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Kalton’s definition of K¨othe function space is not the classic one. A K¨othe N function space X on N is a linear subspace of C with a norm . X which makes it into a complete Banach space such that (for x CN X we makek k the convention ∈ \ that x X = ): k k ∞ N (1) BX is closed in C , where BX is the closed unit ball of X; N (2) For every x, y C , if y X and x y then x X and x X y X ; (3) There are h, k∈ CN strictly∈ positive| | ≤ such | | that for∈ every x k kCN ≤kwe havek ∈ ∈ xh 1 x X xk . k k ≤k k ≤k k∞ As a comparison, in the standard definition one asks that the characteristic function of every finite set is in X. Kalton’s definition implies that χ n X for every n N, since by (3) there is a positive element in X. { } ∈ Let us∈ consider the unit circle T together with normalized Haar measure λ. Definition 2.1 (Strongly admissible family). A strongly admissible family is a family = Xw w T of K¨othe function spaces on N such that H { } ∈ N (1) The map (w, x) x w from T C into [0, ] is Borel; (2) There are h, k 7→CN k strictlyk positive× such that∞ for every x CN and a. e. ∈ ∈ w T we have xh 1 x w xk ; ∈ k k ≤k k ≤kN k∞ (3) There is a linear subspace V of C of countable dimension such that V BXw is CN-dense in B for a. e. w T. ∩ Xw ∈ A family that satisfies only (1) and (2) is called admissible. We will be interested in finite families:
Definition 2.2. An admissible family = Xw w T is finite if there are n N,a family Xj :1 j n of K¨othe functionH spaces{ } and∈ a function f : T 1,∈ ..., n { ≤ ≤f(w}) 1 →{ } such that Xw = X for every w T and f − (j) is an arc for 1 j n. 1 ∈ j j ≤ ≤k If we denote Aj = f − (j), then we also write = X ; Aj . If X = X for 1 j < k n, we say that is a family of n spacesH . { } 6 ≤ ≤ H Let D denote the open unit disk.
+ Definition 2.3. Let = Xw w T be an admissible family. The space = +( ) is the space ofH functions{ }F∈: D CN such that F (.)(n): D C isN in the NSmirnovH class N + for every n N and → → ∈
F + = esssup F (w) Xw < k kN w T k k ∞ ∈ where F (w) is the radial limit of F at w in CN. For z D the K¨othe sequence ∈ space Xz is X = F (z): F + z { ∈ N } with the quotient norm. Let B : X + be a homogeneous function for which z → N there is C 1 such that B(x) + C x Xz and B(x)(z)= x for every x Xz. ≥ k N kN ≤ k k ∈ The derivation Ω : X C is defined as Ω (x)= B(x)′(z). z z → z In most cases, Kalton’s method of interpolation agrees with the classical complex method for families from [2] (see [1]). Kalton showed that the derivations induced by complex interpolation of K¨othe function spaces are centralizers (see the definition in the Introduction). Notice that the exact expression of Ωz depends on the choice of B, but another choice induces the same centralizer up to bounded equivalence: COMPLEX INTERPOLATION OF FAMILIES OF ORLICZ SEQUENCE SPACES 3
Definition 2.4. Let X be a K¨othe function space on N. Two centralizers Ω and Ψ on X are called equivalent if there is a linear map L : X CN such that Ω Ψ L : X X is bounded, in the sense that there is a constant→ C > 0 such that− − → Ω(x) Ψ(x) L(x) C x k − − kX ≤ k kX for every x X. If we may take L = 0 then Ω and Ψ are said boundedly equivalent, and we write∈ Ω Ψ. If there is λ C, λ = 0 such that Ω is equivalent to λΨ then Ω and Ψ are∼ said projectively equivalent∈ 6. If Ω is boundedly equivalent to λΨ, then Ω and Ψ are boundedly projectively equivalent. Ω is said to be trivial if it is equivalent to 0.
We shall use an alternative description of Xz that is akin to Lovanozkii factor- ization.
