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Surjective isometries on rearrangement-invariant spaces

arXiv:math/9211208v1 [math.FA] 5 Nov 1992Surjective isometries on rearrangement-invariant spacesN.J. Kalton and Beata Randrianantoanina1Department of MathematicsUniversity of MissouriColumbiaMo.

65211Abstract. We prove that if X is a real rearrangement-invariant function space on [0, 1],which is not isometrically isomorphic to L2, then every surjective isometry T : X →Xis of the form Tf(s) = a(s)f(σ(s)) for a Borel function a and an invertible Borel mapσ : [0, 1] →[0, 1].

If X is not equal to Lp, up to renorming, for some 1 ≤p ≤∞then inaddition |a| = 1 a.e. and σ must be measure-preserving.1.

Introduction.The main result of this paper is the following theorem, which combines the statementsof Theorems 6.4 and 7.2 below. We denote Lebesgue measure on [0, 1] by λ and use theterm rearrangement-invariant Banach function space in the sense of Lindenstrauss-Tzafriri[19].Theorem 1.1.

Let X be a (real) rearrangement-invariant Banach function space on [0, 1].Suppose X is not (isometrically) equal to L2[0, 1]. Let T : X →X be a surjective isometry.Then(1) There exists a nonvanishing Borel function a : [0, 1] →R and an invertible Borel mapσ : [0, 1] →[0, 1] such that, for any Borel set B ⊂[0, 1], we have λ(σ−1B) = 0 if and onlyif λ(B) = 0 and so that Tf(s) = a(s)f(σ(s)) a.e.

for any f ∈X.1Both authors were supported by NSF-grant DMS-9201357This research will form part of the Ph.D. thesis of the second author currently underpreparation at the University of MisssouriAMS Classification 46B041

(2) If X is not equal to Lp for some 1 ≤p ≤∞up to renorming, then |a| = 1 a.e. and σis measure-preserving.The first part of this theorem is known for the case of spaces of complex functionsand is due to Zaidenberg [32], [33]; we discuss the relationship between our result andZaidenberg’s below.

The second part also holds for the complex case, and is apparentlynew even in this case.The study of isometries on classical function spaces goes back to Banach [1] who provedthat the isometries of Lp[0, 1] are disjointness-preserving when p ̸= 2 (see [1] p. 175).Lamperti [18] later characterized the isometries on Lp. Since then there has developed anextensive literature on isometries of particular function spaces; see [3], [4], [5] and [13] forexample.In the case of (not necessarily rearrangement-invariant) complex function spaces, atechnique developed by Lumer [20], [21], [22] has proved particularly effective.

This tech-nique was used by Lumer [21], [22] to study isometries on reflexive Orlicz spaces and laterby Zaidenberg [32] and [33] to study isometries on general r.i. spaces, X. The idea is tocharacterize first the hermitian operators H : X →X.

H is hermitian if exp(itH) is anisometry for every real t. One shows that hermitian operators are simply multiplicationoperators by real functions, unless X = L2. Then, if U is a surjective isometry on X wehave that UHU −1 is hermitian for every hermitian H. Combining these ideas leads toTheorem 1.1 (1) in the complex case.

See also, for example, [6], [7], [10] and [31]. For afuller discussion of the existing literature we refer the reader to the forthcoming survey ofFleming and Jamison [8].This line of argument simply does not work for real spaces, and most of the knownresults use geometric techniques (e.g.

extreme point arguments) for special spaces. In thispaper, we follow a line of reasoning which is distantly related to the Lumer technique.We use the notion of a numerically positive operator [26]; this is an operator T such that∥exp(−tT)∥≤1 for all t ≥0 (see also [23] where −T is called dissipative).

Unfortunately,this is far too weak a notion to allow us to characterize such operators on an r.i. space,but by studying rank-one numerically positive operators and using results of Flinn (see[26]) we are able to prove a representation theorem for surjective isometries (Proposition6.3), which is a partial step towards our main result. Then by a probabilistic techniquewe obtain Theorem 6.4 which is equivalent to Theorem 1.1 (1).

Finally in Theorem 7.2 weshow that if X is not Lp up to renorming then the representation in Theorem 6.4 can befurther narrowed to the trivial case as in Theorem 1.1 (2).Some remarks on the nature of our results are in order. First notice that we mustrestrict ourselves (as do Lumer and Zaidenberg) to surjective isometries.

Many of thespecial results quoted above apply equally to isometries which are not surjective. Secondly,2

it will be seen that there appear to be obstacles to extending the main result to r.i. spaceson [0, ∞), (see, for example, [11], [12]). However our results do apply equally to separableand nonseparable r.i. spaces on [0, 1].

The proof can be simplified a little in the separablecase and we indicate such simplifications as various points in the paper.Acknowledgement: We wish to thank Jim Jamison and Anna Kaminska for bringingthis question to our attention and for providing copies of both [8] and a translation of [33].2. Introductory remarks on K¨othe function spaces.Let us suppose that Ωis a Polish space and that µ is a σ−finite Borel measure on Ω.We use the term K¨othe space in the sense of [19] p. 28.

Thus a K¨othe function space Xon (Ω, µ) is a Banach space of (equivalence classes of) locally integrable Borel functions fon Ωsuch that:(1) If |f| ≤|g| a.e. and g ∈X then f ∈X with ∥f∥X ≤∥g∥X.

(2) If A is a Borel set of finite measure then χA ∈X.We say that X is order-continuous if whenever fn ∈X with fn ↓0 a.e. then ∥fn∥X ↓0.X has the Fatou property if whenever 0 ≤fn ∈X with sup ∥fn∥X < ∞and fn ↑f a.e.then f ∈X with ∥f∥X = sup ∥fn∥X.The K¨othe dual of X is denoted X′; Thus X′ is the K¨othe space of all g such thatR|f||g| dµ < ∞for every f ∈X equipped with the norm ∥g∥X′ = sup∥f∥X≤1R|f||g| dµ.Then X′ can be regarded as a closed subspace of the dual X∗of X.

If X is order-continuousthen X′ = X∗; if X has the Fatou property then X′ is a norming subspace of X∗.A rearrangement-invariant function space (r.i. space) is a K¨othe function space on([0, 1], λ) where λ is Lebesgue measure which satisfies the conditions:(1) Either X is order-continuous or X has the the Fatou property. (2) If τ : [0, 1] →[0, 1] is any measure-preserving invertible Borel automorphism thenf ∈X if and only if f ◦τ ∈X and ∥f∥X = ∥f ◦τ∥X.

(3) ∥χ[0,1]∥X = 1.In this section we will make some introductory remarks about operators and isometrieson K¨othe function spaces. Let us suppose that X is a K¨othe function space on (Ω, µ).

Wefirst consider those operators T : X →X which are continuous for the topology σ(X, X′).This is equivalent to requiring the existence of adjoint T ′ : X′ →X′. Of course, if X isorder-continuous, every operator T : X →X is σ(X, X′)−continuous.Our first result is well-known, but we know no explicit reference.Proposition 2.1.

Let X be a K¨othe function space on (Ω, µ). The following conditions3

on T : X →X are equivalent:(1) Tis σ(X, X′)−continuous. (2) If 0 ≤fn ∈X and fn ↑f a.e.

then limn→∞R|h||Tfn −Tf|dµ = 0 for every h ∈X′. (3) If 0 ≤g ∈X and |fn| ≤g with fn →f a.e.

then limn→∞R|h||Tfn −Tf|dµ = 0 forevery h ∈X′.Remark: Note that (2) says that T : X →L1(|h|dµ) is an order-continuous operator forevery h ∈X′; see Weis [29].Proof: (1) →(3) : Consider the operator S : L∞(µ) →L1(µ) defined by Sφ = h(Tφg).Then S is σ(L∞, L1) →σ(L1, L∞)-continuous and hence weakly compact. Now we maychoose φn →φ a.e.

such that φng = fn and φg = f. Consider the adjoint S′ : L∞→L1.Then S′(BL∞) is weakly compact and hence uniformly integrable in L1 ([30], p. 137).Thuslimn→∞sup∥h∥∞≤1ZφnS′h dµ = 0which quickly gives (3). (3) →(2) : Obvious.

