The canonical injection of the Hardy-Orlicz space HΨ into the Bergman–Orlicz space BΨ

arXiv:1005.1996v1 [math.FA] 12 May 2010 The canonical injection of the Hardy-Orlicz space HΨinto the Bergman-Orlicz space BΨ Pascal Lefèvre, Daniel Li, Hervé Queffélec, Luis Rodríguez-Piazza May 13, 2010 Abstract. We study the canonical injection from the Hardy-Orlicz space HΨ into the Bergman-Orlicz space BΨ. Mathematics Subject Classification. Primary: 46E30 – Secondary: 30D55; 30H05; 32A35; 32A36; 42B30 Key-words. absolutely summing operator – Bergman-Orlicz space – compactness – Dunford-Pettis operator – Hardy-Orlicz space – weak compactness 1 Introduction and notation 1.1 Introduction There are two natural Orlicz spaces of analytic functions on the unit disk Dof the complex plane: the Hardy-Orlicz space HΨand the Bergman-Orlicz space BΨ. It is well-known that in the classical case Ψ(x) = xp,Hp⊆Bpand the canonical injection Jpfrom Hpto Bpis bounded, and even compact. In fact, for any Orlicz function Ψ, one has HΨ⊆BΨand the canonical injection JΨ:HΨ→BΨis bounded, but we shall see in this paper that its compactness requires that Ψdoes not grow too fast. We actually characterize in Section 2 the compactness: JΨis compact if and only if limx→+∞Ψ(Ax)/[Ψ(x)]2= 0 for every A > 1, and the weak compactness: JΨis weakly compact if and only if lim supx→+∞Ψ(Ax)/[Ψ(x)]2<+∞for every A > 1. We show that, if these two properties are “often” equivalent (this happens for example if Ψ(x)/x is non-decreasing for xlarge enough), it is not always the case. We actually show a stronger result in Section 4: there is an Orlicz function Ψsuch that JΨis weakly compact and is Dunford-Pettis, but such that JΨis not compact. 1.2 Notation An Orlicz function is a non-decreasing convex function Ψ: [0,+∞[→[0,+∞[ such that Ψ(0) = 0 and Ψ(∞) = ∞. One says that the Orlicz function Ψhas 1 property ∆2(Ψ∈∆2) if Ψ(2x)≤CΨ(x)for some constant C > 0and xlarge enough. It is equivalent to say that, for every β > 1,Ψ(βx)≤CβΨ(x). It is known that if Ψ∈∆2, then Ψ(x) = O(xp)for some 1≤p < +∞. One says (see [6], [7]) that Ψsatisfies the condition ∆0if, for some β > 1, one has lim x→∞ Ψ(βx)/Ψ(x) = +∞. If Ψ∈∆0, then Ψ(x)/xp−→ x→∞ +∞for every 1≤p < ∞. Indeed, let 1≤p < ∞. For every β > 1one can find x0>0such that Ψ(βx)/Ψ(x)≥βpfor x≥x0; then Ψ(βnx0)≥βnpΨ(x0)for every n≥1. That implies that Ψ(x)≥Cpxpfor every x > 0large enough. Since p≥1is arbitrary, we get xp=o[Ψ(x)]. We say that Ψ∈ ∇0(1) if, for every A > 1,Ψ(Ax)/Ψ(x)is non-decreasing for xlarge enough. This is equivalent to say (see [7], Proposition 4.7) that log Ψ(ex)is convex. When Ψ∈ ∇0(1), one has either Ψ∈∆2, or Ψ∈∆0. If (S, S, µ)is a finite measure space, one defines the Orlicz space LΨ(µ)as the set of all (classes of) measurable functions f:S→Cfor which there is a C > 0such that RSΨ(|f|/C)dµ is finite. The norm kfkΨis the infimum of all C > 0for which the above integral is ≤1. The Morse-Transue space MΨ(µ)is the subspace of f∈LΨ(µ)for which RSΨ(|f|/C)dµ is finite for all C > 0; it is the closure of L∞(µ)in LΨ(µ). One has MΨ(µ) = LΨ(µ)if and only if Ψ∈∆2. If Ψ(x)/x −→ x→+∞+∞, the conjugate function Φof Ψis defined by Φ(y) = supx>0xy −Ψ(x). It is an Orlicz function and [MΨ(µ)]∗=LΦ(µ), isomorphically. We may note that if Ψ(x)/x does not converges to infinity, we must have Ψ(x)≤ax for some a≥1and xlarge enough. Then