S. V. Astashkin and F. A. Sukochev- Best Constants in Rosenthal-Type Inequalities and the Kruglov...

8/3/2019 S. V. Astashkin and F. A. Sukochev- Best Constants in Rosenthal-Type Inequalities and the Kruglov Operator

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BEST CONSTANTS IN ROSENTHALS TYPE INEQUALITIES 1987

holds for every sequence f X of independent random variables (i.r.v.s) (resp.,of mean zero i.r.v.s) and for every n N. The inequalities above can be viewedas a fundamental generalization of the famous Khintchine inequality and, in the

case when X = Lp (resp., X Lp), 1 p < , they may be found in [23] (resp.,[12]). The original proof of Rosenthal in [23], as well as a subsequent proof of amore general result by Burkholder in [7] yielded only constants CLp in (2) and (3)which grow exponentially in p, as p . The sharp result that CLp pln(p+1) ,that is, there are universal constants 0 < < < such that the ratio betweenCLp and

pln(p+1) lies in the interval [, ] for all p [1, ) was obtained in [13]

(see also subsequent alternative proofs in [14, 18]).The main purpose of this paper is to provide a sharp estimate on the constant

CX in (2) in the more general setting of symmetric spaces X. At the same time,

we also believe that our methods shed additional light on the well-studied caseX = Lp. Indeed, the methods exploited in [13, 14, 18] have a distinct Lp-flavorand do not appear to extend to other symmetric function spaces for which (2) and(3) hold. Our approach here is linked with the so-called Kruglov property (see thedefinition in Section 2.2). Consider the special case of Rosenthals inequalities (2)and (3) when i.r.v.s fk , k = 1, 2, . . . , n, satisfy the additional assumption that

nk=1

({fk = 0}) 1, n N(4)

[in this case, the right-hand sides of (2) and (3) become equal]. In this special case,it was first established by Braverman [5] that ifX is a symmetric space with theFatou property (see Section 2.1 below), then X has the Kruglov property if andonly if (2) holds. Recently, in [24] we have developed a novel approach to thestudy of spaces with the Kruglov property that involves defining a positive lin-ear operator K : L1[0, 1] L1[0, 1] (see details in Section 2.3) which is boundedin a symmetric function space X with the Fatou property if and only ifX has theKruglov property. Furthermore, we have shown in those papers that in this case the(Kruglov) operator K is bounded in X if and only if(2) and (3) hold in full gener-ality. The following key fact is an immediate consequence of Prohorovs familiar

inequality [22] (see also the proof of Theorem 3.5 from [2]): if X is a symmet-ric space, then for every sequence {fk}k=1 L1[0, 1] of symmetrically distributedi.r.v.s satisfying assumption (4) we have

nk=1

fk

16K(F) (n N),

where the function F is defined by (1) (the definition of the decreasing rearrange-ment f of a measurable function f is given in the next section). This observationhas naturally led us to the conjecture that the best constant CX in (2) and (3) should

be equivalent to the norm of the operator K in X. We prove this conjecture in Sec-tion 3 and present computations of the norm K in various classes of symmetric


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1988 S. V. ASTASHKIN AND F. A. SUKOCHEV

spaces in Section 4. In the case ofLp-spaces, 1 p < , our results, of course,yield the same estimates as in [13, 14, 18]. In the case of symmetric Lorentz andMarcinkiewicz spaces (and other classes of symmetric spaces in which we are

able to compute the norm of the operator K) our results are new and appear to beunattainable by methods used in [13, 14, 18]. In the final section of this paper weprovide two complements to Rosenthals inequality (2).

2. Preliminaries.

2.1. Symmetric function spaces and interpolation of operators. In this subsec-tion we present some definitions from the theory of symmetric spaces and interpo-lation of operators. For more details on the latter theory we refer to [16, 20].

We will denote by S(,P) the linear space of all measurable finite a.e. func-

tions on a given measure space (,P) equipped with the topology of convergencelocally in measure.

