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Lower estimates of random unconditional constants of

arXiv:math/9202203v1 [math.FA] 28 Feb 1992Lower estimates of random unconditional constants ofWalsh-Paley martingales with values in Banach spacesStefan Geiss ∗Mathematisches Institut der Friedrich Schiller Universit¨at JenaUHH 17.OG, D O-6900 Jena, GermanyOctober 17, 2018Abstract. For a Banach space X we define RUMDn(X) to be the infimum of all c > 0 such that AVεk=±1∥nX1εk(Mk −Mk−1)∥2LX2!1/2≤c∥Mn∥LX2holds for all Walsh-Paley martingales {Mk}n0 ⊂LX2 with M0 = 0.

We relate the asymptotic behaviourof the sequence {RUMDn(X)}∞n=1 to geomertrical properties of the Banach space X such as K-convexityand superreflexivity.1980 Mathematics Subject Classification (1985 Revision): 46B10, 46B200IntroductionA Banach space X is said to be an UMD-space if for all 1 < p < ∞there is a constant cp = cp(X) > 0such thatsupεk=±1∥nX1εk(Mk −Mk−1)∥LXp ≤cp∥nX1(Mk −Mk−1)∥LXpfor all n = 1, 2, ... and all martingales {Mk}n0 ⊂LXp with values in X. It turns out that this definition isequivalent to the modified one if we replace ”for all 1 < p < ∞” by ”for some 1 < p < ∞”, and ”for allmartingales” by ”for all Walsh-Paley-martingales” (see [3] for a survey).

Motivated by these definitions weinvestigate Banach spaces X by means of the sequences {RUMDn(X)}∞n=1 whereas RUMDn(X) := inf csuch that AVεk=±1∥nX1εk(Mk −Mk−1)∥2LX2!1/2≤c∥Mn∥LX2holds for all Walsh-Paley martingales {Mk}n0 ⊂LX2 with the starting point M0 = 0. ”RUMD” standsfor ”random unconditional constants of martingale differences”.

We consider ”random” unconditionalconstants instead of the usual one, where supε=±1 is taken in place of AVε=±1, since they naturallyappear in the lower estimates we are interested in. These lower estimates are of course lower estimates∗The author is supported by the DFG (Ko 962/3-1).1

for the non-random case, too. The paper is organized in the following way.

Using a technique of Maureywe show that the exponent 2 in the definition of RUMDn(X) can be replaced by any 1 < p < ∞(seeTheorem 2.4). Then we observe (see Theorem 3.5)Xis not K-convex⇐⇒RUMDn(x) ≍n.In the case of superreflexive Banach spaces this turns intoXis not superreflexive=⇒RUMDn(X) ⪰n1/2,and, under the assumption X is of type 2,Xis not superreflexive⇐⇒RUMDn(x) ≍n1/2(see Theorems 4.3 and 4.4).

Using an example due to Bourgain we see that the type 2 condition isnecessary. In fact, for all 1 < p < 2 < q < ∞there is a superreflexive Banach space X of type p andcotype q such that RUMDn(X) ≍n1p −1q (see Corollary 5.4).

According to a result of James a non-superreflexive Banach spaces X is characterized by the existence of large ”James-trees” in the unit ballBX of X. We can identify these trees with Walsh-Paley martingales {Mk}n0 which only take values in theunit ball BX and which satisfy infω ∥Mk(ω) −Mk−1(ω)∥≥θ for some fixed 0 < θ < 1.

In this way wecan additionally show that the martingales, which give the lower estimates of our random unconditionalconstants, are even James trees (see Theorems 3.5(2) and 4.3).1PreliminariesThe standard notation of the Banach space theory is used (cf.[10]). Throughout this paper IK standsfor the real or complex scalars.

BX is the closed unit ball of the Banach space X, L(X, Y ) is the spaceof all linear and continuous operators from a Banach space X into a Banach space Y equipped withthe usual operator norm. We consider martingales over the probability space [Ωn, µn] which is givenby Ωn := {ω = (ω1, ..., ωn) ∈{−1, 1}n} and µn(ω) :=12n for all ω ∈Ωn.The minimal σ-algebrasFk, such that the coordinate functioinals ω = (ω1, ..., ωn) →ωi ∈IK are measurable for i = 1, ..., k,and F0 := {∅, Ωn} form a natural filtration {Fk} on Ωn.

A martingale {Mk}n0 with values in a Banachspace X over [Ωn, µn] with respect to this filtration {F}n0 is called Walsh-Paley martingale. As usualwe put dM0 := M0, dMk := Mk −Mk−1 for k ≥1 and M ∗k(ω) = sup0≤l≤k ∥Ml(ω)∥.

