ENSAYO DE MURETES DE MAMPOSTERÍA

Theorem 49. Supposef : Λ → [0,∞) is measurable. Then there is aσ ∈ _G(R)such that, for 1≤n≤ ∞, the mappingt7→(f◦σ) (ω, t)is nonincreasing onJna.e. (µn).

Proof. ChooseR >kfk_∞. Suppose1≤n≤ ∞. Let

X ={(h, σ)∈L∞(δn)×_MP(Jn) : 0≤h≤R, h◦σ is nonincreasing onJn},

where {f : 0≤f ≤R} is given the k·k_2,δ

n-topology, MP(Jn) is given the weak topology, and

to the set of n×npermutation matrices and has the discrete topology.) Since k·k₂ convergence

implies subsequential convergence almost everywhere, it follows that X is a complete separable

metric space. Since every measurablehhas a nonincreasing rearrangement, the map

π1 :X → {h: 0≤h≤R}

is onto, so, by Lemma 27, there is an absolutely measurable cross-sectionγn :Y → X forπ1. Let

ηn=π2◦γn:Y →MP(Jn).

We now definesn: Ωn→MP(Jn)by

sn(ω) =ηn(fω)∈MP(Jn).

It is clear from the construction that thatfω◦sn(ω)is a nonincreasing function oft, i.e.,f(ω, sn(ω) (t))

is a nonincreasing function oftfor eachω ∈Ωn.

We define

σn(ω, t) = (ω, sn(ω) (t)).

Thenσ ={σn}1≤n≤∞∈G(R)has the desired properties.

Note that the functionσ is not necessarily unique, but the functionf ◦σis unique. It is called thenonincreasing rearrangement functionforf, and we denote it bysf. Iffandhare nonnegative

measurable functions onΛ, we say thatf andhare_G(R)-equivalentif and only if sf = sh a.e.

(λ). This holds if and only if there is aσ1 ∈G(R)such thath=f ◦σ1.

For eachω∈Ωnandt ∈Jn,sf(ω, t)is callthetths-number of f atω. Definition 50. SupposeT ∈ R. We can writeT = P

1≤n≤∞ R⊕ ΩnT (ω)dµn(ω). We definesT ∈ L∞(Λ, λ)by sT (ω, t) =sT(ω)(t) when1≤n≤ ∞, ω ∈Ωnandt ∈Jn.

Definition 51. Supposef ∈ L∞(Λ, λ)and0 ≤ f. For each 1 ≤ n ≤ ∞, and each ω ∈ Ωn, we definefω ∈L∞(Jn, δn)by fω(t) = f(ω, t). We view f = ⊕ X 1≤n≤∞ Z ⊕ Ωn fωdµn(ω). We then definesf ∈L∞(Λ, λ)by sf(ω, t) =sfω(t).

Lemma 52. Suppose0≤f ∈L∞(Λ, λ). Then there is aσ∈_Gsuch that,f◦σ =sf.

Proof. For1≤n≤ ∞, the mapω 7→fωfromΩntoL∞(Jn, δn)is measurable. For eachω∈Ωn,

there is a σω ∈ MP(Jn, δn)such thatfω ◦σω = sfω. Using measurable cross-sections, we can choose theσω’s so that{σω :ω∈Ω}is measurable. Thusσ =

P

1≤n≤∞ R⊕

Ωnσω ∈Gand

(f ◦σ) (ω, t) = f(ω, σω(t)) = (fω◦σω) (t) = sfω(t) = sf(ω, t).

Lemma 53. Suppose T ∈ R, A is a masa in R, |T| ∈ A, πA : L∞(Λ, λ) → A is a tracial embedding as in Theorem 33, andf ∈L∞(Λ, λ)satisfiesπA(f) =|T|. ThensT =sf.

Proof. We can write

A= X

1≤n≤∞ Z ⊕

Ωn

Aωdµn(ω),

where, for1≤n≤ ∞andω ∈Ωn,Aωis a masa inRω. We can also write

πA = X 1≤n≤∞ Z ∞ Ωn πωdµn(ω),

where, for eachω∈Ωn,πω :L∞(Jn, δn)→ Aωis a tracial embedding. IfπA(f) = |T|, then, for

almost everyω,

Thus, for almost everyω ∈Ω,

sfω =sTω . Thussf =sT.

Lemma 54. Suppose A1, A2 are masas in R, 0 ≤ Ak ∈ Ak, πk : L∞(Λ, λ) → Ak are the isomorphisms in Theorem 33 and f1, f2 ∈ L∞(Λ, λ) satisfy πk(fk) = Ak for k = 1,2. The following are equivalent:

1. sf1 =sf2

2. There is aγ ∈_G(R)such thatf2 =f1◦γ

3. There is a sequence{Un}of unitary operators inRsuch that

kUnA1Un∗−A2k →0.

4. For every unitarily invariant normαonR

α(A1) = α(A2)

5. For every rational numbert∈(0,1]KFt(A1) =KFt(A2).

Proof. (1) ⇒ (2). There areγ1, γ2 ∈ G(R) suchsfk = fk◦γk fork = 1,2. By(1) we have

f2 =f1◦ γ1◦γ2−1

.

(2) ⇒(3). Defineπ3 :L∞(Λ, λ)→ A2by

π3(f) =π2(f ◦γ).

Thus π3(f1) = A2. By Theorem 23, π1 ∼a π3. Thus there is a net (sequence) {Ui}of unitary

operators inRsuch that

lim i kUiA1U ∗ i −A2k= lim i kUiπ1(f1)U ∗ i −π3(f1)k= 0.

Hence, for everyn∈_N, there is a unitaryUnsuch that

kUnA1Un∗−A2k<1/n.

(3)⇒(4),(4)⇒(5)are trivial.

(5) ⇒(1). We know thatKFt(A1) = KFt(sf1)andKFt(sf2). Let

Et={ω∈Ω :KFt(sf1) (ω)6=KFt(sf2) (ω)},

and let E = ∪Et, then λ(E) = 0. Therefore

Rt

0 f1(x)dx =

Rt

0f2(x)dxfor every 0 < t ≤ 1.

Thusf1(x) =f2(x)except on a countable set. Thereforef1 =f2 a.e.(δ∞).

Corollary 55. Suppose A1, A2 are masas in R, 0 ≤ A ∈ Ak, πk : L∞(Λ, λ) → Ak are the isomorphisms in Theorem 33 and f1, f2 ∈ L∞(Λ, λ) satisfy πk(fk) = A for k = 1,2. Then

sf1 =sf2.

IfT ∈ R, we define

KFt(T) =KFt(s(fT))

We need to definetth_{Ky Fan functionKF}

t(T)solely in terms ofT andR. (See Lemma 17)

Note that whenn = ∞, KFt is defined onL∞(Jn, δn) for all0 < t ≤ 1.For 1 ≤ n < ∞,

KFtis only defined whent∈

₁

n, . . . , n

n . The next definition extends this concept.

Definition 56. Suppose1≤n <∞and0< t≤1. We choose an integerk,1≤k≤nsuch that

k−1 n < t≤ k n. We defineKFtonL∞(Jn, δn)by KFt =KFk n.

Forf ∈L∞(Λ)and1≤n≤ ∞andω ∈Ωnandt ∈Jn, we define

KFt(f) (ω, t) = KFt(sfω),

and we define, forT ∈ R,

KFt(T) =KFt(sT).

We easily have that forS, T ∈ R

KFt(S+T)≤KFt(S) +KFt(T)

always holds.