Independent Random Variables and Rearrangement Invariant Spaces (London Mathematical Society Lecture Note Series)

-1/q

n

at,Xk

I-]. F I k=1

I

>x

< vx-g.

(15)

Rearrangement Invariant Spaces

58

In addition, the upper estimate in (4) together with the other conditions of the lemma imply the upper estimate in (15). We break the proof into several steps. Without loss of generality we may assume n

Elakl9=1.

(16)

k=1

Let f (t) be the characteristic function corresponding to the r.v. X1. Then the sum Ek=1 cakXk has the characteristic function n

g(t) _ II f(akt).

(17)

k=1

We show that (7) holds for g(t) with constants independent of ak and n, which together with Lemma 2 implies (15).

Proposition 2. Let ak, 13k E R and Ial:l < 1 (1 < k < n). Then

1-H(ak+if3k)

+exp 1:.1

k=1

(i:

IQkI

Proof: We have n

n

1-II (ak+0k)

n

n

k=1

k=1

H (Yk +

k=1

k=1

Expanding the parentheses, we get n

II (ak + ij k) k=1

n-2

n-1

n qq

- II ak = On II (ak + i/ k) + iani n-1 11 (ak + iqqk) qq

k=1

k=1

k=1

n-3 + iancxn_lfn-2 II (ak + i k) + ... + ianan_1 ... Q1. k=1

Since lak.I < 1, it follows that n

n

k.=1

k.=1

n-k

n-1

11 ((.Yk+zlk)-Iak.

<:IQn-k+11 11 (1 + IQs 1) + 1911 k=1

j=1

=ll(1+lakl)-1. k=1

As 1 + IxI < exp(IxI), then n

n

fl (1 + IQk.1) < exp (t 1,6kI k=1

This implies the needed inequality. El

k=1

3. Linear combinations of independent random variables

59

Proposition 3. Let (7) and (12) be fulfilled. Then there are positive constants o' i )31 and yl such that if I1 < -y1, then

exp (-)61 It I') < Ref(t) < If(t)I < exp (-al It I') .

(18)

In addition, the upper estimate in (7) implies the lower estimate in (18).

Proof: It follows from (7) and the well known inequality 1- x < exp(-x) < 1 - x/2, where 0 < x < 1, that there is b > 0 such that exp (-2,Q It I') < Ref (t) < exp (-a It I')

for Itl < b, where a and 3 are the constants from (7). This together with (12) gives us the estimate

If MI < (exp (-2a

ItI')+/-,' ItI2q) 1/2

where ItI < min 1b, v}.

If h > 0, then

<exp(-4)

exp(-x) +hr;2<1- 2+h.x2<1-

for small enough positive x. Therefore there exist4positive constants a1 and b1 such that exp (-2a ItIl) +{p2 ItI2q < exp (-2a1 Itlq) for ItI < 61. This and the above implies the needed inequalities. One can see that the lower estimate in (18) follows from the upper estimate in (7).

Proof of Lemma 4: According to Lemmas 2 and 3, from the conditions of Lemma 4 the estimates (7) and (12) follow. Suppose that g(t) is defined by the formula (17) and (16) holds. Then Iakl < 1 and we obtain, using (18), n

1-Ref(t)> 1-Ig(t)I=1-

f(akt) k=1

n

> 1 - exp

(_ai

E Iaktlq = 1 - exp (-a1 Itlq) ,

(19)

k=1

where It I < yl.

Denote c(t) = Ref(t) and d(t) = Imf(t). According to (17) and Proposition 2, n

1 - Ref(t) < I1- g(t) I =

1 - II (c(akt) + id(akt))

k=1

c(akt) + exp

E Id(akt) I (k="

1

1.

Rearrangement Irtvariautt Spaces

60

Since Ic(t)I < 1, then from (18) and (16) n

tb

0 < 1 - fl c(akt) < 1 - fl exp (-,31 IaktI') = 1 - exp (-)1 It1) k=i

.

k=1

Applying (12), we get for ItI < v n

n

EId(ak.t)I pEIaktI'=it It I' k=1

L=1

The last estimates and (19) yield that there exist positive constants a2, /32 and 72, independent of n and cak, such that a2 It I' < 1 - Reg(t) < Q2 It 1'

(20)

if Iti < 72. These estimates and Lemma 2 give us (15). According to Lemma

2 and Proposition 3, the upper estimate in (4) implies the same estimate in (7) and the lower bound in (18). Reasoning as above, we get the righthand side inequality in (20). Using Lemma. 2 once more, we obtain the same estimate in (15).

