Composition operators on Hardy-Orlicz spaces
Abstract
We investigate composition operators on Hardy-Orlicz spaces when the Orlicz function Ψ grows rapidly: compactness, weak compactness, to be p-summing, order bounded,..., and show how these notions behave according to the growth of Ψ. We introduce an adapted version of Carleson measure. We construct various examples showing that our results are essentially sharp. In the last part, we study the case of Bergman-Orlicz spaces.
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arXiv:math/0610905v1 [math.FA] 29 Oct 2006 Comp osition op erators on Hardy-Orliz spaes Pasal Lefèvre, Daniel Li, Hervé Queéle, Luis Ro dríguez-Piazza 2nd February 2008 Abstrat. We investigate omposition operators on Hardy-Orliz spaes when the Orliz funtion Ψ grows rapid ly: ompatness, weak ompatness, to be p - summing, order bounded, . . . , and show how these notions behave aording to the growth of Ψ . We introdue an adapted version of Carleson measure. We onstrut various examples showing that our results are essential ly sharp. In the last part, we study the ase of Bergman-Orliz spaes. Mathematis Sub jet Classiation. Primary: 47 B 33 46 E 30; Se- ondary: Key-words. Bergman-Orliz spae Carleson measure omposition operator Hardy-Orliz spae 1 Intro dution. Comp osition op erators on the lassial Hardy spaes Hp have b een widely studied (see [36℄, and [11℄, and referenes therein; see also [19℄ and [20℄, and [7℄, [10℄, [18℄, [28℄, [33℄, [37℄, [38℄ for some more reent works), but it seems that one has not paid muh attention to the Hardy-Orliz spaes (in [40℄ and [41℄, J.-O. Strömb erg studied Hardy-Orliz spaes in the ase when the Orliz funtion Ψ inreases smo othly; see also [26℄ for omp osition op erators). We shall investigate what happ ens when the Orliz funtion grows more rapidly than a p ower funtion. Reall that, given an analyti self-map φ:D→D of the unit disk D , the omposition operator asso iated to φ is the map Cφ:f7→ f◦φ . This map may op erate on various Banah spaes X of analyti funtions on D (Hardy spaes, Bergman spaes, . . . , and their weighted versions (see [42℄ for instane), Blo h spaes B and B0 , BMOA and V MOA , Dirihlet spaes (see [1℄), or some more general spaes as Nevanlinna or Smirnov lasses: [8℄, [9℄, [17℄, [21℄; see also [3℄, [4℄, [14℄ for omp osition op erators on Hp spaes of Dirihlet series, though they are not indued by an analyti self-map of D ). The goal is to link prop erties of the omp osition op erator Cφ:X→X (ompatness, strong or weak, for example) to prop erties of the symbol φ (essentially its b ehaviour near the frontier of D ). For that study, one an roughly sp eaking (see [11℄, Chapter 4, though their notions are dierent from ours) distinguish two kind of spaes. 1
1) The smal l spaes X ; those spaes are in a sense lose to the Hardy spae H∞ : the ompatness of Cφ:X→X is very restritive and it imp oses severe restritions on φ . For example, if X=H∞ , a theorem of J. Shwartz ([34℄) implies that Cφ:H∞→H∞ is ompat if and only if kφk∞<1 (weakly ompat sues: see [23℄), whih in turn implies reinfored ompatness properties for Cφ . For example, Cφ:H∞→H∞ is nulear and 1 -summing as soon as it is ompat. 2) The large spaes X ; those spaes are in a sense lose to the Hardy spae H1 : the ompatness of Cφ:X→X an take plae fairly often, and in general implies no self-improvement. For example, for X=H2 , Cφ:H2→H2 an b e ompat without b eing Hilb ert-Shmidt, even if φ is injetive ([39℄, Theorem 6.2). Another formulation (whih lends b etter to generalizations in the non- Hilb ertian ase) is that Cφ an b e non-order b ounded (see [16℄, and our Setion 3) and yet ompat. In this pap er, we shall rather b e on the small spae side, sine we shall work in spaes asso iated to a very large Orliz funtion Ψ (typially: Ψ(x) = Ψ2(x) = ex2−1 ), and the previous situation will not take plae: our op erators will b e e.g. order-b ounded as so on as they are (weakly) ompat, even if the situation is not so extreme as for H∞ . However, for slightly smaller Orliz funtions (for instane Ψ(x) = exp log(x+ 1)2−1 ), the situation is loser to the H2 ase: the omp osition operators may be ompat on HΨ , but not order-b ounded (Theorem 4.22). This pap er is divided into ve parts. Setion 1 is this Intro dution. In Se- tion 2, whih is essentially notational, we reall some more or less standard fats on Orliz funtions Ψ , on asso iated Orliz spaes LΨ , and the little Orliz spae MΨ , and their banahi properties, asso iated with various slow growth onditions (indiated by subsripts: ∆1 , ∆2 , . . . ) or fast growth onditions (indiated by sup ersripts: ∆0 , ∆1 , ∆2 , . . . ). In Setion 3, we intro due the Hardy-Orliz spae HΨ , and its (or his) little brother HMΨ . These spaes have already been studied (see [15℄, [31℄), but rather for slowly growing funtions Ψ (having ∆2 most of the time), and their denition is not so learly outlined, so we give a detailed exp osition of the equivalene of the two natural denitions that one has in mind (if one wants to extend the ase of Hardy spaes Hp asso iated to Ψ(x) = xp ), as well as of the automati b oundedness (through the Littlewoo d subordination priniple and the ase of inner self-maps of the disk) of omp osition op erators on those spaes. Two of the main theorems are Theorem 3.24 and Theorem 3.27. Roughly sp eaking, Theorem 3.24 says the following: if Ψ is very fast growing (having ∆2 more preisely), HΨ is a small spae, the (weak) ompatness of Cφ is very restritive, and even if the situation is not so extreme as for H∞ ( kφk∞<1 ), φ has to tend to the boundary very slowly, and Cφ is automatially order-bounded into MΨ . However, Theorem 3.27 shows the limits of this self-improvement: Cφ may b e order-b ounded into MΨ (and hene ompat), but p -summing for no nite p . We also show that, when Ψ has ∆2 growth, there are always symb ols 2
φ induing ompat omp osition op erators on H2 (even Hilb ert-Shmidt), but not ompat on HΨ . Setion 4 is devoted to the use of Carleson measures. The usefulness of those measures in the study of omp osition op erators is well-known (see [12℄, [11℄, [6℄) and, to our knowledge, rst expliitly used for ompatness in [27℄. In partiular, we reall the following neessary and suient ondition for Cφ:H2→H2 to b e ompat: if h > 0 and w∈∂D , onsider the Carleson window : W(w, h) = {z∈D;|z| ≥ 1−h and |arg(z¯w)| ≤ h}. If φ is an analyti self-map of D with b oundary values φ∗ , and µφ=φ∗(m) denotes the image under φ∗ of the normalized Leb esgue measure (Haar measure) on T=∂D , the measure µφ is always a Carleson measure, i.e. : sup w∈∂D µφW(w, h)=O(h). Now, we an state: Theorem 1.1 (B. MaCluer [27℄) The omposition operator Cφ:H2→H2 is ompat if and only if µφ satises the little-oh ondition, i.e. if and only if: (1.1) sup w∈∂D µφW(w, h)=o(h) as h→0. There is another famous neessary and suient ompatness ondition, due to J. Shapiro ([35℄): Let us denote by Nφ the Nevanlinna ounting funtion of φ , i.e. : Nφ(w) = (P φ(z)=w log 1 |z| if w6=φ(0) and w∈φ(D) 0 if w /∈φ(D). By Littlewoo d's inequality, one always has (see [11℄, page 33): Nφ(w) = O(1 −|w|). Now, Shapiro's Theorem reads: Theorem 1.2 (J. Shapiro [35℄) The omposition operator Cφ:H2→H2 is ompat if and only if Nφ satises the little-oh ondition, i.e.: (1.2) Nφ(w) = o(1 −|w|) as |w|< →1. Theorem 1.2 is very elegant, and probably more popular than Theorem 1.1. Yet, it is diult to apply beause the assumption (1.2) is diult to hek. Here, we shall app eal to Theorem 1.1 to prove (Theorem 4.1) that the ompatness of Cφ:H2→H2 annot b e read on |φ∗| when φ is not nitely valent; more preisely, there are two analyti self-maps φ1 and φ2:D→D 3
suh that: |φ∗ 1|=|φ∗ 2|m - a.e. , but Cφ1:H2→H2 is not ompat, though Cφ2:H2→H2 is ompat. We show then that every omposition op erator whih is ompat on HΨ is neessarily ompat on Hp for all p < ∞ . However, there exist (see ab ove, or Setion 3) symbols φ induing ompat omp osition op erators on H2 but whih are not ompat on HΨ , when Ψ has ∆2 growth. Hene ondition (1.1) do es not sue to haraterize the ompat omposition op erators on HΨ . We have to replae Carleson measures and ondition (1.1) by what we may all Ψ - Carleson measures, and an adaptated little-oh ondition, whih allows us to haraterize ompatness for omp osition operators. It follows that if Ψ∈∆0 , then the weak ompatness of Cφ:HΨ→HΨ implies its ompatness. We also show that the ab ove example φ2 indues, for an Orliz funtion Ψ whih does not satisfy ∆2 , but whih satises ∆1 , a omp osition op erator on HΨ whih is ompat, but not order b ounded into MΨ(T) (Theorem 4.22), showing that the assumption that Ψ∈∆2 in Theorem 3.24 is not only a tehnial assumption. In Setion 5, we intro due the Bergman-Orliz spaes. Let us remind that, in the Hilb ertian ase, the study of ompatness of omp osition op erators is simpler for the Bergman spae B2 than for the Hardy spae H2 . For example, we have the following: Theorem 1.3 (see [36℄) i) Cφ:B2→B2 is ompat if and only if (1.3) lim |z|< →1 1−|φ(z)| 1−|z|= +∞. ii) (1.3) is always neessary for Cφ:H2→H2 to be ompat, and it is suient when φ is injetive, or only nitely valent. iii) There are Blashke produts φ satisfying (1.3) for whih Cφ:H2→H2 is (in an obvious manner) non-ompat. We p erform here a similar study for the Bergman-Orliz spae BΨ , and ompare the situation with that of the Hardy-Orliz spae HΨ . We are naturally led to a reinforement of (1.3) under the form: (1.4) Ψ−11 (1 −|φ(a)|)2 Ψ−11 (1 −|a|)2−→ |a|< →1 0 (always neessary, and suient when Ψ is ∆2 ), whih reads, in the ase Ψ(x) = Ψ2(x) = ex2−1 : (1.5) 1−|φ(z)| ≥ cε(1 −|z|)ε for all ε > 0. 4