Given a functionM ∈LXp (Ωn) we can set Mk := IE(M|Fk) for k = 0, ..., n. Consequently, for each M ∈LXp (Ωn) there isa unique Walsh-Paley martingale {Mk}n0 with Mn = M. In this paper we consider a further probabilityspace [IDn, Pn] with IDn = {ε = (ε1, ..., εn) ∈{−1, 1}n} and Pn(ε) =12n for all ε ∈IDn. IEε,ω meansthat we take the expectation with respect to the product measure Pn × µn.

To estimate the randomunconditional constants of Walsh-Paley martingales from above we use the notion of the type.For1 ≤p ≤2 an operator T ∈L(X, Y ) is of type p if IEε∥XkT εkxk∥2!1/2≤c Xk∥xk∥p!1/pfor some constant c > 0 and all finite sequences {xk} ⊂X. The infimum of all possible constants c > 0is denoted by Tp(T ).

Considering the above inequality for sequences {xk}nk=1 ⊂X of a fixed length nonly we obtain the corresponding constant T np (T ) which can be defined for each operator T ∈L(X, Y ).In the case T = IX is the identity of a Banach space X we write Tp(X) and T np (X) instead of Tp(IX)and T np (IX), and say ”X is of type p” in place of ”IX is of type p” (see [17] for more information).2

2Basic definitionLet T ∈L(X, Y ) and 1 ≤q < ∞. Then RUMDqn(T ) = inf c, where the infimum is taken over all c > 0such that IEε,ω∥nX1εkT dMk(ω)∥q!1/q≤c (IEω∥Mn(ω)∥q)1/qholds for all Walsh-Paley martingales {Mk}n0 with values in X and M0 = 0.Especially, we setRUMDqn(X) := RUMDqn(IX) for a Banach space X with the identity IX.

It is clear that RUMDqn(T ) ≤2n∥T ∥since∥nX1εkT dMk∥LXq ≤nX1∥T ∥∥dMk∥LXq ≤∥T ∥nX1∥Mk∥LXq + ∥Mk−1∥LXq≤2n∥T ∥∥Mn∥LXq .In the case X is an UMD-space we have supn RUMDqn(X) < ∞whenever 1 < q < ∞(the converse seemsto be open). q = 1 yields a”singularity” since RUMD1n(X) ≍RUMD1n(IK) ≍n for any Banach space X(see Corollary 5.2) therefore we restrict our consideration on 1 < q < ∞.

Here we show that the quantitiesRUMDqn(T ) are equivalent for 1 < q < ∞. In [5](Thm.4.1) it is stated that supn RUMDqn(X) < ∞iffsupn RUMDrn(X) < ∞for all 1 < q, r < ∞.

Using Lemma 2.2, which slightly extends [11](Thm.II.1), weprove a more precise result in Theorem 2.4.Let us start with a general martingale transform. Assuming T1, ..., Tn ∈L(X, Y ) we defineφ = φ(T1, ..., Tn) : LX0 (Ωn) −→LY0 (Ωn)byφ(M)(ω) :=nX1TkdMk(ω),where Mk = IE(M|Fk).

The following duality is standard.Lemma 2.1 Let 1 < p < ∞, T1, ..., Tn ∈L(X, Y ) and φ = φ(T1, ..., Tn) : LXp −→LYp . Thenφ′(F) =nX1T ′kdFkfor allF ∈LY ′p′ .Proof.

Using the known formula< M, M ′ >=nX0< dMk, dM ′k >M ∈LZs (Ωn), M ′ ∈LZ′s′ (Ωn), 1 < s < ∞,for M ∈LXp (Ωn) and F ∈LY ′p′ (Ωn) we obtain< M, φ′F >==nX1< TkdMk, dFk >=nX1< dMk, T ′kdFk >=< M,nX1T ′kdFk > .✷3

Now we recall [11](Thm.II.1) in a more general form. Although the proof is the same we repeat some ofthe details for the convenience of the reader.Lemma 2.2 Let 1 < p < r < ∞, T1, ..., Tn ∈L(X, Y ) and φ = φ(T1, ..., Tn).

Then∥φ : LXp →LYp ∥≤6r2(p −1)(r −1)∥φ : LXr →LYr ∥.Proof. We define 1 < q < ∞and 0 < α < 1 with 1p = 1r + 1q and pr = 1−α and obtain αq = (1−α)r = p.Let {Mk}n0 be a Walsh-Paley martingale in X with dM1 ̸= 0.

