Theorem: let the function f(z) be continuous on a domain \mathcal{D}. If for every simple closed contour C contained within \mathcal{D}, the following condition holds:
\oint_{C} f(z)\mathrm{d}z = 0
then the function f(z) is analytic in the domain \mathcal{D}.
The proof proceeds by explicitly constructing a primitive (or antiderivative) for the function f(z) and then leveraging the properties of this primitive to demonstrate the analyticity of f(z) itself.
Proof: let’s define a function F(z) by an integral. Fix a point z_0 \in \mathcal{D} and for any other point z \in \mathcal{D}, we define:
F(z) = \int_{z_0}^z f(\zeta) \mathrm{d}\zeta
The hypothesis of the theorem—that the integral of f(z) over any closed loop is zero ensures that F(z) is a well-defined, single-valued function.
To verify this, we consider two distinct paths, C_1 and C_2, from z_0 to z within \mathcal{D}.
The concatenation of C_1 with the reverse of C_2 forms a closed loop C = C_1 \cup (-C_2). By hypothesis, \oint_C f(\zeta)\mathrm{d}\zeta = 0.
This implies:
\begin{aligned} & \int_{C_1} f(\zeta)\mathrm{d}\zeta - \int_{C_2} f(\zeta)\mathrm{d}\zeta = 0 \\ & \int_{C_1} f(\zeta)\mathrm{d}\zeta = \int_{C_2} f(\zeta)\mathrm{d}\zeta \end{aligned}
The value of the integral defining F(z) is independent of the path of integration chosen between z_0 and z, making F(z) uniquely defined for each z \in \mathcal{D}.
The next step is to demonstrate that F(z) is analytic and that its derivative is precisely f(z). We employ the definition of the derivative:
F^\prime(z) = \lim_{\Delta z \to 0} \frac{F(z+\Delta z) - F(z)}{\Delta z}
From the definition of F(z) and its path independence, we can write:
F(z+\Delta z) - F(z) = \int_{z_0}^{z+\Delta z} f(\zeta)\mathrm{d}\zeta - \int_{z_0}^z f(\zeta)\mathrm{d}\zeta = \int_z^{z+\Delta z} f(\zeta)\mathrm{d}\zeta
As it is path independent, we can choose the path of integration for this last integral to be the straight line segment connecting z to z+\Delta z. The difference quotient then becomes:
\frac{F(z+\Delta z) - F(z)}{\Delta z} = \frac{1}{\Delta z} \int_z^{z+\Delta z} f(\zeta)\mathrm{d}\zeta
To show this expression approaches f(z) as \Delta z \to 0, we subtract f(z) and analyze the magnitude of the difference. Note that f(z) = \frac{1}{\Delta z} \int_z^{z+\Delta z} f(z)\mathrm{d}\zeta, as f(z) is a constant with respect to the integration variable \zeta:
\begin{aligned} \frac{F(z+\Delta z) - F(z)}{\Delta z} - f(z) = & \frac{1}{\Delta z} \left( \int_z^{z+\Delta z} f(\zeta)\mathrm{d}\zeta - \int_z^{z+\Delta z} f(z)\mathrm{d}\zeta \right) \\ = & \frac{1}{\Delta z} \int_z^{z+\Delta z} [f(\zeta) - f(z)]\mathrm{d}\zeta \end{aligned}
We can now apply the ML-bound to the integral on the right-hand side. The length of the straight-line path is L = |\Delta z|. The maximum value of the integrand’s modulus is M = \max_{\zeta \in [z, z+\Delta z]} |f(\zeta) - f(z)|:
\begin{aligned} & \left| \frac{F(z+\Delta z) - F(z)}{\Delta z} - f(z) \right| = \left| \frac{1}{\Delta z} \int_z^{z+\Delta z} [f(\zeta) - f(z)]\mathrm{d}\zeta \right| \\ & \quad \quad \le \frac{1}{|\Delta z|} (L \cdot M) = \frac{|\Delta z|}{|\Delta z|} \max_{\zeta \in [z, z+\Delta z]} |f(\zeta) - f(z)| \end{aligned}
This inequality then holds:
\left| \frac{F(z+\Delta z) - F(z)}{\Delta z} - f(z) \right| \le \max_{\zeta \in [z, z+\Delta z]} |f(\zeta) - f(z)|
By the initial hypothesis, the function f(z) is continuous in \mathcal{D}. This means that for any given \varepsilon > 0, there exists a \delta > 0 such that for any \zeta satisfying |\zeta - z| < \delta, we have |f(\zeta) - f(z)| < \varepsilon. If we choose |\Delta z| < \delta, then every point \zeta on the segment between z and z+\Delta z satisfies |\zeta - z| \le |\Delta z| < \delta. Consequently, the maximum value of the difference is also bounded by \varepsilon.
This demonstrates that for any \varepsilon > 0, there exists a \delta > 0 such that for |\Delta z| < \delta:
\left|\frac{F(z+\Delta z) - F(z)}{\Delta z} - f(z) \right| < \varepsilon
This is precisely the definition of the limit:
\lim_{\Delta z \to 0} \left|\frac{F(z+\Delta z) - F(z)}{\Delta z} - f(z) \right| = 0
and we have therefore shown that F^\prime(z) exists and is equal to f(z) for all z \in \mathcal{D}.
So, have demonstrated that F(z) is an analytic function in \mathcal{D}, with its derivative being F^\prime(z) = f(z).
We demonstrated here that if a function (F(z) in this case) is analytic in a domain, then its derivative must also be analytic in that same domain.
Therefore, f(z), being the derivative of an analytic function, is itself analytic in \mathcal{D}. This completes the proof.
Morera’s theorem provides a partial converse to Cauchy’s Integral Theorem. While Cauchy’s theorem asserts that if a function is analytic in a simply connected domain, its integral around any closed loop within that domain is zero, Morera’s theorem establishes that continuity and the universal vanishing of loop integrals are sufficient conditions to guarantee analyticity.