Proposition 27.2.2.label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, and $f \in H(U; E)$, then the following are equivalent:
- (1)
$f = 0$.
- (2)
There exists $z_{0} \in U$ such that $D^{n}f(z_{0}) = 0$ for all $n \in \natp$.
- (3)
There exists $z_{0} \in U$ such that $z_{0} \in \ol{\bracs{f = 0} \setminus \bracs{z_0}}$.
Proof. (3) $\Rightarrow$ (2): Let $r > 0$ and $\bracs{a_n}_{0}^{\infty} \subset E$ such that $f(z) = \sum_{n = 0}^{\infty} a_{n} (z - z_{0})^{n}$ for all $z \in B(z_{0}, r)$, where the series has a radius of convergence of at least $r$. By continuity of $f$, $a_{0} = f(0) = 0$.
Suppose inductively that $a_{k} = 0$ for each $0 \le k \le n$. Let $[\cdot]_{E}: E \to [0, \infty)$ be a continuous seminorm, then for each $z \in B(z_{0}, r) \cap \bracs{f = 0}\setminus \bracs{z_0}$,
Since the series $\sum_{n = 0}^{\infty} a_{n} (z - z_{0})^{n}$ has a radius of convergence of at least $r$, $\limsup_{n \to \infty}[a_{n}]_{E}^{1/n}\le 1/r$, so there exists $N \ge n+2$ such that $[a_{k}]_{E}^{1/k}\le 2/r$ for all $k \ge N$. For each $s \in (0, r/4)$, $B(z_{0}, s) \cap \bracs{f = 0}\setminus \bracs{z_0}\ne \emptyset$, thus
As the above holds for all $s \in (0, r/4)$, $[a_{n+1}]_{E} = 0$. Therefore $a_{n} = 0$ for all $n \in \natp$.
(2) $\Rightarrow$ (1): By continuity, $\bracs{f = 0}$ is closed. By local power series expansion and (3), $\bracs{f = 0}$ is open. As $U$ is connected, $U = \bracs{f = 0}$.$\square$
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