5. Convergence of the norms. Lemma 5. Let the condition (3) be fulfilled, 1 < q < 2 and r.v.s Yn be weakly convergent to the r.v. Y

.

Suppose there exists a constant b > 0 such

that f'oralla:>0andit EN P {I1,1 I > x:} < bx:-q.

(21)

Then Y,Yn EEanditYnIIE -'IIYIIE'

Proof: The assumptions of the lemma imply that the r.v. Y satisfies (21) with the same constant. Since Zq E E, Proposition 1.2 gives that Yn and Y are contained in the space E. Let a > 0. The estimate (21) and Proposition 1.2 give us IIY"1{IYnI>a}IIE

B

IIZ9I{jzqj>a}IIE

where B is independent of a and it. From here and (3) lim sup -, n

a

0,IIYnIflYnl>a} (I

E-

So, for fixed e > 0 there exists a > 0 such that IIYn'flYYI>a}IIE < e

0.

3. Linear combinations of independent random variables

61

for all it E N and we may conclude that for all integers n IIYnIIE - E < IIYnr{IYni
Since the r.v. Y satisfies (21), the same inequality is true for Y.

The condition (3) implies E OL,,(S2). Hence ¢E(t) -; 0 as t -> 0 (see [48]). Using this, it is not difficult to verify that nlim IIYnr{isrni
From here and the previous IIYIIE - e < hm IIYn IIE << n_Oo lim IIYnIIE 5 IIYIIE + E.

n00

Putting f --+ 0, we obtain the needed relation. p

6. Some more inequalities for characteristic functions. Here we show that (1) implies (7).

Lemma 6. Let {Xk}k 1 be i.i.d.r.v.s, f(i) be the corresponding characteristic function and let 0 < p < q < 2 be fixed. Suppose p

C},,q = sup

nil

< 00.

(22)

n

Then (12) and the upper estimate in (7) are true. First we prove some auxiliary results. Since f (t) is continuous and f (0) 1, there exists log f (t) near zero. We consider the branch of the logarithm for which log(1) = 0 and put

-logf(t) ItI4

Therefore, there is b > 0 such that

f(t) = exp((- Itlg0 (L))

(23)

for ItI < S, t # 0. To prove Lemma. 6 we have to set bounds for 0(t) near zero.

Let's denote

X,

Sn = k=1

(24)

62

Rearrangement Invariant Spaces

This sum has the characteristic function fn (t) = (f (tn-lfq))n

(25)

.

The condition (22) yields that the sequence {Sn},, 1 is weakly compact (see [35]). Let's denote the collection of all characteristic functions related to the limit distributions of {Sn};'_1 by A. Put °° =A U {fn}n=1

By Yh we denote a r.v. with the characteristic function h(t). If h E ', then (22) and the well-known properties of absolute moments (see [35]) imply that E IYhI < Cj,q Therefore the collection {Yh : h E ti'} is also weakly compact.

Proposition 4. Let (22) be fulfilled. Then there is e > 0 for which v(e) = inf {Reh(t) : ItI < c, h E'1 } > 1/2.

Proof: Suppose the contrary holds. Then there are tk 'N 0 and hk E I such that Rehk(tk) < 1/2 (k E N). By virtue of the compactness we may choose integers k(n) / oo with the property h.k(n)(t) g(t) for every t E R, where g(t) is a characteristic function. From here (see [35]) lim hk(n)(tk(n)) = g(0) = 1.

n

oo

This relation contradicts the above. Choosing e > 0 under the condition v(e) > 1/2 we have Reh(t) > 1/2 for all h E T and t E (-E, E). So, for every h E ' the function log(h(t)) ItIl

is defined on (-e, e)\ {0} and on this set the formula h(t) = eXp (- It I' Oh (t))

(26)

takes place.

