Question

Determine which of the following functions are of bounded variation on[0,1]
a. f(x)= x^2sin(1/X) if x#0 ,f(0)=0
b. f(x)= √x sinx,if x#0,f(0)=0

Answer 1

$f(x)=x^2sin(\frac1x)$

$f(0)=0;f(1)=sin(1)$

In the interval (0,1):

$0\leq sin(\frac1x)\leq 1$

$0\leq x^2sin(\frac1x)\leq x^2$

So,the function doesn't blow up to infinity,thus, f(x) is bounded.

(b)$f(x)=\sqrt{x}sin x$

$f(0)=0;f(1)=sin(1)$

$0\leq sinx \leq1$

$0\leq \sqrt{x}sinx \leq\sqrt x$

So,the function doesn't blow up to infinity,thus, f(x) is bounded.