Symmetric Functionals and Singular Traces

1 2 3,† 1, P.G. DODDS , B. DE PAGTER , E.M. SEMENOV and F.A. SUKOCHEV ∗ 1Department of Mathematics and Statistics, The Flinders University of South Australia, GPO Box 2100, Adelaide, SA 5001, Australia; 2Department of Mathematics, Delft University of Technology, Mekelweg 4, 2628 CD Delft, The Netherlands; 3Department of Mathematics, Voronezh State University, Universitetskaya pl.1, Voronezh 394693, Russia

(Received: 11 November 1997; Accepted: 4 December 1997) Abstract. We study the construction and properties of positive linear functionals on symmetric spaces of measurable functions which are monotone with respect to submajorization. The construction of such functionals may be lifted to yield the existence of singular traces on certain non- commutative Marcinkiewicz spaces which generalize the notion of Dixmier trace.

Mathematics Subject Classiﬁcations (1991): 46E30, 46L50

Key words: symmetric functional, singular trace, Marcinkiewicz space, symmetric operator space

1. Introduction and Preliminaries The purpose of this paper is to study the properties of positive linear functionals on symmetric Banach function spaces which are monotone with respect to submajorization, in both the commutative and non-commutative setting. Such functionals are called symmetric (see Definition 2.1 below). This notion finds its roots and motivation in the work of Dixmier [7] on the existence of non-trivial traces on the factor B(H) which are singular in the sense that they vanish on all finite rank operators. Such traces have subsequently found important application in non- commutative geometry and quantum field theory [4]. The methods used by Dixmier in his construction of singular traces are based on ideas familiar in the theory of rearrangement invariant spaces; in fact, his approach can be viewed as first constructing a symmetric functional on a certain Marcinkiewicz sequence space and then “lifting” this functional to the ideal of compact operators whose singular value sequences lie in this space. Our objective therefore is to study the existence and properties of symmetric functionals, first in the setting of symmetric spaces E on the positive half-line (see sections 2,3 below) and then to lift the results (see section 4) to the more general setting of symmetric spaces of measurable operators affiliated with some semi-finite von Neumann algebra.

† Research supported by RFFI, Grant 95-01-00135 ∗ Research supported by the Australian Research Council.

PIPS No.: 159135 (postkap:mathfam) v.1.15

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Since any normal symmetric functional is a scalar multiple of integration with respect to Lebesgue measure (see Proposition 2.6), our principal results concern the existence of singular symmetric functionals. The familiar Lebesgue spaces p 1 1 L 0, ), 1

x∗(t) inf s 0 m( x >s ) t ,t>0. = { ≥ : {| | } ≤ } For basic properties of symmetric spaces and rearrangements, we refer to the monographs [14], [3]. We note that for any symmetric space E,

1 1 L 0, ) L∞ 0, ) E L 0, ) L∞ 0, ), [ ∞ ∩ [ ∞ ⊂ ⊂ [ ∞ + [ ∞

post56.tex; 23/04/1998; 10:32; p.2 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 49 with continuous embeddings. We will denote by E0 the symmetric space associated with E. The space E0 consists of all measurable functions y for which

∞ y E sup x(t)y(t) dt x E, x E 1 < . k k 0 := | | : ∈ k k ≤ ∞ Z0 If E∗ denotes the Banach dual of E, it is known that E0 E∗ and E0 E∗ if ⊂ = and only if the norm E is order-continuous, i.e. from xn E,xn n 0, it k·k { }⊆ ↓ follows that xn E 0. We note that the norm E on the symmetric space E k k → k·k is order-continuous if and only if E is separable, in which case limt x∗(t) 0 for all x E. →∞ = ∈ 1 If x,y L 0, ) L∞ 0, ), we will say that x is submajorized by y and write x y∈if and[ only∞ + if [ ∞ ≺≺ t t x∗(s)ds y∗(s)ds for all t>0. ≤ Z0 Z0 The symmetric space E on 0, ) will be called fully symmetric if and only if [ ∞ x y, y E, implies x E and x E yE. It is well known (see, for ≺≺ ∈ ∈ k k ≤kk example, [14]) that (E, E)is fully symmetric if and only if E is an interpolation 1 k·k space for the pair (L 0, ), L∞ 0, )), with interpolation constant 1. If the norm on E is order continuous,[ ∞ then E[is fully∞ symmetric.

2. Symmetric Functionals on 0, ). [ ∞ Throughout this section, E will denote a fully symmetric Banach function space on 0, ). [ ∞ DEFINITION 2.1. A linear functional 0 φ E∗ will be called symmetric if and only if 0 f, g E and f g imply that≤ φ(f)∈ φ(g). ≤ ∈ ≺≺ ≤ If E L1 0, ), then the functional φ deﬁned by setting φ(f) ∞ f(t)dt for ⊆ [ ∞ = 0 all f E is clearly symmetric. Recall that a functional 0 φ E∗ is called ∈ ≤ ∈R normal (or order continuous)iffn n 0a.e.inEimplies that φ(fn) n 0. For example, see [25]. ↓ ↓

1 PROPOSITION 2.2. If 0 <φ E∗is normal and symmetric, then E L 0, ) ∈ ⊆ [ ∞ and there exists 0

φ(χA) w(t)dt αm(A) < βm(B) w(t)dt φ(χB), = ≤ ≤ = ZA ZB

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and this is a contradiction. 2

Before presenting some non-trivial examples of symmetric functionals, we ﬁrst show that symmetric functionals are dilation invariant in the sense of of the following proposition. For s>0, the dilation operator Ds is deﬁned by setting Dsf(t) f(t/s)for all t>0 and functions f on 0, ). The linear mapping Ds := [ ∞ is bounded on E; moreover, Ds E max 1,s ([14], Corollary 1 to Theorem II k k ≤ { } 4.5). Note that (Dsf)∗ Dsf∗. = PROPOSITION 2.3. If 0 <φ E∗ is symmetric, then φ(Dsf) sφ(f ) for all f E and all s>0. ∈ = ∈

Proof. Let 0 f E.Forn N , there exist disjointly supported 0 ≤ ∈ ∈ ≤ f ,...,fn Eand such that f ∗ f ∗ for all k 1,...,n.Leth f fn. 1 ∈ k = = := 1 +···+ It is easy to see that h∗(t) f ∗(t/n) for all t>0, that is, h∗ Dnf ∗. Hence, = = φ(Dnf) φ((Dnf)∗) φ(Dnf∗) φ(h∗) = n = n =

φ(h) φ(fk) φ(f∗) = = = k k 1 k 1 X= X= nφ(f ∗) nφ(f ). = = Via a standard argument, it now follows that φ(Drf) rφ(f ) for all rationals =

1 = 1 ≤ 2 = 2 Q Given 0

~~for all n 1, 2,... which implies that φ(Dsf) sφ(f ). 2 = = Remark 2.4. It is clear from the above that non-trivial symmetric functionals fail to exist on any of the spaces E Lp 0, ), 1~~

0(where1(t) 1, for all t>0). = = We now present some examples of non-trivial symmetric functionals. Further constructions will be discussed in the next section. EXAMPLE 2.5. (i) (cf. [19],[1]) Let ψ 0, ) 0, )be concave and :[ ∞ →[ ∞ increasing with ψ( ) limt ψ(t) and ∞ := →∞ =∞ ψ(2t) lim 1. (2.1) t ψ(t) →∞ =

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For example, we may take ψ(t) log(t 2), t 0. Let Mψ be the corresponding Marcinkiewicz space consisting:= of all measurable+ ≥ functions f on 0, ) for which [ ∞ 1 t sup f ∗(s)ds < ψ(t) ∞ t>0 Z0 equipped with the norm 1 t f M sup f ∗(s)ds. k k ψ := ψ(t) t>0 Z0 For basic properties of Marcinkiewicz spaces, we refer to [14]. The space (Mψ ,

Mψ)is a fully symmetric Banach function space. Note that ψ0 Mψ ,butthat k·k 1 1 ∈ ψ0 L 0, ) since ψ( ) . Consequently, Mψ L 0, ).For0 f 6∈ [ ∞ ∞ =∞ 6⊆ [ ∞ ≤ ∈ Mψ, we deﬁne the function Φ(f) by setting 1 t Φ(f )(t) f ∗(s)ds, t > 0. (2.2) := ψ(t) Z0 It follows that 0 Φ(f) Cb(0, ),whereCb(0, )denotes the space of all bounded (real-valued)≤ continuous∈ functions∞ on (0, )∞. We now observe that ∞ Φ(f g) Φ(f) Φ(g) wD / Φ(f g) (2.3) + ≤ + ≤ 1 2 + for all 0 f, g Mψ ,wherew(t) ψ(2t)/ψ(t), t > 0. In fact, the ﬁrst ≤ ∈ = inequality is just the well known submajorization f g f ∗ g∗. The second inequality follows by noting that + ≺≺ +

t t 2t f ∗(s)ds g∗(s)ds (f g)∗(s)ds + ≤ + Z0 Z0 Z0 which follows from [14], II.2.2. Now let L be a linear functional on the space Cb(0, ) with the properties ∞ ∞ (a) L (u) 0forall0 u Cb(0, ). ∞ ≥ ≤ ∈ ∞ (b) L (u) limt u(t), for all u Cb(0, ) for which lim u(t) exists. ∞ →∞ t = ∈ ∞ →∞ (c) L (Dsu) L (u) for all s>0, u Cb(0, ). ∞ = ∞ ∈ ∞ Such a functional is easily obtained from a translation invariant generalized limit

at on Cb( R ). Note that if u, v Cb(0, ) and limt v(t) exists, then +∞ ∈ ∞ →∞ L (vu) (limt v(t))L (u).For0 f Mψ,wenowdeﬁneφ(f) L∞(Φ(f )).= From→∞(2.3) it follows∞ that ≤ ∈ := ∞

