Problem in walter rudin RAC

Question

The function $f(x)=x^2\sin(\frac{1}{x})$ if $x\neq 0$ , $f(0)=0$.Then $f$ is differentiable at every point, but $\int_0^1|f^{'}(x)|dx=\infty.$

I proved $f$ is differentiable at every point. To prove $f$ is not integrable i integrate the derivative of $f$ but it was difficult to conclude.

If you know something about absolutely continuous functions, you can equivalently prove that $f$ is not AC... — Romeo, Jan 29 '14 at 13:39
So you want to prove that $f$ is not in $L^1$, or is it that $f'$ is not in $L^1$? — i like xkcd, Jan 29 '14 at 14:26

score 0 · Answer 1 · answered Jan 29 '14 at 15:16

Your derivative is $f' = 2x\sin(1/x) - \cos(1/x)$. You want to know whether

$$\int_0^\infty |f'(x)|dx<\infty$$

There would be two areas where integrability may fail : at 0 (because of 1/x) and at $\infty$.

-At $x\sim 0$ your |f'| is equivalent to cos(1/x)... which is integrable near 0!

-At $x\sim\infty$, $2x\sin(1/x)$ is equivalent to 2 and $\cos(1/x)$ is equivalent to 1. What do you conclude about integrability of $|f'|$?

1 Answers1