Let I denote either [0, 1] or (0, ) with Lebesgue measure . Iff S(I,) wedenote by f the decreasing rearrangement off, that is,

f(t) = inf(A)=t

supsI\A

|f(s)|.

A Banach function space X on I is said to be symmetric if the conditions f Xand g f imply that g X and gX fX. We will assume always thenormalization that

(0,1)

X

=1, where A is the characteristic function of the

set A I. Let X(t) = (0,t )X be the fundamental function ofX. A symmetricspace X is said to have the Fatou property if for every sequence (fn)n=1 Xof nonnegative functions such that fn f a.e. and limn fnX < we havef X and fX = limn fnX.

Let us recall some classical examples of symmetric spaces on [0, 1].Let M(t) be an increasing convex function on [0, ) such that M(0) = 0. By

LM we denote the Orlicz space on [0, 1] (see, e.g., [16, 20]) endowed with thenorm

xLM =

inf > 0 : 10

M|x(t)|/dt 1.We suppose that is a positive concave function on [0, 1] with (0+) = 0.

The Lorentz space () is the space of all measurable functions f on the interval[0, 1] such that

f() =1

0f(s)d(s) < .

The Marcinkiewicz space M() is the space of all measurable functions f on theinterval [0, 1] such that

fM() = sup0


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It is easy to check that ()(t) = M()(t ) = (t). In this paper, we mainly workwith the case when (t) = t1/p , 1 p < .

Let X=

(X0, X1) be a Banach couple and X be a Banach space such that

X0 X1 X X0 + X1. We say that X is an interpolation space between X0and X1 if any bounded linear operator A : X0 + X1 X0 + X1 which maps Xiboundedly into Xi (i = 0, 1) also maps X boundedly into X. Then AXX C(AX0X0 , AX1X1 ) for some C 1. If the last inequality holds with C = 1we will refer X to an 1-interpolation space between X0 and X1.

In what follows supp f is the support of a function f defined on , that is,supp f := { : f() = 0}, F is the distribution function of a random vari-able , and [z] is the integral part of a real number z.

2.2. The Kruglov property of symmetric function spaces. Let f be a measur-able function (a random variable) on [0, 1]. By (f) we denote the random vari-able

Ni=1 fi , where fi s are independent copies off, and N is a Poisson random

variable with parameter 1 independent of the sequence {fi}.

DEFINITION. A symmetric function space X is said to have the Kruglov prop-erty if and only iff X (f) X.

This property has been studied by Braverman [5], using some probabilistic con-structions of Kruglov [15] and by the authors in [24] via an operator approach.

We refer to the latter papers for various equivalent characterizations of the Kruglovproperty. Note that only the implication f X (f) X is nontrivial, sincethe implication (f) X f X is always satisfied [5], page 11. Moreover,a symmetric space X has the Kruglov property ifX Lp for some p < [5],Theorem 1.2, and [2], Corollaries 5.4, 5.6. At the same time, some exponentialOrlicz spaces which do not contain Lq for any q < also possess this property(see [2, 5]).

2.3. The Kruglov operator in symmetric function spaces. Let {Bn}n=1 be asequence of pairwise disjoint measurable subsets of[0, 1] and let (Bn) = 1en! . Iff L1[0, 1], then we set

Kf (0, 1, . . . ) =

n=1

nk=1

f (k)Bn (0).(5)

Then K : L1[0, 1] L1(,P) is a positive linear operator. Here (,P) =n=0([0, 1], n), where n is the Lebesgue measure on [0, 1].For convenience, by Kf we also denote another random variable defined on

[0, 1

]and having the same distribution as the variable introduced in (5). If f

L1[0, 1], {Bn} is the same sequence of subsets of[0, 1] as above, and, for each n N fn,1, fn,2, . . . , f n,n and Bn form a set of independent functions such that f

n,k =


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f for every k = 1, . . . , n, then Kf(t) is defined as the decreasing rearrangementof the function

n=1

nk=1

fn,k (t)Bn (t) (0 t 1).(6)As we have pointed out above K is a linear operator from L1[0, 1] to L1(,P).By saying that K maps boundedly a symmetric space X on [0, 1] into symmet-ric space Y, we mean that K is bounded as a linear mapping from X[0, 1] intoY(,P). The representation ofKf given by (6) allows us, without any ambiguity,also speak about K as a bounded map from X[0, 1] into Y[0, 1]. A direct compu-tation (see, e.g., [2]) yields the following equality for the characteristic functionKf ofKf:

Kf(t) = exp

(eit x 1) dFf(x)

(7)

= expf(t ) 1= (f)(t), t R.Therefore, FKf = F(f), and we can treat Kf as an explicit representation of(f). In particular, a symmetric space X has the Kruglov property if and only ifK is bounded in X.

It follows from the definition of the operator K that for any symmetric spacesX and Y

K

X

Y

1/e provided that

[0,1

]X

=

[0,1

]Y

=1 (see also [5],

page 11). It is shown in [24] that the operator K plays an important role in es-timating the norm of sums of i.r.v.s through the norm of sums of their disjointcopies. In particular, in [2] the well-known results of Johnson and Schechtmanfrom [12] have been strengthened. In the next section, we shall improve the mainresults of [2] and explain the role of this improvement in obtaining sharp constantsin Rosenthal-type inequalities studied earlier in some special cases [12, 13, 18].Subsequent sections contain explicit computation of the norm of the operator Kin various classes of symmetric spaces X and further modifications of Rosenthalsinequality (2).

3. Kruglov operator and Rosenthals inequalities. The main objective ofthe present section is the strengthening of [2], Theorem 3.5. For an arbitrary sym-metric space X on [0, 1] and an arbitrary p [1, ], we defined in [2] a functionspace ZpX on [0, ) by

ZpX := {f L1[0, ) + L[0, ) :fZpX < },

where

fZpX :=f[0,1]X + f[1,)p f[0,1]X +

k=1f(k)p

1/p

.


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Clearly, Z

pX

is a quasi-norm. It is easy to see that ZpX equipped with the equiv-

alent norm

fZpX := f[0,1]X + f(L1+Lp)(0,), f ZpX,is a symmetric space on [0, ). The spaces Z2X were introduced in [11], and thespaces Z1X and Z

2X were used in [12] in the study of Rosenthal-type inequalities.

Following [20], page 46, we define the space X(lp) as the set of all sequencesf= {fk ()}k=1, fk X (k 1) such that

fX(lp) := supn=1,2,...

n

k=1|fk|p

1/pX

<

(with an obvious modification for p = ). The closed subspace ofX(lp) gener-ated by all eventually vanishing sequences fX(lp) is denoted by X(lp).

Let X and Y be symmetric spaces on [0, 1] such that X Y. The main focus of[24] is on inequalities of the type

ni=1

fi

Y

C

ni=1

fi

ZX

(8)

and

fY (lp) C

ni=1

fi

Z

pX

,(9)

where the sequence f:= {fk}k=1 X consists of i.r.v.s, and the sequence {fk}k=1is a disjointly supported sequence of equimeasurable copies of the elements fromthe sequence f.

Our first result in this section strengthens [2], Theorems 3.5 and 6.1, by es-tablishing sharp estimates on the constant C in (8). Let F be the disjointification

function related to the sequence f:= {fk}k=1 [see (1)].THEOREM 1. Let X and Y be symmetric spaces on [0, 1] such that X Y

andY has the Fatou property.

(i) If there exists a constant C such that the estimate

ni=1

fi

Y

CFZX= C

ni=1

fi

X

,(10)

holds for every sequence {fk}k=1 X of i.r.v.s satisfying the assumption (4) forall n N, then the operatorK acts boundedly from X into Y andKXY C.


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(ii) If the operator K acts boundedly from X into Y, then for every sequence{fk}k=1 X of independent random variables, we have

n

i=1fi

Y

KXYFZX ,(11)

where > 0 is a universal constant which does not depend on X andY.