~~φ(f g) φ(f) φ(g) L (wD1/2Φ(f g)) + ≤ + ≤ ∞ + for all 0 f, g Mψ.By(2.1), limt w(t) 1, and so the given properties of the functional≤ L∈ imply that →∞ = ∞~~

~~L (wD1/2Φ(f g)) L (Φ(f g)) φ(f g) ∞ + = ∞ + = +~~

post56.tex; 23/04/1998; 10:32; p.5 52 P.G. DODDS ET AL. and consequently φ(f g) φ(f) φ(g) for all 0 f, g Mψ . Moreover, it + = + ≤ ∈ is clear that φ(λf) λφ(f ) for all 0 f Mψ , 0 λ R , and therefore φ = ≤ ∈ ≤ ∈ extends to a positive linear functional on Mψ , which we continue to denote by φ. From the deﬁnition it is clear that φ is symmetric; further, ψ(t) ψ(0 ) Φ(ψ0)(t) − + 1 = ψ(t) → 1 as t . Consequently, φ(ψ0) 1andso φ 1. If 0 f Mψ L 0, ), →∞ = k k= ≤ ∈ ∩ [ ∞ then limt Φ(f )(t) 0andsoφ(f) 0. This shows that φ 0onMψ 1 →∞ 1= = = ∩ L 0, ). Note that L 0, ) Mψ whenever ψ(0 )>0. [Finally,∞ we observe[ that∞ if the⊆ above construction+ is carried out in the case that 1 ψ( )< ,thenMψ L 0, )and the functional φ obtained is given by ∞ ∞ 1 ⊆ [ ∞ φ(f) ψ( )− ∞ f(s)ds since = ∞ 0 R 1 ∞ lim Φ(f )(t) f∗(s)ds t = ψ( ) →∞ ∞ Z0 for all 0 f Mψ in this case. (ii) Assume≤ ∈ now that ψ 0, ) 0, )is concave and increasing with ψ(0 ) 0, and that :[ ∞ →[ ∞ + = ψ(2t) lim 1. (2.4) t 0 ψ(t) = ↓ 1 1 For example, we may take ψ to be given by ψ(t) 2(1 log t)− for 0 0 ≤ ψ(t) k k∞ and since (2.4) in combination with ψ(0 ) 0 implies that limt 0 t/ψ(t) 0, it + = ↓ = follows that φ(f) 0. Consequently, φ 0onMψ L∞ 0, ). (iii) If the concave= increasing function= ψ on 0∩, ) [with∞ψ(0 ) 0and ψ( ) satisﬁes both (2.1) and (2.4), then we[ can∞ construct non-trivial+ = sym- ∞ =∞ metric functionals φ1,φ2 on Mψ via the methods given in (i) and (ii) above respectively. If we then set φ φ1 φ2,thenφis a symmetric functional which vanishes 1 := + on (L L∞) 0, ). ∩ [ ∞

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Our next objective is to obtain decomposition theorems for symmetric functionals which will give considerable insight into their structure. We recall that a functional 0 φ E∗ is called singular if whenever 0 φ0 E∗ is order ≤ ∈ ≤ ∈ continuous and satisﬁes 0 φ0 φ, it follows that φ0 0 (equivalently, φ is disjoint from all order continuous≤ ≤ functionals on E; see, for= example [25], Section 87). As is well known, a functional 0 φ E∗ is singular if and only if φ ≤ ∈ vanishes on some order dense ideal, [25] Section 90. Moreover, any 0 φ E∗ ≤ ∈ has a unique decomposition φ φn φs , with φn order continuous (normal) and = + φs singular. In the proposition which follows, we will show that if φ is symmetric, then both φn and φs are symmetric. We denote by Eb the norm closure in E of the 1 (order dense) ideal (L L∞) 0, ); equivalently, Eb is the norm closure in E of the space of all stepfunctions.∩ [ ∞

~~PROPOSITION 2.6. If 0 φ E∗ is a symmetric functional, then there exist ≤ ∈ symmetric functionals 0 φ ,φ E∗ for which φ φ φ and such that ≤ 1 2 ∈ = 1 + 2~~

∞ R (i) φ1(f ) c 0 f(t)dt for some 0 c , with c>0only possible if E L1=0, ). ≤ ∈ ⊆ [ ∞R (ii) φ 0 on Eb (in particular, φ is singular). 2 = 2 The decomposition φ φ1 φ2 is the unique decomposition of φ into its normal and singular parts. = + Proof. Observe ﬁrst that if φ(χA) 0 for all measurable A 0, )with = ⊆[ ∞ m(A) < ,thenφ 0onEb and so we may take φ 0andφ φ ∞ = 1 = = 2 in this case. Now assume that there exists a measurable set A0 0, )such χ χ χ ⊆[ ∞ that m(A0)< and φ( A0)>0. Since A∗ Dm(A0) 0,1 , it follows that χ ∞χ χ 0 = χ [ ] 0 φ( A0) φ( A∗ ) m(A0)φ( 0,1 ),andsoφ( 0,1 )>0. Without loss 6= = 0 = [ ] [ ] of generality, we may assume that φ(χ 0,1 ) 1. It then follows that φ(χA) [ ] = = χ ∞ m(A)φ( 0,1 ) m(A) for all measurable A 0, )so that φ(f) 0 f(t)dt for all stepfunctions[ ] = f .If0 f E, there⊆[ exists∞ a sequence of= stepfunctions ≤ ∈ R fn n∞ 1 such that 0 fn n f a.e. on 0, ).Since { } = ≤ ↑ [ ∞ ∞ 0 fn(t)dt φ(fn) φ(f), n 1,2,..., (2.5) ≤ = ≤ = Z0 it follows from the Beppo–Levi theorem that f L1 0, ). Consequently, E 1 ∈ [ ∞ ⊆ L 0, ).Ifφ(f ) ∞ f(s)ds for all f E,thenφ E∗ and from (2.5) [ ∞ 1 := 0 ∈ 1 ∈ it follows that 0 φ1 φ.Sinceφ(f) φ1(f ) for all stepfunctions f ,itis ≤ R≤ = clear that φ φ on Eb. Putting φ φ φ , it follows that 0 φ E∗ = 1 2 = − 1 ≤ 2 ∈ and φ 0onEb. It remains to show that φ is symmetric. To this end, suppose 2 = 2 that 0 0, and let f1 χ ∈ χ ≺≺ := f ∗ 0,t0 and g1 g∗ 0,t0 . It is clear that f1 g1. Moreover, φ2(f ) φ2(f ∗) [ χ] := χ[ ] ≺≺ χ = = φ2(f ∗ 0,t0 ) φ2(f ∗ (t0, )) φ2(f1),sincef∗ (t0, ) E L∞ 0, ) 1 [ ] + ∞ = ∞ ∈ ∩ [ ∞ = (L L∞) 0, ) Eb. Similarly, φ (g) φ (g ). Now take t t such that ∩ [ ∞ ⊆ 2 = 2 1 1 ≥ 0 (t t )f (t ) ∞ g (t)dt ∞ f (t)dt. 1 − 0 1 0 = 1 − 1 Z0 Z0

post56.tex; 23/04/1998; 10:32; p.7 54 P.G. DODDS ET AL. χ If f2 f1 f1(t0) (t0,t1 ,thenf2 g1,φ1(f2) φ1(g1) and φ2(f2) φ2(f1). Since:=φ is symmetric,+ f ] g implies≺≺ that φ(f ) =φ(g ), hence = 2 ≺≺ 1 2 ≤ 1 φ (f ) φ (f ) φ(f ) φ (f ) 2 = 2 2 = 2 − 1 2 φ(g ) φ (g ) φ (g ) φ (g). ≤ 1 − 1 1 = 2 1 = 2 This shows that g is symmetric. Since the last assertion of the proposition is obvi-

~~ous, this completes the proof of the proposition. 2~~

It follows in particular from the preceding proposition that any singular symmetric 1 functional vanishes on the order dense ideal Eb. Further, if E L 0, ),then any symmetric functional on E is singular. Let us note that a6⊆ result[ similar∞ to Proposition 2.6 is given in [12] Lemma 2.5, although there appears to be a gap in the proof given there. We will now give a further decomposition of singular symmetric functionals. We shall need the following terminology.

DEFINITION 2.7. A symmetric functional 0 φ E∗ is said to be (i) supported ≤ ∈ 1 at 0 if φ 0 on E L∞ 0, ); (ii) supported at if φ 0 on E L 0, ). = ∩ [ ∞ ∞ = ∩ [ ∞ Remark 2.8. (i) A symmetric functional supported either at 0 or at is clearly ∞ singular. (ii) If 0 φ ,φ E∗ are symmetric functionals such that φ is supported ≤ 1 2 ∈ 1 at 0 and φ2 at ,thenφ1 φ2 0. Indeed , φ1 φ2 0onE L∞ 0, )and on 1 ∞ ∧ = 1 ∧ = ∩ [ ∞ E L 0, ), and hence on E since E (L L∞) 0, ). ∩ [ ∞ ⊆ + [ ∞ The terminology introduced in Deﬁnition 2.7 is in fact motivated by the alternative characterizations given in the following lemma. We leave the simple proof to the reader.

