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An exercise requires to show that if $f$ is a real function of bounded variation over $[0,1]$, so is $|f|^r$, where $r \geq 1$.

It is trivial to prove that $|f| \in \operatorname{BV}[0,1]$, and, should one notice that $$\operatorname{V}_0^1|f|^p \leq p \sup |f| \operatorname{V}_0^1 |f| < \infty, \quad \text{for all } p \in \mathbb{N},$$ it is straightforward that the statment is true for integer $r \geq 1$.

(The last result is obtained with the help of the inequality $\operatorname{V}_a^b (gh) \leq \sup |g| \operatorname{V}_a^b h + \sup |h| \operatorname{V}_a^b g,$ which holds for any functions $g,h : [a,b] \to \mathbb{R}$).

We have proven the case in which $r \in \mathbb{N}$. But what if $r$ is rational or even irrational?

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Use the Mean Value Theorem: $$ \bigl|\,|f(x)|^r-|f(y)|^r\,\bigr|=r\,\xi^{r-1}\bigl|\,|f(x)|-|f(y)|\,\bigr|\le r\,(\sup|f|)^{r-1}|f(x)-f(y)|. $$ In the above, $\xi$ is between $|f(x)|$ and $|f(y)|$.

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    That's great! But I couldn't possibly think of the Mean Value Theorem myself... I was looking for a more 'standard' way to derive it. Do you happen to know any alternatives? – 2017-02-17
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    It is the standard way when you have to estimate differences of the form $g(x)-g(y)$ for some function $g$. – 2017-02-17
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    OK then, I'll keep it in mind, in case I need it elsewhere. Thank you! – 2017-02-17