Prove that a product of functions of bounded variation is a function of bounded variation

Question

We consider functions defined on an interval $[a,b]$. I have to prove that a product of functions of bounded variation is a function of bounded variation. I have to also show that this isn't true for quotient in general and tell which additional assumption guarantees that quotient IS of bounded variation.

Answer 1

Hint:

$|(fg)(x)-(fg)(y)|\leq |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|$. Again $f,g$ are bounded so what will you get from here??

Answer 2

This came to my mind first.

$$fg = \frac{(f+g)^2 - (f-g)^2}{4}$$

Since sum of functions of bounded variation have bounded variation, it suffices to show that $f^2$ is of bounded variation when $f$ is. To show the latter, use

$$|f(x)^2-f(y)^2| = |f(x)-f(y)||f(x)+f(y)| \leq |f(x)-f(y)|(|f(x)|+|f(y)|)$$

and the fact that $f$ is bounded.