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.

File:Morera's Theorem.png

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.

Source: http://www.math.ucla.edu/~mhofmann/math246a-3a.pdf

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