Proposition 5. Let h E 'I' and fk(n)(t) , h(t) for each t E R ¢(tk(n)`11) - Oh(t) uniformly on each set It : 0 < a < Itl G Q G E} 0(t) is the function from (23).

.

,

Then where

Proof: It follows from (23) and (25) that f" (t) = exp (- I w m (n-1/qt))

(27)

3. Linear combinations of independent random variables

63

if 0 < ItI < E. From here 10

(n-1/qt)

-q

log f(t) - log fni(t)I < Ifn(t) - fm(t)I (m-1/gt) I _

ItI° Idm,n(t)I

ItI°

where d,,, n(t) is a point contained in the linear segment joining the points f,n(t) and f, (t). According to Proposition 2, If,, (t) I > 1/2 if ItI < c (n E N). (t)I > 2. Hence It is well known that. the convergence of characteristic functions is uniform

on each finite segment (see [35]). Therefore the last estimates yield the needed assertion. Proof of Lemma 6: According to (23), the function q5(t) is defined on (-E, e)\ {0}. Proposition 5 and Arzela.-Ascoli's theorem (see [16]) give us C

sup 11

(tn-119)

E2-114 < ItI < E, n E N} < oo.

We show that I¢(t) I < C for every i E

{0}. It is not difficult to verify

that for such t there is an integer n with the property c2- 11q < I tn1/Q I < From this and the above I0(t)I = 10 (n-'/q (tn1/4)) I < C

This bound and (23) give us 1 - f (t) = 0 (ItI') as t

0, which implies

Lemma. 6.

Lemma 7. Let one of the following conditions be fulfilled: (a) 1 < q < 2 and (3) holds;

(b)0
1 C E be a sequence of i.i.d.r.v.s such that (1) holds and, if

E=L1,(1l)andp
(Xzx. - X2k-1) k..1

E

Then for the characteristic function f (t) corresponding to X1 the estimates (7) and (12) are true. If follows from (1) and Proposition 1.1 that in the case (a) the condition (22) holds, where 1 = p < q < 2. This condition follows from (1) in the case (b). According to Lemma. 6, the upper estimates in (7) and (12) are true. Hence, we have only to prove the lower estimate in (7).

Rearrangement Invariant Spaces

64

Proposition 6. Suppose It,, E IF and Yh , are weakly convergent to Y Then Y E E and lim IIYh,,, IIE nl-w

= IIYIIE

Proof: The upper estimates in (7) and (12) give us the conditions of Lemma 4 (see Lemmas 2 and 3). So, for the sum (24) the inequality (21) holds with a constant independent of n. According to the definition of the collection 'Ir, the same estimate is true for every r.v. Yh, h E T. Suppose (a) holds. Then the needed assertion follows directly from Lemma 5. Let (b) be fulfilled. According to (21)

supElYhI< oo

(29)

hET for every r E (p, q).

From here and the theorem on the convergence of

moments (see [35]) the needed relation follows.

Proposition 7. Let h E T and {Yk } _1 be independent r.v.s equidistributed with Yh . Then these r.v.s satisfy the condition of Lemma 7.

Proof: For each h E' there are integers n(j) / oo such that S,,(j) weakly converges to Yh, where S,, is determined by the formula (24). Let's fix an integer in and consider nx

(Jnx = M-

11q

EYk.

(30)

k=1

We have n(j) S,nxn(j) =

n(j)-11v

7711-1h9

k=1

Hence S,,,,(j) IlSrnn(j)IIE

E i=1X(k-1)n(j)+i

U,,, weakly as j =- oo, and according to Proposition 6 IIUn2IIE. From (1)

C1 < IjSrnn(j)IIE < C2

for all j. So, the same estimate holds for Un,, which is equivalent to (1) for Yk.

In the case p < q = I it is similarly proved that (28) holds for the sequence in question.

As usual we call a, r.v. X and the corresponding characteristic function degenerate if P {x = a} = 1 for some a E R.

