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 is a connected open set, then consists of isolated points if is analytic on . (Oops, I started writing this up and realized that I need to trivially throw out the case where .

Proof: Let and let . Reformulating the setup we see that 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 is just the regions that f has non-isolated zeros.

It is straightforward to check that both and are open (just choose ‘s sufficiently small to stay inside the declared sets). Also we have that and I now claim .

This seems obvious, but should be pinned down in some sort of argument. Let . We claim that there is a punctured disk about such that either on the disk or anywhere on the disk. By analyticity, we have a power series convergent on some radius about , i.e. on .

Suppose is the first nonzero coefficient (by not being equivalently zero, this must exist). Then . So since the series converges in and since is continuous we can choose small enough so that . This clearly shows that on else we’d have . 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 proving the claim.

The properties and (along with both sets being open) combine to give that either or by the connectedness of . 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 is to continuous on . In fact, even infinitely differentiable functions on . 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.

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I am a mathematics graduate student fascinated in how all my interests fit together.

September 23, 2010 at 10:01 am

i liked your proof very much. specially the use of connectedness of omega. thank you