aa r X i v : . [ m a t h . OA ] A ug TRACES ON SYMMETRICALLY NORMED OPERATOR IDEALS
F. SUKOCHEV AND D. ZANIN
We dedicate this paper to the memory of Nigel Kalton whose influence on us both has beenprofound. Without his collaboration this paper would never have been written.
Abstract.
For every symmetrically normed ideal E of compact operators, wegive a criterion for the existence of a continuous singular trace on E . We alsogive a criterion for the existence of a continuous singular trace on E whichrespects Hardy-Littlewood majorization. We prove that the class of all contin-uous singular traces on E is strictly wider than the class of continuous singulartraces which respect Hardy-Littlewood majorization. We establish a canonicalbijection between the set of all traces on E and the set of all symmetric func-tionals on the corresponding sequence ideal. Similar results are also proved inthe setting of semifinite von Neumann algebras. Introduction
In his groundbreaking paper [6], J. Dixmier proved the existence of positivesingular traces (that is, linear positive unitarily invariant functionals which vanishon all finite dimensional operators) on the algebra B ( H ) of all bounded linearoperators acting on infinite-dimensional separable Hilbert space H. Namely, if ψ : R + → R + is a concave increasing function such that(1) lim t →∞ ψ (2 t ) ψ ( t ) = 1 , then there is a singular trace τ ω , defined for every positive compact operator A ∈ B ( H ) by setting(2) τ ω ( A ) = ω ( 1 ψ ( n ) n X k =1 s k ( A )) . Here, { s k ( A ) } k ∈ N is the sequence of singular values of the compact operator A ∈ B ( H ) taken in the descending order and ω is an arbitrary dilation invariant gen-eralised limit on the algebra l ∞ of all bounded sequences. This trace is finiteon 0 ≤ A ∈ B ( H ) if and only if A belongs to the Marcinkiewicz ideal (see e.g.[14],[15],[27]) M ψ := { A ∈ B ( H ) : sup n ∈ N ψ ( n ) n X k =1 s k ( A ) < ∞} . Mathematics Subject Classification.
Key words and phrases.
Symmetric functionals, singular traces.
In [18], Dixmier’s result was extended to an arbitrary Marcinkiewicz ideal M ψ withthe following condition on ψ (3) lim inf t →∞ ψ (2 t ) ψ ( t ) = 1 . All the traces defined above by formula (2) vanish on the ideal L consisting ofall compact operators A ∈ B ( H ) such that P ∞ k =1 s k ( A ) < ∞ .An ideal E of algebra B ( H ) is said to be symmetrically normed if { s k ( B ) } k ∈ N ≤{ s k ( A ) } k ∈ N and A ∈ E implies that k B k E ≤ k A k E (see [14], [15], [29] , [28], [20]).Since the ideal M ψ is just a special example of symmetrically normed operatorideal, the following question (suggested in [18], [16], [17], [7]) arises naturally. Question 1.
Which symmetrically normed operator ideals admit a nontrivial sin-gular trace ? In analyzing Dixmier’s proof of the linearity of τ ω given by (1), it was observedin [18] (see also [3]) that τ ω possesses the following fundamental property, namelyif 0 ≤ A, B ∈ M ψ are such that(4) n X k =1 s k ( B ) ≤ n X k =1 s k ( A ) , ∀ n ∈ N , then τ ω ( B ) ≤ τ ω ( A ) . Such a class of traces was termed “fully symmetric”in [20],[30] (see also earlier papers [8],[25], where the term “symmetric”was used). It isnatural to consider such traces only on fully symmetrically normed operator ideals E (that is, on symmetrically normed operator ideals E satisfying the condition:if A, B satisfy (4) and A ∈ E , then B ∈ E and k B k E ≤ k A k E ). In fact, it wasestablished in [8] that every Marcinkiewicz ideal M ψ with ψ satisfying the condition(3) possesses fully symmetric traces. Furthermore, in the recent paper [18], thefollowing unexpected result was established. If ψ satisfies the condition (3), thenevery fully symmetric trace on M ψ is a Dixmier trace τ ω for some ω. The following question ( also suggested in [18], [7], [16], [17]) arises naturally.
Question 2.
Which fully symmetrically normed operator ideals admit a nontrivialsingular trace which is fully symmetric?
In papers [16],[17] the following two problems (closely related to Question 1 andQuestion 2) were also suggested.
Question 3.
Which fully symmetrically normed operator ideals admit a trace whichis not fully symmetric?
Let us fix an orthonormal basis { e n } n ∈ N in H. An operator A ∈ B ( H ) is calleddiagonal if ( Ae n , e m ) = 0 for every n = m. Question 4.
Let the mapping ϕ : E → C be unitarily invariant. Suppose that ϕ is linear on the subset of all diagonal operators from E . Does it imply that ϕ is atrace on E ? We have to caution the reader that in Theorem 1.16 of [29] the assertion ( b ) does not hold forthe norm of an arbitrary symmetrically normed ideal E (see e.g. corresponding counterexamplesin [19, p. 83]). In this paper, we exclusively deal with positive traces
RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 3
In some very special cases (for principal ideals contained in L , which are, strictlyspeaking, not symmetrically normed ideals), Question 3 was answered in the affir-mative in [33]. In [20], question 3 was answered in the affirmative for the specialcase of Marcinkiewicz ideals under the assumption (1). It should be pointed outthat the method used in [20] cannot be extended to an arbitrary Marcinkiewicz ideal M ψ and, furthermore, cannot be extended to a general symmetrically normed op-erator ideal. Question 4 was answered in [20] in full generality using deep resultsfrom [11, 10] (see also [9]).The following theorem is the main result of this paper. It yields answers toQuestions 1–3. In the course of the proof of Theorem 5, we also present a new (andvery simple) proof answering Question 4. Prior to stating Theorem 5, we make afew preliminary observations, for which we are grateful to the referee.Any trace ϕ : E → C obeys the condition1 m ϕ ( A ⊕ m ) = ϕ ( A ) , A ∈ E , m ≥ . Here, the direct sum A ⊕ m is formed with respect to some arbitrary Hilbert spaceisomorphism H ⊕ m ≃ H. Thus, traces are closely related to the following convex(see Lemma 11 below) functional on E .π : A → lim m →∞ m k A ⊕ m k E , A ∈ E . The non-triviality of the functional π : E → R is an obvious necessary condition forthe existence of a trace. Theorem 5.
Let E be a symmetrically normed operator ideal. Consider the fol-lowing conditions.(1) There exist nontrivial singular traces on E . (2) There exist nontrivial singular traces on E , which are fully symmetric.(3) There exist nontrivial singular traces on E , which are not fully symmetric.(4) E 6 = L and there exist an operator A ∈ E such that (5) lim m →∞ m k A ⊕ m k E > . (i) The conditions (1) and (4) are equivalent for every symmetrically normedoperator ideal E . (ii) The conditions (1) , (2) and (4) are equivalent for every fully symmetricallynormed operator ideal E . (iii) The conditions (1) − (4) are equivalent for every fully symmetrically normedoperator ideal E equipped with a Fatou norm. Recall that the norm on a symmetrically normed operator ideal E is called aFatou norm if the unit ball of E is closed with respect to strong (or, equivalently,weak) operator convergence. Observe that classical ideals (such as Schatten-vonNeumann ideals L p , Marcinkiewicz, Orlicz and Lorentz ideals [14], [15], [29]) havea Fatou norm. In fact, in some standard references on the subject (e.g. Simon’sbook [29]), the requirement that symmetrically normed operator ideal has a Fatounorm appears to be a part of the definition. Similarly, in the book [24], devotedto the study of symmetric function spaces (which are a commutative counterpart We are grateful to the referee for this remark. termed there “rearrangement invariant”. F. SUKOCHEV AND D. ZANIN of symmetrically normed operator ideals), an assumption that the norm is a Fatounorm is incorporated into the definition [24, p. 118].The proof of Theorem 5 is given in Section 7. In fact, in this paper we willprove a more general result for symmetric spaces associated with semifinite vonNeumann algebras. The precise statements are given in Section 4 (see Theorems23, 28, 29), Section 5 (see Theorems 33, 35, 36) and Section 6 (see Theorems 47,48). The appendix contains the proof of important technical results for which wewere unable to find a suitable reference. We also present a new and short proof ofthe Figiel-Kalton theorem from [13].Finally, we say a few words about our proof and its relation to the previousresults in the literature. Our strategy is based on the approach from recent papers[30] and [21], where condition (5) was connected to the geometry of E (see also [2]).The condition (5) is easy to verify in concrete situations. For example, the followingcorollary of Theorem 5 strengthens the main result of [20] and complements earlierresults of J. Varga [32]. Corollary 6.
Every Marcinkiewicz ideal M ψ with ψ satisfying the condition (3) admits a trace which is not fully symmetric. Indeed, it is proved in [1, Proposition 2.3] that the condition (4) of Theorem 5is equivalent to the condition (3) for the Marcinkiewicz ideal M ψ . Some examplesof symmetrically normed operator ideals, which are not Marcinkiewicz ideals, pos-sessing symmetric traces were presented in [7]. These results are also an immediatecorollary of Theorem 5.For completeness, we note that the assertion ( ii ) in Theorem 5 holds for a widerclass of relatively fully symmetrically normed operator ideals. The latter class isdefined as follows: if A, B ∈ E are such that (4) holds, then k B k E ≤ k A k E . Itcoincides with the class of all symmetrically normed subspaces of a fully symmetricoperator ideal (see [19])2.