Definition 2.5. Let = Xw w T be an admissible family. The space = ( ) is the space of all functionsH { (up} to∈ almost everywhere equivalence) φ : NE TE HC such that × →
φ = esssup φ(., w) Xw < k kE k k ∞ 1 r2 Let P (r, t) denote the Poisson kernel, that is, P (r, t)= 1 2r cos(− t)+r2 . − Definition 2.6 ([7], Theorem 3.3). Let = Xw w T be an admissible family. H {N } ∈ Then Xz coincides with the space of all x C such that there is φ , φ 0, with ∈ ∈ E ≥ 1 π (2.1) x(n) = exp( P (r, θ t) log φ(n,eit)dt) | | 2π Z π − − for every n N. Furthermore, x Xz = inf φ , where the infimum is over all φ satisfying (2.1).∈ k k k kE We remark that [7, Theorem 3.3] is stated only for strongly admissible families, but the result is valid for admissible families in general. We shall also use an alternative description for the derivation Ωz that may be easily derived from the proof of [7, Theorem 3.3].
Definition 2.7. Let = Xw w T be an admissible family and z D. A map H { } ∈ ∈ Bz : Xz is a factorization map for at z if there is C > 0 such that for every x X and→Eλ C H ∈ z ∈ (1) B (x) 0; z ≥ (2) Bz(λx)= λ Bz(x) | | 1 π it (3) x(n) = exp( 2π π P (r, θ t) log Bz(n,e )dt); | | − − (4) Bz(x) C xR Xz . k kE ≤ k k Proposition 2.8. Let be an admissible family and Bz a factorization map for at z. Then H H π it x(n) e it Ωz(x)(n)= it 2 log Bz(x)(n,e )dt π Z π (e z) − − Definition 2.9. A centralizer Ω on a K¨othe function space X is real if Ω(f) is a real function for every real function f X. ∈ In [6, 7] Kalton establishes the following: 4 WILLIAN HANS GOES CORREAˆ
Theorem 2.10. (1) Let be a family of two K¨othe function spaces and z D. H ∈ Then Ωz is boundedly projectively equivalent to a real centralizer. (2) If Ω is a (real) centralizer on a superreflexive K¨othe function space X then there is a family of three (two) K¨othe function spaces such that X0 = X and Ω0 is boundedly projectively equivalent to Ω. N The K¨othe dual X′ of a K¨othe space X is defined as the space of all y C such that ∈ ∞ y X′ = sup y(n)x(n) < k k x BX | | ∞ ∈ nX=1 Then X′ is a subspace of X∗. We have the following duality result:
Theorem 2.11. [7, Lemma 3.2 and Corollary 4.8] Let = Xw w T be a strongly H { } ∈ admissible family. Then the family ′ = Yw w T where Yw = Xw′ is admissible H { } ∈ and Yz = Xz′ isometrically.
3. An abstract counterexample In this section we show that Kalton’s theorem 2.10 is optimal, in the sense that two spaces are not enough to induce a given centralizer. If X is a K¨othe function space and Ω is a centralizer on X then there are real centralizers Ω , Ω on X such that Ω Ω + iΩ [7, Lemma 7.1]. The following 1 2 ∼ 1 2 result characterizes when Ω1 + iΩ2 is a real centralizer. If f : N C, we let M : CN CN be given by M (x)(n)= f(n)x(n). → f → f Lemma 3.1. Let Ω Ω1 + iΩ2 be a centralizer on X with Ω1, Ω2 real centralizers. Then Ω is projectively∼ equivalent to a real centralizer on X if and only if either
(1) Ω1 or Ω2 is trivial; (2) Ω1 and Ω2 are projectively equivalent. Proof. Let Ψ be a real centralizer on X, and suppose that Ω (α + iβ)Ψ, with α, β R. That is, there are f ,f : N R such that ∼ ∈ 1 2 → Ω + iΩ (α + iβ)Ψ M iM 1 2 − − f1 − f2 is bounded (and complex homogeneous).
This happens if and only if Ω1 αΨ Mf1 = B1 and Ω2 βΨ Mf2 = B2 are − − 1 − − bounded. Now, if α =0 then Ω1 is trivial. Otherwise Ψ = α (Ω1 Mf1 B1), and substituting we get − − β β Ω Ω = B + M (M + B ) 2 − α 1 2 f2 − α f1 1 β that is, Ω2 is equivalent to α Ω1. If β = 0 then Ω2 is trivial. Otherwise Ω1 and Ω2 are projectively equivalent. The sufficiency is easy.