PROOF. (i) The claim follows from the inspection of the first part of the proofof [2], Theorem 3.5. For the convenience of the reader, we include details of theargument. Fix f X and n N and choose h X such that Fh = Ff and suchthat h and [0,1/n] are independent. Set hn := h[0,1/n], and let {[0,1/n], hn,k}nk=1be a set of(n

+1) independent random variables such that Fhn,k

=Fhn for all 1

k n. Since the functions |nk=1 hn,k| and |h| have the same distribution function,we conclude that the functions |nk=1 hn,k| and |f| are equidistributed. Observingnow that the sequence {hn,k}nk=1 satisfies (4), we obtain

nk=1

hn,k

Y

C

nk=1

hn,k

X

= CfX.(12)

A direct computation shows that hn (t) = n1f(t ) + (1 n1) for all t R.Hence, the characteristic function of the sum Hn :=

nk=1 hn,k is given by

Hn (t) = n1f(t) 1+ 1n tR.Since limn Hn (t) = exp(f(t ) 1) = (f)(t), for all t R, we see that Hnconverges weakly to Kf. Combining this with (12), (7), [5], Proposition 1.5, andwith the fact that Y has the Fatou property, we conclude that

KfY CfX.This completes the proof of the first assertion.

(ii) Firstly, let us assume that a sequence {fk}k=1 X consists of independentrandom variables satisfying assumption (4). Denote by

{hk

}nk

=1 a sequence of in-

dependent random variables such that Fhk =F(fk ) for all k = 1, 2, . . . . Considerthe following two cases.

(a) fks are symmetrically distributed r.v.s. In [22], Prokhorov proved that inthis case we have

n

k=1fk

x

8

nk=1

hk

x2

(x > 0).

From this inequality (see, e.g., [16], Corollary II.4.2) it follows that

n

k=1fk

Y

16n

k=1hk

Y

.(13)


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(b) fk = ak Ak , where ak 0 and Ak are arbitrary independent subsets of[0, 1](k = 1, 2, . . . ). Without loss of generality, we may assume that fk s are defined onthe measure space

n

=0([0, 1], n) by the formula fk (t) = ak [0,pk](tk ), where

pk = (Ak), k 1. From the definition of the Kruglov operator [see (6)] it fol-lows that ({Kfk = ak}) = pk /e pk/3. Hence, by (7), we may assume thathk ak[0,pk /3], and so

nk=1

fk

Y

3

nk=1

ak [0,pk /3](tk )

Y

3

nk=1

hk

Y

.(14)

Next, from fk fm = 0 (k = m) it follows that

eit F

1

=

n

k=1

(eitfk

1).

Therefore,

F(t) 1 =

(eit F 1) =n

k=1

(eitfk 1) =

nk=1

fk (t ) 1

and

K(F) = exp(F 1) =n

k=1exp(fk 1) =

nk=1

hk = nk=1 hk .

Thus, the sum

nk=1 hk is equidistributed with K(F). Therefore, by inequalities

(13) and (14), we haven

k=1fk

Y

16

nk=1

hk

Y

= 16K(F)Y 16KXYFX(15)

in the case (a) and analogously

n

k=1fk

Y

3KXYFX(16)

in the case (b).Now, we consider the case when a sequence {fk}k=1 X consists of mean zero

i.r.v.s satisfying assumption (4) with 1/2 instead of 1. We shall use the standardsymmetrization trick. Let {fk}nk=1 be a sequence of i.r.v.s such that the sequence{fk, fk}nk=1 consists of independent random variables and Ffk =Ffk for all k 1.

Setting gk := fk fk , we obtain a sequence {gk}nk=1 of symmetrically distrib-uted independent random variables satisfying assumption (4), and, by (15), wehave

n

k=1gk

Y

16KXYGX,(17)


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where G := nk=1 gk . Let B be the -subalgebra generated by the sequence{fk}nk=1 and let EB be the corresponding conditional expectation operator. Thanksto our assumption Y is an 1-interpolation space for the couple (L1, L). Hence,

since EB is bounded in L1 and L (with constant 1) [20], Theorem 2.a.4, we haveEBYY = 1. Therefore, due to the independence offks and fk s, we haven

k=1fk

Y

=EB

n

k=1gk

Y

nk=1

gk

Y

.(18)

On the other hand, it is obvious that GX 2FX . Combining (18) and (17),we see that (11) holds with = 32.

