Martingales Junge

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arXiv:math/0206062v1

[math.O

A]6Jun2002

DOOBS INEQUALITY FOR NON-COMMUTATIVEMARTINGALES

MARIUS JUNGE

Abstract. Let 1 p < and (xn)nN be a sequence of positive elements in a non-commutative Lp space and (En)nN be an increasing sequence of conditional expectations,

then n

En(xn)

p

cpn

xn

p

.

This inequality is due to Burkholder, Davis and Gundy in the commutative case. By

duality, we obtain a version of Doobs maximal inequality for 1 < p .

Introduction:

Inspired by quantum mechanics and probability, non-commutative probability has be-

come an independent field of mathematical research. We refer to P.A. Meyers exposition

[Me], the successive conferences on quantum probability [AvW], the lecture notes by Jajte

[Ja1, Ja2] on almost sure and uniform convergence and finally the work of Voiculescu,

Dykema, Nica [VDN] and of Biane, Speicher [BS] concerning the recent progress in free

probability and free Brownian motion. Doobs inequality is a classical tool in probability

and analysis. Transferring classical inequalities into the non-commutative setting theory

often requires an additional insight. Pisier, Xu [PX, Ps3] use functional analytic and com-

binatorial methods to establish the non-commutative versions of the Burkholder-Gundy

square function inequality. The absence of stopping time arguments, at least until the

time of this writing, imposes one of the main difficulties in this recent branch of martin-

gale theory.

The formulation of Doobs inequality for non-commutative martingales faces the following

problem. For an increasing sequence of conditional expectations (En)nN and a positiveoperator x in Lp, there is no reason why supn En(x) or supn |En(x)| should be an elementin Lp or represent a (possibly unbounded) operator at all. Using Pisiers non-commutative

vector-valued Lp-space Lp(N; ) we can overcome this problem and at least guess the right

formulation of Doobs inequality. However, Pisiers definition is restricted to von Neumann

algebras with a -weakly dense net of finite dimensional subalgebras, so-called hyperfinite

Junge is partially supported by the National Science Foundation.1
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2 MARIUS JUNGE

von Neumann algebras. But maximal inequalities are also interesting for free stochastic

processes where the underlying von Neumann algebra is genuinely not hyperfinite. All

these obstacles disappear for the so-called dual version of Doobs inequality: For everysequence (xn)nN of positive operators

n

En(xn)

p

cpn

xn

p

. (DDp)

In the commutative case this inequality is due to Burkholder, Davis and Gundy [ BDG]

(even in the more general setting of Orlicz norms). Since it is crucial to understand

our approach to Doobs inequality, let us indicate the duality argument relating (DDp)

and Doobs inequality in the commutative case. Indeed, (DDp) implies that T(xn) =n En(xn) defines a continuous linear map between Lp(1) and Lp. The norm ofT

yields the best constant in Doobs inequality for the conjugate index p = p

p1.sup

n

|En(x)|p

= T(x)Lp () T xp = cp xp .

Personally, I learned this argument after reading Dilworths paper [Di]. But I am sure it

is known to experts in the field, see Garcias [Ga] for the general theory (and [AMS] for

the explicit equivalence). (DDp) admits an entirely elementary proof in the commutative

case (see again [AMS]). This elementary proof still works in the non-commutative case

for p = 2, see Lemma 3.1. It is the starting point of our investigation. We recommend

the reader (not familiar with modular theory) to start in section 3 where interpolation

is used to extend (DDp) to 1 p 2 and suitable norms are introduced to make theabove duality argument work in the non-commutative case. In section 4, we establish

the dual version (DDp) in the range 2 p < using duality arguments which rely onPisier/Xus version of Steins inequality in combination with techniques from Hilbert C-

modules. By duality, we obtain the non-commutative Doob inequality in the more delicate

range 1 < p 2. The heart of our arguments rely on the (apparently new) connectionbetween Hilbert C-modules and non-commutative Lp spaces presented in section 2. These

duality techniques are necessary because p > 1 and 0

a

b implies 0

ap

bp only for

commuting operators. This is very often used in the elementary approach to commutativemartingales inequalities as in Garsias book [Ga].

Let us formulate our main results for finite von Neumann algebras. If : N C isa normal, tracial state, i.e. (xy) = (yx), then the space Lp(N, ) is defined by the

completion of N with respect to the norm

xp = ((xx)p2 )

1p .


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NON-COMMUTATIVE DOOB INEQUALITY 3

We refer to the first section for more precise definitions and references. Given a subalgebra

M N, the embedding : L1(M, ) L1(N, ) is isometric because |x| =

xx M forall x M. The dual map E = : N M yields a conditional expectation satisfying

E(axb) = aE(x)b

for all a, b M and x N, see [Tk, Theorem 3.4.]. Since E is trace preserving, E extendsto a contraction E : Lp(N, ) Lp(N, ) with range Lp(M, ). In the following, we con-sider an increasing sequence (Nn)nN N of von Neumann subalgebras with conditionalexpectations (En)nN. We recall that an element x is positive if it is of the form x = y

y.

Theorem 0.1. Let1 p < , then there exists a constant cp depending only on p such

that for every sequence of positive elements (xn)nN Lp(N, )

n

En(xn)

p

cpn

xn

p

. (DDp)

Note the close relation to the non-commutative Stein inequality, see [ PX, Theorem 2.3.],n

En(xn)En(xn)

p

22pn

xnxn

p

.

Using Kadisons inequality En(xn)En(xn) En(xnxn), it turns out that (DDp) is stronger

than Steins inequality. However, Steins inequality combined with the theory of HilbertC-modules yields one of the fundamental inequalities in the proof of Theorem 0.1. Using a

Hahn-Banach separation argument a la Grothendieck-Pietsch, we deduce Doobs maximal

inequality.

Theorem 0.2. [Doobs maximal inequality] Let 1 < p and x Lp(N, ), thenthere exist a, b L2p(N, ) and a sequence of contractions (yn) N such that

En(x) = aynb and a2p b2p cp xp

Here cp

is the constant in Theorem 0.1 for the conjugate index p

=

p

p1. In particular,for every positive x Lp(N, ), there exists a positive b Lp(N, ) such thatEn(x) b

for all n N.

In the case of hyperfinite von Neumann algebras this is equivalent to the corresponding

vector-valued inequality, see [Ps2], and therefore justifies the name maximal inequality.


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4 MARIUS JUNGE

In the semi-commutative case where Nn = L(, n, ) M and n are increasing -algebras this inequality is stronger than the vector-valued Doob inequality, see Remark

5.5. In particular, this applies for random matrices. The inequality can be extendedto a continuous index set under suitable density assumptions, for examples for Clifford

martingales or free stochastical processes. Since these modifications are rather obvious,

we omit the details.