3. Linear combinations of independent random variables

65

Proposition S. Suppose that for some 0 < p < v and for the functions fn determined by (25)

sup{Ifn(t)I:p
Proof: There are tk E [p, v] and integers n(k) / oo such that I fn(k)(tk)I 1. As mentioned above, (29) yields the weak compactness of {Yh : h E f}. Passing to a subsequence, we may assume that tk -> to E [p, v] and fn(k) (t)

h(t) for all y E R and some h E F. From here (see [35]) Ih(t0)I = klim Ifn(k)(tk)I = 1. Suppose Yh is degenerate and consider independent r.v.s {Yk Q 1 equidis-

tributed with Yh. Then Yk equals a constant almost surely. But it is not difficult to verify that then (1) does not hold if q # 1, and (28) is not true if q = 1. This contradicts Proposition 7. Let's denote

Jm = [(2,n)-11q, m-119]

(31)

Proposition 9. There is an integer m such that

sup {Ifn(t)I:tEJ.,nENJ =y<1. Proof: Suppose the contrary holds. Then, according to Proposition 8, for each m there are non-degenerate hm E IF and tm E Jm such that Ihm(tm)I = 1.

Hence hm corresponds to a lattice distribution. If am is the maximal step of this distribution, then (see [43], Ch.1) tm > 2ir/am and am > 21r/tm > 2rmt/4. As the collection {Yh : h E AF } is weakly compact, there are integers

m(k) / oo for which hm(k)(t) , h(t) E IF for every real t. It is not difficult to prove that am -. oo implies Ih(t)I = 1. So, the r.v. Yh is degenerate. As in the proof of Proposition 8, a contradiction follows. The next assertion is easily verified.

Proposition 10. For every 0 < t < v and integer j there is e > 0 such that (0 e) C I

I

n>>j

[µn-1/4 vn-tk]

Rearrangement Invariant Spaces

66

Proof of Lemma 7: Let's denote u(t) = Ref (t)

.

According to (25) and

Proposition 9, for some integer m and every t E Jm j

(to-'I9)

u (t,- 11q) I n< If

In

= If.(t)I C 'Y < 1.

Since u(0) = 1, there is b > 0 such that u(x) > 0 if Ixl < b. In addition, for some integer j the inequality to-1/Q < b holds if n > j and t E Jm, which yields nlog (u (tn-1/g)) < log(7) < 0 for such n and t. From here and (31)

-

(tn-119)-q

log (u (to-1/9)) > -t-Q log(7) > -mlog(ry) = a > 0 .

Therefore u (tn-1/0 < exp (-atQn-1). Proposition 10 implies that the union of the segments J,,. (n > j) contains some interval (0, e). If t E (0, e), then tn1/9 E Jm for some n > j . Taking into account the previous, we get

u(t) =

u n- 11q (t7z11q))

< exp

(_a (tn1/) n

9

= exp (-atQ) .

It follows from (9) that u(t) is even. Hence the last inequality yields the lower estimate in (7).

7. Proof of Theorems 2 and 3. Let the conditions of Theorem 2 be fulfilled. According to Lemma 7, the estimates (7) and (12) are true. From Lemmas 2 and 3, EX1 = 0 and (4) holds. If the right-hand side of (1) holds, then Proposition 1.1 implies (22), where p = 1 < q < 2. Using Lemmas 2 and 3 we get the second part of Theorem 2. Theorem 3 follows immediately from Lemma 4.

8. Proof of Theorem 4. Let {Yk}k 1 be symmetric i.i.d.r.v.s with the q-stable distribution. It follows from (1.18) that Y1 satisfies the estimate (4)

for some positive constants a, b and c. We choose numbers a /3 and xj from (c, oo) under the conditions

xj >

4j1-1/9a

(32)

2bj)1/4 Xj,

fli lim

'l' = 0.

j-.oo aj+l

(33)

(34)

Let's put 00

Xk = EYklf.9:5lY,,l
(35)

3. Linear combinations of independent random variables

67

It is clear that {Xk}k°__1 are symmetric i.i.d.r.v.s contained in the space Lq,cx (S2).

