Disks extremal with respect to interpolation constants

Publicacions Matem`atiques, Vol. 44 (2000), 119–133 DISKS EXTREMAL WITH RESPECT TO INTERPOLATION CONSTANTS Nguyen Van Trao Abstract We define a function µfrom the set of sequences in the unit ball to R∗ +by taking the greatest lower bound of the reciprocal of the interpolating constant of the sequences of the disk which get mapped to the given sequence by a holomorphic mapping from the disk to the ball. Its properties are studied in the spirit of the work of Amar and Thomas. 0. Introduction Much attention has been given to the notion of interpolating sequences since L. Carleson introduced the concept in [C] and used it to establish a generalization of the Pick-Schwarz theorem (see e.g., [C], [V], [B], [B-C-L]). In [A-T] E. Amar and P. J. Thomas gave a new approach to the study of interpolating sequences in the unit ball Bnof Cn, by considering maps from the unit disk to the ball, constrained to reach the points in the sequence, and extremal in the sense that the preimages in the disk should minimize the constant used by L. Carleson in his characterization of interpolating sequences in the disk [C]. The aim of this paper is to continue this approach using the interpolation constant of the sequence of preimages. Although more technical to handle (there is no explicit formula, unlike in the work of Amar and Thomas), it should be more meaningful for the original problem of interpolation. We now present more precisely the content of the paper. Let a= {ak}k∈Na sequence of points in a domain Ω in Cn, we say that ais an interpolating sequence if, for any bounded sequence v={vk}k∈N, there is a bounded holomorphic function on Ω, fv, such that fv(ak)=vk. The constant of interpolation of ais the smallest number MΩ awith ∀v∈∞(N), fv∞≤MΩ av∞. There always is a (finite) constant of interpolation when the sequence is interpolating. In particular, this makes sense for any finite sequence. 120 N. Van Trao E. Amar and P. J. Thomas defined for a={ak}k∈N⊂Bn, δBn(a) := inf k j:j=k dBn G(ak,a j), where dBn Gis the Gleason distance in Bn(in dimension 1, we denote δD(a) instead of δB1(a)), and ρ(a) := inf δD(α):α={αk}⊂D,∃ϕ∈Hol(D,Bn) such that ϕ(αk)=ak,∀k∈N, where Dis the open unit ball in C. We study the following analogue. Definition. µ(a) := inf 1 MD α :ααjj∈N⊂D,∃ϕ∈Hol(D,Bn) such that ϕ(αk)=ak,∀k∈N, where MD αis the interpolation constant of α⊂D. Of course, this also applies to the special case of finite sequences, with 1 ≤j≤N. It is easy to see that µ(a)≥1 MBn a . Maximizing MD αamong the sequences α⊂Dwhich are preimages of a⊂Bnyields mappings —if there are any— that are “the tightest” in the sense that it is more difficult to hit the points of awith a map ϕfrom the disk to the ball when the preimage sequence αallows the interpolation by H∞functions of fewer values, which is the intuitive meaning of a larger value of MD α. We also know by Carleson [C], that for α⊂D,MD α≥1 δD(α),thus µ(a)≤ρ(a). We want to study the function a−→ µ(a) in the same way that ρwas studied in [A-T] and [T1]. Namely, we will prove the following. Theorem 2.7 (monotonicity). Let a=aj1≤j≤Nbe a sequence of points in Bnand abe a subsequence of a. Then µ(a)≥µ(a). Theorem 2.8 (continuity). µ(a)depends continuously on ain the set of finite