Clearly, maximal inequalities immediately imply almost sure convergence. Therefore it is

not surprising that Theorem 0.2 implies almost uniform convergence convergence of the

martingale truncations (En(x)) for p 2 and bilateral convergence for 1 < p 2 incase of a tracial state. We refer to [Ja1, Ja2] for the definition of these notions and more

details. In the tracial case the bilateral convergence of the martingale truncations is knownby a result of Cuculescu [Cu] even for martingales in L1(N, ). Therefore Theorem 0.2

provides a alternative approach to these results but only for p > 1. However, the maximal

inequality discussed in [Ja1] cannot easily be interpolated to obtain Theorem 0.2 as in the

real case. In a subsequent paper [DJ2], we will apply the maximal inequality of Theorem

0.2 in Haagerup Lp spaces in order to obtain (bilateral) almost sure convergence for all

states thus underlining the strength of these maximal inequalities.

Preliminary results and notation are contained in section 1. Section 5 contains immedi-

ate applications to submartingales and conditional expectations associated to actions of

groups.

I am indebted to Q. Xu for many discussion and support. I want to thank A. Defant for

the discussion about almost everywhere convergence. The knowledge of similar results for

almost sure convergence of unconditional sequences, see [DJ1], have been very encouraging.

I would like to thank Stanislaw Goldstein for initiating a correction in the proofs of Lemma

2.3 and Lemma 3.2.

1. Notation and preliminary results

As a shortcut, we use (xn), (xnk) instead of (xn)nN, (xnk)n,kN for sequences indexed by

the natural numbers N or its cartesian product N2. We use standard notation in operator

algebras, as in [Tk, KR]. In particular, B(H) denotes the algebra of bounded operators

on a Hilbert space H and K(H), K denote the subalgebra of compact operators on H, 2,respectively. The letters N, M will be used for von Neumann algebras, i.e. subalgebras

of some B(H) which are closed with respect to the -weak operator topology. We refer

to [Dx, Tk] for the different locally convex topologies relevant to operator algebras. For


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n N we denote by Mn(N) the von Neumann algebra of n n matrices with values inN. We will briefly use Mn for Mn(C). Given C

-algebras A, B, we denote by A Btheminimal tensor product. For von Neumann algebras N B(H1), M B(H2), we useNM for the closure of N M B(H1 H2) in the -weak operator topology. Let usrecall that a von Neumann algebra is semifinite if there exists a normal, semifinite faithful

trace. A trace is a positive homogeneous, additive function on N+ = {xx |x N}, thecone of positive elements ofN, such that for all increasing nets (xi)i with supremum in N

and for all x Nn) (supi xi) = supi (xi).

s) For every 0 < x there exists 0 < y < x such that 0 < (y) < .f) (x) = 0 implies x = 0.

t) For all unitaries u N: (uxu) = (x).A positive homogeneous, additive function w : N+ [0, ] satisfying n), s), f), but notthe last property t), is called a n.s.f. (normal, semifinite, faithful) weight.

It will be worthwhile to clarify the different notions of non-commutative Lp-spaces. If

is a trace then

m() =

n

i=1

yixi

n N,n

i=1

[(yi yi) + (xi xi)] <

is the definition ideal on which there exists a unique linear extension : m() Csatisfying (xy) = (yx). The Lp-(quasi)-norm is defined for x m() by

xp = ((xx)p2 )

1p .

Then Lp(N, ) is the completion of m() with respect to the Lp-norm. (For p < 1 smaller

ideal is needed in order to guarantee that (|x|p) is finite.) We refer to [Ne, Te, KF, Ye] formore on this and the fact that Lp(N, ) can be realized as unbounded operators affiliated

to N.

The starting point of Kosakis [Ko] definition of an Lp-space is a normal faithful state on a von Neumann algebra N. Then N acts on the Hilbert space L2(N, ) obtained by

completing N with respect to the norm

xL2(N,) = (xx)12 .

The modular operator is an (unbounded) operator obtained from the polar decomposi-

tion S = J12 of the antilinear operator S(x) = x on L2(N, ) , see [KR, Section 9.2.]. We


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6 MARIUS JUNGE

denote by t : N N the modular automorphism group defined by t (x) = itxit.Let us recall the standard notation

x.(y) = (xy) .

For each t there is a natural map It : N N (N the unique predual of N) defined by

It(x) = t(x). .

According to [Ko, Theorem 2.5.], there is a unique extension Iz : N N such that forfixed x the function fx : {z| 1 Im(z) 0} N, fx(z) = Iz(x) is analytic andsatisfies

fx(t)(y) = (y

t (x)) and fx(i + t)(y) = (

t (x)y) .

The density of the algebra of analytic elements shows that for 0 1 the map Ii isinjective. By complex interpolation, the Banach space

Lp(N , , ) = [Ii(N), N] 1p

is defined by specifying x0 =I1i(x)N and x1 = xN . We refer to [Te, Fi] for

further information.

Haagerups abstract Lp space [Ha1, Te] is defined for every von Neumann algebra N using

the crossed product Nw R with respect to the modular automorphism group of a n.s.f.

weight w. (In our applications, we can assume w = for a n.f. state.) IfN acts faithfully

on a Hilbert space H, then the crossed product Nw R is defined as the von Neumann

algebra defined on L2(R, H) and generated by

(x)((t)) = wt((t)) and (s)(t) = (t s) .

Then Nw R, see [PTA], is semifinite and admits a unique trace such that the dual

action

s(x) = W(s)xW(s)

satisfies (s(x)) = es(x). Here W(s) is defined by the phase shift

W(s)(t) = eist(t) .

The dual action satisfies s((x)) = (x) and moreover

(N) = {x Nw R | s(x) = x , for all s R}(1.1)


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Let us agree to identify N with (N) in the following. Lp(N) is defined to be the space

of unbounded, -measurable operators affiliated to Nw R such that for all s Rs(x) = e

sp x .

Note that the intersection Lp(N) Lq(N) is {0} for different values p = q. There is anatural isomorphism between N and L1(N) such that for every normal functional Nthere is a unique a L1(N) associated satisfying

(ax) = (

R

s(x))

for all positive x Nwt R. The key point in this construction is the definition of thetrace function tr : L1(N)

C (corresponding to the integral in the commutative case)

given by

tr(a) = (1) .

Let 1p

+ 1p

= 1 and x Lp(N), y Lp(N). Then we have the trace propertytr(xy) = tr(yx) .

The polar decomposition x = u|x| of x Lp(N) satisfies u N andxp = tr(|x|p)

1p .

