I know there is a theorem saying if $f$ defined on $[a,b]$ is of bounded variation, then it is differentiable on $(a,b)$ a.e and $f'$ is integrable over $[a,b]$.
I wonder whether the converse is true, say, if $f$ defined on $[a,b]$ is differentiable on $(a,b)$ a.e and $f'$ is integrable, must $f$ have bounded variation?
In fact, I run into such concrete question:
$$
f=x^\alpha \text{sin}\left(\frac{1}{x^\beta}\right) \text{ for $x\in(0,1]$}
$$
and $f(0)=0$.
The hint says that we can prove that when $\alpha > \beta$, $f$ is of bounded variation by showing $f'$ is integrable.
So here comes my question above, is such argument true?
Best Answer
We know, under the conditions in your problem, that the total variation of $f$ is given by $$V_a^b(f)=\int_a^b|f'(x)|dx$$ and since you're given $f'$ integrable then yes: $f$ is of b.v.
Added to the OP's request:
1) Under the condition of your question, in any subinterval $\,[\alpha,\beta]\subset [a,b]\,$, we have that for some $\,c\in (\alpha,\beta)\,,\,f(\beta)-f(\alpha)=f'(c)(\beta-\alpha)$
2) If $\,\mathcal P:=\{a=x_1<x_2<...<x_{n_p}=b\}\,$ is a partition of $\,[a,b]\,$, then the total variation of $f$ on this interval is defined to be $$V_a^b(f):=\sup_{\mathcal P}\sum_{k=1}^{n_p-1}\left|f(x_{i+1}-f(x_i)\right|$$
3) Can you see now from where the integral of $f'$ comes?