(2) →(1) : We must show that the adjoint T ∗: X∗→X∗maps X′ into X′. Considerany h ∈X′.

It follows quickly that the set-function A →RhTχAdµ is countably additivewhen restricted to any Borel set A of finite measure. Thus there exists φ ∈L0 so thatRh(Tf)dµ =Rfφ dµ for any simple function f supported on a set of finite measure.

Nowfor general positive f ∈X we find a sequence fn of such simple functions so that fn ↑fa.e. and thenZh(Tf) = limn→∞Zh(Tfn)dµ = limn→∞Zφfndµ =Zφfdµ.We conclude that T ∗h = φ ∈X′.An operator T : X →X will be called elementary if there is a Borel function a anda Borel map σ : Ω→Ωsuch that Tf(s) = a(s)f(σ(s)) a.e.

for every f ∈X. Observethat a necessary condition on a and σ is that if B is a Borel set with µ(B) = 0 thenµ(σ−1B ∩{|a| > 0}) = 0.

T is called disjointness-preserving if min(|f|, |g|) = 0 a.e. impliesmin(|Tf|, |Tg|) = 0, a.e.Lemma 2.2.

T is elementary if and only if T is disjointness preserving and σ(X, X′)−con-tinuous.Proof: It is trivial that an elementary operator is disjointness-preserving and continuousfor σ(X, X′). For the converse we check that if 0 ≤f ≤g ∈X then 0 ≤|Tf| ≤|Tg|.It suffices by a density argument to establish this when f, g are both countably simple.4

Pick a maximal family of Borel sets {Ai : i ∈I} of finite positive measure such that|T(fχAi)| ≤|T(gχAi)| a.e. This family is countable and so its union B is a Borel set.If A is a Borel set of positive measure disjoint from B then we may find a further set ofpositive measure A′ ⊂A so that fχA′ = αgχA′ for some 0 ≤α ≤1; thus |TfχA′| ≤|TgχA′|contrary to our maximality assumption.

Hence µ(Ω\ B) = 0. Now by Proposition 2.1,Tf = Pi∈I T(fχAi) and Tg = Pi∈I T(fχAi) in L1(hdµ) where h is any strictly positivefunction in X′.

Thus |Tf| ≤|Tg|. It follows that T is regular and order-continuous as anoperator from X into L1(hdµ).

As shown by Weis [29] this means that T is elementary.Lemma 2.3. Suppose Ωis uncountable.

If X is a K¨othe function space on (Ω, µ) andT : X →X is an invertible elementary operator then T −1 is elementary and T can berepresented in the form Tf(s) = a(s)f(σ(s)) where a is a nonvanishing Borel function andσ : Ω→Ωis an invertible Borel map such that µ(B) > 0 if and only if µ(σ−1B) > 0.Proof: As noted above, T can be represented in the form Tf = af ◦σ0 where a is aBorel function and σ0 : Ω→Ωis a Borel map. Since T is onto it is clear that a canvanish only on a set of measure zero and so we may assume that it is nonvanishing.

Thenfor any f, supp Tf = σ−10 (supp f.) Thus T −1 is disjointness-preserving. Now suppose0 ≤gn ↑g a.e.

; we will verify condition (2) of Proposition 2.1 for T −1. We can supposegn = afn◦σ0 and g = af ◦σ0.

Then T −1gn = fn; we will show that, almost everywhere, wehave both fn(ω) →f(ω) and |fn(ω)| ≤|f(ω)| for all n. Once this is established then theDominated Convergence Theorem establishes Proposition 2.1 (2). Suppose E is any Borelset of finite measure such that for every ω ∈E we have supn |fn(ω)| > |fn(ω)| or fn(ω)does not converge to f(ω).

Then σ−10 E is contained in the set where gn(ω) fails to convergemonotonically to g(ω) and so has measure zero. This implies that TχE = 0 (a.e.) and soµ(E) = 0.

Hence T −1 satisfies (2) of Proposition 2.1 and hence is σ(X, X′)−continuous. Itfollows that T −1 is elementary and so can be represented in the form T −1f = bf ◦τ whereb is a nonvanishing Borel function and τ : Ω→Ωis a Borel map.

Thus the identity mapcan be written in the form f →a(b ◦σ0)f ◦τ ◦σ0 and so τσ0s = s a.e. ; similarly σ0τs = sa.e.Let E = {s : τσ0(s) = s}.

By Lusin’s theorem there is an increasing sequence ofcompact subsets Kn of E so that σ0 is continuous on Kn and µ(Ω\K) = 0 where K = ∪Kn.Then σ0 is a Borel isomorphism of K onto σ0(K) and both sets are Fσ’s. Let F be anuncountable compact subset of K of measure zero.

Then we define σ = σ0 on K \ F andσ = ρ on F ∪(Ω\K) where ρ is any Borel isomorphism between the two uncountable Borelsets F ∪(Ω\ K) and σ0(F) ∪(Ω\ σ(K)). Then σ = σ0 a.e.

and is a Borel automorphism.We thus can replace σ0 by σ and assume that σ is a Borel automorphism. Finally to showthe measure properties of σ note that µ(B) = 0 if and only if TχB = 0 a.e.

i.e. if and only5

if µ(σ−1B) = 0.Lemma 2.4. If T : X →X is an invertible elementary operator then T ′ : X′ →X′ is anelementary operator.Proof: We can represent T in the form Tf = af ◦σ where a is nonvanishing and σ isan invertible Borel map with µ(σ−1B) = 0 if and only if µ(B) = 0.

Let w be the Radon-Nikodym derivative of the σ−finite measure ν(B) = µ(σ−1B). Then for f ∈X, g ∈X′ wehaveZ(T ′g)fdµ =Zgaf ◦σdµ=Z(g ◦σ−1)(a ◦σ−1)fdν=Z(g ◦σ−1)(a ◦σ−1)fwdµ.Thus T ′g = a ◦σ−1wg ◦σ−1 a.e.

and thus is elementary.Of course if X is order-continuous every operator T : X →X is σ(X, X′)−continuous.However, for isometries we can prove a similar result even without this assumption.Proposition 2.5. Let X be a K¨othe function space with the Fatou property and supposeT : X →X is a surjective isometry.

Then T is σ(X, X′)−continuous.Proof: We will use ideas developed in [9]. We recall that the ball topology on X is theweakest topology bX for which every closed ball (with any center and radius) is closed.Then T : (X, bX) →(X, bX) is continuous.

The topology is not a Hausdorfftopology, but([9], Theorem 3.3) its restriction to any absolutely convex Rosenthal set is Hausdorff. Herea set is a Rosenthal set if every sequence contains a weakly Cauchy subsequence.Now suppose h is any strictly positive function in X′.We show that if 0 ≤fn ∈X and fn ↑f a.e.

where f ∈X then Tfn convergesto Tf in L1(hdµ). In fact, setting f0 = 0, Pn≥1(fn −fn−1) is weakly unconditionallyCauchy in X and so Pn≥1(Tfn −Tfn−1) is weakly unconditionally Cauchy in X. ThusPn≥1(Tfn −Tfn−1) converges unconditionally to some g in L1(hdµ).In particular Tfn converges in L1(hµ) to g. Since (Tfn) is bounded in X and X hasthe Fatou property (i.e.

BX is L0-closed) it follows that g ∈X.Now consider the absolutely convex hull of (Tfn) together with the points g and Tf.This is a Rosenthal set. Since bX is weaker than the L0−topology it follows that Tfnconverges to g in bX.

