# The theorem

We are finally ready to prove “the theorem” that I’ve been talking about for so long:

Theorem 1 Let ${f:U\rightarrow{\mathbb C}}$ for ${U}$ open in ${{\mathbb C}}$. Then the following are equivalent:

1. ${f}$ is holomorphic.
2. ${f}$ is infinitely differentiable.
3. ${f}$ is analytic.

Proof:

(2) implies (1). Duh.

(3) implies (2). Check it out here.

(1) implies (3). Finally, here we go:

Pick any point ${z_0}$ in ${U}$. There is a small open ball ${B}$ around ${z_0}$ contained entirely in ${U}$. We’ll show that ${f}$ is given in this ball by a power series centered at ${z_0}$. Pick a small circle around ${z_0}$ of radius ${r}$ contained in ${B}$, and call it ${\beta_r}$. We’ll show that ${f}$ is given by a power series centered at ${z_0}$ (at least within a ball of radius ${r}$. We’ll use this and the Cauchy integral formula to compute what the coefficients of the power series have to be. Actually, I already know what they are, so let

$\displaystyle a_k=\frac1{2\pi i}\int_\gamma\frac{f(w)}{(w-z_0)^{k+1}}dw.$

Can you see why this might work? If I had such a power series ${f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots}$, and I plugged in ${z_0}$, all the terms would go away, except ${a_0}$. By my definition,

$\displaystyle a_0=\displaystyle\frac{1}{2\pi i}\int_\gamma\frac{f(w)}{w-z_0}dw,$

which by the Cauchy integral formula, is just ${f(z_0)}$. Pretty cool, huh?

Okay, so here goes. By the Cauchy integral formula, we can take a small counter-clockwise circle ${\gamma}$ centered at ${z_0}$ and going around ${z}$ (for ${z}$ in the ball ${B}$) to get:

$\displaystyle \begin{array}{rcl} f(z) &=& \displaystyle\frac{1}{2\pi i}\int_\beta\frac{f(w)}{w-z}dw\\ &=& \displaystyle\frac{1}{2\pi i}\int_\beta\frac{1}{w-z_0}\cdot\frac{w-z_0}{(w-z_0)-(z-z_0)}f(w)dw\\ &=& \displaystyle\frac{1}{2\pi i}\int_\beta\frac{1}{w-z_0}\cdot\frac{1}{1-\frac{z-z_0}{w-z_0}}f(w)dw\\ &=& \displaystyle\frac{1}{2\pi i}\int_\beta\frac{1}{w-z_0}\cdot\sum_{k=0}^\infty\left(\frac{z-z_0}{w-z_0}\right)^kf(w)dw\\ &=& \displaystyle\frac{1}{2\pi i}\int_\beta\sum_{k=0}^\infty\frac{(z-z_0)^k}{(w-z_0)^{k+1}}f(w)dw \end{array}$

We’re getting very close. Can you see how this might work? The ${(z-z_0)^k}$ will have the coefficient involving an integral that has ${(w-z_0)^{k+1}}$ in the denominator, just as ${a_k}$ does.

The next step is to interchange the sum and the integral. If the sum was finite, we’d be allowed to do this, but because integrals are really limiting operators, and infinite sums are as well, we don’t just get to exchange them willy-nilly. I had a professor in undergrad that claimed “more than 80 percent of analysis is figuring out when you can interchange limits.” I thought that sounded lame at the time, but it’s definitely helped me remember to always justify such interchanges. Anyway, we can use the Weierstrass M-test. Recall how we came up with the bound ${M_r}$ last time (by compactness of the circle). We’ll use this again. So we know ${|f(w)/(w-z_0)|\le M_r}$. Now we’re left with just

$\displaystyle \left|\frac{(z-z_0)^k}{(w-z_0)^{k+1}}f(w)\right|\le M_r\cdot \left|\frac{z-z_0}{w-z_0}\right|^k.$

Recall that ${w}$ is always chosen on ${\beta_r}$, so ${|w-z_0|=r}$, and ${z}$ is chosen inside the circle, so ${|z-z_0|. Thus, the entire fraction is less than 1 in absolute value. The sum over all ${k}$ of these is a geometric series which converges, and so the Weiersrass ${M}$-test tells us that the the sum of the integrals is the integral of the sum. Thus

$\displaystyle \begin{array}{rcl} f(z) &=& \displaystyle\frac{1}{2\pi i}\int_\beta\sum_{k=0}^\infty\frac{(z-z_0)^k}{(w-z_0)^{k+1}}f(w)dw\\ &=& \displaystyle\sum_{k=0}^\infty\frac{1}{2\pi i}\int_\beta\frac{(z-z_0)^k}{(w-z_0)^{k+1}}f(w)dw\\ &=& \displaystyle\sum_{k=0}^\infty(z-z_0)^k\frac{1}{2\pi i}\int_\beta\frac{f(w)}{(w-z_0)^{k+1}}dw\\ &=& \displaystyle\sum_{k=0}^\infty(z-z_0)^k\cdot a_k. \end{array}$

$\Box$