The canonical injection of the Hardy-Orlicz space $H^\Psi$ into the Bergman-Orlicz space ${\mathfrak B}^\Psi$

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The canonical injection of the Hardy-Orlicz space H ^Ψ into the Bergman-Orlicz space B ^Ψ

Pascal Lefèvre, Daniel Li, Hervé Queffélec, Luis Rodriguez-Piazza

To cite this version:

Pascal Lefèvre, Daniel Li, Hervé Queffélec, Luis Rodriguez-Piazza. The canonical injection of the

Hardy-Orlicz space

into the Bergman-Orlicz space

. 2010. �hal-00482846�

The canonical injection of the Hardy-Orlicz space H ^Ψ into the

Bergman-Orlicz space B ^Ψ

Pascal Lef` evre, Daniel Li,

Herv´ e Queff´ elec, Luis Rodr´ıguez-Piazza

May 11, 2010

Abstract. We study the canonical injection from the Hardy-Orlicz space H^Ψ into the Bergman-Orlicz spaceB^Ψ.

Mathematics Subject Classification. Primary: 46E30 – Secondary: 30D55;

30H05; 32A35; 32A36; 42B30

Key-words. absolutely summing operator – Bergman-Orlicz space – compact- ness – 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 H^Ψ and the Bergman-Orlicz spaceB^Ψ. It is well-known that in the classical case Ψ(x) =x^p,H^p⊆B^p and the canonical injection Jp from H^p to B^p is bounded, and even compact. In fact, for any Orlicz function Ψ, one hasH^Ψ ⊆B^Ψ and the canonical injection JΨ:H^Ψ→B^Ψ is bounded, but we shall see in this paper that its compactness requires that Ψ does not grow too fast. We actually characterize in Section 2 the compactness: JΨ is compact if and only if limx→+∞Ψ(Ax)/[Ψ(x)]²= 0 for everyA >1, and the weak compactness: JΨ is weakly compact if and only if lim sup_x→+∞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 forxlarge enough), it is not always the case. We actually show a stronger result in Section 4: there is an Orlicz function Ψ such thatJΨ is weakly compact and is Dunford-Pettis, but such thatJΨ is 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 Ψ has

property ∆2 (Ψ∈∆2) if Ψ(2x)≤CΨ(x) for some constantC >0 andxlarge 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(x^p) for some 1 ≤ p < +∞. One says (see [6], [7]) that Ψ satisfies the condition ∆⁰ if, for some β > 1, one has lim

x→∞Ψ(βx)/Ψ(x) = +∞. If Ψ ∈ ∆⁰, then Ψ(x)/x^p_x→∞−→ +∞ for every 1≤p <∞. Indeed, let 1≤p <∞. For everyβ >1 one can findx0>0 such that Ψ(βx)/Ψ(x)≥β^p forx≥x0; then Ψ(βⁿx0)≥β^npΨ(x0) for everyn≥1.

That implies that Ψ(x) ≥Cpx^p for every x >0 large enough. Since p≥1 is arbitrary, we getx^p=o[Ψ(x)].

We say that Ψ∈ ∇⁰(1) if, for every A >1, Ψ(Ax)/Ψ(x) is non-decreasing for x large enough. This is equivalent to say (see [7], Proposition 4.7) that log Ψ(e^x) 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 L^Ψ(µ) as the set of all (classes of) measurable functions f:S →C for which there is a C >0 such thatR

SΨ(|f|/C)dµis finite. The normkfk^Ψ is the infimum of all C >0 for which the above integral is≤1. The Morse-Transue spaceM^Ψ(µ) is the subspace off ∈L^Ψ(µ) for whichR

SΨ(|f|/C)dµis finite for allC >0; it is the closure ofL^∞(µ) inL^Ψ(µ). One hasM^Ψ(µ) =L^Ψ(µ) if and only if Ψ∈∆2.

If Ψ(x)/x −→

x→+∞+∞, the conjugate function Φ of Ψ is defined by Φ(y) = sup_x>0 xy−Ψ(x)

. It is an Orlicz function and [M^Ψ(µ)]^∗ =L^Φ(µ), isomorphi- cally.

We may note that if Ψ(x)/x does not converges to infinity, we must have Ψ(x)≤axfor somea≥1 andxlarge enough. ThenL^Ψ(µ) =L¹(µ) isomorphi- cally and then Φ(y) = +∞for y > a (givingL^Φ(µ) = L^∞(µ) isomorphically).

We denote byDthe open unit disk ofCand byT=∂Dthe unit circle. The normalized area-measure on D is denoted by A and the normalized Lebesgue measure onTis denoted bym.

The Hardy-Orlicz space H^Ψ is defined as {f ∈ H¹; f^∗ ∈ L^Ψ(m)}, where f^∗ is the boundary values function of f, and HM^Ψ = H^Ψ ∩M^Ψ(m) is the closure ofH^∞inH^Ψ. The Bergman-Orlicz spaceB^Ψis the subspace of analytic f ∈L^Ψ(A), andBM^Ψ =B^Ψ∩M^Ψ(A) is the closure ofH^∞in B^Ψ. Since, for f ∈ H^Ψ, kfkH^Ψ = sup0<r<1kfrkH^Ψ (see [7], Proposition 3.1), where fr(z) = f(rz), one has:

Z 2π 0

Ψ

|f(re^it)| kfkH^Ψ

dt 2π ≤

Z 2π 0

Ψ

|f(re^it)| kfrkH^Ψ

dt 2π ≤1 ; hence:

Z

D

Ψ

|f(re^it)| kfkH^Ψ

dA=

Z 1 0

Z 2π 0

Ψ

|f(re^it)| kfkH^Ψ

dt 2π

2r dr≤1,

sof ∈B^Ψ andkfk^BΨ ≤ kfkH^Ψ. It follows thatH^Ψ ⊆B^Ψ and the canonical injection JΨ:H^Ψ → B^Ψ is bounded, and has norm 1. Let us point out that

the boundedness also follows from [7], Theorem 4.10, 2), sinceJΨ is a Carleson embeddingJΨ:H^Ψ→B^Ψ⊆L^Ψ(A).

