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.
$\blacksquare$