Average radial integrability spaces of analytic functions

Journal of Functional Analysis 282 (2022) 109262 Contents lists available at ScienceDirect Journal of Functional Analysis www.elsevier.com/locate/jfa Average radial integrability spaces of analytic functions ✩ Tanausú Aguilar-Hernández a, Manuel D. Contreras a,∗, Luis Rodríguez-Piazza b aDepartamento de Matemática Aplicada II and IMUS, Escuela Técnica Superior de Ingeniería, Universidad de Sevilla, Camino de los Descubrimientos, s/n 41092, Sevilla, Spain bDepartamento de Análisis Matemático and IMUS, Facultad de Matemáticas, Universidad de Sevilla, Calle Tarfia, s/n 41012, Sevilla, Spain a r t i c l e i n f o a b s t r a c t Article history: Received 27 February 2020 Accepted 28 September 2021 Available online 12 October 2021 Communicated by K. Seip MSC: primary 30H20, 47B33, 47D06 secondary 46E15, 47G10 Keywords: Mixed norm spaces Radial integrability Bergman projection In this paper we introduce the family of spaces RM(p, q), 1 ≤p, q≤+∞. They are spaces of holomorphic functions in the unit disc with average radial integrability. This family contains the classical Hardy spaces (when p =∞) and Bergman spaces (when p =q). We characterize the inclusion between RM(p1, q1)and RM(p2, q2) depending on the parameters. For 1 <p, q<∞, our main result provides a characterization of the dual spaces of RM(p, q)by means of the boundedness of the Bergman projection. We show that RM(p, q)is separable if and only if q<+∞. In fact, we provide a method to build isomorphic copies of ∞in RM(p, ∞). © 2021 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). ✩This research was supported in part by Ministerio de Economía y Competitividad, Spain, and the European Union (FEDER), project PGC2018-094215-B-100, and Junta de Andalucía, FQM133 and FQM- 104. *Corresponding author. E-mail addresses: [email protected] (T. Aguilar-Hernández), con[email protected] (M.D. Contreras), [email protected] (L. Rodríguez-Piazza). https://doi.org/10.1016/j.jfa.2021.109262 0022-1236/© 2021 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). 2T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 Contents 1. Introduction ....................................................... 2 2. Definition and first properties ........................................... 3 2.1. First examples................................................. 5 2.2. Point-evaluation functionals........................................ 9 2.3. Density of polynomials and separability ................................ 12 3. Containment relationships.............................................. 18 3.1. Inclusions .................................................... 18 3.2. Compactness of the inclusions ...................................... 21 4. Bergman projection .................................................. 25 References ............................................................. 33 1. Introduction In 1923, the classical Hardy spaces Hpwere introduced by F. Riesz [16]. He named those spaces after the article of G.H. Hardy [13]. Subsequently, Bergman developed his theory of reproducing kernels in A2spaces of general domains [4]and the theory of Ap spaces continued developing throughout the 20th century by the works of many authors; see monographs [8]and [14]. Since then, great progress has been made in the study of these and other spaces of analytic functions in the unit disc. In most of the cases, the belonging to the space is given in terms of boundedness (or integrability) of a certain average of the function on circles centered at the origin or in terms of the integrability with respect to the Lebesgue area, maybe with a certain weight. There are many other good books about these spaces, but we single out [7,10,15]. In other less studied cases, the belonging is determined by the average radial integrability. Maybe the most well-known space in this situation is the space of analytic functions of bounded radial variation BRV, a topic that goes back to Zygmund and where many different authors have worked (see, e.g., the papers of Bourgain [5], Rudin [17], and Zygmund [20]). The space BRV of analytic functions with bounded radial variation consists of those holomorphic functions g∈H(D)such that sup θ 1 ˆ 0|g(teiθ)|dt < ∞. Other different situation where the radial integrability plays an important role is in the Riesz-Féjer Theorem which says that there is a constant Cp>0such that if fbelongs to the Hardy space Hpthen sup θ⎛ ⎝ 1 ˆ 0|f(reiθ)|pdr⎞ ⎠ 1/p ≤Cp||f||Hp.