Positivity https://doi.org/10.1007/s11117-018-0563-8

Positivity

The minimal and maximal operator ideals associated to (n + 1)-tensor norms of Michor’s type J. A. López Molina1

Received: 5 December 2017 / Accepted: 1 February 2018 © Springer International Publishing AG, part of Springer Nature 2018

Abstract We study an (n + 1)-tensor norm αrC extending to (n + 1)-fold tensor products a tensor norm defined by Michor when n = 1 by convexification of a certain s-norm. We characterize the maps of the minimal and the maximal multilinear operator ideals related to αrC in the sense of Defant, Floret and Hunfeld. Keywords (n + 1)-fold tensor products · αrC -nuclear and αrC -integral n-linear operators · Ultraproducts Mathematics Subject Classification Primary 46M05 · 46A32

1 Introduction The systematic study of the relation between (n + 1)-fold tensor norms and ideals of multilinear operators was initiated with the works [6,7] of Floret, giving an important impulse to the development of the program proposed by Pietsch in [19]. Since then, several concrete examples of multi-tensor norms has been appeared in the literature to check and enhance the general theory. In [12,13] we have studied two (n + 1)-tensor norms generalizing those of Saphar and Lapresté for the case n = 1. The multiple tensor norms in [13] are strongly dependent of equalities of type 1 1 1 1 1 := + +  + ··· +  s r0 r1 r2 rn+1

B 1

J. A. López Molina [email protected] E. T. S. Ingeniería Agronómica y del Medio Natural, Camino de Vera, 46072 Valencia, Spain

J. A. López Molina

with s = 1. The purpose of this paper is to study the situation in the case s < 1. This question was briefly considered many years ago by Michor in [18] when n = 1 and this is the motivation for the name we have assigned to the (n + 1)-tensor norms we shall deal with. Michor’s work does not deals with the characterization of operator ideals naturally associated to these tensor products from the point of view of the general theory of multilinear operator ideals. So, this will be our main goal. The paper is organized as follows. In this section we introduce  the notation to be used. In Sect. 2 we introduce the (n +1)-fold tensor product ⊗αrC E 1 , E 2 , . . . , E n , F a convexification procedure of Banach spaces E j , 1 ≤ j ≤ n, and F by means of as well as the multilinear αrC -nuclear operators from nj=1 E j into F, which are characterized by means of suitable factorizations in Theorem 3. The next step is the characterization of the so called αrC -integral multilinear maps, i.e. the maps in the ideal associated in the sense of Defant, Floret and Hunfeld to the αrC -tensor norm. To do this we need a technical preparation concerning to concrete factorizations of ultraproducts of some special αrC -nuclear maps. This goal is accomplished in Sect. 3. Finally, in Sect. 4 the αrC -integral operators are characterized. Here important differences with the case of Lapresté type (n + 1)-tensor norms arise (Theorem 23). Notation of the paper is standard in general. We shall deal always with vector spaces defined over the field R of real numbers. ek (resp. ekh ) will denote the sequence (αm ) (resp. the double sequence ((αmv ))) such that αk = 1 and αm = 0 if m = k (resp. αkh = 1 and αmv = 0 if m = k or v = h). Given a Banach space E and k ∈ N ∪ {∞}, for every {xm }km=1 ⊂ E and every p ∈ [1, ∞[, we      1 k   p p . We will write  p (E) :=  define ε p (xm )km=1 := sup x  E  ≤1 m=1 x m , x     N  ε (x )∞ p q (x j )∞ p j j=1 < ∞ . I p,q will denote the inclusion map  ⊂  j=1 ∈ E for every 0 < p ≤ q < ∞. All maps defined by continuous extension of a map T previously defined and continuous on a dense subset will be denoted again by T. In the same way, given a measure space (, M, μ) and A ∈ M, the σ -algebra induced on A by M will be denoted  by M  A while the induced measure still will be designed by μ. Symbols such as E or xi i∈A will denote the linear span of a subset E or  xi , i ∈ A in a larger linear space X. Concerning Banach lattices we refer the reader to [2]. If Y is a subset of a lattice X we put Y ⊥ = {x ∈ X | |x| ∧ |y| = 0 ∀ y ∈ Y }. If Y = {y} is a singleton we simply write y ⊥ instead of {y}⊥ . C(u) will be the set of all components of an element u ≥ 0 in X and B(u) will denote the band generated by u in X.   If F is a quasi-Banach space we shall denote by Ln nj=1 E j , F the space of all  n−linear continuous maps from the product nj=1 E j of normed spaces E j , 1 ≤ j ≤ n, into F. Given A j ∈ L(E j , F j ) between  normed spaces  E j and F j , 1 ≤ j ≤ n, we write (A j )nj=1 := (A1 , A2 , . . . , An ) : nj=1 E j −→ nj=1 F j to denote the continu   ous linear map sending every (x j )nj=1 ∈ nj=1 E j to A1 (x1 ), A2 (x2 ), . . . , An (xn ) ∈ n j=1 F j . As customary, for p ∈ [1, ∞], p  will be the conjugate extended real number such that 1/ p + 1/ p  = 1. Given n ≥ 1, in all the paper we denote by r an (n + 2)-tuple of real numbers r = (r0 , r1 , r2 , . . . , rn , rn+1 ) such that 1 ≤ r0 ≤ ∞, 1 < r j < ∞, 1 ≤ j ≤ n + 1, and

The minimal and maximal operator ideals associated to (n + 1)…

1≤

1 1 1 1 1 := +  +  + ··· +  . sr r0 r1 r2 rn+1

(1)

Moreover, we define tr and w such that

1 1 1 1 1 1 := −  = + ≥ tr sr rn+1 r0 r j rn+1 n

1 1 1 1 :=  +  + · · · +  . (2) w r1 r2 rn

and

j=1

It is important to note that 1 1 + , w rj n

n=

r 0 > tr ,

tr ≤ rn+1 .

and

(3)

j=1

Associated with r, given a measure spaces (i , Ai , μi ), i = 1, 2 and g(t, x) ∈   1 L [1 , L r0 (2 )] we have a canonical n-linear map Sg : nj=1 L ∞ [1 , L r j (2 )] −→    L 1 [1 , L tr (2 )] defined by Sg ( f j (t, x))nj=1 (t, x) = g(t, x) nj=1 f j (t, x) verify   

ing  Sg  ≤ g  L 1 [ ,L r0 ( )] because of the generalized Hölder’s Inequality and (1), 1 2 gives us    Sg ( f j (t, x))n

j=1



≤

1

  g(t, x)

  ≤ g 

 

L 1 [1 ,L tr (2 )] n 

   f j (t, x)

L r0 (2 ) j=1

L 1 [1 ,L r0 (2 )]

n     f j j=1

r

dμ1 (t)

r

.

L j (2 )

L ∞ [1 ,L j (2 )]

(4)

Moreover we have the next general and relevant fact: Lemma 1 If (, A, ν) and (, M, μ) are measure spaces, g ∈ L 1 [, L r0 ()] and M ∈ L L 1 [, L tr ()], L 1 [, L rn+1 ()] satisfies Supp(M( f )) ⊂ Supp( f ) for every f ∈ L 1 [, L tr ()], the map M ◦ Sg can be factorized through an isometric composition map with the same properties but using finite measure spaces (, A, ν) and (, M, μ). Proof Consider the new measures ν and μ defined on (, A) and (, M) respectively setting for A ∈ A and M ∈ M  ν(A) =

  g(t, x)

A

μ(M) = M

L r0 (,μ)

  g(t, x)r01

L (,ν)

dν(t), dμ(x).

J. A. López Molina

Clearly ν() < ∞ and by the Integral Minkowski’s Inequality [4, p. 499]   μ() =





 ≤



  g(t, x) dν(t)

  g(t, x)

L r0 (,μ)

r0 dμ(x) r0 dν(t) < ∞.

It is straightforward to check (by Radon–Nikodým Theorem) that the linear maps  H j , 1 ≤ j ≤ n, sending each f j ∈ L ∞ [, ν, L r j (, μ)] to h j ( f j ) := r0 − r j     f j (t, x) g(t, x) L 1 (,ν) ∈ L ∞ [, ν, L r j (, , μ)] and L j , j = tr , rn+1 , sending − rj0    g(t, x)−1r f j ∈ L 1 [, ν, L j (, μ)] to L j ( f j ) := f j (t, x) g(t, x) L 1 (,ν) ∈ L 0 (,μ) −1  1 j L [, ν, L (, μ)] are well defined onto isometries. Let g(t, x) := g(t, x) L r0 (,μ)       1 , ν, L r0 (, μ) , g  g(t, x)−11 g(t, x). Then g ∈ L = L (,ν) L 1 [,ν,L r0 (, μ)]   −1 g  1 and taking M := L ◦ M ◦ L by (1) we obtain r r t L [,ν,L 0 (, μ)]

∀ ( f j (t, x))nj=1 ∈

n+1

n 

r



L ∞ [, ν, L r j (, μ)] L r−1 ◦ M ◦ Sg ◦ (H j ( f j ))nj=1 n+1

j=1

 r0  r0 n g(t, x) tr1  − r j  L (,ν)    =g  f j g(t, x) L 1 (,ν) g(t, x) 1 L (,ν) j=1 =g

n 

f j = Sg



f j )nj=1 ,

j=1

 

giving the desired factorization with spaces of finite measure. n+1



L ∞ [, L r j  n+1 ()] associated to T := M◦Sg is continuous. Consider for every fixed z j (t, x) j=1 ∈ n+1 ∞ r j j=1 L [, L ()] and each 1 ≤ j ≤ n + 1, the linear form ϕ j sending each   z(t, x) ∈ L ∞ [, L r j ()] to T z 1 (t, x), . . . , z j−1 (t, x), z(t, x), z j+1 (t, x), . . . ,     z n+1 (t, x) . Clearly ϕ j ∈ (L ∞ [, L r j ()]) and hence T is σ L ∞ [, L r j ()], (L ∞    r j ∞ [, L r j ()]) -separately continuous, 1 ≤ j ≤ n + 1, on n+1 j=1 L [, L ()]. It follows from (4) that the canonical (n + 1)-linear form T on

j=1

2 αrC -tensor products and αrC -nuclear maps We assume the readier is familiar with the main properties of (n + 1)-tensor norms, n ≥ 1, and ideals of n-linear operators in the way they are developed in the pioneer

The minimal and maximal operator ideals associated to (n + 1)…

 works [6,7]. If it is needed to emphasize α z; ⊗n+1 j=1 M j or similar notations will denote the value of the (n + 1)-tensor norm α in z ∈ ⊗n+1 j=1 M j . Let E j , 1 ≤ j ≤ n + 1, be normed spaces. Given an (n + 2)-tuple r we define   n+1  j h    h   αr z; ⊗n+1 j=1 εr j (x k )k=1 , taking the inf over all reprej=1 E j := inf (λk )k=1 r0 h j j sentations of z of type z = k=1 λk ⊗n+1 xk with xk ∈ E j , 1 ≤ j ≤ n + 1, 1 ≤  j=1 n+1  E j [18, k ≤ h, h ∈ N. If sr > 1 one has αr z; ⊗ j=1 E j = 0 for every z ∈ ⊗n+1  j=1 n+1  Remark 7.8 (a)]. If sr ≤ 1, using standard methods it can be shown that αr z; ⊗ j=1 E j is an sr -norm in ⊗n+1 j=1 E j . If sr = 1, αr is the (n + 1)-tensor norm of Lapresté studied < 1 the Minkowski’s functional of the absolutely convex cover of the set in [13]. If s r  E | α (z) ≤ 1 is a norm αrC on ⊗n+1 z ∈ ⊗n+1 j r j=1 j=1 E j which can be computed as ∀ z ∈ ⊗n+1 j=1 E j

αrC (z) = inf

s

 (λmk )h

 

n+1 

k=1 r0

m=1

 j h  εr j xmk k=1

j=1

  h taking the infimum over all representations of z of type z = sm=1 k=1 λmk ⊗n+1 j=1 j  j xmk where xmk ∈ E j , 1 ≤ j ≤ n + 1, 1 ≤ k ≤ h, 1 ≤ m ≤ s, h ∈ N, s ∈ N. This     C normed tensor product will be denoted by ⊗αrC E 1 , E 2 , . . . , E n+1 or ⊗n+1 j=1 E j , αr . It can be proved that αrC is a (n + 1)-tensor norm in the class of all normed spaces. The minimal operator ideal (of multi-linear maps) associated to the (n + 1)-tensor norm αrC in the sense of Defant, Floret and Hunfeld is the ideal of so called αrC -nuclear n-linear maps. In order to reduce the length of the paper as far as possible, we give a direct definition suitable for our purposes.   Definition 2 Given Banach spaces E j , 1 ≤ j ≤ n+1, a map T ∈ Ln nj=1 E j , E n+1 is said to be αrC -nuclear if for every m ∈ N there are sequences (ϕmk )∞ k=1 ∈ j

r j



rn+1 r0  (E j ), 1 ≤ j ≤ n, (z mk )∞ (E n+1 ) and (λmk )∞ k=1 ∈  k=1 ∈  such that ∞

  (λmk )∞ 

k=1 r0

m=1

n    j ∞   (z mk )∞ εrn+1 εri ϕmk k=1 < ∞ k=1

(5)

i=1

  n  j   n ∞ ∞ and T (x j )nj=1 = m=1 k=1 λmk j=1 ϕmk , x j z mk for each x j j=1 ∈ n j=1 E j .   It can be shown that the set NαrC nj=1 E j , E n+1 of all n-linear αrC -nuclear maps n from j=1 E j into E n+1 becomes a Banach space under the αrC -nuclear norm NαrC (T )  j ∞ ∞ defined as the infimum of the numbers (5) over all sequences ϕmk k=1 m=1 , 1 ≤ ∞    ∞ ∞ j ≤ n, (z mk )∞ k=1 m=1 and (λmk )k=1 m=1 satisfying the conditions of Definition 2. αrC -nuclear maps can be characterized by means of suitable factorizations as follows.