However fn converges to f in bX and Tfn also converges to Tf. Weconclude that Tf = g and so (Tfn) converges to Tf in L1(hdµ).We now can conclude the argument by appealing to Proposition 2.1.6

Remark: This result will only be needed to prove the main result for nonseparable r.i.spaces. The reader who is only concerned with the separable case can observe that if Xis separable it must be order-continous and then any isometry T is σ(X, X′)−continuous.Also its adjoint T ′ : X′ →X′ can be shown directly to be σ(X′, X′′) continuous byidentifying X′′ as the sequential closure of X in X∗∗.3.

Flinn elements.Let X be a real Banach space and suppose T : X →X is a linear operator. We defineΠ(X) to be the subset of X × X∗of all (x, x∗) such that ∥x∥= ∥x∗∥= x∗(x) = 1.

Werecall that an operator T : X →X is numerically positive (Rosenthal [26]) if x∗(Tx) ≥0whenever (x, x∗) ∈Π(X). This is equivalent to requiring the slightly weaker condition thatgiven x with ∥x∥= 1 there exists x∗so that (x, x∗) ∈Π(X) and x∗(Tx) ≥0 (see Lumer[20], [2]).

By results of Lumer [20] and Lumer and Phillips [23] (see also [2]) it is equivalentto the requirement that ∥exp(−αT)∥≤1 for α ≥0. In the case when T is a projection itis easily seen that T is numerically positive if and only if ∥I −T∥= 1.We next introduce an idea which is a real analogue of the notion of hermitian elements[17].

Based on ideas of P.H. Flinn [26] we say that u ∈X is a Flinn element if there isa numerically positive projection P : X →[u].

The set of Flinn elements will be denotedF(X). Note that 0 ∈F(X) and that u ∈F(X) and α ∈R imply αu ∈F(X).

If 0 ̸= u ∈F(X) then there exists f ∈X∗so that f ⊗u is numerically positive projection onto [u].We say then that (u, f) is a Flinn pair. Clearly (u, f) is a Flinn pair if and only if f(u) = 1and f(x)x∗(u) ≥0 for (x, x∗) ∈Π(X).Proposition 3.1.

The set F(X) is closed.Proof: Suppose un ∈F(X) and lim ∥un −u∥= 0. It suffices to consider the case when∥un∦= 0 and ∥u∦= 0.

Then there exist fn ∈X∗so that fn ⊗un is a numerically positiveprojection. Thus ∥fn ⊗un∥= ∥fn∥∥un∥≤2.

Thus ∥fn∥≤2 sup(1/∥un∥). By Alaoglu’stheorem (fn) has a weak∗-cluster point f and clearly (u, f) is a Flinn pair.The next proposition is trivial, but we record it for future use.Proposition 3.2.

Suppose U : X →Y is a surjective isometry. Then U(F(X)) = F(Y );furthermore if (u, f) is a Flinn pair then (U(u), (U ∗)−1f) is a Flinn pair.The next theorem is due to Flinn (see [26], Theorem 1.1).Theorem 3.3.

Let X be a Banach space and π be a contractive projection on X withrange Y. Suppose (u, f) is a Flinn pair in X.

Suppose f /∈Y ⊥. Then π(u) ∈F(Y ).7

Proof: Let g be the restriction of f to Y. We may assume π(u) ̸= 0.

Then let S =g ⊗πu be a rank one operator on Y. If (y, y∗) ∈Π(Y ) then (y, y∗◦π) ∈Π(X).

Nowf(y)y∗(πu) ≥0 and so S is numerically positive. But S2 = βS where β = g(πu).

Byconsidering y = πu/∥πu∥and choosing y∗to norm πu it is immediately clear that β ≥0.If β = 0 then exp(−αS) = I −αS; since by assumption S is non-zero this contradicts∥exp(−αS)∥≤1 for all α ≥0. Hence β > 0 and (πu, β−1g) is a Flinn pair.4.

Flinn elements in lattices.Now suppose that Ωis a Polish space and that µ is a σ−finite Borel measure on Ω.Proposition 4.1. Let X be an order-continuous K¨othe function space on Ω.

(a) Suppose that (u, f) is a Flinn pair with u ∈X and f ∈X′ = X∗. Then fu ≥0 a.e.

(b) Suppose u ∈F(X). Then there exists f ≥0 such that (|u|, f) is a Flinn pair.Proof: Let A be a Borel subset of the set {f > 0} ∩{u < 0} of finite measure.

Supposeµ(A) > 0 and let x = χA/∥χA∥. Pick x∗so that (x, x∗) ∈Π(X) and supp x∗⊂A.

Thenx∗≥0 (a.e.) andRux∗dµ < 0 butRfx dµ > 0.

This contradiction shows that µ(A) = 0and so the set {f > 0} ∩{u < 0} has measure zero. Similar reasoning shows that the set{f < 0} ∩{u > 0} has measure zero.

(b) There is an isometry of X onto X which carries u to |u| so that |u| ∈F(X) byProposition 3.2. Now suppose (|u|, f) is a Flinn element.

Let A = {f < 0} and considerthe isometry Ux = x −2χAx. Clearly by (a), U(|u|) = |u| and of course (U ∗)−1f = |f| sothat (|u|, |f|) is a Flinn pair.Lemma 4.2.

Suppose µ is nonatomic and suppose f, g ∈L1(µ) withR|f| dµ > 0 satisfythe criterion thatZhf dµ Zhg dµ≥0whenever |h| = 1 a.e. Then there is a nonnegative constant c so that g = cf a.e.Proof: Consider the subset Γ of R2 of all (a, b) such that for some h ∈L∞(µ) with|h| = 1 a.e.

we haveRhf dµ = a andRhg dµ = b. Then it is an immediate consequence ofLiapunoff’s theorem [27] that Γ is closed and convex.

However Γ = −Γ and the criterionis that Γ is contained entirely in the union of the first and third quadrants. This triviallyimplies that Γ is contained in a line through the origin; the hypothesis on f implies thisline is not the y-axis and so we deduce the existence of c ≥0 so thatRhg dµ = cRhf dµfor all such h and the lemma follows.We now establish the analogue of Theorem 6.5 of [17].8

Theorem 4.3. Suppose µ is nonatomic and suppose X is an order-continuous K¨othefunction space on (Ω, µ).

Then u ∈X is a Flinn element if and only if there is a nonnegativefunction w ∈L0(µ) with supp w = supp u = B, so that:(a) If x ∈X(B) then ∥x∥= (R|x|2w dµ)1/2.and(b) If v ∈X(Ω\ B) and x, y ∈X(B) satisfy ∥x∥= ∥y∥then ∥v + x∥= ∥v + y∥.Proof: Assume first that 0 ̸= u ∈F(X). We can assume there exists f ∈X∗so that(u, f) is a Flinn pair.

Suppose first that (x, x∗) ∈Π(X). Then if |h| = 1 a.e.

we also have(hx, hx∗) ∈Π(X) and so (Ruhx∗dµ)(Rfhx dµ) ≥0. By Lemma 4.2, there is a constantkx > 0 so that ux∗= kxfx almost everywhere.

It follows immediately that we must havefχΩ\B = 0 almost everywhere. Thus we can define a function w by w = f/u on B andw = 0 otherwise.

Then if (x, x∗) ∈Π(X) we have x∗χB = kxwxχB.Next let us suppose that e1, e2 ∈X(B) satisfy the conditionsRe21w dµ =Re22w dµ = 1andRe1e2w dµ = 0. Consider the function F(ϕ) = e1 cos ϕ + e2 sin ϕ for 0 ≤ϕ ≤2π.Suppose v ∈X(Ω\ B) and consider the function H(ϕ) = ∥v + F(ϕ)∥.We note that the function H is Lipschitz on [0, 2π].

