Prove Morera's Theorem in circles cases.

Step 1. Assume first that $f \in C^1$. Then writing $f = u + iv$ for $u, v : \Bbb{C} \to \Bbb{R}$ shows that for any $z_0 \in \Bbb{C}$ and $r > 0$,

$$ 0 = \oint_{\partial B_r(z_0)} f(z) \, dz = \oint_{\partial B_r(z_0)} (u \, dx - v \, dy) + i \oint_{\partial B_r(z_0)} (u \, dy + v \, dx)$$

and hence the real part and the imaginary part vanish simultaneously. Now by the Green's theorem,

$$ 0 = -\oint_{\partial B_r(z_0)} (u \, dx - v \, dy) = \iint_{B_r(z_0)} \left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\right) \, dxdy $$ $$ 0 = \oint_{\partial B_r(z_0)} (u \, dy + v \, dx) = \iint_{B_r(z_0)} \left( \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right) \, dxdy $$

Since this is true for any ball $B_r(z_0)$, dividing both equations by $|B_r(z_0)| = \pi r^2$ and taking $r \to 0$ yields the Cauchy-Riemann equation

$$ \frac{\partial u}{\partial x} -= \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}. $$

This shows that $f$ is holomorphic.

Step 2. Now we only impose the condition that $f$ is continuous. In order to utilize the previous step, let $\varphi : \Bbb{C} \to \Bbb{R}$ be a compactly supported smooth function such that

$$ \iint_{\Bbb{C}} \varphi(\mathrm{x}) \, d^2\mathrm{x} = 1. $$

Then it is not hard to check that $f_n$ defined by

$$ f_n(z) = \iint_{\Bbb{C}} f(\mathrm{x})\varphi_n(z-\mathrm{x}) \, d^2\mathrm{x} = \iint_{\Bbb{C}} f(z-\mathrm{x})\varphi_n(\mathrm{x}) \, d^2\mathrm{x}, \qquad \varphi_n(\mathrm{x}) = n^2 \varphi(n\mathrm{x}) $$

are smooth and the sequence $(f_n)$ converges locally uniformly to $f$. Moreover,

\begin{align*} \oint_{\partial B_r(z_0)} f_n(z) \, dz &= \oint_{\partial B_r(z_0)} \iint_{\Bbb{C}} f(z-\mathrm{x})\varphi_n(\mathrm{x}) \, d^2\mathrm{x} dz \\ &= \iint_{\Bbb{C}} \bigg( \oint_{\partial B_r(z_0)} f(z-\mathrm{x}) \, dz \bigg) \varphi_n(\mathrm{x}) \, d^2\mathrm{x} \\ &= \iint_{\Bbb{C}} \bigg( \oint_{\partial B_r(z_0-\mathrm{x})} f(z) \, dz \bigg) \varphi_n(\mathrm{x}) \, d^2\mathrm{x} \\ &= 0. \end{align*}

Therefore $(f_n)$ is a sequence of holomorphic functions by Step 1. Since $f$ is a locally uniform limit of holomorphic functions, $f$ is holomorphic as well. (Cauchy integration formula guarantees this.)

Addendum. Locally uniform convergence of $f_n$:

Let $\bar{B}(0, R)$ be any compact ball in $\Bbb{C}$ and $\epsilon > 0$ be given. Pick $\delta > 0$ and $N \in \Bbb{N}$ as follows:

• Since $f$ is uniformly continuous on the compact set $\bar{B}(0, R+1)$, there exists $\delta > 0$ such that $|f(z) - f(w)| < \epsilon$ whenever $z, w \in \bar{B}(0, R+1)$ and $|z-w| < \delta$. May assume that $\delta < 1$.

• Since $\varphi$ is compactly supported, there exists $N > 0$ such that $\operatorname{supp} \varphi_n \subset B(0, \delta)$ for all $n \geq N$.

Now for $z \in \bar{B}(0, R)$ and for $n \geq N$,

\begin{align*} | f_n(z) - f(z) | &\leq \iint_{\Bbb{C}} |f(z-\mathrm{x}) - f(z)| |\varphi_n(\mathrm{x})| \, d^2\mathrm{x} \\ &= \iint_{B(0, \delta)} |f(z-\mathrm{x}) - f(z)| |\varphi_n(\mathrm{x})| \, d^2\mathrm{x} \\ &\leq \iint_{B(0, \delta)} \epsilon |\varphi_n(\mathrm{x})| \, d^2\mathrm{x} \\ &= C\epsilon, \end{align*}

where $C = \iint_{\Bbb{C}}|\varphi| = \iint_{\Bbb{C}}|\varphi_n|$ is an absolute constant. This shows that $f_n \to f$ uniformly on $\bar{B}(0, R)$.