From (34) and (35) P {IXk l > fl j} = P {IXkI > aj+1}. The formula (35) gives us the estimate P { I Xk I > x} P { IYk I > x}, which holds for every

x > 0. From here and (4)

P{IXkI>/3j}aj+l}
as j --+ oo. Hence Xk does not satisfy the lower estimate in (4). Since Xk satisfies the upper estimate in (4), then according to Theorem 3, the upper estimate in (5) takes place, where E = Lp,, (S2) . We prove the lower estimate. Let's put for fixed n and k E N Zk,n = (Yk - Xk) I{IYkI-Xk}
Let a collection of real numbers {ak}k=1 satisfy (16) and denote by S; (j = 1,2,3) the linear combinations with the coefficients {ak }k=1 of the r.v.s {Xk}, {Zk,n} and {Uk,n} respectively. Then n

Eakyk =S1+S2+S3k=1

Y1. This and the lower estimate in (4)

From (1.20) and (16) S1 + S2 + S3 for the r.v. Y1 imply that

3

ax-q < P{IY1l > x} < EP{1S,I >- 3} i-1

(36)

for x > c. Using (16), (32) and Holder's inequality, we get n

E lakl < n1-1/q k=1

1/q

En ( k=

lakl q

21n4an

l

It follows from the definition that IZk,n I < an, which yields 1S2l < x,,14 and P f1521 > x,,13} = 0. From (35), (33) and (4) we obtain

P{Uk,n#0}=P{IYk-XkI>an}/3n}
.

Rearrangement Invariant Spaces

68

Hence

Pllssl> 31 <EP{Uk,n#0}
These results for S1 and S2 and the estimate (36) yield P {IS,I > xn/3} > ax;-9/2. From here and (1.2), under the condition (16), n

1 a 3 21k

EakXk k=1

P,00

and we get the lower estimate in (5).

9. The estimates for the exponents. Here we show that the conditions of Theorem 5 imply 0 < p < q < 2. The following result is proved in [47].

Lemma 8. Let {Xk}

1 be i.i.d.r.v.s, 0 < p < 2 and E

IX1IP

< co Suppose

EX 1 = 0 if p > 1. Then P

n

E

EXk

= o(n)

(37)

k=1

asn->00. Proof: First let 1 < p < 2 and r.v.s Xk be symmetric. Put for a > 0 Xk,a = XkI{IXkI
Zk,a = Xk - Xk,a.

,

(38)

These r.v.s are symmetric. Since p < 2, then n

E

E Xk,a

2

P

p/2

<E k=1 Xk

k=1

=

a/

From here

n

(EE(Xa))

PI2

< (an)P.

k=1

P

E n-1/P

Xk,a

< apnp/2-1 -> 0.

k=1

According to von Bahr-Esseen's inequality [2], n

E 1: Zk,a

n

P

< 2 E E IZk,aIP = 2nE IZ1,aIP .

k=1

k=1

Using the well-known inequality

EIU+VIP
(39)

3. Linear combinations of independent random variables

69

where b(p) =max 11, 2P-11, we obtain n

P

lim E n-i/P > Xk n-oo

< 2b(p)E I Z1,a I P .

k=1

If a , oo, then E IZi,a lp 0 and we get (37). Now we eliminate the assumption of symmetry. Let's put Yk = X2k X2k_1. Since EX1 = 0 then from Proposition 1.11 P

n

E

P

n

<E

Xk

E Yk

k=1

k=1

which yields (37).

In the the case 0 < p < 1 we have n

E

/ n

P

<E

E Xk,a

P

IXk,aI I k=1

k=1

Asp < 1, then

< (an)P.

/

n

E n- 1/P E Xk,a < aPnP-1

0.

k=1

From (39) n

P

E E Zk,a k=I

n

< E E I Zk,a lp = nE IZi,a lp k=1

and (37) follows.

Lemma 9. If the conditions of Theorem 5 hold, then 0 < p < q < 2.

1/ (p

Proof: As in the proof of Lemma. 1, we get that the estimate (6) holds for the spaces Lp(S2) (p < 1). From here and (1) q < 2. Since q # 2, then we have q < 2. Now we show that p < 2. Suppose p > 2. Applying Theorem 1, we get q > 1. Let q > 1. Then by the same theorem EX, = 0. The lower estimate in (1) and Rosenthal's inequality give us

/n

n

Cinilq < E Xk ko1

C(p) max

EIIXkIIp) k_1

1/p

n

,

k=1

= C(p) max { ni/P IIX1IIp , n1/2 IIXI II2}

which contradicts the strong inequality q < 2.