~~LEMMA 2.9. Let 0 φ E∗ be a symmetric functional. ≤ ∈ (a) φ is supported at 0 if and only if φ(f ∗χ(s, )) 0 for all f E and all ∞ = ∈~~

THEOREM 2.10. If 0 φ E∗ is a singular symmetric functional, then φ has a ≤ ∈ unique decomposition φ φ1 φ2 such that 0 φ1,φ2 E∗ are symmetric, φ1 is supported at 0 and φ is= supported+ at . ≤ ∈ 2 ∞ Proof. Note ﬁrst that any f E can be written as f f1 f2 with f1 1 ∈ = + ∈ E L 0, ) and f2 E L∞ 0, ) (with f1,f2 0iff 0). Suppose that f f ∩ f[ is∞ another such∈ decomposition.∩ [ ∞ Then f f≥ f ≥f (L1 L ) 0, =) ˜1 ˜2 1 ˜1 ˜2 2 ∞ and+ so φ(f f ) φ(f f ) 0. This yields− that=φ(f−) ∈φ(f )∩and φ(f[ )∞ 1 ˜1 ˜2 2 1 ˜1 2 φ(f ). We can− therefore= deﬁne− φ=,φ by setting φ (f ) φ(f= ), φ (f ) φ(f =). ˜2 1 2 1 := 1 2 := 2 It is clear that 0 φ ,φ E∗ and that φ φ φ . Moreover, φ 0on ≤ 1 2 ∈ = 1 + 2 1 =

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1 E L∞ 0, ) and φ 0onE L 0, ). It remains to be shown that φ ,φ ∩ [ ∞ 2 = ∩ [ ∞ 1 2 are symmetric. Suppose then that 0 f, g E and that f g.Setg g1 g2 1 ≤ ∈ ≺≺ = + with g E L 0, ) and g E L∞ 0, ). From [3] Section 3.7, there exist 1 ∈ ∩ [ ∞ 2 ∈ ∩ [ ∞ 0 f1,f2 E with f1 g1,f2 g2 and f f1 f2.Sinceφis symmetric, it follows≤ that∈φ (f ) φ(f≺≺) φ(g≺≺) φ(g),= and similarly,+ φ (f ) φ (g),andit 1 = 1 ≤ 1 = 2 ≤ 2 follows that φ1,φ2 are symmetric. 2

Note that the functional constructed in Example 2.5 (i) is supported at ,andthat in Example 2.5 (ii) is supported at 0. The above decomposition is illustrated∞ in part (iii) of the same example. Note further that if E L1 0, ), then necessarily any ⊆ [ ∞ 1 singular functional is supported at 0, since E L∞ 0, ) (L L∞) 0, ) in this case. ∩ [ ∞ = ∩ [ ∞

Remark 2.11. (Symmetric functionals on 0, 1)). Given a fully symmetric Banach function space on 0, ), we may consider[ the corresponding Banach function space E on 0, 1) consisting[ ∞ of those measurable functions f on 0, 1) for which [ [ f ∗ E, with norm given by f E 0,1) f ∗ E,forallf E0,1).Itis easy∈ to see that every fully symmetrick k [ Banach:= k functionk space on∈0, 1[) is of this form. The notion of a symmetric functional on E 0, 1) is obtained[ by the obvious modiﬁcation of Deﬁnition 2.1. [ Let 0 φ E 0, 1)∗ be a symmetric functional. For 0

~~φ0(f g) φ (f g)∗χ 0,1) + = + [ φ(f ∗χ 0,1) g∗χ 0,1)) φ0(f ) φ0(g). ≤ [ + [ = +~~

~~For the converse inequality, note that f ∗ g∗ 2D / (f g)∗,andso + ≺≺ 1 2 +~~

~~f∗χ0,1/2) g∗χ0,1/2) 2D1/2 (f g)∗χ 0,1) . [ + [ ≺≺ ˜ + [~~

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~~Using the fact that φ vanishes on L∞ 0, 1),thisimpliesthat [~~

~~φ0(f ) φ0(g) φ(f ∗χ 0,1/2) g∗χ 0,1/2)) + = [ + [ 2φ(D1/2 (f g)∗χ 0,1) ) ≤ ˜ + [ φ (f g)∗χ 0,1) . = + [ This concludes the proof of the claim.~~

Finally, we note conversely that if 0 φ E∗ is a symmetric functional and if ≤ 0 ∈ φ(f) φ (f ∗) for all 0 f E 0, 1),thenφextends to a symmetric functional := 0 ≤ ∈ [ 0 φ E 0, 1)∗ (see Theorem 4.2 below). Of course, if φ0 is supported at , then≤ φ ∈ 0.[ The preceding observations reduce the study of symmetric functionals∞ on 0, 1=) to the corresponding study on 0, ). [We conclude this section with some[ remarks∞ concerning the structure of the space of symmetric functionals on E. For this purpose, we will denote by (E∗ )sym + the collection of all symmetric functionals on E. It is clear that (E∗ )sym is a cone, + that is, if φ1,φ2 (E∗ )sym and if 0 α, β R ,thenαφ1 βφ2 (E∗ )sym. ∈ + ≤ ∈ + ∈ + PROPOSITION 2.12. With respect to the order structure induced by E∗, the cone (E∗ )sym is a lattice. More precisely: + (i) if φ1,φ2 (E∗ )sym,thenφ1 φ2 (E∗ )sym,whereφ1 φ2 denotes the ∈ + ∧ ∈ + ∧ inﬁmum of φ1,φ2 in the vector lattice ReE∗;

~~(ii) if φ1,φ2 (E∗ )sym, then the supremum φ1 s φ2 of φ1 and φ2 in (E∗ )sym exists and∈ is given+ by the formula ∨ +~~

φ s φ (f ) sup φ (u) φ (v) 0 u, v E,u v f (2.6) 1 ∨ 2 = { 1 + 2 : ≤ ∈ + ≺≺ } for all 0 f E. ≤ ∈ Proof. (i) We recall that the inﬁmum of φ1,φ2 in the vector lattice ReE∗ is given by the formula

(φ φ )(f ) inf φ (u) φ (v) 0 u, v E,u v f 1 ∧ 2 = { 1 + 2 : ≤ ∈ + = } for all 0 f E. Now suppose that 0 f, g E and f g.If0 u, v E ≤ ∈ ≤ ∈ ≺≺ ≤ ∈ with u v g, then there exist 0 u1,u2 E with u1 v1 f and u1 u, v +v.Since= φ ,φ are symmetric,≤ it follows∈ that + = ≺≺ 1 ≺≺ 1 2 (φ φ )(f ) φ (u ) φ (v ) φ (u) φ (v), 1 ∧ 2 ≤ 1 1 + 2 1 ≤ 1 + 2 and hence (φ φ )(f ) (φ φ )(g). Consequently, φ φ is symmetric. 1 ∧ 2 ≤ 1 ∧ 2 1 ∧ 2 (ii) If the right hand side of (2.6) is denoted by φ0(f ), then it clearly sufﬁces to show that φ0 is additive on E+. To this end, let 0 f, g E and suppose that 0 u, v E satisfy u v f g.There exist≤ 0 ∈f ,g E such ≤ ∈ + ≺≺ + ≤ 1 1 ∈ that u v f1 g1 and f1 f, g1 g respectively. Using the well-known Riesz decomposition+ = + property for≺≺ vector≺≺ lattices, we may write u u u and = 1 + 2

post56.tex; 23/04/1998; 10:32; p.10 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 57 v v1 v2 with 0 u1,u2,v1,v2 Eand u1 v1 f1 and u2 v2 g1.Since u= v+ f and u ≤ v g, it follows∈ that + = + = 1 + 1≺≺ 2 + 2 ≺≺ φ (u) φ (v) (φ (u ) φ (v )) (φ (u ) φ (v )) 1 + 2 = 1 1 + 2 1 + 1 2 + 2 2 φ (f ) φ (g) ≤ 0 + 0 and so φ (f g) φ (f ) φ (g). To prove the converse inequality, let >0be 0 + ≤ 0 + 0 given and assume that 0 ui ,vi E,i 1,2 satisfy u1 v1 f and u2 v2 g and ≤ ∈ = + ≺≺ + ≺≺

φ (u ) φ (v )>φ(f ) , φ (u ) φ (v )>φ(g) . (2.7) 1 1 + 2 1 0 − 1 2 + 2 2 0 − If u 1/2D (u u ) and v 1/2D (v v ),then := 2 1 + 2 := 2 1 + 2 1 1 u v D(u u ) D (v v ) + = 2 2 1 + 2 + 2 2 1 + 2 1 1 D f ∗ D g∗ ≺≺ 2 2 + 2 2 1 1 ∗ 2D / D f D g (f g)∗, ≺≺ 1 2 2 2 + 2 2 = + and hence φ1(u) φ2(v) φ0(f g). On the other hand, using Proposition 2.3 and (2.7), it follows+ that ≤ +

~~φ (u) φ (v) φ (u u ) φ (v v ) 1 + 2 = 1 1 + 2 + 2 1 + 2 (φ (u ) φ (v )) (φ (u ) φ (v )) = 1 1 + 2 1 + 1 2 + 2 2 >φ(f ) φ (g) 2. 0 + 0 −~~

~~Consequently, φ (f ) φ (g) φ (f g), and this sufﬁces to complete the proof.2 0 + 0 ≤ 0 +~~

3. Existence of Symmetric Functionals As in the previous section, E will continue to denote a fully symmetric Banach function space on 0, ). In this section, we consider the existence and non- existence of (non -trivial)[ ∞ symmetric functionals on certain classes of Banach function spaces. As noted in Proposition 2.6, non-zero symmetric normal functionals exist if and only if E L1 0, ), and so the primary interest lies in the existence of singular symmetric⊆ functionals.[ ∞ We note immediately that non-zero symmetric 1 1 functionals do not exist on (L L∞) 0, ) and (L L∞) 0, ). Indeed, this 1 ∩ [ ∞ + [ ∞ is clear for (L L∞) 0, ) from Proposition 2.6. Now suppose that 0 φ is a ∩ [ ∞1 ≤ symmetric functional on (L L∞) 0, ).Sinceφis singular, it follows that φ 0 1 + [ ∞ 1 1 = on (L L∞) 0, ) and hence φ 0onL 0, )as L 0, ) is contained in ∩ [ ∞ 1 = [ ∞ [ ∞ the norm closure of L L∞ 0, ). Moreover, it follows from Proposition 2.3 that ∩ [ ∞ 1 φ 0onL∞ 0, )and hence φ 0on(L L∞) 0, ). We will show ﬁrst = [ ∞ = + [ ∞

post56.tex; 23/04/1998; 10:32; p.11 58 P.G. DODDS ET AL. that non-zero singular symmetric functionals do not exist on Orlicz and Lorentz spaces. For the convenience of the reader we recall some of the definitions, but for the details, we refer to any of the books [3,14,25]. In fact, we will treat the Orlicz and Lorentz spaces by considering the (so-called) Orlicz–Lorentz spaces. For more details on such spaces, we refer to [13]. Let Φ 0, ) 0, be a Young’s function, that is, Φ is convex, Φ(0) 0 and 0 <Φ(t)<:[ ∞ →[for some∞] t>0, and let ψ 0, ) 0, )be increasing,= ∞ :[ ∞ →[ ∞ concave and ψ(0) 0. We assume that if Φ jumps to at some t0 > 0, then = 0 ∞ Φ(t ) Φ(t ). The modular MΦ,ψ is defined on L 0, ) by setting 0 = 0− [ ∞ ∞ MΦ,ψ(f ) Φ(f ∗(t))dψ(t) := Z0 ∞ Φ(f ∗(t))ψ0(t)dt ψ(0 )Φ( f ). (3.1) = + + k k∞ Z0 The Orlicz–Lorentz space LΦ,ψ is defined by setting