Definitions and preliminaries
The theory of singular traces on symmetric operator ideals rests on some classicalanalysis which we now review for completeness.As usual, L ∞ (0 , ∞ ) is the set of all bounded Lebesgue measurable functions onthe semi-axis equipped with the uniform norm. Given a function x ∈ L ∞ (0 , ∞ ) , one defines its decreasing rearrangement t → µ ( t, x ) by the formula (see e.g. [22]) µ ( t, x ) = inf { s ≥ m ( { x > s } ) ≤ t } . Let H be a Hilbert space and let B ( H ) be the algebra of all bounded operators on H equipped with the uniform norm.Let M ⊂ B ( H ) be a semi-finite von Neumann algebra equipped with a fixedfaithful and normal semi-finite trace τ. M is said to be atomic (see [31, Definition5.9]) if every nonzero projection in M contains a nonzero minimal projection. M is said to be atomless if there is no minimal projections in M . For every A ∈ M , the generalised singular value function t → µ ( t, A ) is definedby the formula (see e.g. [12]) µ ( t, A ) = inf {k Ap k : τ (1 − p ) ≤ t } . If, in particular, M = B ( H ) , then µ ( A ) is a step function and, therefore, can beidentified with the sequence of singular numbers of the operators A (the singular RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 5 values are the eigenvalues of the operator | A | = ( A ∗ A ) / arranged with multiplicityin decreasing order).Equivalently, µ ( A ) can be defined in terms of the distribution function d A of A. That is, setting d A ( s ) = τ ( E | A | ( s, ∞ )) , s ≥ , we obtain µ ( t, A ) = inf { s ≥ d A ( s ) ≤ t } , t > . Here, E | A | denotes the spectral measure of the operator | A | . Using the Jordan decomposition, every operator A ∈ B ( H ) can be uniquelywritten as A = ( ℜ A ) + − ( ℜ A ) − + i ( ℑ A ) + − i ( ℑ A ) − . Here, ℜ A := 1 / A + A ∗ ) (respectively, ℑ A := 1 / i ( A − A ∗ )) for any operator A ∈ B ( H ) and B + = BE B (0 , ∞ ) ( respectively, B = − BE B ( −∞ , B ∈ B ( H ) . Recall that ℜ A, ℑ A ∈ M for every A ∈ M and B + , B − ∈ M for every self-adjoint B ∈ M . Further, we need to recall the important notion of Hardy–Littlewood majoriza-tion. Let
A, B ∈ ( L + L ∞ )( M ) . The operator B is said to be majorized by A andwritten B ≺≺ A if and only if Z t µ ( s, B ) ds ≤ Z t µ ( s, A ) ds, t ≥ . We have (see [12]) A + B ≺≺ µ ( A ) + µ ( B ) ≺≺ σ / µ ( A + B )for every positive operators A, B ∈ ( L + L ∞ )( M ) . If s > , the dilation operator σ s is defined by setting( σ s ( x ))( t ) = x ( ts ) , t > , , the operator σ s isdefined by ( σ s x )( t ) = ( x ( t/s ) , t ≤ min { , s } , s < t ≤ . Similarly, in the sequence case, we define an operator σ n by setting σ n ( a , a , · · · ) = ( a , · · · , a | {z } n times , a , · · · , a | {z } n times , · · · )and an operator σ / by setting σ / : ( a , a , a , a , · · · ) → ( a + a , a + a , · · · ) . Definition 7.
The Banach space E ( M , τ ) ⊂ ( L + L ∞ )( M ) is said to be a sym-metric operator space if the following conditions hold.(1) Given A ∈ E ( M , τ ) and B ∈ ( L + L ∞ )( M ) with µ ( B ) = µ ( A ) , we have B ∈ E ( M , τ ) and k B k E = k A k E . (2) Given ≤ A ∈ E ( M , τ ) and ≤ B ∈ ( L + L ∞ )( M ) with B ≤ A, we have B ∈ E ( M , τ ) and k B k E ≤ k A k E . F. SUKOCHEV AND D. ZANIN
The space E ( M , τ ) is called fully symmetric if for every A ∈ E ( M , τ ) and every B ∈ ( L + L ∞ )( M ) with B ≺≺ A, we have B ∈ E ( M , τ ) and k B k E ≤ k A k E . The norm on a symmetric space E ( M , τ ) is a Fatou norm if the unit ball of E ( M , τ ) is closed with respect to strong (or, equivalently, weak) operator conver-gence. Every symmetric space equipped with a Fatou norm is necessarily fullysymmetric.A linear functional ϕ : E ( M , τ ) → C is said to be symmetric if ϕ ( B ) = ϕ ( A )for every positive A, B ∈ E ( M , τ ) such that µ ( B ) = µ ( A ) . A linear functional ϕ : E ( M , τ ) → C is said to be fully symmetric if ϕ ( B ) ≤ ϕ ( A ) for every positive A, B ∈ E ( M , τ ) such that B ≺≺ A. Every fully symmetric functional is symmetricand bounded. The converse fails [20].A functional ϕ : E ( M , τ ) → C is called singular if ϕ = 0 on ( L ∩ L ∞ )( M ) . If E ( M , τ ) L ( M ) , then every symmetric functional is singular.If E = E (0 , ∞ ) and if ϕ : E → R is a symmetric functional, then sϕ ( x ) = ϕ ( σ s x )for every x ∈ E. If E = E (0 ,
1) and if ϕ : E → R is a singular symmetric functional,then sϕ ( x ) = ϕ ( σ s x ) for every x = µ ( x ) ∈ E. Let E be a fully symmetric Banach space either on the interval (0 ,
1) or on thesemi-axis. We need the notion of an expectation operator (see [2]).Let A = { A k } be a (finite or infinite) sequence of disjoint sets of finite measureand denote by A the collection of all such sequences. Denote by A ∞ the complementof ∪ k A k . The expectation operator E ( ·|A ) : L + L ∞ → L + L ∞ is defined by setting E ( x |A ) = X k m ( A k ) ( Z A k x ( s ) ds ) χ A k . Note that we do not require A ∞ to have finite measure.Every expectation operator is a contraction both in L and L ∞ . Therefore, E ( x |A ) ≺≺ x, x ∈ L + L ∞ . It follows that E ( ·|A ) is also contraction in E. It will be convenient to introduce the following notation. If A is a discretesubset of the semi-axis (i.e. a subset without limit points inside (0 , ∞ )), then theelements of A ∪ { } partition the semi-axis. This partition consists of a (finiteor infinite) sequence of sets of finite measure. We identify this partition with theset A . Elements of A will be called nodes of the partition A . The correspondingaveraging operator will be denoted by E ( ·|A ) . Let E be a symmetric Banach space either on the interval (0 ,
1) or on the semi-axis. Define the sets D E = Lin( { x ∈ E : x = µ ( x ) } ) = { µ ( a ) − µ ( b ) , a, b ∈ E } ,Z E = Lin( { x − x : 0 ≤ x , x ∈ E, µ ( x ) = µ ( x ) } ) . Let C be a Hardy operator defined by setting( Cx )( t ) = 1 t Z t x ( s ) ds. The following theorem was proved in [13]. For convenience of the reader, we givea new and simple proof in the appendix.
RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 7
Theorem 8.
Let E be a symmetric space on the semi-axis and let x ∈ D E . Wehave x ∈ Z E if and only if Cx ∈ E. A similar assertion is also valid for the interval (0 , provided that R x ( s ) ds = 0 . The following uniform submajorization was introduced by Kalton and Sukochevin [19].Let x, y ∈ L (0 ,
1) (or x, y ∈ ( L + L ∞ )(0 , ∞ )). We say that y ⊳ x if there exists m ∈ N such that(6) Z bma µ ( s, y ) ds ≤ Z ba µ ( s, x ) ds, ∀ ma ≤ b. Let x, y ∈ l ∞ . We say that y ⊳ x if there exists m ∈ N such that(7) b X k = ma +1 µ ( k, y ) ≤ b X k = a +1 µ ( k, x ) ∀ ma + 1 ≤ b. The following important theorem was proved in [19] (see Theorem 5.4 and The-orem 6.3 there).
Theorem 9.
Let x, y ∈ L (0 , or x, y ∈ ( L + L ∞ )(0 , ∞ ) or x, y ∈ l ∞ be suchthat y ⊳ x. For every ε > , the function (1 − ε ) y belongs to a convex hull of the set { z : µ ( z ) ≤ µ ( x ) } . This theorem led to the following fundamental result (see [19]).
Theorem 10.
Let E = E (0 , (or E = E (0 , ∞ ) or E = E ( N ) ) be a symmetricBanach space either on the interval (0 , or on the semi-axis or on N . It followsthat the corresponding set E ( M , τ ) is a symmetric Banach space. Also, the uniform submajorization permits us to prove the convexity of thefunctional π : E → R defined in Section 1. Lemma 11.
The functional π : E → R is convex on every symmetrically normedoperator ideal E . Proof.
Let E be the corresponding symmetrically normed ideal of l ∞ . For every
A, B ∈ E , it follows from Proposition 8.6 of [19] that µ ( A + B ) ⊳ µ ( A ) + µ ( B ) . Hence, σ m µ ( A + B ) ⊳ σ m ( µ ( A ) + µ ( B )) . By Theorem 9, we have k σ m µ ( A + B ) k E ≤ k σ m ( µ ( A ) + µ ( B )) k E ≤ k σ m µ ( A ) k E + k σ m µ ( B ) k E . Note that k A ⊕ m k E = k σ m µ ( A ) k E . Dividing by m and letting m → ∞ , we obtain π ( A + B ) ≤ π ( A ) + π ( B ) . (cid:3) Lifting of symmetric functionals
In this section, we explain a canonical bijection between symmetric functionalsand traces. In what follows, we require that a semifinite von Neumann algebra M be either atomless or atomic with traces of all atoms being 1 . For an atomless von Neumann algebra M , we have (see e.g. [12]) Z t µ ( s, A ) ds = sup { τ ( p | A | ) : p ∈ P ( M ) , τ ( p ) = t } , A ∈ M . F. SUKOCHEV AND D. ZANIN
For a atomic von Neumann algebra M , we have (see e.g. [12]) m X k =1 µ ( k, A ) = sup { τ ( p | A | ) : p ∈ P ( M ) , τ ( p ) = m } , A ∈ M . In either case, this implies a remarkable inequality (see e.g. [12])(8) µ ( A + B ) ≺≺ µ ( A ) + µ ( B ) ≺≺ σ / µ ( A + B ) , ≤ A, B ∈ ( L + L ∞ )( M ) . Lemma 12.
Let E = E (0 , (or E = E (0 , ∞ ) or E = E ( N ) ) be a symmetricBanach space either on the interval (0 , or on the semi-axis or on N . If x, y ∈ E + are such that y ⊳ x, then ϕ ( y ) ≤ ϕ ( x ) for every positive symmetric functional ϕ on E. Proof.
Fix ε > . By Theorem 9, there exist z k ∈ E, ≤ k ≤ n, and positivenumbers λ k , ≤ k ≤ n, such that µ ( z k ) ≤ µ ( x ) for every 1 ≤ k ≤ n and(1 − ε ) y = n X k =1 λ k z k , n X k =1 λ k = 1 . Since ϕ is positive and symmetric, it follows that ϕ ( z k ) ≤ ϕ ( | z k | ) = ϕ ( µ ( z k )) ≤ ϕ ( µ ( x )) = ϕ ( x ) . Therefore, (1 − ε ) ϕ ( y ) ≤ ϕ ( x ) . Since ε > (cid:3)
The following assertion is essentially known. However, we provide the full prooffor readers convenience.
Lemma 13.
Let M be a semifinite atomless von Neumann algebra and let A, B ∈ ( L + L ∞ )( M , τ ) be positive operators. Z b a µ ( s, A + B ) ds ≤ Z ba ( µ ( s, A ) + µ ( s, B )) ds, ∀ a ≤ b, Z b a ( µ ( s, A ) + µ ( s, B )) ds ≤ Z b a µ ( s, A + B ) ds, ∀ a ≤ b. Similar assertion is valid for atomic von Neumann algebra M . Proof.
Applying inequality (8) to the operators
A, B, we obtain that Z b µ ( s, A + B ) ds ≤ Z b ( µ ( s, A ) + µ ( s, B )) ds and Z a µ ( s, A + B ) ds ≥ Z a ( µ ( s, A ) + µ ( s, B )) ds. Subtracting this inequalities, we obtain Z b a µ ( s, A + B ) ds ≤ Z ba ( µ ( s, A ) + µ ( s, B )) ds. Proof of the second inequality is identical. (cid:3)
The following theorem answers Question 4 in the affirmative, as also does [20,Theorem 5.2]. The proof below is very simple and based on a completely differentapproach.
RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 9
Theorem 14.
Let E = E (0 , (or E = E (0 , ∞ ) or E = E ( N ) ) be a symmetricBanach space either on the interval (0 , or on the semi-axis or on N and let E ( M , τ ) be the corresponding symmetric Banach operator space.(1) If ϕ is a positive symmetric functional on E, then there exists a positivesymmetric functional L ( ϕ ) on E ( M , τ ) such that ϕ ( x ) = L ( ϕ )( A ) for allpositive x ∈ E and A ∈ E ( M , τ ) such that µ ( A ) = µ ( x ) . (2) If ϕ is a positive symmetric functional on E ( M , τ ) , then there exists apositive symmetric functional L − ( ϕ ) on E such that ϕ ( A ) = L − ( ϕ )( x ) for all positive x ∈ E and A ∈ E ( M , τ ) such that µ ( A ) = µ ( x ) . Proof.
We will only prove (1). Proof of (2) is identical.Let
A, B ∈ E + ( M , τ ) . It follows from Lemma 13 that µ ( A + B ) ⊳ µ ( A ) + µ ( B ) ⊳ σ / µ ( A + B ) . It follows from Lemma 12 that ϕ ( µ ( A + B )) ≤ ϕ ( µ ( A ) + µ ( B )) ≤ ϕ (2 σ / µ ( A + B )) = ϕ ( µ ( A + B )) . It follows that L ( ϕ ) is additive on E + ( M , τ ) . We than extend it to E ( M , τ ) bylinearity. (cid:3) Theorem 14 provides a very natural bijection between the set of all symmetricfunctionals on E and that on E ( M , τ ) , observed first for the case of fully symmetricfunctionals in [8]. Next corollary follows immediately. Corollary 15.
Let E and E ( M , τ ) be as in Theorem 14. The functional ϕ is fullysymmetric on E if and only if L ( ϕ ) is a fully symmetric functional on E ( M , τ ) . We also need a lifting between sequence and function spaces. The following spacewas introduced in [21].Let A = { [ n − , n ] } n ∈ N be a partition of the semi-axis. Clearly, E ( ·|A ) maps L + L ∞ into the set of step functions which can be identified with sequences. Proposition 16.
Let E be a symmetric Banach sequence space and let F be thelinear space of all such functions x ∈ L ∞ for which E ( µ ( x ) |A ) ∈ E. The space F equipped with the norm k x k F = k x k ∞ + k E ( µ ( x ) |A ) k E is a symmetric Banach function space. The fact that the space F is a Banach space is non-trivial. Proof of this fact wasmissing in both [19] and [21]. We include it in the appendix.Below, we assume that E is embedded into F. Theorem 17.
Let E = E ( N ) be a symmetric Banach sequence space and let F bethe corresponding function space.(1) If ϕ is a positive symmetric functional on E, then there exists a positivesymmetric functional L ( ϕ ) on F such that ϕ ( E ( µ ( x ) |A )) = L ( ϕ )( x ) for allpositive x ∈ F. (2) If ϕ is a positive symmetric functional on F, then its restriction on E is apositive symmetric functional. This restriction is an inverse operation forthe L in (1) . Proof.
Let us prove (1) ϕ ( σ / a ) = 1 / ϕ ( a , a , · · · ) + 1 / ϕ ( a , a , · · · ) == 1 / ϕ ( a , , a , , · · · ) + 1 / ϕ (0 , a , , a , · · · ) = 1 / ϕ ( a )for every a ∈ E. Let x, y ∈ F be positive. It follows from Lemma 50 that E ( µ ( x + y ) |A ) ⊳ E ( µ ( x ) + µ ( y ) |A ) ⊳ σ / E ( µ ( x + y ) |A ) . It follows from Lemma 12 that ϕ ( E ( µ ( x + y ) |A )) = ϕ ( E ( µ ( x ) + µ ( y ) |A ))and (1) follows.The first assertion of (2) is trivial. Clearly, µ ( x ) − E ( µ ( x ) |A ) ∈ ( L ∩ L ∞ )(0 , ∞ ) . If E = l , then ϕ ( y ) = 0 for every y ∈ ( L ∩ L ∞ )(0 , ∞ ) and every symmetricfunctional ϕ on F. If E = l , then F = ( L ∩ L ∞ )(0 , ∞ ) and the only symmetricfunctional on both spaces is an integral. The second assertion of (2) follows. (cid:3) Existence of symmetric functionals
In this section, we present results concerning existence of symmetric functionalson symmetric function spaces. The main results of this section are Theorem 23,Theorem 28 and Theorem 29.We need the following variation of the Hahn-Banach theorem.
Lemma 18.
Let E be a partially ordered linear space and let p : E → R be convexand monotone functional. For every x ∈ E, there exists a positive linear functional ϕ : E → R such that ϕ ≤ p and ϕ ( x ) = p ( x ) . Proof.
The existence of ϕ follows from the Hahn-Banach theorem. We only haveto prove that ϕ ≥ . If z ≥ , then ϕ ( x − z ) ≤ p ( x − z ) . Therefore, ϕ ( z ) ≥ ϕ ( x ) − p ( x − z ) = p ( x ) − p ( x − z ) ≥ z ≥ p is monotone. (cid:3) Define operators M m : ( L + L ∞ )(0 , ∞ ) → ( L + L ∞ )(0 , ∞ ) (or, M m : L (0 , → L (0 , M m x )( t ) = 1 t log( m ) Z tt/m x ( s ) , m ≥ . Lemma 19. If ≤ x ∈ L + L ∞ (or, ≤ x ∈ L (0 , ), then Z b/ma x ( s ) ds ≤ Z ba ( M m x )( s ) ds ≤ Z ba/m x ( s ) ds provided that ma ≤ b. In particular, m − σ m x ⊳ M m x ⊳ x provided that x = µ ( x ) . Proof.
Clearly, Z ba ( M m x )( s ) ds = 1log( m ) Z ba Z tt/m x ( s ) ds dtt == 1log( m ) Z ba/m Z min { ms,b } max { a,s } dtt x ( s ) ds = 1log( m ) Z ba/m x ( s ) log( min { ms, b } max { a, s } ) ds. RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 11
The integrand does not exceed x ( s ) log( m ) and the second inequality follows im-mediately. The integrand is positive and is equal to x ( s ) log( m ) for s ∈ ( a, b/m ) . The first inequality follows. (cid:3)
Corollary 20. If E is a symmetric Banach function space either on the interval (0 , or on the semi-axis, then M m : E → E is a contraction for m ∈ N . Proof.
Let x = µ ( x ) ∈ E. It follows from Lemma 19 that M m x ⊳ x. It follows fromtheorem 9 that, for every ε > , the function (1 − ε ) M m x belongs to a convex hullof the set { z : µ ( z ) ≤ µ ( x ) } . Therefore, M m x ∈ E and (1 − ε ) k M m x k E ≤ k x k E . Since ε is arbitrarily small, the assertion follows. (cid:3) Lemma 21.
Let E be a symmetric Banach space either on the interval (0 , oron the semi-axis. Let p : D E → R be convex and monotone functional. If p = 0 on Z E ∩ D E , then p extends to a convex monotone functional p : E → R by setting p ( x ) = p ( µ ( x + ) − µ ( x − )) . Also, p ( x ) = 0 for every x ∈ Z E . Proof. If x ∈ D E , then x − µ ( x + ) + µ ( x − ) ∈ Z E ∩ D E . Therefore, p ( x − µ ( x + ) + µ ( x − )) = 0 and, due to the convexity of p, p ( x ) = p ( µ ( x + ) − µ ( x − )) . This provesthe correctness of the definition.For x, y ∈ E, we have µ (( x + y ) + ) − µ (( x + y ) − ) − µ ( x + ) + µ ( x − ) − µ ( y + ) + µ ( y − ) ∈ Z E ∩ D E . It follows that p ( µ (( x + y ) + ) − µ (( x + y ) − ) − µ ( x + ) + µ ( x − ) − µ ( y + ) + µ ( y − )) = 0and p ( x + y ) = p ( µ (( x + y ) + ) − µ (( x + y ) − )) == p ( µ ( x + ) − µ ( x − ) + µ ( y + ) − µ ( y − )) ≤ p ( x ) + p ( y ) . Since p is monotone on D E , then p ( y ) ≤ ≥ y ∈ D E . It follows that p ( y ) = p ( − µ ( y )) ≤ ≥ y ∈ E. Therefore, p ( x + y ) ≤ p ( x ) + p ( y ) ≤ p ( x ) forevery 0 ≥ y ∈ E. (cid:3) Lemma 22.
Let E be a symmetric Banach space either on the interval (0 , oron the semi-axis. The functional p : x → lim sup m →∞ k ( M m x ) + k E , x ∈ D E satisfies the assumptions of Lemma 21. Also, for every x ∈ D E , we have p ( x ) ≤k x k E . Proof.
It follows from Corollary 20 that k ( M m x ) + k E ≤ k M m x k E ≤ k x k E , x ∈ E. It follows that p ( x ) = lim sup m →∞ k ( M m x ) + k E ≤ k x k E , x ∈ D E . Clearly, the mappings x → ( M m x ) + are convex and monotone. So are the mappings x → k ( M m x ) + k E . Therefore, p : D E → R is a convex and monotone functional. If x ∈ Z E ∩ D E , then by Theorem 8 | Cx | ∈ E. Therefore,( M m x )( t ) ≤ m ) ( | t Z t/m x ( s ) ds | + | t Z t x ( s ) ds | ) ≤≤ m ) ( 1 m σ m | Cx | + | Cx | )( t ) . Since k σ m k E → E ≤ m (see [22, Theorem II.4.5]), it follows that k ( M m x ) + k E ≤ m ) k Cx k E and p ( x ) = 0 . (cid:3) Theorem 23.
Let E = E (0 , ∞ ) be a symmetric Banach space on the semi-axis.For a given ≤ x ∈ E, there exists a symmetric linear functional ϕ : E → R suchthat ϕ ( x ) = lim m →∞ m k σ m ( µ ( x )) k E . Proof.
Without loss of generality, x = µ ( x ) . Let p be the convex monotone func-tional constructed in Lemma 22. It follows from Lemma 18 that there exist apositive linear functional ϕ on E such that ϕ ≤ p and ϕ ( x ) = p ( x ) . Since p ( z ) = 0for every z ∈ Z E , it follows that ϕ ( z ) = 0 for every z ∈ Z E . Therefore, ϕ is asymmetric functional.Since ϕ ( z ) ≤ p ( z ) ≤ k z k E for every z = µ ( z ) ∈ E, it follows that k ϕ k E ∗ ≤ . Therefore, ϕ ( x ) = ϕ ( 1 m σ m x ) ≤ m k σ m x k E . Passing m → ∞ , we obtain ϕ ( x ) ≤ lim m →∞ m k σ m µ ( x ) k E . On the other hand, It follows from Lemma 19 that m − σ m x ⊳ M m x. Therefore, p ( x ) = lim sup m →∞ k M m x k E ≥ lim m →∞ m k σ m µ ( x ) k E . The assertion follows immediately. (cid:3)
Consider the functional π : E → E (identical to the one defined in Section 1).(9) π ( x ) = lim m →∞ m k x ⊕ m k E , x ∈ E. Note that π ( − x ) = π ( x ) for every x ∈ E. If p is a functionals defined in Lemma 22,then p ( − x ) = 0 for positive x ∈ E. Therefore, p = π. However, the assertion belowfollows from Theorem 23.