Next, let us consider an arbitrary sequence {fk}k=1 X of i.r.v.s satisfyingassumption (4) with 1/2 instead of 1. In this case, set uk = fk vk , where

vk := 1({supp fk})

1

0fk(t)dt supp fk (k 1).

Clearly, {uk}k1 is a mean zero sequence of i.r.v.s satisfying assumption (4) with1/2 instead of 1. Thus, the preceding argument yields

nk=1

uk

Y

32KXYUX,(19)

where U :=nk=1 uk . Moreover, by (16), we have

nk=1

fk

Y

nk=1

uk

Y

+

nk=1

vk

Y

32KXYUX + 3KXYVX.

Let C be the -algebra generated by the supports of fk s. It is clear that V =EC(F ), where V :=nk=1 vk . Therefore, as above, we have VX FX . More-over, since U = F V, we also have UX FX + VX 2FX . Thus,

n

k=1fk

Y

67KXYFX.

If{fk}k=1 X is an arbitrary sequence of i.r.v.s satisfying assumption (4), thenwe may represent fk = fk +fk , where each of the sequences {fk}k=1 and {fk }k=1consists of i.r.v.s satisfying assumption (4) with 1/2 instead of 1 and moreover|F| |F| and |F| |F|. In this case, using the preceding formulas, we obtainthat

nk=1

fk

Y

nk=1

fk

Y

+

nk=1

fk

Y

134KXYFX.

Finally, repeating verbatim the proof of [2], Theorem 6.1, we obtain (11) for

arbitrary sequences of i.r.v.s [which do not necessarily satisfy assumption (4)] asa corollary of already considered special case when (4) holds.


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Our next result strengthens [2], Theorem 6.7, by establishing sharp estimates onthe constant C in (9). Before proceeding, we recall the following construction dueto Calderon [8]. Let X0 and X1 be two Banach lattices of measurable functions on

the same measure space (M, m) and let (0, 1). The space X1

0 X1 consists ofall measurable functions f on (M, m) such that for some > 0 and fi Xi withfiXi 1, i = 0, 1, we have

|f(x)| |f0(x)|1 |f1(x)| , x M.This space is equipped with the norm given by the greatest lower bound of all num-bers taken over all possible such representations. Even though this constructionis not an interpolation functor on general couples of Banach lattices (see [19]),it is still a convenient tool of interpolation theory. Indeed, if (X0, X1) is a Ba-

nach couple and if (Y0, Y1) is another Banach couple of lattices of measurablefunctions on some measure space (M, m), then any positive operator A fromS(M, m) into S(M, m), which acts boundedly from the couple (X0, X1) intothe couple (Y0, Y1) also maps boundedly X

10 X

1 into Y

10 Y

1 and, in addition,

AX10 X

1Y10 Y1 A

1X0Y0AX1Y1 for all (0, 1). The proof of the

latter claim follows by inspection of the standard arguments from [20], Proposi-tion 1.d.2(i), page 43.

THEOREM 2. LetX andY be symmetric spaces on

[0, 1

]such thatX

Y. If

K acts boundedly from X into Y, then there exists a universal constant > 0 suchthat for every sequence {fk}k=1 X of i.r.v.s and for every q [1, ], we have

fY (lq ) K1/qXYFZqX .