J. A. López Molina

Theorem 3 Let F and E i , 1 ≤ i ≤ n, be Banach spaces. A map T ∈ Ln

n i=1



Ei , F



is αrC -nuclear if and only if there are maps A j ∈ L(E j , ∞ [r j ]), 1 ≤ j ≤ n, ∞  1 r0 C ∈ L(1 [rn+1 ], F) and λ := (λkm )∞ m=1 k=1 ∈  [ ] such that T factorizes in the way n j=1

T

Ej

- F 6

(A j )nj=1

C ?

n

 ∞ rj j=1  [ ]

- 1 [tr ] Sλ

Moreover, NαrC (T ) = inf torizations of type (6).

n j=1

I1,1 ⊗ Itr ,rn+1

- 1 [rn+1 ]. (6)

       A j   Sλ  C  taking the infimum over all fac-

   ∞  ∞ Proof (a) Given ε > 0, assume that T x = λkm ⊗nj=1 ϕmk j , x ymk for k=1 m=1   every x := (x j )nj=1 ∈ nj=1 E j in such a way that δ := ∞ m=1 δm ≤ N C (T ) + ε,     n  j ∞  αr ∞ ∞  where δm := πr0 (λmk )k=1 εrn+1 (ymk )k=1 m ∈ N. j=1 εr j ϕmk k=1 , Taking j

j ϕ mk

=

ϕmk

δm  λmk ,  j ∞  , λmk :=    (λmk )∞ εr j (ϕmk )k=1 k=1 r0

y mk :=

ε

 rn+1

ymk   (ymk )∞ k=1

for every 1 ≤ j ≤ n, m ∈ N and k ∈ N we obtain the new representation T = ∞ ∞ j n m=1 k=1 λmk ⊗ j=1 ϕ mk ⊗ y mk which allows us the desired factorization  j ∞ taking A j (x j ) = (ϕ mk , x j )∞ k=1 m=1 for every x j ∈ E j , 1 ≤ j ≤ n, and defining  ∞ 1 rn+1 ∀ (βmk )∞ ] k=1 m=1 ∈  [

C



(βmk )∞ k=1

∞

m=1

=

∞ ∞

βmk y mk .

m=1 k=1

       We find that  A j  = 1 for each 1 ≤ j ≤ n and  Sλ  ≤ λ1 [r0 ] ≤ ∞ m=1 δm . 1 r n+1 Let B be the closed unit ball in  [ ]. By Hölder’s inequality   C  ≤ =

sup

∞

  (βmk )∞ 

sup

∞

  (βmk )∞ 

(βmk )∈B m=1

(βmk )∈B m=1

 k=1 rn+1 εrn+1

k=1 rn+1

  (y mk )∞ k=1

= 1.

The minimal and maximal operator ideals associated to (n + 1)…

n      ∞    S  C  ≤ Then j=1 A j m=1 δm ≤ NαrC (T ) + ε. λ (b) Conversely, let T factorizes as in (6). We define ymk = C(em ⊗ ek ) ∈ E n+1 and ∞ ∞  j j ϕmk (x) := A j (x), em ⊗ ek k=1 m=1 , x ∈ E j , 1 ≤ j ≤ n. Then ϕmk ∈ E j  j ∞ ∞ and, putting A j (x) = βmk (x) k=1 m=1 we obtain  j ∞  εr j ϕmk k=1 =

   sup (βmk (x))∞ k=1

r j



x E j ≤1

≤

  sup  A j (x)

x E j ≤1

r ∞ [ j ]

  ≤  A j .



Analogously, given y  ∈ F  we have C  (y  ) ∈ ∞ [rn+1 ]. Then, given m ∈ N  ∞   (ymk )∞ εrn+1 k=1 m=1 =

 ∞  sup  (em ⊗ ek ), C  (y  ) k=1 

y  F  ≤1

 rn+1



  ≤ C .

    j ∞  ∞ On the other hand T (x j )nj=1 = λmk nj=1 ϕmk , x j ymk and so m=1 k=1 n       λC . As ε > 0 is arbitrary, the proof is complete.   NαrC (T ) ≤ j=1 A j

3 Factorization of ultraproducts of αrC -nuclear maps This section has a technical preparatory character in order to find a characterization of αrC -integral multilinear maps in the following Sect. 4. We will deal with ultraproducts of Banach (resp. quasi-Banach) spaces. For this topic our main reference is [9] (resp. [21]). In this section G will be always an arbitrary non void index set and U a non trivial ultrafilter defined on G. The ultraproduct by U of a family {E γ , γ ∈ G} of quasi-Banach spaces is denoted by (E γ )U (or simply by (E)U if E γ = E for every γ ∈ G and then it is called an ultrapower) and provided with its natural quasi-norm.  Analogously, (xγ )U will be the class in (E γ )U of the element (xγ )γ ∈G ∈ γ ∈G E γ . If every E γ , γ ∈ G, is a Banach lattice, the ultraproduct (E γ )U is a Banach lattice with the canonical order defined by the relation (xγ )U ≤ (yγ )U if and only if there exist (x γ ) ∈ (xγ )U and (y γ ) ∈ (yγ )U such that x γ ≤ y γ for every γ ∈ G. Moreover, one has (xγ )U ∧ (yγ )U = (xγ ∧ yγ )U and (xγ )U ∨ (yγ )U = (xγ ∨ yγ )U .   j j Given a family of maps {Tγ ∈ Ln nj=1 E γ , Fγ  γ ∈ G}, E γ , 1 ≤ j ≤ n, being   Banach spaces, Fγ , γ ∈ G, being quasi-Banach spaces, satisfying supγ ∈G Tγ  < ∞     j we can define the canonical ultraproduct map Tγ U sending every x := (xγ )nj=1 U ∈ 

       j  j n to Tγ U (x) = Tγ (xγ )nj=1 U ∈ (Fγ )U . One has  Tγ U  ≤ j=1 E γ  U limγ ,U Tγ . Specifically, we shall deal with ultrapowers ( p [q ])U of Bochner quasi-Banach spaces  p [q ], 0 < p ≤ ∞, 0 <  state several fundamental notations to  qγ< ∞. We be used in this setting. Let X = (xmk ) U ∈  p [q ] U . ∞  γ p q 1. We shall write Xγ := (xmk )∞ k=1 m=1 ∈  [ ] for every γ ∈ G. For each m ∈ N, γ γ ∞ q we put Xm := (xmk )k=1 ∈  . Then we can write X = (Xγ )U . γ 2. Denote by P(Z ) theset of all subsets of a given set Z . Given W = ((wmk ))U ∈ ( p [q ])U and a set G γ , γ ∈ G ⊂ P(N2 ) we define W(G γ ) = (WG γ )U where

J. A. López Molina

 γ  γ γ γ WG γ = (xmk ) is defined by xmk = wmk if (m, k) ∈ G γ and xmk = 0 if (m, k) ∈ / G γ . Note that this definition is independent of  the used representative for W. W(G γ ) has a similar meaning if W ∈ (1 )U and G γ , γ ∈ G ⊂ P(N).  3. Let G γ , γ ∈ G ⊂ P(N) and assume that for every γ ∈ G we have another →

γ

set F γ := {Fm , m ∈ G γ } ⊂ P(N). We define X γ

γ

γ Gγ

→ ,F γ

γ

= ((z mk )) ∈  p [q ] γ

γ

γ

such that z mk = 0 if m ∈ / G γ , z mk = 0 if m ∈ G γ , k ∈ / Fm and z mk = xmk if γ be γ ∈ G and (m, k) ∈ N2 . Moreover, we put m ∈ G γ and k ∈ Fm whatever  p q  γ γ → = X X → U ∈  [ ] U . If every set Fm , m ∈ G γ is the same set γ (G γ ),( F )

Gγ ,F γ

→ . Clearly F independent of m and γ we shall write X(G γ ),(F) instead of X (G γ ),( F γ )  p q   p q  → : X ∈  [ ] → is a projection in  [ ] −→ X the map P U U. γ γ (G γ ),( F ) (G γ),( F ) 4. Let N := N U be the set theoretic ultrapower of N by U. Every element of N is a family k = (kγ )U with kγ ∈ N for each γ ∈ G. If (k, h) ∈ N2 , Xkh will denote

→

the element X

(G γ

→ ),( F γ )

where G γ = {kγ } and F γ = {h γ } for every γ ∈ G, while

Xk := X(G γ ),(N) . Note that   Xk 

( p [q ])U

   γ   γ  = lim Xkγ q ≤ lim  (xmk )  p [q ] = X( p [q ]) . γ ,U

U

γ ,U

(7)

  Lemma 4 Let p ∈]0, ∞] , q ∈]0, ∞[ and let 0 ≤ W ∈  p [q ] U . Then  C(W) = W

(G γ

Proof As W

γ

→

G γ , Fγ

→ ),( F γ )

  G γ ⊂ N,

 →  γ F γ = Fm ⊂ N, m ∈ G γ ∀γ ∈ G .

∈ C(Wγ ), γ ∈ G, it is clear that W

→

(G γ ),( F γ )

∈ C(W). Conversely,

if X ∈ C(W), there are representatives W = (Wγ )U and X = (Xγ )U such that Xγ ∈ C(Wγ ) for every γ ∈ G (see [14], Remark 2). Now it is clear that for every →  γ γ ∈ G there exist a set G γ ⊂ N and a family F γ = Fm , m ∈ G γ of subsets of N γ → . such that Xγ = W → . Then X = W   Gγ ,F γ

(G γ ),( F γ )

To simplify notation, from now on we write ∀q>0

  X q∞ := ∞ [q ] U ,

  X q := 1 [q ] U ,

  Yq := q [q ] U .

(8)

Note that, if 1 ≤ q < ∞, by [8, Proposition 4.6] and [5, Theorem 11.12] the Banach lattices X q and Yq are order continuous (and Dedekind order complete). pose now the main problem we shall deal with in this section. Let 0 < g := We γ ∞ ∞  γ gmk k=1 m=1 U ∈ X r0 , gmk ≥ 0 for every (m, k) ∈ N2 and γ ∈ G. For every    γ ∈ G we consider the canonical n-linear map Sgγ ∈ Ln nj=1 r j , tr . Our purpose in this section is to find a factorization of the composition of ultraproduct maps S := (I1,1 ⊗ Itr ,rn+1 )U ◦ (Sgγ )U through concrete Lebesgue–Bochner spaces. The bad

The minimal and maximal operator ideals associated to (n + 1)…

order properties of ultraproducts X q∞ are a serious obstacle in the search of a direct factorization. To circumvent this problem we need a previous auxiliary factorization of (Sgγ )U through spaces with better order properties. This will be made in Sect. 3.1 with he introduction of some suitable quotient spaces Qq∞ of X q∞ . Finally in Sect. 3.2 we will obtain the desired factorization of S. 3.1 Auxiliary factorization of (Sgγ )U  γ − r0  γ  r0 qγ Put ahk := gh  r q ghk  q for every (h, k) ∈ N2 , γ ∈ G and 0 < q < ∞. Consider 0 the elements  qγ ∞  ∞ Aq := (ahk )∞ k=1 h=1 U ∈ X q ,

  r0  r γ 1− q  γ  q0 ∞ ∞ gm g Uq := ∈ Xq . (9) mk

r0

k=1 m=1 U

We note that by (2)  r  (Sgγ )U A j

(G γ

→ ),( F γ )

n



j=1





 −r  1 − 1  1 1 γ  0 tr r0  γ r0 tr − r0 +1   γ gmk gm r = k∈F 0

=U

tr

→

(G γ ),( F γ )

m



m∈G γ U

.

(10)

Assume 1 < q < ∞. As X q  is order complete by [2, Theorems 3.5 and 3.7] one       has X q  = (Uq )⊥⊥ ⊕ (Uq )⊥ , (Uq )⊥⊥ = B(Uq ) and (Uq )⊥ = B((Uq )⊥ ). Let   S1 : B(Uq ) −→ X q  and S2 : B((Uq )⊥ ) −→ X q  be the canonical injections. The      dual space is a canonical direct sum X q  = B(Uq ) ⊕ B((Uq )⊥ ) and B(Uq ) =                  S1 X q  and B((Uq )⊥ ) = S2 X q  . If (Uq )◦ := ϕ ∈ X q   ϕ, Uq = 0 and  q  ⊥ ◦       (U ) (11) := ϕ ∈ X q   ϕ, Y = 0 ∀ Y ∈ (Uq )⊥ , ◦      we obtain (Uq )◦ = B((Uq )⊥ ) and (Uq )⊥ = B(Uq ) . ⊥  Lemma 5 Let 1 < q < ∞. Then (Aq ⊂ (Uq )◦ .  γ −1 γ   ⊥ γ Proof Write amk := gm r0 gmk , γ ∈ G, (m, k) ∈ N2 . Let X ∈ (Aq , i.e. X ∧   γ Aq = 0. By [14, Remark 1] there are representatives X = (xmk ) U and A j =  γ  γ γ (αmk ) U such that αmk = 0 if xmk = 0 for every γ ∈ G and (m, k) ∈ N2 . Then it is  γ   γ  q   enough to show that Uq = gm  r α  q  since in such a case we will have 

0

mk

U

∞ ∞  q    γ  γ  γ    X, Uq   = lim xmk gm r0 αmk  q   = 0, γ ,U

m=1 k=1

J. A. López Molina 

i.e. X ∈ (Uq )◦ and (A j

⊥



⊂ (Us j )◦ . To do this, we recall the elementary inequalities  1 1 1 |x| q − |y| q  ≤ |x − y| q ,  q    |x| − |y|q  ≤ q |x − y| |x|q−1 + |y|q−1 .

