# Analytic functions (part 2)

July 10, 2012 1 Comment

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

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 1Let (with be a power series centered at with radius of convergence . Let , the open ball on which converges. Then 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 on the interval . For every value except , when you plug it in, the result gets closer and closer to zero. Thus, the limit should be zero. However, if you plug in , the result is always 1, and so the pointwise limit of these polynomials is

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 , the limit function is continuous. The reason for this is that the sequence of functions , each on the interval is not uniformly continuous, whereas, on 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 . 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 centered at can be “exhausted” by these smaller balls, by using uniformity on balls of radius larger and larger (approaching ).

So for any point with , pick in between and . We’ll show that on the ball , the power series is a uniform limit of polynomials. Let

Clearly , but we want to show that this limit is uniform. Since , it follows that there are only finitely many for which . Pick to be larger than all of these finitely many . Then, for any , we have

because , for large enough , we can make this as small as we like. Moreover, to make it less than some specified , our choice of only depends and . This means the same choice of works for all points in , proving that is the uniform limit of the , at least on . On this set is a uniform limit of continuous functions, and hence continuous. In particular, is continuous at . Since was arbitrary, is continuous at every with .

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

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