This injection is not onto, since there are functions f ∈B^Ψ with no radial limit on a subset ofTof positive measure (the proof is the same as in B^p: see [4],§3.2, Lemma 2, page 81). Note that JΨ is not an into-isomorphism: take fn(z) =zⁿ, for everyn∈N; it is easy to see that{fn}ⁿ tends to 0 inB^Ψ, but not inH^Ψ.

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 An´alisis Matem´atico 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 ofJΨ, we introduce the following quantityQA,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: H^Ψ → B^Ψ is 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 thatM := sup_A>1QA <+∞. Let 0 < ε≤1 and A > 1; we can find xA = xA(ε) > 0 such that Ψ(Ax/ε)/[Ψ(x)]² ≤ 2M for x ≥xA. By convexity, one has Ψ(Ax)≤ εΨ(Ax/ε), and hence Ψ(Ax)/[Ψ(x)]² ≤2εM for x≥xA. We getQA= 0.

2) It is clear that condition (2.2) is satisfied whenever Ψ∈∆2, but Ψ(x) = e^[log(x+1)]2 −1 satisfies (2.2) without being in ∆2. However, condition (2.2) implies that Ψ cannot grow too fast. More precisely, we must have

Ψ(x) =o(e^xα) for everyα >0.

Indeed, one has Ψ(At) ≤ [Ψ(t)]² for t ≥ tA, and, by iteration, Ψ(AⁿtA) ≤ [Ψ(tA)]²ⁿ for every n ≥1. For every x > 0 large enough, taking n≥ 1 such thatAⁿtA≤x < Aⁿ⁺¹tA, we get Ψ(x)≤C1e^C2xα, withα= log 2/logA. Since A > 1 is arbitrary, α may be any positive number. The little-oh condition follows from the fact that the inequality is true for allα >0.

Proof of Theorem 2.1. By definition, B^Ψ is a subspace of L^Ψ(D,A); hence we can seeJΨas a Carleson embeddingJΨ:H^Ψ→L^Ψ(D,A). IfS(ξ, h) ={z∈ D; |z−ξ|< h}, the compactness ofJΨ implies, by [7], Theorem 4.11, that, for everyA >1, everyε >0, andh >0 small enough:

h²≤4A[S(ξ, h)]≤ 4ε Ψ[AΨ⁻¹(1/h)],

that is, settingx= 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 iso (1/h)/Ψ[AΨ⁻¹(1/h)]

for everyA >1, if (2.2) holds; hence, by [7],

Theorem 4.11, again,JΨ is compact.

We now turn ourself to the weak compactness.

Theorem 2.2 The following assertions are equivalent:

(a) JΨ:H^Ψ→B^Ψ is weakly compact;

(b)JΨ fixes no copy of c0; (c) JΨ fixes no copy of ℓ∞; (d)QA<+∞, for every A >1;

(e) H^Ψ⊆BM^Ψ;

(f) JΨ is strictly singular.

Recall that an operatorT:X →Y between two Banach spaces is said to be strictly singular if there is no infinite-dimensional subspaceX0 of X on which T is 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 everyA >1 implies condition (2.2). Indeed, if lim

x→+∞

Ψ(βx)

Ψ(x) = +∞, we get, for everyA >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 someA > 1, Ψ(Ax)/Ψ(x) is non-decreasing forxlarge enough (in particular if Ψ∈ ∇0(1)), one has the dichotomy: either Ψ∈∆2, and then JΨ

is compact; or Ψ∈∆⁰ and hence the weak compactness ofJΨ implies, by the two above theorems, its compactness. Hence:

Proposition 2.3 If, for some A > 1, Ψ(Ax)/Ψ(x) is non-decreasing, for x large enough, then the weak compactness ofJΨ is equivalent to its compactness.

However, it is easy to construct an Orlicz function Ψ which satisfies condi- tion (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 defineΨ1(t) = [Ψ(t)]²,t≥0, thenΨ1is an Orlicz function for which H^Ψ⊆B^Ψ1 and the canonical injection ofH^Ψ intoB^Ψ1 is continuous.

Proof. It is enough to see that H^Ψ continuously embeds intoL^Ψ1(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 =hKA(h).

Using part 2) of Theorem 4.10 in [7], the lemma follows.

Lemma 2.5 LetM > δ >0 and{fn}nbe a sequence inH^Ψ∩BM^Ψ such that:

(a){fn}n tends to 0uniformly on compact subsets of D; (b)kfnk^BΨ ≥δ, for everyn≥1;

(c) kfnkH^Ψ≤M, for everyn≥1.

Then there exists a subsequence {fnk}k such that P

k|fnk(z)| < +∞, for every z ∈ D, and for every α = (αk)k ∈ ℓ∞, one has, writing T α(z) = P∞

k=1αkfnk(z):

(2.5) T α∈B^Ψ and (δ/2)kαk^∞≤ kT αk^BΨ ≤2Mkαk^∞.