In [36℄, the onstrution of a Blashke pro dut satisfying (1.3) is fairly deliate, and app eals to Frostman's Lemma and Julia-Caratheo dory's Theorem on nonangular derivatives at the boundary. Here, we an no longer use these tools for the reinforement (1.5), so we do make a diret onstrution, using the Parseval formula for nite groups. In passing, the onstrution gives a simpler proof for (1.3). Otherwise, the theorem whih we obain is similar to Shapiro's one, if one ignores some tehnial diulties due to the non-separability of BΨ : we have to transit by the smaller Bergman-Morse-Transue spae BMΨ , whih is the losure of H∞ in BΨ , is separable, and has BΨ as its bidual. 2 Notation Let D b e the op en unit disk of the omplex plane, that is the set of omplex numbers with modulus stritly less than 1 , and T the unit irle, i.e. the set of omplex numb ers with mo dulus 1 . We shall onsider in this pap er Orliz spaes dened on a probability spae (Ω,P) , whih will be the unit irle T , with its (normalized) Haar measure m (most often identied with the normalized Leb esgue measure dx/2π on the interval [0,2π] ), or the op en unit disk D , provided with the normalized area measure A . By an Orliz funtion, we shall understand that Ψ: [0,∞]→[0,∞] is a non-dereasing onvex funtion suh that Ψ(0) = 0 and Ψ(∞) = ∞ . To avoid pathologies, we shall assume that we work with an Orliz funtion Ψ having the following additional prop erties: Ψ is ontinuous at 0 , stritly onvex (hene inreasing), and suh that Ψ(x) x−→ x→∞∞. This is essentially to exlude the ase of Ψ(x) = ax . If Ψ′ is the left (or instead, if one prefers, the right) derivative of Ψ , one has Ψ(x) = Rx 0Ψ′(t)dt for every x > 0 . The Orliz spae LΨ(Ω) is the spae of all (equivalene lasses of ) measurable funtions f: Ω →C for whih there is a onstant C > 0 suh that ZΩ Ψ|f(t)| CdP(t)<+∞ and then kfkΨ (the Luxemburg norm ) is the inmum of all p ossible onstants C suh that this integral is ≤1 . The Morse-Transue spae MΨ(Ω) is the subspae generated by L∞(Ω) , or, equivalently, the subspae of all funtions f for whih the above integral is nite for all C > 0 . To every Orliz funtion is asso iated the omplementary Orliz funtion Φ = Ψ∗: [0,∞]→[0,∞] dened by: Φ(x) = sup y≥0xy −Ψ(y), 5
The extra assumptions on Ψ ensure that Φ is itself stritly onvex. When Φ satises the ∆2 ondition (see the denition b elow), LΨ is (isomorphially, if LΦ is itself normed by the Luxemburg norm) the dual spae of LΦ , whih is, in turn, the dual of MΨ . 2.1 Growth onditions We shall have to use various growth onditions for the Orliz funtion Ψ . These onditions are usually denoted as ∆ -onditions. Our interest is in Orliz funtions whih have a somewhat fast growth. Usually, some of these onditions are dened through a mo derate growth ondition on the omplementary funtion Φ of Ψ , and the ondition ∆ for the Orliz funtion is translated as a ∇ -ondition for the omplementary funtion. So we shall distinguish b etween moderate growth onditions, that we shall dene for the omplementary Orliz funtion, and fast growth onditions. To emphasize this distintion, we shall denote, sometimes in hanging the usual notation (see [22, 30℄), the mo derate growth onditions with a subsript, and the fast growth onditions with a sup ersript. Mo derate growth onditions • The Orliz funtion Φ saties the ∆1 -ondition ( Φ∈∆1 ) if, for some onstant c > 0 , one has: Φ(xy)≤cΦ(x)Φ(y) for x, y large enough. This is equivalent to say that Φ(axy)≤Φ(x)Φ(y) for some onstant a > 0 and x, y large enough. This ondition is usually denoted by ∆′ (see [30℄, page 28). •Φ satises the ∆2 -ondition ( Φ∈∆2 ) if Φ(2x)≤KΦ(x) for some onstant K > 1 and x large enough. One has: Φ∈∆1⇒Φ∈∆2. 6
Fast growth onditions • The Orliz funtion Ψ satises the ∆0 -ondition ( Ψ∈∆0 ) if (see [25℄), for some β > 1 : lim x→+∞ Ψ(βx) Ψ(x)= +∞. A typial example is Ψ(x) = exp log(x+ 2) log log(x+ 2)−2log log 2 ; another is Ψ(x) = exp log(x+ 1)3/2−1 . • The Orliz funtion Ψ satises the ∆1 -ondition ( Ψ∈∆1 ) if there is some β > 1 suh that: xΨ(x)≤Ψ(βx) for x large enough. Note that this latter ondition is usually written as ∆3 -ondition, with a subsript (see [30℄, 2.5). This notation ts b etter with our onvention, and the sup ersript 1 agrees with the fat that this ∆1 -ondition is b etween the ∆0 - ondition and the following ∆2 -ondition. Ψ∈∆1 implies that Ψ(x)≥exp α(log x)2 for some α > 0 and x large enough (see [30℄, Prop osition 2, page 37). A typial example is Ψ(x) = e(log(x+1))2−1 . • The Orliz funtion Ψ: [0,∞)→[0,∞) is said to satisfy the ∆2 -ondition ( Ψ∈∆2 ) if there exists some α > 1 suh that: Ψ(x)2≤Ψ(αx) for x large enough. This implies that Ψ(x)≥exp(xα) for some α > 0 and x large enough ([30℄, Prop osition 6, page 40). A typial example is Ψ(x) = Ψ2(x) = ex2−1 . Conditions of regularity • The Orliz funtion Ψ satises the ∇2 -ondition ( Ψ∈ ∇2 ) if its omplementary funtion Φ satises the ∆2 -ondition. This is equivalent to say that for some onstant β > 1 and some x0>0 , one has Ψ(βx)≥2βΨ(x) for x≥x0 , and that implies that Ψ(x) x−→ x→∞∞ . In partiular, this exludes the ase LΨ=L1 . 7
• The Orliz funtion Ψ satises the ∇1 -ondition ( Ψ∈ ∇1 ) if its omplementary funtion Φ satises the ∆1 -ondition. This is equivalent to say that Ψ(x)Ψ(y)≤Ψ(bxy) for some onstant b > 0 and x, y large enough. All p ower funtions Ψ(x) = xp satisfy ∇1 , but Ψ(x) = xplog(x+ 1) do es not. One has (see [30℄, page 43): Ψ∈∆1+3Ψ∈∆0 & F F F F F F F F F F F F Ψ∈∆2 7? x x x x x x x x x x x x x x %- S S S S S S S S S S S S S S S S S S S S S SΨ∈ ∇2 Ψ∈ ∇1 2: l l l l l l l l l l l l l l l l l l l l But ∆1 do es not imply ∇1 . That ∇1 do es not even imply ∆0 is lear sine any power funtion Ψ(x) = xp ( p≥1 ) is in ∇1 . 2.2 Some sp ei funtions In this pap er, we shall make a repeated use of the following funtions: • If Ψ is an Orliz funtion, we set, for every K > 0 : (2.1) χK(x) = ΨKΨ−1(x), x > 0. For example, if Ψ(x) = ex−1 , then Ψ−1(x) = log(1 + x) , and χK(x) = (1 + x)K−1 . Note that: Ψ∈∆0 means that χβ(u) u−→ u→∞+∞ , for some β > 1 . Ψ∈∆1 means that χβ(u)≥uΨ−1(u) for u large enough, for some β > 1 . Ψ∈∆2 means that χα(u)≥u2 for u large enough, for some α > 1 . Ψ∈ ∇1 means that χA(u)≥(Ψ(A)/b)u for u large enough and for every A large enough, for some b > 0 . • For |a|= 1 and 0≤r < 1 , ua,r is the funtion dened on the unit disk D by: (2.2) ua,r(z) = 1−r 1−¯arz 2,|z|<1. Note that kua,rk∞= 1 and kua,rk1≤1−r . 8
3 Comp osition op erators on Hardy-Orliz spaes 3.1 Hardy-Orliz spaes It is well-known that the lassial Hp spaes ( 1≤p≤ ∞ ) an b e dened in two equivalent ways: 1) Hp is the spae of analyti funtions f:D→C for whih, setting fr(t) = f(reit) : kfkHp= sup 0≤r<1kfrkp is nite (reall that the numbers kfrkp inrease with r ). When f∈Hp , the Fatou-Riesz Theorem asserts that the b oundary limits f∗(t) = limr< →1fr(t) exist almost everywhere and kfkHp=kf∗kp . One has f∗∈Lp([0,2π]) , and its Fourier o eients b f∗(n) vanish for n < 0 . 2) Conversely, for every funtion g∈Lp([0,2π]) whose Fourier o eients ˆg(n) vanish for n < 0 , the analyti extension P[g]: D→C dened by P[g](z) = Pn≥0ˆg(n)zn is in Hp and g is the b oundary limit (P[g])∗ of P[g] . Hardy-Orliz spaes HΨ are dened in a similar way. However, we did not nd very satisfatory referenes, and, though the reasonings are essentially the same as in the lassial ase, the lak of homogeneity of Ψ and the presene of the two spaes MΨ and LΨ gives pro ofs whih are not so obvious and therefore we shall give some details. It should b e noted that our denition is not exatly the same as the one given in [30℄, 9.1. We shall b egin with the following prop osition. Prop osition 3.1 Let f:D→C be an analyti funtion. For every Orliz funtion Ψ , the fol lowing assertions are equivalent: 1) sup0≤r<1kfrkΨ<+∞ , where fr(t) = f(reit) ; 2) there exists f∗∈LΨ([0,2π]) suh that b f∗(n) = 0 for n < 0 and for whih f(z) = Pn≥0b f∗(n)zn , z∈D . When these onditions are satised, one has kf∗kΨ= sup0≤r<1kfrkΨ . Let us note that, sine Ψ is onvex and inreasing, Ψ(a|f|) is subharmoni on D , and hene the numb ers RTΨ(a|fr|)dm inrease with r , for every a > 0 . This proposition leads to the following denition. Denition 3.2 Given an Orliz funtion Ψ , the Hardy-Orliz spae HΨ asso- iated to Ψ is the spae of analyti funtions f:D→C suh that one of the equivalent onditions of the above proposition is satised. The norm of f is de- ned by kfkHΨ=kf∗kΨ . We shal l denote by HMΨ the Hardy-Morse-Transue spae , i.e. the subspae {f∈HΨ;f∗∈MΨ(T)} . In the sequel, we shall make no distintion between f and f∗ , unless there may b e some ambiguity, and shall write f instead of f∗ for the b oundary limit. 9
