Space of Bounded Sequences with Supremum Norm forms Banach Space

Theorem
Let $\ell^\infty$ be a space of bounded sequences.

Let $\norm {\, \cdot \,}_\infty$ be a supremum norm.

Then $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ 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 $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$ converges.

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

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 $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$ with the limit $\mathbf x$.

Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\ell^\infty, \norm{\, \cdot \,}_\infty}$.

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:


 * $\displaystyle \exists N \in \N : \forall m,n \in \N : m,n > N : \norm {\mathbf x_n - \mathbf x_m}_\infty < \epsilon$

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 $\displaystyle \lim_{n \mathop \to \infty} \sequence {x_n^{\paren k}}_{n \mathop \in \N} = x^{\paren k}$.

Denote $\sequence {x^{\paren k}}_{k \mathop \in \N} = \mathbf x$.

$\mathbf x$ belongs to $\ell^\infty$
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:


 * $\displaystyle \forall m > N : \size {x_n^{\paren k} - x_m^{\paren k}} < \epsilon$

Take the limit $m \to \infty$:


 * $\size {x_n^{\paren k} - x^{\paren k}} < \epsilon$

Since $k$ was arbitrary:

By definition, $\mathbf x_n - \mathbf x \in \ell^{\infty}$.

By assumption, $\mathbf x_n \in \ell^\infty$.

Hence, $\mathbf x \in \ell^\infty$.

$\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ to $\mathbf x$
From previous lemma we have that for some $\epsilon > 0$, $N \in \N$ and fixed $n > N$:


 * $\displaystyle \norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$

Repeat the same argument for all $\epsilon \in \R_{>0}$ and note that $n$ was arbitrary:


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

Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^\infty, \norm{\, \cdot \,}_\infty}$.

Also see

 * P-Sequence Space with P-Norm forms Banach Space