(1.1) T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 3 The left-hand side of (1.1), considered as a function in the variables θand r, is the norm of fin the space L∞(T, Lp[0, 1]). This paper is devoted to introducing and studying the family of spaces RM(p, q)of analytic functions on the disk Dsuch that f∈Lq(T, Lp[0, 1]) (Definition 2.1). This family of spaces contains the Bergman spaces (when p =q) and Hardy spaces (when p =∞). As far as we know, there is no systematic study of spaces of average radial integrability. A second part of this research will appear in [1] where Littlewood-Paley type inequalities and integration operators are analyzed in the setting of these spaces. In Section 2, we introduce the family of spaces RM(p, q)and show a range of examples. Among them, we point out Proposition 2.5 where we characterize lacunary series belonging to RM(p, q). We analyze other properties such as boundedness of point-evaluation functionals and separability. We show that RM(p, q)is separable if and only if q<+∞ (see Proposition 2.13 and Theorem 2.18). In fact, RM(p, ∞)always contains a subspace isomorphic to ∞(Theorem 2.18). The main results of the paper appear in Section 3and 4. In Section 3we provide a complete characterization of when one of such spaces is included in another one (Theorem 3.3) and, in such a case, we characterize when the inclusion mapping is compact (Theorem 3.8). As a byproduct of such characterization, we see that the converse of (1.1)does not hold, that is, there are holomorphic functions fin Dsuch that supθ´1 0|f(reiθ)|pdr1/p <+∞but f/∈Hp. In the last section of this article we show the boundedness of the Bergman projection from Lq(T, Lp[0, 1]) onto our spaces RM(p, q)when 1 <p, q<∞. This allows us to identify the dual space of RM(p, q)for 1 <p, q<+∞(Corollary 4.8). The proof of the boundedness of the Bergman projection depends on techniques and tools coming from Harmonic Analysis. In particular, we use a classical result of C. Fefferman and E. Stein. The case p =qgives the well-known boundedness of the Bergman projection from Lp(D) onto the Bergman space Ap, which is usually proved by different techniques that do not work in our situation. Throughout the paper the letter C=C(·) will denote an absolute constant whose value depends on the parameters indicated in the parenthesis, and may change from one occurrence to another. We will use the notation a bif there exists a constant C=C(·) >0such that a ≤Cb, and a bis understood in an analogous manner. In particular, if a band a b, then we will write a b. Acknowledgment. The authors are grateful to the anonymous referee for a very careful reading of the manuscript and a number of helpful comments. 2. Definition and first properties We start this section introducing the spaces which are the goals of our study and providing some different kind of functions that belongs to them. In addition, we deal with some properties of such spaces, as for instance the separability. 4T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 Definition 2.1. Let 0 <p, q≤+∞. We define the spaces of analytic functions RM(p, q)={f∈H(D):ρp,q(f)<+∞} where ρp,q(f)=⎛ ⎜ ⎝ 1 2π 2π ˆ 0 ⎛ ⎝ 1 ˆ 0|f(reit)|pdr⎞ ⎠ q/p dt⎞ ⎟ ⎠ 1/q ,if p, q < +∞, ρp,∞(f)=esssup t∈[0,2π)⎛ ⎝ 1 ˆ 0|f(reit)|pdr⎞ ⎠ 1/p ,if p<+∞, ρ∞,q(f)=⎛ ⎝ 1 2π 2π ˆ 0sup r∈[0,1) |f(reit)|q dt⎞ ⎠ 1/q ,if q<+∞, ρ∞,∞(f)=fH∞. Remark 2.2. In the definition of ρp,∞the essential supremum can be replaced by the supremum. Fix θ∈[0, 2π]. Since the set of t ∈[0, 2π]such that ´1 0|f(reit)|pdr1/p ≤ ρp,∞(f)is dense in [0, 2π], we can extract a sequence {tn}in this set such that tn→θ. Using Fatou’s lemma it follows that ´1 0|f(reiθ)|pdr1/p ≤ρp,∞(f). One can easily check that if 1 ≤p, q≤+∞, then RM(p, q)is a