∀ q ≥ 1, ∀ x, y ∈ R ∀ q ≥ 1, ∀ x, y ∈ R

By (12), (13) and Hölder’s Inequality, since q  =

q q−1

(13)

we obtain

  ∞ 

  γ  γ  r0  γ  q        q q lim − αmk  gm r0 amk γ ,U

   

k∈N q 

m=1

∞

 γ gm  ≤ lim γ ,U

(12)

r0

m=1 ∞

≤ q lim γ ,U

 1 ∞ q

  γ r    q γ  a  0 − α   mk mk k=1 ∞ 

 γ gm 

   γ  r 0  γ   a  q − α  mk   mk

r0

m=1

k=1

  1  γ  r0 (q−1)  γ q−1 q × amk  q + αmk  ≤ q lim γ ,U

∞

 γ gm 

r0

  1  γ rq0 q γ   |amk | − |αmk | k∈N  q 

m=1

1     γ r0   γ × |amk | q  + |αmk |q−1 k∈N  qq     1 r0  γ q γ  ≤ q lim sup  |amk | q − |αmk | k∈N  q

γ ,U m∈N

×



∞

 γ gm 

r0

m=1

1 ∞ q 2

 γ γ q r0 |amk | + |αmk | k=1

  1

 γ r0 q γ  ≤ q lim sup  |amk | q − |αmk | k∈N  q 

γ ,U m∈N



∞   γ q 12  γ gm  r × 1 + sup (αmk )k∈N q q 0 m∈N

 ≤ q lim sup

γ ,U m∈N

m=1

  1

 γ rq0 q γ   |amk | − |αmk | k∈N 

   12   × 1 + Aq  X ∞ q g X q

q

r0

=0

 γ r0   γ    γ  γ q   (because Aq = (|amk | q ) U = (αmk ) U ) and by (9) Uq = (gm r0 αmk  q  ) U .   We are now ready to introduce the announced quotient spaces. Let 1 < q < ∞.  Clearly (Aq )⊥ is a solid set in X q∞ . On the other hand, as Uq is a weak order unit

The minimal and maximal operator ideals associated to (n + 1)… 

in the order complete Banach lattice (Uq )⊥⊥ , by Spectral Theorem [2,   Freudenthal’s   Theorem 6.8] the equality (Uq )◦ = ϕ ∈ X q   ϕ, Y = 0 ∀ Y ∈ (Uq )⊥⊥ holds.   p. 166] and the quotient Banach lattices It follows that (Uq )◦ is a solid set in (X q  ) q[2,  q ◦ ∞ ∞ ϕ the class in Dq Dq := (X q ) /(U ) and Qq := X q (A )⊥ are well defined. Let  of ϕ ∈ (X q ) and  X the class in Qq∞ of X ∈ X q∞ . However, some times it will be more useful to set K q : X q∞ −→ Qq∞ for the canonical quotient map.  q ) =  X, X ∈ C(Aq ) . Lemma 6 For each 0 < q < ∞ C(A Proof K q being an order homomorphism one has  X ∈ C( Aq ) if X ∈ C(Aq ). Conγ γ q q −  X ∧ (A X) = 0. versely, assume that A = ((amk ))U and X = ((xmk ))U verify  γ γ γ γ q ⊥ q −  As A X ≥ 0 there exists C = ((cmk ))U ∈ (A ) such that amk − xmk ≥ cmk for γ γ γ γ γ ∈ G and (m, k) ∈ N2 . If xmk := xmk + cmk it turns out that X := ((xmk ))U satisfies γ γ  =   ∧ (  ) = 0. Then X X, amk ≥ xmk , γ ∈ G, (m, k) ∈ N2 , and necessarily X Aq − X γ γ 2 we can assume amk − xmk ≥ 0 for every γ ∈ G and (m, k) ∈ N . Since X ∧ (Aq − X) ∈ (Aq )⊥ , one has X ∧ Aq ∧ (Aq − X) = 0 and    γ γ γ γ  lim xmk ∧ amk ∧ (amk − xmk )  ∞ s j = 0. (14) γ ,U

 [ ]

 γ  γ  γ γ  Let Dγ := (m, k) ∈ N2  xmk ∧ amk > amk − xmk  and E γ := N2 \Dγ . Define γ γ γ γ γ X = (x mk )U where x mk := amk if (m, k) ∈ Dγ and x mk := xmk if (m, k) ∈ E γ . γ γ  γ γ γ γ   Note that if (m, k) ∈ Dγ we have amk − xmk = xmk ∧ amk ∧ amk − xmk  ≥ 0. It   follows from (14) that X − X ∧ Aq = 0. Then K q (X) =  X. On the other hand, since γ γ γ γ γ γ γ γ xmk ≤ amk , if (m, k) ∈ E γ we have x mk ∧ (amk − x mk ) = xmk ∧ (amk − xmk ) = γ γ γ γ xmk ∧ amk ∧ (amk − xmk ) and by (14) X ∧ (Aq − X) = 0. Then X ∈ C(Aq ) and the result follows.   q is a weak order unit in Qq∞ and so Qq∞ = B(A q ). Lemma 7 If 1 ≤ q < ∞, A q ∧  Y = 0 in Qq∞ . Then  Y ≥ 0 and so there is T ∈ (Aq )⊥ such that Proof Let A   ∞ Y ≥ T in X q . We have K q Y − T =  Y and Y − T ≥ 0 and hence we can take γ γ γ γ q 2 A = (amk )U ,Y = (ymk )U with amk ≥ 0, ymk ≥ 0 for (m, k) ∈ N G.   γand γ ∈ γ γ γ 2 2   Let Dγ := (m, k) ∈ N amk ≥ ymk , E γ := (m, k) ∈ N amk < ymk , γ ∈ q q q   q , Y ≤ ≤ Aq and Y ≤ Y we have A ≤A Y and 0 ≤ A ∧ G. As A (Dγ )

(Dγ )

(Dγ )

(Dγ )

(Dγ )

q q   q ∧     Y Y = 0, i. e. A (Dγ ) ≤ A (Dγ ) ∧ Y(Dγ ) = 0. Analogously A(E γ ) ∧ Y(E γ ) = 0. On

q q q    the other hand A(Dγ ) ≥ Y(Dγ ) and A(E γ ) ≤ Y(E γ ) . Then Y (Dγ ) = A(Dγ ) ∧ Y (Dγ ) = 0   q q q q A(E γ ) ∧  Y(E γ ) = K q A(E γ ) ∧ Y(E γ ) = 0. That means A(E γ ) ∧ Y(E γ ) ∈ and  A(E γ ) = 

(Aq )⊥ , i. e. 0 = A(E γ ) ∧ Y(E γ ) ∧ Aq = A(E γ ) ∧ Y(E γ ) . But A(Dγ ) ∧ Y(E γ ) = 0 and q

q

q

 Y(E γ ) = 0. Then  Y=Y Y(E γ ) = 0. so Y(E γ ) ∈ (Aq )⊥ and  (Dγ ) + 

 

X ∈ Dq  is a surjective Lemma 8 Let 1 < q < ∞. The map ι :  X ∈ Qq∞ −→  ∞ topological and order isomorphism and Q j is Dedekind complete.

J. A. López Molina

Proof Keeping in mind [9, Sect. 7] it is easy to check that X q∞ is a sublattice of (X q  ) . By Lemma 5 we see that ι ∈ L(Qq∞ , Dq  ) and it is an order monomorphism. Let ϕ ∈  Dq  = (X q  ) /(Uq )◦ . By Kürsten’s result [10] about local duality of ultraproducts, there is an order isomorphism Tϕ from the sublattice generated by ϕ in (X q  ) onto        a sublattice of X ∞ such that Tϕ (ϕ), Us j = ϕ, Uq . Then  ϕ − Tϕ (ϕ), Uq  = 0. q

 ∞  That means ϕ − Tϕ (ϕ) ∈ (Uq )◦ and hence  ϕ=T ϕ (ϕ). As Tϕ (ϕ) ∈ X q we obtain   ϕ and ι turns out to be surjective. Then it is enough to apply the ι(T ϕ (ϕ)) = T ϕ (ϕ) =  open mapping theorem.   On the other hand, by the structure of X q  as a direct sum one has Dq  =      S1 (X q  ) = B(Uq ) , which being a dual Banach lattice is order complete by [2,   Corollary 12.5 and Theorem 1.13]. Then Qq∞ is order complete.

 q ) is dense in Qq∞ . Corollary 9 If 1 < q < ∞ the linear span C(A Proof Apply Lemmata 6, 7 and 8 and Freudenthal’s Spectral Theorem.    n     j Lemma 10 The map (S ∈ nj=1 Qr∞ −→ (Sgγ )U (X j )nj=1 ∈ X tr g γ )U : X j=1 j n  ∞ tr  is a well defined and continuous n-linear map and (S g γ )U j=1 Qr  ⊂ B(U ). j

   j = Y j ∈ Q∞ , 1 ≤ j ≤ n. As X j − Y j  ∧ Ar j = 0 by [14, Remark Proof Let X rj    jγ   jγ   1 ], there are representatives (z mk ) U ∈ Y j − X j  and (αmk ) U ∈ Ar j such that  jγ   jγ  jγ jγ (z mk ) ∧ (αmk ) = 0 for every γ ∈ G. Then z mk = 0 if αmk = 0 and putting  jγ  r j γ    jγ Dm := k ∈ N  αmk = 0 necessarily limγ ,U supm∈N  amk jγ   ≤ k∈Dm r j  r j γ jγ ∞  − αmk k=1  r j = 0. Hence, given ε > 0 there is U ∈ U such limγ ,U supm∈N  amk  that for each m ∈ N and γ ∈ U  γ   g

 jγ 

mk k∈Dm

r0

 γ = gm 

r0

r j

 r j γ

 a mk

r



rj  r0j  γ   ≤ ε r0 gm r . jγ r 0

k∈Dm



j

r j

  ≤ε = ε r0 g X and ε > 0 being m=1 r0 r0 arbitrary, g(N),(D jγ ) = 0. Writing Wh := (X1 , . . . , Xh−1 , Yh − Xh , Yh+1 , . . . , Yn ),  mγ  γ 1 ≤ h ≤ n, (g mk ) U := g(N),(D jγ ) and g γ = ((g mk )), γ ∈ G, by the definition  Then g

 

jγ (N),(Dm ) X r0

hγ

r0

 γ  limγ U ∞ gm 

m

of (Sgγ )U and Dm and (4) it is obtained (Sgγ )U (Wh ) = (Sgγ )U (Wh ) = 0 and  n     consequently (Sgγ )U Y j j=1 − (Sgγ )U X j )nj=1 = nh=1 (Sgγ )U (Wh ) = 0 and  (S g γ )U is well defined.  n  j j ∈ nj=1 Qr∞ , for every ε > 0 there are X j ∈ X On the other hand, given X j=1 j     X j  + ε, 1 ≤ j ≤ n. From the arbitrariness of ε > 0 we obtain such that X j  ≤    n    n    ≤ g     j easily (S g γ )U X j=1 X j and (Sg γ )U turns out to be continuous. j=1

The minimal and maximal operator ideals associated to (n + 1)…

As B(Utr ) is closed [2, Theorem 11.2], by continuity, (10), Lemma 7, Corollary 9 and n  ∞ tr    Lemma 4 we obtain (S g γ )U j=1 Qr  ⊂ B(U ). j

3.2 Factorization of S The main result of this section is Theorem 11 There are measure spaces (R, A, ν) and (T, M, μ), a function g ∈    L 1 R, L r0 (T) and a map M ∈ L L 1 [R, L tr (T)], L 1 [R, L rn+1 (T)] verifying Supp(M( f )) ⊂ Supp( f ) for each f ∈ L 1 [R, L tr (T)] such that S factorizes as 

n j=1

 (∞ [r j ])

Sgγ

 U

(I1,1 ⊗ Itr ,rn+1 )U

- (1 [tr ])U

U

- (1 [rn+1 ])U 6

(X j )nj=1 n

C

?