2) One has: kCφk ≥ 1 8Ψ−1(1)Ψ−11 + |φ(0)| 1−|φ(0)|. 3) When Ψ∈ ∇1 global ly: Ψ(x)Ψ(y)≤Ψ(bxy) for al l x, y ≥0 , we also have: kCφk ≤ bΨ−11 + |φ(0)| 1−|φ(0)|. 4) Moreover, Cφ maps HMΨ into HMΨ . Hene, if Ψ∈ ∇2 , then Cφ:HΨ→ HΨ is the bi-adjoint of the omposition operator Cφ:HMΨ→HMΨ . Note that when Ψ(x) = xp for 1≤p < ∞ , then Ψ∈ ∇1 globally, with b= 1 . Pro of. 1) Assume rst that φ(0) = 0 . Let f∈HΨ , with kfkΨ= 1 . Sine Ψ is onvex and inreasing, the funtion u= Ψ ◦ |f| is subharmoni on D , thanks to Jensen's inequality. The ondition φ(0) = 0 allows to use Littlewo o d's sub ordination priniple ([12℄, Theorem 1.7); for r < 1 , one has: Z2π 0 Ψ|(f◦φ)(reit)|dt 2π≤Z2π 0 Ψ|f(reit)|dt 2π≤1. Hene f◦φ∈HΨ and kf◦φkΨ≤1 . Assume now that φ is an inner funtion, and let a=φ(0) . It is known that (see [29℄, Theorem 1) that φ∗(m) = Pa.m , where φ∗(m) is the image under φ∗ (the b oundary limit of φ ) of the normalized Leb esgue measure m , and Pa.m is the measure of density Pa , the Poisson kernel at a . Therefore, for every f∈HΨ with kfkΨ= 1 , one has for 0≤r < 1 , in setting Ka=kPak∞=1+|a| 1−|a| : Z2π 0 Ψ|(f◦φ)(reit)| Kadt 2π≤ZT Ψ|(f◦φ)∗| Kadm =ZT Ψ|f∗◦φ∗| Kadm (reall that |φ∗|= 1 ) =ZT Ψ|f∗| Kadφ∗(m) = ZT Ψ|f∗| KaPadm (3.1) ≤ZT 1 Ka Ψ(|f∗|)Padm , sine Ka>1 ≤ZT 1 Ka Ψ(|f∗|)kPak∞dm =ZT Ψ(|f∗|)dm ≤1. Hene k(f◦φ)rkΨ≤Ka , and therefore kf◦φkΨ≤Ka . 16
Then, for an arbitrary φ , let a=φ(0) again, and let φa b e the automorphism z7→ z−a 1−¯az , whose inverse is φ−a . Sine φ=φ−a◦(φa◦φ) , one has Cφ= Cφa◦φ◦Cφ−a . But φ−a is inner and, on the other hand, (φa◦φ)(0) = 0 ; hene parts a) and b) of the pro of give: kCφk ≤ kCφ−ak ≤ Ka=1 + |a| 1−|a| , whih gives the rst part of the proof. 2) By Lemma 3.11, we have for every f∈HΨ with kfkΨ≤1 : |(f◦φ)(0)| ≤ kδ0k(HΨ)∗kf◦φkΨ≤2Ψ−1(1) kCφk. In other words: |fφ(0)| ≤ 2Ψ−1(1)kCφk for every suh f∈HΨ . Hene: kδφ(0)k(HΨ)∗≤2Ψ−1(1) kCφk, giving kCφk ≥ 1 8Ψ−1(1)Ψ−11 + |φ(0)| 1−|φ(0)|, by using Lemma 3.11 again, but the minoration. 3) When Ψ∈ ∇1 globally, we go bak to the pro of of 1) . We have only to mo dify inequalities (3.1) in b). Setting K′ a= Ψ−1(Ka) , and writing Pa= ΨΨ−1(Pa) , we get, for every f∈HΨ with kfkΨ= 1 : Z2π 0 Ψ|(f◦φ)(reit)| bK′ adt 2π≤ZT Ψ|f∗| bK′ aPadm =ZT Ψ|f∗| bK′ aΨΨ−1(Pa)dm ≤ZT Ψ|f∗| K′ a Ψ−1(Pa)dm ≤ZT Ψ(|f∗|)dm ≤1, sine Ψ−1(Pa)≤Ψ−1(kPak∞) = K′ a , giving kf◦φkΨ≤bK′ a . 4) Supp ose now that f∈HMΨ . As b efore, when φ(0) = 0 , Littlewo o d's sub ordination priniple gives, for every C > 0 : Z2π 0 ΨC|(f◦φ)(eit)|dt 2π= sup r<1Z2π 0 ΨC|(f◦φ)(reit)|dt 2π ≤sup r<1Z2π 0 ΨC|f(reit)|dt 2π=Z2π 0 ΨC|f(eit)|dt 2π<+∞; 17
hene f◦φ∈HMΨ . When φ is inner, the same omputations as in 1) b) ab ove, using that φ∗(m) = Pa.m , where a=φ(0) , give, for every C > 0 : Z2π 0 ΨC|(f◦φ)(reit)|dt 2π≤ZT Ψ(C|f∗|)kPak∞dm =KaZT Ψ(C|f∗|)dm < +∞, and f◦φ∈HMΨ again. The general ase follows, as in 1) ) ab ove, sine f∈HMΨ implies f◦φ−a∈HMΨ , b eause φ−a is inner, and then f◦φ= (f◦φ−a)◦(φa◦φ)∈HMΨ sine (φa◦φ)(0) = 0 . 3.4 Order b ounded omp osition op erators Reall that an op erator T:X→Z from a Banah spae X into a Banah subspae Z of a Banah lattie Y is order bounded if there is some p ositive y∈Y suh that |T x| ≤ y for every x in the unit ball of X . Before studying order b ounded omp osition op erators, we shall reall the following, ertainly well-known, result, whih says that order boundedness an b e seen as stronger than ompatness. Prop osition 3.13 Let T:L2(µ)→L2(µ) be a ontinuous linear operator. Then T is order bounded if and only if it is a Hilbert-Shmidt operator. The pro of is straightforward: if B is the unit ball of L2(µ) , and (ei)i is an orthonormal basis, one has supf∈B|T f|=Pi|T ei|21/2 . Hene supf∈B|T f| ∈ L2(µ) if and only if RPi|T ei|2dµ =PikT eik2<+∞ , i.e. if and only if T is Hilb ert-Shmidt. J. H. Shapiro and P. D. Taylor proved in [39℄ that there exist omposition op erators on H2 whih are ompat but not Hilb ert-Shmidt. We are going to see that, when the Orliz funtion Ψ grows fast enough, the ompatness of omp osition op erators on HΨ is equivalent to their order b oundedness. Prop osition 3.14 The omposition operator Cφ:HΨ→HΨ is order bounded (resp. order bounded into MΨ(T) ) if and only if Ψ−11 1−|φ|∈LΨ(T) (resp. Ψ−11 1−|φ|∈MΨ(T) ). Equivalently, if and only if (OB1) χA1 1−|φ|∈L1(T) for some A > 0, respetively: (OB2) χA1 1−|φ|∈L1(T) for every A > 0, 18
In other words (reall that χA(x) = ΨAΨ−1(x) ), if and only if: ZT ΨAΨ−11 1−|φ|dm < +∞ for some (resp. every) A > 0 . Remark For a sequene (gn)n of elements of LΨ(Ω) , one has kgnkΨ−→ n→+∞0 if and only if (3.2) ZΩ Ψ|gn| εdP−→ n→+∞0 for every ε > 0. In fat, if (3.2) holds, then for n≥nε , the ab ove integrals are ≤1 , and hene kgnkΨ≤ε . Conversely, assume that kgnkΨ−→ n→+∞0 , and let ε > 0 b e given. Let 0< δ ≤1 . Sine kgn/(εδ)kΨ−→ n→+∞0 , one has kgn/(εδ)kΨ≤1 , and hene RΩΨ|gn| εδ dP≤1 , for n large enough. Then, using the onvexity of Ψ : ZΩ Ψ|gn| εdP=ZΩ Ψδ|gn| εδ dP≤δZΩ Ψ|gn| εδ dP≤δ , for n large enough. Therefore, using Leb esgue's dominated onvergene Theorem, it follows that if Cφ:HΨ→HΨ is order b ounded into MΨ(T) , then the omp osition op erator Cφ:HΨ→HΨ is ompat. Pro of of Prop osition 3.14. As HMΨ is separable, there exists a sequene (fn)n≥1 in the unit ball of HMΨ suh that sup nfn◦φ(reiθ)=kδφ(reiθ)k(HMΨ)∗. Now, supp ose that Cφ is order b ounded into LΨ(T) (resp. into MΨ(T) ). Then there exists some g in LΨ(T) (resp. in MΨ(T) ) suh that g≥ |Cφ(f)| a.e. , for every f in the unit ball of HΨ . Using Lemma 3.11, we have a.e. Ψ−11 1−|φ(reiθ)|≤4kδφ(reiθ)k(HΨ)∗= 4 sup nfn◦φ(reiθ)≤4g. The result hene follows letting r↑1 . The onverse is obvious. Theorem 3.15 If the omposition operator Cφ:HΨ→HΨ is order bounded, then: (OB3) m(1 −|φ|< λ) = O1 χA(1/λ), as λ→0, for some A > 0. and if it is order bounded into MΨ(T) , then: (OB4) m(1 −|φ|< λ) = O1 χA(1/λ), as λ→0, for every A > 0. Under the hypothesis Ψ∈∆1 , the onverse holds. 19
Pro of. The neessary ondition follows from Prop osition 3.14 and Markov's inequality: m1 1−|φ|> t≤1 χA(t)ZT χA1 1−|φ|dm. For the onverse, we shall prove a stronger result, and for that, we dene the weak- LΨ spae as follows: Denition 3.16 The weak- LΨ spae LΨ,∞(Ω) is the spae of measurable fun- tions f: Ω →C suh that, for some onstant c > 0 , one has, for every t > 0 : P(|f|> t)≤1 Ψ(ct)· For subsequent referenes, we shall state separately the following elementary result. Lemma 3.17 For every f∈LΨ(Ω) , one has, for every t > 0 : kfkΨ≥t Ψ−11 P(|f|>t)· Pro of. By Markov's inequality, one has, for t > 0 : Ψt kfkΨP(|f|> t)≤ZΩ Ψ|f| kfkΨdP≤1; and that gives the lemma. Sine Lemma 3.17 an be read: P(|f|> t)≤1 Ψ(t/kfkΨ) , we get that LΨ(Ω) ⊆LΨ,∞(Ω) . The onverse of Theorem 3.15 is now an immediate onsequene of the following prop osition. Prop osition 3.18 1) a) If Ψ∈∆1 , then LΨ(Ω) = LΨ,∞(Ω) . b) If LΨ(Ω) = LΨ,∞(Ω) , then Ψ∈∆0 . 2) If LΨ(T) = LΨ,∞(T) , then ondition (OB3) implies that Cφ:HΨ→HΨ is order bounded, and ondition (OB4) implies that it is order bounded into MΨ(T) . Lemma 3.19 The fol lowing assertions are equivalent i) LΨ(Ω) = LΨ,∞(Ω) . ii) Z∞ 1 Ψ′(u) Ψ(Bu)du ≡Z+∞ Ψ(1) 1 χB(x)dx < +∞ , for some B > 1 . 20
Pro of of the Lemma. Assume rst that 1/χB is integrable on (Ψ(1),∞) . For every f∈LΨ,∞(Ω) , there is a c > 0 suh that P(|f|> t)≤1/Ψ(ct) . Then ZΩ Ψc|f| BdP=Z+∞ 0 P(|f|> Bt/c) Ψ′(t)dt ≤Ψ(1) + Z+∞ 1 P(|f|> Bt/c) Ψ′(t)dt ≤Ψ(1) + Z+∞ 1 Ψ′(t) Ψ(Bt)dt = Ψ(1) + Z+∞ Ψ(1) du χB(u)<+∞, so that f∈LΨ(Ω) . Conversely, assume that LΨ(Ω) = LΨ,∞(Ω) . Sine 1/Ψ is dereasing, there is a measurable funtion f: Ω →C suh that P(|f|> t) = 1/Ψ(t) , when t≥Ψ−1(1) . Suh a funtion is in LΨ,∞(Ω) ; hene it is in LΨ(Ω) , by our hyp othesis. Therefore, there is a B > 1 suh that ZΩ Ψ(|f|/B)dP<+∞. But ZΩ Ψ(|f|/B)dP=Z+∞ 0 P(|f|> Bt) Ψ′(t)dt ≥Z+∞ 1 BΨ−1(1) Ψ′(t) Ψ(Bt)dt ≥Z+∞ Ψ−1(1) Ψ′(t) Ψ(Bt)dt =Z+∞ 1 du χB(u) , and hene 1/χB is integrable on (1,∞) . Pro of of Prop osition 3.18. 1)a) We rst remark that for every Orliz funtion Ψ , one has Ψ(2x)≥xΨ′(x) for every x > 0 , b eause, sine Ψ′ is p ositive and inreasing: Ψ(2x) = Z2x 0 Ψ′(t)dt ≥Z2x x Ψ′(t)dt ≥xΨ′(x). Assume now that Ψ∈∆1 : xΨ(x)≤Ψ(βx) for some β > 0 and for x≥x0>0 . By Lemma 3.19, it sues to show that 1/χ2β is integrable on (Ψ(x0),+∞) . But Z+∞ Ψ(x0) dx χ2β(x)=Z+∞ x0 Ψ′(u) Ψ(2βu)du ≤Z+∞ x0 Ψ(2u)/u 2uΨ(2u)du =Z+∞ x0 du 2u2<+∞. 1)b) Supp ose now that LΨ(Ω) = LΨ,∞(Ω) . By the preeding lemma, there exists some B > 1 suh that lim x→+∞Z2x x Ψ′(u) Ψ(Bu)du = 0. 21
By onvexity, Ψ(2x)≥2Ψ(x) and Ψ′ is nonnegative, so that Z2x x Ψ′(u) Ψ(Bu)du ≥1 Ψ(2Bx)Z2x x Ψ′(u)du ≥Ψ(2x)−Ψ(x) Ψ(2Bx)≥Ψ(x) Ψ(2Bx)· Therefore Ψ satises ∆0 . 2) Assume that LΨ(T) = LΨ,∞(T) and that ondition (OB3) (resp. (OB4)) holds. By Lemma 3.19, there is a B > 0 suh that 1/χB is integrable on (1,+∞) . We get, using ondition (OB3) (resp. (OB4)) and setting C=A/B : ZT χC1 1−|φ|dm =Z+∞ 0 m1 1−|φ|> tχ′ C(t)dt =Z+∞ 0 m(1 −|φ|<1/t)χ′ C(t)dt ≤χC(1) + KZ+∞ 1 χ′ C(t) χA(t)dt. But, if we set u=χC(t) , i.e. u= ΨCΨ−1(t) , then Ψ−1(u) = CΨ−1(t) , and hene χA(t) = ΨAΨ−1(t)= ΨBΨ−1(u)=χB(u). Therefore: ZT χC1 1−|φ|dm ≤χC(1) + KZ+∞ χC(1) du χB(u)du < +∞. It follows from Prop osition 3.14 that Cφ is order b ounded (resp. into MΨ(T) ). Remark. The ondition Ψ∈∆1 is not equivalent to LΨ(Ω) = LΨ,∞(Ω) . For example, we an take: Ψ(x) = exp log(x+ 1)3/2−1. Then, as x tends to innity, Ψ(Kx)∼Ψ(x) exp 3 2(log K)(log x)1/2, and hene Ψ6∈ ∆1 . On the other hand, Ψ′(x)∼3 2 (log x)1/2 xΨ(x) ; hene: Z+∞ Ψ(1) du χK(u)=Z+∞ 1 Ψ′(t) Ψ(Kt)dt ∼Z+∞ 0 3u2 exp(3 2(log K)u)du =Z+∞ 0 3u2 K3u/2du < +∞ for K > 1 . Therefore LΨ(Ω) = LΨ,∞(Ω) by Lemma 3.19. 22
