I want to prove that
$f(t)=t^{1+\epsilon}sin(1/t), f(0)=0$ has bounded first variation on [0,1].
First order variation is defined as: $_{||||→0}∑^{−1}_{=0}|Δ_|$, where $p$ is a partition of $[0,1]$ and $||p||$ is the mesh of $p$
My first attempt was to show that the function has bounded first derivatives (because I already showed that such functions have bounded first variation in the previous part.) But, it is not the case for this function.
Any idea how to proceed? Any kind of help would be much appreciated!
With $\epsilon > 0$, the derivative is $f'(t) = (1+\epsilon)t^\epsilon \sin(1/t) - t^{-1+\epsilon}\cos(1/t)$ for $0 < t \leqslant 1$ and $f'(0) = 0$.
For $ \epsilon \geqslant 1$ derivative is bounded on $[0,1]$ and the function is of bounded variation.
For $0 < \epsilon \leqslant 1$ we see that
$$|f'(t)| = | (1+\epsilon)t^\epsilon \sin(1/t) - t^{-1+\epsilon}\cos(1/t)| \leqslant (1+\epsilon)t^{\epsilon}+ t^{-1+\epsilon},$$
and
$$\int_0^1|f'(t)| \, dt \leqslant (1+\epsilon)\int_0^1 t^{\epsilon} \, dt + \int_0^1 t^{-1+\epsilon} \, dt = \left.t^{1+\epsilon}\right|_0^1 + \left.\frac{t^\epsilon}{\epsilon}\right|_0^1 = 1 + \frac{1}{\epsilon} $$
Thus, $f$ is of bounded variation in this case as well, since the derivative is absolutely integrable (as discussed here).