- L 1 [R, L tr (T)]



rj ∞ j=1 L [R, L (T)]

M

Sg

- L 1 [R, L rn+1 (T)]. (15)

Proof The (long) proof will be made through some steps and previous lemmata. Step 1. Background for the construction of the desired measure spaces.  γj   γ  1−r0  γ  r0 ∞ ∞  gm  q g  q 1.a) Given q ∈]0, ∞[ let Vq = (vmk ) U := ∈ mk k=1 m=1 U r0  1  q  q q     Yq . Remark that V = V for 0 < q < ∞. Every C(V ) → → Y Y (G γ ),( F γ )

1

(G γ ),( F γ )

q

is a Boolean algebra with the natural operations of join, meet and complement given by ∨, ∧ and Vq − x, x ∈ C(Vq ) respectively. By Stone’s Representation Theorem there are a totally disconnected compact topological space  and a Boolean algebra isomorphism ϕ1 from C(V1 ) onto the Boole algebra B  of the clopen  sets of . Then by q q ) −→ ϕ V1 ∈ C(V is a Boolean algebra Lemma 4 the map ϕq : V → → 1 (G γ ),( F γ )

(G γ ),( F γ )

isomorphism from C(Vq ) onto B. As Y1 is order Dedekind complete, by [2, Theorem 3.15]C(V1 ) is a Dedekind complete Boolean algebra as well as C(Vq ), 0 < q < ∞, (since ϕ1−1 ◦ ϕq is an isomorphism from C(Vq ) onto C(V1 )). As in the proof of Kakutani–Bohnenblust–Nakano’s theorem [2, Theorem 12.26], the real function μ on B ∀ A∈B

  q  μ(A) = ϕ1−1 (A)Y = ϕq−1 (A)Y , 0 < q < ∞ 1

q

(16)

is a measure in (, B) that can be extended by Caratheodory’s method to a measure (again denoted by μ) defined in the σ -algebra M of μ-measurable sets in . Once again from [2, Theorem 3.15] we obtain that ϕ1 C(V1 ) = M = B. Moreover, the map Jq : w ∈ C(Vq ) −→ Jq (w) = χϕq (w) can be extended to an isometry (again denoted

J. A. López Molina

by Jq ) from the band B(Vq ) generated by Vq in Yq onto the space L q (, M, μ). Note that Jq (Vq ) = χ and ∀0
   1   1 χ  q q = Vq  = g Xq r < ∞.  L () = μ() Y q

0

(17)

 γ  ∞  1.b) Let Z = gm r0 m=1 U . As in Lemma 4, it can be shown that the set of   components of Z in (1 )U is C(Z) = Z(G γ )  G γ ⊂ N ∀ γ ∈ G . Arguing as in part 1.a) there are a totally disconnected compact topological space  and a Boolean algebra isomorphism ψ from C(Z) onto the Boolean algebra C ofthe clopen sets of  . Moreover, the real function ν0 : A ∈ C −→ ν(A) = ψ −1 (A)(1 ) is a measure U in (, C) which can be extended to a measure, again denoted by ν0 , defined in the σ -algebra A of ν-measurable sets of . It follows from [8, Proposition 4.6] and [2, Theorem 3.15] that ψ(C(Z)) = A = C. It turns out that the map I : w ∈ C(Z) −→ I(w) = χψ(w) can be extended to an isometry (again denoted by I) from B(Z) onto the space L 1 (, A, ν). Remark that for every 0 < q < ∞  q  q    ν0 ψ(Z(G γ ) ) = Z(G γ ) (1 ) = V(G γ ),(N) Y q U

 γ   gm  r ≤ g . = lim (18) 0 X γ ,U

r0

m∈G γ

    It follows that ν0 () = Z(1 ) = g X < ∞. r0 U In the sequel the discrete and the continuous parts of (, M, μ) and (, A, ν0 ) will need to be handled in a completely different way. 1.c) The atomic parts of (, M, μ) and (, A, ν0 ). It is clear that ϕq , 0 < q < ∞, sends the components of Vq which are discrete elements in Yq onto the atoms of (, M, μ). As μ() < ∞, by [22, Theorem 3.2]  the set  of atoms of M must be a numerable set of elements of type ekm ,hk := ϕ1 Vk1 m ,hk where (km , hk ) ∈ N2 and (m, k) ∈ P, a certain subset of N2 . Since for every (m, k) ∈ P one as Vk1 m ,hk = 0 and    γ 1−r  γ r  μ(ekm hk ) = Vk1 m hk  = lim gk m r0 0 gk m h k  0 γ ,U

γ

γ

1−r   γ r   = limgk m h k  0 gkm  0 = 0, γ ,U

γ

γ

(19)

γ

 γ  γ −1 noting that gk m h k gk m r0 ≤ 1 for every γ ∈ G, we have necessarily γ

γ

∀ (m, k) ∈ P

γ

    gkm  = limgγm  γ ,U

kγ

r0

= 0 and

 γ  γ −1 limgk m h k gk m r0 = 0. γ ,U

γ

γ

γ

(20) q

If follows Ukm ,hk = 0 for 0 < q < ∞ and (m, k) ∈ P.  Let 1 := (m,k)∈P ekm ,hk the purely atomic part of  and 0 := \1 its purely non atomic part. As μ() < ∞ given a set A ⊂ P, if A = ∪(m,v)∈A ekm ,hk and F(A)

The minimal and maximal operator ideals associated to (n + 1)…

is the set of finite parts of A, one has χ A = lim F∈F (A) q (1 , μ).



(i,v)∈F

C(Vq )

χe

ki ,hv

in the topology

of As is Dedekind complete and Jq is a topological isomorphism, by [1, Theorem 5.6] we obtain ϕq−1 (A) =

lim

F∈F (A)



=

F∈F (A)



q

(i,v)∈F q Vki ,hv

Vki ,hv = lim

h→∞

F∈F (A)

q

Vki ,hv

∈ C(V ). q

(21)

→

Hence by Lemma 4, there are sets Aγ ⊂ N, Bγ ⊂ P(N), γ ∈ G, such that ϕq−1 (A) = V

→

q

→ (Aγ ),( Bγ )

. In particular, we define sets Pγ ⊂ N, Q γ ⊂ P(N), γ ∈ G, such that

ϕq−1 (1 ) = V

q

→

(Pγ ),( Q γ )

, which will be used in the sequel.

 q On the other hand, let X q1 be the closure in X q of the linear span Ukm ,hk (m,k)∈P .     q q put A := (m, v)  ekm ,hv ⊂ If 1 ≤ q < ∞, given X := U → ∈ C U → γ γ (Aγ ),( B ) (Pγ ),( Q ) ϕq (X) . Since X q is order continuous and Dedekind complete, there exists Sq =  q   q 1 . → (m,k)∈A Ukm ,hk ∈ X q ⊂ B U (Pγ ),( Q γ )

 q  q = Lemma 12 If 1 ≤ q < ∞ the equalities Sq = U → and B U → (Aγ ),( B γ ) (Aγ ),( B γ )       q → → P B(Uq ) hold. In particular B U =P B(Uq ) = X q1 . → γ γ (Aγ ),( B )

(Pγ ),( Q )

(Pγ ),( Q γ )

 γ  q q Proof Assume 0 ≤ T = (tmk ) U ∈ X q such that Ukm ,hv ≤ T ≤ U → for every (Aγ ),( B γ )  γ    γ (m, v) ∈ A. Fixed (m, v) ∈ A there are (αik ) ∈ 1 [q ]G and (βik ) ∈ 1 [q ]G such  γ 1− r0  γ  r0  γ   γ  that limγ ,U  (αik ) 1 [q ] = limγ ,U  (βik ) 1 [q ] = 0 and gk m r0 q gk m h v  q γ γ γ  γ 1− r0  γ  r0 γ γ γ γ γ q q     gha + αk m h v ≤ tk m h v , γ ∈ G, and tha ≤ gh r + βha for every a ∈ Bh , h ∈ γ

γ

γ

γ

0

Aγ , γ ∈ G. Then, with respect the order in 1 [q ] one has  γ  1 −1  γ 1− r0  γ  r0 g m  q g m  q g γ γ  q ek m h v kγ r0 kγ r0 γ γ km h v  γ  1 −1 γ + gk m  qr αk m h v ekγm h vγ γ

0

γ

 γ  1−r0  γ = g m  q g m

γ

 1  r0  q ek m h v + gγγ  q −1 αk m h v kγ r0 γ γ γ γ km r0  γ  1 −1 γ  × ekγm h vγ ≤ gh  qr thk 0    γ  1 −1  γ 1− r0  γ  r0  γ ≤ g  q g  q g  q kγ h vγ

h

h

r0

hk

 γ  1 −1 γ  + g m  q β kγ

r0

hk

k∈Q h h∈Pγ

J. A. López Molina

=

 γ  1−r0  γ  r0  g  q g  q h

hk

r0

γ





k∈Q h h∈Pγ

 γ  1 −1 γ  + gk m  qr βhk . γ

(22)

0

 γ  1 −1  γ   γ  1 −1  γ  By (20) one has limγ ,U gk m  qr  (αik ) 1 [q ] = limγ ,U gk m  qr  (βik ) 1 [q ] = γ 0 γ 0  γ  1 −1 γ  q q q   gh r thk U ≤ V 0 and hence (22) and (21) imply Vkm ,hv ≤ = → (Aγ ),( B γ )

0



 γ  1 −1 γ  As (m, v) ∈ A is arbitrary we deduce gh  qr thk U = 0  γ 1− 1  γ  1 −1 γ   γ 1− 1  γ  1−r0 q q       gh r gh r q gh  r q gh r thk U = . Then T =

q (m,v)∈A Vkm ,hv .

V

q

→

(Aγ ),( B γ )  γ  r0    g  q γ hk k∈Bh h∈Aγ U

0

=U

q

→

(Pγ ),( Q γ )

0

0

. It follows that Sq = U

q

→

(Aγ ),( B γ )

0

. As X is a weak

order  q unit in B(X) [2, p. 36] and B(X) is order continuous (a closed subspace of ), Freudenthal’s Spectral Theorem gives the density of the linear span B U → γ γ ),( B )    (A  q  q C U in B U .   → → (Aγ ),( B γ )

(Aγ ),( B γ )

 As above we find that the set of atoms of (, A, ν0 ) is a numerable set 1 := m∈N ekm , where we have defined ekm = ψ(Zkm ), m ∈ N, and the purely non atomic part of (, ν) is 0 := \1 . Let ν1 be the measure on 1 such that ν1 (ekm ) = 1 for each m ∈ N and for every 0 < q < ∞ let Pq be the natural projection from 1 [1 , ν1 , q (1 , μ)] onto the closure of the linear span  Hq := χe m (t)χe m v (x) (m,v)∈P . The motivation for the use of the measure ν1 is k k ,h the following result. Lemma 13 Let 0 < q < ∞. There is an isometry Wq from X q1 onto the complemented subspace Hq of 1 [1 , ν1 , q (1 , μ)].  q h vi q Proof For every X = i=1 v=1 αiv Ukm i ,hkv ∈ Ukm ,hv (m,v)∈P we define v1 h

  1   g m 1− q αiv χe m (t)χe m Wq (S) := (x) ∈ 1 1 , ν1 , q (1 , μ) . k i k i k i ,hkv i=1 v=1

q

q

As Ukm ,hv1 ∧ Ukm ,hv2 for v1 = v2 in N we have vi h

1     1   q Wq (X) = g m 1− q αiv  μ(e m i kv ) q k ,h k i i=1

=

h

 g

km i

i=1

=

vi h 

 

i=1 v=1

1− 1  q

v=1 vi  v=1

 q  q  1 αiv  V m k q q k i ,h v

    q αiv Ukm i ,hkv  = X X

q

The minimal and maximal operator ideals associated to (n + 1)…

and hence Wq can be continuously extended to an isometry from X q1 onto Hq .     γ Lemma 14 For 1 < q < ∞, the map Xq1 sending X = (xmk ) U ∈ X q∞ to Xq1 (X) =



  r0 −1    r0 g m  q limγ U x γm v g γm v − q ∈ ∞ [1 , ν1 , q (1 , μ)] k k h k h γ

γ

γ

γ

ekm hv ⊂1 ekm ⊂1

∞ 1 ∞ q 1 :  is continuous. Moreover, the map X q X ∈ Qq −→ Xq (X) ∈  [1 , ν1 ,  (1 , μ)] is well defined and continuous.  Proof Fixed m ∈ N, for every u ∈ N there is Uu ∈ U such that the numbers h vγ , 1 ≤   v ≤ u are pairwise different whatever be γ ∈ Uu . Define Pm := w ∈ N  (m, w) ∈ P . Then by (19)

⎛

∀ u ∈ Pm

 γ q  ⎞ q1 x m v  μ(ekm ,hv ) kγ h γ ιmu (X) := ⎝ lim  γ r0  1−r ⎠ 0 γ ,U g m v  g  kγ h γ 1≤v∈Pm km ⎛ ⎞1 q u

 γ q   ⎝ ⎠ = lim xk m h v u

1≤v∈Pm

γU

γ

γ

∞ 1

 γ q q  γ  x m   ≤ lim ≤ lim sup (xas )∞ s=1 q k b γ ,U

b=1

  = X X ∞ .

γ

γ U a∈N

(23)

q

  As m and u are arbitrary it follows that Xq1 (X) = supm∈N supu∈Pm ιmu (X) ≤     X ∞ and X1 ∈ L X ∞ , ν1 , ∞ [1 , q (1 , μ)] . q q X q

To finish it is enough to prove that if X ∈ (Aq )⊥ one has Xq1 (X) = 0. As   γ   γ γ limγ ,U ((|x hk | ∧ ahk )) ∞ q = 0, given (m, v) ∈ P we have γ ∈ G  |xk m h v | > γ γ  [ ] γ q / U since in the contrary we would have Akm ,hv = 0 and by (20) and ak m h v ∈ γ

γ

the definition of V1 we would have Vk1 m ,hv = 0, a contradiction. It follows that  γ  γ γ ∈ G  |xk m h v | ≤ ak m h v ∈ U and so Xq1 (X)(ekm ,hv ) = 0. By the arbitrariness of γ

γ

γ

γ

(m, v) ∈ P we obtain Xq1 (X) = 0.

