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 is a continuous, complex-valued function defined on an open set in the complex plane, satisfying

for every triangle in , then must be holomorphic on .

The assumption of Morera’s theorem is equivalent to that has an anti-derivative on . 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 is continuous on , and

for every circle . Prove is holomorphic.

*Proof*. If is a smooth bounded function with

let

so is an approximate identity.

Define

,

that is, the convolution of and . Then by Folland Prop. 8.10 is smooth (actually smooth on compact subsets which implies smooth everywhere). By Folland Thm. 8.14 uniformly on compact subset as and

.

By a change of variables

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

.

Thus if the partial derivatives all exist and are continuous and

which implies

or equivalently,

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

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

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

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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