0 1 LΦ,ψ f L 0, ) MΦ,ψ f < for some k>0 := ∈ [ ∞ : k ∞ equipped with the norm Φ,ψ given by k·k 1 f Φ,ψ inf k>0 MΦ,ψ f 1 . (3.2) k k := : k ≤ The Orlicz–Lorentz space (LΦ,ψ, Φ,ψ ) is a fully symmetric Banach function space on 0, ). In the special casek·k that ψ(t) t for all t 0, we obtain the [ ∞ = ≥ Orlicz space LΦ (with the Luxemburg norm Φ). In the case that Φ(t) t for k·k = all t 0, we obtain the Lorentz space associated with ψ, usually denoted 3ψ . ≥ THEOREM 3.1. If φ is a symmetric singular functional on the Orlicz–Lorentz space LΦ,ψ,thenφ 0. Proof. Suppose, on= the contrary, that 0 <φis a non-trivial symmetric singular functional on LΦ,ψ. We may assume that φ 1. Let 0 1. Since φ is symmetric, it may be assumed that f f ∗.From Theorem 2.10, it sufﬁces to consider only the cases that φ is supported= at 0 or at 1 . We assume ﬁrst that φ is supported at 0, that is, φ 0onL L∞ 0, ). ∞ = ∩ [ ∞ Since ψ(0 )>0 implies that LΦ,ψ L∞ 0, ), it follows that ψ(0 ) 0. + ⊆ [ ∞ + = Set fn f χ 1 ,n 1,2,....Forτ>0, it follows from Proposition 2.3 that 0, n ) 1 = [ = φ(τ− Dτfn) φ(fn)>1, and hence = 1 τ − Dτ fn Φ,ψ > 1. k k The expression for the norm given by (3.1) and (3.2) now implies that

~~∞ 1 Φ D f (s) ψ0(s)ds > 1, τ τ n Z0 ~~

~~post56.tex; 23/04/1998; 10:32; p.12 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 59 or equivalently~~

∞ 1 1 Φ f (s) ψ0(τs)ds > , τ n τ Z0 for all τ>0andalln 1,2,....Sincef fn n 0on 0, )this implies that = ≥ ↓ [ ∞ ∞ 1 Φ f(s) ψ0(τs)ds τ =∞ Z0 for all τ>0. Since ψ 0 is decreasing, it follows that

∞ 1 Φ f(s) ψ0(s)ds τ =∞ Z0 for all τ 1, and hence for all τ>0. This contradicts the fact that f LΦ,ψ. ≥ ∈1 Suppose now that φ is supported at ,thatis,thatφ 0onLΦ,ψ L 0, ). If ∞ = ∩ [ ∞ 1 LΦ,ψ then L∞ LΦ,ψ. As noted in Remark 2.4, φ 0onL∞ 0, ).Sinceφ is∈ supported at , it⊆ then follows that φ 0, contrary to= hypothesis.[ Consequently,∞ ∞ = 1 LΦ,ψ. In particular, it follows that f(t) 0ast .Iffn (f χ n, ))∗ 6∈ → →∞ := [ ∞ for n 1, 2,...,thenf fn 0 pointwise on 0, ); in fact, fn n 0. As in the preceding= paragraph,≥ but using↓ now the assumption[ ∞ that φ isk supportedk∞ ↓ at , 1 ∞ we obtain that τ − Dτ fn Φ,ψ > 1 and hence k k ∞ 1 1 Φ Dτ fn(s) ψ0(s)ds ψ(0 )Φ Dτ fn > 1 τ + + τ Z0 ∞ for all τ>0. The last term is only present if ψ( 0 )>0, in which case LΦ,ψ + ⊆ L∞ 0, ). Again via a substitution it follows that [ ∞ ∞ 1 1 1 1 Φ fn(s) ψ0(τs)ds ψ(0 )Φ fn > τ + τ + τ τ Z0 ∞ for all τ>0andn 1,2,.... For each ﬁxed τ, it follows from fn n 0that 1 = k k∞ ↓ Φ fn n 0, and so f fn n 0on 0, ), implies that τk k∞ ↓ ≥ ↓ [ ∞

~~∞ 1 Φ f(s) ψ0(τs)ds τ =∞ Z0 for all τ>0. This again contradicts the fact that f LΦ,ψ and sufﬁces to complete ∈ the proof of the theorem. 2~~

~~COROLLARY 3.2. (i) Orlicz spaces LΦ on 0, ) admit no non-trivial singular symmetric functionals. [ ∞ (ii) Lorentz spaces 3ψ on 0, ) admit no non-trivial singular symmetric functionals. [ ∞~~

~~post56.tex; 23/04/1998; 10:32; p.13 60 P.G. DODDS ET AL.~~

We observe that part (i) of the preceding Corollary establishes a special case of the conjecture in [22] concerning the non-existence of (so-called) additive functionals on non-separable Orlicz spaces. We now characterise those Marcinkiewicz spaces on 0, ) which do admit non-trivial singular symmetric functionals. We shall need[ the∞ following proposition, which is of interest in its own right. PROPOSITION 3.3. Let E be a fully symmetric Banach function space on 0, ) and let 0

t 0 f ∗(s)ds lim sup t 1 (3.5) t g (s)ds ≤ →∞ R0 ∗ implies that φ(f)R φ(g).Given>0, it follows from (3.5) that there exists a> t ≤ t 0 such that f ∗(s)ds (1 ) g∗(s)ds for all t>a.Setg1 g∗ f ∗χ 0,a). 0 ≤ + 0 := + [ Since φ is supported at , it follows that φ(f∗χ 0,a)) 0andsoφ(g) φ(g∗) R ∞ R [ = = = φ(g1). It is easy to see that f (1 )g1 and so φ(f) (1 )φ(g1) (1 )φ(g). Since this holds for≺ every≺ >+ 0, we may conclude≤ that φ(f)+ φ(g)=. +A similar argument establishes the assertion of (ii), and this sufﬁces to complete≤

~~the proof of the Proposition. 2~~

Let ψ 0, ) 0, )be an increasing, concave function. The concavity of ψ implies:[ that∞ the→[ function∞t ψ(t)/t is decreasing on (0, ), from which it follows that → ∞ ψ(2t) 1 2,t>0. ≤ ψ(t) ≤

As will become clear below, those Marcinkiewicz spaces Mψ for which the quo- tient ψ(2 )/ψ( ) has limit 1 at 0 or play an important role with respect to symmetric functionals.· · In particular, it has∞ been already observed in Example 2.5 that if ψ satisﬁes either of the conditions limt ψ(2t)/ψ(t) 1(andψ( ) ) →∞ = ∞ =∞ or limt 0 ψ(2t)/ψ(t) 1(andψ(0 ) 0), then non-trivial singular symmetric ↓ = + = functionals exist on Mψ .

~~post56.tex; 23/04/1998; 10:32; p.14 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 61~~

~~THEOREM 3.4. (i) There exists a non-zero symmetric functional on Mψ supported at if and only if ∞ ψ(2t) lim inf 1 and lim ψ(t) . t ψ(t) t →∞ = →∞ =∞~~

~~(ii) There exists a non-zero symmetric functional on Mψ supported at 0 if and only if ψ(2t) lim inf 1andψ(0 ) 0. t 0 ψ(t) = + = ↓~~

Proof. (i) Suppose that 0 <φ Mψ∗ is a symmetric functional which is sup- ∈ 1 ported at so that φ 0onMψ L 0, ). Since limt ψ(t) < implies ∞ 1 = ∩ [ ∞ →∞ ∞ that Mψ L 0, ), the non-triviality of φ implies that limt ψ(t) .Since ⊆ [ ∞ t →∞ =∞ the derivative ψ0 of ψ is decreasing and satisﬁes ψ0(s)ds ψ(t) ψ(0 ) for 0 = − + all t>0, it is clear that ψ0 Mψ and ψ 0 Mψ 1. Moreover, if 0

~~t t 0 f ∗(s)ds 1 lim sup t lim sup f∗(s)ds f Mψ. t ψ (s)ds = t ψ(t) ψ(0 ) 0 ≤k k →∞ R0 0 →∞ − + Z~~

~~By PropositionR 3.3 (i), it follows that φ(f) f M φ(ψ0). In particular, it follows ≤k k ψ that φ(ψ0)>0. From Proposition 3.3(i), we obtain that~~

