📂Complex Anaylsis

Proof of Morera's Theorem

Theorem ¹

A complex function $f : \mathbb{C} \to \mathbb{C}$ is analytic in a simply connected domain $\mathscr{R}$ if it is continuous in $\mathscr{R}$ and satisfies $\displaystyle \int_{\mathscr{C}} f(z) dz = 0$ for all closed paths $\mathscr{C} \subset \mathscr{R}$ contained in $\mathscr{R}$.

Explanation

This can be thought of as roughly the converse of Cauchy’s theorem. Interestingly, it’s common knowledge in analysis that ‘if it’s differentiable, then it’s continuous, and if it’s continuous, then it’s integrable’. However, Morera’s theorem intriguingly uses integration to determine the differentiability of a function, which is truly a remarkable theorem.

Proof ²

$$F(z) := \int_{z_{0}}^{z} f(w) dw$$ Define the complex path integral of $f$ from a fixed point $z_{0} \in \mathscr{R}$ to any point $z \in \mathscr{R}$ as a function of $z \in \mathscr{R}$ as shown above. First, let’s check if it’s well-defined. Assuming the premise that $\displaystyle \int_{\mathscr{C}} f(z) dz = 0$ for all closed paths $\mathscr{C}$, regardless of which path $w_{0} : z \to z_{0}$ is fixed, $$\int_{z_{0}}^{z} f(w) dw + \int_{w_{0}} f(u) du = 0$$ it’s evident that integrating $F(z)$ over any path from $z_{0}$ to $z$ always yields the same value. Thus, $F$ is confirmed to be a function determined solely by the choice of $z$.

By the fundamental property of complex path integrals, since ${{F(z+h) - F(z)}\over{h}} = {{1} \over {h}} \int_{z}^{z+h} f(w) dw$, $$\begin{align*} & \left| {{F(z+h) - F(z)}\over{h}} - f(z) \right| \\ =& \left| {{1} \over {h}} \int_{z}^{z+h} f(w) dw - {{ 1 } \over { h }} h f(z) \right| \\ =& \left| {{1} \over {h}} \int_{z}^{z+h} f(w) dw - {{ 1 } \over { h }} \int_{z}^{z+h} f(z) dw \right| \\ =& \left| {{1} \over {h}} \int_{z}^{z+h} (f(w) - f(z)) dw \right| \end{align*}$$ Given that $f$ is continuous, for a given $\varepsilon >0$, $$|h|< \delta \implies |f(z+h) - f(z)| < \varepsilon$$ there exists a $\delta$ that satisfies.

ML Lemma: For a positive number $M$ that satisfies $|f(z)| \le M$ and the length &VariableDoubleVerticalBar;$\mathscr{C}$&VariableDoubleVerticalBar; of $L$, $$\left| \int_{\mathscr{C}} f(z) dz \right| \le ML$$

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, $$f(z) = \lim_{h \to 0} {{F(z+h) - F(z)} \over {h}} = F ' (z)$$ Thus, $f$ is the derivative of some function $F$. In complex analysis, if it’s differentiable once, it’s infinitely differentiable; hence, if $F$ is differentiable, so is $f$.

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