# Showing a function is not of a bounded variation.

$V_a^b(P,f):=\sum_{i=1}^k|f(x_i)-f(x_{i-1})|$, where $P$ is a partition.

$$f(x)= \begin{cases} x^2\sin(\frac{1}{x^2}), &\text{if } x\neq0, \\ 0, &\text{if } x=0 \end{cases}$$ does not have bounded variation over $[-1,1]$. I'm trying to show this without using the fact that it may not have bounded variation over some subset of $[-1,1]$ .

I want to find a partition $\lbrace x_0,\dotsc,x_n \rbrace$ of $[-1,1]$ for which $$\sum_{i=1}^n|f(x_i)-f(x_{i-1})|$$ gives a partial sum of the Harmonic Series. My partition is $P=\lbrace -1,0,\frac{1}{\sqrt{\frac{\pi}{2}+\pi n}},\frac{1}{\sqrt{\frac{\pi}{2}+\pi(n-1)}},\dotsc,\frac{1}{\sqrt{\frac{\pi}{2}}},1\rbrace$.

\begin{align} V_{-1} ^1(P,f) &=|f(0)-f(-1)| + \left|(f\Bigl(\frac1{\sqrt{\smash[b]{\frac\pi2}+\pi n}}\Bigr)-f(0)\right|+\left|f(1)-f\Bigl(\frac1{\sqrt{\smash[b]{\frac\pi2}}}\Bigr)\right| \\ &\mathrel{\phantom=} +\sum_{i=0}^{n}\left|f\Bigl(\frac1{\sqrt{\smash[b]{\frac\pi2}+\pi (n+1)}}\Bigr)-f\Bigl(\frac1{\sqrt{\smash[b]{\frac\pi2}+\pi n}}\Bigr)\right| \\ &= |f(0)-f(-1)|+\left|f\Bigl(\frac1{\sqrt{\smash[b]{\frac\pi2}+\pi n}}\Bigr)-f(0)\right| \\ &\mathrel{\phantom=}+\left|f(1)-f\Bigl(\frac1{\sqrt{\smash[b]{\frac\pi2}}}\Bigr)\right|+\sum_{i=0}^n\left(\frac1{\smash[b]{\frac\pi2}+\pi n}+\frac1{\smash[b]{\frac\pi2}+\pi (n-1)}\right). \end{align}

I don't know what to do next with the summation. Am I on to something?

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I think you might have a good idea here, but where are the $i$'s in your sums...? – icurays1 Nov 23 '12 at 3:40
You do not want to know how much time it took to get your math massaged into proper shape. The horrendous misuse of cases at the start was easy to fix, but please take a careful look at what I did to your huge equation starting with "$V_{-1}^1(P,f)$". – kahen Nov 23 '12 at 3:48
That is nice. Didn't know I can do that. – cap Nov 23 '12 at 5:24

I think the problem is that your sum is not right. It should be \begin{align} \sum_{k=0}^n\left(\frac1{\frac\pi2+\pi(k+1)}+\frac1{\frac\pi2+\pi k}\right) &=\frac2\pi+\frac2\pi\sum_{k=1}^n\frac1{k+\frac12}\\ &\ge\frac2\pi+\frac2\pi\sum_{k=1}^n\frac1{k+1}\\ &=\frac2\pi\sum_{k=1}^n\frac1k \end{align}