The problem reads:
Prove that if $ f:[0,1]\rightarrow(0,\infty) $ is absolutely continuous $ \sqrt{f} $ may not be.
I am having trouble figuring out how to show this. I found that $x^2\sin\left(\frac{1}{x^2}\right)$ is not absolutely continuous, but then I need to show that $\left[x^2\sin\left(\frac{1}{x^2}\right)\right]^2$ is absolutely continuous and I don't think that it is. Is there a more general way to show this or is there a counterexample that works?
Best Answer
As stated, the conclusion is false, since $f([0,1])$ would then be an interval $[a,b]$ with $a>0.$ In this case the MVT shows
$$|\sqrt {f(y)} - \sqrt {f(x)}| \le \frac{1}{2\sqrt a}|f(y)-f(x)|$$
and the absolute continuity of $\sqrt f$ follows.
So the hypothesis should probably be $f:[0,1]\to [0,\infty).$ Here it looks like $x^2\sin(1/x^2)+x$ would be an example.