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:


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

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

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

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


 * $\displaystyle \sum_{k \mathop = 1}^K \size {x_n^{\paren k} - x^{\paren k}}^p < \infty$

Since $K$ was arbitrary, we can take the limit $K \to \infty$.

By definition, $\forall k \in \N : x_n^{\paren k} - x^{\paren k} \in \ell^p$.

In other words, $\mathbf x_n - \mathbf x \in \ell^p$.

But, by assumption, $\mathbf x_n \in \ell^p$.

Therefore:


 * $\paren {\mathbf x - \mathbf x_n} + \mathbf x_n = \mathbf x \in \ell^p$

Let $p = \infty$.

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, $\mathbf x_n - \mathbf x \in \ell^{\infty}$, and so does $\mathbf x$.

$\sequence {\mathbf x_n}_{n \mathop \in \N}$ 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:


 * $\displaystyle \forall n > N : \norm {\mathbf x_n - \mathbf x_m}_p < \epsilon$

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 it converges.

Let $p = \infty$.

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, it converges.