We will show that H′(ϕ) = 0 a.e.and deduce that H is constant.Let us suppose that θ is a point of differentiability of H. Let g ∈X∗be a normingfunction for v + F(θ). Then H(ϕ) −⟨v + F(ϕ), g⟩has a minimum at ϕ = θ and so we candeduce tht H′(θ) = ⟨F ′(θ), g⟩.Since g norms v+F(θ) we conclude that gχB = cwF(θ) for some nonnegative constantc.

Thus⟨F ′(θ), g⟩= cZwF(θ)F ′(θ)dµ= c cos 2θZwe1e2dµ −c2 sin 2θZw(e21 −e22)dµ= 0.Thus H is constant as promised.It follows immediately that if x, y ∈X(B) satisfyRx2w dµ =Ry2w dµ = 1 then∥v + x∥= ∥v + y∥; simply determine e2 so thatRe2xw dµ = 0,Re22wdµ = 1 and y =x cos ϕ + e2 sin ϕ.Taking the special case v = 0 this leads easily to (a). (b) is then the general case.The converse is easy.First note that (b) easily implies that if x, y ∈X(B) with∥x∥≤∥y∥then for v ∈X(Ω\ B) we have ∥v +x∥≤∥v +y∥.

Suppose ∥u∥= 1 and (a) and(b) hold. We show that the pair (u, uw) is Flinn.

Clearly ⟨u, uw⟩= 1. Suppose x ∈X;then I −uw ⊗u(x) = xχΩ\B + y where ∥y∥≤∥xχB∥and so ∥I −uw ⊗u∥≤1.We now apply this theorem to the case when X is a separable r.i. space on [0, 1].9

Theorem 4.4. Suppose X is a separable r.i. space on [0, 1].

If F(X) ̸= {0} then X =L2[0, 1].Proof: If F(X) ̸= 0 then Theorem 4.3 shows that there is a Borel set B ⊂[0, 1] ofpositive measure and a weight function w ∈L0 such that if supp f ⊂B then ∥f∥X =(R|f|2wdλ)1/2. Further if gχB = 0 and f1, f2 ∈X(B) then ∥g + f1∥X = ∥g + f2∥X.

Itfollows immediately from re-arrangement invariance that w is constant and we obtain theexistence of c, δ > 0 so that if λ(supp f) ≤δ then ∥f∥X = c∥f∥2.Now pick an integer N so that 1/N < δ. It follows easily from condition (b) of theprevious theorem that there is a constant a > 0 so that if f1, f2, .

. ., fN are disjointfunctions satisfying λ(supp fk) ≤1/N and ∥fk∥2 = 1 then ∥f∥X = a.

Consider then anysimple function f and write f = f1 + · · · + fN where fk are identically distributed anddisjointly supported. Then ∥f∥X = a∥f1∥2 = aN −1/2∥f∥2.

By considering χ[0,1] it is clearthat aN −1/2 = 1 and the theorem follows easily.5. Flinn elements of finite-dimensional r.i. spaces.Suppose N is a natural number.Let eNi= χ((i−1)2−N ,i2−N] for 1 ≤i ≤2N.

LetXN = [eNi: 1 ≤i ≤2N]. We denote the averaging projection (conditional expectationoperator) of X onto XN by EN.

Notice that X∗N can be identified naturally with X′N. Wewill also let X−N = [eNi : 1 ≤i ≤2N −1].Lemma 5.1.

Suppose X is an r.i. space on [0, 1] so that X ̸= L2. Then there exists N ∈Nso that Pi<2N eNi/∈F(X−N).Proof: Suppose for every n ∈N we have χn = Pi<2n eni = χ[0,1−2−n] ∈F(X−n ).

Since(X−n )∗can be identified with (X′n)−there exists fn = Pi<2n anieni so that (χn, fn) is aFlinn pair for X−n i.e.Rfn dλ = 1 and ∥I −fn ⊗χn∥= 1. Then for every permutationσ of 1, 2, .

. ., 2n −1 we have that (χn, f σn) is a Flinn pair where f σn = Pi<2n anσ(i)eni .

Byaveraging we conclude that (χn, (1 −2−n)−1χn) is a Flinn pair.Now suppose x ∈X. We conclude that∥En(xχn) −(1 −2−n)−1(Z 1−2−n0x(t) dt)χn∥X ≤∥En(xχn)∥X.Letting n →∞we obtain (by the Fatou property of the norm when X is not separable)that∥x −(Z 10x(t) dt)χ[0,1]∥X ≤∥x∥X10

and so (χ[0,1], χ[0,1]) is a Flinn pair in X ×X′. Now if X is separable (i.e.

order-continuous)Theorem 4.4 gives the conclusion that X is isometric to L2. If not we consider X0, theclosure of the simple functions in X; it is immediate that χ[0,1] is Flinn in X0 and so if X0 isseparable, we can again apply Theorem 4.4 to get the conclusion that X is isometric to L2.There remains one case, when X0 is not order-continuous and so ([19]) X0 = L∞[0, 1] upto renorming.

But then we conclude that χ[0,1] is Flinn in X′ which is L1 up to renormingand get a contradiction.We now need to introduce a technical definition. We will say that an r.i. space X hasproperty (P) if for every t > 0,∥e11∥X < ∥e11 + te12∥X.We say that X has property (P ′) if X′ has property (P).Lemma 5.2.

Any r.i. space X has at least one of the properties (P) or (P ′).Proof: Assume X fails both (P) and (P ′). Then for small enough η > 0 we have ∥e11 +ηe12∥X = ∥e11∥X and ∥e11 + ηe12∥X′ = ∥e11∥X′.

But then12(1 + η2) =Z(e11 + ηe12)2dλ≤∥e11∥X∥e11∥X′= 12.This contradiction establishes the lemma.Remark: If X is strictly convex then it has property (P).Lemma 5.3. Assume X has property (P).

Suppose (eNj , u) is a Flinn pair in XN × X′N.Then u = 2NeNj .Proof: It suffices to consider the case j = 1. We can write u = 2NeN1 + Pj>1 ajeNj .

Byusing Proposition 3.2 it follows that (eN1 , |u|) is also a Flinn pair. Then by an averagingprocedure as in the preceding Lemma 5.1 we can show that (eN1 , v) is a Flinn pair wherev = 2NeN1 + η Pj≥2 eNj where (2N −1)η = Pj≥2 |aj|.

We now project by E1 onto X1. ByTheorem 3.3, (E1eN1 , w) is a Flinn pair where w is a multiple of E1v.

Thus (e11, 2(e11 + τe12))is a Flinn pair for some τ > 0.Now consider g = 12τe11 −e12 ∈X1 and suppose this is normed by h = αe11 −βe12 ∈X′1,where α, β ≥0. Thus ∥h∥X′ = 1 and 14τα+ 12β = ∥g∥X.

NowR2g(e11 +τe12)dλ = −12τ < 0,and henceRhe11dλ ≤0 i.e. α ≤0.

hence α = 0 and so h = −∥e12∥−1X′ e12 and ∥g∥X = ∥e12∥Xwhich contradicts property (P).11

Lemma 5.4. Suppose N is a natural number and N = lm.

Suppose d1 ≥d2 ≥· · · ≥dN ≥0. Then there is a permutation σ of {1, 2, .

. ., N} so that ifbj =mXi=1dσ((j−1)m+i)for 1 ≤j ≤l then maxi,j |bi −bj| ≤d1.Proof: The construction is inductive.We will define σ((j −1)m + k) in blocks fork = 1, 2, .

. ., m. For k = 1 we define σ((j −1)m + 1) = j.

