MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 – Mar 29 '13 at 0:27
Related: – Andrés 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
up vote 12 down vote accepted

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

share|cite|improve this answer
Thank you very much – Apr 17 '13 at 22:24
For what it's worth, I have a writeup of the Joris argument at – Terry Tao Apr 18 '13 at 4:17
@Terry Tao,I do not konw your article until now. – 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.

share|cite|improve this answer
@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 (But 2.5 is wrong - corrected in – 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 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.