LΨ(µ) = L1(µ)isomorphically and then Φ(y) = +∞for y > a (giving LΦ(µ) = L∞(µ)isomorphically). We denote by Dthe open unit disk of Cand by T=∂Dthe unit circle. The normalized area-measure on Dis denoted by Aand the normalized Lebesgue measure on Tis denoted by m. The Hardy-Orlicz space HΨis defined as {f∈H1;f∗∈LΨ(m)}, where f∗is the boundary values function of f, and HMΨ=HΨ∩MΨ(m)is the closure of H∞in HΨ. The Bergman-Orlicz space BΨis the subspace of analytic f∈LΨ(A), and BMΨ=BΨ∩MΨ(A)is the closure of H∞in BΨ. Since, for f∈HΨ,kfkHΨ= sup0<r<1kfrkHΨ(see [7], Proposition 3.1), where fr(z) = f(rz), one has: Z2π 0 Ψ|f(reit)| kfkHΨdt 2π≤Z2π 0 Ψ|f(reit)| kfrkHΨdt 2π≤1 ; hence: ZD Ψ|f(reit)| kfkHΨdA=Z1 0Z2π 0 Ψ|f(reit)| kfkHΨdt 2π2r dr ≤1, so f∈BΨand kfkBΨ≤ kfkHΨ. It follows that HΨ⊆BΨand the canonical injection JΨ:HΨ→BΨis bounded, and has norm 1. Let us point out that 2 the boundedness also follows from [7], Theorem 4.10, 2), since JΨis a Carleson embedding JΨ:HΨ→BΨ⊆LΨ(A). This injection is not onto, since there are functions f∈BΨwith no radial limit on a subset of Tof positive measure (the proof is the same as in Bp: see [4], § 3.2, Lemma 2, page 81). Note that JΨis not an into-isomorphism: take fn(z) = zn, for every n∈N; it is easy to see that {fn}ntends to 0in BΨ, but not in HΨ. Acknowledgment. This work is partially supported by a Spanish research project MTM 2009-08934. Part of this paper was made during an invitation of the second-named author by the Departamento de Análisis Matemático of the Universidad de Sevilla. It is a pleasure to thanks the members of this department for their warm hospitality. 2 Compactness and weak-compactness In order to characterize the compactness and the weak-compactness of JΨ, we introduce the following quantity QA,A > 1: (2.1) QA= lim sup x→+∞ Ψ(Ax) [Ψ(x)]2, which will turn out to be essential. We are going to start with the compactness. Theorem 2.1 The canonical injection JΨ:HΨ→BΨis compact if and only if (2.2) lim x→+∞ Ψ(Ax) [Ψ(x)]2= 0 for every A > 1. Remarks. 1) Condition (2.2) means that QA= 0 for every A > 1. It is equivalent to say that: (2.3) sup A>1 QA<+∞. Indeed, assume that M:= supA>1QA<+∞. Let 0< ε ≤1and A > 1; we can find xA=xA(ε)>0such that Ψ(Ax/ε)/[Ψ(x)]2≤2Mfor x≥xA. By convexity, one has Ψ(Ax)≤εΨ(Ax/ε), and hence Ψ(Ax)/[Ψ(x)]2≤2εM for x≥xA. We get QA= 0. 2) It is clear that condition (2.2) is satisfied whenever Ψ∈∆2, but Ψ(x) = e[log(x+1)]2−1satisfies (2.2) without being in ∆2. However, condition (2.2) implies that Ψcannot grow too fast. More precisely, we must have Ψ(x) = o(exα)for every α > 0. 3 Indeed, one has Ψ(At)≤[Ψ(t)]2for t≥tA, and, by iteration, Ψ(AntA)≤ [Ψ(tA)]2nfor every n≥1. For every x > 0large enough, taking n≥1such that AntA≤x < An+1tA, we get Ψ(x)≤C1eC2xα, with α= log 2/log A. Since A > 1is arbitrary, αmay be any positive number. The little-oh condition follows from the fact that the inequality is true for all α > 0. Proof of Theorem 2.1. By definition, BΨis a subspace of LΨ(D,A); hence we can see JΨas a Carleson embedding JΨ:HΨ→LΨ(D,A). If S(ξ, h) = {z∈ D;|z−ξ|< h}, the compactness of JΨimplies, by [7], Theorem 4.11, that, for every A > 1, every ε > 0, and h > 0small enough: h2≤4A[S(ξ, h)] ≤4ε Ψ[AΨ−1(1/h)] , that is, setting x= Ψ−1(1/h),Ψ(Ax)≤4ε[Ψ(x)]2, and (2.2) is satisfied. Conversely, one has: sup 0<t≤h sup |ξ|=1 A[S(ξ, t)] t≤sup 0<t≤h t2 t=h , which is o(1/h)/Ψ[AΨ−1(1/h)]for every A > 1, if (2.2) holds; hence, by [7], Theorem 4.11, again, JΨis compact.  