We set∗Mk(ω) := M ∗k−1(ω) + sup0≤l≤k∥dMl(ω)∥fork ≥1and obtain an Fωk−1 -measurable random variable with0 < ∗M1(ω) ≤... ≤∗Mn(ω)andM ∗k(ω) ≤∗M k(ω) ≤3M ∗k(ω).Using [11](L.II.B) and Doob’s inequality we obtain∥φ(M)∥p= IEω∥nX1TkdMk(ω)∗M αk (ω)∗M αk(ω)∥p!1/p≤2 IEω sup1≤k≤n∥kX1TldMl(ω)∗M αl (ω) ∥p∗M αpn (ω)!1/p≤2 IEω sup1≤k≤n∥kX1TldMl(ω)∗M αl (ω) ∥r!1/r(IEω∗M αqn (ω))1/q≤2rr −1 IEω∥nX1TkdMk(ω)∗M αk (ω) ∥r!1/r(IEω∗M pn(ω))1/q≤2rr −1∥φ∥r IEω∥nX1dMk(ω)∗M αk (ω)∥r!1/r(IEω∗M pn(ω))1/q .Applying [11](L.II.A) in the situation ∥Pl1 dMi(ω)∥≤(∗Ml(ω)α)1/α yields∥kX1dMl(ω)∗M αl (ω)∥≤1/α1/α −1∗Mk(ω)α(1/α−1) ≤rp∗Mn(ω)p/rsuch that∥φ(M)∥p≤2r2(r −1)p∥φ∥r (IE∗ωMn(ω)p)1/p ≤6r2(r −1)p∥φ∥r (IEωM ∗n(ω)p)1/p≤6r2(r −1)(p −1)∥φ∥r∥Mn∥p.✷4

We deduceLemma 2.3 Let T1, ..., Tn : X →LY1 (IDn) and φ := φ(T1, ..., Tn).For i = 1, ..., n assume thatTi(X) ⊆{f : f = Pn1 εkyk, yk ∈Y }. Then1cpr∥φ : LXp →LLYpp ∥≤∥φ : LXr →LLYrr∥≤cpr∥φ : LXp →LLYpp ∥for 1 < p < r < ∞, where the constant cpr > 0 is independent from X,Y ,(T1, ..., Tn) and n.Proof.

The left-hand inequality follows from Lemma 2.2 and∥φ : LXp →LLYpp ∥≤6r2(p −1)(r −1)∥φ : LXr →LLYpr∥≤6r2(p −1)(r −1)∥φ : LXr →LLYrr∥.The right-hand inequality is a consequence of Lemma 2.1, Lemma 2.2 and∥φ : LXr →LLYrr∥=∥φ′ : LLY ′r′r′→LX′r′ ∥≤6r′2(p′ −1)(r′ −1)∥φ′ : LLY ′r′p′→LX′p′ ∥=6r′2(p′ −1)(r′ −1)∥φ : LXp →LLYrp ∥≤6r′2(p′ −1)(r′ −1)Krp∥φ : LXp →LLYpp ∥,where we use Kahane’s inequality (cf. [10] (II.1.e.13)) in the last step.✷If we apply Lemma 2.3 in the situation Tkx := εkT x and exploitRUMDpn(T ) ≤∥φ(T1, ..., Tn) : LXp →LLYpp ∥≤2RUMDpn(T )then we arrive atTheorem 2.4 Let 1 < p < r < ∞and T ∈L(X, Y ).

Then1cprRUMDpn(T ) ≤RUMDrn(T ) ≤cprRUMDpn(T ),where the constant cpr > 0 is independent from X,Y ,T and n.The above consideration justifiesRUMDn(T ) := RUMD2n(T )forT ∈L(X, Y )and RUMDn(X) := RUMDn(IX) for a Banach space X.5

3K–convexityWe show that RUMDn(X) ≍n if and only if X is not K-convex, that is, if and only if X uniformlycontains ln1 . To do this some additional notation is required.

For x1, ..., xn ∈X we set|x1 ∧... ∧xn|X := sup{|det(⟨xi, aj⟩)ni,j=1| : a1, ..., an ∈BX′}.Furthermore, for fixed n we define the bijectioni : {−1, 1}n →{1, ..., 2n}asi(ω) = i(ω1, ...ωn) := 1 + 1 −ωn2+ 1 −ωn−122 + ... + 1 −ω122n−1and the corresponding sets I0 := {1, ..., 2n}, I(ω1, ..., ωn) := {i(ω1, ..., ωn)},I(ω1, ..., ωk) := {i(ω1, ..., ωn) : ωk+1 = ±1, ..., ωn = ±1}fork = 1, ..., n −1.It is clear thatI(ω1, ..., ωk−1) = I(ω1, ..., ωk−1, 1) ∪I(ω1, ..., ωk−1, −1)andI0 = I(1) ∪I(−1).Our first lemma is technical.Lemma 3.1 Let {Mk}n0 be a Walsh-Paley-martingale in X and let xk := Mn(i−1(k)) ∈X fork=1, ..., 2n.Then, for all ω∈{−1, 1}n and 1≤k≤n , there exist natural numbers1 ≤r0 ≤s0 < r1 ≤s1 < ... < rk ≤sk ≤2n with|M0(ω) ∧... ∧Mk(ω)| = 12kxr0 + ... + xs0s0 −r0 + 1∧... ∧xrk + ... + xsksk −rk + 1Proof. Let us fix ω ∈{−1, 1}n. Since Ml(ω1, ..., ωl) =12n−lPi∈I(ω1,...,ωl) xi we have for l = 0, ..., n −1,Ml(ω1, ..., ωl) −12Ml+1(ω1, ..., ωl+1) =12n−lXI(ω1,...,ωl,−ωl+1)xi =12#I(ω1, ..., ωl, −ωl+1)XI(ω1,...,ωl,−ωl+1)xifor l = 0, ..., n−1.

