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|<R\}}, 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|<R}, 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|<R}.


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!”).


One Response to Analytic functions (part 2)

  1. Pingback: Analytic functions (part 3) « Andy Soffer

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s