Problem 11070
(American Mathematical Monthly, Vol.111, March 2004) Proposed by R. Tauraso (Italy).
Let f and g be two commuting analytic maps from a non-empty open connected set D ⊂ C into D. Suppose that z0 ∈ D be a fixed point of both f and g, and that neither f′(z0) nor g′(z0) is a root of unity. Suppose also there exists an integerN ≥ 1 such that f(k)(z0) = g(k)(z0) = 0 for 1 ≤ k ≤ N − 1, while f(N )(z0) = g(N )(z0) 6= 0. Prove that the restriction of f and g to D are equal.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We will use the Fa`a di Bruno’s formula for the nth-derivative of the composition of two functions:
letting fn= f(n)(z0) and gn = g(n)(z0) for n ≥ 1 then
(f ◦ g)(n)(z0) = Bn(f1, g1, · · · , fn, gn) where Bn is the Bell polynomial of degree n
Bn(x1, y1, · · · , xn, yn) =X
π(n)
n!
r1!r2! · · · rn!(1!)r1(2!)r2· · · (n!)rn · xry1r1y2r2· · · ynrn.
The sum runs over all partitions π(n) of the integer n: n = r1+ 2r2+ · · · + nrn, rk denotes the number of parts of size k, and r = r1+ r2+ · · · + rn is the total number of parts.
By hypothesis fk= gk for k = 1, . . . , N and deriving the identity f ◦ g = g ◦ f we get useful relations that will help us to prove by induction that fn= gn for all n that is f ≡ g:
(f ◦ g)(n)(z0) = Bn(f1, g1, · · · , fn, gn) = Bn(g1, f1, · · · , gn, fn) = (g ◦ f )(n)(z0).
First we consider the case N = 1 and assume that fk = gk for k = 1, . . . , n − 1 and n ≥ 2. It easy to see that the monomials in Bn(f1, g1, · · · , fn, gn) which depend on fn or gn are precisely
f1gn and fng1n.
Gathering all the other monomials in the polynomial Pn(f1, g1, · · · , fn−1, gn−1) we have that (f ◦ g)(n)(z0) = f1gn+ Pn(f1, g1, · · · , fn−1, gn−1) + fng1n
and similarly by swapping f and g we obtain
(g ◦ f )(n)(z0) = g1fn+ Pn(g1, f1, · · · , gn−1, fn−1) + gnf1n. Therefore
f1gn+ Pn(f1, g1, · · · , fn−1, gn−1) + fngn1 = g1fn+ Pn(g1, f1, · · · , gn−1, fn−1) + gnf1n, and since fk= gk for k = 1, . . . , n − 1 we get
g1(g1n−1− 1)fn= f1(f1n−1− 1)gn and we conclude that fn= gn because f1= g16= 0 is not a root of unity.
Now suppose that N > 1 and assume that fk = gkfor k = 1, . . . , N +m−1 with m ≥ 1. Let’s consider the Bell polynomial Bn(f1, g1, · · · , fn, gn) for n = N2+ m. Since fk = gk= 0 for 1 ≤ k ≤ N − 1 it suffices to take into account the monomials
frg1r1g2r2· · · gnrn
with r1= r2= · · · = rN −1= 0 and r ≥ N . Then Xn
k=1
rk = Xn k=N
rk= r and Xn k=1
krk= Xn k=N
krk = n.
and by subtracting N times the first sum from the second one we get
0 ≤ Xn k=N
(k − N )rk =
n−NX
k=1
krN+k= n − rN ≤ n − N2= m
This implies that rN+k= 0 for k > m and r ≤ n/N = N + m/N < N + m. Moreover, if rN+m6= 0 then rN+m = 1, rN+k= 0 for k = 1, · · · , m − 1, rN = N − 1, and r = rN + rN+m= N . Thus we have selected the monomial
fNgNN −1gN+m.
On the other hand, if rN+m = 0 then the corresponding monomials depend on fk and gk for k = N, · · · , N +m−1. We gather these monomials in the polynomial Qn(fN, gN, · · · , fN+m−1, gN+m−1).
Thus
(f ◦ g)(n)(z0) = cfNgNN −1gN+m+ Qn(fN, gN, · · · , fN+m−1, gN+m−1) and in the same way
(g ◦ f )(n)(z0) = cgNfNN −1fN+m+ Qn(gN, fN, · · · , gN+m−1, fN+m−1)
where c is a positive coefficient. Since fk = gk for k = 1, . . . , N + m − 1, after equating we get fNgN −1N gN+m= gNfNN −1fN+m
and we conclude that fN+m= gN+mbecause fN = gN 6= 0.
Remark. The condition that the first derivatives are not roots of unity is necessary: the maps f (z) = zn+1+ ωz and g(z) = ωz with ωn= 1 commute, f (0) = g(0) = 0 and f′(0) = g′(0) = ω 6= 0 but f 6≡ g
If |f′(z0)| = |g′(z0)| 6= 1 then the statement can be proved also by the existence of a common local conjugation due to Schr¨oder and B¨ottcher theorems. This approach is much more difficult when z0
is an irrationally indifferent fixed point.