Rearrangement Invariant Spaces

70

In the case q = 1 we have to apply Rosenthal's inequality to the r.v.s {X2k - X2k_1}' 1, which together with (28) leads to a contradiction.

So, p < 2. Now we prove that p < q. Let, at first, q # 1. If p > 1, then from Theorem 1, q > 1 and EX1 = 0. If p < 1 then b(p) = 1 in (39). Using von Bahr-Esseen's inequality for p > 1 and (39) for p < 1 and applying (1), we get P

n

G+1 P/q <E E Xk

n

< 2 1: E IXk IP = 2nE IXL IP . k=1

k-1

From here p < q. Let q = 1. Considering the r.v.s {X2k - X2k_1}k 1, we obtain as above,

p
Lastly we show that p # q. If p = q > 1, then EX 1 = 0, which follows from Theorem 1 . Using Lemma 8, we get a contradiction to the left hand-side of (1). In the case p = q = 1 we apply Lemma 8 to the r.v.s {X2k - X2k_1}k 1 and obtain a contradiction to (28). Lemma 8 shows that in the case p = q < 1 the lower estimate in (1) is not true.

10. Proof of Theorems 5 and 6. Let the conditions of Theorem 5 be fulfilled. Then Lemma 9 yields 0 < p < q < 2. Lemma 7 gives us the estimates (7) and (12). Applying Lemmas 2 and 3, we get the assertions 2)-4). Theorem 6 follows directly from Lemma. 4.

2. 12-estimates 1. Results. In this section we consider estimates of the types (1) and (5) for q = 2. From Lemma 1, in this case the left-hand side of (5) holds. Hence, we have to study the right-hand side of (5) only, i.e. the estimate n


akXk k=1

1/2

E a2

(k=1

E

where D is a constant independent of n and ak. Theorem 1 and (40) yield EX 1 = 0. Lemma 2.6 and Proposition 1.1 give

us EXi < 00. Putting ak = n-1/2 (1 < k < n), we obtain sup n


n-1/2 E X k k-1

E

It follows from the Central Limit Theorem that the normed sums Sn = n-1/2 Ek_1 Xk are weakly convergent to a, r.v. with the normal distribution.

3. Linear combinations of independent random variables

71

Using Proposition 1.5 we conclude that the space E" contains a r.v. with such a distribution. So, the necessary conditions for (40) to be fulfilled are the following: 1) the space E" contains a r.v. with the normal distribution;

2) EXi < oo ;

3)EX1=0. The question of the sufficiency of these conditions is open. But under some additional assumptions the answer is positive. Let L2(fl) C E. Then 1IXIIE < C IIXII2 for every X E L2(1) and some constant C, which together with 2) and 3) yields (40). If E has von BahrEsseen's 2-property, then (40) also holds. Applying Theorem 2.7, we get the estimate (40) for the spaces E = Lr ,(fl), where p > 2 and q > 2. It was proved in [15] that (40) is true in the case p > 2 and q < 2. The results of the section 2.4 permit us to conclude that the considered estimate holds for the Orlicz spaces LN, (fl), where 0 < p < 2. We introduce a new notation. For a = {ak}k 1 E 12 and a r.v. X we put CO

P { IakX > x} .

Qa(x) =

(41)

k=1

Suppose X E L2(S2). Then P {JX > x} < Cx-2, where C is a constant. So, 00

Qa(x) < li t ak2)

x-2 < 00

A1=1

for allx>0. Definition 1. We say that a r.v. X has the A2(E)-property (X E A2(E)) if for all a E 12 the r.i. space E contains all r.v.s Y such that P { IY I > x} < CQa(x) for every x > 0 and a constant C. It is obvious that X E A2(E) implies X E E. We describe some conditions which yield the A2(E)-property. Suppose that there are positive constants b and c such that p 11XI > xy} < cy-2P {IX I > x} (42) for x > b and all positive y. From (42)

Qa(x)x}. k_1

This and Proposition 1.2 imply that if X E E, then X E A2(E). Let P { IX I > x} = 4(x)x-2 and ¢(x) be decreasing for large enough x. Then (42) holds. It is easy to show that if P {IX (> x} is a function with regular variation at infinity and the exponent p > 2 (see [20]), then the estimate (42) is also true. If E satisfies the upper 2-estimate, then X E E implies that X E A2(E). The main result of this section is the following.