We now turn ourself to the weak compactness. Theorem 2.2 The following assertions are equivalent: (a) JΨ:HΨ→BΨis weakly compact; (b) JΨfixes no copy of c0; (c) JΨfixes no copy of ℓ∞; (d) QA<+∞, for every A > 1; (e) HΨ⊆BMΨ; (f) JΨis strictly singular. Recall that an operator T:X→Ybetween two Banach spaces is said to be strictly singular if there is no infinite-dimensional subspace X0of Xon which Tis an into-isomorphism. The proof will be somewhat long, and before beginning it, we shall remark that if Ψ∈∆0, then condition (2.4) QA<+∞for every A > 1 implies condition (2.2). Indeed, if lim x→+∞ Ψ(βx) Ψ(x)= +∞, we get, for every A > 1: lim sup x→+∞ Ψ(Ax) [Ψ(x)]2= lim sup x→+∞ Ψ(Ax) Ψ(βAx) Ψ(βAx) [Ψ(x)]2≤lim sup x→+∞ Ψ(Ax) Ψ(βAx)QβA = 0 . Now, if, for some A > 1,Ψ(Ax)/Ψ(x)is non-decreasing for xlarge enough (in particular if Ψ∈ ∇0(1)), one has the dichotomy: either Ψ∈∆2, and then JΨ is compact; or Ψ∈∆0and hence the weak compactness of JΨimplies, by the two above theorems, its compactness. Hence: 4 Proposition 2.3 If, for some A > 1,Ψ(Ax)/Ψ(x)is non-decreasing, for x large enough, then the weak compactness of JΨis equivalent to its compactness. However, it is easy to construct an Orlicz function Ψwhich satisfies condition (2.4), but not condition (2.2). We do not give an axample here because we have a stronger result in Section 4. In order to prove Theorem 2.2, we shall need several lemmas. Lemma 2.4 Let Ψbe any Orlicz function. If we define Ψ1(t) = [Ψ(t)]2,t≥0, then Ψ1is an Orlicz function for which HΨ⊆BΨ1and the canonical injection of HΨinto BΨ1is continuous. Proof. It is enough to see that HΨcontinuously embeds into LΨ1(A), and for this we can use Theorem 4.10 in [7]. Following the notation of that theorem for the measure µ=A, it is easy to see that, as h→0+,ρA(h)≈h2, and KA(h)≈h. Observe that, for t > 1, we have Ψ1[Ψ−1(t)] = t2, and so, for h∈(0,1), 1/h Ψ1[Ψ−1(1/h)] =1/h 1/h2=hKA(h). Using part 2) of Theorem 4.10 in [7], the lemma follows.  Lemma 2.5 Let M > δ > 0and {fn}nbe a sequence in HΨ∩BMΨsuch that: (a) {fn}ntends to 0uniformly on compact subsets of D; (b) kfnkBΨ≥δ, for every n≥1; (c) kfnkHΨ≤M, for every n≥1. Then there exists a subsequence {fnk}ksuch that Pk|fnk(z)|<+∞, for every z∈D, and for every α= (αk)k∈ℓ∞, one has, writing T α(z) = P∞ k=1 αkfnk(z): (2.5) T α ∈BΨand (δ/2)kαk∞≤ kT αkBΨ≤2Mkαk∞. Remark. It is clear that, by (2.5), we are defining an operator Tfrom ℓ∞ into BΨwhich is an isomorphism between ℓ∞and its image. In particular, the subsequence {fnk}kis equivalent, in BΨ, to the canonical basis of c0. Proof. First we are going to construct, inductively, a subsequence {fnk}kof {fn}, and an increasing sequence {rk}kin (0,1), such that limk→∞ rk= 1 and, setting Dk={z∈D;|z| ≤ rk},for k≥1, and C1=D1, Ck=Dk\Dk−1={z∈D;rk−1<|z| ≤ rk}, k ≥2, we have: (2.6) |fnk(z)| ≤ 2−k,for every z∈Dk−1,and every k≥2 ; 5 and (2.7) kfnk1ID\CkkLΨ< δ2−k−2,for every k≥1. Start the construction by taking n1= 1. It is a known fact that, for every function fin the Morse-Transue space MΨ(A), we have (2.8) lim A(A)→0kf1IAkLΨ= 0. Now, using (2.8), with f=fn1and considering sets Aof the form A={z∈ D;r < |z|<1}, we get r1∈(0,1) so that, for C1=D1={z∈D;|z| ≤ r1}, we have kf11ID\C1kLΨ< δ2−3. By the uniform convergence of {fn}nto 