Remark. It is clear that, by (2.5), we are defining an operatorT from ℓ∞

intoB^Ψ which is an isomorphism betweenℓ∞and its image. In particular, the subsequence{fnk}k is equivalent, inB^Ψ, to the canonical basis of c0.

Proof. First we are going to construct, inductively, a subsequence {fnk}k of {fn}, and an increasing sequence{rk}^k in (0,1), such that limk→∞rk = 1 and, setting

Dk={z∈D; |z| ≤rk}, fork≥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 everyz∈Dk−1, and everyk≥2 ;

and

(2.7) kfnk1I_D\CkkL^Ψ< δ2^−k−2, for everyk≥1.

Start the construction by taking n1= 1. It is a known fact that, for every functionf in the Morse-Transue spaceM^Ψ(A), we have

(2.8) lim

A(A)→0kf1IAkL^Ψ= 0.

Now, using (2.8), with f = fn1 and considering sets A of the form A= {z ∈ D; r <|z|<1}, we getr1∈(0,1) so that, for C1 =D1={z ∈D; |z| ≤r1}, we have

kf11I_D\C1kL^Ψ< δ2⁻³.

By the uniform convergence of{fn}nto 0 onD1, we can findn2> n1such that

|fn2(z)| ≤1/4, for everyz∈D1, and kfn21ID1kL^Ψ< δ2⁻⁵. Using this last inequality and (2.8) again (for f = fn2), we get r2 ∈ (r1,1), r2>1−1/2, such that, settingC2={z∈D; r1<|z| ≤r2}, we have

kfn21I_D\C2kL^Ψ< δ2⁻⁴.

Now that we have (2.6) and (2.7) fork= 1 andk= 2, it is clear how we must iterate the inductive construction. At the time of choosing rk ∈ (rk−1,1), we also impose the conditionrk>1−1/kin 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 setK inDandz∈D, there existslK ∈Nsuch that:

|fnk(z)| ≤2^−k, for everyz∈K, and everyk≥lK. This yields two facts. First,P

k|fnk(z)|<+∞, for everyz∈D, and secondly:

for every bounded complex sequence α = (αk)k ∈ ℓ∞, the series P

kαkfnk

converges uniformly on compact subsets ofD, and its sum, the functionT α, is analytic onD.

It remains to prove the estimates in (2.5) about the norm ofT αinL^Ψ(A).

By homogeneity, we may assume thatkαk∞= 1. Let us writegk =fnk1ICkand hk =fnk1I_D\Ck, for everyk≥1,

g= X∞

k=1

αkgk and h= X∞

k=1

αkhk.

We have T α = g+h. By (2.7) and the fact that |αk| ≤ 1, we have that h∈L^Ψ(A) andkhkL^Ψ≤δ/4.

By the condition (c) in the statement and the definition of the norm inH^Ψ we have, for everynand everyr∈(0,1):

(2.9) 1

2π Z 2π

0

Ψ |fn(re^it)|/M dt≤1.

The functiongk is 0 outside ofCk, and the sequence{Ck}k is a partition ofD. Therefore:

Z

D

Ψ(|g|/M)dA= X∞ k=1

Z

Ck

Ψ(|g|/M)dA= X∞ k=1

Z

Ck

Ψ(|αk| |fnk|/M)dA

≤ X∞

k=1

Z

Ck

Ψ(|fnk|/M)dA.

Integrating in polar coordinates, settingr0= 0, and using (2.9), we get:

Z

D

Ψ(|g|/M)dA ≤ X∞

k=1

Z rk

rk−1

2r 1 2π

Z 2π 0

Ψ(|fnk(re^it)|/M)dt dr

≤ X∞

k=1

Z rk

r_k−1

2r dr= 1,

and thereforekgkL^Ψ≤M, andkT αkL^Ψ≤δ/4 +M ≤2M. On the other hand, for everyk, we have:

kgkL^Ψ≥ kg1ICkkL^Ψ=|αk|kfnk−hkkL^Ψ≥ |αk|(δ−δ/2^2+k)≥ 3δ 4|αk|. Taking the supremum onk, we getkgkL^Ψ≥(3δ/4)kαk^∞= 3δ/4. Consequently,

kT αkL^Ψ≥ kgkL^Ψ− khkL^Ψ≥(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 functionΨ is such that, for someA >1,

(2.10) lim sup

x→+∞

Ψ(Ax) [Ψ(x)]² = +∞ Then the injectionJΨ:H^Ψ→B^Ψ fixes a copy ofℓ∞.

Proof.Let us take a sequence of positive numbers{dn}n, and a sequence{ξn}n

inT, such that the disks{D(ξn, dn)}n are pairwise disjoint inD. In particular, we should have limn→∞dn = 0.

The convexity of Ψ implies the existence of somec >0 such that Ψ(x)≥cx for everyx≥1. Given a sequence {βn}ⁿ in (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 everyn∈N.

Defineyn as 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 thereforehn∈(0,1/2). Define

un(z) = hn

1−rnξnz ²

, and fn(z) =ynun(z). It is easy to see thatkunk^∞= 1, and thatkunkH¹ ≤hn.

The first condition imposed to βn is βn > 16/d²_n. That gives [Ψ(yn)]² >

16/d²_n andhn < dn/4. Let us write Dn for the disk D(ξn, dn). Observe that, forz∈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 conditionβn >16An²/cd²_n, and so we have:

(2.13) |fn(z)| ≤ 1

n², forz∈D\Dn.

