Problem 10991
(American Mathematical Monthly, Vol.110, February 2003) Proposed by R. Mortini (France).
For complex a, z ∈ D = {s : |s| < 1}, let F (a, z) = (a + z)/(1 + az) be a map of D onto D. Let ρ(a, b) = |(a − b)/(1 − ab)| be the pseudohyperbolic distance.
(a) Prove that there exists a function C : D → R+ so that ρ(F (a, z), F (b, z)) ≤ C(z)ρ(a, b) for every a, b, z ∈ D.
(b) Find the minimal value of C(z) for which this bound holds.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
It is well known that the map F (a, z) is a ρ-isometry of the unit disc D and ρ(z, a)2= |F (−a, z)|2= 1 − (1 − |z|2)(1 − |a|2)
|1 − az|2 . Moreover
F (−b, F (a, z)) = F (a, z) − b
1 − bF (a, z) = z + a − b(1 + az) (1 + az) − b(z + a)
= (1 − ab)z + (a − b) 1 − ab + (a − b)z
= eiθF (c, z)
where c = (a − b)/(1 − ab) and eiθ= (1 − ab)/(1 − ab). In particular, when a = b, this implies that F−1(b, z) = F (−b, z).
We claim that it is possible to take C(z) = (1 + |z|)/(1 − |z|) and this function is the best one.
Indeed, we will show that for every z ∈ D
C(z) := sup ρ(F (a, z), F (b, z))
ρ(a, b) : a 6= b ∈ D
= 1 + |z|
1 − |z|. (a) We first prove that the inequality ≤ holds. Note that
ρ(F (a, z), F (b, z)) = ρ(F (−b, F (a, z)), F (−b, F (b, z)))
= ρ(eiθF (c, z), z) = ρ(F (c, z), e−iθz)
≤ ρ(F (c, z), z) + ρ(z, e−iθz).
We consider the first term ρ(F (c, z), z):
ρ(F (c, z), z)2 = 1 −(1 − |z|2)(1 − |F (c, z)|2)
|1 − zF (c, z)|2 = 1 − (1 − |z|2)2(1 − |c|2)
|1 − zF (c, z)|2|1 + cz|2
= 1 −(1 − |z|2)2(1 − |c|2)
|1 + cz − z(z + c)|2 = 1 − (1 − |z|2)2(1 − |c|2)
|1 − |z|2+ 2iIm(cz)|2
= 1 − (1 − |z|2)2(1 − |c|2)
(1 − |z|2)2+ 4|Im(cz)|2 ≤ 1 − (1 − |z|2)2(1 − |c|2) (1 − |z|2)2+ 4|c|2|z|2
≤ (1 − |z|2)2+ 4|c|2|z|2− (1 − |z|2)2(1 − |c|2) (1 − |z|2)2+ 4|c|2|z|2
≤ (1 + |z|2)2|c|2
(1 − |z|2)2+ 4|c|2|z|2 ≤ 1 + |z|2 1 − |z|2
2
· |c|2.
We consider the second term ρ(z, e−iθz):
ρ(z, e−iθz) =
z − e−iθz 1 − eiθ|z|2
≤ |z|
1 − |z|2 ·
1 − e−iθ
≤ |z|
1 − |z|2 ·
1 − 1 − ab 1 − ab
≤ 2|z|
1 − |z|2 ·|Im(ab)|
|a − b| · |c| ≤ 2|z|
1 − |z|2 · |c|
The last inequality holds because
|Im(ab)| = |Im(ab − |b|2)| = |Im(b(a − b))| ≤ |b(a − b)| ≤ |b| · |a − b| ≤ |a − b|.
Therefore, summing up the inequalities involving the two terms, we have that ρ(F (a, z), F (b, z)) ≤ 1 + |z|2
1 − |z|2 · |c| + 2|z|
1 − |z|2· |c| ≤ 1 + |z|
1 − |z|· |c|.
Since |c| = ρ(a, b), we obtain
C(z) ≤1 + |z|
1 − |z|.
(b) We now prove that also the inequality ≥ holds. Since rotations around 0 are ρ-isometries then C(z) = C(|z|) and therefore we can assume that z ∈ [0, 1). If b = a then
ρ(F (a, z), F (a, z))2 = 1 −(1 − |F (a, z)|2)(1 − |F (a, z)|2)
|1 − F (a, z)F (a, z)|2
= 1 − (1 − |a|2)2(1 − z2)2
|(1 + az)2− (z + a)2|2 and
ρ(a, a)2=
a − a 1 − a2
2
=4|Im(a)|2
|1 − a2|2.
Let a = (1 − s) + is2 with s ∈ (0, 1) then a, b ∈ D and they converge to 1 along two symmetric parabolic arcs as s goes to 0+. Moreover
|1 − a2|2 = |2s − s2+ s4− 2is2(1 − s)|2= 4s2− 4s3+ 5s4+ o(s4), (1 − |a|2)2 = (2s − s2− s4)2= 4s2− 4s3+ s4+ o(s4)
and after some easy calculations
|(1 + az)2− (z + a)2|2 = |(1 + az) + (z + a)|2· |(1 + az) − (z + a)|2
= |(2 − s)(1 + z) − is2(1 − z)|2· s2|(1 − z) + is(1 + z)|2
= (4s2− 4s3+ s4)(1 − z2)2+ 4s4(1 + z)4+ o(s4).
Therefore
ρ(F (a, z), F (a, z))2
ρ(a, a)2 =
1 − (1 − |a|2)2(1 − z2)2
|(1 + az)2− (z + a)2|2
· |1 − a2|2 4|Im(a)|2
= 4s4(1 + z)4+ o(s4)
4s2(1 − z2)2+ o(s2)·4s2+ o(s2)
4s4 =(1 + z)2
(1 − z)2 + o(1).
Hence for all z ∈ D
C(z) = C(|z|) ≥ lim
s→0+
ρ(F (a, |z|), F (a, |z|))
ρ(a, a) = 1 + |z|
1 − |z|.
