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Let $f:[a,b]\to\mathbb{R}$ be of bounded variation. Then it follows that $f$ is differentiable almost everywhere in $[a,b]$, and $$ \int_a^b|f'(x)|\ dx<\infty. $$

Conversely, suppose $f:[a,b]\to\mathbb{R}$ is differentiable almost everywhere and $$ \int_a^b|f'(x)|\ dx<\infty. $$ Can one conclude that $f$ is of bounded variation on $[a,b]$?

I'm not sure if the stronger condition of "differentiable everywhere" is needed, but I don't have a counterexample.

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    Play a bit with the [Cantor function](https://en.wikipedia.org/wiki/Cantor_function). – 2017-01-02
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    @DanielFischer: I think this is a classical example that is BV but not absolutely continuous? Would you elaborate your point? – 2017-01-02
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    Side remark: So far, you don't require $f$ to be continuous. Discontinuous $f$ with derivative $0$ almost everywhere and unbounded variation are easy. If you want a continuous $f$, the Cantor function gives you an a.e. differentiable function with $f' = 0$ a.e. Of course, since it is monotonic, the Cantor function has bounded variation. But if you modify it a bit, letting it oscillate, you may be able to construct a continuous function with $f' = 0$ a.e. and unbounded variation. (Note: I haven't fully explored it, so I don't _know_ that it works. But it looks like it ought to.) – 2017-01-02
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    @DanielFischer: Thanks for your comment. What would be an "easy example" of "Discontinuous $f$ with derivative $0$ almost everywhere and unbounded variation" you mentioned? – 2017-01-02
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    The characteristic function of an infinite closed set with empty interior, for example $\{0\} \cup \{2^{-n} : n \in \mathbb{N}\}$ or the Cantor set. – 2017-01-02

1 Answers 1

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There is a continuous function $f : [0, 1] \to \Bbb{R}$ with infinite variation such that $f'(x)$ exists and is zero for a.e. $x \in [0, 1]$. For instance, let $\phi : [0, 1] \to \Bbb{R}$ be the Cantor-Lebesgue function and let $a_n = 1 - \frac{1}{2} + \frac{1}{3} + \cdots + (-1)^{n-1}\frac{1}{n}$ with $a_0 = 0$. Then we define $f$ piecewise by

$$ f(x) = a_{n-1} + \frac{(-1)^{n-1}}{n} \phi \left(1 - \tfrac{1}{2^{n-1}} + \tfrac{1}{2^n}x\right) \quad \text{if} \quad x \in [1 - \tfrac{1}{2^{n-1}}, 1 - \tfrac{1}{2^n}]$$

for each $n = 1, 2, \cdots$ and $f(1) = \log 2$.

$\hspace{7em}$enter image description here

This function is continuous and has infinite variation. But since $\phi'$ is zero a.e., it follows that $f'$ is also zero a.e.

On the other hand, if we assume that $f:[a,b]\to\Bbb{R}$ is everywhere differentiable on $[a, b]$ and $f'$ is integrable, then $f$ is of bounded variation. This is because the following FToC-type result holds:

$$ f(x) - f(a) = \int_{a}^{x} f'(t) \, dt, \qquad x \in [a, b] $$

The proof of this result can be found, for instance, in Rudin RCA, Theorem 7.25.

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    Thank you for your answer. Do you have a reference for the example or a quick justification why it works? – 2017-01-02
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    @Jack, If you don't mind, would you be a more specific about which part you want me to justify? – 2017-01-02
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    Oh, I mean the function $f$ you constructed using the Cantor function. I don't see how it has infinite variation. Would you elaborate? – 2017-01-02
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    @Jack, Oh, I see. Notice that $$V(f, [0, 1]) \geq \sum_{k=1}^{n} \left| f\left( 1 - \tfrac{1}{2^{k-1}} \right) - f\left( 1 - \tfrac{1}{2^{k}} \right) \right| = \sum_{k=1}^{n} \frac{1}{k} \to \infty \qquad \text{as } n \to \infty. $$ – 2017-01-02
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    Thanks for your elaboration! What tool did you use to give the graph? – 2017-01-02
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    @Jack I use Mathematica 11. It has a native function representing the Cantor-Lebesgue function, so it wass not hard to produce the graph. – 2017-01-02