From (2.13) we deduce that {fn}ⁿ converges to 0 uniformly on compact subsets ofD. Moreover (2.13) yields that, for every bounded sequence{αn}ⁿ of complex numbers, the seriesP

n≥1αnfn is uniformly convergent on compact subsets ofD. Let us writef_n^∗for the boundary value (onT=∂D) of the function fn. We claim that :

(2.14) S=

X∞ n=1

|f_n^∗| ∈L^Ψ(T, m).

From this, it is not difficult to deduce that, for every bounded sequence{αn}n

of complex numbers, the functionP∞

n=1αnfnis inH^Ψand, forM =kSkL^Ψ(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 constantγ∈(0,1) such thatA(An)≥γh²_n, and, for everyz∈An, we have:

|1−rnξnz| ≤ |1−rn|+|rnξnξn−rnξnz|=hn+rn|z−ξn| ≤2hn,

and consequently|un(z)| ≥1/4. If δ=γ/4A, we have, for everyn, Z

D

Ψ|fn| δ

dA ≥ Z

An

Ψyn

4δ

dA ≥γh²_nΨ1 γAyn

≥h²_nΨ(Ayn)> h²_nΨ(Axn) = 1.

Thuskfnk^BΨ ≥δ, for everyn∈N. We can apply Lemma 2.5. Using this lemma and (2.15), we get a subsequence{fnk}k such that, for everyα= (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 thatJΨ fixes a copy ofℓ∞.

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/p βn ≤1.

Let us set gn = |f_n^∗|1IDn. Thanks to (2.13), S −P∞

n=1gn is a bounded function. Thus we just need to prove thatG=P∞

n=1gn is inL^Ψ(T). We have kGkL^Ψ(T)≤A. Indeed, recalling that the Dn’s are pairwise disjoint, and that eachgn is 0 out ofDn, we have:

Z

T

ΨG A

dm= X∞ n=1

Z

Dn∩T

ΨG A

dm= X∞ n=1

Z

Dn∩T

Ψ|f_n^∗| A

dm

≤ X∞

n=1

Z

T

Ψyn|u^∗_n| A

dm

and by the convexity of Ψ, and the fact that|un| ≤1,

≤ X∞ n=1

Z

T

|u^∗_n|Ψyn

A

dm= X∞ n=1

kunkH1Ψyn

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).

The implications (a) =⇒(b) =⇒(c) and (f) =⇒(b) are trivial, and we have seen in Lemma 2.6 that (c) =⇒(d).

(d) =⇒ (e). By Lemma 2.4, there exists a constant C > 0 such that, for everyf in the unit ball ofH^Ψ, we have:

(2.17)

Z

D

[Ψ(|f|/C)]²dA ≤1.

For everyA >0, there existxA, such that Ψ(Ax)≤(QA+ 1)[Ψ(x)]², for every x≥xA. Thus for every x≥0 we have Ψ(Ax)≤(QA+ 1)[Ψ(x)]²+ Ψ(AxA).

Then, by (2.17), we have Z

D

Ψ(A|f|/C)dA<+∞, for everyA >0.

Thereforef ∈BM^Ψ, for everyf in the unit ball of H^Ψ, and thus for everyf inH^Ψ.

(e) =⇒ (a). Let {fn}ⁿ be in the unit ball of H^Ψ. We have to prove that {fn}ⁿ has a subsequence which converges in the weak topology of B^Ψ. By Montel’s Theorem{fn}n has a subsequence converging uniformly on compact subsets ofD, to a functiongwhich, by Fatou’s lemma, also belongs to the unit ball ofH^Ψ. If this subsequence converges togin the norm ofB^Ψ we are done.

If not, after perhaps a new extraction of subsequence, there existδ >0 and a subsequence{fnk}k, such that

kfnk−gk^BΨ≥δ, and kfnk−gkH^Ψ ≤2.

Since moreover{fn_k−g}^k converges to 0 uniformly on compact subsets of D and, by condition (e),fn_k−g ∈BM^Ψ, we may apply Lemma 2.5 and we get that {fnk−g}^k has a subsequence equivalent to the canonical basis of c0 in B^Ψ, and is therefore weakly null. This yields that {fn}ⁿ has a subsequence converging tog in the weak topology ofB^Ψ.

(b) =⇒(f). Suppose there exists an infinite-dimensional subspaceX ofH^Ψ on which the normsk · k^BΨ andk · kH^Ψ are equivalent. We shall have finished if we prove thatX contains a subspace isomorphic toc0 because thenJΨ will fix a copy ofc0.

We can assume thatX is contained inBM^Ψ because we already know that (b) implies (e). X being infinite-dimensional, there exists, for every n ∈ N, fn ∈ X, such that kfnkH^Ψ = 1, and cfn(k) = 0, for k = 0,1, . . . , n. By the equivalence of the norms in X, there exists δ > 0 such thatkfnk^BΨ ≥ δ, for everyn. The unit ball ofH^Ψ is compact in the topology ofH(D). Since

n→∞lim cfn(k) = 0, for everyk≥0,

the only possible limit of a subsequence of {fn}ⁿ is the function 0. So {fn}ⁿ converges to 0 uniformly on compact subsets of D. As fn ∈ X ⊆ BM^Ψ, for

everyn, we can apply Lemma 2.5, and we get that {fn}n has a subsequence generating an spaceY isomorphic toc0inB^Ψ. This spaceY is contained inX, where the norms are equivalent, soY is also isomorphic toc0 for the norm of

H^Ψ.

