**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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