F´elix Cabello S´anchez presented the following result to us: Theorem 3.2. There is a centralizer which is projectively equivalent to a centralizer induced by complex interpolation of a family of three K¨othe function spaces but non projectively equivalent to one induced by complex interpolation of two K¨othe function spaces. COMPLEX INTERPOLATION OF FAMILIES OF ORLICZ SEQUENCE SPACES 5
Proof. Let Ω1 and Ω2 be two real centralizers on ℓ2, both nontrivial, which are not projectively equivalent. By Kalton’s Theorem 2.10 there is a family of three K¨othe function spaces which induces a centralizer projectively equivalent to Ω1 + iΩ2, which by Lemma 3.1 is not projectively equivalent to a centralizer induced by complex interpolation of a family of two K¨othe function spaces.
4. Complex interpolation of Orlicz spaces For general facts on Orlicz spaces we refer to [8, Chapter 4] and [9]. Recall that φ : [0, ) [0, ) is an Orlicz function if it is convex, non-decreasing, φ(0) = 0 ∞ → ∞ and limt φ(t)= . The function φ is said non-degenerate if φ(t) > 0 for t> 0, →∞ ∞ φ(2t) and satisfies the ∆2-condition at 0 if lim supt 0 φ(t) < . We deal only with → ∞ non-degenerate Orlicz functions, unless otherwise stated. Given an Orlicz function φ we have the Orlicz space
N x(n) ℓφ = x C : φ | | < for some ρ> 0 { ∈ X ρ ∞ } endowed with the complete norm x(n) x φ = inf ρ> 0 : φ | | 1 k k { X ρ ≤ } The ∆2-condition at 0 is equivalent to the separability of ℓφ. Of course, the most famous examples of Orlicz spaces are the ℓp spaces. φ(t) An N-function is a nondegenerate Orlicz function φ such that limt 0 t = 0 φ(t) → and limt t = . For a nondegenerate Orlicz function φ which is not an N- function→∞ℓ ℓ . If∞φ is an N-function satisfying the ∆ -condition at 0 then there φ ≃ 1 2 is an N-function φ∗ such that ℓ∗ ℓ ∗ , where the action of an element x∗ ℓ ∗ on φ ≃ φ ∈ φ x ℓ is given by ∞ x∗(n)x(n). In fact, ∈ φ n=1 P (4.1) x∗ ℓ ∗ x∗ ℓ∗ 2 x∗ ℓ ∗ k k φ ≤k k φ ≤ k k φ The function φ∗ is given by φ∗(y) = sup xy φ(x):0
Definition 4.1. A family φw w T of Orlicz functions will be called admissible if { } ∈ (1) For every t [0, ) the function w φw(t) is Borel; ∈ ∞ 7→ 1 (2) There is t0 > 0 such that for every t (0,t0) the function w log φw− (t) is integrable. ∈ 7→ (3) There are h, k ℓ strictly positive such that for every x ℓ and a. e. ∈ 0 ∈ 0 w T we have xh 1 x ℓ xk . ∈ k k ≤k k φw ≤k k∞ If, furthermore, the functions φw satisfy the ∆2-condition at 0, the family will be called strongly admissible.