 

−1 1.d) The continuous part of (, M, μ). Define V1 → := ϕ1 (0 ). Clearly (Nγ ),( M γ )   the measure algebra C(Z(Nγ ) ), ν is isometric to the measure sub-σ -algebra (Z, μ)    1 . By Maharam’s results where Z := V(G γ ),(N) , G γ ⊂ Nγ , γ ∈ G of C V1 → (Nγ ),( M γ )

[16, Theorem 1] and [15, Definition 10.2, Theorem 10.6 and Section 17] there are  , a σ -algebra E in V1 ∈ C V1 such that a component V1 → → → (Vγ ),(W γ ) (Nγ ),( M γ ) (Vγ ),(W γ )   1   1 ⊂C V , a measure ρ0 in E and an isometry W from the E⊂C V → → (Vγ ),(W γ ) (Nγ ),( M γ )   tr , μ) onto the product measure algebra (Z, ν)×(E, ρ0 ). measure algebra (C V → (Nγ ),( M γ )

J. A. López Molina

Moreover, for every V1 because V1

(G γ

→

(G γ ),( F γ )

 ∈ E one has W V1

→ ),( F γ )

= V1

→

(G γ ),( F γ )

∧ V1

(G γ

→

(G γ ),( N )

 → ),( F γ )

1 = V(G ×Vtr γ ),(N)

→

(G γ ),( F γ )

in C(V1 ). If P1 (resp. P2 ) denote the

canonical projectionsfrom the cartesian product Z × E onto Z (resp. E), from now on,  1 we shall put given V1 → ∈ C V → (E γ ),( D γ )

(Vγ ),(W γ )

  E := ψ ◦ P1 ◦ W V1 → , (E γ ),( N )   . D := ϕ1 ◦ P2 ◦ W V1 →

(24)

(E γ ),( D γ )

 Note that μ V1



  = ν0 (E)ρ0 (D). We define  := ϕ1 V1 ⊂ 0 → (E γ (Vγ ),(W γ )   provided with the σ -algebra M := ϕ1 E and the measure ρ 0 (ϕ1 (A)) = ρ0 (A) if A ∈ E. → ),( D γ )

Lemma 15 There is an isometry  from L 1 (0 , M0 , μ) onto the Lebesgue– Bochner space L 1 [0 , A0 , ν0 , L 1 (, M, ρ 0 )]. Proof Let S(0 ) be the set of simple functions in L 1 (0 , M0 , μ). For every   h  αkv χ Ahv (x) ∈ S(0 ), where Ahv = ϕ1 V1 with pairwise S = kh=1 vv=1 → γ (E γh ),( Dhv )   1 disjoint components V → , 1 ≤ v ≤ vh and Z(E h ) , 1 ≤ h ≤ k , using (24) γ k

γ

(E h ),( Dhv )

vγh

define (S) = h=1 v=1 αkv χ Eh (t)χ Dhv (x) ∈ L 1 [0 , A0 , ν0 , L 1 (, M, ρ 0 )]. As W is a Boolean algebra isomorphism it is easy to check that  is well defined. Then vh k

    αhv ρ 0 (Dhv ) (S) = ν0 (E h ) v=1

h=1

=

vh k

    αhv μ(Ahv ) =  S 

h=1 v=1

L 1 (0 ,μ)

  and  is an isometry from S(0 ) into L 1 0 , ν0 , L 1 (, ρ 0 ) . By density it can be continuously extended to L 1 (0 , μ). On the other hand, W being an isomorphism, for every E × D ∈ A0 × M there is V1 such that (24) holds. If → (E γ ),( D γ )  1  L := ϕ1 V we have (χ L ) = χ E (t)χ D (x). Since L 1 [0 , ν0 , L 1 (, ρ 0 )] = → (E γ ),( D γ )

π L 1 (, ρ 0 ) (a theorem of Grothendieck) it turns out that  is surjective L 1 (0 , ν0 )⊗ finishing the proof.    q Lemma 16 If 1 ≤ q < ∞ there is a linear isometry q from B U → ) onto (Nγ ),( M γ )   1 L 1 [0 , A0 , ν0 , L q (, M, ρ 0 )] such that for every V1 one → ∈ C V → has

(E γ ),( D γ )

(Nγ ),( M γ )

The minimal and maximal operator ideals associated to (n + 1)… 1−q

q−1 (χ E (t)χ D (x)) = ρ 0 (D) q 

q−1

   γ 1−r0  γ r0 q  γ  rq0 gmh gmk × gm



γ k∈Dm

γ

h∈Dm



m∈E γ U

.

 q  q Proof Let Lq : h ∈ L q (, ρ0 ) −→ h +  − h −  ∈ L 1 (, ρ 0 ). Lq is a well defined bijective, disjointness preserving non linear map. After an easy computation   q  we see that Lq (h) L 1 () = h  L q () for h ∈ L q (, ρ 0 ). Analogously, the map  1  1     Bq : h ∈ L 1 0 , ν0 , L 1 (, ρ 0 ) −→ h +  q − h − q ∈ Lq 0 , ν0 , L q (, ρ 0 ) is non linear, bijective, preserves disjointness and Bq (h) L q [ ,ν ,L q (,ρ )] = 0 0 0   q1 h  . L 1 [,ν0 ,L 1 (,ρ 0 )]  γ   γ  On the other hand, given X = (xmk ) U ∈ X q put Nqm (Xγ ) = (xmk )k∈N q for every γ ∈ G and m ∈ N. Consider the non linear maps θq : X = −→

 γ   q (xmk ) U ∈ B U



→

(Nγ ),( M γ )

1

Nqm (Xγ ) q

−1

γ

xmk

)



U

 q ∈B V

→

(Nγ ),( M γ )

)

and q−1    q : h(t, x) ∈ L q [0 , ν0 , L q (, ρ 0 )] −→ h(t, x) q

L (,ρ 0 )

×h(t, x) ∈ L 1 [0 , ν0 , L q (, ρ 0 )].  γ  θq and q are bijective maps with inverses θq−1 (Y) = Nqm (Xγ )q−1 ymk U ∈  q1 −1   q  q γ −1   B U → ) if Y = ((ymk ))U ∈ B V → ) and q (h) = h(t, x) L q (,ρ 0 ) (Nγ ),( M γ ) (Nγ ),( M γ )   h(t, x) ∈ L q 0 , ν0 , L q (, ρ 0 ) if h(t, x) ∈ L 1 [0 , ν0 , L q (, ρ 0 )]. Note that  1        θq (X) = X q for each X ∈ B Uq q (h) = h q for every → ) and (Nγ ),( M γ )

h∈ 0 , ν0 0 )].  q Consider the map q := q ◦ Bq ◦  ◦ Lq ◦ Jq ◦ θq from B U → ) into (Nγ ),( M γ )  q L 1 [0 , ν0 , L q (, ρ 0 )]. As q and Jq are isometries, for X ∈ B U → ) we L 1 [

, L q (, ρ

(Nγ ),( M γ )

obtain      q (X) = Bq ◦  ◦ Lq ◦ Jq ◦ θq (X)q    q   = Lq ◦ Jq ◦ θq (X) = θq (X) = X.

 Since q (0) = 0, by a result of Mazur and Ulam [3,17], Chapter VII, §1, page 142 , the isometry q turns out to be linear.

J. A. López Molina

  q Now we will prove that q B U

(Nγ

→ ),( M γ )

 ) contains the set of products of charac-

teristic functions χ E (t)χ D (x) corresponding to arbitrary components V1

→

(E γ ),( D γ )

. As

this set is total in L 1 [0 , ν0 , L q (, ρ 0 )] (the quoted Grothendieck’s Theorem) the surjective character of q will follows immediately. First note that for each γ ∈ G  Nq

  1−r0  r γ  q  γ  q0  gm g r 

γ k∈Dm

mk

0

q−1

 m∈E γ

 γ  1−r0 (q−1)   γ r0 q−1 q g  = gm r0q . mk γ

k∈Dm

Then it can be easily checked that  q−1 (χ E (t)χ D (x)) = θq−1 ◦ Jq−1 ◦ Lq−1 ◦ −1 ◦ Bq−1 ρ 0 (D)  = θq−1 ◦ Jq−1 ◦ Lq−1 ◦ −1 ρ 0 (D)  ρ 0 (D)

=

θq−1

◦ Jq−1

◦ Lq−1

=

θq−1

◦ Jq−1

 ρ 0 (D)

 = θq−1 ρ 0 (D)





1 q −1

V

q

(E γ

1 q −1

χ D (x)

χ D (x)



1 1 q −1 q







1 1 q −1 q



1 1 q −1 q







χ E (t)χ D (x)

χ E (t)χ D (x)









 → ),( D γ )

  1−r0 



1 1    r0   γ −1 q −1 q gm  r q g γ  q γ = θq ρ 0 (D) mk 0 k∈D m∈E m

= ρ 0 (D)

1−q q

γ

U

   

 q−1   r0   γ 1−r0 gm g γ r0 q g γ  q γ r 0 mh mk  k∈D m∈E

 q ∈B U

γ

m

γ

h∈Dm →

(Nγ ),( M γ )

).

    q is complemented in X q (X q is order complete). If 1 ≤ q < ∞ the band B U → Mγ )   q (Nγ ),(  So the dual space B U ⊂ (X q ) and next lemma is meaningful. → (Nγ ),( M γ )

Lemma 17 Let 1 < q < ∞.

  (a) Let 0 ≤ h(t, x) ∈ L ∞ [0 , ν0 , L q (, ρ 0 ) . Put Y = q (h(t, x)) ∈   q   = ι−1 ( B U and let H ∈ X q∞ such that H Y). Assume that for some → (Nγ ),( M γ )

→ = 0. the equality χG (t)χ F (x)h(t, x) = 0 holds. Then H (G γ ),( Fγ )  q    → (b) Let X ∈ B U . Then χ(×)\(E×D) (t, x) q X (t, x) = 0. → γ) γ (E ),( D γ (E γ ),( D )    −1 X (t, x) = 0. → = 0 then χ (c) If X ∈ X q∞ and X E×D (t, x)( q ) γ  q (E γ ),( D ) → = 0 then χ (d) Let X ∈ B U → ). If X E×D (t, x) q (X)(t, x) = 0. γ

V1

→

(G γ ),( F γ )

(Nγ ),( M γ )

(E γ ),( D )

U

The minimal and maximal operator ideals associated to (n + 1)…

Proof (a) Given V1 with A ∈ A0 and B ∈ M, define Wq → (Aγ ),( B γ )     q q q−1 χ A (t)χ B (x) ∈ B U . By Lemma 16 one has Wq = W → (Nγ ),( M γ )

and so Wq = W

→

(Aγ ),( B γ )

q

−→ (Aγ ∩G γ ),( B γ ∩Fγ )

+W

=

q

−→

+W

(Aγ \G γ ),( B γ ∩F γ )

q

−→

(Aγ ∩G γ ),( B γ \Fγ ) q

+W

−→

(Aγ \G γ ),( B γ \F γ )

.

 q → that W → = ,H It follows from the definition of H −→ γ γ (G γ ),( Fγ ) (G γ ),( Fγ ) γ ∩G γ ),( B \F )  q  (A q → → 0, W = 0 and W = ,H ,H −→ −→ (Aγ \G γ ),( B γ ∩F γ )

(G γ ),( Fγ )

(Aγ \G γ ),( B γ \F γ )

(G γ ),( Fγ )

0. On the other hand, by (11) and the hypothesis about h(t, x), since q is a positive map  q 0≤ W 

(Aγ ∩G γ q

= W

(Aγ ∩G γ

 q = W = ≤

−→ ),( B γ ∩F γ ) −→ ),( B γ ∩F γ ) −→

,H

→



(G γ ),( Fγ )

 q ,H = W

(Aγ ∩G γ

  , q h(t, x)

−→ ),( B γ ∩F γ )

,Y



(Aγ ∩G γ ),( B γ ∩F γ )  q  , h(t, x) q W −→ (Aγ ∩G γ ),( B γ ∩F γ )     q q W −→ , h(t, x) = χG (t)χ F (x), h(t, x) γ (G γ ),( F )



 Then Wq , H

→



(G γ ),( Fγ )

= 0.

= 0. As q is an isomorphism (Lemma 16), by den-

sity of the linear span of the respective simple functions in L 1 (0 , A0 , ν0 ) and L q (, M, ρ 0 ) and the already used Grothendieck’s Theorem we obtain → = 0. H (G γ ),( Fγ )



(b) Let h(t, x) ∈ L ∞ [0 , ν0 , L q (, ρ 0 )] verifying χ E (t)χ D (x)h(t, x) = 0 and take         = ι−1 → H ∈ X q∞ such that H ,h = q (h) . By part a) and (11) q X(E ),( D γ) γ  → ,H → X = 0 and the result follows. γ γ (E γ ),( D ) →

(E γ ),( D )

 γ γ (c) Define D γ = D m := N\Dm , m ∈ E γ for every γ ∈ G. Consider sets A ⊂ E, B ⊂ D such that  A ∈ A0  and B ∈ M and h := χ A (t)χ B (x). q and hence, ϕtr being an isomorphism By Lemma 16 q−1 (h) ∈ B U → (Aγ ),( B γ )   −1     −1      → ( q ) X , h = ( q ) X(N\E γ ),(N) , h + ( q )−1 X ,h = (E γ ),( D γ )   −1 → , X(N\E γ ),(N) , q−1 (h) + X q (h) = 0. The result follows by density as in part a). (d) As X = X

(E γ ),( D γ )

→

(Nγ \E γ ),( D γ )

+X

−→

(Nγ ),(\D γ )

the result follows from part b).