Now suppose we have completedthe construction up to k −1 < m. We calculate bk−1j= Pk−1i=1 dσ((j−1)m+i. We then defineσ((j −1)m+k) ∈[(k −1)l+1, kl] in such a way that bk−1i< bk−1jimplies σ((i−1)m+k) <σ((j −1)m + k).

This describes the construction of σ.Now by induction we have maxi,j |bki −bkj | ≤d1. For k = 1 this is obvious.

Supposewe have the result for k −1. Suppose σ((i−1)m+k) ≤σ((j −1)m+k).

Then bk−1i≤bk−1j;furthermore bki = bk−1i+ x and bkj = bk−1j= y where 0 ≤y ≤x. Hence |bki −bkj | ≤max(x −y, bk−1j−bk−1i) ≤d1.Now if we let k = l, the lemma is proved.Proposition 5.5.

Suppose X is an r.i. space on [0, 1] with property (P ′) and such thatX ̸= L2. Then for any 0 < p < ∞there is a constant Ap = Ap(X) so that for every n ∈Nand every u = P2ni=1 aieni ∈F(Xn) we have 2nXi=1|ai|p!1/p≤Ap max1≤i≤2n |ai|.Proof: We start with the simple observation that if P aieni is Flinn then so is P |ai|eniand so it suffices to consider only the case when u ≥0.

Similarly we are free to permutethe (ai). We therefore consider the case when a1 ≥a2 ≥· · · ≥a2n ≥0.Now according to Lemma 5.1 there exists m so that Pi<2m emi/∈F(X−m).

In factby Proposition 3.1, this means that there exists δ > 0 so that if w ∈F(X−m)) then∥w −(2m −1)−1 Pi<2m emi ∥∞≥δ/2. This implies that if w = Pi<2m biemiand P bi = 1then maxi,j |bi −bj| ≥δ.Now let us suppose n > m and that u = P2nj=1 ajenj is Flinn in Xn where a1 ≥a2 ≥· · · ≥a2n ≥0.

Let us set Sk = Pj≤k aj for 1 ≤k ≤2n. Let S0 = 0 and S = S2n−2m.Fix 1 ≤k ≤2n−m.

We consider a permutation σ of {1, 2, . .

., 2n} so that σ{2n −2n−m + 1, . .

., 2n} = {i : i < k} ∪{i : i ≥2n −2n−m + k} and such that ifbj =2n−mXi=1aσ((j−1)2n−m+i)12

for 1 ≤j ≤2m −1 then max |bi −bj| ≤ak. Such a permutation exists by Lemma 5.4.Now we argue that if v = P2nj=1 aσ(j)enj then v ∈F(Xn) and so Em(v) ∈F(Xm).To see this observe that there exists g ∈X′n with g ≥0 so that (v, g) is a Flinn pairby Proposition 4.1; clearly Em(g) ̸= 0 and so by Theorem 3.3, Em(v) ∈F(Xm).

Thusw = Pi≤2m biemi ∈F(Xm).Next we claim that w0 = Pi<2m biemi ∈F(X−m). If w0 = 0 this is trivial.

If not, selecth ≥0 in X′m so that (w, h) is a Flinn pair. If h = Pi≤2m ciemiwe argue that there existsi < 2m so that ci > 0.

For, if not, h is a multiple of em2m and by Lemma 5.3, since X′ has(P), we get that bi = 0 for i < 2m, i.e. w0 = 0.

Now we can apply Theorem 3.3 to deducethat w0 ∈F(X−m).Recalling the original choice of δ this implies that:maxi,j<2m |bi −bj| ≥δ2m−1Xj=1bj.In view of the selection of σ we haveak ≥δ(S2n−2n−m+k−1 −Sk−1) ≥δ(S −Sk−1)and this holds for 1 ≤k ≤2n−m. For convenience, let us put α = 1 −δ.

Then, for1 ≤k ≤2n−m we have(S −Sk) ≤α(S −Sk−1).By induction, we have(S −Sk) ≤αkSfor 1 ≤k ≤2n−m. This gives an estimate on ak, i.e.ak ≤S −Sk−1 ≤αk−1S ≤δ−1αk−1a1,for 1 ≤k ≤2n−m.If 0 < p < ∞, this implies that2nXi=1api ≤2m2n−mXi=1api≤2map1δ−p(1 −αp)−1= ap1Bpp,say.

This estimate holds if n > m. If we take Ap = max(2m/p, Bp) we obtain the Proposi-tion as stated.13

6. Isometries on r.i. spaces.Theorem 6.1.

Let X be an r.i. space on [0, 1] with X ̸= L2. Suppose X has property(P).

Then for any 0 < p ≤1 there is a constant Cp = Cp(X) with the following property.Suppose Y is any K¨othe function space on some Polish space (Ω, µ), for which Y ′ isnorming. Suppose T : X →Y is an isometric isomorphism of X onto Y. Thensupn ∥(2nXi=1|Teni |p)1/p∥Y ≤Cp.Remark: The sequence (P2ni=1 |Teni |p)1/p is increasing.

If Y has the Fatou property itwill follow that supn(P2ni=1 |Teni |p)1/p ∈Y.Proof: We note first that by Proposition 2.5, T −1 is σ(X, X′)−continuous and so hasan adjoint S = (T −1)′ : X′ →X′. We define f ni = Teni and gni = Seni .

Suppose (x, x∗) ∈Π(Xn) where x = P aieni and x∗= P a∗i eni . Then (Tx, Sx∗) ∈Π(Y ) and this implies that(*)(2nXi=1aif ni (ω))(nXi=1a∗i gni (ω)) ≥0for µ−a.e.

ω ∈Ω.Using the fact that Π(Xn) is separable it follows that there is a set of measure zeroΩn0 so that if ω /∈Ωn0, (*) holds for every (x, x∗) ∈Π(Xn). Let Ω0 = ∪n≥1Ωn0.Now define Fn(ω) = P2ni=1 f ni (ω)eni ∈X′n and Gn(ω) = P2ni=1 gni (ω)eni ∈Xn.

Theabove remarks show the operator Gn(ω) ⊗Fn(ω) is numerically positive on X′n if ω /∈Ω0.Now let Bn = {ω : Gn(ω) = 0}. Clearly (Bn) is a descending sequence of Borelsets.

Let B = ∩Bn. If µ(B) > 0 then there exists a nonzero h ∈Y supported on B and⟨h, Sx′⟩= 0 for every x′ ∈X′.

Thus T −1h = 0, which is absurd.Let Dn = Ω\ (Ω0 ∪Bn). If ω ∈Dn then Gn(ω) ̸= 0 and so it follows that Fn(ω) ∈F(X′n).

We recall that X has property (P) and so X′ has property (P ′). Hence lettingAp = Ap(X′) be the constant from Proposition 5.5(2nXi=1|f ni (ω)|p)1/p ≤Ap max1≤i≤n |f ni (ω)|.14

Hence∥χDn(2nXi=1|f ni |p)1/p∥Y ≤Ap∥max1≤i≤2n |f ni |∥X≤Ap∥(2nXi=1|f ni |2)1/2∥X≤KGAp∥(2nXi=1|eni |2)1/2∥X= KGApby Krivine’s theorem ([19] 1.f.14, p.93.) Now the sequence χDn(P2ni=1 |f ni |p)1/p is increas-ing, as 0 < p ≤1.

If g ≥0 and ∥g∥Y ′ ≤1, we haveZDng(2nXi=1|f ni |p)1/pdµ ≤KGApand soZΩg(supn (2nXi=1|f ni |p)1/p)dµ ≤KGAp.We now quickly obtain the Theorem since Y ′ is norming.Let M = M[0, 1] = C[0, 1]∗denote the space of regular Borel measures on [0, 1]. If0 < p ≤1 and µ ∈M we define the p-variation of µ by∥µ∥p = sup{(nXk=1|µ(Bk)|p)1/p : n ∈N, B1, .

. .

