Is Differentiation of total variation function $V(f, [0,x])$ possible in this condition? I am having difficulty of determining whether differentiation of total variation function $V(f, [0,x])$ is possible on $x = 0$ when $f:[0,1]→R$ is defined as:
$ f(x) = 0$ for $x = 0$ and $f(x) = x^2sin(1/x)$ for $x≠0$
I think it would not be possible intuitively, but I cannot come up with precise proof. 
 A: To compute the derivative of the total variation of $f$ at $x = 0$ as a definitional limit, take a partition with subintervals 
$\left[((k+1)\pi )^{-1},(k\pi)^{-1}\right].$  Since $f(x) = x^2 \sin(1/x)$ vanishes at the endpoints, there is an extremum at some point $\xi_k$ where
$$\frac{1}{(k+1)\pi} \leqslant \xi_k \leqslant \frac{1}{k\pi}.$$
The total variation of $f$ on the subinterval is
$$V\left[(k\pi)^{-1}\right] - V\left[((k+1)\pi)^{-1}\right] = 2|f(\xi_k)|,$$
since $f$ is alternately monotone increasing and decreasing in the subintervals on either side of $\xi_k$.
Observing where $f$ takes the extreme value in the interval we find
$$\frac{1}{((k+1/2)\pi)^2} = |f(\, ((k+1/2)\pi)^{-1})| < |f(\xi_k)| < \xi_k^2 < \frac{1}{(k\pi)^2}.$$
Hence,
$$\frac{2}{\pi^2}\frac{1}{(k+1/2)^2} \leqslant   V\left[(k\pi)^{-1}\right] - V\left[((k+1)\pi)^{-1}\right]  \leqslant \frac{2}{\pi^2}\frac{1}{k^2}$$
Summing from $k = n$ to $\infty$ we get
$$\frac{2}{\pi^2}\sum_{k=n}^{\infty}\frac{1}{(k+1/2)^2}  < V[(n\pi)^{-1}] <  \frac{2}{\pi^2}\sum_{k=n}^{\infty}\frac{1}{k^2}.$$
Using the following bounds for the sums,
$$\sum_{k=n}^{\infty}\frac{1}{k^2} < \int_{n-1}^{\infty}\frac{dx}{x^2}= \frac1{n-1},\\\sum_{k=n}^{\infty}\frac{1}{(k+1/2)^2} > \int_{n+1/2}^{\infty}\frac{dx}{x^2}= \frac1{n+ 1/2},$$
it follows that,
$$\frac{2}{\pi^2}\frac1{n + 1/2} <  V[(n\pi)^{-1}] < \frac{2}{\pi^2}\frac1{n-1}.$$
Take $h$ such that $(n\pi)^{-1} < h < ((n-1)\pi)^{-1},$ and note that $h \to 0$ as $n \to \infty$.
Since $V(x)$ is increasing, we have
$$\frac{2}{\pi^2}\frac{1}{n + 1/2}<V[(n\pi)^{-1}] < V(h) < V[((n-1)\pi)^{-1}]< \frac{2}{\pi^2}\frac{1}{n-2}.$$
Since, $ (n-1)\pi < 1/h < n\pi,$ we have
$$ \frac{2}{\pi} \frac{1 - 1/n}{1 + 1/(2n)} = \frac{2}{\pi^2}\frac{(n-1)\pi}{n + 1/2} < \frac{V(h)}{h} < \frac{2}{\pi^2}\frac{n\pi}{n-2} = \frac{2}{\pi} \frac{1}{1 - 2/n}
$$
Taking limits as  $n \rightarrow \infty$ and applying the squeeze theorem, we find the right-hand derivative of the total variation
$$V'(0) = \lim_{h \rightarrow 0} \frac{V(h)}{h } = \frac{2}{\pi}.$$