0on D1, we can find n2> n1such that |fn2(z)| ≤ 1/4,for every z∈D1,and kfn21ID1kLΨ< δ2−5. Using this last inequality and (2.8) again (for f=fn2), we get r2∈(r1,1), r2>1−1/2, such that, setting C2={z∈D;r1<|z| ≤ r2}, we have kfn21ID\C2kLΨ< δ2−4. Now that we have (2.6) and (2.7) for k= 1 and k= 2, it is clear how we must iterate the inductive construction. At the time of choosing rk∈(rk−1,1), we also impose the condition rk>1−1/k in order to get limk→∞ rk= 1. Once the construction is achieved, let us see why the subsequence {fnk}k works. The condition (2.6) and the fact that limk→∞ rk= 1 imply that, for every compact set Kin Dand z∈D, there exists lK∈Nsuch that: |fnk(z)| ≤ 2−k,for every z∈K, and every k≥lK. This yields two facts. First, Pk|fnk(z)|<+∞, for every z∈D, and secondly: for every bounded complex sequence α= (αk)k∈ℓ∞, the series Pkαkfnk converges uniformly on compact subsets of D, and its sum, the function T α, is analytic on D. It remains to prove the estimates in (2.5) about the norm of T α in LΨ(A). By homogeneity, we may assume that kαk∞= 1. Let us write gk=fnk1ICkand hk=fnk1ID\Ck, for every k≥1, g= ∞ X k=1 αkgkand h= ∞ X k=1 αkhk. We have T α =g+h. By (2.7) and the fact that |αk| ≤ 1, we have that h∈LΨ(A)and khkLΨ≤δ/4. By the condition (c) in the statement and the definition of the norm in HΨ we have, for every nand every r∈(0,1): (2.9) 1 2πZ2π 0 Ψ|fn(reit)|/Mdt ≤1. 6 The function gkis 0outside of Ck, and the sequence {Ck}kis a partition of D. Therefore: ZD Ψ(|g|/M)dA= ∞ X k=1 ZCk Ψ(|g|/M)dA= ∞ X k=1 ZCk Ψ(|αk||fnk|/M)dA ≤ ∞ X k=1 ZCk Ψ(|fnk|/M)dA. Integrating in polar coordinates, setting r0= 0, and using (2.9), we get: ZD Ψ(|g|/M)dA ≤ ∞ X k=1 Zrk rk−1 2r1 2πZ2π 0 Ψ(|fnk(reit)|/M)dt dr ≤ ∞ X k=1 Zrk rk−1 2r dr = 1 , and therefore kgkLΨ≤M, and kT αkLΨ≤δ/4 + M≤2M. On the other hand, for every k, we have: kgkLΨ≥ kg1ICkkLΨ=|αk|kfnk−hkkLΨ≥ |αk|(δ−δ/22+k)≥3δ 4|αk|. Taking the supremum on k, we get kgkLΨ≥(3δ/4) kαk∞= 3δ/4. Consequently, kT αkLΨ≥ kgkLΨ−khkLΨ≥(3δ/4) −δ/4≥δ/2, and Lemma 2.5 is fully proved.  In the following lemma we isolate the proof of the implication (c) =⇒(d) in the statement of Theorem 2.2. Lemma 2.6 Assume that the Orlicz function Ψis such that, for some A > 1, (2.10) lim sup x→+∞ Ψ(Ax) [Ψ(x)]2= +∞ Then the injection JΨ:HΨ→BΨfixes a copy of ℓ∞. Proof. Let us take a sequence of positive numbers {dn}n, and a sequence {ξn}n in T, such that the disks {D(ξn, dn)}nare pairwise disjoint in D. In particular, we should have limn→∞ dn= 0. The convexity of Ψimplies the existence of some c > 0such that Ψ(x)≥cx for every x≥1. Given a sequence {βn}nin (4,+∞)to be fixed later, we can find, thanks to (2.10), an increasing sequence {xn}satisfying: (2.11) xn>1,Ψ(xn)>1,Ψ(Axn)> βn[Ψ(xn)]2,for every n∈N. 7 Define ynas the point in the interval (xn, Axn)such that (2.12) [Ψ(yn)]2= Ψ(Axn). Put now hn= 1/Ψ(yn)and rn= 1 −hn. By (2.11) and (2.12), we have [Ψ(yn)]2> βn>4, and therefore hn∈(0,1/2). Define un(z) = hn 1−rnξnz2,and fn(z) = ynun(z). It is easy to see that kunk∞= 1, and that kunkH1≤hn. The first condition imposed to βnis βn>16/d2 n. That gives [Ψ(yn)]2> 16/d2 nand hn< dn/4. Let us write Dnfor the disk D(ξn, dn). Observe that, for z∈D\Dn, we have |1−rnξnz|=|1−rn+rnξnξn−rnξnz| ≥ rn|ξn−z|−hn≥(1/2)dn−hn≥dn/4, and therefore, since [Ψ(xn)]2≥Ψ(xn)≥c xn, |fn(z)| ≤ yn4hn dn2=16yn d2 n[Ψ(yn)]2≤16Axn d2 