# Analytic functions (part 2)

Last time we saw that power series have a well-defined radius of convergence, given by

$\displaystyle R=\left(\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.$

The question is: Within that radius of convergence, must a power series be holomorphic? The answer isn’t just “yes.” It’s “HELL YES!” Our proof will take two days. First, we require a fact that you probably have already taken for granted:

Theorem 1 Let ${f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots}$ (with ${a_n\in{\mathbb C})}$ be a power series centered at ${z_0}$ with radius of convergence ${R}$. Let ${B=\{z\mid |z-z_0|, the open ball on which ${f}$ converges. Then ${f|_B}$ is continuous.

You’re thinking: Of course it’s continuous. It’s like a “super polynomial,” right? Unfortunately, if we’re careful about it, the result isn’t so obvious. By definition, it’s a limit of polynomials, and limits of polynomials aren’t always continuous. Consider the sequence of functions ${1, x, x^2, x^3, x^4,\dots}$ on the interval ${[0,1]}$. For every value except ${1}$, when you plug it in, the result gets closer and closer to zero. Thus, the limit should be zero. However, if you plug in ${1}$, the result is always 1, and so the pointwise limit of these polynomials is

$\displaystyle x\mapsto\left\{\begin{array}{cl}0 & \mbox{if }x\in[0,1)\\ 1 & \mbox{if }x=1\end{array}\right.$

As we’ll see, the problem isn’t with the sequence of functions that we have. It’s with their domain. If we restrict the domain to just the open interval ${(0,1)}$, the limit function is continuous. The reason for this is that the sequence of functions ${1,x,x^2,\dots}$, each on the interval ${[0,1]}$ is not uniformly continuous, whereas, on ${(0,1)}$ it is. We’ll use this uniformity condition in an important way.

Proof: The approach for the first part is to show that power series are a uniform limit of polynomials on a ball of radius smaller than ${R}$. Then, since polynomials are continuous, and a uniform limit of continuous functions is continuous, the power series on that smaller ball must be continuous. We then see that the open ball of radius ${R}$ centered at ${z_0}$ can be “exhausted” by these smaller balls, by using uniformity on balls of radius larger and larger (approaching ${R}$).

So for any point ${w}$ with ${|w-z_0|, pick ${\hat R}$ in between ${|w-z_0|}$ and ${R}$. We’ll show that on the ball ${B_{\hat R}(z_0)=\{z\mid |z-z_0|<\hat R\}}$, the power series ${f}$ is a uniform limit of polynomials. Let

$\displaystyle f_n(z)=a_0+a_1(z-z_0)+\cdots+a_n(z-z_0)^n.$

Clearly ${0=\displaystyle\lim_{n\rightarrow\infty}f-f_n}$, but we want to show that this limit is uniform. Since ${\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=1/R}$, it follows that there are only finitely many ${n}$ for which ${\sqrt[n]{|a_n|\hat R^n}>\hat R/R}$. Pick ${N}$ to be larger than all of these finitely many ${n}$. Then, for any ${m>N}$, we have

$\displaystyle \begin{array}{rcl} f(w)-f_m(w) &=& \displaystyle\sum_{k=m+1}^\infty a_k(w-z_0)^k\\ |f(w)-f_m(w)| &\le& \displaystyle\sum_{k=m+1}^\infty |a_k|\cdot|w-z_0|^k\\ &\le& \displaystyle\sum_{k=m+1}^\infty \left(\sqrt[k]{|a_k}\hat R\right)^k\\ &\le& \displaystyle\sum_{k=m+1}^\infty \left(\frac{\hat R}{R}\right)^k\\ &=& \left(\frac{\hat R}{R}\right)^{m+1}\cdot\left(\frac{1}{1-\left(\frac{\hat R}{R}\right)}\right) \end{array}$

because ${\hat R/R<1}$, for large enough ${m}$, we can make this as small as we like. Moreover, to make it less than some specified ${\varepsilon}$, our choice of ${m}$ only depends ${R'}$ and ${R}$. This means the same choice of ${m}$ works for all points in ${B_{\hat R}(z_0)}$, proving that ${f}$ is the uniform limit of the ${f_n}$, at least on ${B_{\hat R}(z-z_0)}$. On this set ${f}$ is a uniform limit of continuous functions, and hence continuous. In particular, ${f}$ is continuous at ${w}$. Since ${w}$ was arbitrary, ${f}$ is continuous at every ${w}$ with ${|w-z_0|.

$\Box$

Next time we’ll show that ${f'}$ is also a power series, and also has radius of convergence ${R}$. Then we’ll argue that this implies ${f}$ is not only holomorphic (giving the answer “yes”) but also infinitely differentiable (giving the answer “HELL YES!”).

Advertisements