~~t 0 ψ 0(s)ds φ(ψ0) lim sup t φ(2D1/2ψ0). ≤ t 2D ψ (s)ds →∞ 0 R 1/2 0 !~~

Since φ(ψ0) φ(2D / Rψ0) by Proposition 2.3, and since = 1 2 t 0 ψ 0(s)ds ψ(t) ψ(0 ) lim sup t lim sup − + , t 2D1/2ψ (s)ds = t ψ(2t) ψ(0 ) →∞ 0 R 0 →∞ − + it follows thatR ψ(t) ψ(0 ) lim sup − + 1 t ψ(2t) ψ(0 ) ≥ →∞ − + and hence also lim supt ψ(t)/ψ( 2t) 1sinceψ( ) .Asψ(2t)/ψ(t) →∞ ≥ ∞ =∞ ≥ 1forallt>0, we obtain that lim inft ψ(2t)/ψ(t) 1. For the proof of the converse implication,→∞ we assume= that ψ( ) and that ∞ =∞ lim inft ψ(2t)/ψ(t) 1. We make ﬁrst the following observation. Since ψ is concave,→∞ the function τ = ψ(τt)/ψ(t) is concave and increasing on (0, ) for → ∞ each ﬁxed t>0. If we set α(τ) lim inft ψ(τt)/ψ(t) for τ>0, the function α is concave and increasing on :=(0, ).Since→∞ α(1) 1, it is clear that α(2) 1 ∞ = =

post56.tex; 23/04/1998; 10:32; p.15 62 P.G. DODDS ET AL. if and only if α(τ) 1forallτ 1. Consequently, the assumption on ψ implies that = ≥ ψ(2nt) lim inf 1,n 1,2,3,... t ψ(t) →∞ = = and so there exists a sequence 0

~~We deﬁne Φ M + l ∞ ( N ) by : ψ →~~

~~n 1 2kt 1 − 1 n ∞ Φf f (s)ds k ∗ := n ψ(2 tn) 0 ( k 0 Z )n 1 X= =~~

for all 0 f Mψ . It is clear that Φ(af) aΦ(f ) for all 0 a R , while ≤ ∈ = ≤ ∈ the inequality Φ(f g) Φf Φg for all 0 f, g Mψ follows immediately + ≤ + ≤ ∈ from the submajorization (f g)∗ f ∗ g∗. We will now show that Φf Φg + ≺≺ + + ≤ Φ(f g) u, for some null sequence u l∞( N ),forall0 f, g Mψ .Tothis end, we+ use+ the estimate (see [14]. II 2.2) ∈ ≤ ∈

~~k k k 1 2 tn 2 tn 2 + tn f ∗(s)ds g∗(s)ds (f g)∗(s)ds + ≤ + Z0 Z0 Z0~~

for all n N ,k 0,1,2,...,n 1andall0 f, g Mψ to obtain ∈ = − ≤ ∈ n 1 2k 1t 1 − 1 + n Φf (n) Φg(n) (f g)∗(s)ds + ≤ n ψ(2kt ) + k 0 n 0 X= Z n 1 k 1 1 − 1 2 + tn (f g)∗(s)ds u (n) = n ψ(2k 1t ) + + 1 k 0 + n 0 = Z Xn k 1 1 2 tn (f g)∗(s)ds u (n) = n ψ(2kt ) + + 1 k 1 n 0 X= Z n 1 2kt 1 − 1 n (f g) (s)ds u (n) u (n) k ∗ 1 2 ≤ n ψ(2 tn) + + + k 0 0 X= Z Φ(f g)(n) u (n) u (n) = + + 1 + 2 where

~~n 1 k 1 1 − 1 1 2 + tn u (n) (f g) (s)ds 1 k k 1 ∗ = n ψ(2 tn) − ψ(2 tn) + k 0 + 0 X= Z~~

post56.tex; 23/04/1998; 10:32; p.16 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 63 and n 1 2 tn u (n) (f g)∗(s)ds, 2 = nψ(2nt ) + n Z0 for all n 1, 2,....Since = n 1 k 1 k 1 2 + tn 1 − ψ(2 + tn) 1 0 u (n) 1 (f g)∗(s)ds ≤ 1 = n ψ(2kt ) − ψ(2k 1t ) + k 0 n + n 0 X= Z n 1 n 1 − ψ(2 tn) 1 f g M ≤ n ψ(t ) − k + k ψ k 0 n X= n ψ(2 tn) 1 f g M , = ψ(t ) − k + k ψ n it follows from the choice of the sequence tn n∞ 1 that u1(n) 0asn . Further, since { } = → →∞ 1 0 u (n) f g M ≤ 2 ≤ nk + k ψ it follows also that u (n) 0asn .NowletLbe any generalized limit

2 → →∞ N on l∞( N ),thatis,Lis a positive linear functional on l∞( ) such that L(u) = limn u(n) whenever u(n) n∞ 1 is a convergent sequence. For 0 f Mψ ,we deﬁne→∞φ(f) L(Φf ).{ It is clear} = that φ is positively homogeneous,≤ and∈ from the := argument above, it follows that φ is additive on Mψ+. Consequently, φ has a unique extension to a positive linear functional on Mψ , which we continue to denote by φ. Note that φ, being positive, is necessarily continuous([25]). If 0 f, g Mψ ≤ ∈ satisfy f g, then the inequality Φf Φg holds in l∞( N ) and so φ(f) φ(g), and hence≺≺φ is symmetric. To show that≤φ is non-trivial, observe that ≤

n 1 1 − ψ(2kt ) ψ(0 ) Φ(ψ )(n) n , 0 −k + = n ψ(2 tn) k 0 X= and so Φ(ψ0)(n) 1asn . Consequently, φ(ψ0) 1 and this shows that φ → →∞ = 1 is non-trivial since ψ0 Mψ . Finally, suppose that 0 f Mψ L 0, ), and observe that ∈ ≤ ∈ ∩ [ ∞

~~n 1 k 1 − 1 2 tn Φf (n) f∗(s)ds = n ψ(2kt ) k 0 n 0 X= Z n 1 1 − 1 ∞ f ∗(s)ds ≤ n ψ(tn) k 0 0 X= Z 1 ∞ f ∗(s)ds. = ψ(t ) n Z0~~

~~post56.tex; 23/04/1998; 10:32; p.17 64 P.G. DODDS ET AL.~~

~~It follows that Φf (n) 0asn and hence φ(f) 0. Consequently, φ is supported at and this→ completes→∞ the proof of (i). The proof= of (ii) is similar and ∞ is therefore omitted. 2~~

Remark 3.5. If the concave function ψ satisﬁes the conditions of (ii) in the preceding theorem, then via Remark 2.11, there exist non-trivial symmetric functionals on Mψ 0, 1) which vanish on L∞ 0, 1). We recall that a functional 0 φ Mψ 0, 1)∗ [ [ ≤ ∈ [ is called anormal if φ(f) 0forallf L∞0,1). Moreover, a functional = ∈ [ 0 φ Mψ 0, 1)∗ is called localizable if for every >0, there exists a ≤ ∈ [ measurable subset A 0,1)such that A <and φ(χEf) φ(f) for all ⊆[ | | = f Mψ 0, 1). It is clear that any symmetric functional is not localisable in the above∈ sense.[ Hence, the above theorem yields the existence of anormal nonlocaliz- able functionals on Mψ 0, 1), and hence Theorem 3.4 (ii) can be considered as an extension of a result of G.[ Ya. Lozanovskii [15–17].

Remark 3.6. Let ψ be as in Theorem 3.4 (i) (or (ii)). We denote by Nψ the norm 0 closure in Mψ of the (order) ideal N f Mψ f ∗ kDkψ0 for some k ψ := { ∈ : ≤ ∈ N . Clearly, Nψ is a Banach function space in its own right and is symmetric. } However, under the present assumptions on ψ, the space Nψ is not fully symmetric, and is different from Mψ . For these facts, we refer the reader to [2], [14] II.5.7 and [20]. Of course, the notion of symmetric functional on Nψ has an obvious sense and it is clear that the restriction of an non-trivial symmetric functional on Mψ to Nψ is symmetric. For the construction of symmetric functionals on Nψ , a method different to that employed in the proof of Theorem 3.4 can be employed. We will brieﬂy indicate this alternative construction under the hypothesis 1 that limt ψ(2t)/ψ(t) 1. For any function f (L L∞) 0, ), there exists →∞ f = ∈ + [ ∞ a measurable set A 0, )such that t ⊆[ ∞ t f ∗(s)ds f(s)ds = f | | Z0 ZAt f and At t(see, for example [14] II.2.2.(7)). By the observations made in the proof| of|= Theorem 3.4, there exists a sequence 0

~~post56.tex; 23/04/1998; 10:32; p.18 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 65~~

Indeed, let f, g N0 and let ∈ ψ f g f g Rn A n A n A n+ := 3 tn ∪ 3 tn ∪ 3 tn for n 1, 2,.... It follows that = Ψf (n) Ψ g(n) Ψ(f g)(n) | + − + | 1 1 f(s)ds f(s)ds n n f ≤ ψ(3 tn) Rn − ψ(3 tn) A n Z Z 3 tn

~~1 1 g(s)ds g(s)ds n n g + ψ(3 tn) Rn − ψ(3 tn) A n Z Z 3 tn~~

~~1 1 (f g)(s)ds (f g)(s)ds . n n f g + ψ(3 tn) Rn + − ψ(3 tn) A n+ + Z Z 3 tn~~

It sufﬁces to show that each of the three terms on the right hand side of the above inequality tends to 0 as n . We consider only the ﬁrst term; the remaining →∞ n 1 terms can be treated similarly. Using the fact that Rn 3 + tn, it follows that | |≤ 1 1 f(s)ds f(s)ds n n f ψ(3 tn) Rn − ψ(3 tn) A n Z Z 3 tn