N acts as a left and right module on Lp(N) and more generally Holders inequality

xyr xp yq(1.2)holds whenever 1

p+ 1

q= 1

r. As for semifinite von Neumann algebras, there is a positive

cone Lp(N)+ in Lp(N) consisting of elements in Lp(N) which are positive as unbounded

operators affiliated to Nw R. Following [Te, Proposition 33, Theorem 32], we deduce

for 0 x y Lp(N) and 1p + 1p = 1xp = sup

zLp(N)+,zp1

tr(zx) supzLp(N)+,zp1

tr(zy) = yp .(1.3)

In the sequel, we will often use the following simple observation.

Lemma 1.1. Let 0 < p and x, y Lp(N) such that xx yy. If p is the leftsupport projection of y, then xy1p is a well-defined element in N of norm less than one.

Proof: We note that a = xy1p is affiliated with Nw R and

p(y1)xxy1p p(y1)yyy1p p


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8 MARIUS JUNGE

shows that a is a contraction and in particular -measurable. Moreover, we have

p = s(p) = s(yy1p) = e

sp ys(y

1p)

and hence

s(y1p) = e

sp y1p .(1.4)

Therefore, the equality

s(a) = s(x)s(y1p) = e

sp xe

sp y1p = a

shows with (1.1) that a is a contraction in N.

In the -finite case, Kosakis Lp-space is isomorphic to Haagerups Lp space, see [Ko,

section 8]. Indeed, given a n. f. state with corresponding density D in L1(N) = N,then

t (x) = DitxDit

and for all x NIi(x)[Ii(N),N] 1

p

=D p xD 1p

Lp(N).

We recall that an element N is analytic, ift t (x) extends to an analytic function withvalues in N. The -closed subalgebra of analytic elements will be denoted by

A. The

following Lemma is probably well-known, see [Ko], [JX, Lemma 1.1.]. We add a shortproof for the convenience of the reader.

Lemma 1.2. Let0 < p < , then D 12pA+D 12p is dense in Lp(N)+ and N D1p is dense in

Lp(N) and for 1 p the map Jp : Lp(N) L1(N), Jp(x) = xD11p is injective.

Proof: D is a -measurable operator with support projection 1. Therefore, xD11p = 0

implies that x = xD11p D

1p1 is a well-defined -measurable operator and equals 0. Hence,

Jp is injective for 1

p

. Let 1

p+ 1

p= 1. We show the density ofD

12p

A+D

12p in Lp(N)+

for 1 p < . If this is not the case, the Hahn-Banach theorem implies the existenceof x Lp(N)+ and y Lp(N)sa such that tr(yD

12p aD

12p ) = tr(D

12p yD

12p a) 0 for all

a A+ and tr(xy) > 0. Using the -strong density ofA+ in N+, see [PTA], we deduce thaty = D

12p D

12p yD

12p D

12p is negative and hence tr(xy) 0, a contradiction. Let 1

2 p 1

and y Lp(N)+, then we can approximate y 12 by an element x = D14p aD

14p , a A+ and

hence Holders inequality implies that x2 y = x(x y 12 ) + (x y 12 )y 12 approximates y.Since a is analytic, we observe that x2 = D

14p aD

14p D

14p aD

14p = D

12p i

4p(a) i

4p(a)D

12p


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is in D12pA+D 12p . By induction, we deduce the density of D 12pA+D 12p in Lp(N)+ for all

0 < p k

tr(En(xn)Ek(xk))

=nk

tr(Ek(En(xn)xk)) +n>k

tr(En(xnEk(xk)))

=nk

tr(En(xn)xk) +n>k

tr(xnEk(xk))

=k

tr(

nk

En(xn)

xk) +

n

tr(xn

k


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Lemma 3.2. If (DD2) holds with constant c2 and 1 p 2, then for all sequences (xn)and (yn) in L2p(N)

n

En(xnyn)

p

c2(p1)

p

2

n

xnxn

12

p

n

ynyn

12

p

.

Proof: Let us first prove the assertion for finite sequences and p = 2 or p = 1. We start

with (DD1). Using Lemma 2.1, we get

n

En(xnxn)

1

= tr(n

En(xnxn)) =

n

tr(En(xnxn))

=n

tr(xnxn) =

n

xnxn1

.

By the density of elements of the form xn = anD1p , the Cauchy-Schwarz inequality, see

Theorem 2.17, implies with Lemma 3.1 for p {1, 2}n

En(xnyn)

p

n

En(xnxn)

12

p

n

En(ynyn)

12

p

cpn

xnxn

12

p

n

ynyn

12

p

,

(3.1)

where c2 is the constant given in the assumption and c1 = 1. To use interpolation, we

consider finite sequences (xn) and (yn) such thatn

xnxn

p

= 1 =

n

ynyn

p

.

We define X =

n xnxn, Y =

n y

nyn. Their support projections are denoted by qX

and qY and are in N, see [Te, Proposition 4. 2) c), Proposition 12]. Since X1

2 qX, qYY1

2

are well-defined unbounded operators, we can define

vn = xnX 1

2 qX , wn = ynY 1

2 qY

Note that xnxn X and ynyn Y implies xn = xnqX, yn = ynqY, respectively andaccording to Lemma 1.1 we have vn N, wn N. Then, we observe

n

vnvn = qXX1

2 XX12 qX qX 1 ,

n

wnwn = qYY1

2 Y Y12 qY qY 1 .


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Let be determined by 1p

= 11

+ 2

. According to Kosakis interpolation [Ko]

[L2,L(N, ), L4,L(N, )] = L2p,L(N, ) and [L2,R(N, ), L4,R(N, )] = L2p,R(N, )

with respect to the state (x) = tr(Dx). By approximation, we may assume that there are

continuous functions X(z), Y(z) on the strip {0 Re(z) 1} with values in N, analyticin the interior, such that X

12 = D

12p X(), Y

12 = Y()D

12p and

supt

max{D 12 X(it)2, D 14 X(1 + it)4} 1,

supt

max{Y(it)D 122, Y(1 + it)D 144} 1.

Now, we note Kosakis symmetric interpolation result

[L1,sym(N, ), L2,sym(N, )] = Lp,sym(N, ) .

Hence, by (3.1) and Holders inequality we getn

En(X12 vnwnY

12 )

p

=

n

D12p En(X()v

nwnY())D

12p

p

supt

n

D12 En(X(it)v

nwnY(it))D

12

1

1

supt

n

D14 E

n(X(1 + it)v

nw

nY(1 + it))D

14

2

supt

n

D12 X(it)vnvnX(it)

D12

12

1

n

D12 Y(it)wnwnY(it)

D12

12

1

c2 supt

n

D14 X(1 + it)vnvnX(1 + it)

D14

2

1

supt

n

D14 Y(1 + it)wnwnY(1 + it)D

14

2

2

supt

D 12 X(it)X(it)D 12 121

supt

D 12 Y(it)Y(it)D 12 121

supt

D 14 X(1 + it)X(1 + it)D 14 122

c2 supt

D 14 Y(1 + it)Y(1 + it)D 14 122

c2 = c2(p1)

p

2 .