, Bn ∈B disjoint}.If p = 1 this reduces to the usual variation norm. For p < 1 it is easily seen that ∥µ∥p < ∞if and only if µ = P∞n=1 anδ(tn) for some sequence of distinct elements (tn) in [0, 1] andan ∈R such that P |an|p = ∥µ∥pp (see [14], [24]).

The following lemma is standard and weomit the proof.Lemma 6.2. For µ ∈M we have∥µ∥p = supn (2nXk=1|µ(D(n, k))|p)1/pwhere D(n, 1) = [0, 2−n] and D(n, k) = ((k −1)2−n, k2−n] for 2 ≤k ≤2n.We now use the machinery developed in [15].Suppose X is an r.i.space.LetT : X →L0[0, 1] be a continuous linear operator.

We say that T is controllable if there15

exists h ∈L0 so that |Tx| ≤h a.e. when ∥x∥∞≤1.

T is said to be measure-continuous ifit satisfies the criterion that that |Txn| converges to zero in L0 whenever sup ∥xn∥∞≤1and |xn| converges to zero in L0. Thus it follows from Theorem 3.1 of [15] (cf.

Sourour[28]) that T is controllable and measure-continuous if and only if there is a weak∗-Borelmap s →νTs from [0, 1] into M satisfying |νTs |(B) = 0 almost everywhere when B hasmeasure zero, and such that we have for any x ∈L∞Tx(s) =Zx(t) dνTs (t).The map s →νTs is called the representing kernel or representing random measure for Tand it is unique up to sets of measure zero.We further remark that if ∥νTs ∥p < ∞a.e. for some p < 1 then νTs is purely atomicfor almost every s and so (cf.

[14], [29] Theorem 4.1) there is a sequence of Borel mapsσn : [0, 1] →[0, 1] and Borel functions an on [0, 1] so that |an(s)| ≥|an+1(s)| a.e. for everyn and σm(s) ̸= σn(s) whenever m ̸= n and s ∈[0, 1] and for whichνTs =∞Xn=1an(s)δ(σn(s)).We can now summarize our conclusions, restricting attention to surjective isometrieson X.Proposition 6.3.

Let X be an r.i.space on [0, 1] with property (P), and such thatX ̸= L2 (isometrically). Then for any 0 < p ≤1 there is a constant Cp depending only onX such that the following holds.

Suppose T : X →X is a surjective isometry. Then T iscontrollable and further its representing kernel νTs satisfiesZ 10∥νTs ∥ppds ≤Cpp.Proof: This is an almost immediate consequence of Theorem 6.1.We use the samenotation.

If F = supn(P2nk=1 |Tenk|) then we have ∥F∥X′′ ≤C1 and so ∥F∥1 ≤C1. Itis easy to deduce that if ∥x∥∞≤1 then |Tx| ≤F.

To conclude the argument we willneed that T is measure-continuous. This is immediate if X is not equal to L∞with someequivalent renorming, since in this case ∥xn∥∞≤1 and |xn| →0 in measure imply that∥xn∥X →0.

In the exceptional case we use Propositions 2.1 and 2.5 to deduce that T ismeasure-continuous. We conclude that in every case T has a representing random measureνTs .16

Now if 0 < p ≤1, then by Lemma 6.2∥νTs ∥p = supn (2nXk=1|Tenk(s)|p)1/palmost everywhere. Hence(Z∥νTs ∥ppds)1/p ≤∥∥νTs ∥p∥X′′ ≤Cp.Theorem 6.4.

Let X be an r.i. space on [0, 1] which is not isometrically equal to L2[0, 1],and let T : X →X be a surjective isometry. Then there exists a Borel function a on [0, 1]with |a| > 0 and an invertible Borel map σ : [0, 1] →[0, 1] such that λ(σ−1(B)) > 0 if andonly if λ(B) > 0 for B ∈B and so that Tx(s) = a(s)x(σ(s)) a.e.

for every x ∈X.Proof: We start by assuming that X has property (P). According to the previous propo-sition every surjective isometry T is controllable and further for every 0 < p ≤1 there isa constant Cp depending only on X so that(Z 10∥νTs ∥ppds)1/p ≤Cp.Let us define Kp to be the least such constant i.e.Kp = sup{∥(∥νTs ∥p)∥p : T is a surjective isometry}.Suppose T is any fixed isometry.

We can represent νTs = P∞n=1 an(s)δ(σn(s)) wherean is a sequence of Borel functions, and σn : [0, 1] →[0, 1] is a sequence of Borel mapssatisfying σm(s) ̸= σn(s) whenever m ̸= n and 0 ≤s ≤1. In this representation we canassume that σi(s) ̸= 0 for all i, s since the measure of set where νTs ({0}) ̸= 0 is clearlyzero; thus we could simply redefine σi to avoid 0 without changing the kernel except on aset of measure zero.

It follows that∥νTs ∥pp =∞Xn=1|an(s)|p.We define the function Hp(s) = P |an(s)|p.¿From now on we will fix 0 < p ≤1. Let MN(s) be the greatest index such thatσ1(s), .

. ., σM(s) belong to distinct dyadic intervals D(N, k).

Then MN(s) →∞for all s17

and it follows easily that given ǫ > 0 we can find M, N and a Borel subset E of [0, 1] withλ(E) > 1 −ǫ and such that MN(s) ≥M for s ∈E, andZ[0,1]\EHpdt < ǫZE∞Xn=M+1|an(s)|pds < ǫ.For notational convenience we will set P = 2N. Let us identify the circle group T withR/Z = [0, 1) in the natural way.

For θ ∈[0, 1)P we define a measure preserving Borelautomorphism γ = γ(θ1, . .

., θP) given by γ(0) = 0 and thenγ(s) = s + (θk −ρ)2−Nfor (k−1)2−N < s ≤k2−N where ρ = 1 if 2Ns+θk > k and ρ = 0 otherwise. Thus γ leaveseach D(N, k) invariant.

The set of all such γ is a group of automorphisms Γ which weendow with the structure of the topological group TP = [0, 1)P. We denote Haar measureon Γ by dγ. For each k let Γk be the subgroup of all γ(θ) for which θi = 0 when i ̸= k.Thus Γ = Γ1 .

. .

ΓP .We also let the finite permutation group ΠP act on [0, 1] by considering a permutationπ as inducing an automorphism also denoted π by π(0) = 0 and then π(s) = π(k) −k + sfor (k −1)2−N < s ≤k2−N. We again denote normalized Haar measure on ΠP by dπ.Finally note that the set ΓΠP = T also forms a compact group when we endow this withthe product topology and Haar measure dτ = dγ dπ when τ = γπ.We now wish to consider the isometries Vτ : X →X for τ ∈T defined by Vτx = x◦τ.For every τ ∈T the operator S(τ) = TVτT is a surjective isometry and so has an abstractkernel νS(τ)s.Lemma 6.5.

For almost every τ ∈T we have thatZ 10∞Xn=1∞Xj=1|aj(s)||an(τσjs)|ds < ∞(1)∞Xn=1∞Xj=1aj(s)an(τσjs)δ(σnτσis) = νS(τ)sa.e. (2)Proof: Let us prove (1).

Note that for any fixed s and (n, j) we haveZT|an(τσjs)|dτ =Z 10|an(t)|dt18

and so it follows thatZ 10ZT∞Xn=1∞Xj=1|aj(s)||an(τσjs)|dτ ds < ∞.This proves the first assertion. Note, in particular, if (1) holds,∞Xn=1∞Xj=1|aj(s)||an(τσjs)| < ∞for almost every s.To obtain (2) let us suppose that τ is such that (1) holds.

Suppose ∥x∥∞≤1. ThenVτTx may not be bounded but there is an increasing sequence Fm of Borel subsets of [0, 1]with ∪Fm = [0, 1], so that χFmVτTx is bounded.