~~1 f(s)ds n f = ψ(3 tn) Rn A n Z \ 3 tn 3n 1t 1 + n f∗(s)ds n n ≤ ψ(3 tn) 3 tn Z n 1 k 3 + tn s ψ ( )ds n 0 ≤ ψ(3 t ) n k n Z3 tn k2 3n 1t 3nt ψ + n ψ( n ) = ψ(3nt ) k − k n ~~

where k N satisfies f ∗ kDkψ0. Using (*), it is not difficult to see that ∈ ≤ k2 3n 1t 3nt lim ψ + n ψ n 0, n ψ(3nt ) k − k = →∞ n and this concludes the proof of the claim. Note that Ψ 1. If L (l∞/c0)∗,we 0 0 0 k k≤ ∈ 0 define φ (N )∗ by setting φ L Ψ.The unique bounded extension of φ L ∈ ψ L := ◦ L to Nψ will be denoted by φL. We claim that φL is symmetricb whenever L 0. It ≥ clearly suffices to show that b

~~n 1 3 tn ∞ φ (f ) f (s)ds L L n ∗ = ψ(3 tn) 0 ( )n 1! Z =~~

post56.tex; 23/04/1998; 10:32; p.19 66 P.G. DODDS ET AL. for all 0 f Nψ . This however follows immediately by observing that, for ≤0 ∈ 0 g N , 0 f Nψ ,wehave ≤ ∈ ψ ≤ ∈ n 1 1 3 tn g(s)ds f∗(s)ds n g n ψ(3 tn) A n − ψ(3 tn) 0 Z 3 tn Z n n 1 3 tn 1 3 tn g∗(s)ds f∗ (s)ds = ψ(3nt ) − ψ(3nt ) n Z0 n Z0 n 1 3 tn (g f ) (s)ds n ∗ ∗ ∗ ≤ ψ(3 tn) 0 − Z n 1 3 tn (g f)∗(s)ds g f M ≤ ψ(3nt ) − ≤k − k ψ n Z0 for all n 1, 2,..., where we have used [14], II.3.1. = The above construction of symmetric functionals on Nψ essentially follows that of Varga [24] in the setting of trace ideals. A similar construction of symmetric functionals can be found in [12] where only the case of the spaces N 0 on 0, ) ψ [ ∞ is considered. As pointed out in [24], if the sequence tn n∞ 1 satisﬁes the condition (*) and if { } = ψ(3nt ) lim n 0, n 1 n ψ(3 + tn 1) = →∞ + 0 then the mapping Ψ maps the unit ball of Nψ onto the unit ball of l∞/c0. Conse- quently, the mapping L φ is an isometry from (l∞/c )∗ into N ∗ in this case. → L 0 ψ The trace ideal analogueb of Mψ was also considered in [24], using averages similar to those used in the proof of Theorem 3.4. However, we wish to point out that the claim in [24], that the mapping so constructed is an isometry from (l∞/c0)∗ into Mψ∗ , is false. Actually, the mapping T constructed in Remark 6 of [24] is not even injective (due to the fact that the operator on l∞/c0 induced by the Cesaro operator on l∞ does not have dense range in l∞/c0). We omit further details. Our next objective is to show that if ψ 0, ) 0, )is a concave function :[ ∞ →[ ∞ satisfying lim inft ψ(2t)/ψ(t) 1 and limt ψ(t) ,thenMψ admits an uncountable linearly→∞ independent collection= of→∞ symmetric=∞ functionals supported at , together with an analogous result for symmetric functionals supported at 0. For the∞ sake of convenience, the construction will be divided into two lemmas.

LEMMA 3.7. Let ψ 0, ) 0, )be a concave function satisfying lim inft :[ ∞ →[ ∞ →∞ ψ(2t)/ψ(t) 1 and limt ψ(t) . For any sequence tn n for which =n →∞ =∞ ↑ ∞ limn ψ(2 tn)/ψ(tn) 1, there exists a symmetric functional 0 <φ Mψ∗ →∞ = ∈ supported at , satisfying φ(ψ0) 1 and ∞ =tn (i) φ(f) lim infn 1/ψ(tn) f ∗(s)ds for all 0 f Mψ ; ≥ →∞ 0 ≤ ∈ tn (ii) φ(f) 0 for all 0 f MRψ for which limn 1/ψ(tn) f ∗(s)ds 0. = ≤ ∈ →∞ 0 = R

~~post56.tex; 23/04/1998; 10:32; p.20~~

SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 67 R Proof. Let L l∞(N ) be generalized limit, that is, L is a positive : → functional on l∞( N ) which coincides with the limit on the subspace of convergent

~~sequences. For 0 f Mψ and k N ,deﬁne ≤ ∈ ∈ k 1 2 tn ∞ Φ(f ) f (s)ds . k L k ∗ := ψ(2 tn) 0 ( )n 1! Z = From the inequalities~~

k 1 2 tn (f g) (s)ds k ∗ ψ(2 tn) 0 + Z k k 1 2 tn 1 2 tn f (s)ds g (s)ds k ∗ k ∗ ≤ ψ(2 tn) 0 + ψ(2 tn) 0 Z k 1 Z k 1 2 tn ψ(2 + tn) 1 + (f g)∗(s)ds, ≤ ψ(2kt ) ψ(2k 1t ) + n + n Z0 together with the fact that ψ(2k 1t ) lim + n 1 n ψ( kt ) →∞ 2 n = it follows that

Φk(f g) Φk(f ) Φk(g) Φk 1(f g) (3.6) + ≤ + ≤ + + for all 0 f, g Mψ and all k N . This implies that the sequence Φk(f ) k∞ 1 is ≤ ∈ ∈ { } = non-decreasing for every 0 f Mψ .Since0 Φk(f ) f M for all k N , ≤ ∈ ≤ ≤k k ψ ∈ it follows that Φk(f ) k∞ 1 l∞(N ).Wenowdeﬁneφ Mψ+ 0, )by setting { } = ∈ : →[ ∞ φ(f) lim Φk(f ). := k →∞ From (3.6), it follows immediately that φ is additive. Since it is clear that φ is positively homogeneous, it follows that φ extends to a positive linear functional on Mψ which we continue to denote by φ. From the deﬁnition, it is clear that φ 1 is symmetric and that φ(ψ0) 1. Further, if 0 f Mψ L 0, ),then = ≤ ∈ ∩ [ ∞ Φk(f ) 0forallk 1,2,...,and so φ(f) 0. Accordingly, φ is supported at = = = . To show that (i) holds, let 0 f Mψ and observe that ∞ ≤ ∈ k 2 tn tn tn 1 1 ψ(tn) 1 f∗(s)ds f ∗(s)ds f ∗(s)ds ψ(2kt ) ≥ ψ(2kt ) = ψ(2kt ) ψ(t ) n Z0 n Z0 n n Z0 k for all k,n N . Since limn ψ(tn)/ψ(2 tn) 1, it follows easily from the properties of∈ the generalized limit→∞ L that =

~~tn tn ψ(tn) 1 ∞ 1 Φk(f ) L f ∗(s)ds lim inf f ∗(s)ds ψ( kt ) ψ(t ) n ψ(t ) ≥ 2 n n 0 n 1 ≥ →∞ n 0 Z = Z~~

~~post56.tex; 23/04/1998; 10:32; p.21 68 P.G. DODDS ET AL.~~

for all k N and this implies (i). ∈ For the proof of (ii), assume that 0 f Mψ satisﬁes limn 1/ψ(tn) tn ≤ ∈ →∞ f ∗(s)ds 0. We then have 0 = k R 1 2 tn 2k tn 0 f (s)ds f (2ks)ds k ∗ k ∗ ≤ ψ(2 tn) 0 = ψ(2 tn) 0 Z Z tn k ψ(tn) 1 2 f ∗(s)ds ≤ ψ(2kt ) ψ(t ) n n Z0 k for all k,n N . Using again the fact that limn ψ(tn)/ψ(2 tn) 1, this implies that ∈ →∞ = t ψ(t ) 1 n ∞ Φ (f ) k n f (s)ds , k 2 L k ∗ 0 ≤ ψ(2 tn) ψ(tn) 0 n 1 = Z = and so Φk(f ) 0forallk 1,2, .Consequently, φ(f) 0, and this proves = = ··· =

~~(ii). 2~~

~~The construction presented in the following lemma is essentially due to Sedaev [21]. We include the details for the convenience of the reader.~~

~~LEMMA 3.8. Let ψ 0, ) 0, )be a concave function satisfying :[ ∞ →[ ∞ ψ(t) lim ψ(t) , lim 0. (3.7) t t t →∞ =∞ →∞ =~~

~~Assume that tn n∞ 1 is a sequence such that 0~~

~~→∞ + →∞ + →∞ + N Let N nk k∞1 mk k∞1 be a partition of into two disjoint subsets, where 1 ={ }= ∪{ } = = n~~

~~1 tn 1 tn tn tn 1 lim + − ψ(tn 1) − − ψ(tn 1) 0. (3.9) t ψ(tn) tn 1 tn 1 − + tn 1 tn 1 + = →∞ + − − + − − Now deﬁne the function f by setting~~

~~∞ ψ(sk 1) ψ(sk) + χ f − sk,sk 1). := sk sk [ + k 1 1 X= + −~~

~~post56.tex; 23/04/1998; 10:32; p.22 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 69~~

Since ψ is concave, it is clear that f is decreasing so that f f ∗. Further, = sk f(t)dt ψ(sk) ψ(0) (3.10) = − Z0 for all k 1, 2,...,and sf(t)dt ψ(s) ψ(0) for all s 0. This implies = 0 ≤ − ≥ in particular that f Mψ and f M 1. It is clear that (3.10) implies (i). It

~~∈ R k k ψ = N remains to show that f satisﬁes (ii). For each k N , there exists j such that ∈ ∈ sj~~

rk sj 1 rk rk sj f(t)dt + − ψ(sj ) − ψ(sj 1) ψ(0) 0 = sj 1 sj + sj 1 sj + − Z + − + − tmk 1 tmk tmk tmk 1 + − ψ(tmk 1) − − ψ(tmk 1) ψ(0). ≤ tmk 1 tmk 1 − + tmk 1 tmk 1 + − + − − + − − Since mk as j , property (ii) is now an immediate consequence of →∞ →∞

