aa r X i v : . [ m a t h . F A ] J a n SUBREARRANGEMENT-INVARIANT FUNCTION SPACES
BEN WALLIS
Abstract.
Rearrangement-invariance in function spaces can be viewed asa kind of generalization of 1-symmetry for Schauder bases. We definesubrearrangement-invariance in function spaces as an analogous generaliza-tion of 1-subsymmetry. It is then shown that every rearrangement-invariantfunction space is also subrearrangement-invariant. Examples are given todemonstrate that not every function space on (0 , ∞ ) admits an equiva-lent subrearrangement-invariant norm, and that not every subrearrangement-invariant function space on (0 , ∞ ) admits an equivalent rearrangement-invariant norm. The latter involves constructing a new family of functionspaces inspired by D.J.H. Garling, and we further study them by showing thatthey are Banach spaces containing copies of ℓ p . Introduction
A 1-symmetric basis can be viewed as a rearrangement-invariant function spaceover N (equipped with the counting measure). Some bases, though, are only 1-subsymmetric, and we would like to explore an analogous property for certainnonatomic function spaces corresponding to this generalization. For instance,in the case where we have a function space X on (0 , ∞ ), we say that it is subrearrangement-invariant whenever the following holds:For every function f ∈ X , every measurable F ⊆ (0 , ∞ ), and everystrictly increasing bijection m : (0 , ∞ ) → F such that m and m − are both measure-preserving, we have k f ◦ m k X = k f F k X .Later, in §
2, we give a broader definition for other nonatomic function spacesbesides (0 , ∞ ).D.J.H. Garling was the first to prove that not every subsymmetric basis issymmetric, by publishing a counterexample in 1968 ([Ga68, § N which is not rearrangement-invariant under any equiva-lent norm. It is also well-known that there exist 1-unconditional bases whichare not subsymmetric, and hence function spaces on N which fail to admit anequivalent subrearrangement-invariant norm. In §
3, we extend these results toa purely nonatomic case, exhibiting an example of a function space on (0 , ∞ )which is not essentially subrearrangement-invariant, and another example whichis subrearrangement-invariant but not essentially rearrangement-invariant. Somegeometric properties of these “Garling function spaces” are then explored in § Mathematics Subject Classification.
Primary 46E30; Secondary 06A05, 28A99.
Key words and phrases.
Rearrangement-invariant function spaces, subsymmetric bases inBanach spaces.
All Banach spaces and function spaces are taken over the real field R , and allmeasure spaces we assume to be countably additive and σ -finite. If θ and φ arereal-valued functions, we use the symbolism θ ( x ) ≈ ε φ ( y ) to mean that for any ε >
0, the arguments x and y can be chosen such that φ ( y ) − ε < θ ( x ) < φ ( y ) + ε. Beyond that, all notation and terminology is either standard (such as appears,for instance, in [LT77]) or defined as encountered. In § Subrearrangement-invariant function spaces
For the following definition, β denotes the Borel measure. Definition 2.1.
Let (Ω , µ ) be a σ -finite measure space, and let M +0 (Ω) denotethe cone of (nonnegative) ( µ, β )-measurable functions f : Ω → [0 , ∞ ]. Suppose ρ : M (Ω) → [0 , ∞ ] satisfies the following properties for all a ∈ (0 , ∞ ) and all f, g ∈ M +0 (Ω):(i) ρ ( f + g ) ρ ( f ) + ρ ( g );(ii) ρ ( af ) = aρ ( f ); and(iii) ρ ( f ) = 0 if and only if f ≡ X, k·k X ) consisting the a.e.-equivalenceclasses of measurable functions f : Ω → [ −∞ , ∞ ] satisfying k f k X := ρ ( | f | ) < ∞ .In this case we say that ρ is a function norm on Ω, and X is a function space on Ω with respect to ρ .Note that our definition differs from other classes of function spaces such asBanach function spaces defined in [BS88, §
1] or K¨othe function spaces. It issuitable for the present purposes, however.
Remark . If ( e i ) ∞ i =1 is a 1-unconditional basis for a Banach space X , we candefine a function norm ρ X by setting, for all f : N → [0 , ∞ ], ρ X ( f ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 f ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X if P ∞ i =1 f ( i ) e i converges, and ∞ otherwise.In this way, X can be viewed as a function space on N , with respect to ( e i ) ∞ i =1 .Let (Ω , µ ) be a σ -finite measure space, and f : Ω → [ −∞ , ∞ ] a ( µ, β )-measurable function (where β is the Borel measure). The distribution function dist f : [0 , ∞ ] → [0 , ∞ ] of f is given by the ruledist f ( s ) = µ { x ∈ Ω : | f ( x ) | > s } . Two such measurable functions f and g are said to be equimeasurable wheneverdist f = dist g . In this case we write f ∼ g . UBREARRANGEMENT-INVARIANT FUNCTION SPACES 3
Definition 2.3.
Let (Ω , µ ) be a σ -finite measure space. A function space X onΩ is called rearrangement-invariant iff k f k X = k g k X for all equimeasurablefunctions f, g ∈ X . It is essentially rearrangement-invariant provided itadmits an equivalent rearrangement-invariant norm. Remark . The above definition follows [BS88] rather than the somewhat morestilted definition of rearrangement-invariance found, for instance, in [LT79].Let (
E, µ E ) and ( F, µ F ) be measure spaces. A map m : E → F is called ( µ E , µ F )-measurable (or, when µ E and µ F are clear from context, simply, measurable ) if whenever A is a measurable subset of F , the set m − ( A ) ismeasurable in E . The map m is called a measure-preserving transformation if whenever A is a measurable subset of F , the set m − ( A ) is measurable with µ E ( m − ( A )) = µ F ( A ). If furthermore m is bijective with m − also measure-preserving, we say that it is a measure-isomorphism . Definition 2.5. If E and F are totally-ordered measure spaces, we denote by MO ( E, F ) the set of all maps m : E → F such that m is both a stictly increasingmeasure-isomorphism. Any such m ∈ MO ( E, F ) is called an MO -isomorphism between E and F .Let us now introduce the main subject under study. Definition 2.6.
