# Can a non-constant analytic function have infinitely many zeros on a closed disk?

I think not, however my proof is quite sketchy so far.. My attempt: Suppose an analytic function f has infinitely many zeros on some closed disk D. Then there exists a sequence of zeros in D with a limit point in D. Thus by the identity theorem (Let D be a domain and f analytic in D. If the set of zeros Z(f) has a limit point in D, then f ≡ 0 in D.), f is identically zero and thus constant.

My main reasons for confusion (other than having a weak understanding of the identity theorem):

-Couldn't such a function f have a finite number of distinct zeros, each with infinite multiplicity? in this case there wouldn't be a convergent sequence of zeros...

-What is the relevance of the fact that D is closed?

Any help in understanding this problem would be greatly appreciated! Thanks

Your proof is correct, you just need to realize that when you say "has infinitely many zeros" you mean "has infinitely many points where it evaluates to $0$", so one is not talking about multiplicities here. The importance of $D$ being a closed disk is that it is then compact, and that implies the existence of a convergent sequences of zeros of the functions, which allows you to invoke the identity theorem (which you certainly do want to understand fully).

• What does multiplicity have to do it? What can go wrong? If there are infinitely many zeros "counting multiplicity" then there are infinitely many distinct zeros, or else there is a zero of infinite multiplicity, right? And in either case, everything is fine? What am I missing? – bof Apr 13 '15 at 11:11
• Multiplicity is irrelevant since the distinctness of the zeros is implied. The other interpretation of "has infinitely many zeros" that I was considering was "has an infinite number of zeros counting multiplicity", in which case there COULD be say exactly 2 zeros in the disk both with infinite multiplicity, in which case the rest of the proof falls apart. – Will Apr 13 '15 at 13:16
• but as quid pointed out, this is an impossible case anyways – Will Apr 13 '15 at 13:17

The relevance of the fact that the set is closed is that the limit point must be in the set. For example $\sin (1/(z+1))$ has infinitely many zeros in the open unit disc, but is not zero. (The sole point of accumulation is $-1$, outside the domain, and the function is not analytic there.)

On the multiplicity: first I think distinct zeros are meant. Second, a non-zero analytic function (on a connected domain) cannot vanish to infinite order anywhere; this would give the series there is $0$ and so the function is zero.

Since the disk $\mathcal{D}$ is bounded and closed it is compact, so, by Heine-Borel it fulfils the Bolzano-Weierstrass property: every infinite set has at least one limit point. So the set of $\mathcal{Z}$ of zeros has a limit point, and you're right to go with the identity theorem. So you can see the reason for closure: it lets you call on the Heine-Borel theorem.

Incidentally, before about 1930 and Pavel Alexandrof's school, "compact" meant by definition having the Bolzano-Weierstrass property.