~~(3.9), by which the lemma is proved. 2~~

THEOREM 3.9. Let ψ 0, ) 0, )be a concave function such that :[ ∞ →[ ∞ ψ(2t) lim inf 1, lim ψ(t) t ψ(t) t →∞ = →∞ =∞ ψ(2t) respectively, lim inf 1, lim ψ(t) 0 . (3.11) t 0 ψ(t) = t 0 = → → + There exists an uncountable collection of linearly independent symmetric functionals supported at (respectively, supported at 0) in the Marcinkiewicz space ∞ Mψ . Proof. We consider only the case of functionals supported at . First observe that (3.11) implies that ψ satisﬁes the conditions (3.7) in Lemma∞ 3.8. Therefore, there exists a sequence tn n∞ 1 satisfying (3.8) and simultaneously the condition in { } =

Lemma 3.7. Let A be an uncountable collection of countable subsets of N with the property that 1 α and α β is ﬁnite for all α, β A with α β. We consider ∈ ∩ ∈ 6= each α A as a sequence α(k) k∞ 1such that 1 α(1)<α(2)< . ∈ { } = = ··· ↑ ∞ For each α A there is a corresponding subsequence tα(k) k∞ 1 of the sequence ∈ { } = tn n∞ 1. From Lemma 3.8, for each α A, there exists a function 0 fα Mψ { } = ∈ ≤ ∈ such that f ∗ fα and α = 1 tα(k) 1 tβ(k) lim fα(t)dt 1, lim fα(t)dt 0,α β. k ψ(t ) = k ψ(t ) = 6= →∞ α(k) Z0 →∞ β(k) Z0 (3.12)

~~For each α A,let0 φα M∗ be a symmetric functional which is supported ∈ ≤ ∈ ψ at with φα 1 and which has the properties (i), (ii) of Lemma 3.7 with ∞ k k=~~

post56.tex; 23/04/1998; 10:32; p.23 70 P.G. DODDS ET AL. respect to the sequence tα(k) k∞ 1. These properties, together with (3.12) imply that { } = φα(fα) 1andφα(fβ) 0forallα, β A with α β. Accordingly, the = = ∈ 6= system φα α A is linearly independent and this completes the proof of the { : ∈ } theorem. 2

4. Symmetric Functionals on Fully Symmetric Operator Spaces In this section, we will extend the ideas of the previous sections to the setting of (noncommutative) spaces of measurable operators. We denote by M asemi- finite von Neumann algebra on the Hilbert space H, with a fixed faithful and normal semifinite trace τ. The identity in M is denoted by 1. A linear operator x:dom(x) H, with domain dom(x) H, is called affiliated with M if ux xu → ⊆ = for all unitary u in the commutant M0 of M. The closed and densely defined operator x, affiliated with M, is called τ-measurable if for every >0there exists an orthogonal projection p M such the p(H) dom(x) and τ(1 p) < . The collection of all τ-measurable∈ operators is denoted⊆ by M. With the− sum and product defined as the respective closures of the algebraic sum and product, M is a *-algebra. For ,δ > 0 we denote by N(,δ) the set of alle x M for which ∈ there exists an orthogonal projection p M such that p(H) dom(x), xp e ∈ ⊆ k k∞ ≤ and τ(1 p) δ,where denotes the usual operator norm.e The sets N(,δ) ,δ− >≤0 form a basek·k at∞ 0 for a metrizable Hausdorff topology in M, which{ is called: the measure} topology. Equipped with this measure topology, M is a complete topological *-algebra. These facts and their proofs can be found in thee papers [18] and [23]. e We next recall the notion of generalized singular value function [10]. Given a self-adjoint operator x in H, we denote by ex( ) the spectral measure of x.Now x · assume that x M.Thene| |(B) M for all Borel sets B R , and there exists ∈ x ∈ ⊆ s>0 such that τ(e| |(s, )) < .Forx Mand t 0wedefine ∞ ∞ ∈ ≥ e x µt(x) inf s 0 τ(e| |(s, )) t . = { ≥ : ∞ ≤ } e The function µ(x) 0, ) 0, is called the generalized singular value :[ ∞ →[ ∞] function (or decreasing rearrangement) of x; note that µt (x) < for all t>0. For the basic properties of this singular value function we refer the∞ reader to [10]. We note that a sequence xn Mconverges to 0 for the measure topology if { }⊆ and only if µt (x) 0forallt>0. If we consider M L∞( R + ,m),where → = mdenotes Lebesgue measure on Re+ , as an abelian von Neumann algebra acting 2 via multiplication on the Hilbert space H L ( R + ,m), with the trace given by integration with respect to m, it is easy to= see that M consists of all measurable

functions on R + which are bounded except on a set of ﬁnite measure, and that for f M, the generalized singular value function µ(fe ) is precisely the decreasing ∈ rearrangement f ∗.IfM L(H)and τ is the standard trace, then it is not difﬁcult = to seee that M M and that the measure topology coincides with the operator = e

~~post56.tex; 23/04/1998; 10:32; p.24 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 71 norm topology. If x M,thenxis compact if and only if limt µt (x) 0; in this case, ∈ →∞ =~~

µn(x) µt (x), t n, n 1), n 0, 1, 2,..., = ∈[ + = and the sequence µn(x) n∞ 0 is just the sequence of eigenvalues of x in non- increasing order and{ counted} = according to multiplicity. | | Given a semiﬁnite von Neumann algebra (M,τ) and a (fully) symmetric Ba-

nach function space (E, E)on ( R + ,m), we deﬁne the corresponding non- commutative space E(M,τ)k·kby setting E(M,τ) x M µ(x) E . ={ ∈ : ∈ } Equipped with the norm x µ(x) , the space (E(M,τ), ) e E(M,τ) E E(M,τ) is a Banach space and isk calledk the:= (non-commutative) k k symmetric operatork·k space associated with (M,τ)corresponding to (E, E). An extensive discussion of the various properties of such spaces can be foundk·k in [5,8,9].

DEFINITION 4.1. A linear functional φ E(M,τ)∗ is called symmetric if φ is positive, (that is, φ(x) 0 whenever 0∈ x E(M,τ)) and φ(x) φ(y) whenever µ(x) µ(y), 0 ≥x,y E(M,τ)≤. ∈ ≤ ≺≺ ≤ ∈ We show ﬁrst that a symmetric functional 0 φ0 E∗ can be lifted to a symmetric functional on E(M,τ). ≤ ∈

THEOREM 4.2. Let 0 φ E∗ be symmetric. If φ(x) φ (µ(x)), for all ≤ 0 ∈ := 0 0 x E(M,τ),thenφextends to a symmetric functional 0 φ E(M,τ)∗. ≤ ∈ ≤ ∈ Proof. It clearly sufﬁces to show that φ is additive on E(M,τ)+.Let0 x,y ≤ ∈ E(M,τ).Sinceµ(x y) µ(x) µ(y) ([10] Theorem 4.4), and since φ0 is symmetric, it follows+ that ≺≺ + φ(x y) φ (µ(x y)) φ (µ(x) µ(y)) φ(x) φ(y). + = 0 + ≤ 0 + = + To prove the converse inequality, we use the easily veriﬁed fact ([12], Proposition 1.10) that t t 2t µs (x)ds µs (y)ds µs (x y)ds. (4.1) + ≤ + Z0 Z0 Z0 Observing that (4.1) is equivalent to the submajorization

µ(x) µ(y) 2D / µ(x y), (4.2) + ≺≺ 1 2 + it follows from Proposition 2.3 that φ(x) φ(y) φ (µ(x) µ(y)) + = 0 + 2φ (D / µ(x y)) φ (µ(x y)) φ(x y). ≤ 0 1 2 + = 0 + = + Thus φ is additive on E(M,τ) and this sufﬁces to complete the proof of the

~~Theorem. 2~~

~~post56.tex; 23/04/1998; 10:32; p.25 72 P.G. DODDS ET AL.~~

Remark 4.3. If 0 φ0 E∗, is a singular symmetric functional, then φ0(f ) 0 1 ≤ ∈ = for all f L 0, ) L∞ 0, ) (see proposition 2.6). Hence, for the corre- ∈ [ ∞ ∩ [ ∞ sponding symmetric functional 0 φ E(M,τ)∗, it follows that φ(x) 0for 1 ≤ ∈ = all x L L∞(M,τ). Similarly, if φ0 is supported at ,thenφ(x) 0for all x ∈ L1(∩M,τ) E(M,τ) and if φ is supported at 0,∞ then φ(x) 0forall= x M∈ E(M,τ).∩It may happen that φ 0. This will be the case, for= example, ∈ ∩ = if φ0 is supported at and τ(1)< ,orifφ0 is supported at 0 and M B(H). It is now clear that∞ Theorem 4.2,∞ in combination with the results of= Section 3, can be used to construct non-trivial symmetric functionals on non-commutative Marcinkiewicz spaces Mψ (M,τ) for appropriate concave functions ψ. The following results show that for a large class of non-commutative spaces E(M,τ), every symmetric functional on E(M,τ)can be obtained via the construction given by Theorem 4.2. THEOREM 4.4. Let (M,τ)be a semi-finite von Neumann algebra without mini- mal projections, and let E be a fully symmetric Banach function space on 0, ). [ ∞ If φ E(M,τ)∗ is a symmetric functional, then there exists a symmetric ≤ ∈ functional 0 φ0 E∗ such that φ(x) φ0(µ(x)) for all 0 x E(M,τ). Proof. In≤ view∈ of Remark 2.11, it is= sufficient to show that≤ there∈ exists a symmetric functional 0 φ E 0,τ(1))∗ satisfying φ(x) φ (µ(x)) for all ≤ 0 ∈ [ = 0 0 x E(M,τ).LetN be the commutative von Neumann algebra L∞ 0,τ(1)), with≤ trace∈ given by integration. The algebra N may be identified with the[ space of all measurable functions on 0,τ(1)) which are bounded except on a set of finite [ measure. Since M does not contain any minimale projections, there exists a positive rearrangement-preserving algebra homomorphism J N M ([6], Lemma 4.1, : → [8], Theorem 3.5). Let 0 φ E(M,τ)∗ be symmetric. For f E 0,τ(1)), ≤ ∈ ∈ [ define φ0(f ) φ(Jf). It is clear that 0 φ Ee 0,τ(e1))∗ is symmetric. Moreover, if 0:= x E(M,τ),thenµ(J (µ(x)))≤ ∈ µ(x)[ and hence φ(x) ≤ ∈ = =

