31.6 Entire Functions
Definition 31.6.1 (Entire Function).label Let $E$ be a complete separated locally convex space over $\complex$, then an entire $E$-valued function is an $E$-valued holomorphic function on $\complex$.
Proposition 31.6.2.label Let $E$ be a complete separated locally convex space over $\complex$, and $f \in H(\complex; E)$, then for any $z_{0} \in \complex$ and $z \in \complex$,
where the radius of convergence of the series is infinite.
Proof. See (4) of Definition 31.1.6.$\square$
Theorem 31.6.3 (Liouville).label Let $E$ be a complete separated locally convex space over $\complex$ and $f \in H(\complex; E)$. If $f$ is bounded, then $f$ is constant.
Proof. Let $x_{0} \in \complex$ and $[\cdot]_{E}: [E, \infty)$ be a continuous seminorm on $E$, then by Cauchy’s Estimate,
for all $r > 0$. Therefore $Df = 0$, and $f$ is constant.$\square$
Theorem 31.6.4 (Fundamental Theorem of Algebra).label Let $p \in \complex[z]$ be a non-constant polynomial, then there exists $z \in \complex$ such that $f(z) = 0$.
Proof. Let $p \in \complex[z]$ such that $p(z) \ne 0$ for all $z \in \complex$. Let $f = 1/p$, then $f \in H(\complex; \complex) \cap C_{0}(\complex; \complex)$, so $f$ is bounded. By Liouville’s Theorem, $f$ and thus $p$ is constant.$\square$
Proposition 31.6.5.label Let $f \in H(\complex; \complex)$, then the following are equivalent:
- (1)
$f(z) \ne 0$ for all $z \in \complex$.
- (2)
There exists $\ell \in H(\complex; \complex)$ such that $f = e^{\ell}$.
Proof. (1) $\Rightarrow$ (2): For each $z \in \complex$, let
then $g \in H(\complex; \complex)$ with $g' = f/f'$, and
Therefore $fe^{-g}$ is a non-zero constant. For any $\lambda \in \complex$ with $e^{\lambda} = fe^{-g}$, let $h = g + \lambda$, then $f = e^{h}$.$\square$
Proposition 31.6.6.label Let $f \in H(\complex; \complex)$ such that:
- (a)
$f(0) = 1$.
- (b)
$f'(0) = 0$.
- (c)
$0 < |f| < |\exp|$.
then $f = 1$.
Proof, [Lemma 4.4, Zhu93]. By (c) and Proposition 31.6.5, there exists $g \in H(\complex; \complex)$ such that $f = e^{g}$. Since $f(0) = 1$, $g(0) = 0$. As $f'(0) = g'(0)f(0) = 0$, $g'(0) = 0$ as well.
From (c), $|g(z)| \le e^{|z|}$ for each $z \in \complex$. Thus for every $r > 0$ and $z \in B_{\complex}(0, r)$, $|g(z)| \le |g(z) - 2r|$, and
extends to a holomorphic function on $B_{\complex}(0, r)$ by Proposition 31.2.3. Since $|g_{r}(z)| \le 1$ for all $z \in \partial B_{\complex}(0, r)$, $g_{r}(z) \le 1$ for all $z \in B_{\complex}(0, r)$ by the Maximum Modulus Theorem.
Finally, since $|g_{r}(z)| \le 1$ for all $z \in \complex$ and $r > |z|$, $g = 0$ and $f = 1$.$\square$
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