Let (Ω , µ ) be a totally-ordered σ -finite measure space satisfy-ing µ (Ω) = ∞ . We say that a function space X on Ω is subrearrangement-invariant if for every measurable F ⊆ Ω, every m ∈ MO (Ω , F ), and every f ∈ X ,we have k f ◦ m k X = k f F k X . We say that X is essentially subrearrangement-invariant whenever it admits an equivalent subrearrangement-invariant norm.Here, the restriction µ (Ω) = ∞ has been included since MO (Ω , F ) would beempty otherwise, whenever µ ( F ) = µ (Ω), and that would make every functionspace on Ω trivially subrearrangement-invariant.The following well-known fact is proved in the appendix. Proposition 2.7.
A 1-unconditional basis ( e i ) ∞ i =1 for a real Banach space X is 1-symmetric if and only if X is rearrangement-invariant as a function spaceon N with respect to ( e i ) ∞ i =1 . It is symmetric if and only if X is essentiallyrearrangement-invariant. Next, we give a result which in some sense justifies our definition ofsubrearrangement-invariance.
Proposition 2.8.
A 1-unconditional basis ( e i ) ∞ i =1 for a real Banach space X is 1-subsymmetric if and only if X is subrearrangement-invariant as a function spaceon N with respect to ( e i ) ∞ i =1 . It is subsymmetric if and only if X is essentiallysubrearrangement-invariant.Proof. ( ⇒ ): Suppose ( e i ) ∞ i =1 is 1-subsymmetric. Let F ⊆ N and m ∈ MO ( N , F ),and select any f ∈ X . By 1-subsymmetry of ( e i ) ∞ i =1 we have BEN WALLIS k f ◦ m k X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 ( f )( m ( i )) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X i ∈ F f ( i ) e m − ( i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X i ∈ F f ( i ) F ( i ) e m − ( i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 f ( i ) F ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = k f F k X . Hence, X is subrearrangement-invariant with respect to ( e i ) ∞ i =1 .( ⇐ ): Suppose now that X is subrearrangement-invariant with respect to( e i ) ∞ i =1 . Let ( e i k ) ∞ k =1 be a subsequence and f ∈ X . Define m ( k ) = i k for k ∈ N ,and F := ( i k ) ∞ k =1 . Clearly, m ∈ MO ( N , F ). Define g : N → [0 , ∞ ] by letting g ( i ) = ( | f | ◦ m − )( i ) if i ∈ F and g ( i ) = 0 otherwise. We will need to check that g ∈ X , but this follows from the facts below, together with the identity g F = g .Now, by subrearrangement-invariance and 1-unconditionality we have (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 f ( k ) e i k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 ( f ◦ m − )( i k ) e i k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X i ∈ F ( f ◦ m − )( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 g ( i ) F ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = k g F k X = k g ◦ m k X = k f ◦ m − ◦ m k X = k f k X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 f ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X . That ( e n ) ∞ n =1 is subsymmetric if and only if it is essentially subrearrangement-invariant follows easily by considering the equivalent norm ||| x ||| =sup ( n k ) ∈ N ↑ k P ∞ k =1 x ∗ k ( x ) x n k k . (cid:3) It is well-known that every 1-symmetric basis is 1-subsymmetric. Similarly, it iseasy to show that rearrangement-invariance implies subrearrangement-invariance.We just need a quick preliminary fact before we do.
Proposition 2.9 ([BS88, Proposition 2.7.2]) . Let m : E → F be a measurepreserving transformation between σ -finite measure spaces ( E, µ E ) and ( F, µ F ) .If f : F → [0 , ∞ ] is a ( µ F , β ) -measurable function on F , then f ◦ m : E → [0 , ∞ ] is a ( µ E , β ) -measurable function on E , and f and f ◦ m are equimeasurable. Proposition 2.10.
Let (Ω , µ ) be a totally-ordered σ -finite measure space satis-fying µ (Ω) = ∞ . If X is a rearrangement-invariant function space on Ω , then itis also subrearrangement-invariant.Proof. Select any f ∈ X , measurable F ⊆ Ω, and m ∈ MO (Ω , F ). Notice that f | F ◦ m = f ◦ m so that, by Proposition 2.9, f ◦ m ∼ f | F . We also clearly have f F ∼ f | F , and hence f ◦ m ∼ f F . By rearrangment-invariance this means k f ◦ m k X = k f F k X . (cid:3) Let us close this section by discussing the nontriviality of essential-subrearrangement invariance. There are, after all, well-known examples of 1-unconditional bases which are not subsymmetric under any renorming, for in-stance the basis for the Tsirelson space. This furnishes us with examples of
UBREARRANGEMENT-INVARIANT FUNCTION SPACES 5 function spaces on N which are not essentially subrearrangement-invariant. Thefollowing example—a simple modification of the Schreier sequence space—givesus a function space on the purely nonatomic measure space (0 , ∞ ) which fails tobe essentially subrearrangement-invariant. Example 2.11.
Denote by A the family of all subsets A of (0 , ∞ ) satisfying λ ( A ) √ inf A . For a nonnegative ( λ, β )-measurable function f : (0 , ∞ ) → [0 , ∞ ], we set ρ Y ( f ) = sup A ∈A Z A f ( t ) dt. Then ρ Y is a function norm, and we can denote by Y the function space itgenerates. Furthermore, Y is a Banach space which fails to be essentiallysubrearrangement-invariant. Proof.
