Derivative of total variation

Question

Problem:

Let $\alpha \in BV[a,b]$ (bounded variation on the interval $[a,b]$) and let $\beta(x) = V_a^x \alpha$ be the total variation of $\alpha(x)$ from $a$ to $x \in [a,b]$.

Prove or disprove: If $\alpha$ is differentiable at a point $c \in (a,b)$ then $\beta$ is differentiable at $c$.

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I have been struggling with this problem for a few hours. Any hints would be appreciated.

Answer 1

If $\alpha$ is differentiable but $\alpha'$ is not continuous at $c$ then it is possible that

$$\lim_{x \to c} \frac{\beta(x) - \beta(c)}{x-c} = \infty.$$

An example where the one-sided derivative of $\beta$ is infinite at $c = 0$ is $\alpha(x) = x^{3/2} \cos(1/x)$ on the interval $(0,1]$ and $\alpha(0) := 0$.

(Note that this can be extended to an example where we consider the two-sided derivative at an interior point $c = 0$, by choosing $\alpha(x) = |x|^{3/2} \cos(1/x)$ for $x \in [-1,0) \cup (0,1]$.)

In this case, $\alpha$ has a bounded derivative and so is of bounded variation.

We have

$$\alpha'(0) = \lim_{x \to 0+} \frac{x^{3/2} \cos(1/x) - 0}{x - 0} = 0,$$

but
$$\beta'(0) = \lim_{x \to 0+} \frac{\beta(x) - 0}{x-0} = \infty.$$

The last result can be shown using a variation on the argument presented here: The derivative of total variation.

In that problem, we considered the function $\alpha(x) = x^2 \cos(1/x)$ and showed that $\beta'(0) = 2/\pi$. With an exponent of $3/2 < 2$, the upper and lower bounds for $\beta(x)/x$ will diverge to $\infty$ as $x \to 0$.

Answer 2

Hint: Suppose not only that $\alpha$ is differentiable but that $\alpha'$ is Riemann-integrable.

Then you can try to show that:

$$ \beta(x) = \int_a^x |\alpha '(t)| dt $$