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In Pugh's Real Mathematical Analysis there is an exercise, marked with three stars (which denotes that the author doesn't know the answer), whether there exist a nonsmooth function $f : \mathbb{R} \to \mathbb{R}$ such that $f^2$ and $f^3$ are both smooth.

My question is not strictly about this exercise, but rather about cases when we weaken the hypotheses when only one of $f^2$ and $f^3$ are smooth.
The fact that the exercise comes with this hypotheses suggest we should be able to find those functions. For the case when $f^2$ need to be smooth we have a function $f(x) = x$ if $x$ is rational and $-x$ if $x$ is irrational, but what about the case when $f^3$ needs to be smooth?

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    $\begingroup$ How about $f(x) = \sqrt[3]{x}$? This is not smooth at $x=0$, at least. $\endgroup$ – copper.hat Oct 2 '12 at 0:54
  • $\begingroup$ The example would suggest multiplication, otherwise the example would have $f \circ f = f$. $\endgroup$ – copper.hat Oct 2 '12 at 1:20
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    $\begingroup$ The example would have $f \circ f$ being the identity map. $\endgroup$ – Nick Alger Oct 2 '12 at 1:53
  • $\begingroup$ Does the function have to be onto? Otherwise something simple like $f(1)=-1$, $f(x\neq 1)=0$ would do the trick $\endgroup$ – Nick Alger Oct 2 '12 at 2:14
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The idea is that it's easy to have a cusp that is "cured" by either squaring or cubing the function (or raising it to any other particular power), but more difficult to think of a case where both operations work. The simplest parametrized family of examples where $f$ is not smooth, but $f^{1+a}$ is, is probably $f(x; a)=\lvert x \rvert^{2/(1+a)}$. Choosing $a=1$ or $a=2$ gives the desired examples where $f^2$ and $f^3$ are smooth.

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For a resolution of the exercise, see

http://www.math.ucla.edu/~tao/whatsnew.html

The Feb 16,2007 comment contains a proof that if $f^2$ and $f^3$ are smooth, $f$ is smooth.

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How about: $$y=\sin^{1/3} x,\;\;x>0$$

(if by $f^3$ you mean $f(x)^3$ of course)

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  • $\begingroup$ No, i thinbk he mean both f^2 ans f^3. $\endgroup$ – Victor Oct 2 '12 at 1:51
  • $\begingroup$ @Victor not according to his second sentence? $\endgroup$ – user39572 Oct 2 '12 at 2:01

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