PROOF. The case q = 1 has been treated in Theorem 1. Ifq = , then it issufficient to observe that

FZX F[0,1]

X,

and that (see, e.g., [10], Proposition 2.1)1

2

F[0,1] > sup

k=1,2,...|fk| >

F[0,1] > ( > 0).(20)

Therefore, for some constant (which does not depend on X and Y) we have

fY (l) fX(l) FZX .The rest of the proof is based on methods from interpolation theory and is verysimilar to the arguments in [2], Theorem 6.7. Let : (,P)

(

[0, 1

], ) be a

measure preserving isomorphism, where (,P) := k=0([0, 1], k) (here, k isthe Lebesgue measure on [0, 1] for every k 0). For every g S(,P), we set


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T (g)(x) := g(1x), x [0, 1]. Note that T is a rearrangement-preserving map-ping between S(,P) and S([0, 1], ). We define the positive linear mapping Qfrom S(0, ) into S(,P)N{0} by setting

Qf (0, 1, . . . ) := {(Qf )k}k=0, f S(0, ),where (Qf )k(0, 1, . . . ) := f (k + k) (k [0, 1]), k 0. The arguments aboveshow that the positive operator Qf := {T (Qf )k}k=0 is bounded from Z1X intoY (l1) with the norm not exceeding KXY (where is a universal constant)and also from ZX into Y (l) with the norm not exceeding . It follows that

Q : (Z1X)1 (ZX )

(Y(l1))1 (Y(l))and

Q Q1

Z1XY (l1) QZX Y (l), (0, 1).

Now, we shall use the following facts: for every (0, 1) we have

ZqX (Z1X)1 (ZX ) , (Y (l1))1 (Y(l)) Y (lq ), q =

1

1 .

To see the first embedding above, fix g = g ZqX , gZqX = 1, and setg1 := g[0,1] + gq [1,), g := g[0,1] + [1,).

Clearly, g

=(g1)

1 (g

) . Moreover, since g(1)

1, then g1 decreases, which

implies that gi ZiX and giZiX 3 (i = 1, ). The second embedding above(in fact, equality) is established in [6], Theorem 3. Now, we are in a position toconclude that

QZqXY (lq ) 1/qK1/qXY 11/q max(1, , )K1/qXY.

The proof is completed by noting that every sequence {fk}k=1 X of i.r.v.s maybe represented in the form {fk} = Q(F ), with some function F which is equidis-tributed with F.

The results presented in this section show that the sharp estimates in the deter-ministic estimates of expressionsn

k=1fk

X

,

nk=1

|fk|p1/p

X

, 1 p < ,

in terms of the sum of disjoint copies of individual terms offare fully determinedby the norm of the Kruglov operator K in X. In the next section, we shall presentsharp estimates of this norm in a number of important cases, including the caseX

=Lp (1

p 0 such that for every finite sequence (xj )nj

=1 X of

pairwise disjoint elements,n

j=1xj

X

C

nj=1

xj pX1/p

.

Recall also that if > 0, the dilation operator is defined by setting

(x)(s) =

x(s/ ), s min{1, },0, < s 1.

The operator , > 0 acts boundedly in every symmetric function space X [16],

Theorem II.4.4.First, we suppose that 1 p < and that X is a symmetric space satisfying

the following two conditions:

(i) X satisfies an upper p-estimate;(ii) tXX Ct1/p, 0 < t < 1.

Conditions (i) and (ii) imply, in particular, that both Boyd indices of X (see, e.g.,[20]) are equal to p.

The value of the constant C varies from line to line in this section.

PROPOSITION 3. If a symmetric space X satisfies the above assumptions, thenthere exists a universal constant > 0 whose value depends only on the constantsin (i) and(ii) above such that

KXX pln(p + 1) , p 1.(21)

PROOF. Let 0 f X, p 1, and n N. Let fn,1, fn,2, . . . , f n,n and Bnhave the same meaning as in Section 2.3. Then, by (6), the random variable Kf is

equimeasurable with the random variablen=1

gnBn where gn =n

k=1fn,k (n = 1, 2,...).

Since gn and Bn are independent, then the assumptions on X imply

KfX C

n=1gnBnpX

1/p= C

n=1

(Bn)gnpX1/p

C n=1

1e n!gnpX

1/p C n=1

np

n!1/pfX.


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It is clear that

n=1

np

n! sup

n

np

pn

n=1

pn

n! =ep sup

n

np

pn

.

The function xp/px takes its maximal value at x = p/ log(p) and this maximumdoes not exceed (p/ log(p))p . Therefore,

n=1

np

n!

