Space of Bounded Sequences with Supremum Norm forms Banach Space

== Theorem ==
Let $\struct {\map {\ell^\infty} \R, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences on $\R$.
Then $\struct {\map {\ell^\infty} \R, \norm {\, \cdot \,}_\infty}$ 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 $\struct {\map {\ell^\infty} \R, \norm {\,\cdot\,}_\infty}$ converges.
We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\map {\ell^\infty} \R, \norm {\,\cdot\,}_\infty}$.
Then we consider the $k$th component and show, that a real Cauchy sequence $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$ with the limit $x^{\paren k}$ and denote the entire set as $\mathbf x$.
Finally, we show that $\sequence {\mathbf x_n}_{n \in \N}$, composed of components $x_n^{\paren k},$ converges in $\struct {\map {\ell^\infty} \R, \norm {\,\cdot\,}_\infty}$ with the limit $\mathbf x$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\map {\ell^\infty} \R, \norm{\, \cdot \,}_\infty}$.
Denote the $k$th component of $\mathbf x_n$ by $x_n^{\paren k}$.
=== $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$=== 
Let $\epsilon >0$.
Then:
:$\exists N \in \N : \forall m, n \in \N : m,n > N : \norm {\mathbf x_n - \mathbf x_m}_\infty < \epsilon$
For same $N, m, n$ consider $\size {x_n^{\paren k} - x_m^{\paren k} } $:
Hence, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$.
From Real Number Line is Complete Metric Space, $\R$ is a complete metric space.
Consequently, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$.
Denote the limit $\ds \lim_{n \mathop \to \infty} \sequence {x_n^{\paren k}}_{n \mathop \in \N} = x^{\paren k}$.
Denote $\sequence {x^{\paren k}}_{k \mathop \in \N} = \mathbf x$.
=== $\mathbf x$ belongs to $\map {\ell^\infty} \R$ ===
From previous lemma, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$.
Fix any $n > N$ and $k \in \N$.
Then:
:$\forall m > N : \size {x_n^{\paren k} - x_m^{\paren k} } < \epsilon$
Take the limit $m \to \infty$:
:$\size {x_n^{\paren k} - x^{\paren k}} < \epsilon$
Since $k$ was arbitrary:
By definition of space of bounded sequences:
:$\mathbf x_n - \mathbf x \in \ell^{\infty}$
By assumption:
:$\mathbf x_n \in \map {\ell^\infty} \R$
Hence:
:$\mathbf x \in \map {\ell^\infty} \R$
=== $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\map {\ell^\infty} \R, \norm {\, \cdot \,}_\infty}$ to $\mathbf x$ ===
From previous lemma we have that for some $\epsilon > 0$, $N \in \N$ and fixed $n > N$:
:$\norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$
Repeat the same argument for all $\epsilon \in \R_{>0}$ and note that $n$ was arbitrary:
:$\forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$
Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\map {\ell^\infty} \R, \norm{\, \cdot \,}_\infty}$.
== Also see ==
* P-Sequence Space with P-Norm forms Banach Space
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Functional Analysis
Category:Banach Spaces
Category:Space of Bounded Sequences