3 Other properties

3.1 Dunford-Pettis

Recall that an operator T: X →Y between two Banach spaces X and Y is said to beDunford-Pettis if{T xn}n converges in norm whenever{xn}n con- verges weakly. Every compact operator is Dunford-Pettis. The next proposition shows that, in “most” of the cases, these two properties are equivalent forJΨ. Proposition 3.1 If the conjugate function of Ψ satisfies condition ∆2, then JΨ:H^Ψ→B^Ψ is Dunford-Pettis if and only if it is compact.

We shall see in Section 4 that without condition ∆2for the conjugate func- tion,Jψ may be Dunford-Pettis without being compact.

Proof. Remark first that speaking of the conjugate function of Ψ implicitly assume that Ψ(x)/x tends to +∞as xgoes to +∞.

Assume that JΨ is not compact. By Theorem 2.1, there are some A > 1 and a sequence{xj}j going to +∞such that Ψ(Axj)≥[Ψ(xj)]². Settingrj = 1−1/Ψ(xj), this is equivalent to say thatAΨ⁻¹ 1/(1−rj)

≥Ψ⁻¹ 1/(1−rj)² . Define:

fj(z) =xj

1−rj

1−rjz ²

·

One has fj ∈ HM^Ψ and kfjkH^Ψ ≤ 1 (see [7], Corollary 3.10). Since {fj}j

converges to 0 uniformly on compact subsets ofD, {fj}j converges to 0 in the weak-star topology ofH^Ψ ([7], Proposition 3.7). But, since the conjugate func- tion of Ψ satisfies condition ∆2,H^Ψ is the bidual of HM^Ψ([7], Corollary 3.3);

hence{fj}j converges weakly to 0 in HM^Ψ.

On the other hand, ifSj =D(1,1−rj)∩D, one has|1−rjz| ≤2(1−rj) forz∈Sj; hence, writingK=kfjk^BΨ, one has:

1 = Z

D

Ψ(|fj|/K)dA ≥ Z

Sj

Ψ(|fj|/K)dA ≥ A(Sj)Ψ(xj/4K).

Since A(Sj) ≥ α(1 −rj)², with 0 < α < 1, we get (since Ψ(αxj/4K) ≤ αΨ(xj/4K), by convexity):

kfjk^BΨ≥(α/4) xj

Ψ⁻¹ 1/(1−rj)² = (α/4) Ψ⁻¹ 1/(1−rj) Ψ⁻¹ 1/(1−rj)² ≥ α

4A·

ThereforeJΨ is not Dunford-Pettis.

On the other hand, one has:

Proposition 3.2 If JΨ is Dunford-Pettis, thenJΨ is weakly compact.

Proof. By Theorem 2.2, if JΨ is not weakly compact, there is a subspaceX0

ofH^Ψ isomorphic to c0 on whichJΨ is an into-isomorphism; hence JΨ cannot

be Dunford-Pettis.

We shall see in the next section that JΨ may be weakly compact without being Dunford-Pettis.

3.2 Absolutely summing

Everyp-summing operator is weakly compact and Dunford-Pettis; so it may be expected thatJΨ isp-summing for somep <∞. The next results show that this is never the case as soon as Ψ grows faster than all the power functions.

Recall that an operatorT:X →Y between two Banach spacesX andY is called (p, q)-summing if there is a constantC >0 such that

Xⁿ

k=1

kT xkk^p1/p

≤C sup

kx^∗kX∗≤1

Xⁿ

k=1

|x^∗(xk)|^q1/q

,

for every finite sequence (x1, . . . , xn) inX. Ifq=p, it is saidp-summing. Every p-summing operator is (p, q)-summing forq≤p.

Theorem 3.3 If JΨ:H^Ψ→B^Ψ isp-summing, then, for everyq > p,Ψ(x) = O(x^q)for xlarge enough. Moreover, ifp <2, thenJΨ is compact.

In order to prove this, we need two lemmas.

Lemma 3.4 If the canonical injection IΨ:A→ B^Ψ is (p,1)-summing, where A=A(D)is the disk algebra, thenΨ(x) =O(x^2p)for xlarge enough.

In particular, Jr:H^r →B^r is (p,1)-summing for no p < r/2, and, if Ψ∈

∆⁰, thenJΨ is(p,1)-summing for nop <∞.

Recall that the disk algebra is the space of continuous functions onDwhich are analytic inD.

We refer to [9] for a detailed study ofr-summing Carleson embeddingsH^r→ L^r(µ). In particular, it follows from these results that Jr: H^r → B^r is 1- summing for 1≤r <2. On the other hand, it is easy to see thatJ2: H²→B² is not Hilbert-Schmidt (i.e. not 2-summing): for the canonical orthonormal basis {zⁿ}n and {√

n+ 1zⁿ}n of H² and B², J2 is the diagonal operator of multiplication by{1/√

n+ 1}n. It also follows from [9] that, for r ≥2, Jr is p-summing for no finitep.