Proposition 4.2. Let φw w T be a (strongly) admissible family of Orlicz func- { } ∈ tions. Then = Xℓw w T is a (strongly) admissible family. H { } ∈ 6 WILLIAN HANS GOES CORREAˆ
Proof. We must show that the map T : (w, x) x ℓφw is Borel. N 7→ k k x(n) Let ψ : T C [0, ] be given by ψ (w, x) = ∞ φ | | and a λ × → ∞ λ n=1 w λ ∈ 1 P [0, ]. We want to check that T − [0,a) is a Borel set. Let (λn) be an enumeration of Q∞ [0,a). We have ∩ 1 ∞ x(n) (w, x) T − [0,a) inf ρ : φ | | 1 ∞ 1 (w, x) ψ− [0, 1] ⇐⇒ ∈ λk k[=1 1 so we just need to check that ψ− [0, 1] is a Borel set for every k. λk N x(k) n n+1 Fix λ and let Ak,n,m = x C : | λ | [ 2m , 2m ) . This is a Borel set, since x(k) { ∈ ∈ N} the function x | | is continuous. Let ξ : T C [0, ] be given by 7→ λ k,n × → ∞ ∞ n ξ (w, x)= χ (x)φ k,m Ak,n,m w 2m nX=0 n The functions (w, x) χAk,n,m (x) and (w, x) φw m are Borel, so that 7→ 7→ 2 x(k) ξk,n is Borel. But ξk,m(w, x) converges in m to φw | | , so that this function is λ Borel. N x(n) This implies that the function ψλ ,N (w, x) = φw | | is Borel, and k n=1 λk therefore so is ψ , proving that (w, x) x is aP Borel function. λk 7→ k kw For strong admissibility, we may take V = c00. Indeed, take x BXw . Then the N-truncations xN V B and converge to x in CN, since x ∈c . ∈ ∩ Xw ∈ 0 Proposition 4.3. Condition (2) of Definition 4.1 is equivalent to the existence of Orlicz functions φ, ψ such that φ φw ψ on a neighborhood of 0 for a. e. w T. Here we allow φ to be degenerate.≤ ≤ ∈ Proof. Suppose φ and ψ are as above. Then for a. e. w T we have continuous inclusions ℓ ℓ ℓ , with uniform bound on their norms.∈ ψ ⊂ φw ⊂ φ Let y ℓ′ be strictly positive of norm at most 1. Then there is a> 0 such that ∈ φ y ℓ with norm at most a for a. e. w T. Take y as h. φ′ w a ∈ ∈ 1 Now take k strictly positive such that ψ(k(n)− ) 1 with k(n) big enough 1 1 ≤ so that φw(k(n)− ) ψ(k(n)− ) for everyPn and a. e. w T. Then for a. e. w T and every x CN we≤ have ∈ ∈ ∈ x(n) 1 φw | | φw(k(n)− ) xk ≤ X k k∞ X 1 ψ(k(n)− ) ≤ X 1 ≤ For the converse, suppose (2). Taking x = e we have for a. e. w T 1 ∈ 1 1 h(1) inf ρ> 0 : φw 1 = 1 k(1) ≤ { ρ ≤ } φw− (1) ≤ COMPLEX INTERPOLATION OF FAMILIES OF ORLICZ SEQUENCE SPACES 7 1 1 1 So k(1)− φw− (1) h(1)− . Now we may take as φ a degenerate function which ≤ 1 ≤ is 0 on [0,h(1)− ], and ψ(t)= k(1)t. iθ Given an admissible family of Orlicz functions φw w T and z = re D we denote { } ∈ ∈ π 1 1 Iz( φw )(t) = exp P (r, θ t) log φe−it (t)dt { } 2π Z π − − Proposition 4.4. Let φw w T be an admissible family of Orlicz functions. Then for every z D { } ∈ ∈ (1) Iz( φw )(0) = 0; { } 1 (2) If t0 > 0 is such that w log φw− (t) is integrable for every t (0,t0), then I ( φ ) is concave and7→ strictly increasing on [0,t) for some∈t Now let fn (respectively, gn) be a monotone sequence of simple positive functions 1 1 such that fn(w) φw− (t0) (respectively, gn(w) φw− (t1)) for