3.5 Weakly ompat omp osition op erators We saw in Lemma 3.10 that kua,rkΨ≤1 Ψ−1(1 1−r)· The next result shows that the weak ompatness of Cφ transforms the big-oh into a little-oh , when Ψ grows fast enough. Theorem 3.20 Assume that Ψ∈∆0 . Then the weak ompatness of the omposition operator Cφ:HΨ→HΨ implies that: (W) sup a∈TkCφ(ua,r)kΨ=o1 Ψ−11 1−r, as r→1. Pro of. Atually, we only need to use that the restrition of Cφ:HΨ→HΨ to HMΨ is weakly ompat. We proved in [25℄, Theorem 4, that, under the hyp othesis Ψ∈∆0 , the op erator Cφ:HMΨ→HMΨ is weakly ompat if and only if for every ε > 0 , we an nd Kε>0 suh that, for every f∈HMΨ , one has: kCφ(f)kΨ≤Kεkfk1+εkfkΨ. Using Corollary 3.10, we get, for every ε > 0 , sine kua,rk1≤1−r : kCφ(ua,r)kΨ≤Kε(1 −r) + ε1 Ψ−1(1 1−r)· But Ψ(x) x−→ x→+∞+∞ . Hene 1−r=o1 Ψ−1(1 1−r) as r→1, and that proves the theorem. We shall see in Setion 4 that the onverse holds when Ψ satises ∆0 and moreover that Cφ is then ompat. That will use some other tehniques. Nevertheless, we an prove, from now on, the following result. Theorem 3.21 If Ψ∈∆2 , then ondition: (W) sup a∈TkCφ(ua,r)kΨ=o1 Ψ−11 1−r, as r→1. in Theorem 3.20 implies ondition: (OB4) m(1 −|φ|< λ) = O1 χA(1/λ), as λ→0, for every A > 0. in Theorem 3.15 . 23
Remark that when Ψ∈∆0 (and in partiular when Ψ∈∆2 ), one has, for any B > βA (where β is given by the denition of ∆0 ): Ψ(Bx) Ψ(Ax)→+∞ as x→+∞ ; hene 1/χA(x) = o1/χB(x) and the big-oh ondition in (OB4) may b e replaed by a little-oh ondition. Before proving this theorem, let us note that: Prop osition 3.22 Condition (W) sup a∈TkCφ(ua,r)kΨ=o1 Ψ−11 1−r as r→1 implies that m(|φ|= 1) = 0 . Pro of. Otherwise, one has m(|φ|= 1) = δ > 0 . Splitting the unit irle T into N parts, we get some a∈T suh that: m(|φ−a| ≤ π/N)≥δ/N. But, the inequality |φ−a| ≤ π/N implies, with r= 1 −1/N (sine |φ| ≤ 1 ): |1−¯arφ| ≤ |1−¯aφ|+ (1 −r)|¯aφ|=|a−φ|+ (1 −r)|φ|<5/N = 5(1 −r). Hene: m|Cφ(ua,r)|>1/25)=m|1−¯arφ|<5(1 −r) ≥m|φ−a| ≤ π(1 −r)≥δ(1 −r), and therefore, by Lemma 3.17: kCφ(ua,r)kΨ≥1/25 Ψ−11/δ(1 −r)≥δ/25 Ψ−11/(1 −r)· Sine r an b e taken arbitrarily lose to 1 , that proves the proposition. Pro of of Theorem 3.21. Assume that ondition (W) is satised, and x A > 1 . Let ε > 0 to b e adjusted later. We an nd rε<1 suh that rε≤r < 1 implies: kCφ(ua,r)kΨ≤ε Ψ−11 1−r , ∀a∈T. Now, Lemma 3.17 also reads: m(|f|> t)≤1 Ψt kfkΨ , so that, if one sets B= 1/9ε : m|1−¯arφ|<3(1 −r)=m|Cφ(ua,r)|>1/9≤1 ΨBΨ−11 1−r· We laim that this implies a go o d upp er b ound on m(|φ|> r) , even if we lo ose a fator 1/(1 −r) , due to the eet of a rotation on φ . For that, we shall use the following lemma. 24
Lemma 3.23 Let φ:D→D be an analyti funtion. Then, for every r with 0< r < 1 , there exists a∈T suh that: m|1−¯ar φ|<3(1 −r)≥1−r 8m(|φ|> r). Admitting this for a while, we are going to nish the pro of. Fix an r suh that rε≤r < 1 , and take an a∈T as in Lemma 3.23. We get, from the preeding, in setting λ= 1 −r : m(1 −|φ|< λ) = m(|φ|> r) ≤8 1−rm|1−¯arφ|<3(1 −r) ≤8 1−r 1 ΨBΨ−11 1−r , i.e. , setting x= Ψ−1(1/1−r) = Ψ−1(1/λ) : m(1 −|φ|< λ)≤8Ψ(x) Ψ(Bx)· But Ψ saties the ∆2 -ondition: Ψ(y)2≤Ψ(αy) for some α > 1 and y large enough. Then, adjusting now ε > 0 as ε= 1/9αA , in order that B=αA , we get, for x large enough, sine A > 1 : Ψ(x)Ψ(Ax)≤[Ψ(Ax)]2≤Ψ(Bx). Therefore, for r lose enough to 1 : m(1 −|φ|< λ)≤8 Ψ(Ax)=8 χA(1/λ)· We hene get ondition (OB4), and that proves Theorem 3.21. Pro of of Lemma 3.23. Let λ= 1 −r , and let δ > 0 b e a numb er whih we shall sp eify later. Consider the set: Cδ={z∈D;|z| ≥ 1−λ and |arg z| ≤ δ} (for δ=λ , Cδ is a losed Carleson window). It is geometrially lear that (1 −λ)Cδ is ontained in the losed disk of enter 1 and whose edge ontains (1 −λ)2eiδ ; hene, for every z∈Cδ , one has: |1−(1 −λ)z|2≤ |1−(1 −λ)2eiδ|2= 2(1 −λ)2(1 −cos δ) + λ2(2 −λ)2 ≤(1 −λ)2δ2+λ2(2 −λ)2≤9λ2 if δ≤λ . 25
Therefore: ZT Ψ|˘g| 2αdm = +∞ X n=1 Z˜ An Ψ|˘g| 2αdm. Sine ˘gj= 0 on ˜ An for j > n , we get: ZT Ψ|˘g| 2αdm = +∞ X n=1 Z˜ An Ψ|˘g1|+···+|˘gn| 2αdm. Now, by the onvexity of Ψ : Ψ|˘g1|+···+|˘gn| 2α≤1 2Ψ|˘g1|+···+|˘gn−1| α+ Ψ|˘gn| α· But: Ψ|˘gn| α≤1 2nΨ(|˘gn|) + Ψ(xn), and Ψ|˘g1|+···+|˘gn−1| α≤Ψt1+···+tn−1 α; therefore, using that RTΨ(|˘gn|)dm ≤RTΨ(|gn|)dm ≤1 : ZT Ψ|˘g| 2αdm ≤ +∞ X n=1 1 2Ψt1+···+tn−1 αm(˜ An) +1 2nZT Ψ(|˘gn|)dm + Ψ(xn)m(˜ An) ≤ +∞ X n=1 1 2hΨt1+···+tn−1 α+ Ψ(xn)i2ntn Ψ(tn)+1 2n ≤ +∞ X n=1 1 2h1 2n+1 2ni= 1 , whih proves that ˘g∈LΨ , and k˘gkΨ≤2α . The pro of is fully ahieved. Remark. In the ab ove pro of, we hose Mn= log(n+ 1) . This hoie was only used to onlude that PnkCφ(gn)kp Ψ= +∞ for every p < ∞ . Therefore, the ab ove pro of shows that, given any inreasing funtion Υ: (0,∞)→(0,∞) tending to ∞ , we an nd, with a suitable hoie of a slowly inreasing sequene (Mn)n≥1 , a symb ol φ and a sequene (gn)n≥1 in HΨ suh that Pn|gn| ∈ LΨ , although PnΥ(kCφ(gn)kΨ) = +∞ . 32
4 Carleson measures 4.1 Intro dution B. MaCluer ([27℄; see also [11℄, Theorem 3.12) has haraterized ompat omp osition op erators on Hardy spaes Hp ( p < ∞ ) in term of Carleson measures. In this setion, we shall give an analogue of this result for Hardy-Orliz spaes HΨ , but in terms of Ψ -Carleson measures. Indeed, Carleson measures do not haraterize the ompatness of omp osition op erators when Ψ grows to o quikly, as it follows from Corollary 3.26. Before that, we shall reall some denitions (see for example [11℄, pages 3738, or [12℄, page 157). Let ξ∈T and h∈(0,1) . Dene (4.1) S(ξ, h) = {z∈D;|ξ−z|< h}. The Carleson window W(ξ, h) is the following subset of D : (4.2) W(ξ, h) = {z∈D; 1 −h < |z| ≤ 1 and |arg(z¯ ξ)|< h}. It is easy to show that we have for every ξ∈T and h∈(0,1) : S(ξ, h/2) ⊆W(ξ, h) and W(ξ, h/2) ⊆S(ξ, h), so that, in the sequel, we may work equivalently with either S(ξ, h) or W(ξ, h) . Reall that a p ositive Borel measure µ on D (or D ) is alled a Carleson measure if there exists some onstant K > 0 suh that: µS(ξ, h)≤Kh , ∀ξ∈T,∀h∈(0,1). Carleson's Theorem (see [11℄, Theorem 2.33, or [12℄, Theorem 9.3) asserts that, for 0< p < ∞ , the Hardy spae Hp is ontinuously embedded into Lp(µ) if and only if µ is a Carleson measure. Given an analyti self-map φ:D→D , we dene the pul lbak measure µφ on the losed unit disk D (whih we shall denote simply µ when this is unambiguous) as the image of the Haar measure m of T=∂D under the map φ∗ (the b oundary limit of φ ): (4.3) µφ(E) = mφ∗−1(E), for every Borel subset E of D . The automati ontinuity of omp osition op erators Cφ on the Hardy spae Hp , ombined with Carleson's Theorem means that µφ is always a Carleson measure. B. MaCluer ([27℄, [11℄, Theorem 3.12) showed that: The omposition operator Cφ is ompat on H2 if and only if: µφS(ξ, h)=o(h) as h→0, uniformly for ξ∈T. (MC) 33
While the Shapiro's ompatness riterion, via the Nevanlinna ounting fun- tion ([36℄), deals with the b ehavior of φ inside the op en unit disk, the hara- terization (MC) deals with its b oundary values φ∗ . It is natural to wonder whether the mo dulus of φ∗ on T=∂D sues to haraterize the ompatness of Cφ . This leads to the following question: if two funtions φ1 and φ2 have the same mo dulus on T , are the ompatness of the two asso iated omp osition op erators equivalent? We have seen in Theorem 3.24 that the answer is p ositive on HΨ when Ψ∈∆2 . However, on H2 it turns out to be negative. We give the following ounterexample. Theorem 4.1 There exist two analyti funtions φ1 and φ2 from D into itself suh that |φ∗ 1|=|φ∗ 2| on T but for whih the omposition operator Cφ2:H2→H2 is ompat, though Cφ1:H2→H2 is not ompat. Remark. Let Ψ b e an Orliz funtion whih satises ∆2 . We shall see in Theorem 4.3 that every omposition op erator Cφ:H2→H2 is ompat as so on as Cφ:HΨ→HΨ is ompat. Hene, in the ab ove theorem, Cφ1:HΨ→HΨ is not ompat. It follows hene from Theorem 3.24, sine φ∗ 1 and φ∗ 2 have the same mo dulus, that Cφ2:HΨ→HΨ is not ompat (and, even, not weakly ompat), though Cφ2:H2→H2 is ompat. We have already seen suh a phenomenon in Corollary 3.26. However, the results of the next subsetion will allow us to onlude (Theorem 4.22) that, when Ψ(x) = exp log(x+ 1)2−1 , whih do es not satisfy ondition ∆2 , but satises onditions ∆1 and ∇1 , the omp osition op erator Cφ2:HΨ→HΨ is ompat, but not order b ounded into MΨ(T) . That will show that our assumption that Ψ∈∆2 in Theorem 3.24 is not only a tehnial one. Pro of. We start simply with φ1(z) = 1+z 2 . It is well known that Cφ1 is not ompat on H2 (this was rst observed in H. J. Shwartz's thesis [34℄: see [39℄, page 471). Now, let: M(z) = exp −1 + z 1−z and φ2(z) = φ1(z)M(z). For simpliity, we shall write φ=φ2 , and we are going to show that Cφ is a ompat op erator on H2 , using the riterion (MC) . Let ξ= eiα ∈T , with |α| ≤ π . We are going to prove that: µφS(ξ, h)=O(h3/2). First, observe that, for h∈(0,1) : S(ξ, h)⊆ {z∈D; 1 −h < |z| ≤ 1 and |arg(¯zξ)| ≤ 2h}. Hene for h small enough, µφS(ξ,h)≤ m{θ∈(−π, π) ; 1 −h < |φ(eiθ)| ≤ 1 and |arg(e−iαφ(eiθ))| ≤ 2h}). 34
