Showing that function is not Lebesgue Integrable in $[0,1]$

Question

Is an exercise of my course of Measure and Integration.

Let $f:[0,1]\rightarrow\mathbb{R}$ such that: $$ f(x)= \left\{ \begin{array}{ll} x^2\sin(\pi/x^2) & \textrm{ if } 0<x\leq 1\\ 0 &\textrm{ if } x=0 \end{array} \right. $$ Show that $f'(x)$ exists for each $x\in[0,1]$ and $f'(x)$ is not Lebesgue Integrable in $[0,1]$

MY ATTEPMT:

Note:

$$ D^+f(0)= \lim_{\varepsilon\rightarrow 0^+}\sup\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0 $$

$$ D_+f(0)= \lim_{\varepsilon\rightarrow 0^+}\inf\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0 $$

$$ D^-f(0)= \lim_{\varepsilon\rightarrow 0^-}\sup\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0 $$

$$ D_-f(0)= \lim_{\varepsilon\rightarrow 0^-}\inf\{h\sin(\pi/h^2): h\in(0,\varepsilon)\}=0 $$ so, $f'(x)$ exists for each $x\in(0,1)$

To show that function $$f'(x)=2x\sin(\pi/x^2)-\frac{2\pi}{x}\cos(\pi/x^2)$$ is not L Integrable I'm trying show that $f'$ is not bounded a.e. but I'm not sure if is it and how to check this.

Answer 1

$f'(x) = \sin ( \frac{\pi }{{x}^{2}} ) x-\frac{2\pi \cos ( \frac{\pi }{{x}^{2}} ) }{x}$.

One can prove this along the following lines. Note that $\cos \theta \ge {1 \over \sqrt{2}}$ when $\theta \in [n-{1 \over 4}, n+{1 \over 4}] \pi$.

If $x \in I_n=[ {1 \over \sqrt{n+{1\over 4}}}, {1 \over \sqrt{n-{1\over 4}}}]$, we have $\int_{I_n} {1 \over x} \cos ( { \pi \over x^2} ) dx \ge {1 \over \sqrt{2}} \sqrt{n+{1\over 4}} ( {1 \over \sqrt{n-{1\over 4}}} - {1 \over \sqrt{n+{1\over 4}}} ) = {1 \over \sqrt{2}} ( \sqrt{{ 4 + { 1 \over n}\over 4 - { 1 \over n} }} -1 )$, and since $\sqrt{{ 4 + x \over 4 - x }} -1 \ge {1 \over 4} x$ for $x \in [0,1]$, we have (for $n$ suitably large), $\int_{I_n} {1 \over x} \cos ( { \pi \over x^2} ) dx \ge {1 \over \sqrt{2}} { 1 \over 4n}$. Since ${ 1 \over n}$ is not summable, we have the desired result.

Answer 2

Showing that $f'(x)$ exists:

If $x\in (0,1]$ then $f'(x)=2x\sin\left(\frac{\pi}{x^2}\right)-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)$

For $x=0$, $f'(x)=\lim\limits_{d\to 0^{+}}\frac{f(d)-f(0)}{d}=0$

So finally:$$f'(x)= \left\{ \begin{array}{ll} 2x\sin\left(\frac{\pi}{x^2}\right)-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right) & 0<x\leqslant 1\\ 0 & x=0 \end{array} \right.$$ What proves its existence.By the triangle inequality we obtain: $$\left|-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)\right|\leqslant\left|2x\sin\left(\frac{\pi}{x^2}\right)-\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)\right|+\left|-2x\sin\left(\frac{\pi}{x^2}\right)\right|$$ $2x\sin\left(\frac{\pi}{x^2}\right)$ is bounded and continuous in the interval $(0,1]$ besides $\lim\limits_{x\to 0^{+}}2x\sin\left(\frac{\pi}{x^2}\right)=0=f(0)$ what means that if $\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)$ is not L-integrable in the interval $(0,1]$ then $f'(x)$ is also not L-integrable. $$\int\limits_{(0,1]}\frac{2\pi}{x}\cos\left(\frac{\pi}{x^2}\right)dx=\frac{1}{\pi}\int\limits_{[\pi,+\infty)}\frac{\cos(v)}{v}dv$$ where $v(x)=\frac{\pi}{x^2}$. According to definitions presented here or here function is L-integrable if and only if is absolutely integrable. So the only thing we have to show is that $\int\limits_{[\pi,+\infty)}\left|\frac{\cos(v)}{v}\right|dv=+\infty$. $$\int\limits_{[\pi,+\infty)}\left|\frac{\cos(v)}{v}\right|dv\geqslant \int\limits_{[\pi,+\infty)}\frac{\cos(v)^2}{v}dv=\frac{1}{2}\int\limits_{[\pi,+\infty)}\frac{1+\cos(2v)}{v}dv$$ $\int\limits_{[\pi,+\infty)}\frac{\cos(2v)}{v}dv$ is convergent by Dirichlet's test for integrals and $\int\limits_{[\pi,+\infty)}\frac{1}{v}dv$ of course diverges what completes the proof.

Note: $f'(x)$ is R-integrable for any interval $[\varepsilon,1]$ where $0<\varepsilon\leqslant 1$. R-integrability implies L-integrability. Quoting from * : "If a real-valued function on [a, b] is Riemann-integrable, it is Lebesgue-integrable".