Proof. Assume that we do not have Ψ(x) =O(x^2p) forxlarge enough. Then lim sup_x→+∞Ψ(x)/x^2p = +∞. Given any K > 0, take y > 0 such that Ψ(y)/y^2p ≥ K and such that h = 1/p

Ψ(y) ≤ 1/2. Let N be the integer part of (1/h) + 1. Writingξj = e^2πij/N, we set:

uj(z) = h²

[1−(1−h)ξjz]²·

We haveuj ∈A(D). By [7], Lemma 5.6, one has, sinceh≥1/N:

NX−1

j=0

|uj(e^it)| ≤N h² 1−(1−h)^2N

[1−(1−h)²][1−(1−h)^N]² ≤ e²

(1−e)² :=C . Hence:

sup

kx^∗k_A∗≤1 NX−1

j=0

|x^∗(uj)| ≤C .

On the other hand, it is easy to see that|uj(z)| ≥1/9 when|z−(1−h)ξj|< h;

hence, ifSj ={z∈D; |z−(1−h)ξj|< h}, one has, sinceA(Sj) =h²: 1 =

Z

D

Ψ |uj(z)| kujk^BΨ

dA(z)≥ Z

Sj

Ψ 1/9 kujk^BΨ

dA ≥h²Ψ 1/9 kujk^BΨ

,

sokujk^BΨ≥1/9Ψ⁻¹(1/h²). Sincey= Ψ⁻¹(1/h²), one gets:

N−1X

j=0

kujk^pB^Ψ≥(1/9)^pN

y^p ≥(1/9)^p Ψ(y)

y^2p 1/2

≥ K^1/2 9^p ·

This yields that the (p,1)-summing norm of IΨ should be greater than K^1/2p/9C, and, asKis arbitrary, thatIΨ is not (p,1)-summing.

Remark. WhenIΨ: A ֒→B^Ψ is p-summing, we have this shorter argument.

By Pietsch’s factorization theorem, thisIΨ factors throughH^p. It follows from [7], Theorem 4.10, that α h² ≤ ρA(h) ≤ 1/Ψ⁻¹(A/h^1/p), for some constants 0< α <1 andA >0, andhsmall enough. That means that Ψ(x)≤C x^2p for xlarge enough.

Lemma 3.5 If the canonical injection IΨ:A→B^Ψ is1-summing, thenJΨ is compact.

Proof. The canonical injectionJ1:H¹→B¹(as well asJΨwhenever Ψ∈∆2) is compact. Hence we may assume thatH^Ψ is not H¹ and hence that Ψ(x)/x tends to +∞as xtends to +∞.

Assume thatJΨ is not compact. Then, as in the proof of Proposition 3.1, there are someA >1 and a sequence{xk}^k going to +∞such that Ψ(Axk)≥ [Ψ(xk)]². Settinghk= 1/Ψ(xk), we define, as in the proof of Proposition 3.4:

uk,j(z) = h²_k

[1−(1−hk)ξk,jz]²,

whereξk,j = e^2πij/Nk, withNkthe integer part of (1/hk) + 1. One hasuk,j ∈A and (see the proofs of the two quoted propositions):

NXk−1 j=0

|uk,j(e^it)| ≤C and kuk,jk^BΨ≥ δα A

1 Ψ⁻¹(1/hk)·

It follows that:

N_k−1

X

j=0

kuk,jk^BΨ≥δα A

Nk

Ψ⁻¹(1/hk)≥ δα A

1/hk

Ψ⁻¹(1/hk) = δα A

Ψ(xk) xk −→

k→∞+∞.

HenceIΨ is not 1-summing.

Proof of Theorem 3.3. Since JΨ: H^Ψ →B^Ψ is p-summing and the canon- ical injection IΨ: A → B^Ψ factors as IΨ:A → H^Ψ → B^Ψ, this injection is p-summing. By Lemma 3.4, Ψ(x) = O(x^2p) for x large enough. Hence we have the factorizationA →H^2p →H^Ψ →B^Ψ. Since the first injection is 2p- summing and the last one isp-summing, the composition is max(1, p1)-summing, with _p¹₁ = _2p¹ +¹_p (see [2], Theorem 2.22),i. e. p1= ²₃p. If p1>1, we can use again Lemma 3.4 withp1instead of 2p; we get that Ψ(x) =O(x^2p1), forxlarge enough, and that the factorizationIΨ:A→H^2p1 →H^Ψ →B^Ψ is max(1, p2)- summing, with _p¹

2 = _2p¹

1 + ¹_p. Going on the same way, we get a decreas- ing sequence {pn}n such that the canonical injection A → B^Ψ is max(1, pn)- summing and _p¹

n+1 =_2p¹

n +¹_p·Writingpn=αnp, we getαn+1= _2α^2αn

n+1; hence pn_n→∞−→ p/2. In particular, Ψ(x) =O(x^q) for everyq > p.

Ifp <2, one has max(1, pn) = 1 fornlarge enough, and Lemma 3.4 implies

thatJΨ is compact.

Remark 1. It is not clear whetherJΨ p-summing, withp≥2, implies thatJΨ

is compact. However, whenr ≥2,Jr:H^r →B^r isp-summing for nop <∞ (see [9]).

Remark 2. An operator T:X →Y between two Banach spaces is said to be finitely strictly singular (orsuperstrictly singular) if for everyε >0, there is an integerNε≥1 such that, for every subspaceX0 ofX of dimension≥Nε, there is anx∈X0such thatkT xk ≤εkxk. Every finitely strictly singular operator is strictly singular. It is not difficult to see that every compact operator is finitely strictly singular and it is shown in [10] (see also [5], Corollary 2.3) that every p-summing operator is finitely strictly singular. We do not know when JΨ is finitely strictly singular.