1/p e p

log(p).

On the other side, if 1 p < 2, then

n=1n

p

n!1/p

n=1n

2

n! = 2e.

It is well known that an Lp-space, 1 p < , satisfies an upper p-estimate andthat tLpLp = t1/p, 0 < t 1. The facts that similarly M(t1/p), 1 < p < ,satisfies an upper p-estimate and tM(t1/p)M(t1/p) t1/p, 0 < t < 1, followfrom a combination of [9], Theorem 3.4(a)(i), and [20], Proposition 1.f.5, andfrom [16], Chapter II, Theorem 4.4. Combining these facts with Proposition 3,we obtain the following corollary.

COROLLARY 4. There exists a constant C > 0 such that for all p 1

KLpLp Cp

ln(p + 1) and KM (t1/p)M(t1/p) Cp

ln(p + 1) .(22)

Although the Lorentz space (t1/p), 1 < p < , does not satisfy the assump-tions of Proposition 3, nevertheless estimates similar to (22) also hold for the normof the operator K : (t1/p) (t1/p), 1 p < . The proof below is based onthe properties of the Kruglov operator K in Lorentz spaces exposed in [2], Sec-tion 5.

PROPOSITION 5. There exists a constant C > 0 such that for all p 1

K(t1/p)(t1/p) Cp

ln(p + 1) , p 1.(23)

PROOF. By [2], Theorem 5.1, we have

K()() 2 supu(0,1)

1(u)

k=1

uk

k!

.


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If(t) = t1/p , then the latter supremum is equal to

n=1

1

n!1/p

2

+

n=3

1

[n/3]3[n/3]/p

=2

+3

k=1

e3k log(k)/p.

Since the latter sequence decreases, we can replace sum with an integral and obtain

Kt1/p t1/p 22 + 6

e

e3s log(s)/p ds.

Substitute t= s log(s). It follows that dtds

= 1 + log(s) 12 log(t). Therefore,

Kt1/p t1/p 22 + 12

e

1

log(t)e3t /p dt.

It is clear that pe

1

log(t )e3t /p dt

pe

1

log(t)dt const p

log(p).

On the other side,p

1

log(t )e3t /p dt 1

log(p)

p

e3t /p dt p3log(p)

.

It follows that

Kt1/p t1/p const plog(p) .

We shall now estimate the norm of the operator K : (t1/p) M(t1/p), 1

1,

with universal constants.

PROOF. The proof follows from combining Proposition 5 and Corollary 4 withLemma 6.

COROLLARY 8. The order of the constant pln(p+1) in Rosenthals inequality(2) with X = Lp is optimal when p .

PROOF. Apply Theorem 1 to the case X

=Y

=Lp and then apply Theorem 7.

REMARK 9. The same argument as above also shows that the order of theconstant pln(p+1) is optimal in variants of Rosenthals inequality (2) for scales ofLorentz spaces (t1/p) (1 < p < ) and Marcinkiewicz spaces M(t1/p) (1

0.


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PROPOSITION 16. LetX be a symmetric space on [0, 1] such that the Kruglovoperator K acts boundedly on X. Then there exists a universal constant > 0such that for every n

N and any sequence

{fk

}nk

=1

X of i.r.v.s the following

inequality holds:SnX KXX

F[0,1]X + SnL1.(33)PROOF. First, we assume that1

0fk (s)ds = 0 (k = 1, 2, . . . , n ) .(34)

In this case, from Proposition 12 (see also [4], Theorem 3.1) we infer that

SnX KXXF[0,1]X + F[1,)L2,where is a universal constant. In addition, for the space L1, we have by [12],Theorem 1,

SnL1 cF[0,1]L1 + F[1,)L2.

Combining these two estimates we obtain (33) under assumption (34).Suppose now that the sequence {fk}nk=1 X is an arbitrary sequence of i.r.v.s.