That ρ Y is a function norm is clear from the definition, and it’s routine(via an argument such as in Proposition 4.1) to show completeness. So we needonly prove that it fails to be essentially subrearrangement-invariant. Select any b ∈ (0 , ∞ ). When selecting A ∈ A to estimate k (0 ,b ] k Y , we may assume withoutloss of generality that inf A b , else R A (0 ,b ] ( t ) dt = 0. Hence, Z A (0 ,b ] ( t ) dt λ ( A ) √ inf A √ b so that k (0 ,b ] k Y √ b . On the other hand, if c > √ b then k ( c,c + b ] k Y = b. It isclear that ( c,c + b ] ◦ m = (0 ,b ] for the shift map m ∈ MO ((0 , ∞ ) , ( c, ∞ )) definedby m ( t ) = t + c . Hence, k ( c,c + b ] k Y k ( c,c + b ] ◦ m k Y = k ( c,c + b ] k Y k (0 ,b ] k Y > b √ b = √ b. As b ∈ (0 , ∞ ) was arbitrary, it follows that Y is not essentially subrearrangement-invariant. (cid:3) Garling function spaces
The converse of Proposition 2.10 fails to hold in general, as can be seen from thefollowing example. If 1 p < ∞ and w = ( w ( k )) ∞ k =1 is a nonincreasing sequenceof positive real numbers satisfying w ∈ c \ ℓ , then we can define the Garlingsequence space g ( w, p ) as the space of all scalar sequences f : N → [ −∞ , ∞ ]satisfying k f k g := sup ( i ( k )) ∞ k =1 ∈ N ↑ ∞ X k =1 | f ( i ( k )) | p w ( k ) ! /p < ∞ , where N ↑ denotes the family of all increasing sequences in N . (We usually alsoimpose the condition that w (1) = 1 but this is not always necessary.) It is knownfrom [AAW18, Proposition 2.4] and [AALW18, Lemma 3.1] that the unit vectorsin g ( w, p ) form a 1-unconditional basis which is 1-subsymmetric but not symmet-ric. In particular, thusly viewed as a function space on N , by Propositions 2.7and 2.8, it is subrearrangement-invariant but fails to be rearrangement-invariant, BEN WALLIS or even just essentially rearrangement-invariant. Nevertheless, it remains to beseen whether essential subrearrangement-invariance is a strictly weaker condi-tion than essential rearrangement-invariance in the nonatomic setting. We de-vote this section, therefore, to exhibiting a function space on (0 , ∞ ) which issubrearrangement-invariant but fails to be essentially rearrangement-invariant.To accomplish this, we shall simply generalize Garling’s construction. In fact,we will use the very same “split into two sums” trick that Garling did in hisoriginal paper [Ga68, § § Proposition 3.1.
Fix a nonincreasing function w : N → (0 , ∞ ) with w ∈ c \ ℓ .For each function f : N → [0 , ∞ ] we define ρ g ( f ) = sup E,F ⊆ N m ∈ MO ( E,F ) X k ∈ E ( f ◦ m )( k ) p w ( k ) ! /p . Then ρ g is a function norm generating the space g ( w, p ) .Proof. Let ( i ( k )) ∞ k =1 ∈ N ↑ . By taking E = N , F = ( i ( k )) ∞ k =1 , and m ( k ) = i ( k ),it is clear that ρ g ( f ) > k f k g . For the reverse inequality, let E, F ⊆ N and m ∈ MO ( E, F ). We may assume without loss of generality that E and F areboth infinite. Thus, there is a unique n ∈ MO ( N , E ), and this satisfies m ◦ n ∈ MO ( N , F ). Since w is nonincreasing, we have X k ∈ E ( f ◦ m )( k ) p w ( k ) = ∞ X j =1 ( f ◦ m ◦ n )( j ) p w ( n ( j )) ∞ X j =1 ( f ◦ m ◦ n )( j ) p w ( j ) k f k pg . That ρ g is a function norm generating g ( w, p ) follows immediately. (cid:3) Definition 3.2.
Let W denote the set of all nonincreasing ( λ, β )-measurablefunctions W : (0 , ∞ ) → (0 , ∞ ) satisfying the following conditions:(W1) lim t →∞ W ( t ) = 0,(W2) R ∞ W ( t ) dt = ∞ , and(W3) R W ( t ) dt < ∞ .We also denote by λ the Lebesgue measure and Λ the family of Lebesgue-measurable subsets of (0 , ∞ ). (Recall that β is the Borel measure.) For each( λ, β )-measurable f : (0 , ∞ ) → [0 , ∞ ], set ρ G ( f ) = sup E,F ∈ Λ m ∈ MO ( E,F ) (cid:18)Z E ( f ◦ m )( t ) p W ( t ) dt (cid:19) /p , UBREARRANGEMENT-INVARIANT FUNCTION SPACES 7 where W ∈ W and 1 p < ∞ . We then define a Garling function space ,denoted G W,p (0 , ∞ ), as the space of all a.e.-equivalence classes of measurablefunctions f : (0 , ∞ ) → [ −∞ , ∞ ] satisfying k f k G := ρ G ( | f | ) < ∞ . Remark . Conditions (W1) and (W2) are the only ones we use in § §
4, condition (W3) isneeded.It is clear that ρ G is a function norm, and hence G W,p (0 , ∞ ) is a function spaceon (0 , ∞ ). We will show later in § Proposition 3.4.
Fix p < ∞ , and let W ∈ W . Then G W,p (0 , ∞ ) issubrearrangement-invariant.Proof. Fix D ∈ Λ and n ∈ MO ((0 , ∞ ) , D ), and f ∈ G W,p (0 , ∞ ). Observe thatthere are E, F ∈ Λ and m ∈ MO ( E, F ) such that k f D k pG ≈ ε Z E (( f D ) ◦ m )( t ) p W ( t ) dt = Z E ( f ◦ m )( t ) p ( D ◦ m )( t ) W ( t ) dt = Z m − ( D ) ∩ E ( f ◦ m )( t ) p W ( t ) dt Z m − ( D ) ( f ◦ m )( t ) p W ( t ) dt = Z m − ( D ) (( f ◦ n ) ◦ ( n − ◦ m ))( t ) p W ( t ) dt k f ◦ n k pG , where the last inequality follows from the fact that n − ◦ m is an MO -isomorphismfrom m − ( D ) onto its image. On the other hand, there are A, B ∈ Λ and ℓ ∈ MO ( A, B ) such that k f ◦ n k pG ≈ ε Z A ( f ◦ n ◦ ℓ )( t ) p W ( t ) dt = Z A ( f ◦ n ◦ ℓ )( t ) p ( D ◦ n ◦ ℓ )( t ) W ( t ) dt = Z A (( f D ) ◦ ( n ◦ ℓ ))( t ) p W ( t ) dt k f D k pG , where the first equality follows due to the fact that D ◦ n ◦ ℓ is the identityfunction on A , and the final inequality follows from the fact that n ◦ ℓ is an MO -isomorphism from A onto its image. (cid:3) To show that a Garling function space fails to admit an equivalentrearrangement-invariant norm, we need the following intuitively obvious lemma.
Lemma 3.5.
Fix p ∈ [1 , ∞ ) and r ∈ (0 , ∞ ) . Let W ∈ W and f : (0 , ∞ ) → [0 , ∞ ] a measurable function which is nondecreasing on (0 , r ) and zero everywhere else.Then there is s ∈ [0 , r ] so that k f k G = (cid:18)Z s f ( t + r − s ) p W ( t ) dt (cid:19) /p Unfortunately, it requires a somewhat technical proof. We begin with some pre-liminaries.