Lemma 24.
Let E = E (0 , ∞ ) be a symmetric Banach space on the semi-axis. Let p and π be the convex functionals on E defined in Lemma 22 and (9) , respectively.For every positive x ∈ E, we have p ( x ) = π ( x ) . Proof.
For every x ∈ E, consider the functional ϕ constructed in Theorem 23. Byconstruction, we have ϕ ( x ) = p ( x ) = π ( x ) . (cid:3) If E L (0 , ∞ ) , then the functional ϕ constructed in Theorem 23 is necessarilysingular. The case E ⊂ L requires more detailed treatment. RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 13
Lemma 25.
Let E be a symmetric (respectively, fully symmetric) Banach functionspace either on the interval (0 , or on the semi-axis. Let { ϕ i } i ∈ I ∈ E ∗ be a netand let ϕ ∈ E ∗ be such that ϕ i → ϕ ∗− weakly.(1) If every ϕ i is symmetric, then ϕ is symmetric.(2) If every ϕ i is fully symmetric, then ϕ is fully symmetric.Proof. Let each ϕ i be symmetric. If 0 ≤ x , x ∈ E are such that µ ( x ) = µ ( x ) , then ϕ ( x ) = lim i ∈ I ϕ i ( x ) = lim i ∈ I ϕ i ( x ) = ϕ ( x ) . Hence, ϕ is symmetric.Let each ϕ i be fully symmetric. Thus, ϕ i ( x ) ≤ x ∈ D E such that Cx ≤ . Therefore, ϕ ( x ) = lim i ∈ I ϕ i ( x ) ≤ x ∈ D E such that Cx ≤ . Let x , x ∈ E be positive elements such that x ≺≺ x . Therefore, z = µ ( x ) − µ ( x ) ∈ D E and Cz ≤ . It follows from above that ϕ ( z ) ≤ . Hence, ϕ is a fullysymmetric functional. (cid:3) Lemma 26.
Let E be a symmetric (respectively, fully symmetric) Banach functionspace either on the interval (0 , or on the semi-axis and let ϕ be a symmetric(respectively, fully symmetric) functional on E. The formula ϕ sing ( x ) = lim n →∞ ϕ ( µ ( x ) χ (0 , /n ) ) , ≤ x ∈ E. defines a singular symmetric (respectivley, fully symmetric) linear functional on E. Proof. If x, y ∈ E are positive functions, then µ ( x + y ) χ (0 , /n ) ⊳ ( µ ( x ) + µ ( y )) χ (0 , /n ) ⊳ σ / µ ( x + y ) χ (0 , /n ) . Taking the limit as n → ∞ , we derive from Lemma 12 that ϕ sing ( µ ( x + y )) = ϕ sing ( µ ( x ) + µ ( y )) . Since ϕ is symmetric, it follows that ϕ sing ( x + y ) = ϕ sing ( µ ( x + y )) = ϕ sing ( µ ( x ) + µ ( y )) = ϕ sing ( x ) + ϕ sing ( y ) . Hence, ϕ sing is an additive functional on E + . Therefore, it extends to a linearfunctional on E. Clearly, ϕ sing is symmetric. Second assertion is trivial. (cid:3) In fact, the construction in Lemma 26 gives a singular part of the functional ϕ as defined by Yosida-Hewitt theorem. Lemma 27.
Let E = E (0 , ∞ ) ⊂ L (0 , ∞ ) be a symmetric Banach function spaceon the semi-axis and let ϕ be a symmetric functional on E. If ϕ sing is a functionalconstructed in Lemma 26, then ϕ − ϕ sing is a normal functional (that is, an integral).Proof. It is clear that0 ≤ ϕ sing ( z ) ≤ k z k ∞ lim n →∞ ϕ ( χ (0 , /n ) ) = 0for every positive z ∈ ( L ∩ L ∞ )(0 , ∞ ) . It follows that ϕ sing ( µ ( x ) χ (1 /n, ∞ ) ) = 0 forevery x ∈ E. Therefore,(10) ( ϕ − ϕ sing )( x ) = lim n →∞ ϕ ( µ ( x ) χ (1 /n, ∞ ) ) = lim n →∞ ( ϕ − ϕ sing )( µ ( x ) χ (1 /n, ∞ ) ) . On the other hand, for every positive z ∈ ( L ∩ L ∞ )(0 , ∞ ) with k z k ∞ = 1 , we have z ≺ χ (0 , k z k ) . It is proved in [30, Theorem 23] that z belongs to the closure (in thetopology of L ∩ L ∞ ) of the set { u ≥ µ ( u ) = χ (0 , k z k ) } . Thus,( ϕ − ϕ sing )( z ) = ( ϕ − ϕ sing )( χ (0 , k z k ) ) = k z k ( ϕ − ϕ sing )( χ (0 , ) . By linearity,(11) ( ϕ − ϕ sing )( z ) = ( ϕ − ϕ sing )( χ (0 , ) · Z ∞ z ( s ) ds, ∀ z ∈ ( L ∩ L ∞ )(0 , ∞ ) . It follows that from (10) and (11) that( ϕ − ϕ sing )( x ) = lim n →∞ Z ∞ /n µ ( s, x ) ds · ( ϕ − ϕ sing )(0 ,
1) = Z x ( s ) ds · ( ϕ − ϕ sing )( χ (0 , )for every positive function x ∈ E. The assertion follows immediately. (cid:3)
Theorem 28.
Let E ⊂ L (0 , ∞ ) be a symmetric Banach space on the semi-axis.For a given ≤ x ∈ E, there exists a singular symmetric linear functional ϕ sing such that ϕ sing ( x ) = lim m →∞ m k σ m ( µ ( x )) χ (0 , k E . Proof.
Apply Theorem 23 to the function µ ( x ) χ (0 , /n ) . It follows that there existsa symmetric linear functional ϕ n such that k ϕ n k E ∗ ≤ ϕ n ( µ ( x ) χ (0 , /n ) ) = lim m →∞ m k σ m ( µ ( x ) χ (0 , /n ) ) k E ≥ lim m →∞ m k σ m ( µ ( x )) χ (0 , k E . Since the unit ball in E ∗ is ∗− weakly compact (Banach-Alaoglu theorem), thereexists a convergent subnet ψ i = ϕ F ( i ) , i ∈ I , of the sequence ϕ n , n ∈ N . Let ψ i → ϕ. It follows from Lemma 25 that ϕ is a symmetric functional.By the definition of a subnet (see [26, Section IV.2]), for every fixed n ∈ N , thereexists i n ∈ I such that F ( i ) > n for every i > i n . Thus, for every i > i n , we have ψ i ( µ ( x ) χ (0 , /n ) ) ≥ ϕ F ( i ) ( µ ( x ) χ (0 , /F ( i )) ) ≥ lim m →∞ m k σ m ( µ ( x )) χ (0 , k E . The subnet ψ i , i n < i ∈ I converges to the same limit ϕ. Therefore, ϕ ( µ ( x ) χ (0 , /n ) ) ≥ lim m →∞ m k σ m ( µ ( x )) χ (0 , k E . Now, taking the limit as n → ∞ , we obtain the inequality ϕ sing ( x ) ≥ lim m →∞ m k σ m ( µ ( x )) χ (0 , k E , where ϕ sing is a singular symmetric functional defined in Lemma 26. The oppositeinequality is trivial. (cid:3) Theorem 29.
Let E be a symmetric Banach space on the interval (0 , . For agiven ≤ x ∈ E, there exists a singular symmetric linear functional ϕ sing suchthat ϕ sing ( x ) = lim m →∞ m k σ m ( µ ( x )) k E . RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 15
Proof.
Let F be a symmetric Banach space on the semi-axis with a norm given bythe formula k x k F = k µ ( x ) χ (0 , k E + k x k , ∀ x ∈ F. Clearly, F ⊂ L (0 , ∞ ) . Applying Theorem 28, we obtain a symmetric singularfunctional ϕ on F such that ϕ ( x ) = lim m →∞ m k σ m ( µ ( x )) χ (0 , k F = lim m →∞ m k σ m ( µ ( x )) k E . (cid:3) Existence of fully symmetric functionals
In this section, we present results concerning existence of fully symmetric func-tionals on fully symmetric function spaces. The main results of this section areTheorem 33, Theorem 35 and Theorem 36.
Lemma 30.
Let E be a symmetric Banach function space either on the interval (0 , or on the semi-axis. If x, z ∈ D E are such that Cx ≤ Cz, then CM m x ≤ CM m z. Proof.
Let x = µ ( a ) − µ ( b ) and z = µ ( c ) − µ ( d ) with a, b, c, d ∈ E. It followsfrom assumption Cx ≤ Cz that C ( µ ( a ) + µ ( d )) ≤ C ( µ ( b ) + µ ( c )) or, equivalently, µ ( a ) + µ ( d ) ≺≺ µ ( b ) + µ ( c ) . Arguing as in Lemma 19, we have Z t ( M m z )( s ) ds = Z t z ( s ) h ( s, t ) ds with h ( s, t ) = , ≤ s ≤ t/m log( t/s )log( m ) , t/m ≤ s ≤ t It is now clear that Z t M m ( µ ( a ) + µ ( d ))( s ) ds = Z t ( µ ( s, a ) + µ ( s, d )) h ( s, t ) ds, Z t M m ( µ ( b ) + µ ( c ))( s ) ds = Z t ( µ ( s, b ) + µ ( s, c )) h ( s, t ) ds. Clearly, h is positive and decreasing with respect to s. It follows from [22, Equality2.36] that M m ( µ ( a ) + µ ( d )) ≺≺ M m ( µ ( b ) + µ ( c ))and the assertion follows. (cid:3) Lemma 31.
Let E be a fully symmetric Banach function space either on the in-terval (0 , or on the semi-axis and let x = µ ( x ) ∈ E. If z ∈ D E is such that Cx ≤ Cz, then p ( x ) ≤ p ( z ) . Proof.
Since M m x is decreasing, it follows from Lemma 30 that Z t µ ( s, M m x ) ds = Z t ( M m x )( s ) ds ≤ Z t ( M m z ) + ( s ) ds ≤ Z t µ ( s, ( M m z ) + ) ds. Therefore, ( M m x ) + = M m x ≺≺ ( M m z ) + . The assertion follows now from thedefinition of the functional p. (cid:3) Lemma 32.
