Answer to Question #216567 in Real Analysis for CNZ345

Question

Prove or disprove: If f and g are both of bounded variation on [a, b], then so is f · g.

Answer 1

Let f and g be two absolutely continuous functions on [a, b]. Then fg are absolutely continuous on [a, b]. If, in addition, there exists a constant C > 0 such that |g(x)| ≥ C for all x ∈ [a, b], then f/g is absolutely continuous on [a, b]. If f is integrable on [a, b], then the function F defined by

Theorem. Let f be an absolutely continuous function on [a, b]. Then f is of bounded

variation on [a, b]. Consequently, f'(x) exists for almost every x ∈ [a, b].

Proof

Since f and g are absolutely continuous functions on [a, b]., there exists some $\delta > 0$ such that $\sum ^n_{i=1}|f(y_i)|<1$ whenever $\{[x_i,y_i]: i=1,...,n \}$ is a finite collection of mutually disjoint sub intervals of [a,b] with $\sum ^n_{i=1}|f(y_i)|<\delta$. Let N be the least integer such that $N> (b-a)/ \delta$ and let $a_j : =a+j(b-a)/N$ for j= 0,1,...,N. Then $a_j -a_{j-1} = (b-a)/N < \delta$. Hence, $V^{a_j}_{a_{j-1}} f< 1; j = 0,1,...,N$. It follows that

$\displaystyle\bigvee_{a}^b f = \sum_{j=1}^N \displaystyle\bigvee_{a_{j-1}}^{a_j} f <N$

This shows that f and g are bounded variations on [a,b]. Consequently f'(x) and g'(x) exist for almost every $x \epsilon [a,b]$