The assertion is proved.


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26 MARIUS JUNGE

Proof of Theorem 0.1 in the case 1 p 2: For 1 p 2 and a sequence ofpositive elements (zn) Lp(N), we can apply Lemma 3.2 to xn = yn = z

12n and deduce

the assertion for the sequence (zn).

The duality argument relies on the following norm for sequences (xn) Lp(N)

(xn)Lp(N;1) = infnj

vnjvnj

12

p

nj

wjnwjn

12

p

.

Here the infimum is taken over all (double indexed) sequences ( vnj) and (wnj) such that

for all n

xn =j

vnjwjn .

We require norm convergence for p < and convergence in the -weak operator topologyfor p = . In fact, we think ofxn being obtained by matrix multiplication of a row with acolumn vector. We denote by Lp(N; 1) the set of all sequences admitting a decompositionas above.

Remark 3.3. This norm is motivated by the following characterization of a normal, de-

composable map T :

N, see [Pa]. Indeed, a normal map is decomposable if and only

if there are sequences (xn) N, (yn) N such that T(en) = ynxn andn

ynyn

N

n

xnxn

N

< .

Lemma 3.4. If (DDp) holds, then

n

En(xnyn)

p cp

n

xnxn12

p

n

ynyn12

p

,

The linear map T : Lp(N; 1) Lp(N; 1), T((xn)) = (En(xn)) satisfies

T cp .

Proof: We have seen in (3.1) that the first inequality is an immediate consequence of

the Cauchy-Schwarz inequality, see Theorem 2.17. As for the second assertion, we can


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assume that N is separable and use the Kasparov maps unp : Lp(N, En) Lp(Nn; C2 )

from Proposition 2.8. Let xn = j vnjwnj , then(En(xn))Lp(N;1) =

(nj

unp(vnj)

unp (wnj))

Lp(N;1)

nj

unp(vnj)

unp(vnj)

12

p

nj

unp (wnj)unp(wnj)

12

p

=

nj

En(vnjvnj)

12

p

nj

En(wnjwnj)

12

p

cpnj

vnjvnj

12

p

nj

wnjwnj

12

p

.

Taking the infimum over all these decompositions, we obtain the assertion.

Let us state some elementary properties of the space Lp(N; 1).

Lemma 3.5. For 1 p the set Lp(N; 1) is a Banach space. For 1 p < , the

set L0

p of elements

xn =j

vnjwjn

such thatcard{(j, n)|vnj = 0 or wjn = 0} < is dense inLp(N; 1). If (xn) is a sequencethen

n

xn

p

(xn)Lp(N;1)

and equality holds if all the xns are positive.

Proof: The proof of the triangle inequality is completely elementary, see also [Ps2], but

essential. Indeed, let > 0 and

xn =j1

vnj1wj1n and yn =j2

vnj2wj2n ,


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28 MARIUS JUNGE

such that

nj1

vnj1vnj1p

=

nj1

wj1nwj1np

(1 + ) (xn)Lp(N;1) ,nj2

vnj2vnj2

p

=

nj2

wj2nwj2n

p

(1 + ) (yn)Lp(N;1) .

We have

xn + yn =j1

vnj1wj1n +j2

vnj2wj2n

and the triangle inequality in Lp(N) impliesnj1

vnj1vnj1

+nj2

vnj2vnj2

p

nj1

vnj1vnj1

p

+

nj2

vnj2vnj2

p

(1 + )((xn)Lp(N;1) + (yn)Lp(N;1)) .

Similarly,

nj1

w

j1nvj1n +nj2

w

j2nwj2np

(1 + )((xn)Lp(N;1) + (yn)Lp(N;1))

and the assertion follows with 0. We consider the spaces of column matrices, rowmatrices, Lp(N;

C2 (N

2)), Lp(N; R2 (N

2)) Lp(B(2(N2))N), respectively. Using,

((xnk) (ynk) ) = (k

xnkynk)nN ,

we deduce that Lp(N; 1) is isomorphic to a quotient space of the projective tensor productLp(N;

R2 (N

2))

Lp(N;

C2 (N

2)), see [DF] for a definition and basic properties of the

projective tensor product. Hence Lp(N; 1) is complete. The image under of pairs offinite sequences generates L0p. According to Corollary 2.4 finite sequences are dense in thecolumn and row spaces and hence L0p is dense in Lp(N; 1). Finally, let (xn) be a sequenceof positive elements. Clearly, xn = x

12nx

12n and hence

(xn)Lp(N;1) n

xn

p

.


29/49


On the other hand if xn =

j vnjwnj, we deduce from Holders inequality

n

xn

p

=

nj

(e1,nj vnj)(enj,1 wnj)p

nj

e1,nj vnj2p

nj

enj,1 wnj2p

=

nj

vnjvnj

12

p

nj

wnjwnj

12

p

.

Taking the infimum yields the assertion.

Inspired by Pisiers vector-valued Lp(N, ; ) space, we define for 0 < p

supn

|xn|p

= (xn)Lp(N;) = inf xn=aynb a2p b2p supn ynN ,

where the infimum is taken over all a, b L2p(N) and all bounded sequences (yk). If Nis a hyperfinite, finite von Neumann algebra, this space coincides with Lp(N, ; ) in thesense of Pisier [Ps2]. The first (formal) notation is suggestive and facilitates understand-

ing our inequalities in view of the commutative theory. For positive elements, we will

drop the absolute value. Let us note that Haagerups work [Ha2] shows that the equal-

ity L1(N; ) = L1(N) (operator space projective tensor product) only holds forinjective von Neumann algebras. However, this does not affect the following factorization

result which is, nowadays, a standard application of the Grothendieck-Pietsch version of

the Hahn-Banach theorem, see [Ps1, Ps2].

Proposition 3.6. Let1 p < . If p = 1, thenL1(N; 1) = 1(L1(N)) holds with equalnorms. If 1 < p , p < satisfy 1

p+ 1

p= 1, then

Lp(N; 1) = Lp(N; )

holds isometrically.