3.3 Order boundedness

Recall that an operator T: X →Y from a Banach spaceX into a Banach latticeY is said to beorder bounded if there isy ∈Y+ such that|T x| ≤y for everyxin the unit ball ofX. Since the Bergman-Orlicz spaceB^Ψis a subspace of the Banach lattice L^Ψ(D,A), we may study the order boundedness of JΨ. Actually, we are going to see that the natural space for the order boundedness ofJΨ is notL^Ψ(D,A), but theweak Orlicz space L^Ψ,∞(D,A), the definition of which we are recalling below (see [7], Definition 3.16).

Definition 3.6 Let (S,S, µ) be a measure space; the weak-L^Ψ space L^Ψ,∞ is the set of the (classes of ) measurable functions f: S →C such that, for some constantc >0, one has, for every t >0:

µ(|f|> t)≤ 1 Ψ(ct)·

One has L^Ψ ⊆ L^Ψ,∞ and ([7], Proposition 3.18) the equality L^Ψ = L^Ψ,∞

implies that Ψ ∈ ∆⁰. On the other hand, this equality holds when Ψ grows sufficiently; for example, if Ψ satisfies the condition ∆¹: xΨ(x) ≤ Ψ(αx), for some constantα >1 andxlarge enough.

Proposition 3.7 JΨ:H^Ψ→B^Ψ is always order bounded intoL^Ψ,∞(D,A).

Proof. Since (see [7], Lemma 3.11):

(3.1) 1

4Ψ⁻¹ 1 1− |z|

≤ sup

kfk_HΨ≤1|f(z)| ≤4Ψ⁻¹ 1 1− |z|

,

one has, denoting byS(z) the supremum in (3.1), fortlarge enough:

A(|S|> t)≤ A {z∈D; |z|>1−1/Ψ(t/4)}

≤ 2

Ψ(t/4) ≤ 1 Ψ(t/8),

and the result follows.

Since we also have, fort large enough:

A(|S|> t)≥ A {z∈D; |z|>1−1/Ψ(4t)}

≥ 1 Ψ(4t), we get:

Corollary 3.8 JΨ is order bounded into L^Ψ(D,A)if and only if L^Ψ =L^Ψ,∞. This is the case if Ψ∈∆¹.

Remark. Contrary to the compactness, or the weak compactness, which re- quires that Ψ does not grow too fast, the order boundedness ofJΨintoL^Ψ(D,A) holds when Ψ grows fast enough. Nevertheless, for Ψ(x) = exp[ log(x+1)2

]−1, JΨ is compact and order bounded intoL^Ψ(D,A).

WhenJΨis weakly compact,JΨmapsH^ΨintoBM^Ψ(Theorem 2.2); hence, we may ask whether JΨ may be order bounded into M^Ψ(D,A); however, we have:

Proposition 3.9 JΨ is never order bounded intoM^Ψ(D,A).

Proof. If it were the case, we should haveS∈M^Ψ(D,A), and hence Z

D

Ψ

4×1

4Ψ⁻¹ 1 1− |z|

dA(z)<+∞,

which is false.

4 An example

Theorem 4.1 There exists an Orlicz functionΨsuch thatJΨis 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 everyA >1, Ψ(Ax)/Ψ(x) is not non-decreasing forxlarge enough, and the conjugate function of Ψ does not satisfies condition ∆2.

The following lemma is undoubtedly well-known, but we have found no ref- erence, so we shall give a proof. Recall that a sublatticeX of L⁰(µ) is solid if

|f| ≤ |g|andg∈X impliesf ∈X andkfk ≤ kgk.

Lemma 4.2 Let (S,S, µ) be a measure space, and let X be a solid Banach sublattice ofL⁰(µ), the space of all measurable functions. Then, for every weakly null sequence{fn}ⁿinX and every sequence{An}ⁿof disjoint measurables sets, the sequence{fn1IAn}ⁿ converges weakly to0 inX.

Proof. If the conclusion does not hold, there are a continuous linear functional σ: X→Cand someδ >0 such that, up to taking a subsequence,|σ(fn1IAn)| ≥ δ. Set, for every measurable setA∈ S:

µn(A) =σ(fn1IA).

Thenµn is 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}k such that:

µn_k

[

l6=k

An_l≤ |µn_k| [

l6=k

An_l

≤δ/2.

Now, ifA=S

l≥1Anl,{fnk1IA}k is weakly null, but:

|σ(fnk1IA)| ≥ |σ(fnk1IA_nk)| − |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}ⁿ of positive numbers in the following way: setx1 = 4 and, for everyn≥1, xn+1 >2xn is the abscissa of the second intersection point of the parabolay =x² with 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, forn≥1:

(4.1) Ψ(xn) =x²_n, Ψ(2xn) =x⁴_n, Ψ affine between xn andxn+1. Then Ψ is an Orlicz function and

(4.2) x²≤Ψ(x)≤x⁴ for x≥4.

For this Orlicz function Ψ,JΨ is not compact, since Ψ(2x)/[Ψ(x)]²does not tend to 0. However, JΨ is weakly compact, because one has the factorization H^Ψ֒→H²֒→B⁴֒→B^Ψ (by (4.2) and Lemma 2.4).

Assume that JΨ is not Dunford-Pettis: there exists a weakly null sequence {fn}n in the unit ball of H^Ψ which does not converges for the norm in B^Ψ. Then{fn}n converges uniformly to 0 on the compact subsets of D(since it is weakly null) and we may assume thatkfnk^BΨ ≥ δ for some δ >0. We may also assume thatkfnk∞_n→∞−→ +∞because if{fn}nwere uniformly bounded, we should havekfnk^BΨ_n→∞−→ 0, by dominated convergence.

