📂Complex Anaylsis

Proof of Morera's Theorem

Theorem ¹

If a complex function $f : \mathbb{C} \to \mathbb{C}$ is continuous on a simply connected region $\mathscr{R}$ and satisfies $\displaystyle \int_{\mathscr{C}} f(z) dz = 0$ for every closed path $\mathscr{C} \subset \mathscr{R}$ contained in $\mathscr{R}$, then $f$ is analytic on $\mathscr{R}$.

Explanation

This can be thought of as roughly the converse of Cauchy’s theorem. What is interesting is the fact that the common wisdom of analysis is originally ‘differentiable implies continuous, continuous implies integrable.’ Yet Morera’s theorem instead determines the differentiability of a function through integration, so it cannot but be a truly astonishing theorem.

Proof ²

$$ F(z) := \int_{z_{0}}^{z} f(w) dw $$ We want to define the complex contour integral of $f$ from a fixed $z_{0} \in \mathscr{R}$ to an arbitrary $z \in \mathscr{R}$ as above, as a function $F : \mathscr{R} \to \mathbb{C}$ of $z \in \mathscr{R}$. First, let us examine whether this is genuinely well-defined. As a premise, we assumed that $\displaystyle \int_{\mathscr{C}} f(z) dz = 0$ on every closed path $\mathscr{C}$. Accordingly, no matter which path $w_{0} : z \to z_{0}$ we fix, $$ \int_{z_{0}}^{z} f(w) dw + \int_{w_{0}} f(u) du = 0 $$ holds, and we can see that $F(z)$ always takes the same value no matter which path from $z_{0}$ to $z$ we integrate along. Accordingly, we have confirmed that $F$ is a function whose value is determined uniquely by the choice of $z$ alone.

By the basic properties of the complex contour integral, ${{F(z+h) - F(z)}\over{h}} = {{1} \over {h}} \int_{z}^{z+h} f(w) dw$, so $$ \begin{align*} & \left| {{F(z+h) - F(z)}\over{h}} - f(z) \right| \\ =& \left| {{1} \over {h}} \int_{z}^{z+h} f(w) dw - {{ 1 } \over { h }} h f(z) \right| \\ =& \left| {{1} \over {h}} \int_{z}^{z+h} f(w) dw - {{ 1 } \over { h }} \int_{z}^{z+h} f(z) dw \right| \\ =& \left| {{1} \over {h}} \int_{z}^{z+h} (f(w) - f(z)) dw \right| \end{align*} $$ Here, since we said that $f$ is continuous, for a given $\varepsilon >0$ there exists a $\delta$ satisfying $$ |h|< \delta \implies |f(z+h) - f(z)| < \varepsilon $$

ML lemma: For a positive number $M$ satisfying $|f(z)| \le M$ and the length $L$ of $\mathscr{C}$, $$ \left| \int_{\mathscr{C}} f(z) dz \right| \le ML $$

By the ML lemma, $$ \left| {{F(z+h) - F(z)}\over{h}} - f(z) \right| = \left| {{1} \over {|h|}} \int_{z}^{z+h} (f(w) - f(z)) dw \right| < {{1} \over {|h|}} \varepsilon |h| = \varepsilon $$ Therefore $$ f(z) = \lim_{h \to 0} {{F(z+h) - F(z)} \over {h}} = F ' (z) $$ That is, $f$ is the derivative of some function $F$. In complex analysis, being differentiable once means being infinitely differentiable, so if $F$ is differentiable, then $f$ is also differentiable.

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