complete metric spaces have banach continuous bounded function spaces

If $X$ is a complete metric space, then the continuous bounded function space $C_{\mathrm{\infty }}(X)$ is a Banach space.

NTS every Cauchy sequence $\{u_n\}\in C_{\mathrm{\infty }}(X)$ has a limit in $C_{\mathrm{\infty }}(X)$.

Let $\{u_n\}$ be a Cauchy sequence in $C_{\mathrm{\infty }}(X)$. Then for all $ϵ>0$ there exists $N\in \mathbb{N}$ such that for all $n,m\ge N$, $||u_n-u_m|{|}_{\mathrm{\infty }}<ϵ$.

In particular, $\mathrm{\exists }{N}_0\in \mathbb{N}$ such that for all $m,n\ge N_0,||u_n-u_m|{|}_{\mathrm{\infty }}<1$. Then for all $n\ge N_0$, we have

$$\begin{array}{rl}||u_n|{|}_{\mathrm{\infty }}& \le ||u_n-u_{N_0}|{|}_{\mathrm{\infty }}+||u_{N_0}|{|}_{\mathrm{\infty }}\\ & <1+||u_{N_0}||\end{array}$$

And therefore for all $n\in \mathbb{N}$, we have

$$\begin{array}{rl}||u_n|{|}_{\mathrm{\infty }}& \le ||u_1|{|}_{\mathrm{\infty }}+||u_2|{|}_{\mathrm{\infty }}+\cdots +||u_{N_0}|{|}_{\mathrm{\infty }}+1\\ & =B\end{array}$$

Since for all $x\in X$ we have $|u_n(x)-u_m(x)|\le ||u_n-u_m|{|}_{\mathrm{\infty }}$, we have that
for each $x\in X$, the sequence$\{u_n(x)\}$ is a Cauchy sequence. And since $\mathbb{C}$ is complete, its limit is in $\mathbb{C}$.

(define a candidate limiting function)
Define $u:X\to \mathbb{C},u(x)=\underset{n\to \mathrm{\infty }}{lim}{u}_n(x)$
Then for all $x\in X$,

$$|u_n(x)|=\underset{n\to \mathrm{\infty }}{lim}|u_n(x)|\le B⟹\underset{x\in X}{sup}|u(x)|\le B$$

thus $u$ is a bounded function.

Finally, we need to show continuity and convergence.

$$||u-u_n|{|}_{\mathrm{\infty }}\to 0$$

Let $ϵ>0$. Since $\{u_n\}$ is Cauchy in $C_{\mathrm{\infty }}(X)$, there exists some $N_1\in \mathbb{N}$ such that $\mathrm{\forall }n,m\ge N$ we have $||u_n-u_m|{|}_{\mathrm{\infty }}<\frac{ϵ}{2}$.

So for any fixed $x\in X$, we have

$$|u_n(x)-u_m(x)|\le ||u_n-u_m|{|}_{\mathrm{\infty }}<\frac{ϵ}{2}$$

Thus as $m\to \mathrm{\infty }$, we have for all $n\ge N_1$ and each $x\in X$,

$$|u_n(x)-u(x)|\le \frac{ϵ}{2}<ϵ$$

Thus we have $||u_n-u|{|}_{\mathrm{\infty }}\to 0$ and this implies that $u_n\to u$ uniformly on $X$. And because each $u_n$ is continuous, this implies that the limit $u$ is also continuous.

Thus, $u\in C_{\mathrm{\infty }}(X)$ and $u_n\to u\in C_{\mathrm{\infty }}(X)$, thus $C_{\mathrm{\infty }}(X)$ is complete and therefore a Banach space.

the approach for this proof is basically the same as any proof to show that something is a Banach space.

• choose a candidate for the limit
• show that the limit is in the space