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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.

Dr Richard Clare
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    The issue is the first term. Start by isolating subintervals where $|\cos(1/x^2)|$ is not too small (say greater than $1/2$), and then consider what happens when you integrate $1/x$ on these subintervals. – Ian Jan 09 '19 at 02:33
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    Could you use the fundamental theorem of calculus, instead of taking the derivative itself? In that case, $\int_0^1 f'(x) dx = f(1) - f(0)$, so you're clearly getting undefined terms. Or is this too naive of an approach? – kkc Jan 09 '19 at 03:27
  • I tried that but I think we can't do that because we need |f'(x)| not $f'(x)$. – Dr Richard Clare Jan 09 '19 at 03:47
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    But f'(x) not integrable => |f'(x)| is not integrable – Dr Richard Clare Jan 09 '19 at 03:48
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    @kkc Hi kkc, unfortunately you can't use the fundamental theorem of calculus because $f$ is not absolutely continuous. Your argument is not even valid. See here. – Sam Wong Dec 03 '20 at 10:06
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    @kkc Or if you want to use the version of F.T.C as listed on Wikipedia(c.f. here), then in order to use it, you will need the Riemann integrability of $f^\prime$ which you don't have and you will never have. – Sam Wong Dec 03 '20 at 10:24

3 Answers3

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$f[0,1]\to \mathbb{R}$ 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.

MathRocks
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  • Correct if you consider Riemann integration (you mention it explicitely) but usually this integral of derivative counter example is introduced in the context of Lebesgue integration. For that, the second answer is better - once corrected in the way I show. – Ulysse Keller Dec 20 '23 at 23:53
  • BTW there is a duplicate of the orig. Q. mentioning explicitely Lebesgue (non-) integrability – Ulysse Keller Dec 21 '23 at 00:23
  • This is said duplicate: https://math.stackexchange.com/questions/836996/prove-the-derivative-of-x2-sin-1-x2-is-not-lebesgue-integrable-on-0-1 – Ulysse Keller Dec 21 '23 at 23:35
  • BTW the tags also show that the question was supposed to concern Lebesgue integration – Ulysse Keller Jan 08 '24 at 12:51
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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. :)

Zixiao_Liu
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  • This is not quite correct. The monotone pieces are not what you say because $x^2$ varies too. But take only those intervals among those you consider where the sine is increasing; in these f is increasing (although such an interval is NOT a complete monotone piece). But this saves your argument - I was looking for a proof myself and was glad to find yours, although it needs a correction - there will be just odd k's or even ones (I have yet to find out which) – Ulysse Keller Dec 20 '23 at 23:44
  • Looking in detail on my modifyed argument, I realize that one must restrict to intervals where the sine is increasing AND $\ge 0$ i.e. take that half of your interval (with incr. sine) where this extra condition is satisfied. This should finally save the argument, I hope – Ulysse Keller Dec 21 '23 at 00:13
  • To make things clearer, one must take intervals with sine decreasing (and positive) because the exponent -2 reverses the order of positive numbers – Ulysse Keller Dec 21 '23 at 00:34
  • I deleted the comment that was in this place. I had not realized that the comparison in your second line is not based on calculating the derivative (which is made in the first answer) but on using the classical fundamental theorem of analysis to each of the intervals mentioned (in which the derivative is continuous) which gives a difference of values of f, one of which is 0 ... this BTW is only the case following my corrections! This lets me think that you tried to reproduce an argument read somewhere (what is that source?) – Ulysse Keller Dec 21 '23 at 07:56
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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.

Ulysse Keller
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