Space of bounded continuous functions complete 
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!
 A: 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.
A: 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:

*

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


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


*$\|T_n-T\|\to 0 $ as $n\to \infty$
Verify all the above claims.
