I know that if $f:[0,\infty)\to\mathbb{R}$ is a function of bounded variation, I can write it as: $$ f(x)=\frac{S(x)+f(x)}{2}-\frac{S(x)-f(x)}{2} $$ where $S(x)$ is $f$'s variation on $[0,x]$, so it is a decomposition of $f$ into two positive monotonoe functions. I read somewhere (And I am truly sorry but I can't remember where) that this composition is minimal in the sense that if I have $f(x)=g(x)-h(x)$ where $g,h$ are positive and monotone, that $\frac{S(x)+f(x)}{2}\leq g(x),\frac{S(x)-f(x)}{2}\leq h(x)$. I was having trouble of proving that last part, and it seems quite simple.
Obviously I tried to write the equations as: $$ \frac{S(x)+f(x)}{2}-\frac{S(x)-f(x)}{2}=g(x)-h(x) \\ \Longrightarrow g(x)-\frac{S(x)+f(x)}{2}=\frac{S(x)-f(x)}{2}-h(x) $$ But even the monotonicity doesn't appear to be useful here.
Thanks for your help.