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30 MARIUS JUNGE

Proof: If zn = aynb and (yn) is a bounded sequence, we deduce from Holders inequality

for all (vnj) L2p(N), (wjn) L2p(N)n

tr(znj

vnjwjn)

=nj

tr(aynbvnjwjn)

=nj

tr(ynbvnjwjna)

sup

n

ynnj

bvnjwjna1

supn

yn

nj

bvnj22 1

2

nj

wjna22 1

2

= supn yn bnj

vnjvjnb

12

1

anj

wnjwjna

12

1

supn

yn b2pnj

vnjvnj

12

p

a2pnj

wjnwjn

12

p

.

Hence, Lp(N; ) Lp(N; 1). Using 1(L1(N)) L1(N; 1), we deduce the equality1(L1(N)) = L1(N; 1). Now we show that for 1 < p < all the functionals are inLp(N; ). Let : Lp(N; 1) C be a norm one functional. Using 1(Lp(N))

Lp(N; 1), we can assume that there exists a sequence (zn)

Lp(N) such that

[(xn)] = (zn)[(xn)] =n

tr(znxn) .

Let us denote by B = B+Lp(N)

the positive part of the unit ball in Lp(N). B is compact

when equipped with the (Lp(N), Lp(N))-topology. The definition of Lp(N; 1) implieswith the geometric/arithmetric mean inequality

nj

tr(znvnjwjn)

=

[(

j

vnjwjn)n]

nj

vnjvnj

12

p

nj

wjnwjn

12

p

12

supc,dB

[nj

tr(vnjvnjc) +

nj

tr(wjnwjnd)] .

Since the right hand side remains unchanged under multiplication with signs nj, we deduce

nj

|tr(znvnjwjn)| 12

supc,dB

[nj

tr(vnjvnjc) +

nj

tr(wjnwjnd)] .


31/49


Following the Grothendieck-Pietsch separation argument as in [Ps1], we observe that the

C given by the functions

fv,w(c, d) =n

[tr(vnvnc) + tr(wnwnd) 2|tr(znvnwn)|]

is disjoint from the cone C = {g| sup g < 0}. Here v = (vn) and w = (wn) are finitesequences and hence fv,w is continuous with respect to the product topology on B B.Since fv,w + fv,w can be obtained by taking the (vn, wn)s to the right of the finite sequence

(vn, wn), we deduce that C is a cone. Hence, there exist a measure on B B and ascalar t such that for all g C and f C

BB

g d < t

BB

f d .

Since we are dealing with cones, it turns out that t = 0 and is positive. Therefore, we

can and will assume that is a probability measure. We define the positive elements c

and d by their projections

a =

BB

c d(c, d) , b =

BB

d d(c, d) .

By convexity of B, we deduce a, b B. Hence, we obtain

n

2 |tr(znvnwn)| BB

n

[tr(vnvnc) + tr(wnv

nd)] d(c, d)

=n

BB

tr(vnvnc) d(c, d) +

BB

tr(wnwnd) d(c, d)

=n

[tr(vnvna) + tr(w

nwnb)] .

Using once more 2st = infr>0(rs)2 + (r1t)2, we get

n

|tr(znvnwn)|

n

tr(vnvna)

12

n

tr(wnwnb)

12

=

n

a 12 vn22

12

n

b12wn22

12

.

(3.2)

Let qa, qb N be the support projections of a, b, respectively. Consider, da = ap + (1 qa)D(1 qa), D the density of . Then da(x) = tr(dax) is a normal, faithful state on Nand according to Lemma 1.2 d

12a N is dense in L2(N). Hence,

qad12a N = a

p

2 N = a12 a

p

2p N a 12 L2p(N)


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32 MARIUS JUNGE

shows that a12 L2p(N) is dense in qaL2(N). Similarly, b

12 L2p(N) is dense in qbL2(N) and

therefore (3.2) implies that for every n

N there is a contraction Tn : qaL2(N)

qbL2(N)

such that for all v, w L2p(N)tr(wznv) = (b

12 w, Tn(a

12 v)) = tr(wb

12 Tn(a

12 v)) .

This means Tn is a bounded extension of the densely defined hermitian form

(b12 qb(h

), zna1

2 qa(h) ) = (b1

2 qbh, zna

12 qah)

Using the density of a12 L2p(N) and b

12L2p(N) it is easily checked that qbTnqa is affiliated

with N. Since Tn is bounded, we deduce qbTnqa N. On the other hand, we have forv L2p(N) and w L2p(N)

|tr(zn(1 qa)vw)| tr(a(1 qa)vv) 12 tr(wwb) 12 = 0and

|tr(znvw(1 qb)| tr(avv) 12 tr(ww(1 qb)b) 12 = 0 .This shows zn = qbznqa and therefore

zn = qbznqa = qbb12 qbb

12 zna

12 qaa

12 qa = b

12 yna

12 .

The assertion is proved because L0p is dense in Lp(N; 1) and hence the functional isuniquely determined by the sequence (zn).

Remark 3.7. Let1 p < and (zn) Lp(N) a sequence of positive elements, thensupn

zn

p

= sup

n

tr(znxn)

xn 0 ,

n

xn

p

1 .

Moreover, there exists a positive element a L2p(N) and a sequence of positive elementsyn such that

zn

= ayn

a and

a2

2p sup

n yn

= supn

znp

.

For positive elements (xn) Lp(N)+, we also haven

xn

p

= sup

n

tr(xnzn)

zn 0 , (zn)Lp(N;) 1

and therefore the cones of positive sequences in Lp(N; 1) respectively Lp(N; ) are induality.


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Proof: For positive elements (zn) satisfying

n

tr(znxn)

nxnp

,

for all sequences of positive elements (xn) Lp(N)+, we deducej,n

|tr(znvnjvnj)| n

tr(zn(j

vnjvnj))

nj

vnjvnj

p

= supcB

j,n

tr(vnjvnjc) .

Using the Hahn-Banach separation argument in the space of continuous functions on B,

we obtain a positive element a in the unit ball of Lp(N) such thatn

|tr(znvnvn)| n

tr(vnvna) .

Since a12 zna

12 is positive, this inequality still ensures that all the yn = a

12 zna

12 are

positive contractions in N. The last equality follows immediately from Lemma 3.5 and

the duality between the positive parts of Lp(N) and Lp(N).

Remark 3.8. Proposition3.6and Remark3.7 can easily be modified for uncountable (or-

dered) index sets by requiring the inequality for all countable (ordered) subsets or for an

essential supremum. This is helpful in the context of continuous filtrations.

The required duality argument is now very simple.

Lemma 3.9. Let1 < p and 1p

+ 1p

= 1 assume that (DDp) holds with constant cp,

then for every y Lp(N)

supn

|En(y)|p

cp yp .

Moreover, for every sequence of positive elements (yn)supn |jn

En(yj)|p

cpn

yn

p

.