nβn[Ψ(xn)]2≤16A c d2 nβn· We also impose the condition βn>16An2/cd2 n, and so we have: (2.13) |fn(z)| ≤ 1 n2,for z∈D\Dn. From (2.13) we deduce that {fn}nconverges to 0uniformly on compact subsets of D. Moreover (2.13) yields that, for every bounded sequence {αn}n of complex numbers, the series Pn≥1αnfnis uniformly convergent on compact subsets of D. Let us write f∗ nfor the boundary value (on T=∂D) of the function fn. We claim that : (2.14) S= ∞ X n=1 |f∗ n| ∈ LΨ(T, m). From this, it is not difficult to deduce that, for every bounded sequence {αn}n of complex numbers, the function P∞ n=1 αnfnis in HΨand, for M=kSkLΨ(T), (2.15)    ∞ X n=1 αnfn  HΨ≤Mk{αn}nk∞. On the other hand, taking An={z∈D;|z−ξn| ≤ hn}, there exists a constant γ∈(0,1) such that A(An)≥γh2 n, and, for every z∈An, we have: |1−rnξnz| ≤ |1−rn|+|rnξnξn−rnξnz|=hn+rn|z−ξn| ≤ 2hn, 8 and consequently |un(z)| ≥ 1/4. If δ=γ/4A, we have, for every n, ZD Ψ|fn| δdA ≥ ZAn Ψyn 4δdA ≥ γh2 nΨ1 γAyn ≥h2 nΨ(Ayn)> h2 nΨ(Axn) = 1 . Thus kfnkBΨ≥δ, for every n∈N. We can apply Lemma 2.5. Using this lemma and (2.15), we get a subsequence {fnk}ksuch that, for every α= (αk)k∈ℓ∞, we have: (δ/2) k{αk}kk∞≤   ∞ X k=1 αkfnk  BΨ≤   ∞ X k=1 αkfnk  HΨ≤Mk{αk}kk∞. This clearly says that JΨfixes a copy of ℓ∞. It remains to prove (2.14). For obtaining this we impose the last condition to the sequence {βn}n. We shall need: (2.16) ∞ X n=1 1/pβn≤1. Let us set gn=|f∗ n|1IDn. Thanks to (2.13), S−P∞ n=1 gnis a bounded function. Thus we just need to prove that G=P∞ n=1 gnis in LΨ(T). We have kGkLΨ(T)≤A. Indeed, recalling that the Dn’s are pairwise disjoint, and that each gnis 0out of Dn, we have: ZT ΨG Adm = ∞ X n=1 ZDn∩T ΨG Adm = ∞ X n=1 ZDn∩T Ψ|f∗ n| Adm ≤ ∞ X n=1 ZT Ψyn|u∗ n| Adm and by the convexity of Ψ, and the fact that |un| ≤ 1, ≤ ∞ X n=1 ZT|u∗ n|Ψyn Adm = ∞ X n=1 kunkH1Ψyn A ≤ ∞ X n=1 Ψ(yn/A) Ψ(yn)≤ ∞ X n=1 Ψ(xn) Ψ(yn)= ∞ X n=1 Ψ(xn) pΨ(Axn)≤ ∞ X n=1 1 √βn≤1, by the required condition (2.16), and that ends the proof of Lemma 2.6.  We are now in position to prove Theorem 2.2. Proof of Theorem 2.2. We shall prove that: (a) =⇒(b) =⇒(c) =⇒(d) =⇒(e) =⇒(a) , and that (b) ⇐⇒ (f). 9 4 An example Theorem 4.1 There exists an Orlicz function Ψsuch that JΨis weakly compact and Dunford-Pettis, but which is not compact. Note that such an Orlicz function is very irregular: Ψ/∈∆2,Ψ/∈∆0, so, for every A > 1,Ψ(Ax)/Ψ(x)is not non-decreasing for xlarge enough, and the conjugate function of Ψdoes not satisfies condition ∆2. The following lemma is undoubtedly well-known, but we have found no reference, so we shall give a proof. Recall that a sublattice Xof L0(µ)is solid if |f| ≤ |g|and g∈Ximplies f∈Xand kfk ≤ kgk. Lemma 4.2 Let (S, S, µ)be a measure space, and let Xbe a solid Banach sublattice of L0(µ), the space of all measurable functions. Then, for every weakly null sequence {fn}nin Xand every sequence {An}nof disjoint measurables sets, the sequence {fn1IAn}nconverges weakly to 0in X. Proof. If the conclusion does not hold, there are a continuous linear functional σ:X→Cand some δ > 0such that, up to taking a subsequence, |σ(fn1IAn)| ≥ δ. Set, for every measurable set A∈ S: µn(A) = σ(fn1IA). Then µnis a finitely additive measure with bounded variation. By Rosenthal’s lemma (see [3], Lemma I.4.1, page 18, or [1], Chapter VII, page 82), there is an increasing sequence of integers {nk}ksuch that: µnk[ l6=k Anl≤ |µnk|[ l6=k Anl≤δ/2. Now, if A=Sl≥1Anl,{fnk1IA}kis weakly null, but: |σ(fnk1IA)| ≥ |σ(fnk1IAnk)|−|µnk|[ l6=k Anl≥δ−δ 2=δ 2, so we get a contradiction.  