# Space of bounded continuous functions complete [closed]

Prove that the space of bounded continuous functions $$C^0(F,X):=\{f: F\rightarrow X : f$$ is continuous and bounded$$\}$$ with $$X$$ Banach space, $$F\subset \mathbb{R}$$ is complete with the sup norm.

I have seen many proves of this for $$X=\mathbb{R}$$ but how do I do this for an arbitrary Banach space? My idea was to show that a Cauchy sequence $$(f_n)_n\in C^0(F,X)$$ is also a Cauchy sequence in $$X$$ and since $$X$$ is complete it follows $$u_k \rightarrow u \in X$$. Now I need to show that $$u$$ is continuous and bounded.

Problem: for $$X=\mathbb{R}$$ I can use uniformly convergence and the uniform limit theorem but I don't know if these theorems apply for any arbitrary Banach space?

How do I prove this statement? Thanks you!

• The same proof works for any Banach space. Commented Apr 20, 2022 at 11:24

## 2 Answers

Carefully analyze the proof in the $$X = \mathbf R$$ case. If you look closely, you will notice that you can use exactly the same proof in the general case. You have - as you noticed - prove the tools use also for the general case:

Lemma (The uniform limit of continuous functions is continuous.) Let $$F, G$$ be metric spaces, $$f_n \colon F \to G$$ continuous functions, $$f \colon F \to G$$. Suppose $$f_n \to f$$ uniformly, that is $$\sup_{x \in F} d\bigl(f_n(x), f(x)\bigr) \to 0$$ Then $$f$$ is continuous.

Proof. As in the case for real functions: Let $$x_0 \in F$$, $$\epsilon > 0$$, choose $$N \in \mathbf N$$ such that $$d(f_n(x), f(x)) < \frac\epsilon 3$$, all $$x \in F$$, all $$n \ge N$$. As $$f_N$$ is continuous at $$x_0$$, there is $$\delta > 0$$ such that $$d(x, x_0) < \delta \implies d(f_N(x), f_N(x_0)) < \frac \epsilon 3$$ Now let $$d(x,x_0)< \delta$$, then $$d(f(x), f(x_0)) \le d(f(x), f_N(x)) + d(f_N(x), f_N(x_0)) + d(f_N(x_0), f(x_0)) < 3 \cdot \frac \epsilon 3 = \epsilon.$$ Hence, $$f$$ is continuous at $$x_0$$, as $$x_0$$ was arbitrary, $$f$$ is continuous.

Proposition. If $$F$$ is any metric space, $$X$$ is a Banach space, then $$C^0(F,X)$$ is complete in the norm $$\|f\|_0 := \sup_{x \in F} \|f(x)\|_X.$$

Proof. Let $$(f_n)$$ be a Cauchy sequence in $$C^0(F,X)$$. Then, by the definition of $$\|\cdot \|_0$$, for each $$x \in F$$: $$\|f_n(x) - f_m(x)\|_X \le \|f_n - f_m\|_0 \to 0$$ hence $$(f_n(x))$$ is Cauchy in $$X$$. As $$X$$ is complete, this sequence is convergent, and we may define $$f \colon F \to X$$ by $$f(x) := \lim_{n \to \infty} f_n(x)$$ We now will show, that $$\|f_n - f\|_0 \to 0$$: Let $$\epsilon > 0$$, choose $$N \in \mathbf N$$ such that $$\|f_n - f_m\|_0 < \epsilon$$ for $$n,m \ge N$$. Now, for arbitrary $$x \in F$$: $$\|f_n(x) - f_m(x)\|_X \le \|f_n - f_m\|_0 < \epsilon, \quad n,m \ge N$$ Now let $$m \to \infty$$, note that $$f_m(x) \to f(x)$$, hence, we have $$\|f_n(x) - f(x)\|_X = \lim_m \|f_n(x) - f_m(x)\|_X \le \epsilon, \quad n\ge N.$$ But this means that $$\|f_n - f\|_0 \to 0$$, and, hence, by the lemma $$f \in C^0(F,X)$$. Therefore $$C^0(F,X)$$ is complete.

• Thank you! Why is it enough to show $||f_n-f||_0\rightarrow 0$? Don't I need to show first that $f$ is bounded?
– Uhmm
Commented Apr 20, 2022 at 11:55
• No, that follows (for example for some $n$: $\|f\|_0 \le \|f_n\| + 1\le\infty$. Commented Apr 20, 2022 at 13:38
• why does $|f_n(x) - f(x)| = \lim_m |f_n(x) - f_m(x)| \le \epsilon, \quad n\ge N$ mean that $\|f_n - f\|_0 \to 0$? And I guess $| . |$ is the norm on $X$?
– Uhmm
Commented Apr 20, 2022 at 17:55
• @Uhmm Fixed the norm $\|\cdot\|_X$ of $X$, and: Because this gives $$\|f_n - f\|_0 = \sup_x \|f_n(x) - f(x)\|_X \le \epsilon, \quad n \ge N$$ and $\epsilon$ was arbitrary. Commented Apr 20, 2022 at 23:28

There is a more general result.

$$X$$ and $$Y$$ be two normed space.

$${\scr{B}}{(X, Y) }=\{T:X\to Y : T \text{ is bounded } \}$$

Claim : $$({\scr{B}}{(X, Y) },\|•\|_{op})$$ is a Banach space whenever $$Y$$ is a Banach space.

$$\|T\|_{op}=\sup\{\|Tx\|:{\|x\|_X\le 1}\}$$

Proof: (For simplicity we will write $$\|•\|$$ and which which should be understood according as the norm on the underlying space.)

Let, $$(T_n)$$ be a cauchy sequence in $${\scr{B}}{(X, Y) }$$

•Then for fixed $$x\in X$$ , $$(T_nx)$$ cauchy in $$Y$$.

\begin{align}\|T_nx -T_m x\|&\le \|T_n-T_m\|\|x\|\\&\to 0 \space \forall m, n\to \infty\end{align}

$$Y$$ is complete.

$$Tx:=\lim_{n\to\infty} T_nx$$

Claim:

1. $$T\in {\scr{L}}{(X, Y) }$$

2. $$T\in { \scr{B}}{(X, Y) }$$

3. $$\|T_n-T\|\to 0$$ as $$n\to \infty$$

Verify all the above claims.