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.