Proof of Theorem 4.1. We begin by defining a sequence {xn}nof positive numbers in the following way: set x1= 4 and, for every n≥1,xn+1 >2xnis the abscissa of the second intersection point of the parabola y=x2with the straight line containing (xn, x2 n)and (2xn, x4 n); we have xn+1 =x3 n−2xn(for example, x2= 56). Define Ψ: [0,+∞)→[0,+∞)by Ψ(x) = 4xfor 0≤x≤4, and, for n≥1: (4.1) Ψ(xn) = x2 n,Ψ(2xn) = x4 n,Ψaffine between xnand xn+1 . Then Ψis an Orlicz function and (4.2) x2≤Ψ(x)≤x4for x≥4. 16 For this Orlicz function Ψ,JΨis not compact, since Ψ(2x)/[Ψ(x)]2does not tend to 0. However, JΨis weakly compact, because one has the factorization HΨ֒→H2֒→B4֒→BΨ(by (4.2) and Lemma 2.4). Assume that JΨis not Dunford-Pettis: there exists a weakly null sequence {fn}nin the unit ball of HΨwhich does not converges for the norm in BΨ. Then {fn}nconverges uniformly to 0on the compact subsets of D(since it is weakly null) and we may assume that kfnkBΨ≥δfor some δ > 0. We may also assume that kfnk∞−→ n→∞ +∞because if {fn}nwere uniformly bounded, we should have kfnkBΨ−→ n→∞ 0, by dominated convergence. We are going to show that there exist a subsequence {fnk}kand pairwise disjoint measurable sets Ak⊆Tsuch that the sequence {fnk1IAk}k⊆LΨ(T, m)is equivalent to the canonical basis of ℓ1, whence a contradiction with Lemma 4.2. It is worth to note from now that the Poisson integral Pmaps boundedly L2(T)into L4(D). Indeed, L2(T) = H2⊕H2 0and the canonical injection is bounded from H2into B4, by Lemma 2.4. We have seen in the proof of Lemma 2.5 that there exist a subsequence {fnk}kand disjoint measurable annuli C1={z∈D;|z| ≤ r1}and Ck= {z∈D;rk−1<|z| ≤ rk},k≥2, with 0< r1< r2<···< rn<···<1, such that kfnk1ICkkLΨ(D)≥δ/2. The assumptions of that lemma are satisfied here: kfnkHΨ≤1,kfnkBΨ≥δ,{fn}nconverges uniformly to 0on the compact subsets of D, and fn∈BMΨbecause HΨ⊆BMΨ, since JΨis weakly compact. Then: Fact 1. There exist two sequences {αk}kand {βk}k, with βn> αn−→ n→∞ +∞ such that, if gk=f∗ nk1I{αk≤|f∗ nk|≤βk}, then: kP(gk)kLΨ(D)≥δ/3, where f∗ nkis the boundary value of fnkon T. Proof. 1) Let αk=δ 12 Ψ−11/A(Ck)and vk=Pf∗ nk1I{|f∗ nk|<αk}1ICk. One has: ZD Ψ|vk|/(δ/12)dA=ZCk Ψ|vk|/(δ/12)dA ≤ Ψαk/(δ/12)A(Ck) = 1 , so kvkkLΨ(D)≤δ/12. Since P(f∗ nk) = fnk, we have kP(f∗ nk) 1ICkkLΨ(D)= kfnk1ICkkLΨ(D)≥δ/2, and we get: kP(f∗ nk1I{|f∗ nk≥αk}) 1ICkkLΨ(D)≥ kfnk1ICkkLΨ(D)−kvkkLΨ(D)≥δ 2−δ 12 =5δ 12 · 2) Let wk=f∗ nk1I{|f∗ nk|≥αk}. Since P(wk1I{|wk|>β})tends to 0uniformly on Ckwhen βgoes to infinity, Lebesgue’s dominated convergence theorem gives: kP(wk1I{|wk|>β}) 1ICkkLΨ(D)≤ kP(wk1I{|wk|>β}) 1ICkkL4(D)−→ β→+∞0, 17 so there is some βk> αksuch that kP(wk1I{|wk|>β}) 1ICkkLΨ(D)≤δ/12. We then have, with gk=f∗ nk1I{αk≤|f∗ nk|≤βk}: kP(gk)kLΨ(D)≥ kP(gk) 1ICkkLΨ(D)≥5δ 12 −δ 12 =δ 3, and that ends the proof of Fact 1.  