Let E be a fully symmetric Banach function space either on the in-terval (0 , or on the semi-axis. Let p be the functional constructed in Lemma 22.The functional q ( x ) = inf { p ( z ) : z ∈ D E , Cx ≤ Cz } , x ∈ D E satisfies the assumptions of Lemma 21.Proof. It is clear from the definition of q that q ≤ p and that q is a positivefunctional.We claim that q is convex on D E . Let x , x ∈ D E . Fix ε > z , z ∈D E such that Cx i ≤ Cz i and p ( z i ) ≤ q ( x i ) + ε for i = 1 , . Thus, C ( x + x ) ≤ C ( z + z ) and q ( x + x ) ≤ p ( z + z ) ≤ p ( z ) + p ( z ) ≤ q ( x ) + q ( x ) + 2 ε. Since ε is arbitrarily small, the claim follows.We claim that q is monotone on D E . Let x , x ∈ D E be such that x ≤ x . Fix ε > z ∈ D E such that Cx ≤ Cz and p ( z ) ≤ q ( x ) + ε. Thus, Cx ≤ Cx ≤ Cz and q ( x ) ≤ p ( z ) ≤ q ( x ) + ε. Since ε is arbitrarily small, theclaim follows.For x ∈ Z E ∩ D E , we have 0 ≤ q ( x ) ≤ p ( x ) = 0 and, therefore, q ( x ) = 0 . s (cid:3) The following theorem is the first main result of this section.
Theorem 33.
Let E = E (0 , ∞ ) be a fully symmetric Banach space on the semi-axis. For a given ≤ x ∈ E, there exists a fully symmetric linear functional ϕ : E → R such that ϕ ( x ) = lim m →∞ m k σ m ( µ ( x )) k E . Proof.
Without loss of generality, x = µ ( x ) . Let q be the convex monotone func-tional constructed in Lemma 32. It follows from Lemma 18 that there exist apositive linear functional ϕ on E such that ϕ ≤ q and ϕ ( x ) = q ( x ) . It is clear that ϕ ≤ q ≤ p. Since p ( z ) = 0 for every z ∈ Z E , it follows that ϕ ( z ) = 0 for every z ∈ Z E . Therefore, ϕ is a symmetric functional. For every z ∈ D E with Cz ≤ , we have ϕ ( z ) ≤ q ( z ) ≤ p (0) = 0 . Let x , x ∈ E be positive elements such that x ≺≺ x . Therefore, z = µ ( x ) − µ ( x ) ∈ D E and Cz ≤ . It follows from above that ϕ ( z ) ≤ . Hence, ϕ is a fullysymmetric functional.Since ϕ ( z ) ≤ q ( z ) ≤ p ( z ) ≤ k z k E for every z = µ ( z ) ∈ E, it follows that k ϕ k E ∗ ≤ . Therefore, ϕ ( x ) = ϕ ( 1 m σ m x ) ≤ m k σ m x k E . Passing m → ∞ , we obtain ϕ ( x ) ≤ lim m →∞ m k σ m µ ( x ) k E . On the other hand, q ( x ) = p ( x ) by Lemma 31. By Lemma 19, we have m − σ m x ⊳ M m x. Therefore, ϕ ( x ) = q ( x ) = p ( x ) = lim sup m →∞ k M m x k E ≥ lim m →∞ m k σ m µ ( x ) k E . The assertion follows immediately. (cid:3)
RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 17 If π : E → E is a convex functional defined in (9), then π ( − x ) = π ( x ) for every x ∈ E. If q is a functional defined in Lemma 32, then q ( − x ) = 0 for positive x ∈ E. Therefore, q = π. However, the assertion below follows from Theorem 33.
Lemma 34.
Let E = E (0 , ∞ ) be a fully symmetric Banach space on the semi-axis. Let q and π be the convex functionals on E defined in Lemma 32 and (9) ,respectively. For every positive x ∈ E, we have q ( x ) = π ( x ) . Proof.
For every x ∈ E, consider the functional ϕ constructed in Theorem 33. Byconstruction, we have ϕ ( x ) = q ( x ) = p ( x ) = π ( x ) . (cid:3) The proofs of the two following theorems are very similar to that of Theorem 28(respectively, Theorem 29) and are, therefore, omitted. The only difference is thatthe reference to Theorem 23 (respectively, Theorem 28) has to be replaced with thereference to Theorem 33 (respectively, Theorem 35).
Theorem 35.
Let E ⊂ L (0 , ∞ ) be a fully symmetric Banach space on the semi-axis. For a given ≤ x ∈ E, there exists a singular fully symmetric linear functional ϕ sing such that ϕ sing ( x ) = lim m →∞ m k σ m ( µ ( x )) χ (0 , k E . Theorem 36.
Let E ⊂ L (0 , be a fully symmetric Banach space on the inter-val (0 , . For a given ≤ x ∈ E, there exists a singular fully symmetric linearfunctional ϕ sing such that ϕ sing ( x ) = lim m →∞ m k σ m ( µ ( x )) k E . The sets of symmetric and fully symmetric functionals aredifferent
In this section, we demonstrate that the sets of symmetric and fully symmetricfunctionals on a given fully symmetric space E are distinct (provided that one ofthese sets is non-empty). The main results are Theorem 47 and Theorem 48.Let x = µ ( x ) ∈ ( L + L ∞ )(0 , ∞ ) (or x = µ ( x ) ∈ L (0 , X ( t ) = R t x ( s ) ds. For every θ > , let a n ( θ ) be such that X ( a n ( θ )) = (3 / n θ for every n ∈ Z such that a n ( θ ) does exist. Given a sequence κ = { κ n } n ∈ Z ∈ ( N ∪ {∞} ) Z , let B κ,θ = { κ n a n ( θ ) , where n ∈ Z is such that κ n a n ( θ ) < a n +1 ( θ ) } . If κ n = m for all n ∈ N , we write B m,θ instead of B κ,θ . Also, set A m = { ma n (1) : m a n (1) < a n +1 (1) , n ∈ Z } . Lemma 37. If x = µ ( x ) ∈ L + L ∞ and if C i , ≤ i ≤ k, are discrete sets, then E ( x | ∪ ki =1 C i ) ≺≺ k X i =1 E ( x |C i ) . Proof.
It is sufficient to verify Z t E ( x | ∪ ki =1 C i )( s ) ds ≤ k X i =1 Z t E ( x |C i )( s ) ds only at the nodes of E ( x | ∪ ki =1 C i ) , that is at the nodes of E ( x |C i ) for every i. However, if t ∈ C i for some i, then Z t E ( x | ∪ ki =1 C i )( s ) ds = X ( t ) = Z t E ( x |C i )( s ) ds and we are done. (cid:3) We will need the following lemma.
Lemma 38. If x = µ ( x ) ∈ L + L ∞ and if κ ≥ κ ′ (that is κ n ≥ κ ′ n for every n ),then (12) E ( x |B κ,θ ) ≺≺ E ( x |B κ ′ ,θ ) . Proof.
Let n ∈ Z be such that κ n a n ( θ ) < a n +1 ( θ ) . It follows that κ ′ n a n ( θ ) . (cid:3) Remark 39.
The inequality (12) holds if κ n ≥ κ ′ n only for such n ∈ Z that satisfythe inequality κ n a n ( θ ) < a n +1 ( θ ) . Lemma 40.
Let E be a fully symmetric Banach function space either on the in-terval (0 , or on the semi-axis. Let x = µ ( x ) ∈ E and y = µ ( y ) ∈ E be suchthat ϕ ( y ) ≤ ϕ ( x ) for every positive symmetric functional ϕ ∈ E ∗ . There exists ≤ u m ∈ E such that u m → in E and Z bma y ( s ) ds ≤ Z mba ( x + u m )( s ) ds, ∀ ma ≤ b. Proof.
Let p be a convex positive functional considered in Lemma 22. By Lemma18, there exists a positive functional ϕ ∈ E ∗ such that ϕ ≤ p and ϕ ( y − x ) = p ( y − x ) . We have p ( z ) = 0 for every z ∈ Z E and, therefore, ϕ ( z ) = 0 for every z ∈ Z E . Therefore, ϕ is a positive symmetric linear functional on E. By the assumption, ϕ ( y − x ) ≤ p ( y − x ) = 0 . Hence, by thedefinition of p, we have u m = ( M m ( y − x )) + → E. Clearly, M m y ≤ M m x + u m . It follows from Lemma 19 that Z bma y ( s ) ds ≤ Z mbma ( M m y )( s ) ds ≤ Z mbma ( M m x + u m )( s ) ds ≤ Z mba ( x + u m )( s ) ds. (cid:3) RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 19
For each sequence κ and λ > , we define the sequence κ λ by setting κ λn = ( κ n , κ n ≥ λ ∞ , κ n < λ. Lemma 41. If m ∈ N , x = µ ( x ) ∈ L + L ∞ and ≤ u ∈ L + L ∞ are such that Z bma E ( x |B κ,θ )( s ) ds ≤ Z mba ( x + u )( s ) ds, ∀ ma ≤ b ∈ R , then (14) m − σ m E ( x |B κ m ,θ ) ≺≺ µ ( u ) . Proof. If κ mn = ∞ for every n ∈ Z , then E ( x |B κ m ,θ ) = 0 and the assertion istrivial.Let n ∈ Z be such that κ n a n ( θ ) < a n +1 ( θ ) and κ n ≥ m. It follows that(15) Z mκ n a n ( θ )0 u ( s ) ds ≥ Z mκ n a n ( θ ) a n ( θ ) ( x + u )( s ) ds − Z mκ n a n ( θ ) a n ( θ ) x ( s ) ds. By the assumption, we have(16) Z mκ n a n ( θ ) a n ( θ ) ( x + u )( s ) ds ≥ Z κ n a n ( θ ) ma n ( θ ) E ( x |B κ,θ )( s ) ds. Note that mκ n a n ( θ ) < a n +1 ( θ ) . It follows from (15) and (16) that(17) Z mκ n a n ( θ )0 u ( s ) ds ≥ Z κ n a n ( θ ) ma n ( θ ) E ( x |B κ,θ )( s ) ds − Z a n +1 ( θ ) a n ( θ ) x ( s ) ds. Let n ′ be the maximal integer number such that n ′ < n and κ n ′ a n ′ ( θ ) < a n ′ +1 ( θ ) . It is clear that κ n ′ a n ′ ( θ ) < a n ′ +1 ( θ ) ≤ a n − ( θ ) < ma n ( θ )and(18) E ( x |B κ,θ ) = X ( κ n a n ( θ )) − X ( κ n ′ a n ′ ( θ )) κ n a n ( θ ) − κ n ′ a n ′ ( θ ) ≥ X ( a n ( θ )) − X ( a n − ( θ )) κ n a n ( θ )on the interval ( ma n ( θ ) , κ n a n ( θ )) . If κ n ′ a n ′ ( θ ) ≥ a n +1 ( θ ) for every n ′ < n, then(19) E ( x |B κ,θ ) = X ( κ n a n ( θ )) κ n a n ( θ ) ≥ X ( a n ( θ )) κ n a n ( θ )on the interval ( ma n ( θ ) , κ n a n ( θ )) . It follows from (17) and (18) (or (19)) that Z mκ n a n ( θ )0 u ( s ) ds ≥ κ n − mκ n · (1 −
49 ) X ( a n ( θ )) − X ( a n ( θ )) . Since κ n ≥ m, it follows that Z mκ n a n ( θ )0 u ( s ) ds ≥ ((1 − −
49 ) −
12 ) X ( a n ( θ )) = 120 X ( a n ( θ )) == 130 X ( a n +1 ( θ )) ≥ X ( κ n a n ( θ )) = 130 Z κ n a n ( θ )0 E ( x |B κ m ,θ )( s ) ds. It follows immediately that(20) Z t E ( x |B κ m ,θ )( s ) ds ≤ Z mt u ( s ) ds ≤ Z mt µ ( s, u ) ds for every t being a node of the partition B κ m ,θ . Therefore, Z t E ( x |B κ m ,θ )( s ) ds ≤ Z mt µ ( s, u ) ds, t > Z t/m E ( x |B κ m ,θ )( s ) ds ≤ Z t µ ( s, u ) ds, t > . The assertion follows immediately. (cid:3)
Lemma 42.