We are going to show that there exist a subsequence{fnk}kand pairwise dis- joint measurable setsAk ⊆Tsuch that the sequence{fnk1IAk}k ⊆L^Ψ(T, m) is equivalent to the canonical basis ofℓ1, whence a contradiction with Lemma 4.2.

It is worth to note from now that the Poisson integral P maps boundedly L²(T) into L⁴(D). Indeed, L²(T) = H² ⊕H₀² and the canonical injection is bounded fromH²into B⁴, by Lemma 2.4.

We have seen in the proof of Lemma 2.5 that there exist a subsequence {fnk}k and 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 thatkfnk1ICkkL^Ψ(D) ≥δ/2. The assumptions of that lemma are satisfied here:kfnkH^Ψ≤1,kfnk^BΨ≥δ,{fn}nconverges uniformly to 0 on the compact subsets ofD, andfn∈BM^ΨbecauseH^Ψ⊆BM^Ψ, sinceJΨis weakly compact.

Then:

Fact 1. There exist two sequences {αk}^k and {βk}^k, with βn > αn_n→∞−→ +∞ such that, ifgk =f_n^∗_k1I{αk≤|f_nk^∗ |≤βk}, then:

kP(gk)kL^Ψ(D)≥δ/3, wheref_n^∗_k is the boundary value offnk onT. Proof. 1) Letαk = ₁₂^δ Ψ⁻¹ 1/A(Ck)

and vk =P f_n^∗_k1I{|f_nk^∗ |<αk}

1ICk. One has:

Z

D

Ψ |vk|/(δ/12) dA=

Z

Ck

Ψ |vk|/(δ/12)

dA ≤Ψ αk/(δ/12)

A(Ck) = 1, so kvkkL^Ψ(D) ≤ δ/12. Since P(f_n^∗_k) = fnk, we have kP(f_n^∗_k) 1ICkkL^Ψ(D) = kfnk1ICkkL^Ψ(D)≥δ/2, and we get:

kP(f_n^∗_k1I{|f_nk^∗ ≥αk}) 1ICkkL^Ψ(D)≥ kfnk1ICkkL^Ψ(D)− kvkkL^Ψ(D)≥δ 2 − δ

12 =5δ 12· 2) Let wk =f_n^∗_k1I{|f_nk^∗ |≥αk}. Since P(wk1I{|wk|>β}) tends to 0 uniformly on Ck whenβ goes to infinity, Lebesgue’s dominated convergence theorem gives:

kP(wk1I{|wk|>β}) 1ICkkL^Ψ(D)≤ kP(wk1I{|wk|>β}) 1ICkkL⁴(D) −→

β→+∞0,

so there is someβk > αk such thatkP(wk1I{|wk|>β}) 1ICkkL^Ψ(D)≤δ/12.

We then have, withgk=f_n^∗_k1I{αk≤|f^∗

nk|≤βk}: kP(gk)kL^Ψ(D)≥ kP(gk) 1ICkkL^Ψ(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 subsetsEk ⊆ {αk≤ |f_n^∗_k| ≤βk}, such that, ifhk=f_n^∗_k1IEk, then:

kP(hk)kL^Ψ(D)≥δ/4.

Proof. First, since gk ∈ L^∞(T) ⊆ M^Ψ(T), there exists εk > 0 such that m(A)≤εk implieskgk1IAkL^Ψ(T)≤δ/(12kPk) (wherekPkstands for the norm ofP: L²(T)→L⁴(D)). Now, P: L^Ψ(T)→L^Ψ(D) is bounded and its norm is

≤ kPk, thanks to the factorizationL^Ψ(T)֒→L²(T)֒→L⁴(D)֒→L^Ψ(D). Hence kP(gk1IA)kL^Ψ(D)≤δ/12 form(A)≤εk.

LetBk ={αk ≤ |f_n^∗_k| ≤βk}. Up to taking a subsequence, we may assume thatP

l>km(Bl)≤εk. Let

Ek =Bk\[

l>k

Bl.

The setsEk,k≥1, are pairwise disjoint, and

kP(gk1IEk)kL^Ψ(D)≥ kP(gk1IBk)kL^Ψ(D)− kP gk1I^S_l>kBl

kL^Ψ(D)≥ δ 3− δ

12 =δ 4; so we get the Fact 2 withhk =gk1IEk=f_n^∗_k1IEk.

Set

Fk={z∈Ek; Ψ |f_n^∗_k(z)

| ≤M|f_n^∗_k(z)|²}. Forz∈Ek\Fk, one has:

Z

E_k\F_k|f_n^∗_k|²dm≤ 1 M

Z

T

Ψ(|f_n^∗_k)|dm≤ 1 M , sokf_n^∗_k1IEk\FkkL²(T)≤1/√

M and:

kP(f_n^∗_k1IE_k\F_k)kL^Ψ(D)≤ kP(f_n^∗_k1IE_k\F_k)kL⁴(D)

≤ kPk k(f_n^∗_k1IEk\Fk)kL²(T)≤ kPk

√M ≤δ 8,

forM large enough. It follows that, for M large enough,kP(f_n^∗_k1IFk)kL^Ψ(D)≥ δ/8 and

(4.3) kf_n^∗_k1IFkkL^Ψ(D)≥δ/(8kPk).