Absolutely continuous on $[-1,1]$ of a function

Question

Show that $f(x)=x^2 \cos\left(\dfrac{\pi}{2x}\right)$ when $x\neq 0$, and $0$ when $x=0$, is absolutely continuous on $[-1,1]$.

I'm honestly not sure how to get this one off the ground. I thought about maybe trying to prove that it's Lipschitz, but 1) I'm not even convinced that's true, and 2) if it is true, I can't get that off the ground either.

Any thoughts would be greatly appreciated.

Thanks in advance.

Answer 1

To prove that $ f $ is Lipschitz at $ [-1,1] $ , You just need to show that the derivative $ f' $ is bounded at $ [-1,1] $.

For $ x=0 $,

$$f'(0)=\lim_{x\to 0}\frac{f(x)-0}{x}$$ $$=\lim_{x\to 0}x\cos(\frac{\pi}{2x})=0$$

For $ x\ne 0$, $$f'(x)=2x\cos(\frac{\pi}{2x})+\frac{\pi}{2}\sin(\frac{\pi}{2x})$$ $$\implies |f'(x)|\le 2|x|+\frac{\pi}{2}\le 2+\frac{\pi}{2}$$

Answer 2

Since [-1, +1] is closed, any continuous function is absolutely continuous, so you only need to show continuity. And all points other than x=0 are obviously continuous, so you only need to care about x=0. And there the absolute value of f(x) doesn’t exceed x^2.

Answer 3

$f$ is differentiable and so, $f'$ is bounded in $[-1,1]$

For $x=0$,

$$\frac{f(h)-f(0)}{h}=h\cos\big(\tfrac{2}{\pi h}\big)\xrightarrow{h\rightarrow0}0$$

For $x\neq0$

$$f'(x)=2x\cos\big(\frac{2}{\pi x}\big)+\frac{2}{\pi}\sin\big(\frac{2}{\pi x}\big)$$

Thus $|f'(x)|\leq 2+\frac{2}{\pi}$ for all $|x|\leq 1$. Then, for any finite collection of closed pairwise disjoint intervals $[a_1,b_1],\ldots, [a_N,b_N]$,

$$ \sum^N_{j=1}|f(b_j)-f(a_j)|\leq K\sum^K_{j=1}(b_j-a_j)$$

for some constant $K>0$. For ay $\varepsilon>0$, let $\delta=\varepsilon/M$. If $\sum^N_{j=1}b_j-a_j\leq \delta$, then $$\sum^n_{j=1}|f(b_j)-f(a_j)|<\varepsilon$$