Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2\sin(\frac{1}{x^2}))$

Question

Show that $f'(x)$ is not integrable on $[0,1]$ if $f(x) = x^2\sin(\frac{1}{x^2}))\chi_{(0,1]}$ and $f(x) = 0$ if $x = 0$.

I took the derivative $$f'(x) = x^2\cos(\frac{1}{x^2})(-2)\frac{1}{x^3} + 2x\sin(\frac{1}{x^2}) \mbox{ if } x \neq 0, f'(x) = 0 \mbox{ if } x = 0$$

I must show that

$$\int_{0}^{1}|f'(x)| = \infty$$

I can't show that the integral is not finite.

Answer 1

$f[0,1]\to \mathbb{R}$ is defined by

f(x) = \begin{cases} x^2sin(\frac{1}{x^2}) & \text{if $x\neq 0$} \\ 0 & \text{if $x=0$} \\ \end{cases}

WTS: $f'$ is not Riemann Integrable on $[0,1]$. Let us do this by showing that $f'(0)$ exists. We do this by $f'(0)= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x^2sin(\frac{1}{x^2})}{x}=\lim_{x\to0} xsin(\frac{1}{x^2})$. Now, $-x\leq xsin(\frac{1}{x^2}) \leq x.$ By Squeeze Theorem, $\lim_{x\to0}(-x)\leq lim_{x\to0} xsin(\frac{1}{x^2})\leq lim_{x\to0}x$. Therefore, $\lim_{x\to0}xsin(\frac{1}{x^2})=0$. Thus, $f'(0)=0$ and does exist.

Now, $f'(x)=x^{2}cos(\frac{1}{x^2}) \cdot \frac{d}{dx}(x^{-2})+ sin(\frac{1}{x^{2}})\cdot2x.$ = $x^2cos(\frac{1}{x^2})(-2x^{-3})+2xsin(\frac{1}{x^2})$ = $\frac{-2}{x}cos(\frac{1}{x^2})+2xsin(\frac{1}{x^2})$. So $f'(x)$ is not bounded on [-1,1]. $|2xsin(\frac{1}{x^2})| \leq2$ for all $x\in[-1,1]$. For ease of notation, Let $a_n = \sqrt{\frac{2}{(2n-1)\pi}}$. Then $cos(\frac{1}{a_n^2})=1$ for all n as $a_n\to 0$ and$\frac{2cos(\frac{1}{a_n^2})}{a_n}$= $\frac{2}{a_n}$= $\sqrt{2(2n-1)\pi} \to \infty$ as $n\to\infty$.

Hence $f'$ is unbounded on $[-1,1]$ and therefore $f'$ is unbounded on $[0,1]\subset[-1,1,]$. As a result, we see that $f'$ is not Riemann integrable on $[0,1]$.

Note: I read your question in bold, I used rather elementary methods. Sorry if this isn't what you are looking for.

Answer 2

Formally, the integral of $|f'(x)|$ means we separte the original function $f(x)$ into monotone picese and add them up. Note that the monotone picese are separeted into $$ \dfrac{1}{x_k^2}:=\dfrac{\pi}{2}+k\pi,\quad\forall k\in\mathbb{N}. $$

This makes the integral $$ \int_0^1|f'(x)|\mathtt{d} x\geq \sum_{k=1}^{\infty}x_k^2 =\sum_{k=1}^{\infty}\dfrac{1}{\dfrac{\pi}{2}+k\pi} $$ which is obviously infinite.

Some constanst in the calculates above may not be rigourous, but the idea is quite clear already. I hope this is right. :)

Answer 3

This is based on the answer given by @Zixiao_Liu plus the corrections I posted for it in comments under it. One gets that $$\int_0^1 |f'(x)|dx \ge \sum_{k=0}^\infty 2((4k+1)\pi)^{-1}= \infty;$$ for details, see in that other answer.