# Proof of Uniqueness Theorem

I have the following question on the theorem below, which I've been working on while I study for quals.

Suppose $\Omega$ is an open and connected set, with $f$ analytic in $\Omega$, and $a \in \Omega$. Then:

(1) If $f(a)=f'(a)=f''(a)=\ldots=0$ then $f$ is identically zero in $\Omega$.

(2) If $z_0$ is an accumulation point of $\lbrace z|f(z)=0\rbrace$, then $f$ is identically zero.

I've had no trouble proving (1), but (2) is giving me some issues. I've tried re-writing $f(z)$ as $g(z)(z-a)^m$, where $m$ is the smallest number such that $f^{(m)}(a)\neq 0$, but when I attempt to work out the details I usually get flipped around.

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Try proving the contrapositive (if $f$ is not identically zero then $z_0$ cannot be an accumulation point of its zeroes). – Qiaochu Yuan Aug 17 '12 at 17:07

(2) is often called or used to prove analytic continuation. I will give a proof below:

Suppose $\alpha$ is an accumulation point of the set $\{z : f(z) = 0\}$. Since $f$ is a holomorphic on $\Omega$, let

$f(z) = \sum a_n(z - \alpha)^n$

by the power series expansion at $\alpha$. The claim is that $f$ is zero in a neighborhood of $\alpha$. Suppose not, then there must exists a smallest $m$ such that $a_m \neq 0$. Hence, the power series takes the form

$f(z) = \sum a_m (z - \alpha)^m (1 + h(z - \alpha)) \ \ \ (*)$

for some holomorphic function $h$ such that $h(z - \alpha) = 0$ as $z \rightarrow \alpha$. ($h$ is clearly what you get when you factor out $a_m(z - \alpha)^m$.)

Now since $\alpha$ is a limit point of the set of zeros. There exists a sequence of zeros of $f$, $(y_n)$ such that $y_n \rightarrow \alpha$. But then by $(*)$, it is clear that $f(y_n) \neq 0$ for any $n$. But you choose $y_n$ to be a sequence of zeros of the function $f$. Contradiction.

Now let $U = \text{Int}(\{z : f(z) = 0\})$. $U$ is nonempty open set by the argument above. $U$ is also closed since if $(y_n)$ is a sequence in $U$ with limit point $y$, then $f(y) = \lim_{n \rightarrow \infty} f(y_n) = 0$ by continuity. By the argument above, applied to the point $y$ in place of $\alpha$, $f$ vanishes in a neighborhood of $y$. Hence $y \in U = \text{Int}(\{z : f(z) = 0\})$. So $U$ is closed. By assumption $\Omega$ is connected, so the only clopen subset is $\emptyset$ or $\Omega$. Since $\alpha \in U$, one has that $U = \Omega$. So $f$ vanishes on all of $\Omega$.

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Presumably you mean "holomorphic function $h$" not $holomorphic function$\alpha$." – Thomas Andrews Aug 17 '12 at 18:47 For the set$U$, is it necessary that$U$be defined as the interior? It seems as if we could get away with just taking the union of the accumulation points. – Frank White Aug 19 '12 at 23:46 Defining it as the interior of the set gives us$U\$ open without needing the argument above it. – Frank White Aug 20 '12 at 0:03