Ex2-6

Question: Let f(x)=x^2sin(1/x)
 for 0<x \leq 1 and f(0)=0. Show that V[ f; 0, 1] <+\infty. (Examine the graph of f, or use Exercise 5 and Corollary 2.10.)
Solution: By the hint given by the Author, we will try to proof it using Ex2.5 and corollary 2.10. 
Firstly, since \sin function is \leq 1, we know that f(x)=x^2sin(1/x) \leq 1, when x \in (0,1] and f(0)=0<1. Thus, f(x) is finite on [0,1].
Next, we need to show f(x)=x^2sin(1/x)
 of bounded variation on every interval [0+\epsilon ,1] , \epsilon >0 with V[ f; a+\epsilon, b] \leq M < +\infty.
By Corollary 2.10, since f'(x)=2xsin(1/x)-cos(1/x) 
 is continuous on [\epsilon ,1], \epsilon >0, then we have
 \begin{align} V & = \int_{\epsilon}^1 |f'(x)|dx \\ & = \int_{\epsilon}^1 |2xsin(1/x)-cos(1/x)|\\
 & \leq \int_{\epsilon}^1 3 dx \\ &= 3(1-\epsilon) \leq 3 \end{align} 
where we use trigo inequality and change the xsin(1/x), cos(1/x) to 1, as x \in [\epsilon, 1] .
Thus, f is of bounded variation on [0+\epsilon,1], \forall \epsilon >0,
and V[ f; a+\epsilon, b] \leq 3 < +\infty. Also f(x)
 is continuous at x=0. By Exercise 2.5, V[f; 0 , 1 ] < +\infty.