For θ∈(−π, π) , one has |φ(eiθ)|=|φ1(eiθ)|= cos(θ/2) , and so the ondition 1−h < |φ(eiθ)| ≤ 1 is equivalent to 1−h < cos(θ/2) <1 , whih implies, sine cos t= 1 −2 sin2(t/2) ≤1−2t2/π2≤1−t2/5 for 0≤t≤π/2 (b eause sin t≥2 πt ), that 1−h < 1−(θ/2)2/5 , i.e. θ2≤20h and so |θ| ≤ 6√h . On the other hand, M(eiθ) = exp −icot(θ/2) ; hene arg φ(eiθ) = θ/2−cot(θ/2) , mo dulo 2π . Therefore, for h small enough: µφS(ξ, h)≤m({|θ| ≤ 6√h;|−α+θ/2−cot(θ/2)| ≤ 2h, mod 2π}) ≤2X n∈Z m({|t| ≤ 3√h;|−α+t−cot t+ 2πn| ≤ 2h}). We have to ma jorize b oth X n∈Z m({0< t ≤3√h;|−α+t−cot t+ 2πn| ≤ 2h}) and X n∈Z m({0< t ≤3√h;|α+t−cot t+ 2πn| ≤ 2h}) . The funtion F(t) = t−cot(t) is inreasing , and we dene an, bn∈(0, π) by: F(an) = α−2π(nh+n)−2h and F(bn) = α−2π(nh+n) + 2h, where the integer nh is given large enough to ensure that a0≤3√h . Of ourse, an< bn< an−1 . Observe that 4h=F(b0)−F(a0)≥b0−a0 , and then b0≤3√h+ 4h≤4√h for h small enough. One has: X n∈Z m({0< t ≤3√h;|−α+t−cot t+ 2πn| ≤ 2h})≤∞ X n=0 (bn−an). Sine F′(t) = 1 + 1 sin2t≥1 t2 , one has, on the one hand: (4.4) 4h=F(bn)−F(an) = Zbn an F′(t)dt ≥bn−an anbn ; hene: (4.5) bn−an≤4hanbn≤4hb2 n, for all n≥0. On the other hand, let us rst p oint out that, for 0≤t≤1 , F′(t) = 1 + 1 sin2t≤1 + π2/4 t2≤4 t2· Hene, for h small enough: 2π=F(bn)−F(bn+1) = Zbn bn+1 F′(t)dt ≤4bn−bn+1 bn+1bn , 35
and we get: b2 n+1 ≤bn+1bn≤2 π(bn−bn+1). Hene, using the fat that (4.4) gives b0−a0≤4ha0b0≤48 h2 , we get, from (4.5): ∞ X n=0 (bn−an)≤(b0−a0) + 8h π ∞ X n=0 (bn−bn+1) ≤48 h2+8h πb0≤48 h2+8h π4√h≤48√h+32 πh3/2≤11 h3/2, for h small enough. In the same way, we have ∞ X n=0 m({0< t ≤3√h;|α+t−cot t+ 2πn| ≤ 2h})≤11 h3/2. We an hene onlude that µφS(ξ, h)≤Ch3/2 , where C is a numerial onstant. Remark. Cφ atually maps ontinuously H2 into H3 , and ompatly H2 into Hp , for any p < 3 (see [16℄ or Theorem 4.10 and 4.11). However, in some ases, the b ehaviour of |φ∗| on the b oundary ∂D sues. Prop osition 4.2 Let φ1 and φ2 be two analyti self-maps of D suh that |φ∗ 1| ≤ |φ∗ 2| on ∂D . Assume that they are both one-to-one on D , and that there exists a∈D suh that φ2(a) = 0 . Then the ompatness of Cφ2:H2→H2 implies that of Cφ1:H2→H2 . Pro of. By omp osing φ1 and φ2 with the automorphism of D whih maps 0 into a , we may assume that a= 0 . We an hene write φ2(z) = zφ(z) , with φ:D→C analyti in D . φ do es not vanish in D b eause of the injetivity of φ2 (this is obvious for z6= 0 , and for z= 0 , follows from the fat that the injetivity of φ2 implies φ′ 2(0) 6= 0 ). Then there is some δ > 0 suh that |φ(z)| ≥ δ for every z∈D . In fat, by ontinuity, there is some α > 0 and some 0< r < 1 suh that |φ(z)| ≥ α and |φ2(z)| ≤ α for |z| ≤ r . But b eing analyti and non onstant, φ2 is an op en map, so there is some ρ > 0 suh that ρD⊆φ2(rD) . Injetivity of φ2 shows that φ2(D\rD)∩ρD=∅ , that is to say that |φ2(z)| ≥ ρ for |z|> r . A fortiori |φ(z)| ≥ ρ for |z|> r . The laim is proved with δ= min(α, ρ) . Then 1 φ∈H∞ , as well as φ1 φ . Sine φ∗ 1 φ∗=φ∗ 1 φ∗ 2≤1 , one has φ1(z) φ(z)≤1 for every z∈D . Hene: 1−|φ1(z)| 1−|z|≥1−|φ(z)| 1−|z|=1−|φ2(z)| 1−|z|−|φ(z)|. 36
Now ([36℄, Theorem 3.5), the ompatness of Cφ2:H2→H2 implies that lim z< →1 1−|φ2(z)| 1−|z|= +∞; we get: lim z< →1 1−|φ1(z)| 1−|z|= +∞, whih implies the ompatness of Cφ1:H2→H2 , thanks to the injetivity of φ1 ([36℄, Theorem 3.2). 4.2 Compatness on HΨ versus ompatness on H2 The equivalene (MC) holds atually for every Hp spae (with p < ∞ ) instead of H2 . We are going to see in this setion that for Hardy-Orliz spaes HΨ , one needs a new notion of Carleson measures, whih one may all Ψ - Carleson measures . Before that, we are going to see that ondition (MC) allows to get that the ompatness of omp osition op erators on HΨ always implies that on Hp for p < ∞ . Reall that, when Ψ∈∆2 , we have seen in Corollary 3.26 that the onverse is not true. Theorem 4.3 Let φ:D→D be an analyti funtion. If one of the fol lowing onditions: i) Cφ is a ompat operator on HΨ ii) Ψ∈∆0 and Cφ is a weakly ompat operator on HΨ is satised, the omposition operator Cφ is ompat on H2 . Note that we have proved in Theorem 3.24 that the weak ompatness of Cφ:HΨ→HΨ implies its ompatness only when Ψ satises the ∆2 ondition. Nevertheless, we shall show in Theorem 4.21 that when Ψ∈∆0 , the weak ompatness of Cφ is equivalent to its ompatness. This is obviously false (in partiular when LΨ is reexive) without any assumption on Ψ . Pro of. We are going to use the haraterization (MC) for ompat omp osition op erators on H2 . Supp ose that the ondition on µφ is not fullled. Then there exist β∈(0,1) , ξn∈T , and hn∈(0,1) , with hn−→ n→+∞0 , suh that: µφS(ξn, hn)≥βhn. We are now going to use the funtion: vn(z) = h2 n (1 −anz)2 , 37
where: an= (1 −hn)ξn. Of ourse, vn is atually nothing but uξn,1−hn . We have by Corollary 3.10: kvnkΨ≤1 Ψ−1(1/hn)· Dene gn= Ψ−1(1/hn)vn , whih is in the unit ball of HMΨ . We have assumed at the b eginning of the pap er that x=oΨ(x) as x→ ∞ ; hene Ψ−1(x) = o(x) as x→ ∞ , and so h2 nΨ−1(1/hn)→0 . Therefore (gn)n onverges uniformly to zero on ompat subsets of D and kgnk1→0 , b eause kgnk1≤hnΨ−1(1/hn) . Then, in b oth ases, we should have kCφ(gn)kΨ→0 . Indeed, in ase i) , this follows from Prop osition 3.6, and in ase ii) , this follows from [25℄, Theorem 4. We are going to show that this is not true. Indeed: ZT Ψ4 β|gn◦φ|dm ≥ZD Ψ4 βΨ−1(1/hn)|vn(z)|dµφ ≥ZS(ξn,hn) Ψ4 βΨ−1(1/hn)|vn(z)|dµφ. But when z∈S(ξn, hn) , one has |vn(z)| ≥ 1/4 , b eause |1−anz| ≤ |1−anξn|+|an(ξn−z)|=hn+ (1 −hn)hn≤2hn. We obtain that by onvexity (sine β < 1 ): ZT Ψ4 β|gn◦φ|dm ≥ZT 1 βΨ4Ψ−1(1/hn)|vn(z)|)dm ≥1 βhn µφS(ξn, hn)≥1. This implies that kCφ(gn)kΨ≥β/4 and proves the theorem. 4.3 General measures We used several times the riterion (MC) for ompatness on H2 via Carleson measures. The fat that this provides suh a useful tool leads to wonder if the b oundedness and the ompatness on Hardy-Orliz spaes an b e expressed in suh a pleasant manner. Theorem 4.10 and Theorem 4.11 are the Orliz version of Carleson's Theorem for Hp spaes ([11℄, Theorem 2.35). The key for our general haraterization is the use of the following funtions (see (4.2) for the denition of the Carleson's window W(ξ, h) ). 38
Denition 4.4 For any positive nite Borel measure µ on the unit disk D (or on D ), we set, for h∈(0,1] : ρµ(h) = sup ξ∈T µW(ξ, h), (4.6) Kµ(h) = sup 0<t≤h ρµ(t) t (4.7) Hene µW(ξ, t)≤Kµ(h)t for t≤h . The measure µ is a Carleson measure if and only if Kµ(h) is b ounded by a onstant K , for h∈(0,1) and this happ ens as so on as Kµ(h0) is nite for some h0∈(0,1) . Denition 4.5 We say that Ψ satises the ∇0 ondition if for some x0>0 , C≥1 and every x0≤x≤y , one has: (4.8) Ψ(2x) Ψ(x)≤Ψ(2Cy) Ψ(y)· This is a ondition on the regularity of Ψ . It is satised if Ψ(2x) Ψ(x)≤CΨ(2y) Ψ(y)· Prop osition 4.6 the fol lowing assertions are equivalent i) Ψ satises the ∇0 ondition. ii) There exists some x0>0 satisfying: for every β > 1 , there exists Cβ≥1 suh that Ψ(βx) Ψ(x)≤Ψ(βCβy) Ψ(y) , for every x0≤x≤y. iii) There exist x0>0 , β > 1 and Cβ≥1 suh that Ψ(βx) Ψ(x)≤Ψ(βCβy) Ψ(y) , for every x0≤x≤y. Pro of. We only have to prove i)⇒ii) , sine iii)⇒i) is similar and ii)⇒iii) is trivial. If β∈(1,2] , it is easy, taking Cβ= 2C/β . Now, if β∈(2b,2b+1] for some integer b≥1 , we write for every x0≤x≤y : Ψ(βx) Ψ(x)≤Ψ(2b+1x) Ψ(x)=Ψ(2b+1x) Ψ(2bx)···Ψ(2x) Ψ(x)· 39
But we have for every integer j≥1 : 2j−1x≤(2C)j−1x≤(2C)j−1y , so: Ψ(2jx) Ψ(2j−1x)≤Ψ((2C)jy) Ψ((2C)j−1y) , and we obtain: Ψ(βx) Ψ(x)≤Ψ((2C)b+1y) Ψ(y)≤Ψ(βCβy) Ψ(y) , where Cβ= 2Cb+1 . Examples. It is immediately seen that the following funtions satisfy ∇0 : Ψ(x) = xp , Ψ(x) = exp log(x+ 1)α−1 , Ψ(x) = exα−1 , α≥1 . Note that when Ψ∈ ∇0 , with onstant C= 1 , i.e. Ψ(βx)/Ψ(x) is inreasing for x large enough, then we have the dihotomy: either Ψ∈∆2 , or Ψ∈∆0 . We shall say that ∇0 is uniformly satised if there exist C≥1 and x0>0 suh that, for every β > 1 : (4.9) Ψ(βx) Ψ(x)≤Ψ(Cβy) Ψ(y) for x0≤x≤y, One has: Prop osition 4.7 1) Condition ∆2 implies ondition ∇0 uniformly. 2) If Ψ∈ ∇0 uniformly, then Ψ∈ ∇1 . 3) The funtion κ(x) = log Ψ(ex) is onvex on (x0,+∞) if and only if ∇0 is satised with onstant C= 1 . We shall say that Ψ is κ -onvex when κ is onvex at innity. Note that Ψ is κ -onvex whenever Ψ is log-onvex. In the ab ove examples Ψ is κ -onvex; it also the ase of Ψ(x) = x2/log x , x≥e ; but, on the other hand, if Ψ(x) = x2log x for x≥e , then Ψ is not κ -onvex. Nevertheless, for β2≤x≤y , one has: Ψ(βx) Ψ(x)=β21 + log β log x≤3β2 2≤3 2 Ψ(βy) Ψ(y) , and hene Ψ∈ ∇0 . We do not know whether Ψ∈ ∇0 uniformly implies that Ψ is equivalent to an Orliz funtion for whih the asso iated funtion κ is onvex. Pro of. 1) Sine Ψ∈∆2 , one has Ψ(u)2≤Ψ(αu) for some α > 1 and x≥x0 . We may assume that Ψ(x0)≥1 . Then, for y≥x≥x0 and every β > 1 : Ψ(βx)Ψ(y)≤Ψ(βy)2≤Ψ(αβy)≤Ψ(αβy)Ψ(x), whih is (4.9). 40
