My problem is the following (exercise 4.14 from L Barreira, C Valls, Dynamical systems, an introduction):
Given $f$: $S^1 \rightarrow \mathcal{R}$ defined as:
$x\sin(\frac{\pi}{x})$ for $x\neq0$ and
$f(x) = 0$ for $x=0$
I am supposed to show that for any n$\in \mathbb{N}$:
$\sum_{i=1}^{n-1} |f(x_i) - f(x_{i-1}) | \quad = \quad 4\sum_{i=1}^n\frac{1}{2i +1}$
for the partition of $[0, 1]$ given by $x_0 = 0, \quad x_1 = \frac{2}{2n+1}, \quad x_2 = \frac{1}{2n-1}, \quad ... ,x_{n-1} = \frac{2}{5}, \quad x_n = \frac{2}{3}, \quad x_{n+1} = 1 $
I know that the sum on the right hand side of the above equation in not convergent as n tends to infinity, hence I can conclude that the above equality shows, by definition of bounded variation, that $f(x)$ is indeed not of bdd variation.
My approach was to use induction to show that, for any n the above equation hold, such that all I need to find is a base case, then show the induction step, and I'm done.
By inspection, the natural choice of basecase is $n=2$, however when doing the calculations we get:
$|f(x_1) - f(x_0)| = |\frac{2}{3}\sin(\frac{3\pi}{2}) - 0| = \frac{2}{3} \neq 4\frac{8}{15} = 4(\frac{1}{3} + \frac{1}{5})$
So the equation above does not hold for $n=2$. My question is then, is there an another approach to showing that $f(x)$ is not of bdd variation?