a. e. w T. Then monotone convergence→ gives → ∈ (Iz( φw )(αt0 + (1 α)t1)) { } 2π − 1 1 = exp log φw− (αt0 + (1 α)t1)Pz(w)dw 2π Z0 − 2π 1 1 1 exp log(αφw− (t0)+(1 α)φw− (t1))Pz (w)dw ≥ 2π Z0 − 1 2π = lim exp log(αfn(w)+(1 α)gn(w))Pz (w)dw n 2π Z0 − 2π 2π lim α exp log f(w)Pz (w)dw + (1 α)exp log g(w)Pz(w)dw ≥ n Z0 − Z0 2π 2π 1 1 = α exp log φw− (t0)Pz(w)dw + (1 α)exp log φw− (t1)Pz(w)dw Z0 − Z0 = α(I ( φ )(t )+(1 α)(I ( φ )(t ) z { w} 0 − z { w} 1 8 WILLIAN HANS GOES CORREAˆ Since I ( φ ) is concave and nonnegative, the only way for (2) to not be satisfied z { w} is if there is 0 < t′ < t0 such that Iz( φw )(t) = 0 for 0 t Now let x Xz of norm 1. Consider the family ∗ = Yw w T = ℓψw w T, ∈ H { } ∈ { } ∈ where ψ = φ∗ . By Lemma 4.6 the family ψ is admissible and by the previous w w { w} calculation, ℓψz Yz with continuous inclusion. The proof now⊂ follows [9, Lemma 6.3.3]. By [9, Proposition 2.1.1(ii)] we have 1 1 1 1 t < φ− (t)(φ∗)− (t) < 2t and t < φ− (t)ψ− (t) < 2t for every t> 0, w T. So if z z w w ∈ t′ is a witness for both φ and ψ then for t If t is such that ψ (t) Now [9, Proposition 3.3.4] tells us that we have a continuous inclusion Xz ℓφz , and that ⊂ 1 a x(n) Bz(x)(n, w)= a x ℓφz φw− (φz | | ), x ℓφz k k x ℓφ ∈ k k z π eit is a factorization map. Since π (eit z)2 dt = 0, we obtain the formula for Ωz. R− − Of course, the previous theorem generalizes the classical results regarding com- plex interpolation of couples of Orlicz spaces [4], stating that (ℓφ0 ,ℓφ1 )θ = ℓφ, with 1 1 1 θ 1 θ φ− = (φ0− ) − (φ1− ) See [3] for a generalization to complex interpolation of couples of Orlicz spaces with respect to a vector measure. 5. The example We have the following characterization of the interpolation spaces Xz for finite families: 10 WILLIAN HANS GOES CORREAˆ Theorem 5.1. [1, Proposition 3.11] Let = Xj; A n be a finite admissible H { j }j=1 family. Let µz denote harmonic measure on T with respect to z D. Then Xz coincides with the space ∈ n n (5.1) (Xj)µz (Aj ) = x ℓ : x = x µz(Aj ) ,f Xj { ∈ 0 | | | j | j ∈ } jY=1 jY=1 and x = inf n x µz(Aj ) , where the infimum is over all x as in (5.1). k kXz { j=1 k j kXj } j Q In the proof of Theorem 4.7 we asked that the functions φw be N-functions satisfying the ∆2-condition at 0 so that we may use duality and the invertibility of φw∗ . The next theorem shows that when we are dealing with finite families we do not need these tools. Its proof should be compared with that of [3, Lemma 4.1]. Theorem 5.2. Let = ℓ ; A n be an admissible family such that the func- H { φj j }j=1 tions φj are non-degenerate. Then Xz = ℓφz with equivalence of norms for every z D, where ∈ n 1 1 µz (Aj ) φz− = (φj− ) jY=1 The induced centralizer is n 1 x(n) Ωz(x)= x ψj′ (z) log φj− φz | | x ℓφ Xj=1 k k z for x ℓφz , where ψj is an analytic function on D which real part agrees with χAj on T.