BEN WALLIS
Proposition 3.6 ([Bo07, Theorem 2.9.3]) . Let (Ω , µ ) be a measure space and f : Ω → [ −∞ , ∞ ] a ( µ, β ) -measurable function. Then the µ -integrability of f isequivalent to the Lebesgue integrability of the function t dist f ( t ) , and Z Ω | f | dµ = Z ∞ dist f ( t ) dt. Corollary 3.7. If E and F are measurable subsets of (0 , ∞ ) , and f : (0 , ∞ ) → [0 , ∞ ] is a (nonnegative) ( λ, β ) -measurable function, then for any measure-preserving transformation m : E → F we have Z E ( f ◦ m )( t ) dt = Z F f ( t ) dt. Proof.
By Proposition 2.9, f and f ◦ m are equimeasurable, which is to say thatdist f = dist f ◦ m . Now by Proposition 3.6 we have Z E ( f ◦ m )( t ) dt = Z ∞ dist f ◦ m ( t ) dt = Z ∞ dist f ( t ) dt = Z F f ( t ) dt. (cid:3) The proof of the following is given in the appendix.
Proposition 3.8.
Let E be a measurable subset of [ −∞ , ∞ ] with λ ( E ) < ∞ .Then there is a measure-zero subset E of E , a measure-zero subset D of [0 , λ ( E )] , and an MO -isomorphism between E \ E and [0 , λ ( E )] \ D .Proof of Lemma 3.5. First, observe that since the map b Z b f ( t + r − b ) p W ( t ) dt is continuous on the compact set [0 , r ], we can find s ∈ [0 , r ] so that Z s f ( t + r − s ) p W ( t ) dt = sup b ∈ [0 ,r ] Z b f ( t + r − b ) p W ( t ) dt. (3.1)Let E, F ∈ Λ and m ∈ MO ( E, F ) be such that k f k pG ≈ ε Z E ( f ◦ m )( t ) p W ( t ) dt. (3.2)Without loss of generality we may assume that F ⊆ (0 , r ), and set b := λ ( F ) r .By Proposition 3.8 we can find measure-zero subsets E of E and D of (0 , r ),and an MO -isomorphism n : (0 , b ) \ D → E \ E . We claim that ( f ◦ m ◦ n )( t ) p f ( t + r − b ) p , (3.3)or, equivalently, b − t r − ( m ◦ n )( t ), for each t ∈ (0 , b ) \ D . Indeed, as m ◦ n is order-preserving, we have, for c ∈ ( t, b ) \ D ,( m ◦ n )(( t, c ) \ D ) ⊆ [( m ◦ n )( t ) , ( m ◦ n )( c )] UBREARRANGEMENT-INVARIANT FUNCTION SPACES 9 and since m ◦ n is a measure isomorphism from (0 , b ) \ D onto its image, we alsohave c − t = λ ( t, c ) = λ (( t, c ) \ D ) = λ (( m ◦ n )(( t, c ) \ D )) λ [( m ◦ n )( t ) , ( m ◦ n )( c )] = ( m ◦ n )( c ) − ( m ◦ n )( t ) . For ε > c ∈ ( t, b ) \ D so that b − c < ε . Hence, b − t < c − t + ε ( m ◦ n )( c ) − ( m ◦ n )( t ) + ε r − ( m ◦ n )( t ) + ε. As ε > b − t r − ( m ◦ n )( t ) as desired.Next we claim that ( W ◦ n )( t ) W ( t ) , (3.4)or, equivalently, t n ( t ), for each t ∈ (0 , b ) \ D . Indeed, for δ ∈ (0 , t ) \ D wehave n (( δ, t ) \ D ) ⊆ [ n ( δ ) , n ( t )]and hence t − δ = λ ( n (( δ, t ) \ D )) λ [ n ( δ ) , n ( t )] = n ( t ) − n ( δ ) n ( t ) . As δ ∈ (0 , t ) \ D can be chosen arbitrarily close to zero, this means t n ( t ) asclaimed.From (3.3) and (3.4) we obtain that( f ◦ m ◦ n )( t ) p ( W ◦ n )( t ) f ( t − r + b ) p W ( t )for all t ∈ (0 , b ) \ D , and hence, by the above together with (3.1), (3.2), andCorollary 3.7, we have k f k pG ≈ ε Z E ( f ◦ m )( t ) p W ( t ) dt = Z E \ E ( f ◦ m )( t ) p W ( t ) dt = Z (0 ,b ) \ D ( f ◦ m ◦ n )( t ) p ( W ◦ n )( t ) dt Z (0 ,b ) \ D f ( t + r − b ) p W ( t ) dt = Z b f ( t + r − b ) p W ( t ) dt Z s f ( t + r − s ) p W ( t ) dt k f k pG . (cid:3) We are now set to prove the main result of this section.