 

J. A. López Molina

Corollary 18 q and and q are order isomorphisms for every 1 < q < ∞. Proof As q is a positive linear map, by Lemma 17, part b) we obtain q (X+ ) = ( q (X))+ for each X ∈ X q . By [2, Theorem 7.2] and Lemma 16 q is an order isomorphism. From this fact and the definition of the positive part of an element of the dual of a Banach lattice [2, Sect. 5] it follows that q (h + ) = ( q (h))+ for every  h ∈ L ∞ [0 , ν0 , L q (, ρ 0 )] and by duality and [2, Theorem 7.2] once again, q is an order isomorphism.      −1 ∞ ∞ q → (X Lemma 19 Aq :  X ∈ Kq P q ) −→ ( q  ) (X) ∈ L [0 , ν0 , L (, γ (Nγ ),( M )

ρ 0 )] is a well defined and continuous linear map for every q ∈]1, ∞[.  γ    ∞ ∞  → (X Proof Let X = (xmk ) U ∈ P q ) verifying X = 0. As X q ⊂ X q  (Nγ ),( M γ )   q ∞ → (X by Lemma 12 we find P and by Lemma 16 → ) q ) ⊂ B(U γ (Nγ ),( M ) →  γ   −1 ( q  ) (X) is defined. Let C γ = Cm := k →  γ γ γ γ Mm , m ∈ Nγ and F γ = Fm := Mm \Cm , m

 q V

→ (N),(C γ )

(Nγ ),( M γ )

  γ  qγ qγ ∈ N  0 ≤ amk = xmk  ∧ amk ∩ ∈ Nγ . As X ∈ Aq⊥ we have

  γ  q1 qγ     = lim gm r a   0 mk k∈C γ m∈N q [q ] γ ,U

m

∞  γ     γ  q1 qγ  gm  r ≤ lim sup  xmk  ∧ amk k∈N q 0 γ ,U m∈N

m=1

 1  γ   qγ  ≤ g q lim sup  xmk  ∧ amk k∈N q = 0, γ ,U m∈N

q

= 0. On the other hand, by [11, Chapter 1, Theorem 1] we obtain i. e. V → γ       γ  (N),(C γ)  x  − (x  ∧ aqγ ) ∧ aqγ − (x γ  ∧ aqγ ) = 0 for all (m, k) ∈ N2 and mk mk mk mk mk mk  γ qγ γ qγ γ γ γ γ ∈ G. Setting a mk = amk − (xmk  ∧ amk ) if k  ∈ / Cm , a mk = 0 if k ∈ Cm for γ γ γ qγ each m ∈ N, γ ∈ G, and x mk = xmk − (xmk  ∧ amk ) for every (m, k) ∈ N2  γ   γ  → and γ ∈ G we have D := (x mk ) U = X, Aq = (a mk ) U and D = 0 γ γ

γ

γ

(Nγ ),( F )

(because for every m ∈ Nγ and k ∈ Fm one has x mk = 0 ⇐⇒ a mk = 0). Then  =         1 = ϕ1 V1 ∪ϕ1 V1 = ϕ1 V1 . So ϕ1 V1 → +V → → → → γ γ γ γ γ ),(C )  F ) (Nγ ),( F ) (Nγ ),(C ) (Nγ ),( F ) (Nγ ),(  (Nγ−1  by Corollary 18 and Lemma 17, part (c) one has ( q  ) (X) = ( q  )−1 (X) = 0. Then Aq is well defined and its continuity follows easily.     j   j n Lemma 20 If tr < 1 we have S (X(N\Pγ ),(N) )nj=1 = 0 and S (X −→ ) j=1 = (Pγ ),(N\Q γ )  0 for every (X j )nj=1 ∈ nj=1 X r∞ . j

Proof We only show the first since  assertion  the  proof of the second one is completely 1 analogous. (L , μ0 ) := ϕ1 V(N\P , μ 0 ⊂ 0 is a purely non atomic measure γ ),(N)

The minimal and maximal operator ideals associated to (n + 1)… 1 ) onto M and the purely space since ϕ1 is a Boolean algebra from  1 isomorphism   tC(V    r . Then B U(N\Pγ ),(N) ⊂ B Utr . atomic part of  is 1 = ϕ1 V → → γ γ (Pγ ),( Q ) (Nγ ),( M )   By (3) we have (I1,1 ⊗ Itr ,rn+1 )U B(Utr ) ⊂ B(Urn+1 ). So if there would exists  j n   (X j )nj=1 ∈ nj=1 X r∞ such that R := S X(N\Pγ ),(N) j=1 = 0 there would exists j      j η ∈ (X rn+1 ) such that R, η = 0 and Y := Sgγ U (X(N\Pγ ),(N) )nj=1 = 0. Jtr being j  an isomorphism, the map  : h ∈ L tr (L , μ0 ) −→ (I1,1 ⊗ Itr ,rn+1 )U ◦ J−1 tr (h), η    t is a continuous linear form  ∈ L r (L , μ0 ) = 0 (because (L , μ0 ) is a purely non atomic measurable space (the proof of this fact is a repetition of Day’s proof for L p ([0, 1]), 0 < p < 1 adapted to our setting using the isometric classification  of tr (by L p (L , μ0 ) for 0 < p < 1 given in [20, Theorem 0]). But Y ∈ B U(N\P γ ),(N)  Lemma 10) and (Jtr (Y)) = R, η = 0, a contradiction.  

Step 2. Construction of the measure spaces (R, M, μ) and (T, A, ν). Now we are ready to define our definitive measure spaces. If tr ≥ 1 we shall take (T, MT , ρ) := ( ∪ 1 , M ∪ M1 , ρ 0 + μ) and (R, AR , ν) := (0 ∪ 1 , A, ν0 + ν1 ), where the meaning of ρ 0 + μ and ν0 + ν1 is self-evident. However, if tr < 1 we shall consider (T, MT , ρ 0 ) := (1 , M1 , μ) and (R, AR , ν) := (1 , A, ν1 ). Jr0n+1 (resp. Jr1n+1 ) will be the natural embedding from L 1 [0 , ν0 , L rn+1 (, ρ 0 )] (resp. 1 [1 , ν1 , rn+1 (1 , μ)]) into L 1 [R, ν, L rn+1 (T, ρ)] and pq0 (resp. pq1 ) will denote the natural projection from L 1 [R, ν, L q (T, ρ)] onto L 1 [0 , ν0 , L q (, ρ 0 )] (resp. 1 [1 , ν1 , q (1 , μ)]) for each q > 0. Step 3. Definition of the linking maps. 2.a) Definition of Xr j , 1 ≤ j ≤ n. If tr < 1 $ 1 ◦K  ◦P 0 → . If t ≥ 1 we define X  = J ◦ A  ◦ K  ◦ we choose X  := X r j

rj

P

→

(Nγ ),( M γ )

rj

$ 1 ◦K  ◦ P + Jr1 ◦ X rj r j

j

r

(Pγ ),( Q γ ) →

(Pγ ),( Q γ )

rj

r j

rj

rj

.

2.b) Definition of g.  −1 n   1. Assume tr ≥ 1. Put Z := tr ◦ (S g γ )U ◦ ( r  ) j=1 and g0 (t, x) := j   Z (χ0 (t)χ (x))nj=1 (t, x). Given E ⊂ 0 , E ∈ Z and D ⊂ , D ∈   M we define v(t, x) := Z (χ E (t)χ D (x))nj=1 (t, x). By Lemmas 10, 16 and 17, part (b) we have v(t, x) = 0 and χ E (t)χ D (x)g  0 (t, x) = 0 almost everywhere on (0 × )\(E × D). But χ E (t)χ D (x) χ0 (t)χ (x) −  χ E (t)χ D (x) = 0. Hence, by Lemma 17, part (a) and 10 we obtain    n   γ j  γj  (S g γ )U r j χ0 (t)χ (x) − χ E (t)χ D (x) j=1 = (z mk ) U such that z mk = γ 0 for every k ∈ Dm , m ∈ E γ , γ ∈ G. It follows from Lemma 17, part (d)     n   that χ E (t)χ D (x) tr (S = 0, g γ )U r j χ0 (t)χ (x) − χ E (t)χ D (x) j=1 i.e. χ E (t)χ D (x)(g0 (t, x) − v(t, x)) = 0. Then finally we obtain v(t, x) = χ E (t)χ D (x)g0 (t, x).  As a consequence, given simple S j ∈ L ∞ (0 , ν0 )⊗L r j (, ρ 0 ), 1 ≤   j functions n j ≤ n it turns out that Z (S )nj=1 = g0 (t, x) j=1 S j (t, x). By continu-

J. A. López Molina   ε L r j (, ρ 0 ) we obtain ity of Z, for every (h j )nj=1 ∈ nj=1 L ∞ (0 , ν0 )⊗    Z (h j )nj=1 = g0 (t, x) nj=1 h j (t, x). It is known that in this setting we have necessarily g0 (t, x) ∈ L 1 [0 , ν0 , L r0 (, ρ 0 )] and hence the diagonal map   Sg0 from nj=1 L ∞ [0 , ν0 , L r j (, ρ 0 )] into L 1 [0 , ν0 , L tr (, ρ 0 )] is well defined. We denote by g0 the function defined in R × T by g0 (t, x) = g0 (t, x) if (t, x) ∈ 0 ×  and g0 (t, x) = 0 in other case. 2. If 0 < tr < ∞ let g1 : 1 × 1 −→ R be the function

 1− 1

g m  r0 χ (t) g1 (t, x) := χekm ,hv (x). e m k k

v∈Pm

m∈N

By (16) and (7) one has   g1 

L 1 [1 ,ν1 ,L r0 (T1 ,μ)]

=

  

 1− 1  r0 r10 0 γ g m  r0 lim gγm 1−r  g m v r k k k h 0 γ ,U

m∈N

≤

γ

γ

γ

 1− 1   0  γ r0  1 g m  r0 lim gγm 1−r g m  r r0 k k k r0 0 γ ,U

m∈N

=

v∈Pm γ

γ

 1− 1   1   g m  r0 g m  r0 ≤ g k k

m∈N

   and the diagonal map Sg1 ∈ Ln nj=1 ∞ [1 , ν1 , r j (1 , μ)], 1 1 , ν1 , tr  (1 , μ) is well defined. It is important to remark that as a consequence of    the definition of g1 actually we have Sg1 nj=1 ∞ [1 , ν1 , r j (1 , μ)] ⊂   Ptr 1 [1 , ν1 , tr (1 , μ)] . As above, we denote by g1 the function defined in R×T by g1 (t, x) = g1 (t, x) if (t, x) ∈ 1 × 1 and g1 (t, x) = 0 in other case. 3. As a consequence the final function g(t, x) ∈ L 1 [R, ν, L r0 (T, ρ)] we shall g0 (t, x) +  g1 (t, x) if consider is g(t, x) := g1 (t, x) if tr < 1 and g(t, x) :=  tr ≥ 1. 2.c) Definition of M. If tr < 1, we take M = Wrn+1 ◦ (I1,1 ⊗ Itr ,rn+1 )U ◦ W−1 tr . In ◦ p0tr + the case tr ≥ 1 we consider M := Jr0n+1 ◦ rn+1 ◦ (I1,1 ⊗ Itr ,rn+1 )U ◦ t−1 r 1 Jr1n+1 ◦ Wrn+1 ◦ (I1,1 ⊗ Itr ,rn+1 )U ◦ W−1 tr ◦ ptr . It follows easily from the definition of the involved maps and Lemmata  13 and 17 that the inclusion Supp(M( f )) ⊂ Supp( f ) holds for every f ∈ L 1 R, ν, L tr (T, ρ) . 2.d) Definition of C. If tr < 1 we define C := Wr−1 ◦ Prn+1 while, for tr ≥ 1 we n+1 0 −1 ◦ P 1 . consider C := r−1 ◦ p + W ◦ p rn+1 rn+1 rn+1 rn+1 n+1 Step 3. Commutativity of the diagram.

X

j

→

(N\Pγ ),( N )  j

S (X

+X

→

(Pγ ),( Q γ )

j

−→

n

∈ X r∞ one has X j = X → + j (Pγ ),( Q γ )   , 1 ≤ j ≤ n, and hence by Lemma 20 S (X j )nj=1 =

(a) Assume first tr < 1. For every (X j )nj=1 ∈ (Pγ ),(N\Q γ )  n ) j=1 . It is easy

j

j=1

to see that by Corollary 9, Lemmata 6 and 10 and

The minimal and maximal operator ideals associated to (n + 1)…

continuity it is enough to check the commutativity of (15) for elements X j = r j

r j

), 1 ≤ j ≤ n. In such a case consider the set F of  k ∈ all finite subsets of  := ϕ1 (V1 → ). If P ∈ F, we define F P := (Aγ ),( B γ )   N  ∃ h ∈ N such that (km , h) ∈ P and the sets Cγ = Aγ \ kγ , k ∈ FP and γ γ  Dkγ := Bkγ \ h vγ ,  (k, h) ∈ P , γ ∈ G. Keeping in mind (3) and (10) one has A

(Aγ

→ ),( B γ )

  r   j S A

(Aγ

∈ C(A

→

(Pγ ),( Q γ )

→ ),( B γ )

−

r

j Ak,h

    t = (I1,1 ⊗ Itr ,rn+1 )U (U r  j=1

n

(Cγ

(k,h)∈P

r  − r0 + r0  r r0  1− 0  γ  r0

 γ  tr rn+1  γ  tr0 − rn+1 γ gm gm  r rn+1 gmv  rn+1 = gmv r





0

 1− r0  r0

 γ  rn+1  γ  rn+1 gm ≤ gmv r0

0

γ

  )

γ

  

v∈Dm m∈Cγ U

    rn+1  = U



→ ),( D γ )

v∈Dm m∈Cγ U

→

(Aγ ),( B γ )

−

(25)  rn+1  Uk,h .

