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