Ex2-1

Solution to Chapter 2, Exercise 1
This proof seems correct!
Let f(x)=x\sin\frac1x for x\in(0,1] and f(0)=0. 
Boundedness: We find that |f(x)|\leq |x|\leq 1 on [0,1], thus f is bounded.
Continuity: For each x_0\in(0,1], we see 1/x_0 is finite, and so the function is continuous. 
Now we consider the case x_0=0. For x>0 we have
 -x\leq x\sin\frac1x\leq x,
hence when x\to 0, f(x)\to 0=f(0) by Squeeze theorem. This proves f is continuous on [0,1].
Not of bounded variation: For each odd natural number N we choose a partition \Gamma=\{x_{N+1},x_N,\ldots,x_1,x_0\} (consists of N+2 elements) of [0,1] as: x_{N+1}=0, x_0=1 , and 
 x_n=\frac2{n\pi} \quad\text{ for }n=1,2,\ldots,N.
Our purpose will be realize the sum S_\Gamma=\sum_{i=1}^{N+1} |f(x_i)-f(x_{i-1})| as a partial sum of harmonic series, thus proving it is unbounded.
We note that for i=1,2,\ldots,N,N+1,
|f(x_i)|=\begin{cases}0&\text{ if i is even,}\\\frac2{i\pi}&\text{ if i is odd.}\end{cases}
This implies 
\begin{align}S_\Gamma&=\sum_{i=1}^{N+1}|f(x_i)-f(x_{i-1})|\\
&\geq \sum_{i=2}^{N+1}|f(x_i)-f(x_{i-1})|\\
&=\frac2\pi +\frac2{3\pi}+\frac2{3\pi} +\frac2{5\pi}+\frac2{5\pi}+\cdots+\frac2{N\pi}+\frac2{N\pi}\\
&>\frac4\pi \sum_{n=1}^{(N-1)/2} \frac1{2n+1}.\end{align}
As N\to \infty, we find that S_\Gamma becomes unbounded, hence f(x) is not of bounded variation.