2) Supp ose that Ψ satises ondition ∇0 uniformly. We may assume that Ψ(x0)≥1 . Let x0≤u≤v ; we an write u=βx0 for some β≥1 . Then ondition (4.9) gives: Ψ(u)Ψ(v) = Ψ(βx0)Ψ(v)≤Ψ(x0)Ψ(Cβv)≤ΨΨ(x0)Cβv= Ψ(buv), with b=CΨ(x0)/x0 . 3) Assume that κ is onvex on (x0,+∞) . For every β > 1 , let κβ(t) = κ(tlog β) = log(Ψ(βt)) , whih is onvex on (x0/log(β),+∞) . Taking y≥x≥ ex0 , write x=βθ and y=βθ′ with θ≤θ′ . Convexity of κ gives, sine θ′≥θ≥ x0/log(β) : κβ(θ+ 1) −κβ(θ)≤κβ(θ′+ 1) −κβ(θ′), whih means that: Ψ(βx) Ψ(x)≤Ψ(βy) Ψ(y)· Assume that (4.9) is fullled for every β > 1 , with C= 1 . Then, taking y=βx , one has: Ψ(βx)2≤Ψ(x)Ψ(β2x). Let u < v b e large enough. Taking x=u2 and β=v/u , we get: Ψ(uv)2≤Ψ(u2)Ψ(v2), whih means that κ is onvex. Remark. The growth and regularity onditions for Ψ an b e expressed in the following form: •Ψ∈∆0 i κ(x+β′)−κ(x)−→ x→∞+∞ , for some β′>0 . •Ψ∈∆1 i for some β′>0 , one has x+κ(x)≤κ(x+β′) , for x large enough. •Ψ∈∆2 i for some α′>0 , one has 2κ(x)≤κ(x+α′) , for x large enough. •Ψ∈ ∇1 i κB(x)+κB(y)≤κB(x+y) for x, y large enough, with B= e−b . •Ψ∈ ∇0 i for some c′≥1 and A > 1 , one has κA(θ+ 1) −κA(θ)≤ κA(θ′+c′)−κA(θ′) for θ≤θ′ large enough. Before proving the main results of this setion, let us ollet some basi fats on the ompatness of the embedding of HΨ1 into LΨ2(µ) . First: 41
Z∞ xA Ψ′ 2(t) Ψ2(2t)Ψ1(2t/A)m({Mf>8t/A})dt ≤Z∞ 0 Ψ1(2t/A) tm({Mf>8t/A})dt ≤2 AZ∞ 0 Ψ′ 1(2t/A)m({Mf>8t/A})dt =Z∞ 0 Ψ′ 1(x)m({Mf>4x})dx =ZT Ψ11 4Mfdm ≤1 4ZT Ψ1(Mf)dm. whih leads to the desired result. For the pro ofs of Theorem 4.10 and Theorem 4.11, we may restrit ourselves to the ase of funtions Ψ1 and Ψ2 satisfying ∇2 . Indeed, supp ose that Ψ1 and Ψ2 are Orliz funtions and dene f Ψj(t) = Ψj(t2) , for j∈ {1,2} . The funtions f Ψ1 and f Ψ2 are Orliz funtions satisfying ∇2 sine, with β= 2 , we have for every t≥0 : f Ψj(βt) = Ψj(4t2)≥4Ψj(t2) = 2βf Ψj(t). Now, we laim that µ satises ( R ), ( K ), ( R0 ) or ( K0 ) for the ouple Ψ1,Ψ2 if and only if µ satises it for the ouple f Ψ1,f Ψ2 . This is simply due to the fat that for every A > 0 and t≥0 , we have f Ψ2Af Ψ1−1(t)= Ψ2A2Ψ−1 1(t). Moreover, notie that, writing f=Bg2 (where B is a Blashke pro dut), we have f∈HΨ if and only if g∈He Ψ ; thus kgkLe Ψ=pkfkLΨ . It is then lear that Id :HΨ1−→ LΨ2(µ)=Id :Hf Ψ1−→ Lf Ψ2(µ) 2, so that the anonial emb edding is b ounded (resp. ompat) for the ouple Ψ1,Ψ2 if and only if it is so for the ouple f Ψ1,f Ψ2 , thanks to Prop osition 4.9. Pro of of Theorem 4.10. 1) Let C b e the norm of the anonial emb edding j:HΨ1֒→LΨ2(µ) , and let ξ∈T and h∈(0,1) . It sues to test the ontinuity of j on f= Ψ−1 1(1/h)uξ,1−h , whih is in the unit ball of HMΨ1 , by Corollary 3.10. But, when z∈W(ξ, h) one has, with a= (1 −h)ξ : |1−¯az| ≤ |1−¯aξ|+|¯aξ −¯az|=h+ (1 −h)hξ−z |z|+z |z|−zi ≤h+ (1 −h)[h+ (1 −|z|)] ≤h+ (1 −h)[h+h]≤3h; 48
hene |uξ,1−h(z)| ≥ 1/9 and |f(z)| ≥ (1/9)Ψ−1 1(1/h) ; therefore: 1≥ZD Ψ2|f| Cdµ ≥Ψ21 9CΨ−1 1(1/h)µW(ξ, h), whih is ( R ). 2) By Prop osition 3.5, the maximal (non-tangential) funtion M is b ounded on LΨ1(T) : there exists a onstant C≥1 suh that kMfkΨ1≤CkfkΨ1 for every f∈LΨ1(T) . We x f in the unit ball of HΨ1 (note that kf/CkΨ1 remains ≤1 ) and use Lemma 4.14, with E=D and f replaed by f/C (here hA= 1 )). Writing ˜ C=π 2+µ(D)Ψ2(xA) , we get: ZD Ψ2A 8C˜ C|f|dµ ≤1 ˜ CZD Ψ2A 8C|f|dµ ≤1 ˜ C µ(D)Ψ2(xA) + π 2ZT Ψ11 CMfdm! ≤1 ˜ Cµ(D)Ψ2(xA) + π 2= 1, whih means that kfkLΨ2(µ)≤8C˜ C A· Pro of of Theorem 4.11. 1) Supp ose that the emb edding is ompat, but that ondition ( R0 ) is not satised. Then there exist ε0∈(0,1) , A > 0 , a sequene of p ositive numbers (hn)n dereasing to 0 , and a sequene of ξn∈T , suh that: µW(ξn, hn)≥ε0 Ψ2AΨ−1 1(1/hn)· Consider the sequene of funtions fn(z) = Ψ1−1(1/hn)h2 n (1 −¯anz)2= Ψ1−1(1/hn)uξn,|an|, where an= (1 −hn)ξn . By Corollary 3.10, fn is in the unit ball of HMΨ1 . Moreover, it is plain that (fn)n onverges to 0 uniformly on every ompat subset of D . By the ompatness riterion (Prop osition 3.6), (fn)n is norm- onverging to 0 in LΨ2(µ) . But, as ab ove (pro of of Theorem 4.10), for every n≥1 , one has |fn(z)| ≥ (1/9)Ψ−1 1(1/hn) when z∈W(ξn, hn) ; hene: ZD Ψ29A ε0|fn|dµ ≥Ψ2A ε0 Ψ−1 1(1/hn)µW(ξn, hn) ≥Ψ2A ε0 Ψ−1 1(1/hn)ε0 Ψ2AΨ−1 1(1/hn)≥1, 49
by the onvexity of Ψ2 . This implies that kfnkLΨ2(µ)≥ε0/9A and gives a ontradition. 2) We have to prove that for every ε > 0 , there exists an r∈(0,1) suh that the norm of the injetion Ir:HΨ1−→ LΨ2(D\rD, µ) is smaller than ε (see Prop osition 4.9). Let C≥1 b e the norm of the maximal op erator, as in the pro of of Theorem 4.10: kMfkΨ1≤CkfkΨ1 for every f∈LΨ1(T) , and set A= 16C/ε . Condition ( K0 ) gives us hA∈(0,1) suh that: Kµ(h)≤1 2 1/h Ψ2AΨ−1 1(1/h) when h≤hA . Let f in the unit ball of HΨ1 and r∈(0,1) . By Lemma 4.14: ZD\rD Ψ2|f| εdµ =ZD\rD Ψ2A 16C|f|dµ ≤1 2ZD\rD Ψ2A 8C|f|dµ ≤1 2 µD\rDΨ2(xA) + π 2ZT Ψ1Mf Cdm! ≤π 4+ Ψ2(xA)µD\rD. As µ(T) = 0 , there exists some r0∈(0,1) suh that π 4+ Ψ2(xA)µD\rD≤1 , for every r∈(r0,1) . This ends the pro of of 2). 3) Assume that Ψ satises ondition ∇0 and that ondition ( R0 ) is fullled: for every A > 0 and every h∈(0, hA) ( hA small enough), we have: ρµ(h)≤1 ΨAΨ−1(1/h)· This implies that: Kµ(h) = sup 0<s≤h ρµ(s) s≤sup 0<s≤h 1/s ΨAΨ−1(1/s)= sup x≥Ψ−1(1/h) Ψ(x) Ψ(A x)· Fix an arbitrary β > 1 and ho ose A=βCβ>1 , where Cβ is given by the ∇0 ondition for Ψ and Prop osition 4.6. We have, for h small enough and x≥Ψ−1(1/h) : ΨβΨ−1(1/h) ΨΨ−1(1/h)≤Ψ(βCβx) Ψ(x)=Ψ(A x) Ψ(x); we get hene, for h small enough: Kµ(h)≤ΨΨ−1(1/h) ΨβΨ−1(1/h)=1/h ΨβΨ−1(1/h) , 50
and ondition ( K0 ) is fullled. With 1) and 2) previously shown, this nishes the pro of. Remark. Atually, the pro of of Theorem 4.10, 1) gives, for every measure µ on D and every Orliz funtion Ψ : 1≥ZD Ψ|uξ,1−h| kuξ,1−hkLΨ(µ)dµ ≥Ψ1 9kuξ,1−hkLΨ(µ)µW(ξ, h), and hene: (4.12) µW(ξ, h)≤1 Ψ1 9kuξ,1−hkLΨ(µ)· In partiular, if µ=µφ is the image of the Haar measure m under a self-map φ of D , one has: (4.13) µW(ξ, h)≤1 Ψ1 9kCφ(uξ,1−h)kΨ· Condition (4.12) allows to have an upp er ontrol for the µ -measure of Carleson windows, with kuξ,1−hkLΨ(µ) . It is p ossible, onversely, to ma jorize these norms. Denition 4.15 We shal l say that a measure µ on D is a Ψ -Carleson measure if there exists some A > 0 suh that µW(ξ, h)≤1 ΨAΨ−1(1/h) , for every ξ∈T and every h∈(0,1). We shal l say that a measure µ on D is a vanishing Ψ -Carleson measure if, for every A > 0 , lim h→0ΨAΨ−1(1/h).ρµ(h) = 0. Equivalently if, for every A > 0 , there exists hA∈(0,1) suh that µW(ξ, h)≤1 ΨAΨ−1(1/h) , for every ξ∈T and every h∈(0, hA). We have the following haraterizations: Prop osition 4.16 1) µ is a Ψ -Carleson measure on D if and only if there exists some onstant C≥1 suh that : kuξ,1−hkLΨ(µ)≤C Ψ−1(1/h) , for every ξ∈T and every h∈(0,1). 2) µ is a vanishing Ψ -Carleson measure on D if and only if lim h→0sup ξ∈T Ψ−1(1/h)kuξ,1−hkLΨ(µ)= 0. 51
Pro of. The suieny (b oth for 1) and 2)) follows easily from (4.12) in the preeding remark. The onverse is an obvious onsequene of the following lemma. Lemma 4.17 Suppose that there exist A > 0 and h0∈(0,1) suh that: ρµ(h)≤1 ΨAΨ−1(1/h) , for every h∈(0, h0). Then there exists h1∈(0,1) suh that: kuξ,1−hkLΨ(µ)≤24 AΨ−1(1/h) , for every ξ∈T and every h∈(0, h1) . Pro of of the lemma. This is inspired from [13℄, Chapter VI, Lemma 3.3, page 239. We may assume that h≤h0/4≤1/4 . First, writing a= (1−h)ξ (where ξ∈T ), we observe that, when |z−ξ| ≥ bh for a b > 0 , we have: |1−¯az|2= 1 + |a|2|z|2−2|a|ℜ(¯ ξz) =|a||ξ−z|2+ (1 −|a|) + |a|2|z|2−|a||z|2 ≥ |a|b2h2+ (1 −|a|)2≥(|a|b2+ 1)h2. So, we have |uξ,1−h(z)| ≤ 1 |a|b2+ 1 ≤min(1,2/b2) , when |z−ξ| ≥ bh . Now, dene, for every n∈N and ξ∈T : Sn=S(ξ, 2n+1h) = {z∈D;|z−ξ|<2n+1h} ⊂ W(ξ, 2.2n+1h). Our observation implies that |uξ,1−h(z)| ≤ min(1,2/4n) , for every z∈D\Sn−1 . For z∈S0 , one has simply |uξ,1−h(z)| ≤ 1 . There exists an integer N suh that 2N+2h≤h0<2N+3h . Let us ompute: ZD ΨM|uξ,1−h|dµ =ZS0 ΨM|uξ,1−h|dµ + N X n=1 ZSn\Sn−1 ΨM|uξ,1−h|dµ +ZD\SN ΨM|uξ,1−h|dµ ≤Ψ(M)µ(S0) + N X n=1 Ψ2M 4nµ(Sn) + Ψ2M 4Nµ(D) ≤ N X n=0 1 2n+1 Ψ4M 2nµ(Sn) + Ψ2M 4Nµ(D). 52
But for n≤N , we have 2.2n+1h≤2N+2h≤h0 , so the hypothesis gives: µ(Sn)≤1 ΨAΨ−1(1/2n+2h)· Take now: (4.14) M=A 24Ψ−11 h· We have, using that Ψ−11 2n+2h≥1 2n+2 Ψ−11 h , ZD ΨM|uξ,1−h|dµ ≤ N X n=0 1 2n+1 +µ(D)ΨA 12.4NΨ−11 h ≤1 2+µ(D)Ψ16A.h2 3h2 0 Ψ−11 h, b eause 1 4N≤8h h02 . We an hoose h1 small enough to have: µ(D)Ψ16A.h2 3h2 0 Ψ−11 h≤1 2 for every h∈(0, h1) , sine lim h→0h2Ψ−11 h= 0 . We get for suh hZD ΨM|uξ,1−h|dµ ≤1, so that kuξ,1−hkLΨ(µ)≤1 M=24 A 1 Ψ−1(1/h) , as it was announed. Examples and ounterexamples. We are going to give some examples showing that we do not have the reverse impliations in general in Theorem 4.10 and Theorem 4.11. 1. Condition ( R ) is not suient in general to have a ontinuous emb edding. Let Ψ(x) = ex−1 (note that this Orliz funtion even fullls the ∆1 ondition !). Note that Ψ(AΨ−1(1/h)) ∼h−A , when h→0 . a. Let ν b e a probability measure on T , supp orted by a ompat set L of Leb esgue measure zero, suh that ν(I)≤ |I|1/2 , for eah I . We an asso iate to ν the measure on D dened by ˜ν(E) = ν(E∩T) . Then the identity map from HΨ to LΨ(˜ν) is not even dened. Nevertheless the ondition ( R ) is learly fullled with A= 1/2 . 53
