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

## Proof

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

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

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

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 $\struct {\map \CC I, \norm {\,\cdot\,}_\infty}$ 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 $\map \CC I$:

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

Fix $t \in I$.

Then:

 $\ds \size {\map {x_n} t - \map {x_m} t}$ $\le$ $\ds \max_{\tau \mathop \in I} \size {\map {x_n} \tau - \map {x_m} \tau}$ $\ds$ $=$ $\ds \norm {x_n - x_m}_\infty$ Definition of Supremum Norm $\ds$ $<$ $\ds \epsilon$

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

$\Box$

### $\sequence {\map {x_n} t}_{n \mathop \in \N}$ 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 $\map x t : I \to \R$:

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

$\Box$

### $\map x t$ 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 $m = N + 1 > N$:

 $\ds \size {\map {x_n} \tau - \map {x_{N + 1} } \tau }$ $\le$ $\ds \norm {x_n - x_{N + 1} }_\infty$ $\ds$ $<$ $\ds \frac \epsilon 3$

Take the limit $n \to \infty$:

 $\ds \size {\map x \tau - \map {x_{N + 1} } \tau}$ $=$ $\ds \lim_{n \mathop \to \infty} \size {\map {x_n} \tau - \map {x_{N + 1} } \tau}$ $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R,\size {\, \cdot \,} }$ $\ds$ $<$ $\ds \frac \epsilon 3$

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

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:

 $\ds \size {\map x \tau - \map x t}$ $=$ $\ds \size {\map x \tau - \map {x_{N + 1} } \tau + \map {x_{N + 1} } \tau - \map {x_{N + 1} } t + \map {x_{N + 1} } t - \map x t}$ $\ds$ $<$ $\ds \size {\map x \tau - \map {x_{N + 1} } \tau} + \size {\map {x_{N + 1} } \tau - \map {x_{N + 1} } t} + \size {\map {x_{N + 1} } t - \map x t}$ Triangle Inequality for Real Numbers $\ds$ $<$ $\ds \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3$ $\ds$ $=$ $\ds \epsilon$

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

$\Box$

### $\sequence {x_n}_{n \mathop \in \N}$ 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, $\forall m > N$:

 $\ds \size {\map {x_n} t - \map {x_m} t}$ $\le$ $\ds \norm {x_n - x_m}_\infty$ $\ds$ $<$ $\ds \epsilon$

Take the limit $m \to \infty$:

 $\ds \lim_{m \mathop \to \infty} \size {\map {x_n} t - \map {x_m} t}$ $=$ $\ds \size {\map {x_n} t - \map x t}$ $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R,\size {\, \cdot \,} }$ $\ds$ $\le$ $\ds \max_{t \mathop \in I} \size {\map {x_n} t - \map x t}$ $\ds$ $=$ $\ds \norm {x_n - x}_\infty$ Definition of Supremum Norm $\ds$ $<$ $\ds \epsilon$

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 $\struct {\map \CC I, \norm {\,\cdot\,}_\infty}$:

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

$\blacksquare$