Banach space when we endow it with the norm ρp,q. In fact in this paper, we will be interested only in these cases. So, we will stand most of our results for 1 ≤p, q≤+∞. Nevertheless, sometimes in the proofs of such results considering other values of pand qwill help us. In these cases, ρp,q is a quasi-norm and RM(p, q)is a quasi-Banach space. For certain parameters p, qthese spaces RM(p, q)are well known spaces. Namely, it is clear that RM(p, p)is nothing but the Bergman space Ap, for 0 <p <∞. Considering the case p =+∞, one can check that RM(∞, q)is contained in the Hardy space Hq. On the other hand, by [18, Theorem 17.11(a), p. 340], there is a constant C=C(q)such that if f∈Hq, then 2π ˆ 0 sup r∈[0,1] |f(reiθ)|qdθ ≤ 2π ˆ 0 sup{|f(z)|q:|eiθ −z|<3(1 −|z|)}dθ ≤Cfq Hq,(2.1) so that we get that RM(∞, q) =Hqfor all q∈(0, +∞]. Another interesting space that fits in this family is the space of analytic functions of bounded radial variation BRV (see, e.g., [5]), that is the space of analytic functions such that f∈RM(1, ∞). T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 5 2.1. First examples Example 2.3. For α>0, consider the function fα(z) =(1 −z)−α, where we are using the main branch of the logarithm to define w−α. Then fαbelongs to RM(p, q)if and only if α<1 p+1 q. Proof. Assume that 0 <p, q<∞. Write I(t) =´1 0|fα(reit)|pdr. Since Iis even and decreasing in [0, π], we have π/4 ˆ 0 I(t)q/p dt ≤ρq p,q(fα)=2 π ˆ 0 I(t)q/p dt ≤8 π/4 ˆ 0 I(t)q/p dt. In addition, for t ∈[0, π/4], we have that 1 −cos(t) t2/2. Therefore, ρq p,q(fα) π/4 ˆ 0 ⎡ ⎣ 1 ˆ 0 1 ((1 −r)2+rt2)αp/2dr⎤ ⎦ q/p dt. With a similar argument, we can reduce the integral in rto the interval [1/2, 1] and using that when rruns through this interval, the function rt2is equivalent to t2we have ρq p,q(fα) π/4 ˆ 0 ⎡ ⎢ ⎣ 1 ˆ 1/2 1 ((1 −r)2+t2)αp/2dr⎤ ⎥ ⎦ q/p dt. (2.2) If α≥1 p+1 q, and t ∈[0, 1/2], then 1 ˆ 1/2 1 ((1 −r)2+t2)αp/2dr ≥ 1 ˆ 1−t 1 ((1 −r)2+t2)αp/2dr ≥ 1 ˆ 1−t 1 (2t2)αp/2dr =1 2αp/2 1 tαp−1. Thus π/4 ˆ 0 ⎡ ⎢ ⎣ 1 ˆ 1/2 1 ((1 −r)2+t2)αp/2dr⎤ ⎥ ⎦ q/p dt ≥1 2αq/2 1/2 ˆ 01 tαp−1q/p dt =+∞, and so, by (2.2), fαdoes not belong to RM(p, q). Assume now that α<1 p+1 q. If αp <1, then 1 ˆ 1/2 1 ((1 −r)2+t2)αp/2dr ≤ 1 ˆ 1/2 1 (1 −r)αp dr < +∞, 6T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 so that, by (2.2), fα∈RM(p, q). If αp =1, then we obtain 1 ˆ 1/2 1 ((1 −r)2+t2)αp 2 dr ≤ 1 ˆ 1−t 1 tdr + 1−t ˆ 1/2 1 1−rdr ≤ln e 2t. Integrating with respect to tit follows that ρp,q(fα) <+∞. It remains to see what happens if 1 <αp <1 +p q. In this case, if t ∈[0, π/4], we have 1 ˆ 1/2 1 ((1 −r)2+t2)αp/2dr ≤ 1−t ˆ 1/2 1 (1 −r)αp dr + 1 ˆ 1−t 1 tαp dr ≤αp αp −1 1 tαp−1. Therefore, by (2.2), ρq p,q(fα)αp αp −1q/p π/4 ˆ 0 1 tαq−q p dt < +∞. Summing up, the result holds if both pand qare finite. For p =∞, since RM(∞, q) = Hq, the result is well-known (see, e.g., [7, Page 13]). For q=∞, arguing as above we have ρp p,∞(fα)sup 0≤t≤π/2 1 ˆ 1/2 1 ((1 −r)2+t2)αp/2dr = 1 ˆ 1/2 dr (1 −r)αp <+∞(2.3) if and only if α<1 p. Finally it is clear that the function fαdoes not belong to H∞. Example 2.4. Let 1 ≤p, q<∞, n ≥1and take αsuch that 1 p+1 q=1 α. The RM(p, q)- norm of the holomorphic function fn,α(z)=n  k=0 zk1/α =1−zn+1 1−z1/α , where we are using the main branch of the logarithm to define w1/α, can be estimated as ρp,q(fn,α)p p−α1/p ln1/q(n+1).(2.4) Proof. Clearly, fn,α is well-defined. Since ρp,q(fn,α) =ρ1/α p/α,q/α(fn,1), the proof of (2.4) is reduced to the case α=1and 1 p+1 q=1. Notice that, in this case, p >1. T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 7 Since π ˆ π/4 ⎛ ⎝ 1 ˆ 0|fn,1(reiθ)|pdr⎞ ⎠ q/p dθ ≤ π ˆ π/4 ⎛ ⎝ 1 ˆ 0|4|pdr⎞ ⎠ q/p dθ =3π 44q, we have 2πρp,q(fn,1)q≤2 π/4 ˆ 0 ⎛ ⎝ 1 ˆ 0|fn,1(reiθ)|pdr⎞ ⎠ q/p dθ +3π 44q. If 1 −θ≤r≤1and θ∈[0, π/4], arguing as in Example 2.3, we obtain |fn,1(reiθ)|2 (1 −r)2+rθ2≤2 θ√1−θ≤2 θ1−π 4 <5 θ. Therefore, there is a constant C>2such that π/4 ˆ 1 n+1 ⎛ ⎝ 1 ˆ 0|fn,1(reiθ)|pdr⎞ ⎠ q/p dθ = ≤ π/4 ˆ 1 n+1 ⎛ ⎝ 1−θ ˆ 0 2p (1 −r)pdr + 1 ˆ 1−θ Cp θpdr⎞ ⎠ q/p dθ ≤Cq π/4 ˆ 1 n+1 p p−1θ−p+1q/p dθ =Cqp p−1q/p (ln(π/4) + ln(n+1)) and 1 n+1 ˆ 0 ⎛ ⎝ 1 ˆ 0|fn,1(reiθ)|pdr⎞ ⎠ q/p dθ = ≤2q 1 n+1 ˆ 0 ⎛ ⎜ ⎝ 1−1 n+1 ˆ 0 1 (1 −r)pdr + 1 ˆ 1−1 n+1 (n+1) pdr⎞ ⎟ ⎠ q/p dθ ≤2qp p−1q/p . Putting altogether, we get the estimation of ρp,q(fn,α).  