Fact 2. There are a further subsequence, denoted yet by {fnk}k, and pairwise disjoint measurable subsets Ek⊆ {αk≤ |f∗ nk| ≤ βk}, such that, if hk=f∗ nk1IEk, then: kP(hk)kLΨ(D)≥δ/4. Proof. First, since gk∈L∞(T)⊆MΨ(T), there exists εk>0such that m(A)≤εkimplies kgk1IAkLΨ(T)≤δ/(12 kPk)(where kPk stands for the norm of P:L2(T)→L4(D)). Now, P:LΨ(T)→LΨ(D)is bounded and its norm is ≤ kPk, thanks to the factorization LΨ(T)֒→L2(T)֒→L4(D)֒→LΨ(D). Hence kP(gk1IA)kLΨ(D)≤δ/12 for m(A)≤εk. Let Bk={αk≤ |f∗ nk| ≤ βk}. Up to taking a subsequence, we may assume that Pl>k m(Bl)≤εk. Let Ek=Bk\[ l>k Bl. The sets Ek,k≥1, are pairwise disjoint, and kP(gk1IEk)kLΨ(D)≥ kP(gk1IBk)kLΨ(D)−kPgk1ISl>k BlkLΨ(D)≥δ 3−δ 12 =δ 4; so we get the Fact 2 with hk=gk1IEk=f∗ nk1IEk. Set Fk={z∈Ek; Ψ|f∗ nk(z)| ≤ M|f∗ nk(z)|2}. For z∈Ek\Fk, one has: ZEk\Fk|f∗ nk|2dm ≤1 MZT Ψ(|f∗ nk)|dm ≤1 M, so kf∗ nk1IEk\FkkL2(T)≤1/√Mand: kP(f∗ nk1IEk\Fk)kLΨ(D)≤ kP(f∗ nk1IEk\Fk)kL4(D) ≤ kPkk(f∗ nk1IEk\Fk)kL2(T)≤kPk √M≤δ 8, for Mlarge enough. It follows that, for Mlarge enough, kP(f∗ nk1IFk)kLΨ(D)≥ δ/8and (4.3) kf∗ nk1IFkkLΨ(D)≥δ/(8 kPk). 18 Now, we may assume that, for some α > 0, ZT|f∗ nk|21IFkdm ≥α , because, if not, there would be a subsequence {f∗ nkj1IFkj}jconverging to 0in L2(T); but then {Pfnkj1IFkj}jwould converge to 0in B4, and hence in BΨ, contrary to (4.3). It follows, using (4.2), that: (4.4) ZFk Ψ(|f∗ nk|)dm ≥α . The following lemma is now the key of the proof. Lemma 4.3 Let δn= 2xn−1/xn= 2/(x2 n−1−2). If Ψ(x)≤Mx2and x≥xn, then, for nlarge enough (n≥N), one has Ψ(εx)≥CMεΨ(x)for δn≤ε≤1. Proof. We may assume that xn≤x < xn+1, because if xk≤x < xk+1 with k≥n, then ε≥δnimplies ε≥δk. Now, remark that: (4.5) Ψ(y) Ψ(x)≤4y x,for 2xn≤x≤y≤xn+1 . Indeed, on the one hand, Ψ(y)−Ψ(xn) Ψ(x)−Ψ(xn)=y−xn x−xn≤y x/2= 2 y x; and, on the other hand, Ψ(y)−Ψ(xn)≥Ψ(y)−Ψ(y/2) ≥Ψ(y)−1 2Ψ(y) = 1 2Ψ(y), so Ψ(y) Ψ(x)≤ Ψ(y) Ψ(x)−Ψ(xn)≤2Ψ(y)−Ψ(xn) Ψ(x)−Ψ(xn)≤4y x· We shall separate three cases: 1) εx ≤xn≤x≤2xn. Then εx ≥εxnand hence Ψ(εx)≥Ψ(εxn). But 2xn−1≤εxn≤xn, since ε≥δn; hence (4.5) implies that Ψ(εx)≥ (ε/4) Ψ(xn) = (ε/4) x2 n. On the other hand, one has, by hypothesis, Ψ(x)≤ Mx2≤M(2xn)2, so we get Ψ(εx)≥(ε/16M)Ψ(x). 2) xn≤εx ≤x≤2xn. Then, since 1≤1/ε: Ψ(x) Ψ(εx)≤Mx2 Ψ(xn)≤M(2xn)2 x2 n = 4M≤4M ε· 3) For x≥2xn, remark that the conditions Ψ(x)≤Mx2and x≥2xnimply that x≥x2 n/√M. Indeed, if x≥2xn, then Ψ(x)≥Ψ(2xn) = x4 n, and the condition Ψ(x)≤Mx2implies x4 n≤Mx2,i.e. x≥x2 n/√M. In this case, one has εx ≥εx2 n/√M≥δnx2 n/√M= 2(xn−1/xn)x2 n/√M= 2xn−1xn/√M≥2xn, if xn−1≥√M. Hence (4.5) gives, for 2xn≤x < xn+1 (since then 2xn≤εx ≤x < xn+1): Ψ(x) Ψ(εx)≤4x εx =4 ε· That ends the proof of Lemma 4.3.  19 Extract now a further subsequence of {fnk}, yet denoted by {fnk}, in order that (see Fact 1) αk≥xN+k. Lemma 4.3 holds, with x= Ψ(|f∗ nk(z)|),z∈Fk, for every k≥1; one has (since, by definition, Ψ(|fnk|)≤M|fnk|2on Fk): ZFk Ψ(ε|f∗ nk|)dm ≥ε C/α := c ε , for δN+k≤ε≤1. The proof of Theorem 4.1 reaches now its end: put uk=f∗ nk1IFk, and take an arbitrary sequence of complex numbers such that Pk≥1|λk|= 1. Let δ0=Pk≥Nδk. One has δ0<1, because we may assume that Nhad been taken large enough. One gets: ZT ΨX k≥1 λkukdm =X k≥1ZFk Ψ(|λkfnk|)dm ≥X |λk|≥δN+k c|λk|+X |λk|<δN+kZFk Ψ(|λkfnk|)dm ≥X |λk|≥δN+k c|λk|=c1−X |λk|<δN+k |λk| ≥c1−X k≥N δk=c(1 −δ0) := c0. Since c0<1, this implies, by convexity, that   X k≥1 λkuk  LΨ(T)≥c0. Hence {uk}kis equivalent to the canonical basis of ℓ1, and that achieves the proof of Theorem 4.1.  Remarks. 