27.2 Zeroes of Holomorphic Functions
Definition 27.2.1 (Zero).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in H(U; E)$, and $z_{0} \in U$, then $z_{0}$ is a zero of $f$ of multiplicity $n \in \natp$ if there exists $g \in H(U; E)$ such that $f(z) = (z - a)^{n} g(z)$ for all $z \in \bracs{g \ne 0}$.
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, $U = \bracs{f = 0}$.$\square$
Proposition 27.2.3.label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be connected, and $f \in H(U; E) \setminus \bracs{0}$, then for each $z_{0} \in \bracs{f = 0}$, there exists $n \ge 1$ and $g \in H(U; \complex)$ such that:
- (1)
$g(a) \ne 0$.
- (2)
For each $z \in U$, $f(z) = (z - z_{0})^{n}g(z)$.
Proof. Since $f \ne 0$, $n = \min\bracs{n \in \natp| D^nf(z_0) \ne 0}< \infty$ by Proposition 27.2.2. Let
then $g$ is continuous on $U$ and holomorphic on $U \setminus \bracs{z_0}$. As $f \in H(U; E)$, there exists $r > 0$ and $\bracs{a_k}_{0}^{\infty} \subset E$ such that $f(z) = \sum_{k = 0}^{\infty} a_{k} (z - z_{0})^{k}$ for all $z \in B(z_{0}, r)$, where the series has a radius of convergence of at least $r$. By assumption on $n$, for each $z \in B(z_{0}, r)$,
where the series $\sum_{k = 0}^{\infty} a_{k+n}(z - z_{0})^{k}$ has the same radius of convergence. Therefore $g$ is holomorphic at $z_{0}$.$\square$
Theorem 27.2.4 (Maximum Modulus Theorem).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, then
- (1)
For each $f \in H(U; E)$ and continuous seminorm $[\cdot]_{E}: E \to [0, \infty)$, if there exists $z_{0} \in U$ with
\[[f(z_{0})]_{E} = \sup_{z \in U}[f(z)]_{E}\]then $[f]_{E}$ is constant.
- (2)
For any $f \in H(U; \complex)$, if there exists $z_{0} \in U$ with
\[|f(z_{0})| = \sup_{z \in U}|f(z)|\]then $f$ is constant.
Proof. (1): Let $M = \sup_{z \in U}[f(z)]_{E}$. Since $[f]_{E}$ is continuous, $\bracsn{[f]_E = M}$ is closed. By Cauchy’s Integral Formula and Proposition 13.4.1, $\bracsn{[f]_E = M}$ is open. As $U$ is connected, $\bracsn{[f]_E = M}= U$.
(2): By (1), $|f|$ is constant. After rescaling, assume without loss of generality that $\ol f = 1/f$. In this case, for each $z \in U$,
and
so $Df = 0$ and $f$ is constant.$\square$
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