variation of $\vert f \vert^{1.5}$

$f$ has a bounded variation. $\vert f\vert ^{1.5}$ also has a bounded variation. $\vert f \vert$ is a bounded variation function, as well as integer powers of $\vert f \vert$. $\vert f \vert ^{1.5}$ is not an integer product, I don't think I can rely on these facts. $\sqrt{\vert f\vert}$ is not always a bounded variation function.

How can the variation of $f$ help me calculate $\sum_{k=1}^n \vert \vert f(x_{k+1}) \vert^{1.5} -\vert f(x_k) \vert ^{1.5}\vert$?

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Hint: In general, it is easy to see that if $f:[a,b]\to[c,d]$ is of bounded variation and $g:[c,d]\to\mathbb{R}$ is Lipschitz, then $g\circ f$ is of bounded variation. For your question, let $g(x)=|x|^{1.5}$, which is Lipschitz on any bounded interval.
Details: Since $f$ is of bounded variation, $|f|\le M$ for some $M>0$. Note that for $g(x)=|x|^{1.5}$, $|g'(x)|=1.5|x|^{0.5}\le 1.5M^{0.5}$ when $|x|\le M$. Then for any $x<y$,
$$\left||f(y)|^{1.5}-|f(x)|^{1.5}\right|=\left|\int^{f(y)}_{f(x)}g'(t)dt\right|\le\left|\int^{f(y)}_{f(x)}|g'(t)|dt\right|\le1.5 M^{0.5} |f(y)-f(x)|.$$ It follows that, for any $x_0<x_1<\cdots<x_n$, we have: $$\sum_{k=1}^n\left||f(x_k)|^{1.5}-|f(x_{k-1})|^{1.5}\right|\le1.5 M^{0.5}\sum_{k=1}^n\left|f(x_k)-f(x_{k-1})\right|\le1.5 M^{0.5}\cdot \mathrm{Var}f<+\infty,$$ where $\mathrm{Var}f$ denotes the total variation of $f$.
Why not? I think it is the natural way to answer your question. If you are not familiar with the concept "Lipschitz", you may replace it with a stronger condition that $g$ is continuously differentiable, i.e. $g'$ exists and continuous on $[c,d]$(and therefore bounded). – 23rd Dec 12 '12 at 6:28
For $g(x)=|x|^{1.5}$, $g'(x)=1.5x^{0.5}$ when $x\ge 0$, and $g'(x)=-1.5|x|^{0.5}$ when $x<0$, of course $g'$ is continuous. – 23rd Dec 12 '12 at 7:31