Theorem 3.9. If W ( t ) = ( t + 1) − / then G W, (0 , ∞ ) fails to admit an equivalentrearrangement-invariant norm.Proof. Fix r ∈ (0 , ∞ ) and let f r : (0 , ∞ ) → [0 , ∞ ] and f ∗ r : (0 , ∞ ) → [0 , ∞ ] bedefined by f r ( t ) = (cid:26) ( r + 1 − t ) − / if 0 < t < r, r t < ∞ and f ∗ r ( t ) = (cid:26) ( t + 1) − / if 0 < t < r, r t < ∞ . We claim that f r and f ∗ r are equimeasurable. Indeed, it is clear that dist f r ( s ) =dist f ∗ r ( s ) = r for all 0 s (1 + r ) − / and dist f r ( s ) = dist f ∗ r ( s ) = 0 for all s ∞ . Now select (1 + r ) − / < s <
1. We have f r ( t ) > s if and only ifboth 0 < t < r and ( r + 1 − t ) − / > s , or, equivalently, r + 1 − s − < t < r . Inthis case we havedist f r ( s ) = λ { t ∈ (0 , ∞ ) : f r ( t ) > s } = λ ( r + 1 − s − , r ) = s − − . Similarly, f ∗ r ( t ) > s if and only if both 0 < t < r and ( t + 1) − / > s , or,equivalently, 0 < t < s − −
1. This gives usdist f ∗ r ( s ) = λ { t ∈ (0 , ∞ ) : f ∗ r ( t ) > s } = λ (0 , s − −
1) = s − − f r and f ∗ r are equimeasurable as claimed.Note that k f ∗ r k G > Z r ( t + 1) − dt = log( r + 1) → ∞ as r → ∞ . Thus, to complete the proof, it is enough to show that k f r k G isbounded by a number not depending on r .Now we apply Garling’s own “split into two sums” trick, except in our casethe “sums” are actually integrals. Since f r is increasing on its support (0 , r ), and W ∈ W , by Lemma 3.5 we must have s ∈ [0 , r ] so that k f r k G = Z s f r ( t − s + r ) W ( t ) dt. = Z s (1 − t + s ) − / ( t + 1) − / dt = Z s/ (1 − t + s ) − / ( t + 1) − / dt + Z ss/ (1 − t + s ) − / ( t + 1) − / dt. Hence, it suffices to show that each of these pieces is bounded by a number notdepending on s . For the first piece, note that if t ∈ (0 , s/
2] then (1 − t + s ) − / ( s/ − / . Hence, Z s/ (1 − t + s ) − / ( t + 1) − / dt ( s/ − / Z s/ ( t + 1) − / dt = ( s/ − / · (cid:2) ( s/ / − (cid:3) . For the second piece, note that if t ∈ [ s/ , s ] then ( t + 1) − / ( s/ − / , sothat Z ss/ (1 − t + s ) − / ( t + 1) − / dt ( s/ − / Z ss/ (1 − t + s ) − / dt = ( s/ − / · (cid:2) ( s/ / − (cid:3) . (cid:3) Geometric properties of Garling function spaces
In this section we show that Garling function spaces are Banach spaces con-taining (1 + ε )-isomorphic copies of ℓ p . As a consequence, the space G W, (0 , ∞ )is nonreflexive. It remains an open question as to whether G W,p (0 , ∞ ) is reflexivewhen 1 < p < ∞ . UBREARRANGEMENT-INVARIANT FUNCTION SPACES 11
Proposition 4.1.
Fix p < ∞ and W ∈ W . Then space G W,p (0 , ∞ ) iscomplete.Proof. Let ( f i ) ∞ i =1 be a Cauchy sequence in G W,p . Let
E, F ∈ Λ and m ∈ MO ( E, F ). Observe that k f i − f j k pG > Z ∞ | f i ( t ) − f j ( t ) | p W ( t ) dt > k| f i | W /p − | f j | W /p k pL p (0 , ∞ ) so that ( | f i | W /p ) ∞ i =1 is Cauchy in L p (0 , ∞ ). As such, it converges a.e.-pointwiseto g ∈ L p (0 , ∞ ). Similarly, k f i − f j k pG > Z E | ( f i ◦ m )( t ) − ( f j ◦ m )( t ) | p W ( t ) dt > k| f i ◦ m | W /p − | f j ◦ m | W /p k pL p ( E ) so that ( | f i ◦ m | W /p ) ∞ i =1 converges both in L p ( E ) and a.e.-pointwise to some g E ∈ L p ( E ). Set f := gW − /p so that ( | f i | ) ∞ i =1 converges a.e.-pointwise to f .As ( | f i ◦ m | W /p ) ∞ i =1 now converges a.e.-pointwise to | f ◦ m | W /p , it follows that | f ◦ m | W /p and g E are a.e.-identical.Since ( f i ) ∞ i =1 is Cauchy, we can find M ∈ (0 , ∞ ) so that k f i k pG M for all i ∈ N . Furthermore, we can find i ∈ N so that k g E − | f i ◦ m | W /p k pL p ( E ) Z E | f ◦ m | ( t ) p W ( t ) dt Z E | ( | f | − | f i | ) ◦ m | ( t ) p W ( t ) dt + Z E | f i ◦ m | ( t ) p W ( t ) dt = k g E − | f i ◦ m | W /p k pL p ( E ) + Z E | f i ◦ m | ( t ) p W ( t ) dt k g E − | f i ◦ m | W /p k pL p ( E ) + k f i k pG M. As E, F, m were arbitrary, we have ρ G ( | f | ) (1 + M ) /p < ∞ so that f ∈ G W,p (0 , ∞ ).Next, select ε > N ∈ N so that k f i − f j k G < ε/ i, j > N .Select j > N so that k g E − | f j ◦ m | W /p k pL p ( E ) < ε/
2. Then for i > N we have Z E | ( f − f i ) ◦ m | ( t ) p W ( t ) dt Z E || f |−| f j || ( t ) p W ( t ) dt + Z E | f i − f j | ( t ) p W ( t ) dt k g E − | f j ◦ m | W /p k pL p ( E ) + k f i − f j k pG < ε. Again as
E, F, m were arbitrary and independent of N , it follows that k f − f i k pG <ε for all i > N . As ε > f i → f in G W,p (0 , ∞ ). (cid:3) To close, we will show that when 1 p < ∞ and W ∈ W , the space G W,p (0 , ∞ )contains a copy of ℓ p . To do this, we will use a basic sequence of characteristicfunctions as an auxiliary structure. Let us gather some facts about it in the nextlemma. In what follows, we denote i = ( i − ,i ] for each i ∈ N . Lemma 4.2.
