Answer to Question #298022 in Real Analysis for Badha

Question

If f is monotonic on (a,b) then show that f is of bounded variation on(a,b)

Answer 1

If $\ V(f,[a,b])< \infty$ , then $f$ is of bounded variation on $[a,b]$ .

$V(f,[a,b])=\sup \big\{\sum\limits_{i=1}^n\big|f(x_i)-f(x_{i-1})\big|, \text{where }a=x_0<x_1<…<x_n=b\}$

1) $f$ is increasing

$\sum\limits_{i=1}^n\big|f(x_i)-f(x_{i-1})\big|= \sum\limits_{i=1}^n\big(f(x_i)-f(x_{i-1})\big)=f(x_1)-f(x_0)+f(x_2)-f(x_1)+…+f(x_n)-f(x_{n-1})=f(x_n)-f(x_0)=f(b)-f(a)$

So, $V(f,[a,b])=f(b)-f(a)<\infty$ .

Therefore $f$ is of bounded variation on $[a,b]$ .

2) $f$ is decreasing $\sum\limits_{i=1}^n\big|f(x_i)-f(x_{i-1})\big|= \sum\limits_{i=1}^n\big(f(x_{i-1})-f(x_{i})\big)=f(x_0)-f(x_1)+f(x_1)-f(x_2)+…+f(x_{n-1})-f(x_{n})=f(x_0)-f(x_n)=f(a)-f(b)$

So, $V(f,[a,b])=f(a)-f(b)<\infty$ .

Therefore $f$ is of bounded variation on $[a,b]$ .