∈ Proof. We begin by noticing that the definition of φz given above coincides with that of Definition 4.5 when the witness is + . If we mimic now the first part of ∞ the proof of Theorem 4.7 for a = 1 we obtain ℓφz Xz and . Xz . ℓφz . For the other inclusion, let x X of norm 1. By⊂ Theoremk k 5.1,≤k we mayk write ∈ z n x = x µz (Aj ) | | | j | jY=1 N xj (n) with x [1, 2]. Let h C be given by h (n) = φ | | , and let j ℓφj j j j 2 kn k ∈ ∈ h = j=1 hj. Then P x x µz(Aj ) | | = | j | 2 Y 2 1 µz (Aj ) = φj− (hj ) Y 1 µ (A ) φ− (h) z j ≤ j Y1 = φz− (h) so that x(k) x (k) φ | | h(k)= φ | j | n +1 z 2 ≤ j 2 ≤ Xk Xk Xk,j That is, x ℓφz , and therefore Xz = ℓφz as sets. By the first part of the proof and the open mapping∈ theorem, the norms are equivalent. We obtain a factorization map and Ωz as in Theorem 4.7. COMPLEX INTERPOLATION OF FAMILIES OF ORLICZ SEQUENCE SPACES 11 We now present the promised example: it 2π it 2π 4π Theorem 5.3. Let A0 = e : 0 t < 3 , A1 = e : 3 t < 3 and it 4π { ≤ } { ≤ } A2 = e : t< 2π . Let = Xw w T, with Xw = ℓφj for w Aj , where { 3 ≤ } H { } ∈ ∈ φ0(t)= t 1 2 2 4 φ (t)=5− t log(t) | | 2 2 2+2√1 t 2 φ (t)=5− e− − (2t +2√1 t 2) − − on a neighborhood of 0. Then X0 = ℓ2 with equivalence of norms, and the induced centralizer Ω0 is not projectively equivalent to any centralizer obtained by complex interpolation from two K¨othe function spaces. We may check that φ1 and φ2 are Orlicz functions on a neighborhood of 0 by differentiation. One may also check that both satisfy the ∆2 condition at 0. We leave the bothersome details to the reader. Recall that two Orlicz functions φ and ψ satisfying the ∆2-condition are equiv- alent, written φ ψ, if there are K > 0, t > 0 such that for every 0 t t ∼ 0 ≤ ≤ 0 1 φ(t) (5.2) K− K ≤ ψ(t) ≤ Equivalently, 1 1 1 φ− (K− ψ(t)) t φ− (Kψ(t)) ≤ ≤ for 0 t t , which happens if and only if ℓ = ℓ with equivalence of norms. ≤ ≤ 0 φ ψ Proposition 5.4. Under the conditions of Theorem 5.3, X0 = ℓ2 with equivalence of norms. Proof. X0 = ℓ2 with equivalence of norms if and only if there are K > 0 and t0 > 0 such that for every 0 1 2 1 1 1 2 1 2 1 1 2 1 K− 3 t 3 ((φ )− (K− t )) 3 ((φ )− (K− t )) 3 t ≤ 1 2 1 1 2 1 2 1 2 1 K− 3 t 3 ((φ )− (Kt )) 3 ((φ )− (Kt )) 3 ≤ that is 1 1 1 2 1 2 1 1 2 1 1 1 1 1 2 1 2 1 2 1 ((φ )− (K− t )) 3 ((φ )− (K− t )) 3 K 3 t 3 ((φ )− (Kt )) 3 ((φ )− (Kt )) 3 ≤ ≤ so 1 1 1 2 1 2 1 1 2 1 1 1 1 1 2 1 2 1 2 1 ((φ )− (K− t )) 2 ((φ )− (K− t )) 2 K 2 t 2 ((φ )− (Kt )) 2 ((φ )− (Kt )) 2 ≤ ≤ 1 1 Taking u = K 2 t 2 1 1 2 4 1 2 1 2 4 1 1 1 4 1 2 1 4 1 (5.3) ((φ )− (K− u )) 2 ((φ )− (K− u )) 2 u ((φ )− (u )) 2 ((φ )− (u )) 2 ≤ ≤ 12 WILLIAN HANS GOES CORREAˆ We must show that (5.3) is satisfied. Take ψ1(t)= t log(t) 2. From now on we will use the notation for the inverse of a function when the| function| is injective on 2 1 1 a neighborhood of 0. Define ψ = (ψ )∗ as if ψ was an Orlicz function. So ψ2(s) = max(ts ψ1(t)) = max t(s log(t)2) t 0 t 0 ≥ − ≥ − 2 