On separated Carleson sequences in the unit disc of ${\mathbb{C}}.$

The interpolating sequences for $H^{\infty}({\mathbb{D}}),$ the bounded holomorphic function in the unit disc ${\mathbb{D}}$ of the complex plane ${\mathbb{C}},$ {\small where characterised by L. Carleson by metric conditions on the points. They are also characterised by"dual boundedness"conditions which imply an infinity of functions. A. Hartmann proved recently that just one function in $H^{\infty}({\mathbb{D}})$ was enough to characterize interpolating sequences for $H^{\infty}({\mathbb{D}}).$ In this work we use the"hard"part of the proof of Carleson for the Corona theorem, to extend Hartman's result and answer a question he asked in his paper.}\ \par

Let D the unit disc in C and S a sequence of points in D.
To say that the sequence S is separated means that there is a δ > 0 such that To say that the sequence S is Blaschke means that a∈S (1 − |a|) < ∞ and this implies that the Blaschke product B S (z) := a∈S a − z 1 − az |a| a converges in D to a function in H ∞ (D), i.e. B S is holomorphic and bounded in D and is zero exactly on S. We shall also need the notion of Carleson measure. Let (ζ, h) ∈ T×(0, 1), and note W (ζ, h) := {z ∈ D :: 1 − ζz < h} the associated Carleson window. If ν is a borelian measure in D, we shall say that ν is Carleson if there is a constant C > 0 such that ∀ζ ∈ T, ∀h ∈ (0, 1), |ν| (W (ζ, h)) ≤ Ch. If ν is Carleson, we have the embedding Carleson theorem : in a continuous way. A sequence S will be a Carleson sequence if the canonical measure associated to it : µ S := a∈S (1 − |a|)δ a is a Carleson measure.
We shall use the following theorem, basis of the duality L. Carleson characterized these sequences for H ∞ (D) by the condition : in particular this sequence S is Blaschke and separated.
One can see easily that this condition is equivalent to the fact that S is dual bounded in H ∞ (D), which means : ∃C > 0, ∀a ∈ S, ∃ρ a ∈ H ∞ (D), ρ a ∞ ≤ C :: ∀b ∈ S, ρ a (b) = δ ab .
We just take ρ b (z) : a . So the metric condition which caracterises the interpolation is equivalent to the existence of an infinity of functions verifying the above conditions. A. Hartmann [3] showed that this can be reduced to a condition on only one function : Theorem 1 Let S be a separated Blaschke sequence in the unit disc D of C. There is a partition (S 1 , S 2 ) of S such that if there is a function f ∈ H ∞ (D) with f = 0 on S 1 and f = 1 on S 2 , then S is interpolating for H ∞ (D).
The aim of this work is to show that we can weakened the condition on the function and suppress the two conditions on the sequence, these conditions will be automatically fulfilled.
We shall extend this in the unit ball B in C n . In this setting N. Varopoulos [5] proved that if the sequence S is H ∞ (B) interpolating, then S is a separated Carleson sequence. P. Thomas [4] proved that if S is H p (B) interpolating, which is weaker than H ∞ (B), then S is a separated Carleson sequence. A better result was obtained in [2] : if the sequence S is just dual bounded in H p (B), then it is separated and Carleson. Again in the ball dual boundedness involves an infinity of functions and in this work we shall show that just one function is enough.
In the sequel d H (a, b) will be the hyperbolic distance in B.

Definition 2
We shall say that the partition We shall show that any discrete sequence in B (in fact in any metric space) admits a good partition (S 1 , S 2 ) and if the partition is not very good then we shall add points to it S ′ j ⊃ S j , j = 1, 2 to make the new pair

Now we can state the main theorem.
Theorem 2 If the sequence S is ultra-separated in the ball B, then it is Carleson and separated. In the unit disc of C the converse is also true.

Proofs.
The following lemma says that there are always good partitions for a discrete sequence.
Proof Take a point O ∈ X and a 1 ∈ S such that d(a 1 , O) is minimal, then b 1 ∈ S a nearest neighbour for the distance d of a 1 and define ϕ(a 1 ) = b 1 . Take a 2 a nearest neighbour of b 1 and define ψ(b 1 ) := a 2 ; if a 2 = a 1 we stop at this "perfect" pair (a 1 , b 1 ) with ψ(b 1 ) := a 1 . If not we continue with b 2 nearest neighbour of a 2 etc... We stop at a perfect pair. This way we get a branch B 1 finite or infinite. We put all the "a" in S 1 and all the "b" in S 2 .
If it remains points in S we have that the points in S\B 1 are far from the points in B 1 by construction. We take a point c in S\B 1 the nearest from O.
A) If all the nearest points from c are in B 1 , which may happen, we take one of them, d, now if d is in S 1 , we put c in S 2 and we set ψ(c) := d. If d is in S 2 , we put c in S 1 and we set ϕ(c) := d. This completes B 1 and we start all again. B) If c has a nearest neighbour which is not in B 1 , we start a new branch B 2 etc... A new point may have its nearest neighbour in B 1 or in B 2 , etc... Then we put it in B 1 or in B 2 , ... as in the step A).
We continue this way in order to exhaust S. The S 1 part is all the "a" and S 2 is all the "b". Then S is a bipartite graph with components S j , j = 1, 2 on which the two applications ϕ, ψ are well defined.

