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Let topological space $X$ and $f:X \to C$. Show that the space

$$X_{f}=\{g \in C(X) \mid \sup|g-f|<\infty\}$$

is a complete metric space with distance function

$$d(g_1,g_2)=\sup_{x \in X}|g_1(x)-g_2(x)|.$$

Can anyone help me to show this?

Paris Lamp
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  • What have you tried so far? You just need to verify that $d$ is a metric and that the space is complete. The first part is pretty much trivial and the second part follows directly from the fact that $C(X)$ is complete, if you equip it with the uniform norm. – Dominik Sep 19 '15 at 22:33
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    @Dominik Yes in spirit. In fact there is no uniform norm on $C(X)$ because functions in $C(X)$ need not be bounded. But the proof of completeness for, say, $C(K)$ where $K$ is compact, or the completeness of the space of bounded continuous functions in the uniform norm on any topological space, is all that's needed here. – David C. Ullrich Sep 20 '15 at 00:04

1 Answers1

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Let $\left(f_n\right)_{n \in \mathbb{N}}$ a Cauchy sequence in $X_{f}$ with metrix $d$; let us remember that $\lim_{n\rightarrow\infty}f_{n}\notin X_{f}$ with metrix $d$ means that

$\sup_{x\in X}\left|\lim_{n\rightarrow\infty}f_{n}-f\right|=\sup_{x\in X}\left|f_{n}-f\right|=\sup_{x\in X}\lim_{n\rightarrow\infty}\left|f_{n}-f\right|>\infty$

by continuity of the norm.

Suppose that $\lim_{n\rightarrow\infty}f_{n}\notin X_{f}$ with metrix $d$, this means that given $k>0$ there is $x_{k}\in X$ such that

$\lim_{n\rightarrow\infty}\left|f_{n}\left(x_{k}\right)-f\left(x_{k}\right)\right|>k$

This is, there is $\xi>0$ such that for all $N \in \mathbb{N}$, $n>N$ and $\left|f_{n}\left(x_{k}\right)-f\left(x_{k}\right)\right|-k>\xi$.

Therefore, let $m$ fixed but arbitrary, so we have

$\xi+k-d\left(f_{m},f\right)<\xi+k-\left|f_{m}\left(x_{k}\right)-f\left(x_{k}\right)\right|<\left|f_{n}\left(x_{k}\right)-f_{m}\left(x_{k}\right)\right|$

and since

$\left\{ \left|f_{n}\left(x_{k}\right)-f_{m}\left(x_{k}\right)\right|\::\: k\in\mathbb{N}\right\} \subseteq\left\{ \left|f_{n}\left(x\right)-f_{m}\left(x\right)\right|\::\: x\in X\right\} $

then

$\xi+k-d\left(f_{m},f\right)\leq\sup\left\{ \left|f_{n}\left(x_{k}\right)-f_{m}\left(x_{k}\right)\right|\::\: k\in\mathbb{N}\right\} \leq d\left(f_{n},f_{m}\right)$

Therefore, $d\left(f_{n},f_{m}\right)\rightarrow\infty$ which contradicts that $\left(f_n\right)_{n \in \mathbb{N}}$ a Cauchy sequence in $X_{f}$ with metrix $d$.

Diego Fonseca
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