Fix p < ∞ and W ∈ W , and set K = R W ( t ) dt .Then the sequence ( i /K ) ∞ i =1 is a normalized, monotone, 1-unconditional and G W,p (0 , ∞ ) which 1-dominates the unit vectorbasis ( g i ) ∞ i =1 of the Garling sequence space g ( w, p ) , where w = ( w ( i )) ∞ i =1 is formedby letting w ( i ) = K − R ii − W ( t ) dt for each i ∈ N . Furthermore, they are isomet-rically equivalent for constant coefficients.Proof. By replacing W with K − W if necessary, we may assume without loss ofgenerality that K = 1.It’s clear that ( i ) ∞ i =1 is normalized. It is also clear that if M < N ∈ N and( a i ) ∞ i =1 is any sequence of scalars then we have (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) M X i =1 a i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) G (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X i =1 a i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) G , which is precisely the criterion for forming a monotone basic sequence.Next we show that it is 1-unconditional. Let ( a i ) ∞ i =1 , ( b i ) ∞ i =1 ∈ c and satisfy | a i | | b i | for all i ∈ N . Then we can find E, F ∈ Λ and m ∈ MO ( E, F ) suchthat, setting U j = m − ( F ∩ ( j − , j ]) for each j ∈ N so that U < U < · · · with E = S ∞ j =1 U j and each j ◦ m = U j | E , (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 a i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X i =1 a i i ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt = Z E ∞ X i =1 | a i | p i ( m ( t )) W ( t ) dt = ∞ X j =1 Z U j ∞ X i =1 | a i | p i ( m ( t )) W ( t ) dt = ∞ X j =1 Z U j ∞ X i =1 | a i | p U i ( t ) W ( t ) dt = ∞ X j =1 Z U j | a j | p U j ( t ) W ( t ) dt ∞ X j =1 Z U j | b j | p U j ( t ) W ( t ) dt By an analogous argument we have ∞ X j =1 Z U j | b j | p U j ( t ) W ( t ) dt = Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X i =1 b i i ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt whence (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 a i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 b i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG so that ( i ) ∞ i =1 is 1-unconditional.Let us show that it is 1-subsymmetric. Indeed, if ( a i ) ∞ i =1 ∈ c and ( i k ) ∞ k =1 issome subsequence, then we can find E, F ∈ Λ and m ∈ MO ( E, F ) such that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 a k i k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k i k ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt. Set E ′ = S ∞ k =1 m − ( F ∩ ( i k − , i k ]), and define an MO -isomorphism ℓ : (0 , ∞ ) → S ∞ k =1 ( i k − , i k ] by gluing together the shift maps ( k − , k ] ( i k − , i k ]. Then ℓ − ◦ m is an MO -isomorphism between E ′ and its image, and for each k ∈ N and UBREARRANGEMENT-INVARIANT FUNCTION SPACES 13 t ∈ E ′ we have i k ( m ( t )) = k ( ℓ − ( m ( t ))). Furthermore, i k ( m ( t )) = 0 for each k ∈ N and t ∈ E \ E ′ . Hence, (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 a k i k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k i k ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt = Z E ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k i k ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt = Z E ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k k ( ℓ − ( m ( t ))) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 a k k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG . To show the reverse inequality, we instead choose
E, F, m so that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 a k k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k k ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt. Define ℓ as before so that ℓ ◦ m is an MO -isomorphism between E and its image,and k ( m ( t )) = i k ( ℓ ( m ( t ))) for each k ∈ N and t ∈ E . Then Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k k ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt = Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k =1 a k ( i k ( ℓ ( m ( t ))) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X k =1 a k i k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG . It follows that ( i ) ∞ i =1 is 1-subsymmetric.To show that it 1-dominates g ( w, p ), we again let ( a i ) ∞ i =1 ∈ c . Select anysubsequence ( a i k ) ∞ k =1 . As before, there is an MO -isomorphism ℓ : (0 , ∞ ) → S ∞ k =1 ( i k − , i k ] defined by gluing together the shift maps ( k − , k ] ( i k − , i k ].Note that i k ◦ ℓ = k for each k ∈ N . We now have ∞ X k =1 | a i k | p w ( k ) = ∞ X k =1 | a i k | p Z kk − W ( t ) dt = ∞ X k =1 | a i k | p Z ∞ k ( t ) W ( t ) dt = Z ∞ ∞ X k =1 | a i k | p k ( t ) W ( t ) dt = Z ∞ ∞ X k =1 | a i k | p i k ( ℓ ( t )) W ( t ) dt = Z ∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X n =1 a i i ( ℓ ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 a i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG By taking the supremum over all subsequences we obtain k ( a i ) ∞ i =1 k g (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 a i i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) G . Finally, we consider the last part of the lemma, about being isometrically equiv-alent for constant coefficients to ( g i ) ∞ i =1 . Indeed, as ( i ) ∞ i =1 already 1-dominatesit as shown above, we need only show the reverse inequality, i.e. that ( g i ) ∞ i =1 i ) ∞ i =1 for constant coefficients. To that end, fix N ∈ N and let E, F ∈ Λ and m ∈ MO ( E, F ) be such that (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X i =1 i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E N X i =1 i ( m ( t )) W ( t ) dt. For each i = 1 , · · · , N , define A i := m − ( F ∩ ( i − , i ]), and then set A := S Ni =1 A i .It is clear that λ ( A ) < ∞ , so by Proposition 3.8 we can find measure-zero subsets D of [0 , λ ( A )] and A of A , and an MO -isomorphism n from D := [0 , λ ( A )] \ D onto A \ A . We claim that t n ( t ) for all t ∈ D . Indeed, if we set b := inf n ( D )then since b > n ( D t ) ⊆ [ b, n ( t )] we have t = λ [0 , t ] = λ ( D t ) = λ ( n ( D t )) λ [ b, n ( t )] = n ( t ) − b n ( t ) . As W is nonincreasing it follows that W ( n ( t )) W ( t ) for all t ∈ D . Note alsothat λ ( A ) N so that D ⊆ [0 , N ]. Furthermore, it is clear that i ( m ( t )) = 0 forall i = 1 , · · · , N and all t ∈ E \ A . Together with Corollary 3.7 we now obtain (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X i =1 i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E N X i =1 i ( m ( t )) W ( t ) dt = Z A N X i =1 i ( m ( t )) W ( t ) dt = N X j =1 Z A j N X i =1 i ( m ( t )) W ( t ) dt = N X j =1 Z A j j ( m ( t )) W ( t ) dt = N X j =1 Z A j W ( t ) dt = Z A W ( t ) dt = Z D W ( n ( t )) dt Z D W ( t ) dt Z N W ( t ) dt = N X k =1 w ( k ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X i =1 g i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pg . As ( g i ) ∞ i =1 and ( i ) ∞ i =1 are both 1-subsymmetric, we are done. (cid:3) Theorem 4.3.
