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?

  • 1
    $\begingroup$ Remember that your examples should be positive. $\endgroup$
    – zhw.
    Jun 13, 2019 at 14:40
  • $\begingroup$ 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. $\endgroup$
    – lurker
    Jun 13, 2019 at 14:46

1 Answer 1


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.


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