It is clear that I(−ω1), I(ω1, −ω2), I(ω1, ω2, −ω3), ..., I(ω1, ..., ωk−1, −ωk), I(ω1, ..., ωk)are disjoint, such that we have|M0(ω) ∧... ∧Mk(ω)|=M0(ω) −12M1(ω)∧... ∧Mk−1(ω) −12Mk(ω)∧Mk(ω)=12kxr0 + ... + xs0s0 −r0 + 1∧... ∧xrk + ... + xsksk −rk + 1after some rearrangement. ✷The second lemma, which is required, is a special case of [6] (Thm.1.1).6

Lemma 3.2 Let u ∈L(ln2 , X) and let {e1, ..., en} be the unit vector basis of ln2 . Then|ue1 ∧... ∧uen|X ≤ 1n!1/2π2(u)n,where π2(u) is the absolutely 2-summing norm of u.Now we apply Lemma 3.1 to a special Walsh-Paley-martingale {M 1k}n0 with values in l2n1whose differencesdM 1k(ω) are closely related to a discret version of the Haar functions from L1[0, 1].

For fixed n thismartingale is given byM 1n(ω1, ..., ωn) := ei(ω1,...,ωn)andM 1k := IE(M 1n|Fk),where {e1, ..., e2n} stands for the unit vector basis of l2n1 .Lemma 3.3 Let n ≥1 be fixed. Then(1) ∥M 1k(ω)∥= ∥dM 1k(ω)∥= 1for k = 0, ...n and all ω ∈Ωn,(2) infω IEε∥Pn1 εkdM 1k(ω)∥≥αnfor some α > 0 independent from n.Proof.

(1) is trivial. We consider (2).

Lemma 3.1 implies12n |f1 ∧... ∧fn+1|ln+11= |M 10(ω) ∧... ∧M 1n(ω)|l2n1for all ω ∈{−1, 1}n, where {f1, ..., fn+1} denotes the unit vector basis of ln+11. We can continue to12|f1 ∧... ∧fn+1|1/nln+11=|M 10 (ω) ∧dM 11 (ω) ∧... ∧dM 1n(ω)|1/nl2n1≤(n + 1)∥M 10(ω)∥|dM 11 (ω) ∧... ∧dM 1n(ω)|1/n≤c|dM 11 (ω) ∧... ∧dM 1n(ω)|1/n.If we define the operator uω : ln2 −→l2n1by uω((ξ1, ..., ξn)) := Pn1 ξidM 1i (ω) and use Lemma 3.2, then weget|dM 11 (ω) ∧... ∧dM 1n(ω)|1/n ≤ 1n!1/2nπ2(uω).Since the l2n1are uniformly of cotype 2 there is a constant c1 > 0, independent from n, such thatπ2(uω) ≤c1IEε∥nX1εkdM 1k(ω)∥(see [17],[12]).

Summerizing the above estimates yields|f1 ∧... ∧fn+1|1/nln+11≤2ce1/2n−1/2c1IEε∥nX1εkdM 1k(ω)∥≤c2n−1/2IEε∥nX1εkdM 1k(ω)∥.7

The known estimate |f1∧...∧fn+1|1/n+1 ≥1c3 (n+1)1/2 concludes the proof (see, for instance, [6](Ex.2.7)).✷Finally, we need the trivialLemma 3.4 Let T ∈L(X, Y ). Then RUMDn(T ) ≤2n1/2T n2 (T ).Proof.

Using the type 2 inequality for each ω ∈Ωn and integrating yield for a martingale {Mk}n0 IEε,ω∥nX1εkT dMk(ω)∥2!1/2≤T n2 (T ) nX1∥dMk∥2LX2!1/2≤2n1/2T n2 (T )∥Mn∥LX2 .✷Now we can proveTheorem 3.5 There exists an absolute constant α > 0 such that for any Banach space X the followingassertions are equivalent. (1) X is not K-convex.

