Showing that $x^2\cos^2(\pi/(x^2))$ is not of bounded variation on $[0,1]$ Problem: Let $f(x) = x^2\cos^2(\pi/(x^2))$.  Prove that $f$ is not of bounded variation on $[0,1]$ even though $f$ is differentiable on
  all of $\mathbb{R}$.


Visual Intuition:

The following is a visual of $f(x)$ that provides intuition for how $f$ could not be of bounded variation on $[0,1]$.  Clearly our attention should be on the behavior of $f(x)$ when $x$ is near $0$.



Attempt:


Recall that $T_0^1(f)$ denotes the total variation from $0$ to $1$ on $f$ and is defined as follows:

$$
T_0^1(f) = \sup \left\{\sum_{k=1}^n \left| f(x_i) - f(x_{i-1}) \right|\right\} \text{ across all subdivisions of $[0,1]$}
$$
My idea is to show that 

$$
T_0^1(f) \ge \sum_{n=1}^{\infty} T_{1 \over n+1}^{1 \over n}(f) = \infty
$$


But I'm not sure how to execute on this.

$\fbox{EDIT: New proof using fgp's comments:}$


We have$$
  f(x) = x^2\cos^2\left(\tfrac{\pi}{x^2}\right) \text{.}
$$
So let $x_k = \sqrt{\frac{1}{k}}$. Then $f(x_k) = \frac{1}{k}\cos(\pi k)$, which means  $$
  f(x_k) = \begin{cases}
    \frac{1}{k} &amp;\text{if $k$ is even (because then $\cos \pi k = 1$)} \\
    -\frac{1}{k} &amp;\text{if $k$ is odd  (because then $\cos \pi k = -1$).} \\
  \end{cases}
$$
Thus we have


$$
  T_0^1(f) \geq \sum_{n=1}^\infty |f(x_{2n}) - f(x_{2n+1})| = \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2(n+1)} \geq {1 \over 2} \sum_{n=1}^\infty \frac{1}{n} = \infty \text{.}
$$

which completes the proof.
As a first step, let's look at $$
  f(x) = x\cos\left(\tfrac{a}{x}\right) \text{.}
$$
Let $x_k = \frac{a}{k\pi}$. Then $f(x_i) = \frac{a}{k\pi}\cos(\pi k)$, which means $$
  f(x_k) = \begin{cases}
    \frac{a}{k\pi} &amp;\text{if $k$ is even (because then $\cos \pi k = 1$)} \\
    -\frac{a}{k\pi} &amp;\text{if $k$ is odd  (because then $\cos \pi k = -1$).} \\
  \end{cases}
$$
Thus (assuming $a\pi \geq 1$, if that isn't the case, just start the sum at some other index. That won't change anything, it'll still diverge)$$
  T_0^1(f) \geq \sum_{n=1}^\infty |f(x_{2n}) - f(x_{2n+1})| = \sum_{n=1}^\infty \frac{a}{2n\pi} + \frac{a}{2(n+1)\pi} \geq \frac{a}{2\pi} \sum_{n=1}^\infty \frac{1}{n} = \infty \text{.}
$$

For $x^2\cos^2\left(\frac{\pi}{x^2}\right)$, just expand $\cos^2(x)$ (it's something like $\frac{1}{2} + \frac{1}{2}\cos(2x)$ or so), and then substitute $u=x^2$.
Just compute the derivative and integrate its absolute value and you'll be able to proof that it (the function) is unbounded.
The latter is easiest done by substituting z=1/x^2 and using a simple estimated lower bound for the contribution of each interval between two integers.