Fix p < ∞ and let W ∈ W . Then for any ε > the ba-sic sequence ( i ) ∞ i =1 admits a normalized constant coefficient block basic sequencewhich is (1 + ε ) -equivalent to ℓ p , and which is 2-complemented in [ i ] ∞ i =1 .Proof. Let g ( w, p ), ( g i ) ∞ i =1 , and K be as in Lemma 4.2, so that ( i /K ) ∞ i =1 is iso-metrically equivalent to ( g i ) ∞ i =1 for constant coefficients. It was shown in [AAW18, §
3] that there exists a constant coefficient block basic sequence of ( g n ) ∞ n =1 whichis (1 + ε )-equivalent to ℓ p , for any ε >
0. In particular, we can select y ′ i = k i +1 − X n = k i g n and y i = y ′ i k y ′ i k g for each i ∈ N , where 1 = k < k < k < · · · ∈ N , so that ( y i ) ∞ i =1 is (1 + ε )-equivalent to ℓ p .Next, write x ′ i = k i +1 − X n = k i n /K and x i := x ′ i k x ′ i k G for each i ∈ N , where 1 = k < k < k < · · · ∈ N .We claim that ( x i ) ∞ i =1 is 1-dominated by the unit vector basis of ℓ p . Indeed, if( a i ) ∞ i =1 ∈ c then we can find E, F ∈ Λ and m ∈ MO ( E, F ) such that
UBREARRANGEMENT-INVARIANT FUNCTION SPACES 15 (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 a i x i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) pG ≈ ε Z E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X i =1 a i x i ( m ( t )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p W ( t ) dt = Z E ∞ X i =1 | a i | p x i ( m ( t )) W ( t ) dt = ∞ X i =1 | a i | p Z E x i ( m ( t )) W ( t ) dt ∞ X i =1 | a i | p k x i k pG = ∞ X i =1 | a i | p so that ( x i ) ∞ i =1 . ℓ p as claimed.By Lemma 4.2, ( i /K ) ∞ i =1 is isometrically equivalent to ( g i ) ∞ i =1 for constantcoefficients, and so k y ′ i k g = k x ′ i k G for each i ∈ N . Again from Lemma 4.2, weknow that ( g i ) ∞ i =1 is 1-dominated by ( i /K ) ∞ i =1 . It follows that ℓ p ≈ ε ( y i ) ∞ i =1 . ( x i ) ∞ i =1 . ℓ p . That ( x i ) ∞ i =1 spans a 2-complemented subspace of [ i ] ∞ i =1 follows from the fact thatconstant-coefficient block basic sequences of a 1-subsymmetric basis are always2-complemented (see, for instance, [LT77, Proposition 3.a.4]). (cid:3) Remark . Although ℓ p is complemented in [ i ] ∞ i =1 , we do not yet know if it iscomplemented in G W,p (0 , ∞ ). Corollary 4.5.
Fix p < ∞ and W ∈ W . Then for every ε > , the space G W,p (0 , ∞ ) contains a subspace which is (1 + ε ) -isomorphic to ℓ p . Hence, inparticular, the space G ( W, is nonreflexive. Appendix
Proposition 5.1.
Let E and F be Lebesgue-measurable subsets of R , and let m : E → F be a bijection which is both order-preserving and measure-preserving.Then m − is also order-preserving and measure-preserving, i.e. m ∈ MO ( E, F ) .Proof. Clearly, it is enough to show that m − is measurable. To that end, let usfix a measurable set A ⊆ E ; we claim that m ( A ) is also measurable, which willcomplete the proof.Denote by B = σ ( τ ) the Borel σ -algebra on R , where τ denotes the usual metrictopology R . Let τ E be the subspace topology on E , i.e. the topology defined by τ E = E ∩ τ := { E ∩ U : U ∈ τ } . Similarly, we denote by τ F the subspace topology for F . It is well-known (andeasy to see) that the set E ∩ B := { E ∩ B : B ∈ B} is a σ -algebra on E , called the trace σ -algebra. Since E ∩ τ ⊂ E ∩ B , we obtain σ ( τ E ) = σ ( E ∩ τ ) ⊆ σ ( E ∩ B ) = E ∩ B . For the reverse inclusion, defineΣ := { Y ⊆ R : E ∩ Y ∈ σ ( τ E ) } . It is routine to verify that Σ is a σ -algebra on R . Also, it is clear that τ ⊆ Σ,since for U ∈ τ we have E ∩ U ∈ τ E ⊆ σ ( τ E ). It follows that σ ( τ ) ⊆ Σ, whencealso by definition of Σ we obtain E ∩ σ ( τ ) ⊆ σ ( τ E ). This gives us the reverse inclusion as desired. We now have the identity σ ( τ E ) = E ∩ B , and an identicalargument shows that σ ( τ F ) = F ∩ B . It’s a well-known fact in real analysis that we can find C ∈ B such that A ⊆ C and λ ( C \ A ) = 0. Now set C ′ = E ∩ C ∈ σ ( τ E ). Since λ ( C \ A ) = 0 there is ameasure-zero set D ∈ B with C ′ \ A ⊆ C \ A ⊆ D . Set D ′ := E ∩ D ∈ σ ( τ E ) sothat C ′ \ A ⊆ D ′ and λ ( D ′ ) = 0.By a standard argument found, for instance, in the proof of [Bo07, Theorem2.1.2], it follows that m ( B ) is Lebesgue-measurable whenever B ∈ σ ( τ E ). Thuswe have λ [ m ( D ′ )] = 0. Observe m ( C ′ ) \ m ( A ) = m ( C ′ \ A ) ⊆ m ( D ′ ) so that (sincesubsets of measure-zero sets are themselves measure-zero) λ [ m ( C ′ ) \ m ( A )] = 0as well. Note also that since C ′ ∈ σ ( τ E ) we have m ( C ′ ) measurable. Since A ⊆ C ′ , we obtain m ( A ) = m ( C ′ ) \ [ m ( C ′ ) \ m ( A )], which shows that m ( A ) ismeasurable. (cid:3) Proposition 5.2.
Let E and F be Lebesgue-measurable subspaces of [ −∞ , ∞ ] ,and let m : F → E be a surjective measure-preserving transformation which isalso order-preserving. Then there is a measure-zero subset F of F such that m is a bijection between F \ F and E .Proof. For each x ∈ E , let I x be an interval containing m − { x } which is min-imal under the relation ⊆ . Since m is order-preserving, the I x ’s are all dis-joint, which means only countably many of them have positive measure. Inparticular, m − { x } is a singleton for all but countably many x ∈ E . Set E := { x ∈ E : m − { x } is not a singleton } . For each x ∈ E , select some f x ∈ m − { x } . Now set F := S x ∈ E ( m − { x } \ { f x } ) . Clearly, m is a bijec-tion between F \ F and E . Observe that each m − { x } \ { f x } has measure zeroand that E is countable. It follows that F has measure zero. (cid:3) Let E be a totally-ordered set. An initial segment of E is any subset of E ′ of E such that E ′ < E \ E ′ . Proposition 5.3.