(2) For all θ > 0 and all n = 1, 2, ... there is a Walsh-Paley martingale {Mk}n0 with values in BX,inf1≤k≤n infω ∥dMk(ω)∥≥1 −θandinfω IEε∥nX1εkdMk(ω)∥≥αn. (3) RUMDn(X) ≥cn for n = 1, 2, ... and some constant c = c(X) > 0.Proof.

Taking α > 0 from Lemma 3.3 the implication (1) ⇒(2) follows. (2) ⇒(3) is trivial.

(3) ⇒(1): Assumig X to be K-convex the space X must be of type p for some p > 1. Consequently,Lemma 3.4 impliesRUMDn(X) ≤2n1/2T n2 (X) ≤2n1/2n1/p−1/2Tp(X) ≤2n1/pTp(X).✷Remark.

One can also deduce (3) ⇒(1) from [4] and [13] in a more direct way (we would obtain thatLX2 ({−1, 1}IN) is not K-convex).4SuperreflexivityA Banach space X is superreflexive if each Banach space, which is finitely representable in X, is reflexive.We will see that RUMDn(X) ≥cn1/2 whenever X is not superreflexive and that the exponent 12 is thebest possible in general. This improves an observation of Aldous and Garling (proofs of [5](Thm.3.2)and [1](Prop.2)) which says that RUMDn(X) ≥cn1/s in the case X is of cotype s (2 ≤s < ∞) and notsuperreflexive.We make use of the summation operatorsσn : l2n1−→l2n∞andσ : l1 −→l∞with{ξk}k −→( kXl=1ξl)k,8

as well as ofΦ : C[0, 1]′ −→l∞([0, 1])withµ −→{t →µ([0, t])} .The operators σn are an important tool in our situation. Assuming X to be not superreflexive, accordingto [7] for all n = 1, 2, ... there are factorizations σn = BnAn with An : l2n1→X, Bn : X →l2n∞andsupn ∥An∥∥Bn∥≤1 + θ (θ > 0).

It turns out that the image-martingale {Mk}n0 ⊂LX2 of {M 1k} (n isfixed, {M 1k} is defined in the previous section), which is given by Mk(ω) := AnM 1k(ω)(k = 1, ..., n),possesses a large random unconditional constant. To see this we setM ∞k (ω) := σnM 1k(ω)(ω ∈Ωn, k = 0, ..., n)and obtain a martingale {M ∞k }n0 with values in l2n∞.

For k = 1, ..., n it is easy to check thatdM ∞k (ω1, ..., ωk) = ωk2k−n−1(0, ..., 0, 1, 2, 3, ..., 2n−k, 2n−k −1, ..., 3, 2, 1, 0, 0, ..., 0)where the block (1, 2, 3, ..., 2n−k) is concentrated on I(ω1, ..., ωk−1, 1) and the block (2n−k −1, ..., 3, 2, 1, 0)is concentrated on I(ω1, ..., ωk−1, −1), that is, the vectors |dM ∞k (ω)| correspond to a discrete Schaudersystem in l2n∞. Furthermore, we haveLemma 4.1 Let n ≥2 be a natural number and let {ei} be the standard basis of l2n1 .

Then there existsa map e : {−1, 1}n −→{e1, ..., e2n} ⊂l2n1such thatµnω : | < dM ∞k (ω), e(ω) > | ≥14≥12fork = 1, ..., n.Proof. First we observe thatinf {| < dM ∞k (ω1, ..., ωk), ei > | : i ∈I(ω1, ..., ωk, −ωk)} = 2k−n−1 min(2n−k−1 + 1, 2n−k −2n−k−1) ≥14for 1 ≤k < n. Then we use the fact that# (∩n1 supp dM ∞k (ω)) = 1for allω ∈{−1, 1}nto define e(ω) as the i-th unit vector, in the case if{i} = ∩n1 supp dM ∞k (ω) ⊆∩n2 I(ω1, ..., ωk−1).For 1 ≤k ≤n −2 we obtainµnω : | < dM ∞k (ω), e(ω) > | ≥14≥µn(ω1, ..., ωn) : | < dM ∞k (ω1, ..., ωk), e(ω1, ..., ωn) > | ≥14,ωk+1 = −ωk≥µn(ω1, ..., ωn) : inf {| < dM ∞k (ω1, ..., ωk), ei > | : i ∈I(ω1, ..., ωk+1)} ≥14,ωk+1 = −ωk= µn{ωk+1 = −ωk} = 12.Since | < dM ∞k (ω), e(ω) > | ≥14 for all ω in the cases k = n −1 and k = n the proof is complete.

✷We deduce9

Lemma 4.2 Let n ≥1 be fixed. Then(1) ∥M ∞k (ω)∥= 1 and ∥dM ∞l (ω)∥= 12for k = 0, ..., n,l = 1, ..., n, and all ω ∈Ωn,(2) µnω : IEε∥Pn1 εkdM ∞k (ω)∥≥αn1/2> βfor some α, β > 0 independent from n.Remark.