φ(J (µ(x))) φ (µ(x)). 2 = 0 THEOREM 4.5. Let M B(H) with standard trace τ and let E be a fully = symmetric Banach function space on 0, ).If0 φ E(M,τ)∗ is a symmetric [ ∞ ≤ ∈ functional, then there exists a symmetric functional 0 φ E∗ such that φ(x) ≤ 0 ∈ = φ0(µ(x)) for all 0 x E(M,τ). Proof. Let 6 denote≤ ∈ the σ -algebra of subsets of 0, ) generated by the in- [ ∞ tervals n, n 1), n 0, 1, 2,..., and let E6 be the corresponding conditional [ + = 1 expectation operator deﬁned on (L L∞) 0, ).SinceE6f f for all f 1 + [ ∞ ≺≺ ∈ (L L∞) 0, ), it follows in particular that E6 deﬁnes a positive contractive + [ ∞ projection in E. The range E6 E6(E) consists of all f E which are constant = ∈ on each of the intervals n, n 1).Let en n∞0 be any orthonormal sequence in H. [ + { } = Let J E6 E(M,τ)be given by : → ∞ (Jf )(ξ) αn ξ,en en,ξH := h i ∈ n 0 X=

post56.tex; 23/04/1998; 10:32; p.26 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 73 χ for all f n∞ 0 αn n,n 1) in E6. It is clear that J is a positive rearrangement- preserving= linear= mapping.[ + For 0 f E,wenowdeﬁne P ≤ ∈ φ (f ) φ(JE6f ∗). 0 := To show that φ is additive on E+, we ﬁrst show that if f, g E satisfy f ∗ g∗, 0 ∈ ≺≺ then E6f ∗ E6g∗. Indeed, since E6f ∗, E6g∗are decreasing and constant on each of the intervals≺≺ n, n 1), this follows immediately by noting that [ + n n n n E6f ∗(s)ds f ∗(s)ds g∗(s)ds E6g∗(s)ds = ≤ = Z0 Z0 Z0 Z0 for n 1, 2,....Iff, g E,then(f g)∗ f ∗ g∗ and so = ∈ + ≺≺ + J E6(f g)∗ JE6f ∗ JE6g∗. + ≺≺ + Hence

~~φ (f g) φ(JE6(f g)∗) 0 + = + φ(JE6f∗ JE6g∗) φ (f ) φ (g). ≤ + = 0 + 0 For the proof of the reverse inequality, we ﬁrst observe that~~

E6D / h∗ D / E6h∗ (4.3) 1 2 ≺≺ 1 2 for all 0 h E. Indeed, for n 1, 2,...,wehavethat ≤ ∈ = n n 1 2n E6D / h∗(s)ds h∗(2s)ds h∗(s)ds 1 2 = = 2 Z0 Z0 Z0 1 2n n E6h∗(s)ds D / E6h∗(s)ds. (4.4) = 2 = 1 2 Z0 Z0 t Since the function t E6D 1 h∗(s)ds is linear on each of the intervals n, n 1) 0 2 → t [ + and since the function t D / E6h∗(s)ds is concave, (4.4) implies that R→ 0 1 2 t R t E6D / h∗(s)ds D / E6h∗(s)ds, t > 0 1 2 ≤ 1 2 Z0 Z0 which is just (4.3). A second observation needed is that

φ(JD E6h∗) 2φ(JE6h∗) (4.5) 2 = for all 0 h E. Note that D2E6h∗ E6, so the left hand side of (4.5) is well ≤ ∈ ∈ χ deﬁned. For the proof of (4.5), write E6h∗ n∞ 0 αn n,n 1).Ifx1,x2 E(M,τ) are deﬁned by setting by = = [ + ∈ P ∞ ∞ x1(ξ) αn ξ,e2n e2n,x2(ξ) αn ξ,e2n 1 e2n 1, := h i := h + i + n 0 n 0 X= X=

~~post56.tex; 23/04/1998; 10:32; p.27 74 P.G. DODDS ET AL. for all ξ H,thenµ(x ) µ(x ) E6h∗ and µ(x x ) D E6h∗. Hence ∈ 1 = 2 = 1 + 2 = 2~~

φ(JD E6h∗) φ(x x ) 2φ(JE6h∗) 2 = 1 + 2 = and this proves (4.5). If now f, g E, then the submajorization f ∗ g∗ ∈ + ≺≺ 2D / (f g)∗ implies that E6(f ∗ g∗) 2E6D / (f g)∗. Hence 1 2 + + ≺≺ 1 2 +

φ (f ) φ (g) φ(JE6f∗) φ(JE6g∗) 0 + 0 = + φ(JE6(f ∗ g∗)) = + 2φ(JE6D / (f g)∗) ≤ 1 2 + φ(JD E6D / (f g)∗), = 2 1 2 + where the last equality follows from (4.5). Now (4.3) implies that

~~D E6D / (f g)∗ E6(f g)∗ 2 1 2 + ≺≺ + and consequently~~

φ(JD E6D / (f g)∗) φ(JE6(f g)∗) φ (f g). 2 1 2 + ≤ + = 0 + We have just shown that φ (f ) φ (g) φ (f g), and from this it now follows 0 + 0 ≤ 0 + that φ0 is additive on E+. This sufﬁces to complete the proof of the theorem. 2

We note ﬁnally that the alternative construction outlined in Remark 3.6 can be carried out directly in the non-commutative setting. However, in the ﬁnal step, it is necessary to appeal to [8] rather than to [14], II.3.1. We leave the details to the interested reader.

References 1. S. Albeverio, D. Guido, A. Ponosov and S. Scarlatti, Singular traces and compact operators, J. Funct. Anal. 137 (1996), 281–302. 2. M. Braverman and A. Mekler, The Hardy-Littlewood property for symmetric spaces, Sib. Mat. Zhur. 18 (1977) 522–540 (Russian). 3. C. Bennett and R. Sharpley, Interpolation of Operators, Academic Press, 1988. 4. A. Connes, Noncommutative Geometry, Academic Press, 1994. 5. V.I. Chilin and F.A. Sukochev, Symmetric spaces over semiﬁnite von Neumann algebras, Soviet Math. Dokl. 42 (1991), 97–101. 6. V.I. Chilin and F.A. Sukochev, Weak convergence in non-commutative symmetric spaces, J. Operator Theory 31 (1994), 35–65. 7. J. Dixmier, Existence de traces non normales, C.R. Acad. Sci. Paris 262 (1966) A1107–A1108. 8. P.G. Dodds, T.K. Dodds and B. de Pagter, Non-commutative Banach function spaces, Math. Z. 201 (1989), 583–597. 9. P.G. Dodds, T.K. Dodds and B. de Pagter, Fully symmetric operator spaces, Integr. Equat. Oper. Th. 15 (1992), 942–972. 10. T. Fack and H. Kosaki, Generalized s-numbers of τ-measurable operators, Paciﬁc J. Math. 123 (1986), 269–300.

~~post56.tex; 23/04/1998; 10:32; p.28 SYMMETRIC FUNCTIONALS AND SINGULAR TRACES 75~~

11. I.C. Gohberg and M.G. Krein, Introduction to the theory of non-selfadjoint operators Transla- tions of Mathematical Monographs, vol.18, AMS (1969). 12. D. Guido and T. Isola, Singular traces on semiﬁnite von Neumann algebras, J. Funct. Anal. 134 (1995), 451–485. 13. A. Kaminska, Some remarks on Orlicz–Lorentz spaces, Math. Nachr. 147 (1990) 29–38. 14. S.G. Krein, Ju. I. Petunin and E.M. Semenov, Interpolation of linear operators, Translations of Mathematical Monographs, Amer. Math. Soc. 54 (1982). 15. G.Ya. Lozanovskii, On localizable functionals in vector lattices, in: Theory of Functions, Functional Analysis and their Applications, 19 (1974) 66–80 (Russian). 16. G.Ya. Lozanovskii, On the representation of linear functionals in Marcinkiewicz spaces, Izvestiya VUZov. Matematika N1 (188) (1978) 43-53. (Russian) 17. G.Ya. Lozanovskii, A supplement to the paper “On localizable functionals in vector lattices”, J. Soviet Math. 14 (1980), 1170–1173. 18. E. Nelson, Notes on non-commutative integration, J. Functional Anal. 15 (1974), 103–116. 19. R. Prinzis, Traces Residuelles et Asymptotique du Spectre d’Operateurs Pseudo-Differentiels, Ph.D. Thesis, Lyon, 1995. 20. G.Russu, Symmetric spaces not having the majorant property, Mat. Issled. 4 (1969), 82–93 (Russian). 1 21. A.A. Sedaev, A symmetric space which is intermediate for the couple L ,L∞ but which is not a subspace of any interpolation space,inStudies in the theory of{ functions} of many real variables, Collection of papers, Yaroslavl (1990), (134–139). 22. K. Sundaresan, Additive functionals on Orlicz spaces, Studia Math. 32 (1969), 269–276. 23. M. Terp, Lp-spaces associated with von Neumann algebras, Notes, Copenhagen Univ. (1981). 24. J.V. Varga, Traces on irregular ideals, Proc. Amer. Math. Soc. 107 (1989) 715–723. 25. A.C. Zaanen, Riesz Spaces II, (North-Holland), Amsterdam-New York-Oxford (1983).

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