Let E be a fully symmetric Banach function space either on the in-terval (0 , or on the semi-axis. If x = µ ( x ) ∈ E is such that ϕ ( y ) ≤ ϕ ( x ) for every positive symmetric functional ϕ on E and every ≤ y ≺≺ x, then λ − σ λ E ( x |B κ λ ,θ ) → as λ → ∞ . Proof.
Since E ( x |B κ,θ ) ≺≺ x, it follows from the assumption and Lemma 40 thatthere exists 0 ≤ u m → Z bma E ( x |B κ,θ )( s ) ds ≤ Z mba ( x + u m )( s ) ds, ∀ ma ≤ b ∈ R . For every λ ≥ m, we have κ m ≤ κ λ . It follows from Lemma 41 that1 λ σ λ E ( x |B κ λ ,θ ) ≺≺ m σ m E ( x |B κ λ ,θ ) Lemma ≺≺ m σ m E ( x |B κ m ,θ ) (14) ≺≺ µ ( u m ) . The assertion now follows immediately. (cid:3)
Proposition 43.
Let E be a fully symmetric Banach function space either on theinterval (0 , or on the semi-axis equipped with a Fatou norm. If x = µ ( x ) ∈ E issuch that ϕ ( y ) ≤ ϕ ( x ) for every positive symmetric functional ϕ on E and every ≤ y ≺≺ x, then m − σ m E ( x |B m,θ ) → as m → ∞ . Proof.
For every m, r ∈ N , set κ m,rn = ( m ≤ | n | < r ∞ r ≤ | n | and κ m,r = { κ m,rn } n ∈ Z . Clearly, E ( x |B κ m,r ,θ ) → E ( x |B m,θ ) almost everywhere when r → ∞ . It follows from the definition of Fatou norm thatlim r →∞ k σ m E ( x |B κ m,r ,θ ) k E = k σ m E ( x |B m,θ ) k E . Select r m so large that(21) 1 m k σ m E ( x |B κ m,rm ,θ ) k E > m k σ m E ( x |B m,θ ) k E . Now define the sequence κ = { κ n } n ∈ Z by setting κ n = inf m ≥ κ m,r m n = inf r m > | n | m, n ∈ Z . RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 21
Clearly, r κ n ≥ | n | and, therefore, κ n → ∞ as | n | → ∞ . In particular, the set { n : κ n < λ } is finite for every λ ∈ N . Set M ( λ ) = max { λ, max κ n <λ ( a n +1 ( θ ) a n ( θ ) ) / } . If m > M ( λ ) , then m a n ( θ ) ≥ a n +1 ( θ ) whenever κ n < λ. Thus, κ n ≥ λ when-ever m a n ( θ ) < a n +1 ( θ ) . Hence, κ λn = κ n whenever ( κ m,r m n ) a n ( θ ) < a n +1 ( θ ) . Therefore, κ λn ≤ κ m,r m n for every n ∈ Z such that ( κ m,r m n ) a n ( θ ) < a n +1 ( θ ) . According to Remark 39, it follows that E ( x |B κ m,rm ,θ ) ≺≺ E ( x |B κ λ ,θ ) . Since m ≥ λ, it follows that(22) 1 m σ m E ( x |B κ m,rm ,θ ) ≺≺ λ σ λ E ( x |B κ λ ,θ ) . By Lemma 42, for every ε > , there exists λ such that(23) 1 λ k σ λ E ( x |B κ λ ,θ ) k E < ε. It follows that1 m k σ m E ( x |B m,θ ) k E (21) ≤ m k σ m E ( x |B κ m,rm ,θ ) k E (22) ≤ λ k σ λ E ( x |B κ λ ,θ ) k E (23) < ε for every m > M ( λ ) . Since ε > (cid:3)
Lemma 44.
Let E be a fully symmetric Banach space either on the interval (0 , or on the semi-axis equipped with a Fatou norm. If x = µ ( x ) ∈ E is such that ϕ ( y ) ≤ ϕ ( x ) for every positive symmetric functional ϕ on E and every ≤ y ≺≺ x, then m − σ m E ( x |A m ) → as m → ∞ . Proof.
It is clear that a k (3 /
2) = a k +1 (1) and a k ((3 / ) = a k +2 (1) for every k ∈ N . It follows that B m, ∪ B m, / ∪ B m, (3 / = A m . Therefore, by Lemma 37, we have(24) E ( x |A m ) ≺≺ E ( x |B m, ) + E ( x |B m, / ) + E ( x |B m, (3 / ) . The assertion follows now from Proposition 43. (cid:3)
Lemma 45.
Let x = µ ( x ) ∈ L + L ∞ (0 , ∞ ) be a function on the semi-axis. If x / ∈ L (0 , ∞ ) , then, for every t > and every m ∈ N , we have (25) X ( t ) ≤ X ( m t ) + 32 Z m t E ( x |A m )( s ) ds. Proof.
For a given t > , there exists n ∈ Z such that t ∈ [ a n (1) , a n +1 (1)] . If a n +1 (1) > m a n (1) , then Z m t E ( x |A m )( s ) ds ≥ Z ma n (1)0 E ( x |A m )( s ) ds = X ( ma n (1)) ≥ X ( t ) . If a n +1 (1) ≤ m a n (1) and a n +2 (1) > m a n +1 (1) , then Z m t E ( x |A m )( s ) ds ≥ Z ma n +1 (1)0 E ( x |A m )( s ) ds = X ( ma n +1 (1)) ≥ X ( t ) . If a n +2 (1) ≤ m a n +1 (1) and a n +1 (1) ≤ m a n (1) , then X ( m t ) ≥ X ( a n +2 (1)) = 32 X ( a n +1 (1)) ≥ X ( t )and the assertion follows. (cid:3) The situation in the case that x ∈ L is slightly more complicated. Lemma 46. If x = µ ( x ) ∈ L (0 , or x ∈ L (0 , ∞ ) , then there exists constant C such that for every t > X ( t ) ≤ X ( m t ) + 32 Z m t E ( x |A m )( s ) ds + C Z m t χ [0 , ( s ) ds. Proof.
Consider first the case of the semi-axis. Fix n such that X ( a n ) ≤ / X ( ∞ ) . For a givne t ∈ [ a, a n ] , there exists n ∈ Z such that n < n and t ∈ [ a n , a n +1 ] . Then, the argument in Lemma 45 applies mutatis mutandi . For every t ≥ a n wehave X ( t ) ≤ X ( ∞ )min { a n , } min { m t, } = X ( ∞ )min { a n , } Z m t χ [0 , ( s ) ds. Setting C = X ( ∞ ) / min { a n , } , we obtain the assertion.The same argument applies in the case of the interval (0 ,
1) by replacing X ( ∞ )by X (1) . (cid:3) The following two theorems are crucial for the proof of the implication (3) ⇔ (4)in Theorem 5. Theorem 47.
Let E be a fully symmetric Banach space either on the interval (0 , or on the semi-axis and let x ∈ E. Suppose that the norm on E is a Fatou norm. If ϕ ( y ) ≤ ϕ ( x ) for every positive symmetric functional on E and every ≤ y ≺≺ x, then (27) lim m →∞ m k σ m ( µ ( x )) k E = 0 provided that one of the following conditions is satisfied(1) E = E (0 , is a space on the interval (0 , . (2) E = E (0 , ∞ ) is a space on the semi-axis and E (0 , ∞ ) L (0 , ∞ ) . Proof.
Without loss of generality, x = µ ( x ) . If x / ∈ L , then by Lemma 45, Z t/m x ( s ) ds ≤ Z t x ( s ) ds + 32 Z t E ( x |A m )( s ) ds, ∀ t > m σ m x ≺≺ x + 32 E ( x |A m ) . Applying m − σ m to the both parts, we obtain1 m σ m x ≺≺
23 1 m σ m x + 32 1 m σ m E ( x |A m ) . Take norms and let m → ∞ . It follows from Lemma 44 thatlim m →∞ m k σ m x k E ≤
23 lim m →∞ m k σ m x k E . This proves (27).
RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 23 If x ∈ L and C are as in Lemma 46, then it follows from Lemma 46 that Z t/m x ( s ) ds ≤ Z t x ( s ) ds + 32 Z t E ( x |A m )( s ) ds + C Z t χ [0 , ( s ) ds, ∀ t > m σ m x ≺≺ x + 32 E ( x |A m ) + Cχ (0 , . Applying m − σ m to the both parts, we obtain1 m σ m x ≺≺
23 1 m σ m x + 32 1 m σ m E ( x |A m ) + C m σ m χ (0 , . Take norms and let m → ∞ . For every symmetric space E on the interval (0 , E on the semi-axis such that E L (0 , ∞ ) we have m − k σ m χ (0 , k E → . It follows from Lemma 44 thatlim m →∞ m k σ m x k E ≤
23 lim m →∞ m k σ m x k E and again (27) follows. (cid:3) Theorem 48.
Let E = E (0 , ∞ ) be a fully symmetric Banach space on the semi-axis equipped with a Fatou norm such that E (0 , ∞ ) ⊂ L (0 , ∞ ) . If ϕ ( y ) ≤ ϕ ( x ) forevery positive symmetric functional on E and every ≤ y ≺≺ x, then (28) lim m →∞ m k σ m ( µ ( x )) χ (0 , k E = 0 . Proof.