2 Let fs(t) = t(s log(t) ). Then fs′(t)= s log(t) 2 log(t), which is 0 if and 1 √1−s − − only if t = e− ± − for 0 s< 1. Then ≤ 2 1+√1 s 2 ψ (s)= e− − (s ( 1+ √1 s) ) − − − for 0 s < 1. This function is increasing on a neighborhood of 0. By definition ts ψ≤1(t)+ ψ2(s), so ≤ 1 1 2 1 (ψ )− (t)(ψ )− (t) 2t ≤ i i 2 i 1 i 1 1 1 1 2 1 1 Let ξ = (ψ ) . Then (ξ )− (t) = (ψ )− (t 2 ) and (ξ )− (t)(ξ )− (t) 2t 2 . It 1 1 4 1 2 1 4 1 u ≤ follows that (ξ )− (t ) 2 (ξ )− (t ) 2 √2t. Write t = , so that ≤ √10 4 4 1 1 u 1 2 1 u 1 u (ξ )− ( ) 2 (ξ )− ( ) 2 100 100 ≤ √5 1 2 ψ (t) 1 For t near 1 from below, the function f(t)= ψ ( 5t ) ψ (t) is negative. This 1 u 1 1 2 1 − implies, taking ψ (t) = u, that 5 (ψ )− (u)(ψ )− (u) for u small enough. In u 1 1 4 1 2 1 ≤4 1 particular (ξ )− (u ) 2 (ξ )− (u ) 2 . So √5 ≤ 4 4 1 1 u 1 2 1 u 1 u 1 1 4 1 2 1 4 1 (ξ )− ( ) 2 (ξ )− ( ) 2 (ξ )− (u ) 2 (ξ )− (u ) 2 100 100 ≤ √5 ≤ Noticing that φi =5 2ξi and writing u = t, we obtain − √5 4 4 1 1 t 1 2 1 t 1 1 1 4 1 2 1 4 1 (5.4) (φ )− ( ) 2 (φ )− ( ) 2 t (φ )− (t ) 2 (φ )− (t ) 2 100 100 ≤ ≤ Let Ω0 Ω1 + iΩ2 with Ω1 and Ω2 real. Kalton shows in [7, Lemma 7.1] that we may take∼ (5.5) Ω1(x)= Re(Ω0(Re(x))) + iRe(Ω0(Im(x))) and (5.6) Ω2(x)= Im(Ω0(Re(x))) + iIm(Ω0(Im(x))) Lemma 5.5. For n N let s ℓ be the unit vectors given by s = 1 n e . ∈ n ∈ 2 n √n j=1 j Under the conditions of Theorem 5.3, if we write Ω0 Ω1 + iΩ2 withPΩ1 and √3 ∼ 3 Ω2 as in (5.5) and (5.6) then Ω1(sn) = 2π Ansn and Ω2(sn) = 2π Bnsn, where − − − (φ0) 1(1/√n)(φ2) 1(1/√n) (φ2) 1(1/√n) An = log 1 −1 2 and Bn = log 0 −1 . ((φ ) (1/√n)) (φ ) (1/√n) Proof. By the proofs of Theorems 4.7 and 5.2, we may take a factorization map B0 1 1 satisfying B (s )(j, w)= φ− (φ ( )) for 1 j n, and 0 otherwise. 0 n w z √n ≤ ≤ COMPLEX INTERPOLATION OF FAMILIES OF ORLICZ SEQUENCE SPACES 13 Then, for 1 j n, ≤ ≤ π sn(j) 1 it Ω0(sn)(j) = it log B0(sn)(j, e )dt π Z π e − 1 0 1 1 1 1 1 = ((√3 3i) log(φ )− ( ) 2√3 log(φ )− ( ) 2π√n − √n − √n 2 1 1 +(√3+3i) log(φ )− ( )) √n 0 1 1 2 1 1 2 1 1 1 (φ )− ( )(φ )− ( ) (φ )− ( ) = (√3 log √n √n +3i log √n ) 2π√n 1 1 1 2 0 1 1 ((φ )− ( √n )) (φ )− ( √n ) and 0 otherwise. The result now follows from (5.5) and (5.6). Since X0 = ℓ2 with equivalence of norms, there are k,K > 0 such that k 1 K φ ( ) n ≤ 0 √n ≤ n Lemma 5.6. (1) If Ω1 is trivial then there is C > 0 such that An C for every n. | | ≤ (2) If there is γ such that Ω1 is equivalent to γΩ2 then there is C > 0 such that γA B C for every n. | n − n|≤ Proof. (1) If Ω is trivial there is a sequence f CN such that Ω M is bounded. 