sequences. Theorem 3.1 (approximation by finite sequences). Let a=ak, k∈Z∗ +⊂Bn. Then µ(a) = lim N→∞ µ{aj}1≤j≤N= inf N∈Z∗ + µ{aj}1≤j≤N. Interpolation constants 121 The next theorem shows that there is a large set of sequences for which our infimum is in fact a minimum. Theorem 3.5. For any N∈Z∗ +there exists an open set Uof sequences awith #a=Nsuch that for every a∈Uthere is α=αj1≤j≤N⊂D such that ϕ(αj)=ajfor 1≤j≤Nand µ(a)= 1 MD α . Finally, we connect our definition with a more classical extremal problem, which shows that some form of uniqueness holds for the extremal mappings: when the αjare given, the map ϕ0is unique. Theorem 3.6. Let a=aj1≤j≤Nbe a sequence of points in Bn. If there exists a holomorphic map ϕ0from Dto Bnand α=αj1≤j≤N⊂D such that ϕ0(αj)=ajfor 1≤j≤Nand µ(a)= 1 MD α , then ϕ0is a solution of the extremal problem inf ψ∞:ψ∈H(D,Cn),ψ(αj)=ajfor 1≤j≤N. We remark that some of the methods used here can be applied (indeed, in a simpler way) to simplify the proofs of [T1]. 1. Definitions and notations For z,w ∈Cn,z·w:= n  j=1 zjwj,|z|2:= z·z Bn:= z∈Cnsuch that |z|<1,D:= B1. Let Ω be a domain in Cnand z,w ∈Ω, the Gleason (or invariant) distance dΩ G(z,w) is given by dΩ G(z,w) = sup{|f(z)|:f∈H∞(Ω) such that f(w)=0,f∞≤1}. We know that 1−dBn G(z,w)2:= (1 −|z|2)(1 −|w|2) |1−z·w|2, in particular, dD G(z,w)= |z−w| |1−zw|. 122 N. Van Trao For λ∈Dlet us denote by Φλthe M¨obius map of Ddefined as follows: Φλ(ζ):= λ−ζ 1−λζ . (See [G1], [R].) Finally, for fa bounded holomorphic function from Dto Cnwe write f∈H∞ n(D),f2 ∞:= sup z∈D|f1(z)|2+···+|fn(z)|2. 2. Continuity of the function µin the finite case For the beginning we need some easy lemma. Lemma 2.1. For any β1,... ,β p∈∂D,δ > 0, there exists a function hδ∈A(D)(i.e continuous up to the boundary) with hδ∞=1, such that (1) hδ(βj)=1,1≤j≤p; (2) |hδ(ζ)|≤δ, ∀ζ∈D\ p  j=1 D(βj,δ). Proof: Remark that every finite set in ∂Dis a peak-interpolation set. Let g∈A(D) be such that g(βj)=1,1≤j≤p, and |g(ζ)|<1, ∀ζ∈D\{βj,1≤j≤p}.ForNlarge enough, gNsatisfies the required properties. Lemma 2.2. With the hypotheses of Lemma 2.1, there exists f∈A(D), f∞≤1such that (1) f(βj)=0,1≤j≤p; (2) |f(ζ)−1|≤δ, ∀ζ∈D\ p  j=1 D(βj,δ). Proof: Take δ1∈(0,δ] and g=hδ1as in Lemma 2.1. Now, for δ2>0, let f(ζ)=Φ1−δ2(1 −δ2)g(ζ)=(1 −δ2)(1 −g(ζ)) 1−(1 −δ2)2g(ζ), then f(βj) = 0 and f∞≤1 by construction. Since f(ζ)−1=−δ21+(1−δ2)g(ζ) 1−(1 −δ2)2g(ζ) for |g(ζ)|≤δ1,|f(ζ)−1|≤δ21+δ1(1 −δ2) 1−δ1(1 −δ2)2≤δ2 1+δ1 1−δ1 <δfor an appropriate choice of δ2. Interpolation constants 123 Lemma 2.3. Let α={αj,j∈J}and J=J1∪J2with J1∩J2=∅, J1=∅,J2=∅. Set α(1) ={αj,j∈J1},α(2) ={αj,j∈J2},M1= MD α(1) ,M2=MD α(2) . Assume further that α(1) ⊂D(0,r)⊂⊂ D, and there exist βj∈∂D, for j∈J2, such that |αj−βj|≤δ,j∈J2. Fix M>0. Then for any g1,g 2∈H∞(D,Cn)and g1∞≤M,g2∞≤ M, and ε>0, there exists δ0such that for any δ<δ 0and αas above, there exists f∈H∞(D,Cn)and f(αj)=g(αj),j∈J,=1,2, and f∞≤max(g1∞,g2∞)+ε. Proof: By Lemma 2.2, there exists a function h1associated to δ1, which we choose smaller than 1−r 2and ε1, where ε1>0 is to be chosen later. Choose δ2<δ 1small enough so that δ2≤ε1and min j|ζ−βj|≤δ2implies that |h1(ζ)|≤ε1. Let h2be the function obtained from Lemma 2.1 applied with δ=δ2. We pick δ0small enough so that min j|ζ−βj|≤δ0 implies that |h2(ζ)−1|≤ε1. Now consider the function