Writing a sequence of bounded real numbers as the difference of two decreasing sequences of bounded real numbers

Question

I am wondering if any sequence of bounded real numbers as the difference of two decreasing sequences of bounded real numbers.

Let $(u_n)_{n\in\mathbb{N}}$ be a bounded sequence of real numbers. I am looking for another sequence $(v_n)_{n\in\mathbb{N}}$ such that for all $n\in \mathbb{N}$ one has $$ u_n = (u_n^+ + v_n) - (u_n^- + v_n) $$ with $u_n = u_n^+ - u_n^-$ the usual decomposition using nonnegative sequences. I also have the additional condition that for a given $\lambda > 1$, one has $$ \|u^+ + v\|_{\infty} + \|u^- + v\|_{\infty} \leq \lambda \|u\|_\infty. $$ The first step seems to set $v_0 = \lambda \|u\|_\infty - |u_0|$ and pursue with $$ -\min(u^+_{n+1},u^-_{n+1}) \leq v_{n+1} \leq \min(u^+_{n} - u^+_{n+1} + v_n,u^-_{n} - u^-_{n+1} + v_n) $$ but it is not clear to me that this choice of $v_{n+1}$ is always possible.

Answer 1

Inspired by the fact that a function is of bounded variation if and only it is the difference of two positive, monotonic increasing functions, we can state the following:

A sequence $(u_n)$ is the difference of two bounded decreasing sequences if and only if it is of bounded variation, i.e. if $\sum_{n=1}^\infty |u_{n+1}-u_n| < \infty$.

Example: The sequence $((-1)^n)_n$ can not be decomposed into a difference of bounded decreasing sequences.

First assume that $u_n = a_n - b_n$ with decreasing bounded sequences $(a_n)$, $(b_n)$. Without loss of generality we can assume that both sequences are non-negative. Then $$ \sum_{n=1}^N|u_{n+1}-u_n| \le \sum_{n=1}^N|a_{n+1}-b_n| + \sum_{n=1}^N|b_{n+1}-b_n| = (a_1-a_{N+1}) + (b_1 - b_{N+1}) \le a_1 + b_1 $$ for all $N$.

Conversely, if $\sum_{n=1}^\infty |u_{n+1}-u_n| < \infty$ then one can define $$ a_n = \sum_{k=n}^\infty |u_{k+1}-u_k| \, , \, b_n = a_n - u_n $$ and show that these are decreasing bounded sequences.

We also have $$ \Vert a \Vert_\infty \le 2 \Vert u \Vert_\infty \\ \Vert b \Vert_\infty \le 3 \Vert u \Vert_\infty $$ by construction.

I do not know if a decomposition $u_n = a_n - b_n$ with $\Vert a \Vert_\infty + \Vert b \Vert_\infty \le \lambda \Vert u \Vert_\infty$ is possible for arbitrary $\lambda > 1$.