Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Banach Space

Theorem
Let $I = \closedint a b$ be a closed real interval.

Let $\map C I$ be the space of real-valued functions, continuous on $I$.

Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on real-valued functions, continuous on $I$.

Then $\struct {\map C I, \norm {\,\cdot\,}_\infty}$ is a Banach space.

Proof
A Banach space is a normed vector space, where a Cauchy sequence converges the supplied norm.

To prove the theorem, we need to show that a Cauchy sequence in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$ converges.

We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$.

Then we fix $t \in I$ and show, that a real Cauchy sequence $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$ with the limit $\map x t$.

Then we prove the continuity of $\map x t$.

Finally, we show that $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$ with the limit $\map x t$.

$\sequence {\map {x_n} t}_{n \mathop \in \N}$ is a Cauchy Sequence
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\map C I$:


 * $\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n, m > N : \norm {x_n − x_m}_\infty < \epsilon$

Fix $t \in I$.

Then:

Hence $\sequence {\map {x_n} t}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$.

$\sequence {\map {x_n} t}_{n \mathop \in \N}$ Converges in $\struct {\R,\size {\, \cdot \,}}$
From Real Number Line is Complete Metric Space, $\R$ is a complete metric space.

Therefore, $\sequence {\map {x_n} t}_{n \mathop \in \N}$ is convergent.

Denote the limit by $\map x t : I \to \R$:


 * $\ds \lim_{n \mathop \to \infty} \map {x_n} t = \map x t$

$\map x t$ is Continuous
Choose $N$ such that:


 * $\forall n, m > N : \norm{x_n - x_m} < \dfrac \epsilon 3$

Let $\tau \in I$.

Then $\forall n > N$ and $m = N + 1 > N$:

Take the limit $n \to \infty$:

By assumption, $\map {x_{N + 1} } t \in \map C I$.

Then:


 * $\forall t, \tau \in I : \exists \delta > 0: \size {\tau - t} < \delta \implies \size {\map {x_{N + 1} } t - \map {x_{N + 1} } \tau} < \dfrac \epsilon 3$

Thus:

Hence, $\map x t$ is continuous in $I$.

$\sequence {x_n}_{n \mathop \in \N}$ Converges to $x$
Let $\epsilon > 0$.

Choose $N$ such that:
 * $\forall n, m > N : \norm {x_n - x_m}_\infty < \epsilon$

Fix $n > N$.

Let $t \in I$.

Then, $\forall m > N$:

Take the limit $m \to \infty$:

This holds for every $n > N$.

Repeat the argument for all $\epsilon > 0$:


 * $\forall \epsilon > 0 : \exists N \in \N : \forall n \in \N : \forall n > N : \norm {x_n - x}_\infty < \epsilon$.

Therefore, $x_n$ converges to $x$ in $\struct {\map C I, \norm {\,\cdot\,}_\infty}$:


 * $\ds \lim_{x \mathop \to \infty} x_n = x$