f1(ζ)=h1(ζ)g1(ζ)+h2(ζ)g2(ζ). For min j|ζ−βj|≤δ2,|h1(ζ)|≤ε1,|h2(ζ)|≤1, so f1(ζ)≤ε1g1∞+g2∞≤(1 + ε1) max(g1∞,g2∞). For min j|ζ−βj|≥δ2,|h2(ζ)|≤δ2≤ε1,|h1(ζ)|≤1, so f1(ζ)≤g1∞+ε1g2∞≤(1 + ε1) max(g1∞,g2∞). On the other hand, for j∈J1, f1(αj)−g1(αj)=(h1(αj)−1)g1(αj)+h2(αj)g2(αj) ≤ε1g1∞+ε1g2∞≤2ε1max(g1∞,g2∞); for j∈J2, f1(αj)−g2(αj)=h1(αj)g1(αj)+(h2(αj)−1)g2(αj) ≤ε1g1∞+ε1g2≤2ε1max(g1∞,g2∞). We can find a map f2such that f2(αj)=g(αj)−f1(αj), ∀j∈J, =1,2, and f2∞≤2ε1√nMD αmax(g1∞,g2∞) (where MD αis the constant of interpolation of α, and it is well known MD αis bounded by  M, which is depending only on M1,M2,and (1−r)/2, supposing as we may that δ≤δ0≤(1 −r)/2). Setting f=f1+f2, we have f(αj)=g(αj), 124 N. Van Trao ∀j∈J,=1,2, and f∞≤(1 + ε1) max(g1∞,g2∞)+2ε1√n Mmax(g1∞,g2∞) = max(g1∞,g2∞)+ε1max(g1∞,g2∞)(1+2 √n M) ≤max(g1∞,g2∞)+ε1M(1+2 √n M). Taking ε1=ε M(1 + 2√n M), the property is proved. The next result follows immediately from Lemma 2.3. Corollary 2.4. Suppose α={αj,j∈J}and J=J1∪J2with J1∩J2= ∅,J1=∅,J2=∅. Set α(1) ={αj,j ∈J1},α(2) ={αj,j ∈J2}, M1=MD α(1) ,M2=MD α(2) . Assume further that α(1) ⊂D(0,r)⊂⊂ D, and there exist βj∈∂D, for j∈J2, such that |αj−βj|≤δ,j∈J2. Then for any ε>0, there exists δ0such that for any δ<δ 0and αas above, max(M1,M 2)≤MD α≤max(M1,M 2)+ε. Proof: Take n= 1; given values {vj,j ∈J}⊂Dto interpolate, we know there exist functions g1,g 2∈H∞(D), g(αj)=vj,j∈Jand g∞≤M,=1,2. Applying Lemma 2.3 we have the property. The following lemma is our main technical tool, inspired to some extent by the work of Globevnik [G2]. Lemma 2.5. Let a=aj1≤j≤Nbe a sequence of points in Bn. Suppose that there exist a sequence α=αj1≤j≤Nin Dand a holomorphic map ϕ:D→Bnsuch that aj=ϕ(αj)for 1≤j≤N.LetaN+1 be a point of Bnand ε>0. Then there exist α=α j1≤j≤N⊂Dand β∈Dand holomorphic map ψfrom Dinto Bnsuch that ψ(α j)=aj,ψ(β)=aN+1 and MD α∪{β}≥MD α−ε. Proof: For all r<1 such that α/r ⊂D,MD α/r ≤MD α, and we know by the continuity of the interpolation constant that MD α/r tends to MD αas rtends to 1. Let max(|α1|,... ,|αN|)<r<1 and s= max |a1|,... ,|aN|,|aN+1|, sup Dr|ϕ|. Apply Lemma 2.3 with g1(ζ)=ϕ(rζ), g1∞= sup Dr|ϕ|≤ s<1; g2(ζ)=aN+1,∀ζ∈D,g2∞=|aN+1|≤s;α j=αj/r ∈D for j∈J1:= {1,... ,N};J2:= {N+1}, and the additional one-point sequence is chosen as α N+1 := 1 −δ(ε1), where ε1=ε1(r):=1−s, and δ(ε1)<δ 0,where δ0is given by Lemma 2.3. Interpolation constants 125 Take Ψ(ε1)=fin Lemma 2.3; then we have Ψ(ε1)(α j)=g1(α j)=aj, 1≤j≤N, and Ψ(ε1)(α N+1)=g2(α N+1)=aN+1;Ψ(ε1)∞≤s+ε1= 1. Now MD α≥MD α/r, which can be made arbitrarily close to MD αby the considerations at the beginning of the proof. Our lemma is completely proved. We now remark that the solution to our extremal problem is not modified if we only require the map ϕto hit only a subset of the sequence {aj}. This would also hold for the original extremal problem in [A-T]. The quantity µ(a) is defined in the introduction. Theorem 2.6. Let a=aj1≤j≤Nbe a sequence of points in Bn. Then µ(a) = inf 1 MD α :∃J⊂{1,2,... ,N},∃ϕ∈H(D,Bn) and α=αjj∈J⊂Ds.t. ϕ(αj)=aj,∀j∈J. Proof: By using Lemma 2.5 repeatedly to add one point at a time to the set {aj,j∈J},wehave µ(a) = inf 1 MD α :α⊂D,∃ϕ∈H(D,Bn) s.t. ϕ(αj)=aj,1≤j≤N ≤inf 1 MD α :∃J⊂{1,2,... ,N},∃ϕ∈H(D,Bn) and α={αj}j∈J⊂Dsuch that ϕ(αj)=aj,∀j∈J}. The converse inequality is obvious and, hence Theorem 2.6 is proved. Applying the above theorem we have the following. Theorem 2.7. Let a=aj1≤i≤Nbe a sequence of points in Bnand abe a subsequence of a. Then µ(a)≥µ(a). Now we consider the continuity of the map a−→ µ(a). Theorem 2.8. µ(a)depends continuously on ain the set of finite sequences. Proof. First of all we prove the upper semi continuity of the map a−→ µ(a). 