Posted by: hilbertthm90 | September 29, 2008

Zeros of Analytic Functions

A strange property of analytic functions is that the zeros are isolated. I don’t remember the proof I originally learned of this fact, but today I saw a really interesting topological way to do it. It makes sense now.

More precise formulation: If \Omega\subset\mathbb{C} is a connected open set, then \{z: f(z)=0\} consists of isolated points if f is analytic on \Omega. (Oops, I started writing this up and realized that I need to trivially throw out the case where f\equiv 0.

Proof: Let U_1=\{a\in\Omega : \exists\delta>0, \ f(z)\neq 0 \ on \ 0<|z-a|<\delta\} and let U_2=\{a\in\Omega : \exists\delta>0, \ f(z)\equiv 0 \ on \ 0<|z-a|<\delta \}. Reformulating the setup we see that U_1 means: if f has a zero, it is isolated since f is nonzero on a punctured disk (meaning the zero must be the punctured part). Also U_2 is just the regions that f has non-isolated zeros.

It is straightforward to check that both U_1 and U_2 are open (just choose \delta’s sufficiently small to stay inside the declared sets). Also we have that U_1\cap U_2=\emptyset and I now claim \Omega=U_1\cup U_2.

This seems obvious, but should be pinned down in some sort of argument. Let z_0\in\Omega. We claim that there is a punctured disk about z_0 such that either f\equiv 0 on the disk or f\neq 0 anywhere on the disk. By analyticity, we have a power series convergent on some radius r>0 about z_0, i.e. f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n on |z-z_0|<r.

Suppose a_k is the first nonzero coefficient (by not being equivalently zero, this must exist). Then f(z)=\sum_{n=0}^\infty a_{n+k}(z-z_0)^n=(z-z_0)^{-k}\sum_{n=k}a_n(z-z_0)^n. So since the series converges in 0<|z-z_0|<r and since f is continuous we can choose 0<\delta<r small enough so that |f(z)-f(z_0)|=|f(z)-a_k|<\frac{|a_k|}{2}. This clearly shows that f(z)\neq 0 on 0<|z-z_0|<\delta else we’d have |a_k|<\frac{|a_k|}{2}. So either there is a punctured disk on which f is non-zero, or the f has no first non-zero coefficient making it zero everywhere on that first disk |z-z_0|<r proving the claim.

The properties U_1\cap U_2=\empty and \Omega=U_1\cup U_2 (along with both sets being open) combine to give that either U_1=\emptyset or U_2=\emptyset by the connectedness of \Omega. This simply means that all the zeros are isolated since we ruled out the alternative of being equivalently zero.

This goes to show how remarkably different analytic on \mathbb{C} is to continuous on \mathbb{R}. In fact, even infinitely differentiable functions on \mathbb{R}. Bump functions play a crucial role in many areas of analysis and they are smooth functions with compact support meaning that outside of a bounded they are zero. An entire class of important functions violates this property that analytic functions are guaranteed to have.

Posted in analysis | Tags: ,

«
»

Leave a response






Your response:

Categories