Next example provides the lacunary series that belong to RM(p, q). For p =∞, that is for Hardy spaces, the characterization is different and it can be seen in [15, 8T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 Theorem 6.2.2] for q<+∞and in [21, Vol. I, p. 247] for q=∞. We will say that a sequence of positive numbers {xk}is a lacunary sequence if there is a constant λsuch that xk+1 xk≥λ >1. Proposition 2.5. Let {nk}∞ k=0 be a lacunary sequence of positive integer numbers, 1 ≤ p <∞and 1 ≤q≤∞. Then f(z)= ∞  k=0 αkznk belongs to RM(p, q)if and only if ∞  k=0 |αk|p nk <+∞. Moreover, it is satisfied that ρp,q(f)∞  k=0 |ak|p nk1/p .(2.5) Remark 2.6. Notice that the term on the right in (2.5)does not depend on q. Proof. Notice that nk+1 pk≥0is also a lacunary sequence. The proof of this result is based on a characterization of bases on Lp[0, 1] due to Gurariˇı and Macaev [12]. Namely they proved that, fixed p ∈[1, +∞), if a sequence {nk}k≥0is lacunary then there exist two positive constants Aand Bsuch that A∞  k=0 |βk|p1/p ≤     ∞  k=0 βkp nk+1/p tnk    Lp≤B∞  k=0 |βk|p1/p ,(2.6) for every {βk} ∈p. Take now f(z) =∞ k=0 αkznka holomorphic function in the unit disc. Fix θ∈[0, 2π] and write βk:= αk p nj+1/p eiθnkif k≥0. By (2.6), A∞  k=0 |αk|p nk+1/p1/p ≤     ∞  k=0 αkrnkeiθnk    Lp =⎛ ⎝ 1 ˆ 0|f(reiθ)|pdr⎞ ⎠ 1/p ≤B∞  k=0 |αk|p nk+1/p1/p . Now, looking at the very definition of ρp,q we get the result.  T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 9 2.2. Point-evaluation functionals This subsection is devoted to the functionals f→ f(z)and f→ f(z). We prove that both of them are bounded and estimate their norms. We will need the following inclusion. Proposition 2.7. Let 0 <s ≤+∞. Then Hs⊂RM(p, q)if and only if 1 p+1 q≥1 s. Proof. The result is clear for s =+∞since H∞is a subspace of RM(p, q)for every p, q. Thus, from now on we consider the case s <+∞. Assume that Hs⊂RM(p, q) and suppose that 1 s>1 p+1 q. Take 1 s>α>1 p+1 q. Then the function fαdefined in Example 2.3 belongs to RM(∞, s) =Hsand not to RM(p, q). A contradiction. Thus 1 p+1 q≥1 s. To see that converse implication we claim that Hs⊂RM(p1, q1) whenever 1 ≤s < +∞and 1 p1+1 q1=1 s. Assume for the moment that the claim holds. Fix pand qsuch that 1 p+1 q≥1 s. We consider p1≥pand q1≥qsuch as 1 p1+1 q1=1 s. By the claim Hs⊂RM(p1, q1). Moreover, it is easy to prove that RM(p1, q1) ⊂RM(p, q)using Hölder’s inequality twice for p1≥pand q1≥q. Thus it remains to prove the claim. By Féjer-Riesz theorem [7, Theorem 3.13, p. 46], we have that for each f∈Hsand θ, 1 ˆ 0|f(reiθ)|sdr ≤1 2fs Hs (notice that, in particular, this implies that Hs⊂RM(s, ∞)). Now, since 1 p1+1 q1=1 s we can take λ ∈[0, 1] such that 1 p1=λ sand 1 q1=1−λ s. Then ρp1,q1(f)=⎛ ⎜ ⎝ 2π ˆ 0 ⎛ ⎝ 1 ˆ 0|f(reiθ)|p1(1−λ)|f(reiθ)|p1λdr⎞ ⎠ q1/p1 dθ 2π⎞ ⎟ ⎠ 1/q1 ≤⎛ ⎜ ⎝ 2π ˆ 0 sup r|f(reiθ)|q1(1−λ)⎛ ⎝ 1 ˆ 0|f(reiθ)|sdr⎞ ⎠ q1λ/s dθ 2π⎞ ⎟ ⎠ 1/q1 ≤⎛ ⎜ ⎝ 2π ˆ 0 sup r|f(reiθ)|s⎛ ⎝ 1 ˆ 0|f(reiθ)|sdr⎞ ⎠ q1λ/s dθ 2π⎞ ⎟ ⎠ 1/q1 fλ Hs⎛ ⎝ 2π ˆ 0 sup r|f(reiθ)|sdθ 2π⎞ ⎠ 1/q1 =fλ Hsρ∞,s(f)1−λfHs, 16 T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 and 1 ˆ ak−rk εp k |ak−r|2pdr =εp k 2p−11 (ak−1)2p−1−1 r2p−1 k=1. In addition we can find a sequence {θk}such that the disks D(akeθki, rk)are pairwise disjoint. For that, we consider θk=arcsin(rk)+2 k−1  n=0 arcsin (rn). It is easy to see that D(akeiθk, rk) ∩D(ak+1eiθk+1 , rk+1) =∅, because θk+1 −θk=arcsin(rk+1)+arcsin(rk). Moreover, it is also obtained that |θk|≤π 2rk+π k−1  n=0 rn<π k  n=0 rn<π. Finally, take fk(z) := εk (ze−iθk−ak)2, z∈C\{akeiθk}. Since fkis bounded in D, it belongs to RM(p, 0). In addition, we have that |fk(z)| ≤εk r2 kif z/∈D(akeiθk, rk). Since ∞ k=0 εk r2 k<∞, it is easy to see that, given a bounded sequence {αk}, the sequence {k n=0 αnfn(z)}converges uniformly on compacta of Dto T({αk}) := ∞ k=0 αkfk, so that T({αk})is holomorphic in D. By construction, every radius Lθ={teiθ :t ∈[0, 1)}only intersects one of the open disks. Let us see that f=T({αk}) ∈RM(p, ∞). On the one hand, if