Fully symmetric Banach space F on the interval (0 ,
1) consists of those z ∈ E supported on the interval (0 , . Let x = µ ( x ) χ (0 , ∈ F. Suppose thatlim m →∞ m k σ m ( µ ( x )) χ (0 , k E > . It clearly follows that lim m →∞ m k σ m ( µ ( x )) k F > . By Theorem 47, there exists 0 ≤ y ≺≺ x and a positive symmetric functional ϕ ∈ F ∗ such that ϕ ( y ) > ϕ ( x ) . Let ϕ sing be a singular part of the functional ϕ constructed in Lemma 26. It follows from Lemma 26 that ϕ sing is symmetric. ByLemma 27, the difference ϕ − ϕ sing is a symmetric normal functional on F (thatis, an integral). Therefore, ϕ sing ( y ) > ϕ sing ( x ) . Now we show that the functional ϕ sing can be extended from F to E by setting ϕ sing ( z ) = lim n →∞ ϕ sing ( µ ( z ) χ (0 , /n ) ) , ≤ z ∈ E. Repeating the argument in Lemma 26, we prove that the extension above is additiveon E + . Thus, the functional ϕ sing ∈ E ∗ is positive and symmetric. Since y ≺≺ x and ϕ sing ( y ) > ϕ sing ( x ) = ϕ sing ( x ) , the assertion follows. (cid:3) Proof of Theorem 5
In this section, we prove an assertion more general then that of Theorem 5. Theassertion of Theorem 5 follows from that of Theorem 49 by setting M = B ( H ) . In what follows, the semifinite von Neumann algebra M is either atomless oratomic so that the trace of every atom is 1 . Theorem 49.
Let E ( M , τ ) be a symmetric operator space. Consider the followingconditions.(1) There exist nontrivial positive singular symmetric functionals on E ( M , τ ) . (2) There exist nontrivial singular fully symmetric functionals on E ( M , τ ) . (3) There exist positive symmetric symmetric functional on E ( M , τ ) which arenot fully symmetric.(4) If E ( M , τ ) L ( M , τ ) , then there exists an operator A ∈ E ( M , τ ) suchthat (29) lim m →∞ m k σ m µ ( A ) k E > . If E ( M , τ ) ⊂ L ( M , τ ) , then there exists an operator A ∈ E ( M , τ ) suchthat (30) lim m →∞ m k ( σ m µ ( A )) χ (0 , k E > . (i) The conditions (1) and (4) are equivalent for every symmetric operator space E ( M , τ ) . (ii) The conditions (1) , (2) and (4) are equivalent for every fully symmetric op-erator space E ( M , τ ) . (iii) The conditions (1) - (4) are equivalent for every fully symmetric operator space E ( M , τ ) equipped with a Fatou norm.Proof. Implications (2) ⇒ (1) and (3) ⇒ (1) are trivial.(1) ⇒ (4) Let E ( M , τ ) be a symmetric operator space with a singular symmetricfunctional ϕ. Let A ∈ E ( M , τ ) be an operator such that ϕ ( A ) = 0 . Without lossof generality, A ≥ . If E ( M , τ ) L ( M , τ ) , then | ϕ ( A ) | = 1 m | ϕ ( A ⊕ · · · ⊕ A | {z } m times ) | ≤ k ϕ k E ∗ ( M ,τ ) · m k σ m µ ( A ) k E . Passing m → ∞ , we obtain the required inequality (29).Let now E ( M , τ ) ⊂ L ( M , τ ) . If M is atomic, then E ( M , τ ) = L ( M , τ ) andthe assertion is trivial. Let M be atomless. Since ϕ is a singular functional and A − AE A ( µ ( 1 m , A ) , ∞ ) ∈ ( L ∩ L ∞ )( M , τ ) , ∀ m ∈ N , we infer that | ϕ ( A ) | = | ϕ ( AE A ( µ ( 1 m , A ) , ∞ )) | == 1 m | ϕ ( AE A ( µ ( 1 m , A ) , ∞ ) ⊕ · · · ⊕ AE A ( µ ( 1 m , A ) , ∞ ) | {z } m times ) | ≤≤ k ϕ k E ∗ ( M ,τ ) · m k σ m µ ( AE A ( µ ( 1 m , A ) , ∞ )) k E ≤ k ϕ k E ∗ ( M ,τ ) · m k ( σ m µ ( A )) χ (0 , k E . RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 25
Passing m → ∞ , we obtain the required inequality (30).(4) ⇒ (1) Firstly, we assume that the algebra M is finite. Without loss ofgenerality, τ (1) = 1 . Let E ( M , τ ) be a symmetric operator space and let E (0 , x = µ ( A ) ∈ E (0 ,
1) such that m − σ m x E (0 , . By Theorem 29,there exists a positive singular symmetric functional 0 = ϕ ∈ E (0 , ∗ . Let L ( ϕ )be a functional on E ( M , τ ) defined in Theorem 14. Clearly, L ( ϕ ) is a nontrivialpositive symmetric functional on E ( M , τ ) . The case when M is an infinite atomless von Neumann algebra can be treatedin a similar manner. The only difference is that the reference to Theorem 29 hasto be replaced with the reference to either Theorem 28 or Theorem 23.Let E ( M , τ ) be a symmetric operator space on a atomic von Neumann algebra M and let E ( N ) be the corresponding symmetric sequence space. It follows fromthe assumption that E ( M , τ ) = L ( M , τ ) or, equivalently, E ( N ) = l . By theassumption, there exists an element x = µ ( A ) ∈ E such that m − σ m x E. Let F (0 , ∞ ) be a symmetric function space constructed in Proposition 16. Since E ( N ) = l , it follows that F (0 , ∞ ) L (0 , ∞ ) . Recall that the space E ( N ) isnaturally embedded into the space F (0 , ∞ ) and that the norms k · k E and k · k F areequivalent on E ( N ) . We have x ∈ F and m − σ m x F (0 , ∞ ) . By Theorem23, there exists a positive symmetric functional 0 ≤ ϕ ∈ F (0 , ∞ ) ∗ . The restrictionof the functional ϕ to E ( N ) is a nontrivial positive symmetric functional on E ( N ) . Let L ( ϕ ) be a functional on E ( M , τ ) defined in Theorem 14. Clearly, L ( ϕ ) is anontrivial positive symmetric functional on E ( M , τ ) . (4) ⇒ (2) The proof is very similar to that of the implication (4) ⇒ (1) and is,therefore, omitted. The only difference is that references to Theorem 29, Theorem28 or Theorem 23 have to be replaced with references to Theorem 36, Theorem 35or Theorem 33, respectively.(4) ⇒ (3) Firstly, we assume that the algebra M is finite. Without loss ofgenerality, τ (1) = 1 . Let E ( M , τ ) be a symmetric operator space and let E (0 , x = µ ( A ) ∈ E (0 ,
1) such that m − σ m x E (0 , . By Theorem 47,there exists a positive symmetric but not fully symmetric functional ϕ ∈ E (0 , ∗ . Let L ( ϕ ) be a functional on E ( M , τ ) defined in Theorem 14. Clearly, L ( ϕ ) is asymmetric but not fully symmetric functional on E ( M , τ ) . The case when M is an infinite atomless von Neumann algebra can be treatedin a similar manner. The only difference is that the reference to Theorem 47 hasto be replaced with the reference to either Theorem 47 or Theorem 48.Let E ( M , τ ) be a symmetric operator space on a atomic von Neumann algebra M and let E ( N ) be the corresponding symmetric sequence space. It follows fromthe assumption that E ( M , τ ) = L ( M , τ ) or, equivalently, E ( N ) = l . By theassumption, there exists an element x = µ ( A ) ∈ E such that m − σ m x E. Let F (0 , ∞ ) be a symmetric function space constructed in Proposition 16. Since E ( N ) = l , it follows that F (0 , ∞ ) L (0 , ∞ ) . Recall that the space E ( N ) isnaturally embedded into the space F (0 , ∞ ) and that the norms k · k E and k · k F areequivalent on E ( N ) . We have x ∈ F and m − σ m x F (0 , ∞ ) . By Theorem 47,there exists a positive symmetric functional ϕ ∈ F (0 , ∞ ) ∗ and a function 0 ≤ y ≺≺ x such that ϕ ( y ) > ϕ ( x ) . Set z = E ( µ ( y ) |{ ( n − , n ) } n ∈ N ) . Clearly, z ∈ E ( N ) and ϕ ( z ) = ϕ ( y ) > ϕ ( x ) . Hence, the restriction of the functional ϕ to E ( N ) is a positive symmetric but not fully symmetric functional on E ( N ) . Let L ( ϕ ) be a functionalon E ( M , τ ) defined in Theorem 14. Clearly, L ( ϕ ) is a positive symmetric but notfully symmetric functional on E ( M , τ ) . (cid:3) Appendix
In this appendix, we set A = { ( n − , n ) } n ∈ N . Lemma 50. If x, y ∈ ( L + L ∞ )(0 , ∞ ) are positive functions, then E ( µ ( x + y ) |A ) ⊳ E ( µ ( x ) |A ) + E ( µ ( y ) |A ) ⊳ σ / E ( µ ( x + y ) |A ) . Proof.
Recall that µ ( x + y ) ≺≺ µ ( x ) + µ ( y ) ≺≺ σ / µ ( x + y ) . It follows that Z b µ ( s, x + y ) ds ≤ Z b ( µ ( s, x ) + µ ( s, y )) ds, Z a µ ( s, x + y ) ds ≥ Z a ( µ ( s, x ) + µ ( s, y )) ds. Let now a, b be positive integers. Subtracting the above inequalities, we obtain Z b a E ( µ ( x + y ) |A )( s ) ds = Z b a µ ( s, x + y ) ds ≤≤ Z ba ( µ ( s, x ) + µ ( s, y )) ds = Z ba E ( µ ( x ) + µ ( y ) |A )( s ) ds. Similarly, we have Z b a E ( µ ( x ) + µ ( y ) |A )( s ) ds ≤ Z b a E ( µ ( x + y ) |A )( s ) ds. (cid:3) Corollary 51.
The quasi-norm in Construction 16 is a norm.Proof.
It follows from Lemma 50 that E ( µ ( x + y ) |A ) ⊳ E ( µ ( x ) + µ ( y ) |A )provided that x, y are positive functions. By Theorem 9, k E ( µ ( x + y ) |A ) k E ≤ k E ( µ ( x ) |A ) k E + k E ( µ ( y ) |A ) k E . (cid:3) Lemma 52.
Let y = µ ( y ) ∈ ( L + L ∞ )(0 , ∞ ) . It follows that Z b − k λa y ( s ) ds ≤ λλ − Z ba ( σ k y )( s ) ds provided that b ≥ λa. Proof.
Let α be the average value of y on the interval [2 − k λa, − k b ] . Clearly, y ≤ α on the interval [2 − k λa, b ] and y ≥ α on the interval [2 − k a, − k b ] . Thus, σ k y ≥ α on the interval [ a, b ] . Therefore, Z b − k λa y ( s ) ds ≤ ( b − − k λa ) α ≤ λλ − b − a ) α ≤ λλ − Z ba ( σ k y )( s ) ds. (cid:3) RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 27
Theorem 53. If { x n } n ∈ N be a Cauchy sequence in F, then there exists x ∈ F suchthat x n → x in F. Proof.
