Cauchy's Integral Formula for Derivatives we know that all derivatives of $F$ exist and that $F'' = f'$ exists on $D(z_0, r)$, i.e., $f$ is analytic on $D(z_0, r)$. Hence $f$ is analytic on $A$. $\blacksquare$
 Corollary 1 (Morera's Analyticity Corollary: Let $A \subseteq \mathbb{C}$ be open, $z_0 \in A$, and let $f : A \to \mathbb{C}$ be analytic on $A \setminus \{ z_0 \}$. Then $f$ is analytic at $z_0$ as well.