b. Now, we exhibit a similar example (less artiial) on the op en disk. Let ν b e as previously. By a standard argument: for every integer n , there exists a funtion gn in the unit ball of the disk algebra suh that |gn|= 1 on L and kgnkΨ≤4−n . As L is ompat, there exists some rn∈(1/2,1) suh that |gn(rnz)| ≥ 1/2 for every z∈L . Now, dene the measure µ by: µ(E) = ∞ X n=1 1 2nνn(E), where: νn(E) = ν{z∈T|rnz∈E}. If W is a Carleson window of size h then, for eah n≥1 , we have: ν{z∈T|rnz∈W}≤ν(W∩T)≤(2h)1/2. Hene, µ(W)≤(2h)1/2 and the ondition ( R ) is fullled. Nevertheless, the identity from HΨ to L1(µ) is not ontinuous: kgnkΨ≤4−n ; but kgnk1≥1 2nZrnT|gn|dνn≥1 2nZL|gn(rnw)|dν(w)≥1 2n+1 · 2. Condition ( K ) is not neessary in general to have a ontinuous embedding. When Ψ satises ∆2 , the identity from HΨ to LΨ(µ) is ontinuous if and only if µ is a Carleson measure. So the onditions ( R ), ( K ) and the ontinuity are equivalent in this ase. Atually, when Ψ do es not satisfy ∆2 , we onstrut b elow a measure µ on D suh that the identity from HΨ to LΨ(µ) is ontinuous and order b ounded , but µ is not a Carleson measure ( a fortiori do es not verify ( K )). Note that the measure µ is then Ψ -Carleson but not Carleson. Here is the example: We have assumed that Ψ do es not satisfy ∆2 ; so there exists an inreasing sequene (an)n≥1 suh that Ψ(an) n is inreasing and Ψ(2an) Ψ(an)≥n2n . Now, dene the disrete measure µ=∞ X n=1 n Ψ(2an)−n+ 1 Ψ(2an+1)δxn, where: xn= 1 −1 Ψ(2an)· As µ[xN,1]=N Ψ(2aN) , the measure µ is not Carleson: it should b e b ounded by c(1 −xN) = c Ψ(2aN) , where c is some onstant. 54
We know that for every f in the unit ball of HΨ and every x∈(0,1) , we have |f(x)| ≤ 4Ψ−11 1−x : see Lemma 3.11. So we only have to see that g∈LΨ(µ) , where g(x) = Ψ−11 1−x . Indeed, we have: ZD Ψ|g| 2dµ =∞ X n=1 n Ψ(2an)−n+ 1 Ψ(2an+1)Ψ|g(xn)| 2 ≤∞ X n=1 n Ψ(2an)Ψ1 2Ψ−11 1−xn ≤∞ X n=1 nΨ(an) Ψ(2an)≤∞ X n=1 1 2n ≤1 so kgkΨ≤2 . 3. Condition ( K0 ) is not neessary in general to have a ompat emb edding. We an nd, for every Orliz funtion Ψ not satisfying ∇0 , a measure µ suh that the identity from HΨ to LΨ(µ) is ompat but ( K0 ) is not satised. Indeed: sine Ψ/∈ ∇0 , we an selet two inreasing sequenes (xn)n≥1 and (yn)n≥1 , with 1≤xn≤yn≤xn+1 and Ψ(xn)>1 , and suh that lim xn= +∞ and: Ψ(2xn) Ψ(xn)≥Ψ(2nyn) Ψ(yn)· Dene rn= 1 −1 Ψ(yn) and the disrete measure: µ=∞ X n=1 1 Ψ(2nyn)δrn. The series onverge sine Ψ(2nyn)≥2n . Thanks to Lemma 3.11, we have, for every f in the unit ball of HΨ and every n≥1 : |f(rn)| ≤ 4Ψ−11 1−rn= 4yn. Given r > r1 , there exists an integer N≥1 suh that rN< r ≤rN+1 . Then, 55
for every f in the unit ball of HΨ , we have kfkLΨ(D\rD,µ)≤2−N+2 , sine: ZD\rD Ψ|f| 2−N+2 dµ =∞ X n=N+1 1 Ψ(2nyn)Ψ|f(rn)| 2−N+2 ≤∞ X n=N+1 Ψ(2Nyn) Ψ(2nyn) ≤∞ X n=N+1 1 2n−N= 1. This implies that lim r→1sup kfkΨ≤1kfkLΨ(D\rD,µ)= 0 . By Prop osition 4.9, the identity from HΨ to LΨ(µ) is ompat. On the other hand, writing hn=1 Ψ(xn) and tn=1 Ψ(yn) , we have: Kµ(hn)≥µ([1 −tn,1]) tn = Ψ(yn)∞ X m=n 1 Ψ(2mym)≥Ψ(yn) Ψ(2nyn) ≥Ψ(xn) Ψ(2xn)=1/hn Ψ2Ψ−1(1/hn) , and this shows that ( K0 ) is not satised. 4. Condition ( R0 ) is not suient in general to have a ompat embedding. We an nd an Orliz funtion Ψ and a vanishing Ψ -Carleson measure ( i.e. ( K0 ) is satised) µ suh that the identity from HΨ to LΨ(µ) is not ompat. We shall use the Orliz funtion introdued in [25℄. The key prop erties of this funtion Ψ are: i) For every x > 0 , Ψ(x)≥x3/3 . ii) For every integer k≥1 , Ψ(k!) ≤(k!)3 . iii) For every integer k≥1 , Ψ(3(k!)) > k.(k!)3 . One again, the job is done by a disrete measure. Dene xk=k! ; yk=(k+ 1)! k1/3 ; rk= 1 −1 Ψ(yk) and ρk= 1 −1 Ψ(xk)· Of ourse, x2< y2< x3<··· . Let ν b e the disrete measure dened by: ν=∞ X k=2 νk, where: νk=1 Ψ(k+ 1)!X ak2=1 δrka. 56
Observe that kνkk ≤ k2 Ψ(k+ 1)!≤3.k2 (k+ 1)!3 so that the series onverges. Note that ν is supp orted in the union of the irles of radii rk and not in a subset of the segment [0,1] as in the preeding ounterexamples. In order to show that ( R0 ) is satised, it is learly suient to prove that, when 1 Ψ(yk)≤h < 1 Ψ(yk−1) (with k≥3 ), we have: ρν(h)≤1 Ψk1/3 2Ψ−1(1/h)· Supp osing then 1 Ψ(yk)≤h < 1 Ψ(yk−1) , we have Ψ−1(1/h)≤yk so Ψk1/3 2Ψ−1(1/h)≤1 2Ψ(k+ 1)!. Therefore, it is suient to establish that ρν(h)≤2 Ψ(k+ 1)!· A Carleson window W(ξ, h) (where ξ∈T ) an ontain at most one k2 -ro ot of the unity, sine 2h < 2 Ψ(yk−1)≤6 y3 k−1≤6(k−1) (k!)3≤2π k2· This implies that νkW(ξ, h)≤1 Ψ(k+ 1)!· Nevertheless, when j < k , the window W(ξ, h) annot meet any irle of radius rj (entered at the origin), so νjW(ξ, h)= 0 . We obtain: νW(ξ, h)=∞ X j=k νjW(ξ, h)≤1 Ψ(k+ 1)!+X j>k j2 Ψ(j+ 1)! ≤1 Ψ(k+ 1)!+∞ X j=k+1 3j2 (j+ 1)!3 ≤1 Ψ(k+ 1)!+3 (k+ 1)!3 ∞ X s=1 1 k+ 1s ≤2 Ψ(k+ 1)!· This proves that ( R0 ) is satised. Let us intro due the funtion fk(z) = xku1,ρkzk2=xk1−ρk 1−ρkzk22 . It lies in the unit ball of HΨ by Corollary 3.10: 57
Let J b e the ar: J={eit ;−ρ < t < ρ}. The map ψ is regular on J , and so there exist two onstants γ1 and γ2>0 suh that for every Borel subset E of J , one has: γ1m(E)≤mψ(E)≤γ2m(E). If now β∈(0,1) , we put: ψβ(z) = (ψ(z))β. Then, it is easy to see that for every Borel subset E of J , one has: mψβ(E)=β mψ(E), and so γ1β m(E)≤mψβ(E)≤γ2β m(E). In order to prove (4.20) , onsider the funtion F:D→C dened by: F(z) = feiδψβ(z), where β= 4δ/π . Then: Re F(0)=u(1 −√2)βeiδ≤Nu(eiδ)≤α. Let us all χ the map: χ(z) = eiδψβ(z). It is lear that I+ is ontained in χ(J) . If A={|f∗|> λ}∩I+ , then E=χ−1(A) is a Borel subset of J , and for every ξ∈E , one has |F∗(ξ)|> λ . Then: m(A) = mχ(E)=mψβ(E)≤γ2β m(E) ≤8γ2 δ 2πm({|F∗|> λ}) = 8γ2m(I+)m({|F∗|> λ}) and using Lemma 4.24 for F , ≤8γ2m(I+)×8c3Re F(0) λ3 ≤64γ2c3α λ3m(I+) = k2α λ3m(I+). The pro of of (4.20) is nished, and Prop osition 4.25 follows. 64
4.4.3 Pro of of Theorem 4.19 Using the fat h7→ µφS(ξ, h) is nondereasing, it is enough to prove that there exist h0>0 , and ε0>0 , suh that (4.15) is true for 0< h < h0(1−|φ(0)|) , and 0< ε < ε0 , b eause hanging the onstant k1 , if neessary, the theorem will follow. We an also supp ose that ξ= 1 . The real part of 1/1−φ(z) is positive, in fat greater than 1/2, for every z∈D . Take 0< h < h0 , and onsider the analyti funtion f dened by: f(z) = h 1−φ(z)1/3 , where the ubi root is taken in order that, for every z∈D , f(z) b elongs to the one C={z∈C:−π/6<Arg (z)< π/6}. Clearly µφS(1, h)=m({|f∗|>1}) and µφS(1, εh)=m({|f∗|>1/3 √ε}) . We also have: |f(0)| ≤ h 1−|φ(0)|1/3 <3 ph0. We shall write f=u+iv where u and v are real-valued harmoni funtions. Observe that: |v(z)|<1 √3u(z), for every z∈D . It is known that f∈Hp , for every p < 3 (see [12℄, Theorem 3.2), and so u and v are the Poisson integrals of u∗ and v∗ (in partiular u, v ∈h1 ). We are lo oking for a ontrol of m({|f∗|>1/3 √ε}) by ε times m({|f∗|>1}) . Prop osition 4.26 provides this ontrol replaing m({|f∗|>1}) by m({Nu > 2}) : m({|f∗|>1/3 √ε})≤8k2εm({Nu > 2}), when m({Nu > 2})<1/4 . As f is valued in C , |f∗| is ontrolled by u∗ . Then what we need in fat is a ontrol of the measure of level sets of Nu by the measure of level sets of u∗ . This will b e done by using Prop osition 4.27. Indeed, by Prop osition 4.27, we have: m({Nu > 2})≤m({Mu∗>2})≤k4m({u∗>1}). We know that ||u∗||1=u(0) ≤ |f(0)| ≤ h1/3 0 . Then, hoosing h0 small enough, we an have k4h1/3 0<1/4 , and so m({Nu > 2})<1/4 ; therefore, we an use Prop osition 4.26. Moreover we also have {u∗>1} ⊆ {|f∗|>1} . Taking ε0<3√3/64 , we have, for 0< ε < ε0 , by Prop osition 4.26: m({|f∗|>1/3 √ε})≤8k2ε m({Nu > 2})≤8k4k2ε m({u∗>1}) ≤k1ε m({|f∗|>1}), taking k1= 8k4k2 . 65
5 Bergman spaes 5.1 Bergman-Orliz spaes Denition 5.1 Let dA(z) = dx dy π ( z=x+iy ) be the normalized Lebesgue measure on D . The Bergman-Orliz spae BΨ denotes the spae of analyti funtions f:D→C whih are in the Orliz spae LΨ(D, dA) . The Bergman- Morse-Transue spae is the subspae BMΨ=BΨ∩MΨ(D, dA). BΨ , equipp ed with the indued norm of LΨ(D, dA) , is a Banah spae, as an obvious onsequene of the following lemma, analogous to Lemma 3.11. Lemma 5.2 For every a∈D , the norm of the evaluation funtional δa , whih maps f∈BΨ to f(a) , is: kδak ≈ Ψ−11 (1 −|a|)2· Pro of. For every analyti funtion g:D→C , the mean-value prop erty gives: g(0) = ZD g(z)dA(z). Hene if φa:D→D denotes the analyti automorphism φa(z) = z−a 1−¯az , whose inverse is φ−1 a=φ−a , one has, for every f∈BΨ , using the hange of variable formula: f(a) = f◦φ−a(0) = ZD f◦φ−a(z)dA(z) = ZD f(w)|φ′ a(w)|2dA(w) =ZD f(w)Ha(w)dA(w), where: Ha(w) = |φ′ a(w)|2=(1 −|a|2)2 |1−¯aw|4· The kernel Ha plays for BΨ the role that the Poisson kernel Pa plays for HΨ : the analyti reproduing kernel Ka for B2 b eing Ka(z) = 1 (1−¯az)2· We therefore have (using [30℄, Prop osition 4, page 61, or [5℄, Theorem 8.14): |f(a)| ≤ 2kfkΨkHakΦ, whih proves the ontinuity of δa . To estimate its norm, we are going to ma jorize kHakΦ , with the help of Lemma 3.9. Let us notie that, on the one hand, kHak1= 1 (take f= 1I in the ab ove identity); and on the other hand: kHak∞=(1 −|a|2)2 (1 −|a|)4=(1 + |a|)2 (1 −|a|)2· 66
We get, setting b=kHak∞ , and using Lemma 3.9 for k kΦ : kHakΦ≤b Φ−1(b)· But b≤Φ−1(b)Ψ−1(b) (see [30℄, Prop osition 1 (ii), page 14). Hene kHakΦ≤ Ψ−1(b) . Now: b≤4 (1 −|a|)2· We have Ψ−1(4t)≤4Ψ−1(t) for all t > 0 . It follows that kHakΦ≤CΨ−11 (1 −|a|)2· Sine kδak ≤ 2kHakΦ , we get the upp er b ound in Lemma 5.2. For the lower bound, we simply observe that Ha=|Ga| , where Ga(z) = (1 −|a|2)2 (1 −az)4 , and by Lemma 3.9: kδak ≥ |Ga(a)| kGakΨ =|Ha(a)| kHakΨ≥ 1 (1 −|a|2)2 b/Ψ−1(b)=Ψ−1(b) b(1 −|a|2)2 =Ψ−1(b) (1 + |a|)4≥1 16Ψ−1(b)≥1 16Ψ−11 (1 −|a|)2, sine b≥1/(1 −|a|)2 . Prop osition 5.3 We have the fol lowing properties i) BMΨ is the losure of H∞(D) in LΨ(D,A) and atual ly the algebrai polynomials are dense in BMΨ . ii) On the unit bal l of BΨ , the weak-star topology σLΨ(D,A), MΦ(D,A) oinides with the topology of onvergene on ompat subsets of D . iii) BΨ is losed in LΨ(D,A) for the weak-star topology. iv) If Ψ∈ ∇2 , BΨ is (isometri to) the bidual of BMΨ . Pro of. For the rst p oint, let us x f∈BMΨ . Setting fr(z) = f(rz) for z∈D and 0≤r < 1 , it sues to show that kfr−fkΨ−→ r→10 , sine, b eing analyti in the disk rD⊂D , fr an b e uniformly approximated on D by its Taylor series. But the norm of MΨ is absolutely ontinuous (see [30℄, Theorem 14, page 84) and therefore, for every ε > 0 , there is some R > 0 , with 1/3≤R < 1 , suh that kf1ID\RDkΨ≤ε ; hene: ZD\RD Ψ|f| 4εdA≤ZD\RD 1 4Ψ|f| εdA≤1 4· 67
