P-Sequence Space with P-Norm forms Banach Space

== Theorem ==
Let $\ell^p$ be a p-sequence space.
Let $\norm {\, \cdot \,}_p$ be a p-norm.
Then $\struct {\ell^p, \norm {\, \cdot \,}_p}$ 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 $\struct {\ell^p, \norm {\,\cdot\,}_p}$ converges.
We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\ell^p, \norm {\,\cdot\,}_p}$.
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 $\struct {\ell^p, \norm {\,\cdot\,}_p}$ with the limit $\mathbf x$.
Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {\ell^p, \norm{\, \cdot \,}_p}$.
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:
:$\ds \exists N \in \N : \forall m,n \in \N : m,n > N : \norm {\mathbf x_n - \mathbf x_m}_p < \epsilon$
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 $\ell^p$ ===
Let $\epsilon > 0$.
Then:
:$\exists N \in \N : \forall n,m \in \N : n,m > N : \norm {\mathbf x_n - \mathbf x_m}_p < \epsilon$.
Let $K \in \N$, $1 \le p < \infty$.
Then:
Take the limit $m \to \infty$: 
:$\ds \sum_{k \mathop = 1}^K \size {x_n^{\paren k} - x^{\paren k}}^p < \ds \epsilon^p < \infty$
Since $K$ was arbitrary, we can take the limit $K \to \infty$.
By definition, $\forall k \in \N : x_n^{\paren k} - x^{\paren k} \in \R$.
In other words, $\mathbf x_n - \mathbf x \in \ell^p$.
But, by assumption, $\mathbf x_n \in \ell^p$.
Therefore:
:$\paren {\mathbf x - \mathbf x_n} + \mathbf x_n = \mathbf x \in \ell^p$
=== $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^p, \norm {\, \cdot \,}_p}$ to $\mathbf x$===
Let $1 \le p < \infty$. 
Let $\epsilon > 0$.
Fix $N \in \N$, $m > N$.
Then:
:$\ds \forall n > N : \norm {\mathbf x_n - \mathbf x_m}_p < \epsilon$
We have that $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$.
Take the limit $m \to \infty$:
So $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^p, \norm {\, \cdot \,}_p}$.
== Also see ==
* Space of Bounded Sequences with Supremum 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