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Let $f:\mathbb{R}\to\mathbb{R}$ be smooth and satisfy $f(-x)=f(x)$ for all x. Define $g:[0,\infty)\to\mathbb{R}$ by $g(x)=f(\sqrt{x})$. Is $g$ necessarily smooth at $0$?

I guess the answer is positive. $f$ being an even function implies that all its derivatives of odd order vanish at $0$, hence when substituting $\sqrt{x}$ in the Taylor series of $f$ around $0$, one obtains a power series in $x$. This is obviously not a proof, as most of the smooth functions are not analytic.

Showing that $f$ is differentiable is easy, and working a little harder one also shows that it is twice differentiable. It seems like it should work the same for derivatives of larger orders, and yet, I haven't found a satisfactory proof.

Amitai Yuval
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  • Well, it turns out that the answer is yes. My advisor showed me a nice nontrivial proof. If it is of interest to anyone, I can add a sketch of the proof. – Amitai Yuval Jul 25 '14 at 10:32
  • $g$ is defined on $[0,\infty)$: all calculations you did regarding "smoothness at $0$ are just in suitable intervals of the formk $[0,\epsilon)$, with $\epsilon>0$, am I right? – Avitus Jul 29 '14 at 10:29
  • Of course. When I say $g$ is smooth, I mean one can differentiate it from the right infinitely many times. – Amitai Yuval Jul 29 '14 at 19:00
  • @AmitaiYuval: I would indeed be interested in seeing the argument. Do you still have it around? :) – PhoemueX Oct 10 '17 at 15:31
  • @PhoemueX This is a challenge! I'll try and see if I can recover something. Will keep you posted. – Amitai Yuval Oct 10 '17 at 18:20
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    @AmitaiYuval: Thank you very much for your effort, I hope I did not trouble you too much. In the meantime I found the following MO thread in which this question is also discussed: https://mathoverflow.net/questions/77120/smoothness-of-f-sqrt-x . There is even a very nice solution using Fourier analysis :) I am sorry for first bothering you, and then finding a solution elsewhere, but maybe the MO thread is also interesting for you :) – PhoemueX Oct 11 '17 at 07:17

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