θ∈[0, 2π]is such that there is k0with Lθ∩D(ak0eiθk0, rk0) =∅, Then ⎛ ⎝ 1 ˆ 0|f(reiθ)|pdr⎞ ⎠ 1/p ≤|αk0|⎛ ⎝ 1 ˆ 0|fk0(reiθ)|pdr⎞ ⎠ 1/p + j=0 |αj|εj r2 j ≤ ||{αk}||∞⎛ ⎜ ⎝⎛ ⎝ 1 ˆ 0|fk0(reiθk0)|pdr⎞ ⎠ 1/p +1 ⎞ ⎟ ⎠ ≤ ||{αk}||∞⎛ ⎝1+(ak0−rk0)εp k0 r2p k01/p +1 ⎞ ⎠≤3||{αk}||∞. On the other hand, if θ∈[0, 2π]is such that eiθ /∈D(akeiθk, rk)for all k, then T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 17 ⎛ ⎝ 1 ˆ 0|f(reiθ)|pdr⎞ ⎠ 1/p ≤ ||{αk}||∞ ∞  k=0 εk r2 k≤ ||{αk}||∞. That is f=T({αk}) ∈RM(p, ∞) and, in particular, T:∞→RM(p, ∞)is bounded. Let us see that Tis open from ∞to T(∞)so that it establishes an isomorphism between ∞and T(∞). For each n, it follows ρp,∞(T({αk})) ≥⎛ ⎝ 1 ˆ 0|T({αk})(reiθn)|pdr⎞ ⎠ 1/p ≥⎛ ⎝ 1 ˆ an−rn |T({αk})(reiθn)|pdr⎞ ⎠ 1/p ≥|αn|−{αk}∞⎛ ⎝ ∞  j=0 εj r2 j⎞ ⎠(1 −an+rn)1/p ≥|αn|−{αk}∞⎛ ⎝ ∞  j=0 εj r2 j⎞ ⎠. Therefore, |αn| ≤{αk}∞∞ j=0 εj r2 j+ρp,∞(T({αk})) and taking supremum in nwe obtain ρp,∞(T({αk})) ≥⎛ ⎝1− ∞  j=0 εj r2 j⎞ ⎠{αk}∞. Since ∞ j=0 εj r2 j<1, we get that Testablishes an isomorphism between ∞and T(∞). To end the proof, we show that T({αk}) ∈RM(p, 0) if and only if {αk} ∈c0. Let T({αk})∈RM(p, 0). Then sup θ⎛ ⎝ 1 ˆ an−rn |T({αk})(reiθ)|pdr⎞ ⎠ 1/p ≥⎛ ⎝ 1 ˆ an−rn |T({αk})(reiθn)|pdr⎞ ⎠ 1/p ≥|αn|−{αk}∞⎛ ⎝ ∞  j=0 εj r2 j⎞ ⎠(1 −an+rn)1/p. Since 1 −an+rn→0and T({αk})∈RM(p, 0), it follows that {αk} ∈c0. Conversely, let α={αk}k∈c0and let us prove that T(α) ∈RM(p, 0). Since fk∈ RM(p, 0), then n k=1 αkfk∈RM(p, 0) for all n ∈N. Moreover, n k=1 αkfk→T(α) because Tis continuous and (α1, ..., αn, 0, 0, ...) →αin ∞. Finally, T(α) ∈RM(p, 0) since RM(p, 0) is a closed subspace of RM(p, ∞).  18 T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 3. Containment relationships 3.1. Inclusions In this section we will give a characterization for the containment relationships between our spaces. To do this, we recall the notion of the Marcinkiewicz spaces Lp,∞, also called the weak Lpspaces. Definition 3.1. Let 0 <p <∞and (X, Σ, μ)a measure space. We define the weak Lp space of measurable functions Lp,∞=#f:X→Cmeasurable : fp,∞:= sup t>0 tλ1/p f(t)<∞$ where λf(t)=μ({x∈X:|f(x)|>t}). Lemma 3.2. [11, Proposition 1.1.14, p. 8] Let f∈Lp0,∞∩Lp1,∞with p0=p1. Then f∈Lpfor 1 p=1−λ p0+λ p1, λ ∈(0, 1). Moreover, there exists a constant C(p0, p1, λ) >0 such that fp≤C(p0,p 1,λ)f1−λ p0,∞fλ p1,∞, for λ ∈(0, 1). Theorem 3.3. Let 1 ≤p0, q0≤∞and set A(p0,q 0)=#(p, q)∈(0,+∞]×(0,+∞]:1 p+1 q≥1 p0 +1 q0 ,p 0≥p$. (1) If p0, q0<+∞, then RM(p0, q0) ⊂RM(p, q)if and only if (p, q) ∈A(p0, q0) \ {(β,∞)}, where β=p0q0 p0+q0. (2) If either p0or q0are +∞, then RM(p0, q0) ⊂RM(p, q)if and only if (p, q) ∈ A(p0, q0). Before proving the result, it is worth showing a picture of the set A(p0, q0). If p0, q0< +∞then the set {(1 p, 1 q) :(p, q) ∈A(p0, q0)}is the grey region (including its boundary) in Fig. 1A while if either p0or q0are +∞, the set {(1 p, 1 q) :(p, q) ∈A(p0, q0)}is the grey region (including its boundary) in Fig. 1B or Fig. 1C, respectively. Proof. Bearing in mind that Hq0=RM(∞, q0), Proposition 2.7 is nothing but the case p0=+∞. Therefore, from now on, we will assume that p0<+∞. To clarify the exposition, we split the proof in several steps. T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 19 Fig. 1. Containment relationships. Step 1. If p0, q0<+∞(Fig. 1A) and (p, q)is such that (1/p, 1/q) belongs to the open segment with end points (1/p0, 1/q0)and (1 p0+1 q0, 0), then RM(p0, q0) ⊂RM(p, q). Write (1 p, 1 q) =λ(1 p0, 1 q0) +(1 −λ)( 1 p0+1 q0, 0) for some λ ∈(0, 1). Take f∈RM(p0, q0) with ρp0,q0(f) ≤1. For each θ, define fθ(r) := f(reiθ). Let us see that fθ∈Lp0,∞([0, 1]) ∩ Lα,∞([0, 1]) for almost every θ∈[0, 2π], where 1 p0+1 q0=1 α. Since f∈RM(p0, q0), by the very definition, we have that fθ∈Lp0([0, 1)) for almost every θand fθp0,∞≤ fθp0. Moreover, by Proposition 2.8, there is a constant C>0such that |f(z)| ≤ C1 (1−|z|2) 1 p0+1 q0 , for all z, and thus fθα,∞=sup t≥0 tm 1({r∈[0,1) : |fθ(r)|>t})1/α ≤sup t≥0 tm 1#r∈[0,1) : 1 −r≤Cα tα$1/α =sup t≥0 min{t, C}≤C, so that fθ∈Lα,∞([0, 1]) for all θ. Hence, applying Lemma 3.2 we have fθp≤C(p0,α,λ)·fθλ p0,∞·fθ1−λ α,∞. Thus ρp,q(f)≤C(p0,α,λ)⎛ ⎝ 