Prove that this function has finite variation

Question

We know that $f$ has finite variation on $[a,b]$.

Prove that $$g(x)= \begin{cases} 0, & x=a\\[8pt] \frac{1}{x-a} \int _{a} ^x f(t) \, dt , & x \in (a,b] \end{cases} $$

has finite variation.

I would really appreciate all your help, because I'm just starting my course in integral calculus.

Thank you.

Answer 1

The function $g$ is absolutely continuous on any interval $[c,b]$ with $a < c < b$. Try to use this fact along with a bound on $|g(c) - g(a)| = |g(c)|$.

Answer 2

Since $f$ is of bounded variation, you can write it $f=f_1-f_2$ where $f_1,f_2$ are non decreasing. It suffices to prove that

$$g_i(x)= \begin{cases} 0, & x=a\\[8pt] \frac{1}{x-a} \int _{a} ^x f_i(t) \, dt , & x \in (a,b] \end{cases}$$

are of bounded variation.

Since $f_i$ is non decreasing, there exists some $c_i$ so that $f_i(x) < 0 \forall x < c_i$ and $f_i(x) \geq 0 \forall x \geq c_i$. Then $g_i$ is either [ decreasing on $[a, c_i]$ and increasing on $[c_i,b]$] or strictly monotonic on $[a,b]$ depending if $c_i \in (a,b)$.

In particular $g_i$ is of bounded variation.