Let E be a Lebesgue-measurable subset of [ −∞ , ∞ ] with λ ( E ) < ∞ . Then for each t ∈ [0 , λ ( E )] there is an initial segment E t of E such that λ ( E t ) = t .Proof. Note that if E t is an initial segment of E \ {−∞ , ∞} then E t ∪ {−∞} is an initial segment of E with the same measure as E t . Hence, without loss ofgenerality, we may assume E ⊂ R . We may also assume that E is bounded, sinceif the result holds in that case then it can be extended to the unbounded case byconsidering the union of sets E n = [ − n, − n + 1) ∩ E ∩ ( n − , n ] . Say E ⊆ [ a, b ] for −∞ < a < b < ∞ . Define f : [ a, b ] → [0 , λ ( E )] by the rule f ( x ) = λ ([ a, x ] ∩ E ) . Observe that if y < x ∈ [ a, b ] then | f ( x ) − f ( y ) | = λ (( y, x ] ∩ E ) | x − y | so that f is Lipschitz, in particular, continuous. As f ( a ) = 0 and f ( b ) = λ ( E ),we may now apply the Intermediate Value Theorem. (cid:3) UBREARRANGEMENT-INVARIANT FUNCTION SPACES 17
Lemma 5.4.
Let E be a measurable subset of R with λ ( E ) < ∞ . For each t ∈ [0 , λ ( E )] , let E t be an initial segment of E (whose existence is guaranteed byProposition 5.3). Define the map m : E → [0 , λ ( E )] by the rule m ( x ) = inf { t ∈ [0 , λ ( E )] : x ∈ E t } . Then m is both measure-preserving and order-preserving. Furthermore, m can beextended to a map m : R → [0 , λ ( E )] defined by m ( x ) = λ (( −∞ , x ] ∩ E ) . Proof.
It is obvious that m is order-preserving, and it is explicitly proved in[BS88, Proposition 2.7.4] that it is also measure-preserving. For x ∈ E we set E x := ( −∞ , x ] ∩ E , and observe that if t > λ ( E x ) then x ∈ E t and if t <λ ( E x ) then x / ∈ E t . It follows that m ( x ) = λ ( E x ) for all x ∈ E . Thus,we can extend m continuously to the function M : R → [0 , λ ( E )] via the rule M ( x ) = λ (( −∞ , x ] ∩ E ) . (cid:3) Proof of Proposition 3.8.
Since {−∞ , ∞} has measure zero, we may assume with-out loss of generality that E ⊂ R . Let m : R → E be as in Lemma 5.4. It isclear (as in, for instance, the proof of Proposition 5.3) that m is Lipschitz, andhence continuous in the usual sense as well. Since λ is inner-regular, we canfind a sequence ( K n ) ∞ n =1 of compact sets and a measure-zero set L such that E = L ∪ S ∞ n =1 K n . It is known that the image of a bounded measure-zero setunder a Lipschitz function is again measure-zero. Furthermore, the continuousimage of a compact set is again compact, and in particular measurable. We nowhave m ( E ) = m ( L ∪ S ∞ n =1 K n ) = m ( L ) ∪ S ∞ n =1 m ( K n ) . It follows that m ( E ) ismeasurable. We can now apply Proposition 5.2 to find a subset E of measure zerosuch that m is a bijection between E \ E and m ( E ). Set D = [0 , λ ( E )] \ m ( E ).We have λ ( D ) = 0, and by Proposition 5.1, m is an MO -isomorphism between E \ E and [0 , λ ( E )] \ D . (cid:3) Proposition 5.5. If f, g : N → [0 , ∞ ) are equimeasurable with lim n →∞ f ( n ) = lim n →∞ g ( n ) = 0 then either they are both identically zero or else there is a measure-isomorphism m : supp ( f ) → supp ( g ) such that g ◦ m = f on supp ( f ) .Proof. Obviously, if one of f and g is identically zero then, since they are equimea-surable, so is the other. So let us assume that neither is identically zero. Let f ∗ ( n ) = inf { λ : dist f ( λ ) n } denote the “decreasing rearrangement” of f . Sincelim n →∞ f ( n ) = 0 we have also lim n →∞ f ∗ ( n ) = 0. Now [BS88, Corollary 7.6]gives us a measure-preserving transformation m f : supp( f ) → supp( f ∗ ) suchthat f = f ∗ ◦ m f on supp( f ), and analogously we get m g : supp( g ) → supp( g ∗ )with g = g ∗ ◦ m g . Since f and g are equimeasurable, f ∗ = g ∗ , and hence g = f ∗ ◦ m g . This means g ◦ m − g = f ∗ and hence, setting m = m − g ◦ m f ,we obtain g ◦ m = g ◦ m − g ◦ m f = f ∗ ◦ m f = f . (cid:3) Proof of Proposition 2.7. ( ⇒ ): Let ( e i ) ∞ i =1 be 1-symmetric, and suppose f and g are equimeasurable sequences in X . Then so are | f | and | g | . If f and g are identically zero then k f k X = 0 = k g k X and we are done. Otherwise byProposition 5.5 there is a bijection m : supp( f ) → supp( g ) with f = g ◦ m onsupp( f ). Now we have, by 1-symmetry and 1-unconditionality k f k X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 f ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X i ∈ supp( f ) | f ( i ) | e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X i ∈ supp( f ) | g ( m ( i )) | e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X i ∈ supp( g ) | g ( i ) | e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X . = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 g ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = k g k X . ( ⇐ ): Suppose that X is rearrangement-invariant with respect to ( e i ) ∞ i =1 , andselect a permutation π of N . Then its inverse π − exists and is a measure-preserving transformation. Select any f ∈ X , and note that | f ( i ) | < ∞ for all i ∈ N . By Proposition 2.9, | f | and | f | ◦ π − are equimeasurable. Now we have,by 1-unconditionality and rearrangement-invariance (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 f ( i ) e π ( i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 | f ( i ) | e π ( i ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 ( | f | ◦ π − )( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = k| f | ◦ π − k X = k| f |k X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 | f ( i ) | e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X i =1 f ( i ) e i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X . That ( e n ) ∞ n =1 is symmetric if and only if it is essentially rearrangement-invariantis clear from considering the equivalent norm ||| x ||| = sup σ ∈ Π N k P ∞ n =1 e ∗ n ( x ) e σ ( n ) k . (cid:3) Acknowledgments.
Thanks to Lukas Geyer and Ramiro Affonso de TadeuGuerreiro for assisting in the proofs of Propositions 5.1 and 3.8, respectively.
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