2π ˆ 0fθλq p0,∞·fθ(1−λ)q α,∞dθ⎞ ⎠ 1/q ≤C(p0,α,λ)C1−λ⎛ ⎝ 2π ˆ 0fθλq p0dθ⎞ ⎠ 1/q =C(p0,α,λ)C1−λ⎛ ⎝ 2π ˆ 0fθq0 p0dθ⎞ ⎠ 1/q =C(p0,α,λ)C1−λρp0,q0(f)λ≤C(p0,α,λ)C1−λ. 20 T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 Step 2. If 1 p>1 p0+1 q0, then RM(p0, q0) ⊂RM(p, ∞). Take f∈RM(p0, q0). By Proposition 2.8, there is C>0such that ρp,∞(f)=esssup θ∈[0,2π)⎛ ⎝ 1 ˆ 0|f(reiθ)|pdr⎞ ⎠ 1/p ≤Cess sup θ∈[0,2π)⎛ ⎝ 1 ˆ 0 ρp0,q0(f)p (1 −r)1 p0+1 q0pdr⎞ ⎠ 1/p <+∞ Step 3. If p0≥pand q0≥qthen RM(p0, q0) ⊂RM(p, q). This inclusion is a direct consequence of Höder’s inequality. Denote by B(p0, q0) ={(p, q)∈R+×R+:p0≥p, q0≥q}and (pλ, qλ)the couple such that 1 pλ,1 qλ=λ 1 p0,1 q0+(1 −λ) 1 p0+1 q0,0. Since A(p0,q 0)\# p0q0 p0+q0 ,∞$=∪λ∈(0,1]B(pλ,q λ)∪#(p, ∞): 1 p>1 p0 +1 q0$, Steps 1, 2 and 3 give that if p0, q0<+∞and (p, q) ∈A(p0, q0) \p0q0 p0+q0,∞then RM(p0, q0) ⊂RM(p, q). Step 4. If RM(p0, q0) ⊂RM(p, q)then p0≥p. By closed graph theorem there is a constant C>0such that ρp,q(f) ≤Cρp0,q0(f)for all f∈RM(p0, q0). Taking fn(z) =znwe obtain ρp,q(fn)=(1+np)−1 p≤C(1 + np0)−1 p0=ρp0,q0(fn) and this inequality holds for all nif and only if p0≥p. Step 5. If 1 p0+1 q0>1 p1+1 q1then RM(p0, q0) RM(p1, q1) We consider a function fαof Example 2.3 such that 1 p1+1 q1<α< 1 p0+1 q0. Hence, we have a function fαsuch that fα∈RM(p0, q0) \RM(p1, q1). Step 6. If p0, q0<+∞, then RM(p0, q0) RM(β, ∞), where β=p0q0 p0+q0. Assume that RM(p0, q0) ⊂RM(β, ∞). By closed graph theorem there is a positive constant C>0such that ρβ,∞(f) ≤Cρp0,q0(f). For each n, consider the function fn,β introduced in Example 2.4. Then ρβ,∞(fn,β)≥⎛ ⎝ 1 ˆ 0 n  k=0 rkdr⎞ ⎠ 1 β =n  k=0 1 k+11 β ≥ln 1 β(n+1). Thus, Example 2.4 would imply ln1/β(n+1)≤Cp0 p0−β1/p0 ln1/q0(n+1), T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 21 what is not possible if nis large enough. So RM(p0, q0) RM(β, ∞). Clearly, Steps 4, 5 and 6 imply that if RM(p0, q0) ⊂RM(p, q)then (p, q) ∈A(p0, q0) \ {(β,∞)}. Therefore, statement (1) and (2) are proved.  A simple argument shows that if q<+∞, the density of the polynomials in RM(p, q) implies that if RM(p, q) ⊂RM(p0, ∞)if and only if RM(p, q) ⊂RM(p0, 0). The situation is not so clear to study when RM(p0, 0) is contained in RM(p, q). To characterize it, we need the following lemma. Lemma 3.4. Let 1 ≤p, q≤+∞. If {fn}is a bounded sequence in RM(p, q)that converges uniformly on compact subsets of the unit disc to f. Then f∈RM(p, q). Proof. Clearly the function fis holomorphic. Assume that p, q<+∞. By Fatou’s Lemma, for each θwe have 1 ˆ 0|f(reiθ)|pdr ≤g(θ) := lim inf ngn(θ), where, for each n, gn(θ) := ´1 0|fn(reiθ)|pdr. Repeating again the argument, we have ρp,q(f)q≤1 2π 2π ˆ 0 (g(θ))q/p dθ ≤lim inf n 1 2π 2π ˆ 0 (gn(θ))q/p dθ = lim inf nρq p,q(fn)≤sup n ρq p,q(fn)<+∞. A similar argument works in the remaining cases, so that we are done.  Proposition 3.5. Let 1 ≤p0≤+∞. Then RM(p0, ∞) ⊂RM(p, q)if and only if RM(p0, 0) ⊂RM(p, q). Proof. Assume that RM(p0, 0) ⊂RM(p, q). Take f∈RM(p0, ∞). For each r<1, the function frbelongs to RM(p0, 0) and then to RM(p, q). Since {fr:r<1}is bounded in RM(p0, 0), it is also bounded in RM(p, q). Since frconverges uniformly on compact subset of Dto f, Lemma 3.4 guarantees that f∈RM(p, q).  3.2. Compactness of the inclusions Once the containment relationships of these spaces have been determined, we study when such inclusions are compact. A standard argument shows the following characterization of compactness. 22 T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 Lemma 3.6. Let 1 ≤p0, q0≤+∞and 1 ≤p, q≤+∞. Then i :RM(p0, q0) →RM(p, q) is compact if and only if every bounded sequence {fn}in RM(p0, q0)that converges to zero uniformly on compact subsets of the unit disc satisfies that limnρp,q(fn) =0. We will use this lemma several times in the proof of the next theorem without explicit reference. We also need the following result. Proposition 3.7. Let 1 ≤p <+∞, f∈RM(p, ∞), and σ∈∂D. Then for the nontangential limit we have ∠ lim z→σf(z)(1 −σz)1/p =0. Proof. Without loss of generality we assume that σ=1. Suppose that ρp,∞(f) ≤1and consider the holomorphic function h(z) =f(z)(1 −z)1/p. Fix R>1and