An inequality IEε∥Pn1 εkdM ∞k (ω)∥≥αn1/2 can not hold for all ω ∈Ωn since, for example,∥nX1εkdM ∞k (1, 1, ..., 1)∥≤∥nX1dM ∞k (1, 1, ..., 1)∥≤∥σn∥∥nX1dM 1k(1, 1, ..., 1)∥≤2.Proof of Lemma 4.2. Assertion (1) is trivial.

We prove (2). For t > 0 we considerµn(ω : IEε∥nX1εkdM ∞k (ω)∥l2n∞> tn1/2)≥µn(ω : ∥IEε|nX1εkdM ∞k (ω)|∥l2n∞> tn1/2)≥µnω : 1co∥ nX1|dM ∞k (ω)|2!1/2∥l2n∞> tn1/2=µn(ω : ∥nX1|dM ∞k (ω)|2∥l2n∞> c2ot2n).Denoting the last mentioned expression by pt the previous lemma yieldsptn + (1 −pt)c2ot2n≥IEω∥nX1|dM ∞k (ω)|2∥≥IEω =IEωnX1| < dM ∞k (ω), e(ω) > |2≥nX1116µω : | < dM ∞k (ω), e(ω) > |2 ≥116≥n32such that pt ≥1/32−c2ot21−c2ot2for c2ot2 < 1.

✷Lemmas 3.2 and 4.2 implyTheorem 4.3 There are α, β > 0 such that for all non-superreflexive Banach spaces X, for all θ > 0,and for all n = 1, 2, ... there exists a Walsh-Paley martingale {Mk}n0 with values in BX,inf1≤k≤n infω ∥dMk(ω)∥≥12(1 + θ),andµn(ω : IEε∥nX1εkdMk(ω)∥≥αn1/2)> β.10

Proof. We choose factorizations σn = BnAn with An : l2n1→X , Bn : X →l2n∞, ∥An∥≤1 and ∥Bn∥≤1+min(1, θ) (see [7](Thm.4)).

Defining Mk(ω) := AnM 1k(ω) ∈X we obtain sup0≤k≤n,ω∈Ωn ∥Mk(ω)∥≤1from Lemma 3.3 as well as inf1≤k≤n,ω∈Ωn ∥dMk(ω)∥≥12(1+θ) andµn(ω : IEε∥nX1εkdMk(ω)∥X > α2 n1/2)≥µn(ω : IEε∥nX1εkdMk(ω)∥X >α∥Bn∥n1/2)≥µn(ω : IEε∥nX1εkdM ∞k (ω)∥l2n∞> αn1/2)≥βaccording to Lemma 4.2.✷For Banach spaces of type 2 we getTheorem 4.4 For any Banach space X of type 2 the following assertions are equivalent. (1) X is not superreflexive.

(2) 1cn1/2 ≤RUMDn(X) ≤cn1/2for n = 1, 2, ... and some c > 0. (3)1c′ n1/2 ≤RUMDn(X)for n = 1, 2, ... and some c′ > 0.Proof.

(1) ⇒(2) follows from Theorem 4.3 and Lemma 3.4. (3) ⇒(1).

We assume X to be superreflexiveand find ([8], cf. [14](Thm1.2,Prop.1.2)) γ > 0 and 2 ≤s < ∞such thatXk≥0∥dMk∥sLX21/s≤γ supk∥Mk∥LX2for all martingales in X.

This martingale cotype implies IEε,ω∥nX1εkdMk(ω)∥2!1/2≤T2(X) IEωnX1∥dMk(ω)∥2!1/2≤T2(X)n1/2−1/sγ∥Mn −M0∥LX2which contradicts RUMDn(X) ≥1c′ n1/2.✷Remark. Corollary 5.4 will demonstrate that the asymptotic behaviour of RUMDn(X) can not charac-terize the superreflexivity of X in the case that X is of type p with p < 2.

Namely, according to Theorem5.4 for all 1 < p < 2 < q < ∞there is a superreflexive Banach space X of type p and of cotype q withRUMDn(X) ≍n1p −1q . On the other hand, if 1p −1q ≥12 then we can find a non-superreflexive Banachspace Y such that RUMDn(Y ) ≍n1p −1q (add a non-superreflexive Banach space of type 2 to X).Finally, we deduce the random unconditional constants of the summation operators σn, σ, and Φ definedin the beginning of this section.

To this end we need the type 2 property of these operators. From[7] and [9] or [16] as well as [18] we know the much stronger results, that σ and the usual summation11

operator from L1[0, 1] into L∞[0, 1] can be factorized through a type 2 space. We want to present a verysimple argument for the type 2 property of the operator Φ which can be extended to some other ”integraloperators” from C[0, 1]′ into l∞([0, 1]).Lemma 4.5 The operator Φ : C[0, 1]′ →l∞([0, 1]) is of type 2 with T2(Φ) ≤2.Proof.

