Bounded Variate function and its total variation has different points of differentiabilty $f(x) = | x |$ is a functional of Bounded Variation but is not differentiable at zero.
However, it's total variation on $[ -1, 1 ]$ is $1 + x$ , which is differentiable at zero.
But I am unable to find a function for which the function is differentiable at a point 'c' but it's total variation is not differentiable at that point.
Any help is appreciated.
 A: The function $f(x) = \begin{cases}x^{3/2} \cos(1/x), &0<x \leqslant 1 \\0,& x= 0 \end{cases}$ is differentiable with
$$f'(0) = \lim_{h \to 0+}\frac{h^{3/2} \cos(1/h) - 0}{h - 0} =  \lim_{h \to 0+}h^{1/2} \cos(1/h) = 0$$
Since $|f'|$ is integrable on $[0,1]$ we have $f \in BV([0,1])$. However, the total variation function $\beta(x) = V_0^x(f)$ is not differentiable at $0$ as $\beta(h) /h \to +\infty$ as $h \to 0+$.
Proof that $\beta(h)/h \to +\infty$.
Consider a partition with subintervals of the form $I_k = \left[(k\pi + \pi/2)^{-1},(k\pi - \pi/2)^{-1}\right].$  Since $f$ vanishes at the endpoints, there is an extremum at some point $\xi_k$ where  $(k\pi)^{-1} \leqslant \xi_k \leqslant (k\pi-\pi/2)^{-1}$. Note that $f$ is monotone on the intervals $((k\pi)^{-1} ,\xi_k)$ and $(\xi_k,(k\pi - \pi/2)^{-1})$ and it follows that the total variation of $f$ on the subinterval $I_k$ is
$$\beta\left[(k\pi-\pi/2)^{-1}\right] - \beta\left[(k\pi+\pi/2)^{-1}\right] = 2|f(\xi_k)|$$
We also have
$$(k\pi)^{-3/2} = |f(1/k\pi)| < |f(\xi_k)| < \xi_k^{3/2} < (k\pi-\pi/2)^{-3/2}$$
Summing from $k = n$ to $\infty$ we get
$$\pi^{-3/2}\sum_{k=n}^{\infty}\frac{1}{k^{3/2}}  < \sum_{k=n}^{\infty}\underbrace{\frac{1}{2}\left(\beta[(k\pi- \pi/2)^{-1}] - (\beta[(k\pi+ \pi/2)^{-1}] \right)}_{|f(\xi_k)|} <  \pi^{-3/2}\sum_{k=n}^{\infty}\frac{1}{(k-1/2)^{3/2}},$$
The middle sum is telescoping and since $\beta(0) = 0$, this reduces to
$$\sum_{k=n}^{\infty}\frac{1}{k^{3/2}}  < \frac{\pi^{3/2}}{2}\beta[(n \pi- \pi/2)^{-1}] <  \sum_{k=n}^{\infty}\frac{1}{(k-1/2)^{3/2}}$$
Using the bounds, 
$$\displaystyle\sum_{k=n}^{\infty}\frac{1}{k^{3/2}} > \int_n^{\infty}\frac{dx}{x^{3/2}}= \frac{2}{\sqrt{n}}, \quad \displaystyle\sum_{k=n}^{\infty}\frac{1}{(k-1/2)^{3/2}} < \int_{n-3/2}^{\infty}\frac{dx}{x^{3/2}}= \frac{2}{\sqrt{n-3/2}},$$
it follows that
$$\frac{2}{\sqrt{n}} <  \frac{\pi^2}{2}\beta[(n\pi- \pi/2)^{-1}] < \frac{2}{\sqrt{n-3/2}}$$
Take $h$ such that  $(n\pi- \pi/2)^{-1} < h < (n\pi- 3\pi/2)^{-1}.$ Since $\beta(x)$ is increasing,
$$\beta(h)>\beta[(n\pi- \pi/2)^{-1}] >  \frac{4}{\pi^2\sqrt{n}}, $$
and 
$$\tag{*}\frac{\beta(h)}{h}  >  \frac{4}{\pi^2\sqrt{n}}(n\pi - 3\pi/2) = \frac{4}{\pi}\frac{n- 3/2}{\sqrt{n}}$$
We have $h \to 0+$ as $n \to \infty$.  Upon taking the limit of both sides of (*) we get
$$\lim_{h \to 0+} \frac{\beta(h)}{h} > \lim_{n \to \infty} \frac{4}{\pi}\frac{n- 3/2}{\sqrt{n}} = + \infty$$
