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POLYA004:  Functions with smooth square and cube (proposed by Xiaodong Zhang, 11/2/01).

Let f(x) be a function defined on an interval such that f^2(x) and f^3(x) are both smooth, i.e., have continuous derivatives of all orders. Is f(x) also smooth?

Discussion. [SCL, 11/12/01]: Is there a reason not to ask for  a generalization? Let k and m be positive integers with gcd(k,m)=1. Let f(x) be a function defined on an interval I such that g=f^k(x) and h=f^m(x) are both smooth, i.e., have continuous derivatives of all orders on I. Is f(x) also smooth?
Obviously, just asking that g be smooth is not enough, since the function x^{1/m} would fail at x=0 if m>1.
Let X={x in I: f(x)=0}, and let s and t be integers such that sk+tm=1. Then, it is not difficult to see that f=g^sh^t is smooth on I-X, since f^{(w)} is a polynomial in the variables g, h, g^{-1}, h^{-1}, g', h', g'', h'', ..., g^{(w)}, h^{(w)}. The problem then is: exactly what happens for a in  X?
Is it of use to think of f as (x-a)^qr(x), where r(a)<>0?

[JPE, 11/21/01]:  A colleague gave me an article (in French):
Henri Joris,  Une C-infini application non-immersive qui possede la propriete universelle des immersions, Arch.Math. 39 (1982) 269-277.
The author gives a long and very technical proof of the following theorem : Let  n, k be relatively primes integers. If f^n and f^k are smooth, then f is smooth.

[SCL, 11/26/01]: I wonder if Xiaodong was looking for a more elementary proof of the paper cited. Once you prove that it holds for the special case he proposed, then I can make it work for all relatively prime pairs as follows.
If f^k and f^m are smooth, then (as in POLYA007) there is some N so that f^i is smooth for all i>N. Now, that means that there is some q so that f^i is smooth for i of the form 2^j3^{q-j}. But then the (2,3) result would imply that f^i is smooth for i of the form 2^j3^{q-j-1}. By induction (going down), we'd then have f is smooth.
Thus, the (2,3) case looks to be at least as powerful as the (k,m) case, with gcd(k,m)=1.

[FR, 11/27/01]: Henri Joris, according to the review, proves the special case (2,3) and concludes that if every sufficiently large power is smooth, then the function is smooth. So he followed Stephen's outline.
I thought it might be fruitful to approach this problem by focusing on the cube. So the hypothesis is that a function and its cube root squared are smooth, and you want to show that the cube root is smooth. I believe I took care of the points where some derivative (possibly the zero-th) of the cube is nonzero, but I never could quite take care of the other points.
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Bibliography.

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