1) It follows from Theorem 3.3 that, for this Ψ,JΨis not p-summing for p < 4. By modifying the definition of Ψ(taking Ψ(xn) = xr/2 nand Ψ(2xn) = xr n), we get, for every 4≤r < ∞, an Orlicz function Ψsuch that JΨis DunfordPettis and weakly compact, without being p-summing for p < r, and without being compact. We do not know whether it is possible to have JΨp-summing for no finite p. 2) Let us point out that the fact that JΨis Dunford-Pettis does not trivially follows from its weak compactness: HΨdoes not have the Dunford-Pettis property. In fact, if it were the case, the weakly compact injection HΨ֒→H2would be Dunford-Pettis, and hence also H4֒→H2(since H4֒→HΨ֒→H2). But it is not the case: the sequence {zn}nconverges weakly to 0in H4, whereas it does not converges in norm to 0in H2. 20 Proposition 4.4 There is an Orlicz function Ψfor which JΨis weakly compact, but not Dunford-Pettis. Proof. Let us call Ψ0the Orlicz function constructed in Theorem 4.1, and let Ψ(x) = Ψ0(x2). Then, with β= 2,Ψ(βx) = Ψ0(4x2)≥4Ψ0(x2) = (2β)Ψ(x); that means that the conjugate function of Ψsatisfies ∆2. JΨis weakly compact (since JΨfactors as HΨ֒→H4֒→B8֒→BΨ), but is not compact, since [Ψ(√xn)]2= Ψ(√2√xn). Since the conjugate function satisfies ∆2,JΨis not Dunford-Pettis, by Proposition 3.1.  References [1] J. Diestel, Sequences and Series in Banach Spaces, Graduate Texts in Mathematics 92, Springer-Verlag, New York (1984). [2] J. Diestel, H. Jarchow, and A. Tonge, Absolutely Summing Operators, Cambridge Studies in Adv. Math. 43, Cambridge Univ. Press (1995). [3] J. Diestel and J. J., Jr. Uhl, Vector Measures, Mathematical Surveys, No. 15, American Mathematical Society, Providence, R.I. (1977). [4] P. Duren and A. Schuster, Bergman Spaces, Math. Surveys and Monographs 100, Amer. Math. Soc. (2004). [5] P. Lefèvre, When strict singularity of operators coincides with weak compactness, to appear in J. Operator Theory. [6] P. Lefèvre, D. Li, H. Queffélec, and L. Rodríguez-Piazza, A criterion of weak compactness for operators on subspaces of Orlicz spaces, J. Funct. Spaces and Applications 6, No. 3 (2008), 277–292. [7] P. Lefèvre, D. Li, H. Queffélec, and L. Rodríguez-Piazza, Composition operators on Hardy-Orlicz spaces, preprint, math.FA/0610905, to appear in Memoirs Amer. Math. Soc. (2010), DOI: 10.1090/S0065-9266-10-00580-6. [8] P. Lefèvre, D. Li, H. Queffélec, and L. Rodríguez-Piazza, Nevanlinna counting function and Carleson function of analytic maps, preprint, arXiv : 0904.2496, hal-00375955. [9] P. Lefèvre and L. Rodríguez-Piazza, Absolutely summing Carleson embeddings, in preparation [10] A. Plichko, Superstrictly singular and superstrictly cosingular operators, Functional analysis and its applications, 239–255, North-Holland Math. Stud., 197, Elsevier, Amsterdam (2004). 21 Pascal Lefèvre, Univ Lille Nord de France F-59 000 LILLE, FRANCE UArtois, Laboratoire de Mathématiques de Lens EA 2462, Fédération CNRS Nord-Pas-de-Calais FR 2956, F-62 300 LENS, FRANCE pascal.lefevr[email protected] Daniel Li, Univ Lille Nord de France F-59 000 LILLE, FRANCE UArtois, Laboratoire de Mathématiques de Lens EA 2462, Fédération CNRS Nord-Pas-de-Calais FR 2956, Faculté des Sciences Jean Perrin, Rue Jean Souvraz, S.P. 18, F-62 300 LENS, FRANCE [email protected] Hervé Queffélec, Univ Lille Nord de France F-59 000 LILLE, FRANCE USTL, Laboratoire Paul Painlevé U.M.R. CNRS 8524, F-59 655 VILLENEUVE D’ASCQ Cedex, FRANCE [email protected]le1.fr Luis Rodríguez-Piazza, Universidad de Sevilla, Facultad de Matemáticas, Departamento de Análisis Matemático, Apartado de Correos 1160, 41 080 SEVILLA, SPAIN [email protected] 22