Continuous Functions on Compact Space form Banach Space

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a compact Hausdorff space.

Let $Y$ be a Banach space.

Let $\CC = \CC \struct {X; Y}$ be the space of continuous functions on $X$ valued in $Y$.

Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on $\CC$.

Then $\struct {\CC, \norm {\,\cdot\,} }$ is a Banach space.

Proof
We first show that $\CC$ is a vector space.

We have that the set of continuous mappings $X \to Y$ is a subset of the set $Y^X$ of all mappings $X \to Y$.

Therefore by Vector Space of All Mappings is Vector Space, we need only show that $\CC$ is a subspace of $Y^X$.

First, define $f : X \to Y$ by:


 * $\map f x = {\mathbf 0}_Y$

for each $x \in X$.

By Constant Function is Continuous, we have that $f$ is continuous, so $\CC \ne \O$.

By the One-Step Vector Subspace Test we now need to show that $\CC$ is closed under linear combinations.

This is shown in Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous.

So $\CC$ is a vector space.

We move to proving that $\CC$ is a normed vector space.

In order to define the supremum norm on $\CC$, we need to establish that:


 * $\set {\norm {\map f x} : x \in X}$

is bounded for each $f \in \CC$.

From Continuous Image of Compact Space is Compact, $f \sqbrk X$ is a compact subset of $Y$.

From Compact Subspace of Topological Vector Space is von Neumann-Bounded, $f \sqbrk X$ is then von Neumann-bounded.

From Characterization of von Neumann-Boundedness in Normed Vector Space, we have that:


 * $\ds \sup_{x \in X} \norm {\map f x}_Y < \infty$

for each $f \in \CC$.

So we can define the supremum norm:


 * $\ds \norm f_\infty = \sup_{x \in X} \norm {\map f x}_Y$

From Supremum Norm is Norm, $\CC$ is a normed vector space.

We now show that $\CC$ is a Banach space.

Let $\sequence {f_n}_{n \in \N}$ be a Cauchy sequence in $\CC$.

Then for each $\epsilon > 0$ there exists $N \in \N$ such that:


 * $\ds \norm {f_n - f_m} = \sup_{x \in X} \norm {\map {f_n} x - \map {f_m} x}_Y < \epsilon$

for each $n, m \ge N$.

In particular, $\sequence {\map {f_n} x}_{n \in \N}$ is a Cauchy sequence in $Y$.

Since $Y$ is Banach, we have $\map {f_n} x \to L_x$ for some $L_x \in Y$.

Define $f : X \to Y$ by:


 * $\map f x = L_x$

for each $x \in X$.

Taking $N, \epsilon$ as defined previously, we have:


 * $\norm {\map {f_n} x - \map {f_m} x}_Y < \epsilon$

for all $x \in X$ and $n, m \ge N$.

Taking $m \to \infty$ and using Modulus of Limit: Normed Vector Space, we obtain:


 * $\norm {\map {f_n} x - \map f x}_Y < \epsilon$

for all $x \in X$, and hence $f_n \to f$ uniformly.

From the Uniform Limit Theorem, it follows that $f$ is continuous, and hence $f \in \CC$.

So we have shown that every Cauchy sequence in $\CC$ converges.

So $\CC$ is Banach as required.