(k,h)∈P

(26) By Lemma 12, (10) and the definition of (Sgγ )U inequality (26) implies that  r S (A j

(Aγ

→ ),( B γ )

  r j n  )nj=1 = lim S Ak,h j=1 P∈F

(k,h)∈P

= lim

P∈F

(k,h)∈P

 r j n  S Ak,h j=1



= lim

P∈F

(k,h)∈P

 tr ). I1,1 ⊗ Itr ,rn+1 U (Uk,h

On the other hand, by (2) and (9) Sg1 ◦



r j

(Xr j (A

(Aγ

→ ),( B γ )

)

n



j=1

  r0 −1   γ −r0  1 − 1  tr r 0 gk  tr limg  χek (t) χek,h (x) = kγ r (k,h)∈

=

γ ,U

0

  r0 −1 − r0 +1 gk  tr tr χ (t) χ (x) ek ek,h

(k,h)∈

= lim

P∈F

 1− 1 gk  tr χ (t) χe (x) k,h ek

(k,h)∈P

  tr and the commutativity of (15) will be proved if we show that I1,1 ⊗Itr ,rn+1 U (Uk,h )   1− 1  = C ◦ M (gk  tr χek (t) χek,h (x) for every (k, h) ∈ . As for (m, v) ∈ P one has

J. A. López Molina

1− 1   C ◦ M gkm  tr χekm (t)χekm ,hv (x) ◦ Prn+1 ◦ Wrn+1 ◦ (I1,1 ⊗ Itr ,rn+1 )U ◦ = Wr−1 n+1 1− 1   × W−1 gkm  tr χe m (t)χe m v (x) tr

k ,h

k

= Wr−1 ◦ Prn+1 ◦ Wrn+1 ◦ (I1,1 ⊗ Itr ,rn+1 )U (Uktrm ,hv ) n+1 = (I1,1 ⊗ Itr ,rn+1 )U (Uktrm ,hv ), the proof is complete. (b) Let now tr ≥ 1. As  = 0 ∪ 1 and 0 ∩ 1 = ∅ it follows that Vtr =   1 tr tr ≤ g1− tr Vtr we deduce Utr = Utr V tr → +V → . As U → + (Nγ ),( M γ ) tr

(Pγ ),( Q γ )

U

→

(since by Lemma 4 the component U

U

→

is disjoint with Utr

(Pγ ),( Q γ ) tr

(Pγ ),( Q γ )  1− 1 tr   tr

→

(Nγ ),( M γ )

+ U tr

tr

→

(Cγ ),( D γ ) →

(Pγ ),( Q γ )

:=

U tr

(Nγ ),( M γ ) tr

−U

and verifies U

→

(Nγ ),( M γ ) tr →

(Cγ ),( D γ )

− ≤

 = 0). Then by Lemma 10 for every (X j )nj=1 ∈ nj=1 X r∞ j (Cγ   j   j   n n + S (X . Using the one has S (X j )nj=1 = S (X ) ) → → j=1 j=1 (Nγ ),( M γ ) (Pγ ),( Q γ )  1 same argumentation of part 3. (a) we obtain C ◦ M ◦ Sg1 Jr  ◦ Xr1 ◦ K r j ◦ j j  j    n j n → (X ) = S (X . By the properties of Z, Lemma 10 P ) → j=1 j=1 γ g

V

→ ),( D γ )

(Pγ ),( Q )

(Pγ ),( Q γ )

and the definition of the involved maps   j n → (X ) Jr0 ◦ Ar j ◦ K r j ◦ P j=1 (Nγ ),( M γ ) j n   0 j → (X ) = C ◦ M ◦ Z Jr  ◦ Ar j ◦ K r j (P j=1 γ (Nγ ),( M ) j  j  n = (I1,1 ⊗ Itr ,rn+1 )U ◦ (Sgγ )U (X → ) j=1

C ◦ M ◦ Sg0



(Nγ ),( M γ )

and the proof follows.

 

4 αrC -integral multilinear maps Let J F : F −→ F  be the canonical inclusion of a normed space F into F  . Definition 21 Given

Banach  spaces F and

E j , 1 ≤ j ≤ n, and a (n + n E , F is called αrC -integral if J F T ∈ j=1 j

2)-tuple r, a map T ∈ Ln     (αC ) E 1 , E 2 , . . . , E n , F  . ⊗ r

as definition of the αrC -integral norm The norm of J F T in this dual  space is taken

n C IαrC (T ) of a map T ∈ IαrC j=1 E j , F , the set of αr -integral multilinear maps n from j=1 E j into F. (IαrC , IαrC ) turns out to be the maximal ideal of multilinear

The minimal and maximal operator ideals associated to (n + 1)…

maps associated to the (n + 1)-tensor norm αrC in the sense of Defant, Floret and Hunfeld (see [4] and Theorem 4.5 in [7]). Theorem 22 Let (, A, ν) and (, M, μ) bemeasure spaces. If g ∈ L 1 [, L r0 ()] and M ∈ L L 1 [, L tr ()], L 1 [, L rn+1 ()] verify Supp(M( f )) ⊂ Supp( f ) for   every f ∈ L 1 [, L tr ()], the composition map T := M◦ Sg : nj=1 L ∞ [, L r j ()]    −→ L 1 [, L rn+1 ()] is αrC -integral and IαrC (T ) ≤ M g . Proof By Lemma 1 and the properties of ideals of operators we can assume that ν() < ∞ and μ() < ∞.        (a) Define E := ⊗nj=1 L ∞ () ⊗ε L r j () ⊗ L ∞ [, L rn+1 () , (αrC ) and let n  ∞ be the canonical linear form on E defined by the restriction to j=1 L () ⊗ε   L r j () of the map JL 1 [,L rn+1 ()] ◦ M ◦ Sg . First we shall prove that ∈ E  . j

j

For 1 ≤ j ≤ n let S (resp. S ) be the linear span of the set of characteristic  j j ε L r j (). functions in  (resp. in ). Clearly every S ⊗ S is dense in L ∞ ()⊗ By the density lemma for (n + 1)-tensor norms it will be enough to show that ∈  n   j j  ⊗ j=1 (S ⊗ S ) ⊗ L ∞ [, L rn+1 ()], (αrC ) . It ie easy to check that every z ∈  n  j j  ⊗ j=1 (S ⊗ S ) ⊗ L ∞ [, L rn+1 ()] can be written as

z :=

h

j

⊗n+1 j=1 Hk (t, x) :=

k=1

v h  u



j ⊗nj=1 χ Am (t) αkms χ Bs (x) ⊗ Hkn+1 (t, x) k=1

m=1

s=1

  where Am , 1 ≤ m ≤ v and Bs , 1 ≤ s ≤ u are finite families of pairwise disjoint j sets in A and M respectively and αkms ∈ R for all used indexes. By (3) we have     j n  n+1 j ⊗n+1 j=1 Hk = M Sg Hk j=1 , Hk % n & v u 



 n+1 j αkms M χ Am (t)χ Bs (x) g , Hk =

∀1≤k≤h

m=1 s=1

j=1

 n

j ) Hk dν dμ (   = M χ Am (t)χ Bs (x) g , Hkn+1  n ν(Am ) μ(Bs ) m=1 s=1 ⎛ ⎞  '  j u v n

 Hk dν dμ M χ Am (t)χ Bs (x) g , Hkn+1 A ×B m s ⎝ ⎠ = . 1 1 (ν(Am ) μ(Bs )) w m=1 s=1 j=1 (ν(Am ) μ(Bs )) r j u v

'

j=1 Am ×Bs

  So, if we put Ams := χ Am (t)χ Bs (x) g  L 1 [,L r0 ()] , 1 ≤ m ≤ v, 1 ≤ s ≤ u, one  has (z) = z, V , where we have defined

J. A. López Molina

V :=

u v

⎛ Ams ⎝⊗nj=1

m=1 s=1

⎞

χ Am (t)χ Bs (x) 1

⎠⊗

(ν(Am )μ(Bs )) r j

  M χ Am (t)χ Bs (x) g 1

Ams (ν(Am )μ(Bs )) w

1 rj ∈ ⊗n+1 j=1 L [, L ()]. 

Then, for every t ∈   there are vts (x), 1 ≤ s ≤ u in the closed unit ball of L r0 ()  u and (βs )s=1 verifying us=1 |βs |r0 ≤ 1 such that, by Fubini’s Theorem and Hölder’s Inequality  u 1 r0  r0 

    u  χ Am (t)χ Bs (x) g(t, x) 1  Ams )s=1  r = r 

0

=

L [,L 0 (0 )]

s=1 u





 g(t, x)vts (x) dμ(x) dν(t)

βs Am

s=1

 u

Am

s=1

 =

 ≤ Am

βs

g(t, x)vtk (x) dμ(x) dν(t) Bs

 u 

   s=1



Bs

r0  r10  g(t, x)vts (x) dμ(x)  dν(t) 

1  u  r0

r0  g(t, x) dμ(x) dν(t)

 ≤ 

Bs



Am

s=1

Bs

  g(t, x)

= Am

dν(t).

L r0 ()

 − 1 j 2. Put ϕms := ν(Am )μ(Bs ) r j χ Am (t)χ Bs (x)

∈ L 1 [, L r j ()], 1 ≤ j ≤ 

n, 1 ≤ s ≤ u and 1 ≤ m ≤ v. If K denotes the closed unit ball in L ∞ [, L r j ()], by the Bipolar Theorem and Hölder’s Inequality we obtain   ∀ 1 ≤ m ≤ v εr j (ϕms )us=1 ⎛ '

r  ⎞ r1 j j u f dν(t)dμ(x) ⎜ Am ×Bs ⎟ = sup ⎝ ⎠ r f ∈K

⎛ u ⎜ ≤ sup ⎝ f ∈K

j

s=1

(ν(Am )μ(Bs )) r j

'

Am ×Bs

s=1

(ν(Am )μ(Bs ))

 = sup

f ∈K





Am ×

f ∈K

| f |r j dν(t)dμ(x)



= sup

Am





| f |r j dν(t)dμ(x)



r j

(ν(Am )μ(Bs ))

r j rj

⎞ 1 ⎟ ⎠

rj

 1

   f (t, x)r j dμ(x)

rj



 1 dν(t)

rj

1 r

≤ ν(Am ) j .

rj

The minimal and maximal operator ideals associated to (n + 1)…

   3. Next, let W be the closed unit ball of L ∞ [, L rn+1 ()] = L 1 [, L rn+1 ()] . By the hypothesis about M, using (1), (2) and the generalized Hölder’s Inequality, as rn+1 > 1 we have  ∀1≤m≤v  = sup

f ∈W

 εrn+1

u



  u  A−1 ms M χ Am (t)χ Bs (x) g 1

(ν(Am )μ(Bs )) w

 ν(A j )μ(Bs )

r − n+1 w

−r  Amsn+1

(

s=1    ) rn+1 M χ Am (t)χ Bs (x) g , f



1  rn+1

s=1

⎛ u ⎜ = sup ⎝ f ∈W

s=1

⎛ ⎜ ≤ sup ⎜ ⎝ f ∈W

u

(

) r  ⎞ r  1   n+1 n+1 M χ Am (t)χ Bs (x) g , χ Am (t)χ Bs (x) f ⎟ ⎠     rn+1 rn+1 ν(A j )μ(Bs ) w Ams

      f M χ (t)χ (x) g  1 Am Bs L [,L tr ()]   ν(A j )μ(Bs )

s=1

⎛

 rn+1

 rn+1 w

 rn+1 r j

 n

r  L ∞ [Am ,L

 rn+1

n+1

(Bs )]

r

= ≤

 1

ν(Am ) w   M 1

ν(Am ) w

sup sup

t∈Am f ∈W

⎞ r 1



  sup  f 

f ∈W

u 

    f (t, x)rn+1 dμ(x) s=1

Bs

L ∞ [,L

 rn+1

()]

=

  M 1

ν(Am ) w

⎟ ⎟ ⎠

1  rn+1

n+1 Ams

 u A ⎟   ⎜   j=1 μ(Bs ) ⎜ ms  f (t, x)rn+1 dμ(x)⎟ ≤ sup sup M ⎜ ⎟  ⎝ ⎠  rn+1   f ∈W t∈Am s=1 ν(A )μ(B ) w Arn+1 Bs m s ms   M

⎞



n+1

1  rn+1

.

4. Finally we obtain αrC (V ) ≤

v

  (Ams )u 

 s=1 r0 εrn+1

n  j u   j  (ψhs )s=1 εr j (ϕhs )us=1

m=1

j=1

v n 1       r j − w1 ≤ M ν(Am ) ν(Am ) m=1

   = Mg  L 1 [,L r0 ()] .

j=1

  g(t, x) Am

L r0 ()

dν(t) (27)

To finish let M j , 1 ≤ j ≤ n + 1, be a finite dimensional subspace of   ε L r j () such that Hkj , 1 ≤ k ≤ h, 1 ≤ m ≤ v ⊂ M j and let L ∞ ()⊗ Q j : L 1 [, L r j ()] −→ M j = L 1 [, L r j ()]/M ⊥ j , 1 ≤ j ≤ n + 1 be the canon

   n+1 ical quotient maps. As z ∈ ⊗ j=1 M j and , z = z, V , by the Metric Mapping Property and (27) we have

J. A. López Molina

         , z  =  z, V  =  z, ⊗n+1 Q j (V )  j=1  n+1  n+1   C ≤ (αrC ) (z; ⊗n+1 j=1 M j ) αr ⊗ j=1 Q j (V ); ⊗ j=1 M j    C   C   g  (α ) (z; ⊗n+1 M j ≤ (αrC ) (z; ⊗n+1 r j=1 M j )αr (V ) ≤ M j=1        , z  ≤ M g  (α C ) (z), and (αrC ) being a finitely generated (n+1)-tensor norm r    1 

r j n ∞ r n+1  () . showing that ∈ IαrC j=1 L ()⊗ε L () , L , L   (b) There is a localizable measure space (, O, ω) such that L 1 (, ω) = L ∞ (, ν) = (L 1 (, ν)) isometrically. Since L r j (), 1 ≤ j ≤ n + 1, has the Radon–Nikodým property and the approximation property, if N denotes the    ε L r j ()  = operator ideal of nuclear linear mappings, one has L ∞ ()⊗  ∞     π L r j () = L 1 , I L (), L r j () = N L ∞ (), L r j () = L 1 ()⊗      ε L r j ()  = L ∞ , L r j () (Grothendieck’s Theorem). Hence L ∞ ()⊗      L r j () . Let us see that L ∞ , L r j () can be identified with a subspace of      L ∞ , L r j () . In fact, each ζ ∈ L 1 , η, L r j () can be written for every ε > ∞ 0 as ζ = k=1 ϕk (u) ⊗ f k (x) where every ϕk ∈ L 1 (), f k ∈ L r j (), k ∈ N and        ∞       fk  r j ≤ ζ  + ε. Then, given z(t, x) ∈ L ∞ , L r j () k=1 ϕk L 1 () L  ()  the linear form Hz on L 1 , L r j () given by the rule Hz (ζ ) =

∞



  ϕk , JL ∞ () z(t, x), f k (x)

k=1

is well defined and satisfies ∞     ϕk   Hz (ζ ) ≤

L 1 ()

k=1

  sup z(t, x) t∈

r

L j ()

   fk 

 r

L j ()

    ≤ z  ζ  + ε .

