Product inequality for BV function:$V(fg) \leq V(f)V(g)$

Question

Let $[a,b]$ be a closed and bounded interval and $V(f)= \sup\sum^n_i |f(t_{i+1})-f(t_i)|$ where supremum is taken over all partition $a=t_1<t_2<\cdots<t_{n+1}=b$. Suppose $f(a)=g(a)=0$, then$$V(fg)\leq V(f)V(g)$$ How to prove it?

Answer 1

Define $F(t_i)=\sum_{j=2}^i |f(t_j)-f(t_{j-1})|$ for all $2\leq i\leq n+1,$ and

\begin{align*} F(t_{i+1})G(t_{i+1})-F(t_{i})G(t_{i})&=(|f(t_{i+1})-f(t_{i})|+F(t_{i}))(|g(t_{i+1})-g(t_{i})|+G(t_{i}))-F(t_{i})G(t_{i})\\ &=|f(t_{i+1})-f(t_{i})||g(t_{i+1})-g(t_{i})|+F(t_i)|g(t_{i+1})-g(t_{i})|\\ &+|G(t_i)|f(t_{i+1})-f(t_{i})|\\ \end{align*} And \begin{align*} |f(t_{i+1})g(t_{i+1})-f(t_{i})g(t_{i})|&\leq|f(t_{i+1})-f(t_{i})||g(t_{i+1})-g(t_{i})|+|f(t_i)||g(t_{i+1})-g(t_{i})|\\ &+|g(t_i)||f(t_{i+1})-f(t_{i})|\\ &\leq|f(t_{i+1})-f(t_{i})||g(t_{i+1})-g(t_{i})|+F(t_i)|g(t_{i+1})-g(t_{i})|\\ &+G(t_i)|f(t_{i+1})-f(t_{i})|)\\ &=F(t_{i+1})G(t_{i+1})-F(t_{i})G(t_{i}) \end{align*} Done by summing up the term

Answer 2

The worst case is when $f$ and $g$ are increasing.

Given a partition $a=t_1<\dots<t_{n+1}=b$ we can define $F(t_1)=0$ and $F(t_i)=\sum_{j=2}^i |f(t_j)-f(t_{j-1})|$ for all $2\leq i\leq n+1,$ and similarly for $G$ in terms of $g.$ Note these are only defined on the vertices of the partition. (You could define a similar function independent of the partition, but this seems easier with your definition.)

Then $|f(t_i)|\leq F(t_i)$ and $|g(t_i)|\leq G(t_i),$ so \begin{align*} &|f(t_{i+1})g(t_{i+1})-f(t_{i})g(t_{i})|\\ &=|(f(t_{i+1})-f(t_i))g(t_{i+1})+f(t_i)(g(t_{i+1})-g(t_i))|\\ &\leq |(F(t_{i+1})-F(t_i))G(t_{i+1})+F(t_i)(G(t_{i+1})-G(t_i))|\\ &=F(t_{i+1})G(t_{i+1})-F(t_{i})G(t_{i}). \end{align*}

This then telescopes when summed over $i$ giving $$\sum_{i=1}^n|f(t_{i+1})g(t_{i+1})-f(t_{i})g(t_{i})|\leq F(b)G(b)\leq V(f)V(g).$$ Taking the supremum over all partitions gives $V(fg)\leq V(f)V(g)$ as required.