# Prove that set of sequences $(x_n)_{n=1}^{\infty}$ such that $\sum_{n}\lvert x_{n+1} - x_n\rvert <\infty$ is Banach space

I have to prove that set of sequences $(x_n)_{n=1}^{\infty}$ in $\mathbb C$ such that $\sum_{n}\lvert x_{n+1} - x_n\rvert <\infty$ is Banach space with norm $\lVert x\rVert=\rvert x_1 \rvert + \sum_{n=1}^\infty\lvert x_{n+1} - x_n\rvert$

Here is my attempt at proof.

Let $(x^m)_{m=1}^{\infty}$ be Cauchy. let $\epsilon >0$ Then we have for every $m_1,m_2 > m_0$ it is true that $\lVert x^{m_1}-x^{m_2}\rVert<\epsilon$ i.e. $\rvert x_1^{m_1} - x_1^{m_2} \rvert + \sum_{n=1}^\infty\lvert x_{n+1}^{m_1} - x_{n+1}^{m_2} - (x_n^{m_1} - x_n^{m_2})\rvert < \epsilon$. From this two things follow.

First is that $\rvert x_1^{m_1} - x_1^{m_2} \rvert$ is Cauchy since $\rvert x_1^{m_1} - x_1^{m_2} \rvert < \lVert x^{m_1}-x^{m_2}\rVert<\epsilon$. Since it is Cauchy, sequence $(x_1^{m})_{m=1}^\infty$ converges in $\mathbb C$ to some $x_1$.

Second thing is that for every $n \in \mathbb N$ we have that $\lvert x_{n+1}^{m_1} - x_{n+1}^{m_2} - (x_n^{m_1} - x_n^{m_2})\rvert = \lvert x_{n+1}^{m_1} - x_{n}^{m_1} - (x_{n+1}^{m_2} - x_n^{m_2})\rvert < \lVert x^{m_1}-x^{m_2}\rVert<\epsilon$ Which means that for every fixed $n \in \mathbb N$ we have that $(x_{n+1}^m - x_n^m)_{m=1}^\infty$ is Cauchy in $\mathbb C$ hence it is convergent to some $a_n$.

Now since i know that ce $(x_1^{m})_{m=1}^\infty$ converges to $x_1$ and that for $n=1$ , $(x_{2}^m - x_1^m)_{m=1}^\infty$ converges to $a_1$ we have that $(x_2^{m})_{m=1}^\infty$ must converge to some $x_2$. Next we do for $n=3$ ... So we get that for every $n \in \mathbb N$ we have that exists $\lim_{m\to \infty}x_n^m=x_n$. Let $x=(x_1,x_2,....)$.

We let $\lim_{m_1\to \infty}$ in $\rvert x_1^{m_1} - x_1^{m_2} \rvert + \sum_{n=1}^k\lvert x_{n+1}^{m_1} - x_{n+1}^{m_2} - (x_n^{m_1} - x_n^{m_2})\rvert$ and we get $\rvert x_1 - x_1^{m_2} \rvert + \sum_{n=1}^k\lvert x_{n+1} - x_{n+1}^{m_2} - (x_n - x_n^{m_2})\rvert < \rvert x_1 - x_1^{m_2} \rvert + \sum_{n=1}^\infty\lvert x_{n+1} - x_{n+1}^{m_2} - (x_n - x_n^{m_2})\rvert = \lVert x-x^{m_2}\rVert<\epsilon$.

Hence $x^m$ converges to $x$.

• You constructed $x=\{x_n\}$ fine, but after it is constructed, you need to show that this sequence is in the space (i.e. $|x_1|+\sum_n|x_{n+1}-x_n|<\infty$) and that $x^m\to x$ in norm. – Aweygan May 15 '17 at 22:23

## 1 Answer

For the record, the space you are dealing with is the space of sequences of bounded variation, often denoted $BV$.

It does not look like you have a proof of $\|x^m - x\|\to 0$; I can't parse the end of the proof. A typical approach is to consider the operator $$Tx = (x_1, x_2-x_1, x_3 - x_2, \dots)$$ and observe that $T:BV\to \ell^1$ is an isometric isomorphism. Since $\ell^1$ is complete (reference), so is $BV$.