Proof: Indeed, as observed in Lemma 3.4 and using Lemma 3.5, we deduce that (DDp)

implies that the linear map T : Lp(N; 1) Lp(N), T((xn)) =

n En(xn) satisfies


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34 MARIUS JUNGE

T cp. By duality and Proposition 3.6, we deduce for all y Lp(N)

supn |En(y)|p = T(y)Lp(N;1) T yp cp yp .Given a sequence of positive elements (yn), we consider y =

j yj and obtainsup

n

|En(y)|p

cp yp = cpj

yj

p

.

However, for positive elements (xn) Lp(N), we deduce by positivity

n tr(En(jn yj)xn) n j tr(En(yj)xn) (En(y))nNLp(N;)n xn

p

.

Hence, Remark 3.7 impliessupnjn

En(yj)

p

sup

n

En(y))nN

p

cp yp = cpj

yj

p

.


Theorem 3.10. For 1 < p there exists a constant cp such that for every sequence(Nn) of von Neumann subalgebras with sequence of -invariant conditional expectations

(En) satisfying EnEm = Emin(n,m) and for every x Lp(N) there exist a, b L2p(N) anda bounded sequence (yn) N such that

En(x) = aynb and a2p b2p supn

yn cp xp .

If x is positive, one can in addition assume that b = a and all the yns are positive.

Proof for 2 p : This follows immediately from Lemma 3.9, Lemma 3.2 and Lemma3.1. Using that En(x) is positive for positive x, the addition follows from Remark 3.7.

4. The dual version of Doobs inequality for 2 p <

In our approach to (DDp) in the range 2 p < , our aim is to obtain the same kindof inequalities for the maximal function as in Garsias book [Ga]. As mentioned in the

introduction, we are forced to use more duality arguments because 0 a b impliesa b only for 0 1 and therefore most of the elementary proofs in Garsias bookare no longer valid in the non-commutative case. We will make the same assumptions


35/49


36/49

36 MARIUS JUNGE

The following proposition is a modification of the Theorem in the appendix of Pisier, Xus

paper [PX] and enables us to apply duality.

Proposition 4.2. Let 1 r < 2 < r , and 1r

+ 1r

= 1, then for all (xj) Lr(N)and (yn) Lr(N, (En); C2 )

n

tr(ynxn)

2 (yn)Lr(N,(En);C2 )j

xjxj

12

r2

.

Moreover,

n

tr(ynxn)

2 (yn)Lr(N,(En);C2 ) supn jn

En(xjxj)

12

r2

.

Proof: Let us assume that both sequences are finite, i.e. xj = 0 = yj for j m. Bydensity, we can moreover assume that yj = ajD

1r with aj N and

mj=1

D1r Ej(a

jaj)D

1r

r

2

< 1 .

By continuity, we can assume that there is an > 0 such that

D 2r +m

j=1

D1r Ej(a

jaj)D

1r

r

2

1 .

Let 1 q such that 1q

+ 2r

= 1. For n N, we define

Sn =

D

2r +

nj=1

D1r Ej(a

jaj)D

1r

r2q

Lq(Nn) Lq(N) .

The support projection of Sn is 1 and D1q Sn. According to Lemma 1.1, we deduce

that wn := D12q S

12

n Nn. HenceynS

12

n = anD1r S

12

n = anD12 D

12q S

12

n = anD12 wn

L2(N) .

In particular, S 1

2n yn L2(N) and

S 1

2n y

nynS

12

n = wnD

12 ananD

12 wn L1(N) .

Moreover,

En(S 1

2n y

nynS

12

n ) = En(wnD

12 ananD

12 wn) = w

nD

12 En(a

nan)D

12 wn

= S1

2n D

1r En(a

nan)D

1r S

12

n .


37/49


This implies with the Cauchy-Schwarz inequality

n

tr(ynxn) =

ntr(xnyn)

= n

tr((xnS

1

2n )(S1

2n yn))

=

n

tr(S1

2n y

nxnS

12n )

n

tr(S1

2n y

nynS

12

n )

12

n

tr(xnxnSn)

12

=

n

tr(En(S 1

2n y

nynS

12

n ))

12n

tr(xnxnSn)

12

=

n

tr(S1

2n D

1r En(a

nan)D

1r S

12

n )

12

n

tr(En(xnxn)Sn)

12

.

To estimate the first term, we define = 2r

[1, 2] and notice that

1 = 1 2r

= 1 2 + 2r

= 1q

.

For fixed n, we define x = Sqn1 and z = Sqn. Since

r

2 1, we have

x =

D2r +

n1j=1

D1r Ej(a

jaj)D

1r

r2

D2r +

nj=1

D1r Ej(a

jaj)D

1r

r2= z .

Then, we note that z12 = z

12q = S

12

n . Hence Lemma 4.1 implies

tr(S 1

2n D

1r En(a

nan)D

1r S

12

n ) = tr(z12 (z x)z12 ) 2tr(z x)

= 2tr(Sqn Sqn1) .Therefore, we obtain

n

tr(S

12

n D

1r

En(a

nan)D

1r

S

12

n ) =n

2tr(Sq

n Sq

n1) = 2tr(Sq

m)

= 2

D 2r +m

j=1

D2r Ej(a

jaj)D

2r

r

2

r

2

2 .

Now, we want to estimate the second term. Let us define

j = Sj Sj1 .


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38 MARIUS JUNGE

and note that j Lq(Nj). As usual, we set S1 = 0. Moreover, r 2q implies that j ispositive. Then, we deduce with EjEn = Emin(j,n)

n

tr(En(xnxn)Sn) =

jn

tr(En(xnxn)j)

=j

tr(nj

En(xnxn)j)

=j

tr(Ej

nj

En(xnxn)

j)

=j

tr(nj

Ej(xnxn)j)

Now, we can continue in two different ways

j

tr(nj

Ej(xnxn)j) supj Ej(nj

xnxn)r2

j

jq

=

supj Ej(nj

xnxn)

r2

Smq

supj Ej(

nj

xnxn)

r2

.

By homogeneity, we obtain the second assertion

n

tr(ynxn)

2

supj Ej(nj

xnxn)

12

r2

(yn)Lr(N,(En);C2 ) .


39/49


Using for all j N that Ej(j) = j , we also get by positivity

j

tr(nj

Ej(xnxn)j) =

j

tr(nj

xnxnEj(j))

=j

tr(nj

xnxnj)

tr(n

xnxnj

j)

n

xnxn

r2

j

j

q

n x

n

xnr2

Sm

q n x

n

xnr2

.

Again by homogeneity, we deducen

tr(ynxn)

2

n

xnxn

12

r2

(yn)Lr(N,(En);C2 ) .

