$V_a^b(P,f):=\sum_{i=1}^k|f(x_i)-f(x_{i-1})|$, 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?
Best Answer
I think the problem is that your sum is not right. It should be $$ \begin{align} \sum_{k=0}^n\left(\frac1{\frac\pi2+\pi(k+1)}+\frac1{\frac\pi2+\pi k}\right) &=\frac2\pi+\frac2\pi\sum_{k=1}^n\frac1{k+\frac12}\\ &\ge\frac2\pi+\frac2\pi\sum_{k=1}^n\frac1{k+1}\\ &=\frac2\pi\sum_{k=1}^n\frac1k \end{align} $$