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在复分析中，Morera 定理，以 Giacinto Morera 的名字命名，是用来证明一个函数全纯的重要准则。

Morera 定理是说，如果函数 \(f \) 定义在复平面的开集 \(D \) 上的连续函数，并且

\[\displaystyle\oint_C f(z) dz = 0\]

对  \(D\) 内的任意三角形  \(C\), 那么 \(f \) 全纯.

Morera 定理的假设等价于 \(f \) 在 \(D\) 中有原函数. Morera 定理的逆一般而言，是不成立的。一个全纯函数在定义域上未必有原函数，除非添加更多的条件。例如， 如果函数的定义域是单连通, Cauchy 积分定理断定全纯函数沿闭曲线的线积分是 \(0\)

现在我们陈述并且证明 Morera 定理的一个推广:

如若 \(f \) 在 \( \mathbb C\) 连续,  并且

\[\displaystyle\int_C f(z) dz =0 \]

对每个圆  \(C\). 证明\(f \) 全纯.

证明一

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

$$ \iint_{\Bbb{C}} \varphi(\mathrm{x}) \, d^2\mathrm{x} = 0. $$

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 $\bar{B}(0, R)$.

 

证明二

Proof. If $\varphi(z)$ is a smooth bounded function with

$$\displaystyle\int_{\mathbb C} \varphi(z)dxdy = 1 $$

let

$$\displaystyle\varphi_\varepsilon(z) = \frac{1}{\varepsilon^2} \varphi(\frac{z}{\varepsilon})$$

so $\varphi_\varepsilon(z)$ is an approximate identity.

Define

$$\displaystyle f_\varepsilon(z) = \int_{\mathbb C} f(z-w) \varphi_\varepsilon(w)dxdy,$$

that is, the convolution of f and $\varphi_\varepsilon$. Then by Folland Prop. 8.10 $f_\varepsilon$ is smooth (actually smooth on compact subsets which implies smooth everywhere). By Folland Thm. 8.14 $f\_\varepsilon \to f $ uniformly on compact subset as
$\varepsilon \to 0$ and

$$\displaystyle\int_C {{f_\varepsilon }} (z)dz = \int_C {\int_\mathbb{C} f } (z – w){\varphi _\varepsilon }(w)dxdydz = \int_\mathbb{C} {{\varphi _\varepsilon }(w)} \int_C {{f_\varepsilon }} (z – w)dzdxdy.$$

By a change of variables

$$\displaystyle \int_C f(z-w)dz = \int_{C-w } f(z)dz = 0 $$

where C-w is the translate of C by w which is still a circle. So

$$\displaystyle\int_C f_\varepsilon (z)dz = 0.$$

Thus if $f_\varepsilon = u_\varepsilon + iv\_\varepsilon $ the partial derivatives all exist and are continuous and

$$\displaystyle 0 = \int_C {{f_\varepsilon }\left( z \right)dz} = \int_C {\left[ {{u_\varepsilon } + i{v_\varepsilon }} \right]dx + \left[ { – {v_\varepsilon } + i{u_\varepsilon }} \right]dy}$$

which implies

$$\displaystyle 0 = \int_{{\rm int } C} {\left[ {\frac{{\partial ( – {v_\varepsilon })}}{{\partial x}} + \frac{{\partial (i{u_\varepsilon })}}{{\partial x}} – \frac{{\partial ({u_\varepsilon })}}{{\partial y}} – \frac{{\partial (i{v_\varepsilon })}}{{\partial y}}} \right]dxdy}$$

or equivalently,

$$\displaystyle 0 = \int_{{\rm int } C} { – \left[ {\frac{{\partial {u_\varepsilon }}}{{\partial y}} + \frac{{\partial {v_\varepsilon }}}{{\partial x}}} \right] + i\left[ {\frac{{\partial {u_\varepsilon }}}{{\partial x}} – \frac{{\partial {v_\varepsilon }}}{{\partial y}}} \right]dxdy}$$

Since this is true for any circle (any translate or dilate) this implies

$$\displaystyle\frac{{\partial {u_\varepsilon }}}{{\partial y}} = – \frac{{\partial {v_\varepsilon }}}{{\partial x}},\quad \frac{{\partial {u_\varepsilon }}}{{\partial x}} = \frac{{\partial {v_\varepsilon }}}{{\partial y}}$$

the C-R equations for $f_\varepsilon $ so $f_\varepsilon$ is holomorphic and since it converges uniformly on compact subsets to $f$, $f$ is also holomorphic.