# $f^3,f^2$ are the cube and quadratic of f respectively and both infinite differentiable on $R$,how to show so is $f$

Let $f$ be a real function with domain R. If $f^2$ and $f^3$ are both infinite differentiable on R, How to prove $f$ be infinite differentiable on R

This problem which I have been thinking it for a long period,but I I can not find a accurate proof .So if somebody can help me, I will be appreciative very deeply.

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I assume that $f^2$ is $f\circ f$. Note that such composition could be constant even if $f$ is not continuous. This question is more suitable for math.stackexchange. –  Misha Mar 28 '13 at 23:47
There is a known "problem" question like this, where $f^2$ and $f^3$ are exponents (repeated multiplication) not compositions. –  Gerald Edgar Mar 29 '13 at 0:10
@Misha，f^2 means the square of f –  bo.gu Mar 29 '13 at 0:27
–  Andres Caicedo Mar 29 '13 at 2:32
Please do not close this question. The related question referred to by Andres indicates that this question is nontrivial. I am not sure what Gerald is referring to ("problem" question from where?). –  Todd Trimble Mar 29 '13 at 14:11

The following papers prove this:

MR0682456 Reviewed Joris, Henri Une C∞-application non-immersive qui possède la propriété universelle des immersions. (French) [A nonimmersive C∞ mapping having the universal property of immersions] Arch. Math. (Basel) 39 (1982), no. 3, 269–277.

MR0833407 Reviewed Duncan, John; Krantz, Steven G.; Parks, Harold R. Nonlinear conditions for differentiability of functions. J. Analyse Math. 45 (1985), 46–68. (Reviewer: Wiesław Pleśniak) 26E10 (58C25)

MR2179865 Reviewed Myers, Robert An elementary proof of Joris's theorem. Amer. Math. Monthly 112 (2005), no. 9, 829–831. (Reviewer: Clifford E. Weil) 26A24

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Thank you very much –  bo.gu Apr 17 '13 at 22:24
For what it's worth, I have a writeup of the Joris argument at math.ucla.edu/~tao/preprints/Expository/squarecube.dvi –  Terry Tao Apr 18 '13 at 4:17
@Terry Tao,I do not konw your article until now. –  bo.gu Apr 18 '13 at 8:37

If $f^2$ has a zero $p$ of finite order, then $f^3$ also has finite order at $p$, and vice versa. If $f^2$ has a zero of infinite order at $p$, then its square root $f$ does, as well, and hence $f$ is smooth at $p$. Otherwise, both $f^2$ and $f^3$ have finite orders, respectively $a$ and $b$, at $p$, and therefore $f=f^3/f^2$ is smooth of finite order $b-a$ at $p$. Therefore $f$ is always smooth.

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@katz Order of vanishing is pretty subtle for smooth functions. For example, what would you say is the order of vanishing of $\exp(-1/x^2) \sin(1/x)$ at $x=0$? –  David Speyer Apr 17 '13 at 17:58
@katz: Your statement "If f2 has a zero of infinite order at p, then its square root f does, as well, and hence f is smooth at p" is wrong. See section 2 of mat.univie.ac.at/~michor/roots.pdf. (But 2.5 is wrong - corrected in mat.univie.ac.at/~michor/roots2.pdf) –  Peter Michor Apr 17 '13 at 18:19
@katz But with that definition, vanishing of infinite order does not imply smooth. EG $\exp(-1/x^2) \sin (\exp(2/x^2))$ has discontinuous first deriviative. –  David Speyer Apr 18 '13 at 0:01
In mathoverflow.net/questions/105438 the explicit example of $f(x) := \sin^2(1/x) e^{-1/x} + e^{-2/x}$ (for $x>0$) and $f(x) := 0$ (for $x \leq 0$) is given for a smooth function vanishing to infinite order at the origin, such that the square root of $f$ fails to be smooth at the origin. –  Terry Tao Apr 18 '13 at 4:24
I don't feel bad! I feel enlightened :-) –  katz Apr 18 '13 at 9:05