Ngô Quốc Anh

July 20, 2009

A generalization of the Morera’s Theorem

In complex analysis, a branch of mathematics, Morera’s theorem, named after Giacinto Morera, gives an important criterion for proving that a function is holomorphic.

Morera’s theorem states that if $f$ is a continuous, complex-valued function defined on an open set $D$ in the complex plane, satisfying $\displaystyle\oint_C f(z) dz = 0$

for every triangle $C$ in $D$, then $f$ must be holomorphic on $D$.

The assumption of Morera’s theorem is equivalent to that $f$ has an anti-derivative on $D$. The converse of the theorem is not true in general. A holomorphic function need not possess an antiderivative on its domain, unless one imposes additional assumptions. For instance, Cauchy’s integral theorem states that the line integral of a holomorphic function along a closed curve is zero, provided that the domain of the function is simply connected.

Now we state and prove the following generalization of the Morera’s theorem: Suppose that $f$ is continuous on $\mathbb C$, and $\displaystyle\int_C f(z) dz = 0$

for every circle $C$. Prove $f$ is holomorphic. 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.

1 Comment »

1. Seems to contradict Zalcman’s famous theorem: If the integral is zero on all circles of radius r and radius R and R/r is not a ratio of the zeroes of the Bessel function J_1(z), then f need not be analytic at any point.

Comment by Jet Foncannon — September 9, 2021 @ 0:12

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