When r≥2R R+1 ≥1/2 , we therefore have: ZD\1+R 2D Ψ|fr| 4εdA≤1, and, by onvexity of Ψ : ZD Ψ|fr−f| 8εdA≤Z1+R 2D Ψ|fr−f| 8εdA +ZD\1+R 2D 1 2hΨ|fr| 4ε+ Ψ|f| 4εidA ≤Z1+R 2D Ψ|fr−f| 8εdA +1 2ZD\1+R 2D Ψ|fr| 4εdA+1 2ZD\RD Ψ|f| 4εdA ≤1, for r lose enough to 1 sine fr−f tends to 0 uniformly on 1+R 2D . Hene, for some r0<1 , one has kfr−fkΨ≤8ε for r0≤r < 1 . This was the laim. ii) It sues to use a sequential argument sine the top ologies are metrizable on balls (the spae MΦ(D,A) is separable). Assume that f∈BΨ (with kfkΨ≤ 1 ) is the weak-star limit of a sequene of analyti funtions fn∈BΨ (with kfnkΨ≤1 ). Testing this with the funtion hk(z) = (k+ 1)zk , we obtain that the kth Taylor o eient ak(n) of fn onverges to the kth Taylor o eient ak of f : ak(n) = ZD fnhkdA−→ ZD fhkdA=ak, for every k≥0. Fix a ompat K⊂D , there exists an r∈(0,1) suh that K⊂rD . We have: sup z∈K|fn(z)−f(z)| ≤ X k≥0|ak(n)−ak|rk−→ 0 by the dominated onvergene Theorem (observe that |ak−ak(n)| ≤ 2(k+ 1) for every k≥0 and every n≥0 ). For the onverse, supp ose now that fn∈BΨ (with kfnkΨ≤1 ) onverges uniformly on every ompat subsets of D to f∈BΨ (with kfkΨ≤1 ). Fixing g∈MΦ(D,A) and ε > 0 , there exists an r∈(0,1) suh that kg1ID\rDkΦ≤ε . We have: ZD (fn−f)gdA≤ZrD (fn−f)gdA+ 2ε≤sup z∈rD|fn(z)−f(z)|.kgk1+ 2ε. By hyp othesis sup z∈rD|fn(z)−f(z)| −→ 0 . The onlusion follows. 68
iii) By the lassial Theorem of Banah-Dieudonné, it is suient to prove that the balls are weak-star losed (equivalently weak-star ompat) and by separability of MΦ(D,A) , the weak-star top ology is metrizable on balls. The previous fat ii) shows that it is equivalent to prove that the unit ball of BΨ is ompat for the top ology of onvergene on ompat subsets. But this is easy: indeed, if fn in the unit ball of BΨ . This is a normal family thanks to Lemma 5.2. A subsequene onverges to an analyti funtion f on ompat subsets of D and the Fatou Lemma implies that f atually lies in the unit ball. iv) Assume now that Ψ satises ∇2 . Sine (MΨ)∗∗ =LΨ(D,A) , we have (BMΨ)∗∗ =BMΨw∗ , in the spae LΨ(D,A) . Hene it sues to show that BMΨw∗ =BΨ . We already know that BΨ is weak-star losed. Now, let f∈BΨ . Obviously, fr∈BMΨ for every r∈(0,1) , where fr(z) = f(rz) . Moreover kfrkΨ≤ kfkΨ . But this is lear that fr is uniformly onvergent to f on ompat subsets of D , when r→1− . By ii), the onlusion follows. In the previous pro of, we an see the p oints ii) and iii) in a slighlty dierent way: the unit ball B of BΨ is ompat for the top ology τ of uniform onvergene on ompat subsets. Then observe that the identity from B , equipp ed with the top ology τ , to B , equipp ed with the weak-star top ology, is ontinuous (the sequential argument is suient by metrizability). This implies, sine the weakstar top ology is separated, that B is weak-star ompat (hene losed) and that the top ologies oinide. Now, by Banah-Dieudonné, the spae BΨ is weak-star losed. 5.2 Compat omp osition op erators on Bergman-Orliz spaes We shall b egin with showing that, as in the Hardy-Orliz ase, every symb ol φ denes a b ounded omp osition op erator. Prop osition 5.4 Every analyti self-map φ:D→D indues a bounded omposition operator Cφ:BΨ→BΨ . Moreover, Cφf∈BMΨ for every f∈BMΨ ; hene, when Ψ∈ ∇2 , the former operator is the bi-transposed of Cφ:BMΨ→ BMΨ . Pro of. It sues to follow the lines of Prop osition 3.12, and to integrate the integrals written there b etween 0 and 1 , with resp et to the measure 2rdr . Before stating and proving the main theorem, we are going to prove the following auxiliary result, interesting in itself, and whih reinfores an example of J. H. Shapiro ([36℄, Example, page 185). Prop osition 5.5 There exists a Blashke produt B having angular derivative at no point of T=∂D , in the fol lowing sense: (5.1) (∀ε > 0) (∃cε>0) 1 −|B(z)| ≥ cε(1 −|z|)ε,∀z∈D. 69
Pro of. We shall take: B(z) = +∞ Y n=1 |zn| zn zn−z 1−¯znz , where: {zn;n≥1}=[ n≥1 An, with: An={rnωj n;ωn= e2πi/pn,0≤j≤pn−1}, where (rn)n≥1 is a (stritly) inreasing sequene with 0< rn<1 , and the integers pn will have to be adjusted, satisfying the Blashke ondition: +∞ X n=1 (1 −|zn|) = +∞ X n=1 pn(1 −rn)<+∞. One has: |B(z)|2= +∞ Y n=1 zn−z 1−¯znz 2= +∞ Y n=1 1−(1 −|z|2)(1 −|zn|2) |1−¯znz|2≤exp −S(z), where: S(z) = +∞ X n=1 (1 −|z|2)(1 −|zn|2) |1−¯znz|2· We now proeed to minorize S(z) . For this purpose, we shall need the following simple lemma, whose pro of will b e temp orarily p ostp oned. Lemma 5.6 For every positive integer p and every a∈D , one has, setting ω= e2πi/p : 1 p p−1 X k=0 1 |1−aωk|2=1−|a|2p 1−|a|2 1 |1−ap|2≥1 4p|a|p. Then, setting r=|z| , we have: S(z)≥(1 −r) +∞ X n=1 (1 −rn) pn−1 X k=0 1 |1−rnω−k nz|2 ≥1−r 4 +∞ X n=1 p2 n(1 −rn)(rnr)pn. We shall take: pn=(1 −rn)εn−1+ 1 , with: εn=1 √n and rn= 1 −1 2n 70
and where [ ] stands for the integer part. More expliitly: pn=2n−√n+ 1. If r≥1/2 , let N≥1 b e suh that rN< r ≤rN+1 . One has: S(z)≥1−r 4(1 −rN)2εN−1r2pN N≥1 8(1 −rN)2εNr2pN N, sine 1−r≥1−rN+1 = (1 −rN)/2 . Moreover: pN≤2(1 −rN)εN−1= 2.2N(1−εN)≤2.2N, so that: S(z)≥1 8(1 −r)2εN(1 −2−N)2N+1 ≥c(1 −r)2εN, where c is a p ositive numerial onstant. Hene, setting: ε(z) = 2εN for |z| ≥ 1/2 and rN< r ≤rN+1 , one has: ε(z)−→ |z|< →1 0, and we get: 1−|B(z)|2≥1−e−S(z)≥1−e−c(1−|z|)ε(z)≥c′(1 −|z|)ε(z), where c′ is another p ositive numerial onstant. This gives ondition (5.1), sine 1−|B(z)| ≥ 1−|B(z)|2 2· Finally, the Blashke ondition is satised, sine: pn(1 −rn)≤2(1 −rn)εn= 2.2−√n. This ends the pro of of Prop osition 5.5. Pro of of Lemma 5.6. Let Gp b e the nite group of pth ro ots of unity, equipp ed with its normalized Haar measure. For u:Gp→C and 0≤k≤p−1 , we denote by ˆu(k) the kth Fourier o eient of u , i.e. : ˆu(k) = 1 pX z∈Gp u(z)z−k. Then, the Planherel-Parseval formula for Gp reads: p−1 X k=0 |ˆu(k)|2=1 pX z∈Gp|u(z)|2. 71
Applying this to u(z) = 1 1−az =X l≥0 0≤k≤p−1 alp+kzk= p−1 X k=0 ˆu(k)zk, with ˆu(k) = X l≥0 alp+k=ak 1−ap , we get: 1 p p−1 X k=0 1 |1−aωk|2= p−1 X k=0 |a|2k |1−ap|2=1−|a|2p (1 −|a|2)|1−ap|2· To nish, we note that |1−ap| ≤ 2 , and that, by the arithmetio-geometri inequality, we have, with x=|a|2 : 1−|a|2p 1−|a|2= 1 + x+···+xp−1 ≥px1+2+···+(p−1)1/p =p xp−1 2≥p xp/2=p|a|p. Theorem 5.7 If the omposition operator Cφ:BΨ→BΨ is ompat, then (5.2) Ψ−11 (1 −|φ(a)|)2 Ψ−11 (1 −|a|)2−→ |a|< →1 0. This ondition is suient if Ψ∈∆2 . Before giving the pro of of this theorem, let us note that in ase where Ψ = Ψ2 , with Ψ2(x) = ex2−1 , it reads: the omposition operator Cφ:BΨ2→BΨ2 is ompat if and only if: (5.3) (∀ε > 0) (∃cε>0) 1 −|φ(z)| ≥ cε(1 −|z|)ε,∀z∈D. Indeed, Ψ2∈∆2 , and: Ψ−1 21 (1 −|φ(a)|)2 Ψ−1 21 (1 −|a|)2=rlog 1 (1 −|φ(a)|)2 rlog 1 (1 −|a|)2 ; hene Cφ is ompat if and only if log 1 (1 −|φ(a)|)2 log 1 (1 −|a|)2 −→ |a|→10. 72
Then, for every ε > 0 , we an nd some cε>0 suh that, for all a∈D : log 1 1−|φ(a)|≤εlog 1 (1 −|a|)2+cε; whih is equivalent to (5.3). That allows to have ompat omposition op erators on BΨ2 whih are not ompat on HΨ2 . However it is very likely that this is the ase for every Ψ∈∆2 , but we have not tried to see this full generality. Theorem 5.8 There exist symbols φ:D→D suh that the omposition operators Cφ is ompat from BΨ2 into itself, but not ompat from HΨ2 into itself, and even Cφ is an isometry onto its image. A similar example is well-known for the Hilb ert spaes B2 and H2 (see [36℄, pages 180186). Pro of. Let B b e a Blashke pro dut verifying the ondition of Prop osition 5.5. We intro due φ(z) = zB(z) . The funtion still veries 1−|φ(z)| ≥ Cε(1 −|z|)ε , sine |φ(z)| ≤ |B(z)| on D . From Theorem 5.7, it follows, sine Ψ2∈∆2 , that Cφ:BΨ2→BΨ2 is ompat. We are now going to see that Cφ:HΨ2→HΨ2 is an isometry. Indeed, reall the following well-known fat, that we already used (see [29℄, Theorem 1): sine φ is an inner funtion, the image φ(m) of the Haar measure m of T under φ is equal to Pa.m , where a=φ(0) and Pa is the Poisson kernel at a . Here φ(0) = 0 so φ(m) = m . It follows that for every f∈HΨ2 , one has, for C > 0 : ZT Ψ2|f◦φ| Cdm =ZT Ψ2|f| Cdm so that kfkΨ2=kf◦φkΨ2 . We shall need also the following lemma, whih ompletes Lemma 5.2. Lemma 5.9 For every f∈BMΨ , one has: f(a) = oΨ−11 (1 −|a|)2 as |a|< −→1. Pro of. This is obvious for the monomials en:z7→ zn sine |en(a)| ≤ 1 , whereas Ψ−11/(1 −|a|)2−→ |a|→1+∞ . Sine the evaluation δa is b ounded on BMΨ and δa/Ψ−1(1/1−|a|2)=O(1) , it sues to use that the p olynomials are dense in BMΨ ; but this was already proved in Prop osition 5.3. Pro of of Theorem 5.7. If Cφ:BΨ→BΨ is ompat, then so is the restrition Cφ:BMΨ→BMΨ and its adjoint C∗ φ=Cφ:BMΨ∗→BMΨ∗ . Sine 73