I have a general query:
If $f$ is a $C^{\infty}$-function and $g$ is a B.V.$([0,2\pi])$ function (i.e. Total Variation of $g$ over $[0,2\pi]$ is finite), then, I want to know (proof or counterexample) if the function $(f \ast g)g$ ( i.e. $[\int_{0}^{2\pi}f(\theta - \theta_{*})g(\theta_{*},t) d\theta_{*}]g(\theta,t)$ )is in B.V.$([0,2\pi])$ ??
My gut feeling says: yes!! $f\ast g$ being $C^{\infty}$ over $[0,2\pi]$ (-a compact set) is in B.V., also $g$ is in B.V. (assumption,). So, our desired product should be in B.V. !!
Isn't it??
Thanks in advance
Yes, your feeling is correct. As you pointed out, $C^{\infty}([0,2\pi]) \subset BV([0,2\pi])$, thus it suffices to prove that $BV([a,b])$ is closed under multiplication. Let $f,g \in BV([a,b])$, first of all note that $f,g$ must be bounded in $[a,b]$; for every $x,y \in [a,b]$ we can write: $$ |(fg)(x)-(fg)(y)| = |f(x)g(x)-f(y)g(y)+f(x)g(y)-f(x)g(y)| $$ thus $$ |(fg)(x)-(fg)(y)| \le |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)| $$
From now it is easy to conclude that $fg$ is $BV$.