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Let $U\subseteq \mathbb C$ be open and $\mathscr A(U)$ consist of all analytic functions on $U$.

I can easily prove that there exists a sequence $K_n$ of compact sets in $U$ so that $U=\cup_{i=1}^nK_n$, $K_n\subseteq K_{n+1}^{\circ}$ and if $K\subseteq U$ is compact then $K\subseteq K_n$ for large $n$.

Let for $f,g\in \mathscr A(U)$,

$$d(f,g)=\sum_{n=1}^{\infty}2^{-n}\frac{\left\|f-g\right\|_{K_n}}{1+\left\|f-g\right\|_{K_n}}$$

Again I can easily prove that $d$ is a metric on $\mathscr A(U)$. I can also prove that compact convergence (i.e. uniform convergence in compact subsets) implies convergence in $d$: If $f_n\to f$ compactly then $f\in\mathscr A(U)$ and let $\epsilon>0$. For large $N$, $\sum_{n=N+1}^{\infty}2^{-n}<\epsilon$ and $n\ge N\implies \left\|f-f_n\right\|_{K_N}<\epsilon$. Therefore, $$d(f,f_n)=\sum_{n=1}^{N}2^{-n}\frac{\left\|f-g\right\|_{K_n}}{1+\left\|f-g\right\|_{K_n}}+\sum_{n=N+1}^{\infty}2^{-n}\frac{\left\|f-g\right\|_{K_n}}{1+\left\|f-g\right\|_{K_n}}\le \\ \frac{\left\|f-g\right\|_{K_N}}{1+\left\|f-g\right\|_{K_N}}\sum_{n=1}^{N}2^{-n}+\sum_{n=N+1}^{\infty}2^{-n}\le \left\|f-g\right\|_{K_N}+\sum_{n=N+1}^{\infty}2^{-n}<2\epsilon$$

My question is how do we prove the converse? If $f_n\to f$ in $d$ let $\epsilon>0$ and $K\subseteq U$ be compact. For large $N$, $K\subseteq K_N$, $n\ge N\implies d(f,f_n)<\epsilon$ and $$\left\|f-g\right\|_K\le \left\|f-g\right\|_{K_N}$$ It remains to show $\left\|f-g\right\|_{K_N}\le d(f,f_n)$ or something in the same line. The inequality $$d_U(f_n,f)\ge \frac{\left\|f-g\right\|_{K_N}}{1+\left\|f-g\right\|_{K_N}}\sum_{n=N}^{\infty}2^{-n}$$ does not really seem to help. Any ideas?

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1 Answers1

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Suppose $f_n \mathop\to\limits^d f$, given $K\subseteq U$ compact and $\epsilon > 0$. Choose $M$ such that $K \subseteq K_M$ and $N$ such that $d(f_n, f) < 2^{-M}\frac{\epsilon}{1+\epsilon}$ for $n \ge N$. Then we have for $n \ge N$ as $x \mapsto x/(1+x)$ is monotone on $[0,\infty)$: $$ \def\norm#1{\left\|#1\right\|} \frac{\norm{f_n-f}_K}{1 + \norm{f_n-f}_K} \le \frac{\norm{f_n-f}_{K_M}}{1 + \norm{f_n-f}_{K_M}} \le 2^M d(f_n, f) < 2^{M-M}\frac{\epsilon}{1+\epsilon} =\frac{\epsilon}{1+\epsilon} $$ As $x \mapsto x/(1+x)$ is strictly monotone, this implies $\norm{f_n-f}_K < \epsilon$ for $n \ge N$.

martini
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