    ε L r j ()  and the map H : z −→ Hz satisfies  H  ≤ 1. Then Hz ∈ L ∞ ()⊗ Moreover, as JL 1 () is an order isomorphism, if for every ϕ(t) ∈ L 1 () and f (x) ∈ L r j () one has  0 = Hz , ϕ(t) f (x) =



 

ϕ(t)JL ∞ ()



 f (x)z(t, x) dμ(x)

dν(t)

 necessarily we obtain z(t, x), f (x) = 0 ν-almost everywhere on . Then z(t, x) = 0 μ-almost everywhere except for every t in a set of ν-measure 0. Then z = 0 and H turns out to be injective.     ε L r j ()) ⊗ L ∞ [, L rn+1 ()] , As a consequence, if G := ⊗nj=1 (L ∞ ()⊗     ∞ , L r j () ⊂ G. By the (αrC ) one has the set theoretic inclusions E ⊂ ⊗n+1 j=1 L Extension Lemma (see [4] §13.2 for n = 1, the proof for n > 1 following in an analogous way), (αrC ) being a finitely generated (n + 1)-tensor norm, for every η ∈ E 

The minimal and maximal operator ideals associated to (n + 1)…

    there exists the so called Arens Extension Z η ∈ G  verifying  Z η G  = ηE  . So we   can define a canonical map β : G −→ E  defining β(z), η = z, Z η for every z ∈ G and η ∈ E  . β is continuous since for every z ∈ G one has   β(z)

E 

        = sup  β(z), η , ηE  ≤ 1 = sup  z, Z η , ηE  ≤ 1       ≤ sup  z, ψ , ψ   ≤ 1 = z  . G

G

Hence, it follows from the Bipolar Theorem and the metric property of mappings for    ∞ , L r j () there is a net {z , η ∈ (n + 1)-tensor norms that, given z ∈ ⊗n+1 η j=1 L 

rj ∞ ∗ A} ⊂ ⊗n+1 j=1 (L () ⊗ε L ()) such that β(z) = lim η∈A z η in the weak topology    n+1 ∞    ∞   rj σ ⊗ j=1 L , L r j () , ⊗n+1 j=1 (L ()⊗ε L ()) and

          ∞ , L r j () . sup z η E ≤ β(z)E  ≤ z G ≤ (αrC ) z ⊗n+1 j=1 L

η∈A

(28)

   ∞ , L r j () defined by the Let now T be the canonical linear form on ⊗n+1 j=1 L 

rj n+1 ∞ map T = JL 1 [,L rn+1 ()] ◦ M ◦ Sg . Given ⊗n+1 j=1 w j ∈ ⊗ j=1 L [, L ()], for  j  ε L r j () such that every 1 ≤ j ≤ n + 1 there are nets wα j , α j ∈ A j ⊂ L ∞ ()⊗      j ε L r j ()  . Moreover, w j = w j = limα j ∈A j wα j in σ L ∞ [, L r j ()], L ∞ ()⊗          j limα j ∈A j wα j in σ L ∞ , L r j () , L ∞ [, L r j ()] because L ∞ () ⊗ε L r j () 

is a subspace of L ∞ [, L r j ()]. By part a) we have ∈ E  and so there exists Z ∈   G  . As T is σ (L ∞ [, L r j ()], L ∞ [, L r j ()] )-separately continuous (recall the result of Sect. 1), 1 ≤ j ≤ n + 1, we obtain   T ⊗n+1 j=1 w j =

  . . . lim T ⊗n+1 wαj j j=1 αn+1 ∈An+1 α1 ∈A1   = lim . . . lim ⊗n+1 wαj j j=1 αn+1 ∈An+1 α1 ∈A1   j = lim . . . lim Z ⊗n+1 j=1 wα j αn+1 ∈An+1 α1 ∈A1   = Z ⊗n+1 j=1 w j . lim



rj ∞ It follows that T (z) = Z (z) for every z ∈ ⊗n+1 j=1 L [, L ()]. Then by (28)

   z, T  =

       z, Z  = lim  z η , Z  = lim  z η ,  η∈A

η∈A

         ∞ ≤   z η E ≤   (αrC ) z, ⊗n+1 , L r j () j=1 L      ∞ , L r j () , (α C )  , finishing the proof. and it turns out that T ∈ ⊗n+1 r j=1 L We are ready to state the main result of the paper.

 

J. A. López Molina

Theorem 23 Let E j , 1 ≤ j ≤ n + 1, be Banach spaces and let T ∈ Ln ( E n+1 ).

n

E j,

j=1

(a) The following are equivalent: 1. T is αrC −integral. 2. There are measurable spaces (, A, ν) and (, M, μ), a function  g ∈ L 1 [, L r0 ()] and a map M ∈ L L 1 [, L tr ()], L 1 [, L rn+1 ()] verify  ing Supp M( f ) ⊂ Supp( f ) for each f ∈ L 1 [, L tr ()] such that J E n+1 ◦T can be factorized as n j=1

T

Ej

J E n+1

- E n+1

- E  n+1 6

(A j )nj=1 n j=1

L∞



C

? 

, L r j ()

 Sg

- L 1 [, L tr ()]

M

- L 1 [, L rn+1 ()]. (29)

3. There exists a factorization as in 2) but with finite measure spaces (, A, ν) and (, M, μ).        Moreover, IαrC (T ) = inf C  M g  nj=1  A j  taking the inf over all such possible factorizations of type used in 2) or 3). (b) If tr < 1, T is αrC −integral if and only if T is αrC −nuclear. Proof (a) 1) ⇒ 2). The proof can be done using standard methods with help of Theorems 3 and 11 (see for instance [12] for a detailed development of the used method in a similar framework). 2) ⇒ 3). Use Lemma 1. 3) ⇒ 1). The proof is immediate by Theorem 22 and the ideal properties of nlinear αrC -integral operators. The rest of the proof is standard using Theorems 11 and 22 and the used isometries  in the proof  of 2) ⇒ 3). (b) If tr < 1 and T ∈ IαrC nj=1 E j , F by application of the quoted standard arguments given in [12] we find that T has a factorization of the type used in Theorem 11 for a suitable ultrafilter the same notations of this the U. We retain    orem. We consider maps R j ∈ L ∞ [1 , ν1 , r j (1 , μ)], ∞ [r j ] , 1 ≤ j ≤ n,  and V ∈ L 1 [rn+1 ], 1 [1 , ν1 , rn+1 (1 , μ)] defined by 1    m  m r )(k,v)∈P km ∈ −→ (αmv μ(ekm ,hv ) j )v∈Pm km ∈ , 1 ≤ j ≤ n, R j : (αkv 1 1

  r0 −1   γ − r 0  r r V : ((αmv )) −→ αmv gkm  n+1 limgk m ,h v  n+1 χe m (t)χe m v (x).

(m,v)∈P

γ ,U

γ

γ

k

k ,h

(30) By (19) it is straightforward to check that R j , 1 ≤ j ≤ n, and V are isometries. 1   1 ∞  m 1− r0 μ(ekm ,hv ) r0 if Finally we define f := ((λmv )∞ v=1 )m=1 such that λmv = gk

The minimal and maximal operator ideals associated to (n + 1)…

  (m, v) ∈ P and λmv = 0 in other case. It turns out that ((λmv )) ∈ 1 r0 because for every (m, w) ∈ N2 , using (19) one has  γ r ⎞ 1 ⎛ r0 m  w m w g a i  0

 r0 r10  1− 1 kγ h γ λai  gka  r0 ⎝ ≤ lim  r0 −1 ⎠   γ ,U a=1 i=1 a=1 i=1 gka m

 γ  g a 

≤ lim γ ,U

kγ

a=1

r0

  ≤ g,

keeping in mind the existence of Um ∈ U such that {kγa , 1 ≤ a ≤ m} is a set of pairwise different elements.   mj  Then for every z j := (αkv )(k,v)∈P km ∈ ∈ ∞ [1 , ν1 , r j (1 , μ)], 1 ≤ j ≤ n, 1 by (2) we obtain  n  n  V ◦ (I1,1 ⊗ Itr ,rn+1 ) ◦ S f ◦ R j j=1 z j j=1 n

  1  1 mj  gkm 1− r0 αmv μ(ekm ,hv ) tr V(emv )

=

km ∈1

(31)

v∈Pm j=1

and n

  1   n  mj  gkm 1− r0 M ◦ Sg1 z j j=1 = αmv M(χe km ∈1

v∈Pm j=1

km

(t) χe

km ,hv

(x)). (32)

Note that by (19) and (30), for each (m, v) ∈ P one has μ(ekm ,hv )

1 tr

 (1−r0 ) V(emv )=gkm 



1 tr

−r

1 n+1



r   γ limgk m ,h v  0 γ ,U

γ



1 tr

−r

1 n+1

γ



χe

km

(t)χe

km ,hv

(x).

On the other hand, by (9), (19) and the definition of ultraproducts, as in (25) we obtain for every (m, h) ∈ P     γ − rtr0 + r r0  γ  r0 − r0 r (I1,1 ⊗ Itr ,rn+1 )U Uktrm ,hv = limgk m r0 n+1 gk m h v  tr rn+1 Ukn+1 m ,hv  − r0 + r0   γ = gkm  tr rn+1 limg m

γ

γ ,U

γ

k ,h

kγ h γ

γ ,U

γ

 r0 − r0 rn+1  tr rn+1 U m v v

and hence M(χe

km

(t) χe

km ,hv

 (1−r0 ) = gkm 



 1 −1   (x)) = Wrn+1 ◦ (I1,1 ⊗ Itr ,rn+1 )U gkm  tr Uktrm ,hv

1 tr

−r

1 n+1



  γ  rt0 − r r0  limgk m h v  r n+1 χe γ ,U

γ

γ

km

(t) χe

km ,hv

(x).

J. A. López Molina

After comparison of (31) and (32) the proof ends by Theorem 3.

 

References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

20. 21. 22.

Aliprantis, C.D., Burkinshaw, O.: Locally Solid Riesz Spaces. Academic, New York (1978) Aliprantis, C.D., Burkinshaw, O.: Positive Operators. Academic, New York (1985) Day, M.M.: Normed Linear Spaces. Springer, Berlin (1973) Defant, A., Floret, K.: Tensor Norms and Operator Ideals, North Holland Math. Studies 176. North Holland, Amsterdam (1993) Diestel, J., Jarchow, H., Tonge, A.: Absolutely Summing Operators, Cambridge Studies in Advanced Maths. 43. Camdridge University Press, Cambridge (1995) Floret, K.: Natural norms on symmetric tensor products of normed spaces. Note Mat. 17, 153–188 (1997) Floret, K., Hunfeld, S.: Ultrastability of ideals of homogeneous polynomials and multilinear mappings on Banach spaces. Proc. Am. Math. Soc. 130(5), 1425–1435 (2001) Haydon, R., Levy, M., Raynaud, Y.: Randomly Normed Spaces. Herman, París (1991) Heinrich, S.: Ultraproducts in Banach space theory. J. Reine Angew. Math. 313, 72–104 (1980) Kürsten, K.D.: Local Duality of Ultraproducts of Banach Lattices, Lecture Notes in Mathematics 991, pp. 137–142. Springer, Berlin (1983) Lacey, H.E.: The Isometric Theory of Banach Spaces. Springer, Berlin (1974) López Molina, J.A.: Multilinear operator ideals associated to Saphar type (n + 1)-tensor norms. Positivity 11, 95–117 (2007) López Molina, J.A.: (n + 1)-tensor norms of Lapresté’s type. Glasg. Math. J. 54, 665–692 (2012) López Molina, J.A., Puerta, M.E., Rivera, M.J.: Ultraproduts of real interpolation spaces between L p -spaces. Bull. Braz. Math. Soc. 37(2), 191–216 (2006) Maharam, D.: The representation of abstract measure functions. Trans. Am. Math. Soc. 65, 279–330 (1949) Maharam, D.: Decompositions of measure algebras and spaces. Trans. Am. Math. Soc. 69, 142–160 (1950) Mazur, S., Ulam, S.: Sur les transformations isométriques d’espaces vectoriels. C. R. Acad. Sci. Paris 194, 946–948 (1932) Michor, P.W.: Functors and Categories of Banach Spaces, Lecture Notes on Mathematics, 651. Springer, Berlin (1978) Pietsch, A.: Ideals of multilinear functionals (designs of a theory). In: Proceeding of the 2nd International Conference on Operator Algebras, Ideals and their applications in Theoretical Physics, Leipzig 1983, pp. 185–199. Teubner-Texte (1983) Popov, M.M.: An isomorphic classification of the L p (μ) for 0< p<1. J. Sov. Math. 48(6), 674–681 (1990) Schreiber, M.: Quelques remarques sur les charactérisations des spaces L p , 0 ≤ p<1. Ann. Inst. Henri Poincaré VIII(1), 83–92 (1972) Wnuk, W., Wiatrowski, B.: When are ultrapowers lattices discrete or continuous? In: Proceedings Positiviy IV—Theory and Applications, pp. 173–182, Dresden (2006)