Stein 3.12

$\textbf{Exercise 12}$ : Consider the function $f(x)=x^2\sin(\frac{1}{x^2})$ for $x \neq 0$ and $f(0)=0$. Show that $f'$ exists for every $x\in[-1,1]$ but $f' \not \in{L^1[-1,1]}$
$\textit{Proof}$ : For $x \neq 0$, $f'(x)=2x\sin(\frac{1}{x^2})-\frac{2}{x}\cos(\frac{1}{x^2})$...an algebraic combination of continuous functions defined for all nonzero $x$. To check $f'(0)$, we may look at the difference quotient directly: $$\lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim_{h \rightarrow 0} \frac{f(h)}{h}=\lim_{h \rightarrow 0} h\sin\left(\frac{1}{h^2}\right)=0$$ So, $f'$ exists everywhere in $[-1,1]$.
Now consider a collection of closed intervals of $(0,1]$ defined by the points $\{\alpha_k\}_{k=1}^\infty$ and $\{\beta_k\}_{k=1}^\infty$ where $\alpha_i \neq \beta_j \quad \forall i,j \in{\mathbb{Z}^+}$ On any one of these closed intervals, $f'$ is continuous, so we may apply the fundamental theorem of calculus which gives $$\int_\alpha^\beta |f'(x)|dx=x^2\sin\left(\frac{1}{x^2}\right)_{\alpha}^{\beta}$$ Let $\alpha_k$ be s.t. $f(\alpha_k)=\alpha_k$ and $\beta_{k}$ s.t. $f(\beta_{k})=0$ i.e. the roots and local maximum values of $f$. Then, $\alpha_k= \pm \sqrt{\frac{2}{\pi (2k-1)}}$ and $\beta_k = \pm \sqrt{\frac{1}{\pi k}}$. Then, $$\int_{-1}^1 |f'(t)|dt \geq \int_0^1 |f'(t)|dt \geq \sum_{k=1}^\infty \int_{\sqrt{\frac{2}{\pi (2k-1)}}}^{\sqrt{\frac{1}{\pi k}}}|f'(t)|dt = \sum_{k=1}^\infty x^2\sin\left(\frac{1}{x^2}\right)_{\sqrt{\frac{2}{\pi (2k-1)}}}^{\sqrt{\frac{1}{\pi k}}}$$ $$\geq \sum_{k=1}^\infty\frac{2}{\pi (2k-1)}= \frac{2}{\pi} \sum_{k=1}^\infty \frac{1}{2k-1} \geq \frac{2}{\pi}\sum_{k=2}^\infty\frac{1}{3k} = \frac{2}{3\pi} \sum_{k=2}^\infty \frac{1}{k} = +\infty$$ $\therefore$ $f' \not \in{L^1}$ $\blacksquare$