# If $f$ is absolutely continuous $\sqrt(f)$ may not be

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?

• Remember that your examples should be positive.
– zhw.
Jun 13, 2019 at 14:40
• There's not a more general way. You just need one counter example under the specified conditions. Of course unless the theorem is false, in which case you'd need to prove $\sqrt{f}$ is absolutely continuous in the general case whenever $f$ is, again under the specified conditions. Intuitively, $\sqrt{}$ seems well behaved enough on the positive Reals that I can't yet think of a potential counter example. Jun 13, 2019 at 14:46

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.