1 ∈ 1 − f For a given Orlicz function space ℓφ the norm of en is independent of n. This implies that we may choose B in such a way that Ω (e ) = ae for some a C 0 0 n n ∈ and every n. Since Ω1 Mf is bounded, f ℓ . So, for every n, − ∈ ∞ Ω (s ) Ω (s ) f(s ) + f(s ) C s k 0 n k2 ≤k 0 n − n k2 k n k2 ≤ k nk2 The result follows from Lemma 5.5. (2) is proved similarly. Theorem 5.3 follows from Lemma 3.1, Lemma 5.6 and the next two results. Lemma 5.7. The sequence ( A ) of Lemma 5.5 is not bounded. | n| n 0 −1 4 2 −1 4 4 1 (φ ) (un)(φ ) (un) Proof. Let un = . We have An = log 1 −1 4 2 . By (5.4) √n ((φ ) (un)) 4 2 unun An log 1 1 4 3 ≥ ((φ )− (un)) The right side is bounded from above if and only if there is C > 0 such that 6 un C 1 −1 4 3 e for n big enough, that is ((φ ) (un)) ≤ 2 1 un 4 φ ( C ) un e 3 ≤ which cannot happen. Lemma 5.8. Let An and Bn be as in Lemma 5.5. For every γ R the sequence ( γA B ) is not bounded. ∈ | n − n| n 14 WILLIAN HANS GOES CORREAˆ Proof. We must consider a number of cases. Case 1: γ =1 2 1 rn Let rn = . Then An Bn = log 1 −1 2 . If An Bn is bounded √n − ((φ ) (rn)) | − | 1 rn there is C such that for n big enough φ ( C ) rn. For n big enough this is the same as ≥ 4 2 rn rn 5− log 1 C2 C ≥ Case 2: γ 1 > 0 4 − un 1 Let 100 = √n . Then by (5.4) 4(γ+1) 2(γ 1) un un − γAn Bn log 4 − ≤ γ+1 1 1 un 3γ 1 100 ((φ )− ( 100 )) − The right side is bounded from below if and only if for some C > 0 we have 2(3γ+1) 4 3γ−1 un φ1 un . For n big enough this is the same as 100 ≤ C −8 2(3γ+1) 4 u 3γ−1 1 u 3γ−1 n 5 2 log n − 2 100 ≤ C C which cannot happen for n big enough. The case γ < 1 must be divided in a number of subcases. Since they are all very similar, we will do one of them as illustration. Case 3: γ 1 < 0 and 3γ 1 > 0 4 −1 − Let un = √n . By (5.4) 4(γ+1) 2(γ 1) un un − γAn Bn log 1 1 4 3γ 1 − ≤ ((φ )− (un)) − Let us show that the sequence on the right has as limit. Were it bounded from −∞ 2(3γ+1) 3γ−1 4 1 un below we should have for some C > 0 that un φ ( C ). For n big enough this is ≤ 2(3γ+1) 4 −8 2 3γ−1 − 5 u u 3γ 1 − log n n 2 ≤ C C which cannot happen when n grows. 1 The other cases one should check are 4. γ = 3 ; 5. 3γ 1 < 0 and 3γ +1 > 0; 6. 1 1 − γ = 3 ; and 7. γ < 3 . − − 6. Acknowledgements We would like to thank F´elix Cabello S´anchez, Jes´us Castillo and Valentin Fer- enczi for discussions regarding this work, and MichalWojciechow ski for suggesting the study of interpolation of families of Orlicz spaces. COMPLEX INTERPOLATION OF FAMILIES OF ORLICZ SEQUENCE SPACES 15 References 1. J. M. F. Castillo, W. H. G. Corrˆea, V. Ferenczi, and M. Gonz´alez, On the stability of the differential process generated by complex interpolation, arXiv e-prints (2019), arXiv:1712.09647. 2. R. R. Coifman, M. Cwikel, R. 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Tzafriri, Classical Banach Spaces I: Sequence Spaces, Ergebnisse der Mathematik und ihrer Grenzgebiete. 2. Folge, Springer Berlin Heidelberg, 2013. 9. M. M. Rao and Z. D. Ren, Theory of Orlicz Spaces, New York: M. Dekker, 1991. Departamento de Matematica,´ Instituto de Matematica´ e Estat´ıstica, Universidade de Sao˜ Paulo, Rua do Matao˜ 1010, 05508-090 Sao˜ Paulo SP, Brazil E-mail address: [email protected]