Answer to Question #170317 in Real Analysis for Prathibha Rose

Question

Show that ,f defined on [a,b] is of bounded variation on [a,b] if and only if f can be expressed as the difference of two increasing functions

Answer 1

Assume that $f \in V[a,b]$ and let $v(x)=V^x_a(f), x \in (a,b]$ and $v(a)=0$. Then clearly $f(x) = v(x)-[v(x)-f(x)]$. We will show $v(x)$ and $v(x)-f(x)$ are increasing.

$\forall$ $x_1<x_2$ : $v(x_2)-v(x_1)=V ^{x_2}_{x_1} (f) ≥ 0 \iff v(x_2) ≥ v(x_1)$ so $v(x)$ is increasing.

$\forall$ $x_1<x_2$ : $f(x_2)-f(x_1) \le |f(x_2)-f(x_1)| \le V ^{x_2}_{x_1} (f) =v(x_2)-v(x_1)$ $\iff v(x_1)-f(x_1) \le v(x_2)-f(x_2)$ and thus $v(x)-f(x)$ is increasing.

Conversely, suppose that $f(x)=g(x)-h(x)$ with $g$ and $h$ increasing. Since $h$ is increasing, $-h$ is decreasing and thus $f(x) = g(x)+(-h(x))$ is the sum of two monotone functions. So theorem

$[$ If $f : [a,b] \to \R$ is monotone on $[a,b]$ then $f \in V[a,b]$ and $V^b_a(f)=|f(b)-f(a)|$ $]$ together with theorem $[$ Let $f,g:[a,b] \to \R$ be of bounded variation on $[a,b]$. Then $(f+g)\in V[a,b]$ and $V^b_a(f+g) \le V^b_a(f) + V^b_a(g)$ $]$ gives that $f \in V[a,b]$.