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