ThusT(χFmVτTx)(s) =∞Xj=1aj(s)χσ−1jFm(s)Tx(τσjs)=∞Xj=1 ∞Xn=1aj(s)χσ−1jFm(s)an(τσjs)x(σnτσjs)!.Now for almost every s since the double series absolutely converges we may obtainlimm→∞T(χFmVτTx)(s) =∞Xn=1∞Xj=1aj(s)an(τσjs)x(σnτσjs).If X is order-continuous of X the left hand side is simply (S(τ)x)(s) a.e. Thus by theuniqueness of the representing random measure we obtain (2).

For the general case we usePropositions 2.1 and 2.5 to give the same conclusion. It follows thatTVτTx(s) =∞Xn=1∞Xj=1aj(s)an(τσjs)x(σnτσjs)for almost every s. Again the uniqueness of the representing random measure gives (2) andcompletes the proof of the Lemma.Let us now define µ(s, τ) ∈M by settingµ(s, τ) =∞Xn=1∞Xj=1aj(s)an(τσjs)δ(σnτσis)19

whenever the series in (1) converges absolutely and setting µ(s, τ) = 0 otherwise. It isnot difficult to see that the map (s, τ) →µ(s, τ) is weak∗-Borel.For a.e.τ we haveµ(s, τ) = νS(τ)sa.e.

and soZ 10∥µ(s, τ)∥ppds ≤Kpp.It follows thatZTZ 10∥µ(s, τ)∥ppds dτ ≤Kpp.Lemma 6.6. For almost every (s, τ) ∈E × T we have∥µ(s, τ)∥pp ≥MXj=1−∞Xj=M+1∞Xn=1|aj(s)|p|an(τσjs)|p.Proof: Assuming (s, τ) belongs to set where (1) holds it is clear the conclusion fails for(s, τ) if and only if there exist two distinct pairs (n, j), (m, i) where m, n ∈N and i, j ≤Mso that σnτσjs = σmτσis and an(τσjs) ̸= 0.Assume then the conclusion of the lemma is false.Then there is a distinct pair(n, j), (m, i) as above and Borel subset B of E × T so thatRB |an(τσjs)|ds dτ > 0 andσnτσjs = σmτσis for (s, τ) ∈B.

Note first that we must have i ̸= j.It will now follow from Fubini’s theorem that there is a Borel subset B′ of Γ and afixed s ∈E and π ∈ΠP for whichRB′ |an(γπσis)|dγ > 0 and so that σnγπσjs = σmγπσisfor γ ∈B′.Now πσjs ∈D(N, k) for some k and πσis ∈D(N, l) where l ̸= k since i, j ≤M ands ∈E. We write Γ = Γk × Γ′ where Γ′ = Qr̸=k Γr.

Again by Fubini’s theorem thereexists a fixed γ′ ∈Γ′ and a Borel subset B0 of Γk so thatRB0 |an(γkγ′πσis)|dγk > 0 andσnγkγ′πσjs = σmγkγ′πσis for γk ∈B0.Now we note that γ′πσis ∈D(N, l) is fixed by every γk and so σnγkγ′πσjs = s′ is fixedfor γk ∈B0. But B0 has positive measure and γ′πσis ∈D(N, k).

Thus there is a subset Aof D(N, k) so thatR|an(t)|dt > 0 and σn(A) = {s′}. This means that |νTt |({s′}) > 0 on aset of positive measure and we have a contradiction.We now complete the proof.

We haveKpp ≥ZEZT∥µ(s, τ)∥ppdτ ds≥ZEZT∞Xj=1−2∞Xj=M+1∞Xn=1|aj(s)|p|an(τσjs)|pdτ ds20

As beforeZT|an(τσjs)|pdτ =Z 10|an(t)|pdt.Thus we obtainKpp ≥ ∞Xn=1|an|pdt! ZE(∞Xn=1−2Xn>M)|an(t)|pdt!and this implies thatKpp ≥Z 10Hpdt ZEHpdt −2ǫ.We finally deduce that J =R 10 Hpdt then J(J −3ǫ) ≤Kpp.

But ǫ > 0 is arbitrary and sowe have J2 ≤Kpp. But as this applies to all such T we have the conclusion K2p ≤Kp i.e.Kp ≤1.

Now this applies to all 0 < p ≤1.Returning to our original T we note thatZ 10∞Xn=1|an(s)|pds ≤1for all p. Let R(s) be the number of points in the support of νTs . ThenR(s) = limp→0X|an(s)|p.By Fatou’s LemmaZ 10R(s)ds ≤1.To deduce that T is elementary we must show R(s) = 1 a.e.

If X is order-continuousthis is obvious since the fact that T is surjective requires R(s) ≥1 a.e.For the general case we again use Proposition 2.1. We first note that R(s) < ∞a.e.We then show that if x ∈X then(3)Tx(s) =∞Xj=1aj(s)x(σjs)almost everywhere.

To see this it suffices to consider the case x ≥0. We first find anincreasing sequence of Borel sets Fn such that xχFn ∈L∞and ∪Fn = [0, 1].

Then byProposition 2.1 TxχFn converges in measure to Tx. However,TxχFn(s) =∞Xj=1aj(s)x(σjs)χFn(σjs)21

and this converges almost everywhere to the right-hand side of (3). Now as in the order-continuous case we can argue that if T is onto we must have R(s) ≥1 a.e.

and henceR(s) = 1 a.e. We conclude that T is elementary in the case when X has property (P).If X fails property (P) then by Lemma 5.2 X′ has property (P).

Further, Proposition2.5 says that the adjoint T ′ : X′ →X′ is a surjective isometry. But then T ′ is elementaryand by Lemma 2.2 T ′′ and hence T is elementary.7.

Isometries in spaces not isomorphic to Lp.We now recall the definition of the Boyd indices of an r.i. space X (cf. [19] p. 129).For 0 < s < ∞define Ds : X →X by Dsf(t) = f(t/s) where we let f(t) = 0 for t > 1.Then the Boyd indices pX and qX are defined by1pX= lims→∞log ∥Ds∥log s1qX= lims→0log ∥Ds∥log s.Proposition 7.1.

Let X be an r.i. space and suppose T : X →X is an elementaryoperator. Suppose pX ≤r ≤qX.

Then T is bounded on Lr[0, 1] and ∥T∥Lr ≤∥T∥X.This Proposition is proved in [16] Theorem 5.1. In fact the hypotheses of [16] Theorem5.1 suppose X is a quasi-Banach space and have an additional unnecessary restrictionr ≤min(1, qX).

This restriction is not used in Theorem 5.1 of [16] but is important in thefollowing Theorem 5.2.We will however show a direct proof under the assumption that Tx = ax◦σ where σ isa Borel automorphism of [0, 1] which is the case we need. For convenience we consider thecase r < ∞, the other case being similar.

Let us assume ∥T∥X = 1. We define a measureBorel measure µ by µ(B) = λ(σ−1B) and it follows from the fact that T is bounded thatµ is continuous with respect to λ and so has a Radon-Nikodym derivative w. Now for anyx we have∥Tx∥rr =Z 10|a(s)|r|x(σ(s))|rds=Z 10|a(σ−1s)|rw(s)|x(s)|rdsand so we need to show that |a(σ−1s)|rw(s) ≤1 a.e.Suppose not.Then there is aBorel set E of positive measure δ and 0 < α, β so that αrβ > 1 and |a(σ−1s)| > α and22

w(s) > β for s ∈E. Then if x is supported in E it quickly follows that ∥Tx∥X ≥α∥Dβx∥Xand so ∥Dβ∥X[0,δ] ≤α−1.

