Suppose $f(x)$ is of bounded variation. Show $F(x) = \frac{1}{x} \int_0^x f(t) \, dt$ is also of bounded variation.

Question

Suppose $f(x)$ is of bounded variation on the interval $[0,a]$. Show $F(x) = \frac{1}{x} \int_0^x f(t) \, dt$ is also of bounded variation on $[0,a]$.

Here is where I am at for context:

Since $f$ is of bounded variation, $f(x) = g(x) - h(x)$, where g(x) and h(x) are monotone increasing on $[0,a]$. In addition the "average" function of $G(x)$ and $H(x)$ are $G(x) = \frac{1}{x} \int_0^x g(t) \, dt$ and $H(x) = \frac{1}{x} \int_0^x h(t) \, dt$. I have previously proven that these "average" functions are increasing on $[0,a]$.\

Now, how can I use this to show that $F(x)$ is of bounded variation on the interval $[0,a]$ provided that $f(x)$ is?

Answer 1

Let $f(t)$ be of bounded variation on the interval $[0,a]$. Let $F(x)$ be defined as

$$ F(x) = \frac{1}{x} \int_0^x f(t) \, dt $$

we show $F(x)$ is also of bounded variation. Now by Jordan's Theorem $f(t)$ is of bounded variation if and only if $f(t)$ is the difference of two increasing functions on $[0,a]$. Choose $g(t)$ and $h(t)$ such that $f(t) = g(t) - h(t)$, where $g(t)$ and $h(t)$ are increasing over the interval $[0,a]$.

Now, consider the difference of the "average" functions $G(x) = \frac{1}{x} \int_0^x g(t) \, dt$ and $H(x) = \frac{1}{x} \int_0^x h(t) \, dt$,

\begin{align*} G(x) - H(x) &= \frac{1}{x} \int_0^x g(t) \, dt - \frac{1}{x} \int_0^x h(t) \, dt \Rightarrow \\ &= \frac{1}{x} \int_0^x g(t) - h(t) \, dt \Rightarrow \\ &= \frac{1}{x} \int_0^x f(t) \, dt \Rightarrow \\ &= F(x) \end{align*}

Thus,

$$ F(x) = G(x) - H(x). $$

Therefore, since $G(x)$ and $H(x)$ are increasing functions, by Jordan's Theorem $F(x)$ is of bounded variation.

Answer 2

Following through on sranthrop's hint, $f(x) = g(x) - h(x)$ implies that $F(x) = G(x)-H(x)$. So, if $G $ and $H$ are both increasing then $G$ is of bounded variation.

One way to prove that $G$ and $H$ are increasing is to differentiate them using the product rule: $$ G'(x) = \frac{1}{x}g(x) - \frac{1}{x^2}\int_0^x g(t)\,dt = \frac{1}{x^2}\int_0^x (g(x)-g(t))\,dt \ge 0 $$ because $g(x)-g(t)\ge 0$. Unfortunately, this computation is justified at the points of continuity of $g$ (where $G$ is differentiable). A monotone function has at most countably many discontinuities, and it's true that having $G$ continuous and $G'\ge 0$ except at countably many points is enough... but this is complicated.

A simpler way is to take $x<y$ and write $$ G(y)-G(x) = \left(\frac{1}{y} - \frac{1}{x}\right)\int_0^x g(t)\,dt + \frac1y \int_x^y g(t)\,dt $$ The term in parentheses is negative, so using $g(t)\le g(x)$ in the first integral and $g(t)\ge g(x)$ in the second we get $$ G(y)-G(x) \ge \left(\frac{1}{y} - \frac{1}{x}\right)x g(x) + \frac1y (y-x) g(x) = 0 $$