The derivative of total variation Define function $u$ on $[0,1]$ such that
$$  
u(x)=
\begin{cases}
x^2\cos\frac{1}{x} &0<x\leq 1\\
0 & x=0
\end{cases}
$$
and by $V(x):= \text{Var}_{[0,x]}u$, i.e., the total variation of $u$ on $[0,x]$. We actually have $u$ is in $W^{1,1}(0,1)$.
Now I want to show
$$ 0<V'(0)<\infty$$
What I can do now is go with definition that 
$$V'(0):=\lim_{h\to0}\frac{V(h)}{h} $$
and what I have so far is that 
$$ \limsup_{h\to0}\frac{V(h)}{h} <\infty$$
But this is still far away from proving $V'(0)$ is actually exist and has a lower bound. Any help would be very welcome!
 A: The derivative of $u$ is 
$$u'(x)=
\begin{cases}
2x\cos\frac{1}{x} + \sin\frac{1}{x}&0<x\leq 1\\
0 & x=0
\end{cases}.$$
In particular, note that since $\left|h\cos\frac{1}{h}\right|\leqslant |h|,$ 
$$u'(0) = \lim_{h \rightarrow 0} \frac{h^2\cos\frac{1}{h}-0}{h}=\lim_{h \rightarrow 0} h\cos\frac{1}{h}=0.$$
So $u$ has a bounded derivative and bounded total variation $V(x)$ on any interval $[0,x] \subset [0,1]$.
Consider a partition with subintervals of the form
$$I_k = \left[(k\pi + \pi/2)^{-1},(k\pi - \pi/2)^{-1}\right].$$
Then $u$ vanishes at the endpoints and there is an extremum at some point $\xi_k$ where
$$\frac{1}{k\pi} \leqslant \xi_k \leqslant \frac{1}{k\pi-\pi/2}.$$
The total variation of $u$ on the subinterval is
$$V\left[(k\pi-\pi/2)^{-1}\right] - V\left[(k\pi+\pi/2)^{-1}\right] = 2|u(\xi_k)|$$
and
$$\frac{1}{(k\pi)^2} \leqslant |u(1/k\pi)| < |u(\xi_k)| < \xi_k^2 < \frac{1}{(k\pi-\pi/2)^2}.$$
Summing from $k = n$ to $\infty$ we get
$$\sum_{k=n}^{\infty}\frac{1}{k^2}  < \frac{\pi^2}{2}V[(n\pi- \pi/2)^{-1}] <  \sum_{k=n}^{\infty}\frac{1}{(k-1/2)^2}.$$
We can find bounds for the sums using integrals:
$$\sum_{k=n}^{\infty}\frac{1}{k^2} > \int_n^{\infty}\frac{dx}{x^2}= \frac1{n},\\\sum_{k=n}^{\infty}\frac{1}{(k-1/2)^2} < \int_{n-3/2}^{\infty}\frac{dx}{x^2}= \frac1{(n-3/2)},$$
and it follows that
$$\frac1{n} <  \frac{\pi^2}{2}V[(n\pi- \pi/2)^{-1}] < \frac1{(n-3/2)}.$$
Finally, if
$$(n\pi- \pi/2)^{-1} < h < (n\pi- 3\pi/2)^{-1},$$
then, since $V(x)$ is increasing,
$$\frac{2}{n\pi^2}<V[(n\pi- \pi/2)^{-1}] < V(h) < V[(n\pi- 3\pi/2)^{-1}]< \frac{2}{(n-5/2)\pi^2},$$
and 
$$\frac{2}{n\pi^2}(n\pi- 3\pi/2) < \frac{V(h)}{h} < \frac{2}{(n-5/2)\pi^2}(n\pi- \pi/2),
\\ \frac{2-3/n}{\pi} < \frac{V(h)}{h} < \frac{2-1/n}{[1-5/(2n)]\pi}.$$
Apply the squeeze principle in taking the limit as $n \rightarrow \infty$ and $h \rightarrow 0$.  This yields:
$$V'(0) = \lim_{h \rightarrow 0} \frac{V(h)}{h } = \frac{2}{\pi}.$$
