31.2 Zeroes of Holomorphic Functions

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, $U = \bracs{f = 0}$.$\square$

Proof. Since $f \ne 0$, $n = \min\bracs{n \in \natp| D^nf(z_0) \ne 0}< \infty$ by Proposition 31.2.2. 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 $z_{0}$.$\square$

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 14.4.1, $\bracsn{[f]_E = M}$ 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.$\square$

Proof. Let $\eps > 0$ and $\phi_{\eps}(z) = e^{\eps(z^2 - 1)}$, then $\phi f \in H(S; E) \cap BC(\ol S; E)$.

For each $R > 0$, let $S_{R} = \bracs{z \in \complex| \text{Re}(z) \in (0, 1), |\text{Im}(z)| < R}$, then by the Maximum Modulus Theorem,

However, since $\phi_{\eps} f(z) \to 0$ as $|\text{Im}(z)| \to \infty$,

Therefore

$\square$

Proof, [Theorem VI.3.7, Con78]. Assume without loss of generality that $M(0), M(1) > 0$. Let

then $g$ is a non-vanishing entire function, and for each $z \in \complex$,

so $|g|^{-1}$ is bounded on $\ol S$ by Proposition 5.16.3. Let

then $h \in H(S; E) \cap BC(\ol S; E)$ with $\norm{h(z)}_{E} \le 1$ for all $z \in \partial S$. By the Maximum Modulus Theorem, $\norm{h(z)}_{E} \le 1$ for all $z \in S$. Thus for every $z \in S$,

Therefore $M(s) \le M(0)^{s} M(1)^{1-s}$ for every $s \in [0, 1]$.$\square$