Remark 4.3. For r = r = 2 the assertion is trivially true because L2(N, (En); C2 ) =

2(L2(N)) and

n

xn2

2

= n

x

n

xn1

= tr(n

x

n

xn) = tr(E1(n

x

n

xn))

=

E1(n

xnxn)

1

supj Ej(

nj

xnxn)

1

.

Proposition 4.2 also shows that the BM OC and HC1 duality from [PX] is still valid in the

non tracial case.

Corollary 4.4. Let 1 q < and 2q the constant in Steins inequality, then for all(xn)

L2q(N)

n

En(xnxn)

q

222qsupn En(

jn

xjxj)

q

.

Proof: We define r = 2q and noten

En(xnxn)

12

r

= (xn)Lp(N,(En);C2 ) .


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40 MARIUS JUNGE

Therefore the assertion follows from Proposition 4.2 and Theorem 2.17.

Proof of (DDp) for 1 < p < : We define r = 2p > 2 Let (zn) Lp(N) be a sequenceof positive elements and define xn = z

12n . Hence by Theorem 2.17 and Proposition 4.2, we

deduce

n

En(zn)

12

p

=

n

En(xnxn)

12

p

r sup(yn)L

r(N,(En);C2 )

1

n

tr(ynxn)

r

2

n

xnxn

12

p

.

The assertion follows with cp 222p.

Proof of Theorem 0.2 and 3.10 for 1 < p 2: This follows from (DDp) via Lemma3.9 and Remark 3.7.

Remark 4.5. Let N be separable and be a functional on Lp(N, (En); C2 ), then there

exists a sequence (xn) such that

((yn)) =n

tr(xnyn) and

supn En(jn

xjxj)

12

p2

d p2Lp (N,(En);C2 ) .

Here d p2

is the constant in Doobs inequality from Theorem 0.2. The assertion yields anextension of the BM OC-H

C1 duality for 2 < p < and fails for p = 2. The proof uses

the Kasparov isomorphism from Proposition 2.15. We leave it to the interested reader.

Answering a question by G. Pisier, we can even produce an asymmetric version of Doobs

inequality. Indeed, let 1 < p and consider 1 r,s < such that 2p

= 1s

+ 1t.

Given x Lp(N) and sequences (vnj) Lr(N), (wjn) Ls(N), we deduce from the


41/49


Cauchy-Schwartz inequality 2.13, (DDs) and (DDt) that

n,j

tr(En(x)vnjwjn) =

n,j

tr(xEn(vnjwjn))

xpnj

En(vnjwjn)

p

xpnj

En(vnjvnj)

12

s

nj

En(wjnwjn)

12

t

c12s c

12t

x

p

nj

vnjvnj

12

s

nj

wjnwjn12

t

.

Similar as in Proposition 3.6, we deduce the existence of bounded a sequence (zn) and

elements a L2s(N), b L2t(N) such that En(x) = bzna and

a2s b2t supn

zn c12s c

12t xp .

Note that 12t

+ 12s

= 1p

and therefore we have proved the following asymmetric version of

Theorem 0.2. (The assumption 2 < q, r is indeed necessary [DJ1].)

Corollary 4.6. Let 1 < p and 2 < q , r such that1

q +1

r =1

p . Then forevery x Lp(N) there exists a sequence (zn) N and a Lq(N), b Lr(N) such thatEn(x) = aznb and

aq br supn

zn c(p, q, r) xp .

5. Applications

In this section, we present first applications of Doobs inequality in terms of submartin-

gales, Doob decomposition. We make the same assumptions about N, (Nn), (En), and

D as in the previous section and start with almost immediate consequences of the dualversion of Doobs inequality.

Corollary 5.1. Let1 < p and (zn) be an adapted sequence of positive elements, i.e.zn Lp(Nn)+. If for all n N

zn En(zn+1) and supm

zmp < ,


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42 MARIUS JUNGE

then there exist a positive element a L2p(N)+ and a sequence of positive contractions(yn) N such that

zn = ayna and a22p cp supm

zmp .

Proof: It suffices to consider p < . We note that for n mzn En(zn+1) En(En+1(zn+2)) = En(zn+2) En(zm) .

Let 1p

+ 1p

= 1. In order to estimate the norm in Lp(N; ), we refer to Remark 3.7. Givena finite sequence (xn) of positive elements such that xn = 0 for n m, we deduce from(DDp)

n

tr(znxn) =

n

tr(En(znxn)) =n

tr(En(zn)En(xn))

n

tr(En(zm)En(xn)) =n

tr(En(zmEn(xn)))

=n

tr(zmEn(xn)) tr(zmn

En(xn))

zmp

n

En(xn)

p

cp zmp

n

xn

p

Hence, the assertion follows from Remark 3.7.

Corollary 5.2. Let2 < p and (zn) be an adapted seqeunce, i.e. zn Lp(Nn). If forall n N

znzn En(zn+1zn+1) ,then there exist a positive element a Lp(N) and a sequence (yn) of contractions suchthat

zn = yna and ap c12p

p2supm

zmp .

Proof: We apply Corollary 5.1 to zn = znzn L p2 (Nn) and obtain positive contractions

(vn) N and a Lp(N) such thatznzn = avna and a2p c pp2 sup

m

zmzm p2

.

If qa is the support projection of a, we see that yn = v12n a

1qa satisfies the assertion.


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Let us mention an immediate application of (DDp) in terms of the Doob decomposition of

the square function. Given a martingale sequence x = k dk(x), where dk(x) = Ek(x) Ek1(x) (and E0(x) = 0), we recall that one of the square functions is given by

sc(x) =k

dk(x)dk(x)

The square function is the discrete analogue of the quadratic variation term see [BS, Ko]

for more details. The Doob decomposition of sc(x) is given by the martingale part

Vn(x) =n

k=1

dk(x)dk(x) Ek1(dk(x)dk(x))

and the predictable part

Wn(x) =n

k=2

Ek1(dk(x)dk(x)) Lp(Mk1) .

Note that

Vn(x) + Wn(x) =n

k=1

dk(x)dk(x) .

Corollary 5.3. Let2 < p < and x Lp(N), then

supn

max{

Wn

(x)

12p

2

,

Vn

(x)

12p

2} p

(1 + c p2

)12

xp

.

Here p is an absolute constant. In particular, there exists an element V(x) in L p2

(N)

such that Vn(x) = En(V(x)).

Proof: Using (DD p2

) for the sequence (Ek1) and the non-commutative Burkholder-

Gundy inequality [PX, JX], we deduce

Wn(x) p2

c p2

nk=2

dk(x)dk(x)

p2

c p2

2p x2p .

