Theorem 27.2.3 (Montel).label Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, and $\cf \subset H(U; E)$, then the following are equivalent:
- (B1)
$\cf$ is equicontinuous, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is bounded.
- (B2)
$\cf$ is bounded in $H(U; E)$.
and the following are equivalent:
- (C1)
$\cf$ is precompact in $H(U; E)$.
- (C2)
$\cf$ is bounded in $H(U; E)$, and for each $x \in U$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact.
Proof. (B1) $\Rightarrow$ (B2): Let $K \subset U$ be compact and $V \in \cn_{E}(0)$ be circled. Since $\cf$ is equicontinuous, for each $x \in K$, there exists $U_{x} \in \cn_{U}(x)$ such that $f(y) - f(x) \in V$ for all $y \in U_{x}$ and $f \in \cf$. By compactness of $K$, there exists $\seqf{x_j}\subset K$ such that $K \subset \bigcup_{j = 1}^{n} U_{x_j}$. Since $B = \bigcup_{j = 1}^{n} \cf(x_{j})$ is bounded, there exists $\lambda > 0$ such that $\lambda V \supset B$. In which case,
(B2) $\Rightarrow$ (B1): By Cauchy’s Estimate, $\bracsn{Df|f \in \cf}$ is also uniformly bounded on every compact set. Thus $\cf$ is equicontinuous.
(C1) $\Leftrightarrow$ (C2): By the Arzelà-Ascoli Theorem.$\square$
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