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 with respect to 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:
\(\ds \size {\map {x_n} t - \map {x_m} t}\) | \(\le\) | \(\ds \max_{\tau \mathop \in I} \size {\map {x_n} \tau - \map {x_m} \tau}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {x_n - x_m}_\infty\) | 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, $\map {x_{N + 1} } t \in \map C I$.
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:
\(\ds \size {\map x \tau - \map x t}\) | \(=\) | \(\ds \size {\map x \tau - \map {x_{N + 1} } \tau + \map {x_{N + 1} } \tau - \map {x_{N + 1} } t + \map {x_{N + 1} } t - \map x t}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \size {\map x \tau - \map {x_{N + 1} } \tau} + \size {\map {x_{N + 1} } \tau - \map {x_{N + 1} } t} + \size {\map {x_{N + 1} } t - \map x t}\) | 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 $\struct {\map C I, \norm {\,\cdot\,}_\infty}$:
- $\ds \lim_{x \mathop \to \infty} x_n = x$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces