complete metric spaces have banach continuous bounded function spaces

[[concept]]

Theorem

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

Proof

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

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

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

\lvert \lvert u_{n} \rvert \rvert _{\infty}&\leq \lvert \lvert u_{n} - u_{N_{0}} \rvert \rvert _{\infty}+ \lvert \lvert u_{N_{0}} \rvert \rvert _{\infty} \\ &< 1 + \lvert \lvert u_{N_{0}} \rvert \rvert \end{align}$$ And therefore for all $n \in \mathbb{N}$, we have $$\begin{align} \lvert \lvert u_{n} \rvert \rvert _{\infty}&\leq \lvert \lvert u_{1} \rvert \rvert_{\infty} + \lvert \lvert u_{2} \rvert \rvert _{\infty}+ \dots + \lvert \lvert u_{N_{0}} \rvert \rvert_{\infty} + 1 \\ &= B \end{align}$$ Since for all $x \in X$ we have $\lvert u_{n}(x)-u_{m}(x) \rvert \leq \lvert \lvert u_{n} - u_{m} \rvert \rvert_{\infty}$, we have that for each $x \in X$, the sequence$\{ u_{n}(x) \}$ is a Cauchy sequence. And since $\mathbb{C}$ is complete, its limit is in $\mathbb{C}$. (*define a candidate limiting function*) Define $u: X \to \mathbb{C}, \quad u(x) = \lim_{ n \to \infty } u_{n}(x)$ Then for all $x \in X$, $$\lvert u_{n}(x) \rvert = \lim_{ n \to \infty } \lvert u_{n}(x) \rvert \leq B \implies \sup_{x \in X}\lvert u(x) \rvert \leq B$$ thus $u$ is a bounded function. Finally, we need to show continuity and convergence. $$\lvert \lvert u -u_{n} \rvert \rvert_{\infty} \to 0$$ Let $\epsilon >0$. Since $\{ u_{n} \}$ is Cauchy in $C_{\infty}(X)$, there exists some $N_{1} \in \mathbb{N}$ such that $\forall n,m \geq N$ we have $\lvert \lvert u_{n} - u_{m} \rvert \rvert_{\infty} < \frac{\epsilon}{2}$. So for any fixed $x \in X$, we have $$\lvert u_{n}(x) -u_{m}(x) \rvert \leq \lvert \lvert u_{n} - u_{m} \rvert \rvert_{\infty} < \frac{\epsilon}{2}$$ Thus as $m \to \infty$, we have for all $n \geq N_{1}$ and each $x \in X$, $$\lvert u_{n}(x) - u(x) \rvert \leq \frac{\epsilon}{2} < \epsilon$$ Thus we have $\lvert \lvert u_{n} - u \rvert \rvert_{\infty} \to 0$ 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, $u \in C_{\infty}(X)$ and $u_{n} \to u \in C_{\infty}(X)$, thus $C_{\infty}(X)$ is complete and therefore a [[Concept Wiki/Banach space]].

NOTE

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

Mentions

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