First we deduce the type 2 inequality for Dirac-measures. Let λ1, ..., λn ∈IK, t1, ..., tn ∈[0, 1],whereas we assume 0 ≤tk1 = ... = tl1 < tk2 = ... = tl2 < ... < tkM = ... = tlM ≤1, and let δt1, ..., δtn thecorresponding Dirac-measures.

Then, using Doob’s inequality, we obtain IEε∥nX1Φεjλjδtj∥2!1/2= IEε supt |MXi=1 liXkiεjλj!δtkj ([0, t])|2!1/2= IEεsup1≤m≤M|mXi=1 liXkiεjλj!|2!1/2≤2 IEε|MXi=1 liXkiεjλj!|2!1/2=2 nX1|λj|2!1/2.Hence IEε∥nX1Φεjλjδtj∥2!1/2≤2 nX1∥λjδtj∥2!1/2.In the next step for any µ ∈C[0, 1]′ we find a sequence of point measures (finite sums of Dirac-measures){µm}∞m=1 ⊂C[0, 1]′ such that supm ∥µm∥≤∥µ∥and limm µm([0, t]) = µ([0, t]) for all t ∈[0, 1] (take,for example, µm := P2mi=1 δ i−12m µ(Imi ) with Im1 := [0,12m ] and Imi:= ( i−12m ,i2m ] for i > 1). Now, assumingµ1, ..., µn ∈C[0, 1]′ we choose for each µj a sequence {µmj }∞m=1 of point measures in the above way andobtain IEε∥nX1Φεjµj∥2!1/2= IEε supt |nX1εjµj([0, t])|2!1/2= IEε supt limm |nX1εjµmj ([0, t])|2!1/2≤lim supm IEε supt |nX1εjµmj ([0, t])|2!1/2.Using the type 2 inequality for Dirac measures and an extreme point argument we may continue to IEε∥nX1Φεjµj∥2!1/2≤2 lim supm nX1∥µmj ∥2!1/2≤2 nX1∥µj∥2!1/2.✷12

As a consequence we obtainTheorem 4.6 There is an absolute constant c > 0 such that for all n = 1, 2, ...1cn1/2 ≤RUMDn(σn) ≤RUMDn(σ) ≤RUMDn(Φ) ≤cn1/2.Proof. 1cn1/2 ≤RUMDn(σn) is a consequence of Lemma 3.3 and 4.2.

RUMDn(σn) ≤RUMDn(σ) ≤RUMDn(Φ) is trivial. Finally, Lemma 4.5 and Lemma 3.4 imply RUMDn(Φ) ≤4n1/2.✷Corollary 4.7 There is an absolute constant c > 0 such that for all n = 1, 2, ...1c n1/2 ≤ IEε,ω∥nX1εkdM ∞k (ω)∥2!1/2= IEε,ω∥nX1εk|dM ∞k (ω)| ∥2!1/2≤cn1/2.Proof.

This immediately follows from Lemma 4.2, Theorem 4.6, and dM ∞k (ω) = ωk|dM ∞k (ω)|.✷5An exampleWe consider an example of Bourgain to demonstrate that for all 0 ≤α < 1 there is a superreflexiveBanach space X with RUMDn(X) ≍nα. Moreover, the general principle of this construction allows usto show that RUMD1n(IK) ≍n mentioned in section 2 of this paper.The definitions concerning upper p- and lower-q estimates of a Banach space as well as the modulus ofconvexity and smoothness, which we will use here, can be found in [10].Let us start with a Banach space X and let us consider the function space XΩn := {f : Ωn →X}equipped with some norm ∥∥= ∥∥XΩn .

For a fixed f ∈XΩn we defineM f : Ωn →XΩnbyM f(ω) := fωwhere fω(ω′) := f(ωω′)(ωω′ := (ω1ω′1, ..., ωnω′n) for ω = (ω1, ..., ωn) and ω′ = (ω′1, ..., ω′n)). SettingM fk := IE(M fn|Fk) we obtain a martingale {M fk }nk=o with values in XΩn generated by the functionf ∈XΩn.

Furthermore, putting fn := f,fk(ω) := IE(f|Fk)(ω) =12n−kXω′k+1=±1...Xω′n=±1fn(ω1, ..., ωk, ω′k+1, ..., ω′n),dfk := fk −fk−1 for k ≥1, and df0 = f0, it yields nX0αkdfk!ω=nX0αkdM fk (ω)for allω ∈Ωnand allα0, ..., αn ∈IK.The following lemma is now evident.13

Lemma 5.1 Let f ∈XΩn and let {M fk }n0 be the corresponding martingale. If ∥∥= ∥∥XΩn is translationinvariant then ∥Pn0 αkdM fk (ω)∥= ∥Pn0 αkdfk∥for all ω ∈Ωn and all α0, ...., αn ∈IK.First we deduceCorollary 5.2 There exists c > 0 such that nc ≤RUMD1n(IK) ≤RUMD1n(X) ≤cn for all n = 1, 2, ...and all Banach spaces X.Proof.