126 N. Van Trao Let a=aj1≤j≤N⊂Bn. We shall prove that ∀ε>0, ∃η> 0 such that µ(a)≤µ(a)+εfor every a=a j1≤j≤N⊂Bnwith max 1≤j≤N|aj−a j|≤η. Let α=αj1≤j≤Nbe a sequence in Dand ϕbe a map from Dto Bnsuch that ϕ(αj)=ajand 1 MD α <µ(a)+ε 2. For r<1, put ϕr(ζ)=ϕ(rζ). Then ϕrαj r=ajand ϕr∞= sup D(0,r)|ϕ|<1. Since MD αdepends continuously on α(see [T2, Lemma 1]), we have 1 MD α/r ≤1 MD α +ε 2for rlarge, where α r=αj r1≤j≤N. Now there exists a map frfrom Dto Bnsuch that frαj r=a j−ajand fr∞≤ η√nMD α/r ≤η√nMD α. Then (fr+ϕr)αj r=a jand fr+ϕr∞≤ϕr∞+η√nMD α<1 for ηsmall enough. It implies that µ(a)≤1 MD α/r ≤1 MD α +ε 2<µ(a)+ε. We now prove the lower semi continuity of the map µ(a). Assume that ak=ak jN j=1 ⊂Bnsuch that ak j→ajas k→∞.We shall show that lim inf k→∞ µ(ak)≥µ(a). Passing to a subsequence, we may assume that lim k→∞ µ(ak) exists, and there exist αk:= {αk j}1≤j≤N⊂D, for k≥1 such that lim k→∞ µ(ak) = lim k→∞ 1 MD αk . It will be enough to prove the following. Proposition 2.9. Suppose that αk=αk j,j∈J,k≥0,Ja finite set, and lim k→∞ MD αk=M>0. Then there exist a sequence {Ψk}⊂Aut D, a subset J⊂J, and an increasing map k→ n(k)from Nto Nsuch that lim k→∞ Ψn(k)αn(k) j=αj∈D,∀j∈J, and MD α= lim k→∞ MD Ψn(k)(αn(k) j),j∈J=M. Interpolation constants 127 End of proof of the Theorem 2.8: Suppose 1 M:= lim inf k→∞ µ(ak)<µ(a). Then {αj,j∈J}provides a sequence mapping by ϕto {aj,j∈J}, and 1 MD {αj,j∈J} <µ(a), which contradicts Theorem 2.6. Proof of Proposition 2.9: We will proceed by downward induction on the set Jof indices. Claim. Given any βk j,j∈Jk≥0as in the proposition, either Case 1: there exists {Ψk},k→ n(k)such that lim k→∞ Ψn(k)βn(k) j= αj∈Dfor all j∈Jand we are done, or Case 2: there exists JJand k→n(k)such that lim k→∞ MD {βn(k) j,j∈J } = M, in which case we apply the inductive step again. To prove the claim, consider = lim sup k→∞ max i,j∈JdGβk i,βk j.If<1 then taking Ψk=Φβk j1 where j1∈Jis fixed, we have that all points remain within a fixed relatively compact disk and we can choose a convergent subsequence, so we are in the first case. Suppose now = 1. Then, passing to a subsequence, we may assume that there are j1=j2∈Jsuch that lim k→∞ dGβk j1,βk j2=1. Enumerate J={j1,... ,j p}so that for 3 ≤i≤p,ji≤ji+1.If lim sup k→∞ dGβk j1,βk ji= 1, by taking a subsequence, we may assume that lim k→∞ dGβk j1,βk ji= 1; we will say that ji/∈J1. If lim sup k→∞ dGβk j1,βk ji< 1, keep the same sequence and say that j∈J1. After a finite number of steps, we have a sequence βn(k) j,j∈Jand a subset J1:= {j∈J: lim sup dG(βk j1,βk j)<1}such that j1∈J1and j2∈J2:= J\J1={j∈J: lim sup dG(βk j1,βk j)=1}. Thus, if we set Ψk=Φβk j1 , we have Ψn(k)βn(k) j,j ∈J1contained in a relatively compact disk, and lim k→∞ Ψn(k)βn(k) j=1,j∈J2. Passing to a further subsequence, we may assume that lim k→∞ Ψn(k)βn(k) j∈D, for all j∈J. Corollary 2.4 now shows that M= lim k→∞ max MD Ψn(k)βn(k) j,j∈J1,M D Ψn(k)βn(k) j,j∈J2,