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 $N \in N$ such that for all $n, m \geq N$ , $| | u_{n} - u_{m} | |_{\infty} < ϵ$ .

In particular, $\exists N_{0} \in N$ such that for all $m, n \geq N_{0}, | | u_{n} - u_{m} | |_{\infty} < 1$ . Then for all $n \geq N_{0}$ , we have

\begin{aligned} | | u_{n} | |_{\infty} & \leq | | u_{n} - u_{N_{0}} | |_{\infty} + | | u_{N_{0}} | |_{\infty} \\ < 1 + | | u_{N_{0}} | | \end{aligned}

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

\begin{aligned} | | u_{n} | |_{\infty} & \leq | | u_{1} | |_{\infty} + | | u_{2} | |_{\infty} + \dots + | | u_{N_{0}} | |_{\infty} + 1 \\ = B \end{aligned}

Since for all $x \in X$ we have $| u_{n} (x) - u_{m} (x) | \leq | | u_{n} - u_{m} | |_{\infty}$ , we have that
for each $x \in X$ , the sequence ${u_{n} (x)}$ is a Cauchy sequence. And since $C$ is complete, its limit is in $C$ .

(define a candidate limiting function)
Define $u : X \to C, u (x) = lim_{n \to \infty} u_{n} (x)$
Then for all $x \in X$ ,

| u_{n} (x) | = lim_{n \to \infty} | u_{n} (x) | \leq B ⟹ sup_{x \in X} | u (x) | \leq B

thus $u$ is a bounded function.

Finally, we need to show continuity and convergence.

| | u - u_{n} | |_{\infty} \to 0

Let $ϵ > 0$ . Since ${u_{n}}$ is Cauchy in $C_{\infty} (X)$ , there exists some $N_{1} \in N$ such that $\forall n, m \geq N$ we have $| | u_{n} - u_{m} | |_{\infty} < \frac{ϵ}{2}$ .

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

| u_{n} (x) - u_{m} (x) | \leq | | u_{n} - u_{m} | |_{\infty} < \frac{ϵ}{2}

Thus as $m \to \infty$ , we have for all $n \geq N_{1}$ and each $x \in X$ ,

| u_{n} (x) - u (x) | \leq \frac{ϵ}{2} < ϵ

Thus we have $| | u_{n} - u | |_{\infty} \to 0$ 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 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

File	Last Modified
bounded linear operator space is banach	2025-11-09
Functional Analysis Lecture 1	2025-11-09
Functional Analysis Lecture 2	2025-11-09

const { dateTime } = await cJS()

return function View() {
	const file = dc.useCurrentFile();
	return <p class="dv-modified">Created {dateTime.getCreated(file)}     ֍     Last Modified {dateTime.getLastMod(file)}</p>
}