27.4 Entire Functions

Definition 27.4.1 (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 27.4.2.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 $z \in \complex$,

\[f(z) = \sum_{k = 0}^{\infty} \frac{1}{k!}D^{k}f(z_{0})(z - z_{0})^{k}\]

where the radius of convergence of the series is infinite.

Proof. See (4) of Definition 27.1.6.$\square$

Theorem 27.4.3 (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,

\[[Df(z_{0})]_{E} \le \frac{1}{r}\sup_{z \in \ol{B(z_0, r)}}[f(z)]_{E}\]

for all $r > 0$. Therefore $Df = 0$, and $f$ is constant.$\square$

Theorem 27.4.4 (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.$\square$

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