Prove $f$ is not of bounded variation function.

Question

Let $f(x)=\dfrac{1}{x}$, for $0<x\leq 1$ and $f(x)=0$, for $x=0$. Prove $f$ is not of bounded variation function.

This is my attemp.

\begin{align} V(f,[0,1])&=\sum\limits_{k=1}^{n} \left\vert f{(x_k)}-f{(x_{k-1})}\right\vert\\ &=\left\vert f{(x_1)}-f{(x_{0})}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert +\ldots +\left\vert f{(x_n)}-f{(x_{n-1})}\right\vert \\ &=\left\vert f{(x_1)}-f{(0)}\right\vert+\left\vert f{(x_2)}-f{(x_{1})}\right\vert +\ldots +\left\vert f{(1)}-f{(x_{n-1})}\right\vert \\ &= \left\vert\dfrac{1}{x_1}-0\right\vert +\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert +\ldots +\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\\ &= 1-0\\ &=1 \end{align} Now I confused to prove $f$ is not of bounded variation function, because $V(f,[0,1])=1<M$, for some $M>0$. So, anyone can correct my answer if my answer is incorrect?

Answer 1

$$\pi=\{a=0\leq x_1<x_2<...<x_n\leq b=1\}$$take $x_1=\epsilon$ now $$ \left\vert\dfrac{1}{x_1}-0\right\vert +\left\vert\dfrac{1}{x_2}-\dfrac{1}{x_{1}}\right\vert +\ldots +\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert=\\ \left\vert\dfrac{1}{\epsilon}-0\right\vert +\left\vert\dfrac{1}{x_2}-\dfrac{1}{\epsilon}\right\vert +\ldots +\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\geq \\ \left\vert\dfrac{1}{\epsilon}-0\right\vert +\left\vert\dfrac{1}{1}-\dfrac{1}{\epsilon}\right\vert +\ldots +\left\vert\dfrac{1}{1}-\dfrac{1}{x_{n-1}}\right\vert\geq \frac2{\epsilon}-1+|\frac1{x_2}-\frac1{x_3}|+...\\ \geq \frac2{\epsilon}-1\\ \to \infty$$

Answer 2

Bounded variation requires the bound to hold for every finite set $x_k$, not just for one set.

To find a set where the total variation is large, you need some $x_i$ to be positive but small. In fact, you can do it with just one intermediate point. Let $M$ be given (with $M>1$) and let $x_1=1/M$

$$|f(0)-f(x_1)|+|f(x_1)-f(1))| = M + (M-1)=2M-1> M$$

Answer 3

$|\frac{1}{x_1}-0|+|\frac{1}{x_2}-\frac{1}{x_1}|+... +|\frac{1}{x_n}-\frac{1}{x_{n-1}}|$ is not equal to $1$. It is $[\frac{1}{x_1}-0]+[\frac{1}{x_1}-\frac{1}{x_2}]+... +[\frac{1}{x_{n-1}}-\frac{1}{x_n}]$. This works out to $\frac 2 {x_1} -1$ which is not bounded.