For every k > , there exists m k such that k x m − x m k k F ≤ − k for m ≥ m k . Set y k = x m k +1 − x m k . Clearly, k y k k F ≤ − k for every k ∈ N . In particular, theseries P ∞ k =1 y k converges in L ∞ (0 , ∞ ) . Set z n = P ∞ k = n σ k µ ( y k ) . We claim that z n ∈ F and z n → F. Indeed, µ ( y k ) ≤ k y k k ∞ χ (0 , + T E ( µ ( y k ) |A ) . Here, T is a shift to the right. It follows that E ( µ ( z n ) |A ) ≤ ∞ X k = n σ k ( k y k k ∞ χ (0 , + T E ( µ ( y k ) |A )) . Therefore, k z n k F ≤ k z n k ∞ + ∞ X k = n k kk y k k ∞ χ (0 , + T E ( µ ( y k ) |A ) k E ≤≤ k z n k ∞ + ∞ X k = n k +1 k y k k F ≤ · − n + 2 − n = o (1) . It follows from Lemma 8.5 of [19] that Z bλa µ ( s, ∞ X k = n y k ) ds ≤ ∞ X k = n Z b − k λa µ ( s, y k ) ds. It follows from Lemma 52 that Z b − k λa µ ( s, y k ) ds ≤ λλ − Z ba ( σ k µ ( y k ))( s ) ds. Therefore, Z bλa µ ( s, ∞ X k = n y k ) ds ≤ λλ − Z ba z n ( s ) ds. Hence, ∞ X k = n y k ⊳ λλ − z n . Since λ > k ∞ X k = n y k k F ≤ k z n k F → . Thus, the series P ∞ k =1 y k does converge in F. The assertion follows immediately. (cid:3) Proof of Figiel-Kalton theorem
The proof of Theorem 8 follows from the combinations of Lemmas below.
Lemma 54.
Let E be a symmetric Banach space either on the interval (0 , oron the semi-axis. If x ∈ Z E , then C ( µ ( x + ) − µ ( x − )) ∈ E. Proof.
Let x = P nk =1 ( x k − y k ) with x k , y k ∈ E + and µ ( x k ) = µ ( y k ) , ≤ k ≤ n. Set z = x + + n X k =1 y k = x − + n X k =1 x k . It follows from the definition of C and (8) that Cµ ( z ) ≤ C ( x + ) + n X k =1 Cµ ( y k ) = C ( µ ( x + ) − µ ( x − )) + Cµ ( x − ) + n X k =1 Cµ ( x k ) . Using the second inequality in (8), we obtain Z t ( µ ( s, x − ) + n X k =1 µ ( s, x k )) ds ≤ Z ( n +1) t µ ( s, z ) ds ≤ Z t µ ( s, z ) ds + ntµ ( t, z ) . Therefore, Cµ ( z ) ≤ Cµ ( z ) + C ( µ ( x + ) − µ ( x − )) + nµ ( z ) . It follows that C ( µ ( x − ) − µ ( x + )) ≤ nµ ( z ) . Similarly, C ( µ ( x + ) − µ ( x − )) ≤ nµ ( z )and the assertion follows. (cid:3) Lemma 55.
Let E be a symmetric Banach space either on the interval (0 , oron the semi-axis. If x ∈ D E , then C ( µ ( x + ) − µ ( x − )) ∈ Cx + E. Proof.
Since x ∈ D E , it follows that x = µ ( a ) − µ ( b ) with a, b ∈ E. Set u = µ ( a ) − x + ≥ . Clearly, µ ( a ) = u + x + and µ ( b ) = u + x − . It follows from thedefinition of C and (8) that Cµ ( a ) ≤ Cµ ( u ) + Cµ ( x + ) = C ( µ ( x + ) − µ ( x − )) + Cµ ( u ) + Cµ ( x − ) . Using the second inequality in (8), we obtain Cµ ( x − ) + Cµ ( u ) ≤ Cµ ( b ) + µ ( b ) . It follows that Cx ≤ C ( µ ( x + ) − µ ( x − )) + µ ( b ) . Similarly, Cx ≥ C ( µ ( x + ) − µ ( x − )) − µ ( a )and the assertion follows. (cid:3) Lemma 56.
Let E = E (0 , ∞ ) be a symmetric space on the semi-axis. If x ∈ D E is such that Cx ∈ E, then x ∈ Z E . Proof.
Define a partition A = { (2 n , n +1 ) } n ∈ Z and set x = E ( x |A ) . If x = µ ( a ) − µ ( b ) with a, b ∈ E, then x = E ( µ ( a ) |A ) − E ( µ ( b ) |A ) . Clearly, E ( µ ( a ) |A ) ≤ σ µ ( a ) ∈ E, E ( µ ( b ) |A ) ≤ σ µ ( b ) ∈ E are decreasing functions. It follows that x ∈ D E . It is easy to see that | Cx − Cx | ≤ σ ( µ ( a ) + µ ( b )) . RACES ON SYMMETRICALLY NORMED OPERATOR IDEALS 29
Therefore, Cx ∈ E. Define a function z ∈ E by setting z ( t ) = ( Cx )(2 n +1 ) , t ∈ (2 n , n +1 ) . Clearly, x = 2 z − σ z ∈ Z E . Consider the function x − x on the interval (2 n , n +1 ) . By Kwapien theorem[23], there exist positive equimeasurable functions y n , y n supported on (2 n , n +1 )such that µ ( y n ) = µ ( y n ) , k y n k ∞ , k y n k ∞ ≤ k ( x − x ) χ (2 n , n +1 ) k ∞ . Set y = P n ∈ N y n and y n = P n ∈ N y n . It follows that y , y ∈ E + . Since x − x = y − y and µ ( y ) = µ ( y ) , it follows that x − x ∈ Z E . The assertion followsimmediately. (cid:3)
References [1] S. Astashkin, F. Sukochev,
Banach-Saks property in Marcinkiewicz spaces.
J. Math. Anal.Appl. (2007) 1231–1258.[2] M. Braverman, A. Mekler,
The Hardy-Littlewood property for symmetric spaces,
SiberianMath. J. (1977), 371–385.[3] A. Carey, F. Sukochev, Dixmier traces and some applications to noncommutative geometry. (Russian) Uspekhi Mat. Nauk (2006), no. 6 (372), 45–110; translation in Russian Math.Surveys (2006), no. 6, 1039–1099.[4] A. Connes, Noncommutative Geometry
Academic Press, San Diego, 1994.[5] A. Connes,
The action functional in noncommutative geometry.
Comm. Math. Phys. (1988), no. 4, 673–683.[6] J. Dixmier,
Existence de traces non normales (French) C. R. Acad. Sci. Paris Ser. A-B
Singular symmetric func-tionals and Banach limits with additional invariance properties. (Russian) Izv. Ross. Akad.Nauk Ser. Mat. (2003), no. 6, 111–136; translation in Izv. Math. (2003), no. 6,1187–1212.[8] P. Dodds, B. de Pagter, E. Semenov, F. Sukochev, Symmetric functionals and singulartraces
Positivity (1998), no. 1, 47–75.[9] K. Dykema, T. Figiel, G. Weiss, M. Wodzicki, Commutator structure of operator ideals.
IMADA preprint PP-1997–22, May 13, 1997, Odense.[10] K. Dykema, T. Figiel, G. Weiss, M. Wodzicki,
Commutator structure of operator ideals.
Adv. Math. (2004), no. 1, 1–79.[11] K. Dykema, N. Kalton,
Sums of commutators in ideals and modules of type II factors.
Ann.Inst. Fourier (Grenoble) (2005), no. 3, 931–971.[12] T. Fack, H. Kosaki, Generalized s -numbers of τ -measurable operators Pacific J. Math. (1986), no. 2, 269–300.[13] T. Figiel, N. Kalton,
Symmetric linear functionals on function spaces
Function Spaces,Interpolation Theory and Related Topics. Lund, 2000[14] I. Gohberg, M. Krein,
Introduction to the theory of linear nonselfadjoint operators.
Trans-lations of Mathematical Monographs, Vol. 18 American Mathematical Society, Providence,R.I. 1969.[15] I. Gohberg, M. Krein,
Theory and applications of Volterra operators in Hilbert space.
Trans-lations of Mathematical Monographs, Vol. 24 American Mathematical Society, Providence,R.I. 1970.[16] D. Guido, T. Isola,
Singular traces on semifinite von Neumann algebras.
J. Funct. Anal. (1995), no. 2, 451–485.[17] D. Guido, T. Isola,
Dimensions and singular traces for spectral triples, with applications tofractals.
J. Funct. Anal. (2003), no. 2, 362–400.[18] N. Kalton, A. Sedaev, F. Sukochev,
Fully symmetric functionals on a Marcinkiewicz spaceare Dixmier traces
Adv. Math. (2011) 3540–3549.[19] N. Kalton, F. Sukochev,
Symmetric norms and spaces of operators
J.reine angew. Math.(2008), 1–41 [20] N. Kalton, F. Sukochev,
Rearrangement-invariant functionals with applications to traces onsymmetrically normed ideals
Canad. Math. Bull. (2008), 67–80.[21] N. Kalton, F. Sukochev, D. Zanin, Orbits in symmetric spaces. II
Studia Math. (2010),no. 3, 257–274.[22] S. Krein, Ju. Petunin, E. Semenov,
Interpolation of linear operators , Nauka, Moscow, 1978(in Russian); English translation in Translations of Math. Monographs, Vol. , Amer.Math. Soc., Providence, RI, 1982.[23] S. Kwapien, Linear functionals invariant under measure preserving transformations
Math.Nachr. (1984), 175–179.[24] J. Lindenstrauss, L. Tzafriri,
Classical Banach Spaces II: Function Spaces , Springer, 1996.[25] S. Lord, A. Sedaev, F. Sukochev,
Dixmier traces as singular symmetric functionals andapplications to measurable operators.
J. Funct. Anal. (2005), no. 1, 72–106.[26] M. Reed, B. Simon,
Methods of modern mathematical physics. I. Functional analysis.
Secondedition. Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York, 1980.[27] A. Pietsch,
About the Banach envelope of l , ∞ . Rev. Mat. Complut. (2009), no. 1, 209–226.[28] R. Schatten, Norm ideals of completely continuous operators.
Second printing. Ergebnisseder Mathematik und ihrer Grenzgebiete, Band 27 Springer-Verlag, Berlin-New York 1970.[29] B. Simon,
Trace ideals and their applications.
Second edition. Mathematical Surveys andMonographs, 120. American Mathematical Society, Providence, RI, 2005.[30] F. Sukochev, D. Zanin,
Orbits in symmetric spaces
J.Funct.Anal. (2009), no.1, 194-218.[31] M. Takesaki,
Theory of operator algebras. I.
Reprint of the first (1979) edition. Encyclopae-dia of Mathematical Sciences, . Operator Algebras and Non-commutative Geometry, 5.Springer-Verlag, Berlin, 2002.[32] J. Varga,
Traces on irregular ideals.
Proc. Amer. Math. Soc. (1989), no. 3, 715–723.[33] M. Wodzicki,
Vestigia investiganda.
Mosc. Math. J. (2002), no. 4, 769–798, 806. School of Mathematics and Statistics, University of New South Wales, Sydney, 2052,Australia.
E-mail address : [email protected] School of Mathematics and Statistics, University of New South Wales, Sydney, 2052,Australia.
E-mail address ::