Hence the triangle implies

Vn(x) p2

Wn(x) p2

+

n

k=1

dk(x)dk(x)

p2

(c p2

2p + 1) x2p .

By uniform convexity of Lp2

(N), we obtain the limit value V(x) = limn Vn(x) with the

desired properties.


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44 MARIUS JUNGE

The next application yields norm estimates for

npnEn(x)qn with respect to a sequence

(pn), (qn) of disjoint projections. This corresponds to a double sided non-adapted stopping

time.

Corollary 5.4. Let1 < p , (vn), (wn) be sequences of bounded elements, then for allx Lp(N)

n

vnEn(x)wn

p

cp xp maxn

vnvn

12

,

n

vnvn

12

maxn

wnwn

12

,n

wnwn12

.

Proof: The case p = is obvious. Hence, we assume 1 < p < . Let y Lp(N) andchoose y1 L2p(N), y2 L2p(N) such that y = y1y2 and

yp = y122p = y222p .

Then, according to Lemma 3.4, we deduce from (DDp):

tr(n

vnEn(x)wny) = n

tr(xEn(wnyvn)) xp n En(wny1y2vn)p xp cp

n

wny1y1w

n

12

p

n

vny2y2wn

12

p

.

To conclude, we use [PX, Lemma 1.1], see also [JX],

n

wny1y1w

n

p

y1y1p max

n

wnwn

,

n

wnwn

= yp maxn

wnwn

,

nwnwn

.

A similar argument applies for the other term and therefore taking the supremum over all

y of norm 1 implies the assertion.

Remark 5.5. Let1 p < . The non-commutative Doob inequality from Theorem 0.2implies the vector-valued Doob inequality in Lp(, , ; Lp(N)).


45/49


Proof: It suffices to treat the discrete case. Let N = L(, , )N and (n)nN bean increasing sequence of -subalgebras with conditional expectations (En). Let Nn =L(, , )N. Then the conditional expectation En onto Nn is given by En = En id. Let f Lp(, , ; Lp(N)) = Lp(N). According to Theorem 0.2 there exist a L2p(N), b L2p(N) and contractions (zn) Nsuch that

En idLp(N)(f) = aznb and a2p b2p cp fp .Hence, for every and n N

En(f)()Lp(N) = a()zn()b()Lp(N) a()L2p(N) z()Nb()L2p(N) a()L2p(N) b()L2p(N) .

Holders inequality implies the assertion

supn

En(f)()pLp(N) d()

1p

a()pL2p(N) b()p

L2p(N)d()

1p

aL2p(,,;L2p(N) bL2p(,,;L2p(N)= a2p b2p cp fp .

In the next application we want to relate group actions with (DDp). To illustrate this, we

consider a finite von Neumann algebra N and an increasing sequence (An) N of finitedimensional subalgebras with 1N An. Let Nn = An be the relative commutant of An inN. If Gn denotes the unitary group ofAn, we have a natural action : Gn B(Lp(N))

n(u)(x) = uxu

such that the conditional expectation on the commutant Nn is given by

En(x) = ENn(x) =

Gn

uxudn(u) .

Let G =

n Gn and the product measure, then

n

En(xn)

p

G

n

n(un)(xn)

p

p

d(u1, u2, ..)

1

p

.

We will show that for a sequence (xn) of positive elements even the right hand side can be

estimated by n xnp. For simplicity let us use the random variables n(x) : G Lp(N),given for = (u1, u2...) by

n(x)() = unxun .


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46 MARIUS JUNGE

For the special case of tensor products of finite dimensional von Neumann algebras the

following theorem implies (DDp).

Theorem 5.6. Let1 < p < and N, (An), (Nn) be as above and (xn) be a sequence ofpositive elements, then

G

n

n(xn)

p

p

d

1p

pn

xn

p

.

Here p is a constant which only depends on p.

Proof: The assertion is obvious for p = 1 and by interpolation as in Lemma 3.2 it suffices

to prove the assertion for p 4. Let (xn) be a finite sequence of positive elements. Then,we observe

G

n

n(xn)

p

p

d

1p

n

En(xn)

p

+

G

n

n(xn) En(xn)p

p

d

1p

cpn

xn

p

+

G

n

n(xn) En(xn)p

p

d

1p

.

Let n be the -algebra generated by the first n coordinates in G =

k Gk and Nn =L(G, n, )N L(G, , )N with the corresponding conditional expectation En.Then, we note

En1(n(xn)) = knGk

unxnun dn(un)dn+1(un+1) = En(xn) .

Hence the n-th martingale difference of

n n(xn) satisfies

dn = En(k

k(xk)) En1(k

k(xk)) = n(xn) En(xn) .


47/49


We apply the non-commutative Rosenthal inequality, see [JX], in this case and obtain

1rp

IE n

n(xn) En(xn)p

p

1p

n

n(xn) En(xn)pp 1

p

+

n

En1(dndn + dndn)

12

p2

2

n

xnpp 1

p

+ 212

n

En(xnxn) + En(x

nxn)

12

p2

2n

xnpp1p

+ 21

2 c

1

2p2

n

xnxn + xnxn

12

p2

Let (n) be a sequence of independent Rademacher variables. Using the triangle inequality

and the orthogonality of the (n)s, we deduce as in [LP]

max

n

xnxn

12

p2

,

n

xnxn

12

p2

IE

n

nxn

2

p

12

.

By interpolation, we have n

xnpp

1p

n

xnxn

12

p2

.

However, since the xn are positive, we deduce for any choice of signs n with positivityn

nxn

p

n=1

xn

p

+

n=1

xn

p

2n

xn

p

.

Hence, we obtain

IE n

n(xn) En(xn)p

p

1p

rp(2 + 4c12p2

)

n

xn

p

.


Remark 5.7. These methods can also be used to show that for every f Lp(G; Lp(N))there exist a, b L2p(; L2p(N)) and a sequence of contractions (yn) L()N such


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48 MARIUS JUNGE

that

Fn(f) = aynb and

a

2p

b

p

c2p

f

p

where

Fn(f)(g1, g2,...) = n(gn)f(g1,...,gn, ..) .

Note that the crossed product N(n)

Gn acts on Lp(; Lp(N)) and Fn somehow removes

the action of Gn on f. Similar results hold for a von Neumann algebra with a faithful

normal state and -invariant, strongly continuous group actions n : Gn Aut(N) ofcompact groups such that the centralizer algebras are increasing or decreasing.

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2000 Mathematics Subject Classification: 46L53, 46L52, 47L25.

Key-words: Non-commutative Lp-spaces, Doobs inequality

Department of Mathematics, University of Illinois at Urbana-Champaign, Urbana, IL

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Martingales Junge