We consider IKΩn with ∥f∥:= Pω |f(ω)| such that IKΩn = l1(Ωn).Defining f ∈l1(Ωn) asf := χ{(1,...,1)} it follows that fω = χ{ω}. It is clear that the isometry I : l1(Ωn) →l2n1with Ifω := ei(ω)(ei(ω) is the i(ω)-th unit vector where i(ω) is defined as in section 3 of this paper) transforms the martingale{M fk } into the martingale {M 1k} from section 3 by IM fk (ω) = M 1k(ω) for all ω ∈Ωn.

Combining Lemma5.1 and Lemma 3.3 yieldsinfω IEε∥nX1εkdfk∥l1(Ωn) ≥αnand∥f −f0∥l1(Ωn) ≤2.Consequently, RUMD1n(X) ≥RUMD1n(IK) ≥α2 n. On the other hand we have RUMD1n(X) ≤2n ingeneral.✷Now, we treat Bourgain’s example [2] .Theorem 5.3 For all 1 < p < q < ∞and n ∈IN there exists a function lattice X2npq = IKΩ2n such that(1) X2npq has an upper p- and a lower q-estimate with the constant 1,(2) there exists a Walsh-Paley martingale {Mk}2n0with values in BX2npq andinfω IEε∥2nX1εkdMk(ω)∥≥c(2n)1p −1qwhere c > 0 is an absolute constant independent from p,q, and n.Proof. In [2] (Lemma 3) it is shown that there is a lattice norm ∥∥= ∥∥IKΩ2n on IKΩ2n which satisfies(1), such that there exists a function φ ∈IKΩ2n with∥φ∥≤ε1−pq(ε = 2n−1p )and∥ 2nX0|dφk|2!1/2∥≥12[2] (Lemma 4 and remarks below, ε = 2n−1/p is taken from the proof of Lemma 4).

Since ∥∥is translationinvariant Lemma 5.1 implies∥M φk (ω)∥= ∥M φk ∥LX2npq2≤∥M φn∥LX2npq2= ∥φ∥≤4(2n)1q −1pand14

IEε∥2nX0εkdM φk (ω)∥= IEε∥2nX0εkdφk∥≥∥IEε|2nX0εkdφk| ∥≥1A∥ 2nX0|dφk|2!1/2∥≥12A.✷As usual, in the following the phrase ”the modulus of convexity (smoothness) of X is of power type r”stands for ”there is some equivalent norm on X with the modulus of convexity (smoothness) of powertype r”. Now, similarly to [2] we apply a standard procedure to the above finite-dimensional result.Corollary 5.4 (1) For all 1 < p < 2 < q < ∞there is a Banach space X with the modulus of convexityof power type q and the modulus of smoothness of power type p, and a constant c > 0 such that1cn1p −1q ≤RUMDn(X) ≤cn1p −1qforn = 1, 2, ...(2) There is a Banach space X with the modulus of convexity of power type q and the modulus of smooth-ness of power type p for all 1 < p < 2 < q < ∞, and RUMDn(X) →n→∞∞.Proof.

For sequences P = {pn} and Q = {qn} with1 < p1 ≤p2 ≤... ≤pn ≤... < 2 < ... ≤qn ≤... ≤q2 ≤q1 < ∞we set XP Q := L2 X2npnqn and obtain that XP Q satisfies an upper pk- and a lower qk-estimate for allk. According to a result of Figiel and Johnson (cf.

[10] (II.1.f.10)) XP Q has the modulus of convexityof power type qk and the modulus of smoothness of power type pk for all k = 1, 2, .... Furthermore,[14](Theorem 2.2) impliessupε1±1,...,εn±1∥nX1εldMl∥LX2≤cpk nX1∥dMl∥pkLX2!1/pk≤cpkn1pk −1qk nX1∥dMl∥qkLX2!1/qk≤cpkdqkn1pk −1qk ∥nX1dMl∥LX2for all martingales {Ml} with values in XP Q such that RUMDn(XP Q) ≤cpkdqkn1pk −1qk .

On the otherhand, from Theorem 5.3 we obtainc(2n)1pn −1qn ≤RUMD2n(X2npnqn) ≤RUMD2n(XP Q).Now, setting pk ≡p and qk ≡q we obtain (1). Choosing the sequences in the way that pk →k→∞2,qk →k→∞2, and n1pn −1qn →n→∞∞assertion (2) follows.✷15

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