Question about bounded variation and continuity

Question

$$f(x) = \begin{cases} \frac{1}{2n} & \text{if} \ x=\frac{1}{2n} \\ 0 & \text{if} \ x=0 \\ 0 & \text{if} \ x=\frac{1}{2n-1} \\ \text{and linear otherwise} \end{cases}$$

can you show that $f$ is continuous but not of bounded variation in $[0,1]$ ?

Answer 1

Hints:For bounded variation, take the partition P:{ 0,1/ 2n, 1/(2n-1),....,1/3,1/2,1}. V(P,f)=|f(1/2n) - f(0)|+|f(1/2n)-f(1/(2n-1))|+......+|f(1/3)-f(1/2)|+|f(1/2)-f(1)| =1/2n +1/2n+1/(2n-2)+......+1/2.
=(1/2)*(1+1/2+1/3+...+1/(n-1)+1/n+1/n). Clearly right hand side term goes to infinity as n goes to infinity.