Setting

gk := fk

1

0fk(s) ds (k = 1, 2, . . . , n )(35)

we obtain a sequence {gk}nk=1 X satisfying (34), and, therefore, by the abovenX KXX

G[0,1]X + nL1,(36)where

n =n

k=1gk , G(t) :=

n

k=1gk(t k + 1)[k1,k](t), t > 0.(37)

On one hand,

G(t) =n

k=1

fk(t k + 1)

10

fk (s)ds

[k1,k](t)

= F(t) n

k=1

10

fk (s)ds[k1,k](t),

which implies

G[0,1](t) F[0,1](t) + maxk=1,2,...,n

fkL1 ,


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and, therefore, in view of the embedding X L1 with the constant 1,

G[0,1]

X

F[0,1]X

+ maxk=1,2,...,n

fkX 2

F[0,1]X

.(38)

On the other hand,

nX SnX 1

0Sn(u)du

SnX SnL1and

nL1 SnL1 +1

0Sn(u)du

2SnL1 .Combining these estimates with (36), we obtain

SnX (2KXX + 1)F[0,1]X + SnL1,and the assertion is established in view of the fact KXX 1/e.

REMARK 17. The converse inequality to (33) fails in general. However, if inaddition fk 0, k = 1, 2, . . . , n, then for any symmetric space X

cF[0,1]X + SnL1 SnX

for some universal constant c > 0.

DenoteUn(t ) := max

k=1,2,...,n|Sk (t)|, t [0, 1].

THEOREM 18. Let us assume that X is an interpolation space for the couple(L1, L) and that the Kruglov operatorK acts boundedly on X. Then there existsa universal constant > 0 such that for all n N and any sequence {fk}nk=1 Xof i.r.v.s the following inequality holds

15F

[0,1]X + UnL1(39) UnX KXX

F[0,1]X + UnL1.PROOF. First, it is obvious that

2Un(t ) Mn(t ) := maxk=1,2,...,n

|fk (t)|.

Appealing to (20), we see that

Mn (t/2) F(t ) (0 < t 1),and so

UnX 14F[0,1]X,


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whence

UnX 15 F[0,1]X

+ UnL1.

Let us prove the right-hand side inequality in (39). It holds if fks are sym-metrically distributed. Indeed, by the well-known Levy theorem (see, e.g., [17],Proposition 1.1.1), we have

{t [0, 1] : Un( t ) > } 2{t [0, 1] : |Sn(t)| > } ( > 0),and so the result follows from (33) (with the universal constant ) and the assump-tion that X is a symmetric space. Moreover, by the standard symmetrization trickand using the assumption that X is an interpolation space for the couple (L1, L),it is not hard to extend this result to all sequences

{fk

}nk

=1

X of i.r.v.s satisfying

condition (34).Finally, let us consider the case of an arbitrary sequence {fk}nk=1 X of i.r.v.s.

Suppose that the sequence {gk}nk=1 is defined by formula (35). Applying the (al-ready established) assertion to this sequence, we obtain

WnX 2KXXG[0,1]X + WnL1,(40)

where G and k are defined as in (37), and Wn := maxk=1,2,...,n |k|. Noting that

Wn(t)

Un(t )

+max

k=1,2,...,n1

0 |Sk(u)

|du

Un(t)

+ Un

L1 ,

we infer WnL1 2UnL1 . Since

Wn(t ) = maxk=1,2,...,n

Sk(t ) 1

0Sk(u)du

Un(t )

10

maxk=1,2,...,n

|Sk(u)| du = Un(t) UnL1 ,

we have

WnX UnX UnL1 .By (40) and (38), this guarantees

UnX (4KXX + 1)F[0,1]X + UnL1,

and the assertion follows in view of the fact KXX 1/e.

REMARK 19. In the case when X is a symmetric space containing Lp forsome finite p (this condition is more restrictive than the boundedness of the

Kruglov operator in X, see [2]), the last result was also obtained in [10], Theo-rem 5. At the same time, it is established in [10] even for quasi-normed spaces.


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Acknowledgments. The authors thank D. Zanin for a number of commentson an earlier version of this paper. Also authors acknowledge support from theARC.

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