Completion of the partition.
If the hyperbolic distance between a and ϕ(a) or between b and ψ(b) can be arbitrarily big, we shall have to add points to S 1 or to S 2 .
Let M 1 := sup a∈S 1 d(a, ϕ(a)), M 2 := sup b∈S 2 d(b, ψ(b)) ; if M := max (M 1 , M 2 ) < ∞ then we have nothing to do. If M = ∞, then we choose an integer m, for instance m = 6, and we shall add points the following way.
For a ∈ S 1 such that d(a, ϕ(a)) > m we add a point b to S 2 such that b ∈ (a, ϕ(a)) and d(a, b) = 1 ; of course we set ϕ(a) = b and we have ψ(b) = a, because there is no point of S nearest b than a. We do the same for b ∈ S 2 if d(b, ψ(b)) > m, we add a point a in S 1 on the segment (b, ψ(b)) such that d(a, b) = 1 and we set ψ(b) := a and of for the same reason than above we have ϕ(a) = b.
By adding these points we have new S 1 and S 2 such that the new M is bounded by m. Let us show now the lemma.
Lemma 2 If the sequence S is ultra-separated in H ∞ (B) then it is separated.
Proof. The existence of the ultra-separating f gives easily that the sequence S is separated. Let a, b two points of S. Suppose that a ∈ S 1 ; then if b is in S 2 , the ultra-separating function f is such that |f (a)| < δ and |f (b)| ≥ 1, hence a and b are separated.
If b ∈ S 1 , let c = ϕ(a) ∈ S 2 , because c is the nearest neighbour of a, we have d(a, b) ≥ d(a, c), hence, because a and c are separated, this is the same for a and b.
On the other hand, taking powers of f, we can manage to have δ as small as we wish.

Proof of the sufficiency.
We have ∀a ∈ S 1 , |f (a)| ≤ δ hence because f ∈ H ∞ (B) we have |f | ≤ 2δ in a hyperbolic neighbourhood of a, for instance in B a (ǫ) := {z ∈ C n :: |Φ a (z)| < ǫ}, where Φ a is the automorphism of the ball exchanging a and 0. Take ǫ small enough in order that these pseudo-balls are disjoint, this is possible because the sequence is separated.
Hence the derivative of f is big in a "tube" linking B a (ǫ) to B b (ǫ). Precisely let γ(t), t ∈ [0, 1] be a smooth curve in B such that γ(0) = α ∈ B a (ǫ) and γ(1) = β ∈ B b (ǫ), we have Let dµ(z) := c αβ d |z| ρ(z) be the measure on the curve γ, with This measure µ is a probability measure and we have by Hölder : We get, with our parametrisation, then A× |γ| ≤ γ dµ(z) ≤ B× |γ| , where |γ| denotes the euclidean length of the curve γ. Finally 3 On the geometry of the unit ball in C n .
Let a ∈ B and define the pseudo-ball associated to it: B a (ǫ) := {z ∈ B :: |Φ a (z)| < ǫ}. We have, by the non-isotropy of the geometry in the ball, the following lema.
Lemma 1 Let S a be the euclidean sphere of center 0 and radius |a| . The area of S a ∩ B a (ǫ) is equivalent to ǫ n (1 − |a|) n .
Proof. Well known.

A construction of pseudo-geodesics.
Let a, b ∈ B be two points in the ball ; we shall connect them by a smooth curve G(a, b)) the following way : • consider the 2-real plane P containing (0, a, b) ; it cut the ball as a real disc, then see it as a complex disc and take the geodesic, in the hyperbolic metric of this disc, passing through a, b. It is the arc of a circle orthogonal to the boundary passing through a, b.
Take G(a, b) to be the part of this arc between a and b. We have two clear facts: • the minimum of the distance ρ(z) to the boundary of the ball, for z ∈ G(a, b), is attained at a or b ; • the euclidean length of G(a, b) is smaller than π times the length of the straight line between a and b. Integrating in the cone C b linking the vertex b to the points in B a (ǫ) by use of these pseudo-geodesics, we get because of the anisotropic geometry of these pseudo-balls The same in the cone C a of vertex b and containing B a (ǫ), sequence. Because S ′ 1 := S ∪ A is separated, S ′ 1 is still an interpolating sequence. The same for S 2 , hence S ′ := S ′ 1 ∪ S ′ 2 is an interpolating sequence. Now we can choose f ∈ H ∞ (D) such that f = 0 on S 1 and f = 1 on S ′ 2 to be done.

Remark 1
In the ball of C n , n ≥ 2, there is a big difference between interpolating sequences and separated Carleson ones. For instance we can take a maximum net S of separated points in the unit disc and consider this sequence S in the unit ball B of C 2 : S is Carleson in B and S is separated [1] but S is not H ∞ (B) interpolating because if so, S would be interpolating also for H ∞ (D)and this is false by the characterization of Carleson.
Remark 2 This result is not completely satisfactory because we have to add points to the good partition to make it a very good one in case where there are points arbitrarily far from there nearest neighbour. So the question : is this result optimal ? I.e. is there a separated sequence S where there are points arbitrarily far from there nearest neighbour, S not interpolating and such that there is a function f ∈ H ∞ (D), |f | S 1 ≤ δ < 1 and |f | S 2 ≥ 1 with (S 1 , S 2 ) a good partition of S ?