the Stolz region S(1, R) ={z∈D:|1 −z| <R(1 −|z|)}. Looking at (2.7)in the proof of Proposition 2.8, we see that there is a constant Csuch that |h(z)|≤R1/p|f(z)|(1 −|z|)1/p ≤CR1/p,z∈S(1,R). That is, the function his bounded on S(1, R). Therefore, by Lindelöf’s Theorem [6, Theorem 1.5.7, p. 26], it is enough to prove that limr→1−|f(r)|(1 −r)1/p =0. Assume by contradiction that there is a constant c1>0and a sequence {rk}where rk→1−such that c1≤|f(rk)|(1 −|rk|)1/p for all k. Write δk:= 1 −rk. By Proposition 2.9, there is a constant Csuch that |f(x)| ≤C (1−x)1+ 1 pfor all x ∈(0, 1). Choose ε <c1 2C. Then, for 1 −(1 +ε)δk<x <1 −δk, |f(x)−f(rk)|≤C|x−rk|1 (1 −rk)1+1/p ≤Cεδk 1 δ1+1/p k =Cε δ1/p k <c1 2δ1/p k . Thus |f(x)|≥|f(rk)|−|f(x)−f(rk)|≥ c1 (1 −rk)1/p −c1 2δ1/p k =c1 2δ1/p k and ⎛ ⎜ ⎝ 1−δk ˆ 1−(1+ε)δk |f(x)|pdx⎞ ⎟ ⎠ 1/p ≥(εδk)1/p c1 2δ1/p k =c1ε1/p 2. Notice that this lower bound does not depend on k. But, this is impossible because ´1 0|f(x)|pdx <+∞. Theorem 3.8. Let 1 ≤p0, q0≤+∞and 1 ≤p, q≤+∞. Then i :RM(p0, q0) →RM(p, q) is compact if and only if 1 p+1 q>1 p0+1 q0and p <p 0. T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 23 Fig. 2. Compactness of the inclusions. As we can see in the Fig. 2, the grey region, removing this time the dotted lines, represents the spaces RM(p, q)such as i :RM(p0, q0) →RM(p, q)is compact when p0<+∞in Fig. 2A and when p0=+∞in Fig. 2B. Proof. Bearing in mind Theorem 3.3, we have to prove that the inclusion is compact if 1 p+1 q>1 p0+1 q0and p <p 0and it is not compact if either 1 p+1 q=1 p0+1 q0or p =p0. Let us start by showing that it is not compact if p =p0. For each n, consider the function fn(z) =(np0+1) 1/p0zn, z∈D. A simple calculation shows that ρp0,q0(fn) =1 and that the sequence {fn}converges uniformly to zero on compact subsets of the unit disc. Assume that i :RM(p0, q0) →RM(p0, q)is compact, then there exists a subsequence {fnk}such that ρp0,q(fnk)must go to 0as kgoes to ∞. But this is not possible because ρp0,q(fn) =1for all n. Now take pand qsuch that 1 p+1 q=1 p0+1 q0. Assume that i :RM(p0, q0) →RM(p, q) is compact. Then i∗:(RM(p, q))∗→(RM(p0, q0))∗is also a compact operator. Assume firstly that q<+∞. Let us see that δz δz(RM(p,q))∗w∗-converges to 0when |z| →1. Taking pa polynomial we obtain |δz(p)| δz(RM(p,q))∗|p(z)|(1 −|z|2)1 p+1 q≤p∞(1 −|z|2)1 p+1 q, which clearly goes to 0when |z| →1. Since q<+∞, by Proposition 2.13, polynomials are dense in RM(p, q)and then δz δz(RM(p,q))∗w∗-converges to 0when |z| →1. Therefore, the compactness of i∗gives lim |z|→1    i∗δz δz(RM(p,q))∗   (RM(p0,q0))∗ =0. However, this is impossible because, as we shall now see, such norm must be greater than a certain positive constant. Indeed, since i∗(δz) =δz, by Proposition 2.8, 24 T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262     i∗δz δz(RM(p,q))∗   (RM(p0,q0))∗≥δz(RM(p0,q0))∗ δz(RM(p,q))∗1. The argument for the case p0=q=∞is the same. However, we consider a sequence {zn}in the Stolz region such that |zn| →1. In this way, we obtain the w∗-convergence bearing in mind Proposition 3.7. Assume now that 1 p1>1 p0+1 q0and take a sequence {fn}in RM(p0, q0)such that ρp0,q0(fn) ≤1for all nand it converges to zero uniformly on compact subsets of D. We claim that limnρp1,∞(fn) =0. Otherwise, there is  >0and a subsequence (that we denote in the same way) such that ρp1,∞(fn) >for all n. Thus, we find {θn}such that 1 ˆ 0|fn(reiθn)|p1dr ≥p1,(3.1) for all n ∈N. For each n, we write gn(r) := fn(reiθn), r∈[0, 1). Since ρp0,q0(fn) ≤1, by Proposition 2.8 there is a constant C>0such that |gn(r)|=|fn(reiθn)|≤ C (1 −r2)1 p0+1 q0 . Since the map r→ C (1−r2) 1 p0+1 q0 belongs to Lp1([0, 1]) and {gn}converges pointwise to zero, we get that it converges to zero in the norm of Lp1([0, 1]) which contradicts (3.1). So that the claim holds. Take now p, qsuch that there is λ ∈(0, 1) with 1 p,1 q=λ 1 p0,1 q0+(1 −λ) 1 p1,0. Then, for each f∈RM(p, q), applying Hölder’s inequality we have 1 ˆ 0|f(reiθ)|pdr = 1 ˆ 0|f(reiθ)|λp|f(reiθ)|(1−λ)pdr ≤⎛ ⎝ 1 ˆ 0|f(reiθ)|p0dr⎞ ⎠ λp/p0⎛ ⎝ 1 ˆ 0|f(reiθ)|p1dr⎞ ⎠ (1−λ)p/p1 . So that ρp,q(f)≤ρp0,q0(f)λρp1,∞(f)1−λ. This inequality, the above claim and Lemma 3.6 show that i:RM(p0, q0) →RM(p, q) is compact whenever 1 p+1 q>1 p0+1 q0and q0<q. Take now p, qsuch that 1 p+1 q>1 p0+1 q0, p <p 0and q0≥q. Fix ˜q<qsuch that 1 p0+1 ˜q<1 p+1 q. By the above argument, the inclusion map ˜ ifrom RM(p0, ˜q)into RM(p, q)is compact. Since i :RM(p0, q0) →RM(p, q) factorizes through ˜ i, we get that iis compact.  