Analyticity of Holomorphic Functions - Proof

Proof

The argument, first given by Cauchy, hinges on Cauchy's integral formula and the power series development of the expression

.

Suppose ƒ is differentiable everywhere within some open disk centered at a. Let z be within that open disk. Let C be a positively oriented (i.e., counterclockwise) circle centered at a, lying within that open disk but farther from a than z is. Starting with Cauchy's integral formula, we have

\begin{align}f(z) &{}= {1 \over 2\pi i}\int_C {f(w) \over w-z}\,\mathrm{d}w \\ &{}= {1 \over 2\pi i}\int_C {f(w) \over (w-a)-(z-a)} \,\mathrm{d}w \\ &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{1 \over 1-{z-a \over w-a}}f(w)\,\mathrm{d}w \\ &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{\sum_{n=0}^\infty\left({z-a \over w-a}\right)^n} f(w)\,\mathrm{d}w \\ &{}=\sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,\mathrm{d}w.\end{align}

To justify the interchange of the sum and the integral, one must notice that in the intersection of |(z − a)/(w − a)| ≤ r < 1 and some closed domain containing C, ƒ(w)/(w − a) is holomorphic and therefore bounded by some positive number M. So we have

The Weierstrass M-test says the series converges uniformly, and thus the interchange of the sum and the integral is justified.

Since the factor (z − a)n does not depend on the variable of integration w, it can be pulled out:

And now the integral and the factor of 1/(2πi) do not depend on z, i.e., as a function of z, that whole expression is a constant cn, so we can write:

and that is the desired power series.

Read more about this topic:  Analyticity Of Holomorphic Functions

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