Question

If F is Lipschitz function and g(x) is monotonically increasing on [a,b] then fog is bounded variation

Answer 1

Let us consider a subdivision $a=x_0 \leq x_1 \leq ... \leq x_n=b$ and the variation of $f\circ g$ on this subdivision.

$V(f\circ g)=\sum_{i\leq n-1} |f\circ g(x_{i+1})-f\circ g(x_i)|$

First of all, as $f$ is a Lipschitz, there is a constant $k$ such that we have

$|f\circ g(x_{i+1})-f\circ g(x_i)| \leq k\cdot |g(x_{i+1})-g(x_i)|$

Now as $g$ is increasing, we have $|g(x_{i+1})-g(x_i)|=g(x_{i+1})-g(x_i)$

From these two relations we deduce

$V(f\circ g)\leq k\cdot \sum|g(x_{i+1})-g(x_i)| =k\cdot \sum (g(x_{i+1})-g(x_i))$

The later sum is telescopic and gives $V(f\circ g)\leq k\cdot (g(b)-g(a))<+\infty$

Therefore, $f\circ g$ is of bounded variation.