T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 25 4. Bergman projection In the theory of Banach spaces of analytic functions, a useful integral operator is the Bergman projection P(f)(z)=ˆ D K(z,w)f(w)dA(w),z∈D, with kernel K(z,w)=(1−zw)−2,z,w∈D,(4.1) which is called the Bergman kernel. Such function is the reproducing kernel for the Bergman space A2. This projection is well-defined on L1(D), mapping each function of L1(D)to an analytic function and mapping each function of the Bergman space A1into itself. Moreover, for 1 <p <∞it is known that the Bergman projection is a bounded operator from Lp(D)onto Ap. These properties allow to describe the dual of Bergman spaces Ap. Theorem 4.1. For 1 <p <+∞, the dual space of Apcan be identified with Ap, where pis the conjugated index, that is, 1 p+1 p=1. Such functional φ ∈(Ap)∗has a unique representation φ(f)=φg(f)=ˆ D f(w)g(w)dA(w),f∈Ap, for some g∈Ap. Mimicking this schedule for Bergman spaces (but with a deeper argument), in this section we prove the boundedness of the Bergman projection from Lq(T, Lp[0, 1]) onto RM(p, q), where 1 <p, q<∞and, as a byproduct, we describe its dual. To study the duality of RM(p, q) spaces, the following theorem will be important because it provides a characterization of the dual space of Lq(T, Lp[0, 1]), for 1 ≤p, q< +∞. Let us recall that a measurable function f:T×[0, 1) →Cbelongs to Lq(T, Lp[0, 1]) if ρp,q(f) <+∞(here, the definition of ρp,q is the same as the one given in Definition 2.1 for holomorphic functions). Theorem 4.2. [3, Theorem 1, p. 304] Let 1 ≤p, q<+∞. J(f)is a continuous functional on the normed space Lq(T, Lp[0, 1]) if and only if it can be represented by J(f)=ˆ D h(w)f(w)dA(w) 32 T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 Given α∈R, consider the function f(reiθ):=%0,re iθ /∈Ω, θαK(1 −θ,re−iθ),re iθ ∈Ω, where, as usual, Kis the Bergman kernel. Taking α=2 −1 p=1 +1 p, we have f∈ L∞(T, Lp[0, 1]). Indeed, for 0 <θ<1/2, 1 ˆ 0|f(reiθ)|pdr =θpα 1−2θ ˆ 0|K(1 −θ,re−iθ)|pdr ≤θpα 1 ˆ 0 dr (1 −(1 −θ)r)2p =θpα 2p−1 θ1−2p−1 1−θ≤2 2p−1θpα+1−2p=2 2p−1<+∞. Now let us see that this function fsatisfies that |P(f)(a)| (1 −a)−1/p for every a ∈'3 4,1(. We have that the Bergman projection of the function f, for a ∈(0,1), is P(f)(a)= 1/2 ˆ 0 θα⎛ ⎝ 1−2θ ˆ 0 rdr (1 −are−iθ)2(1 −(1 −θ)reiθ)2⎞ ⎠ dθ π = 1/2 ˆ 0 θα⎛ ⎝ 1−2θ ˆ 01−(1 −θ)re−iθ 1−are−iθ 2rdr |1−(1 −θ)re−iθ|4⎞ ⎠ dθ π. By Lemma 4.10 (applying first (3) and then (1)), we obtain |P(f)(a)|≥Re [P(f)(a)] ≥3 5 1/2 ˆ 0 θα⎛ ⎝ 1−2θ ˆ 0&&&& 1−(1 −θ)reiθ 1−areiθ &&&& 2rdr |1−(1 −θ)re−iθ|4⎞ ⎠ dθ π  1/2 ˆ 0 θα⎛ ⎝ 1−2θ ˆ 0 rdr (1 −ar)2(1 −(1 −θ)r)2⎞ ⎠dθ ≥ 1−a ˆ 0 θα⎛ ⎝ 1−2θ ˆ 0 rdr (1 −ar)4⎞ ⎠dθ ≥1 12a 1−a ˆ 0 θα'(1 −a(1 −2θ))−3−(1 −a/4)−3(dθ. Using that θ<1 −aand 3/4 ≤a <1we deduce (1 −a(1 −2θ))−3−(1 −a/4)−3≥ (1 −a(1 −2θ))−3/2and 1 −a(1 −2θ) <3(1 −a). Hence T. Aguilar-Hernández et al. / Journal of Functional Analysis 282 (2022) 109262 33 1 12a 1−a ˆ 0 θα'(1 −a(1 −2θ))−3−(1 −a/4)−3(dθ ≥1 24a 1−a ˆ 0 θα(1 −a(1 −2θ))−3dθ ≥1 24a 1 27 1 (1 −a)3 1−a ˆ 0 θαdθ (1 −a)α−2=(1−a)−1/p. Thus, for a >3/4, we have |P(f)(a)| (1 −a)−1/p and the function P(f)does not belong to RM(p, ∞). The remaining cases. For the remaining cases, we use the fact that if the Bergman projection P:Lq(T, Lp[0, 1]) →RM(p, q)is bounded then P:Lq(T, Lp[0, 1]) → RM(p, q)is bounded since, for f∈Lq(T, Lp[0, 1]), ρp,q(P(f)) ρp,q(rP(f))) sup g∈BLq(T,Lp[0,1]) &&&&&& ˆ D P(f)(w)g(w)dA(w)&&&&&& =sup g∈BLq(T,Lp[0,1]) &&&&&& ˆ D f(w)P(g)(w)dA(w)&&&&&& ≤ρp,q(f)sup g∈BLq(T,Lp[0,1]) ρp,q(P(g)) ≤Cρp,q(f), where Cis the norm of the operator P:Lq(T, Lp[0, 1]) →RM(p, q) and, as usual, BLq(T,Lp[0,1]) denotes the unit ball of Lq(T, Lp[0, 1]).  References [1] T. Aguilar-Hernández, M.D. Contreras, L. 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