# A sequence $\{x_n\}$ if of bounded variation. Prove $\exists a_n, b_n$ such that $x_n = a_n - b_n$ where $a_n, b_n$ are bounded and increasing.

Prove that for any sequence $$\{x_n\}$$ of bounded variation there exist increasing bounded sequences $$\{a_n\}$$, $$\{b_n\}$$ such that $$x_n = a_n - b_n$$ for $$n\in \Bbb N$$

This problem appears in the context of theese two problems. Those two posts show some properties related to the bounded variation of the sequences. Now I want to show the last part, namely what's in the problem statement.

I was thinking of the following sketch. Let's suppose $$x_n$$ may be presented as a difference of the two sequences, namely: $$x_n = a_n - b_n$$ Then what is left to show is that both $$a_n$$ and $$b_n$$ are increasing. Since the variation is bounded then it must follow: $$\exists \lim_{n\to\infty}\sigma_n = L\\ \exists \lim_{n\to\infty}x_n = x$$

Then: \lim_{n\to\infty}(\sigma_n - \sigma_{n-1}) = 0 \\ \begin{align} \sigma_n - \sigma_{n-1} &= |x_{n+1} - x_n| \\ &= |(a_{n+1} - b_{n+1}) - (a_n - b_n)| \\ &= |(a_{n+1} - a_n) - (b_{n+1} - b_n)| \end{align}

It follows that: $$\lim_{n\to\infty} |(a_{n+1} - a_n) - (b_{n+1} - b_n)| = 0 \implies \\ \lim_{n\to\infty} (a_{n+1} - a_n) = \lim_{n\to\infty}(b_{n+1} - b_n) = C \tag1$$

But this seems to be a road to nowhere. What would be the way to show $$a_n$$ and $$b_n$$ are increasing and bounded? Because according to $$(1)$$ it looks like $$a_n$$ and $$b_n$$ are not even bounded.

You can just define $$a_n=x_1^{+}+(x_2-x_1)^{+}+\cdots+(x_n-x_{n-1})^{+}$$ and $$b_n=x_1^{-}+(x_2-x_1)^{-}+\cdots+(x_n-x_{n-1})^{-}$$ where $$x^{+}=\max\{x,0\}$$ and $$x^{-}=-\min \{x,0\}$$.