# The theorem

July 19, 2012 Leave a comment

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

Theorem 1Let for open in . Then the following are equivalent:

- is holomorphic.
- is infinitely differentiable.
- is analytic.

*Proof:*

(2) implies (1). Duh.

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

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

Pick any point in . There is a small open ball around contained entirely in . We’ll show that is given in this ball by a power series centered at . Pick a small circle around of radius contained in , and call it . We’ll show that is given by a power series centered at (at least within a ball of radius . 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

Can you see why this might work? If I had such a power series , and I plugged in , all the terms would go away, except . By my definition,

which by the Cauchy integral formula, is just . Pretty cool, huh?

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

We’re getting very close. Can you see how this might work? The will have the coefficient involving an integral that has in the denominator, just as 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 last time (by compactness of the circle). We’ll use this again. So we know . Now we’re left with just

Recall that is always chosen on , so , and is chosen inside the circle, so . Thus, the entire fraction is less than 1 in absolute value. The sum over all of these is a geometric series which converges, and so the Weiersrass -test tells us that the the sum of the integrals is the integral of the sum. Thus