However for any δ > 0 we have the estimate ∥Dβ∥X[0,δ] ≥max(β1/p, β1/q) where p = pX and q = qX. Thus β1/r ≤α−1 contrary to assumption.In [18] Lamperti shows that if 1 < p < ∞then Lp[0, 1] has an equivalent r.i. norm(not equal to the original norm) so that there are isometries of the form Tf = af ◦σ with|a| ̸= 1 on a set of positive measure.

In the next theorem we show that if X is not equalto Lp up to equivalence of norm then the isometries of X can only be of the very simplestform.Theorem 7.2. Suppose X is an r.i. space and that T is a surjective isometry.

Theneither X = Lp[0, 1] up to equivalence of norm for some 1 ≤p ≤∞or there is an invertiblemeasure-preserving Borel map σ : [0, 1] →[0, 1] and a function a ∈L0[0, 1] with |a| = 1a.e. such that Tx = ax ◦σ for x ∈X.Remark: If pX < qX this follows routinely from Proposition 7.1.

The interesting case isthus when pX = qX.Proof: We have that Tx = ax ◦σ where |a| > 0 a.e. and σ : ([0, 1], λ) →([0, 1], λ)is a Borel automorphism by Theorem 6.4.

Suppose pX ≤r ≤qX. By Proposition 6.1,∥Tx∥r ≤∥x∥r whenever x ∈Lr and similarly ∥T −1x∥r ≤∥x∥r.

Thus T also defines anisometry on Lr for pX ≤r ≤qX.Let us consider first the case pX = qX = ∞. Then |a| ≤1 a.e.

In fact if B = {|a| <1 −ǫ} for some ǫ > 0 then TχσB = aχB since σ is invertible and so λ(B) = 0. Hence|a| = 1 a.e.

Again suppose ǫ > 0. Then, assuming X is not isomorphic to L∞, there existsa least δ so that ∥χ[0,δ]∥X = ∥χ[0,ǫ]∥X.

Then if λ(B) = δ we have λ(σ−1(B)) ≥δ. For anarbitrary Borel subset E of [0, 1], we can split E into sets of measure δ and one remainderset to conclude λ(σ−1(E)) ≥λ(E) −δ.

As ǫ was arbitrary λ(σ−1(E)) ≥λ(E) for all E.Since σ is invertible this forces λ(σ−1(E)) = λ(E) for every E i.e. σ is measure-preserving.We turn to the case when pX = p < ∞.

It then follows that if |a| = 1 a.e. we musthave σ measure-preserving.

We thus assume that |a| ̸= 1 on a set of positive measure; wewill prove that the norm ∥∥X is equivalent to ∥∥p. It suffices to consider the case whena > 0.

It follows first that {a > 1} and {a < 1} both must have positive measure.Let us now make an assumption.Assumption. There exist two disjoint closed intervals I1 and I2 contained in (1, ∞) andso that a−1(I1) and a−1(I2) have positive measure.We will proceed under this assumption.

We can then deduce that there is a constantκ > 1 and two disjoint Borel sets A1, A2 of positive measure such that a(s) > κ for s ∈A2,while a(s) > κa(t) whenever s ∈A1, t ∈A2 but a(s) ≤κa(t) whenever s, t are either bothin A1 or both in A2.23

Let δ = min(λ(σ(A1)), λ(σ(A2))). Let us first note that if x is supported in σ(A1) ∪σ(A2) then since a > κ on A1 ∪A2 we will have λ(supp Tx) ≤κ−pλ(supp x).

We alsohave since a is bounded on A1 ∪A2 an estimate λ(supp Tx) ≥cλ(supp x) for some c > 0.Let us consider any nonnegative x ∈X with support E of measure at most δ, andsuch that ∥x∥p = 1. We define the distortion H(x) by settingH(x) = ess sup{x(s)/x(t) : (s, t) ∈E2}.If the distortion H(x) < ∞then it is clear we can define α(x) = ess inf{x(s) : s ∈E}and β(x) = ess sup{x(s) : s ∈E} and then 0 < α(x) < β(x) < ∞and β(x) = H(x)α(x).Further α(x)χE ≤x ≤β(x)χE.We now define a procedure.

Assume H(x) < ∞. Given such x we define x′ with thesame distribution supported on σ(A1) ∪σ(A2) so that x′ ≤(α(x)β(x))1/2 on σ(A1) butx′ ≥(α(x)β(x))1/2 on σ(A2) ∩supp x′.Now compute y = Tx′.

Then y is supported on A1 ∪A2. If y(s), y(t) are both nonzeroand s, t are in the same Aj we have y(s) ≤β(x)1/2α(x)−1/2κy(t).

If s ∈A1 and t ∈A2we have y(s) ≤β(x)α(x)−1κ−1y(t). If s ∈A2 and t ∈A1 we have y(s) ≤κy(t).

It followsthatH(y) ≤max(κH(x)1/2, κ−1H(x)).Notice also that cλ(supp x) ≤λ(supp y) ≤κ−pλ(supp x).If we put y = x1 we can then iterate the procedure to produce a sequence (xn). Letδn = λ(supp xn); then cδn ≤δn+1 ≤κ−pδn and, in particular, limn→∞δn = 0.

SinceH(xn) ≤max(κH(xn−1)1/2, κ−1H(xn−1)),we deduce that lim sup H(xn) < κ5.Fix any n where H(xn) < κ5. Then for suitable α > 0 and a Borel set E of measure δnwe have αχE ≤xn ≤κ5αχE.

However ∥xn∥p = 1 and so we obtain αδ1/pn≤1 ≤κ5αδ1/pnorκ−5δ−1/pn≤α ≤δ−1/pn.Now we introduce the notation φ(t) = ∥χ[0,t]∥X. The above inequalities give usαφ(δn) ≤∥xn∥X = ∥x∥X ≤κ5αφ(δn),and henceκ−5φ(δn)δ−1/pn≤∥x∥X ≤κ5φ(δn)δ−1/pn.Now since δn+1 ≥cδn we have that for δn+1 ≤t ≤δn,24

c1/pκ−5φ(t)t−1/p ≤∥x∥X ≤c−1/pκ5φ(t)t−1/p.As H(xn) ≤κ5 eventually we can conclude that0 < lim inft→0φ(t)t−1/p < lim supt→0φ(t)t−1/p < ∞.In fact if we let K = lim sup φ(t)t−1/p we obtaincκ−5K ≤∥x∥X ≤c−1κ5K.But this estimate is independent of the original choice of x subject to λ(supp x) ≤δ,H(x) < ∞and ∥x∥p = 1. Hence we obtain that ∥x∥X is equivalent to ∥x∥p.Thus our assumption yields the conclusion that X = Lp[0, 1] up to an equivalentnorm.

Clearly it suffices to find one surjective isometry for which the assumption holds togive this conclusion.If the assumption fails for T then a is essentially constant (with value α, say) on{a > 1}. If the assumption fails for T −1 it is easy to see that a is also essentially constant(with value β, say) on {a < 1}.

Now the same reasoning must apply to any surjectiveisometry. However it is now easy to construct an isometry of the form S = TVτ1TVτ2T,where Vτ = x ◦τ for some measure preserving Borel automorphism τ, and so that Sχ[0,1]takes each of the four distinct values α3, α2β, αβ2 and β3 (of which three must be distinctfrom 1) with positive measure.

Thus we can again conclude that X is isomorphic to Lp.Remarks: This theorem can be cast as a statement about maximal norms (cf. [25], [17]).A Banach space X has a maximal norm if no equivalent norm has a strictly bigger groupof invertible isometries.

The above theorem shows immediately that any r.i. space on [0, 1]which is not isomorphic to Lp[0, 1] has a maximal norm; Rolewicz [25] showed that thespaces Lp[0, 1] have maximal norms. However if X is isomorphic but not isometric to Lpits norm cannot be maximal; this follows rather easily from Proposition 7.1 and the almosttransitivity of the norm in Lp.References.1.

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