Rearrangement Invariant Spaces

72

Theorem T. Let E have the Kruglov property and {Xk}' 1 be i.i.d.r.v.s, such that X1 E A2(E), EX' < oo and EX1 = 0. Then (40) holds.

2. On infinite divisible distributions. The proof of Theorem 7 is based on the next result which has an independent interest. We recall that a r.v. X has an infinite divisible distribution if for every integer n there are i.i.d. r.v.s {Xk,n}k=1 such that n

X k=1

The corresponding characteristic function is determined by Levy-Khinchin's formula

f(t) = exp (i7i + J

(eat-

oo\

1x2 x2

2

-1-1+

dG(x)/ I

(43)

where y is a real constant and G(x) is a bounded non-decreasing function on R, which is called the Levy-Khinchin spectral function. We may assume G(-oo) = 0. Let's put

FG(x) = GG

(44)

This is a distribution function. We recall X EC(F) denotes that the r.v. X has the distribution F. Lemma 10. Suppose the r.v. X has an infinite divisible distribution, G(x) is the corresponding Levy-Khinchin function and Y E £(FG). Let a r.i. space E have the Kruglov property. Then the conditions X E E and Y E E are equivalent.

Proof. The implication (Y E E)

(X E E): Let's put

Y(t X) = Ceitx - 1 f1 (t) = exp

(i7t +

1

+ X2)

1

+22

j=I<1 IF(x,t)dG(x)fI

f2(t) = exp (JI-1>1 T(x,t)dG(x)f I

(45)

,

,

(46)

(47)

and denote by X1 and X2 independent r.v.s with the characteristic functions f1(t) and f2(t) respectively . Since f (t) = f, (t) f2 (t) then

X

a

Xl -I- X2.

(48)

3. Linear combinations of independent random variables

73

We show that X1 and X2 belong to E. The following statement is proved in [31]. Let V be a r.v. with an infinite divisible distribution. Suppose that the related spectral function is a constant on each of the intervals (-co, -a) and (a, oo), where a is a positive constant. Then

lim - log P { Ivi > X1 xoo xlog(1+x)

1

a Applying this result to the r.v.s X1 and Z with the Poisson distribution and

the parameter A = 1, we get P {IX1I > x} < CP {JZJ > x} for all x > 0, where C is a constant. As mentioned above, if E E K, then Z E E, which yields X1 E E. Now we show that X2 E E. Let's denote b=

JIxI>1

x-1dG(x)

and define the function H(x) by the equalities fxO, l

H(x) =

if -00 < x < -1, if -1 < x < 1,

dG(y)

H(-1)

H(-1)+j

j dG(y) ifx>1.

Putting g(t) = exp

(eitx

(f _

- 1) dH(x) I

(49)

00

we have from (47) f2(t) = g(t) exp(ibt). Let U be a r.v. with the charactera istic function g(t). We have X2 U + b and show U E E. If H(oo) = 0, then U - 0. So, we may assume H(oo) > 0, which implies G(oo) > 0. Let FH(x) be defined by the formula (44), where G will be replaced by H, and Z E .C(FH). First we prove that the conditions Y E E and Z E E are equivalent, which follows from Proposition 1.2 and the next assertion.

Proposition 11. For x > 1

1P{IYI > x} < P {IZI > x} <

H( ) P

11Y,

> x}

Proof: Since 1 < (1 + y2)y-2 < 2 if y > 1, then for x > 1

PIP > x} = H(oo) 2

H(oo)

f

y,>x 1 y2y2dG(x)

dG(x) =

(oo) P {lYI > x} .

(50)

Rearrangement Invariant Spaces

74

In addition,

P {iZI > x} >

H(oo)

I"I>X dG(x) =

H 0PI

> x} .

Now we reduce the proof to the case H(oo) = 1. Let us put for v > 0

C its (e - 1) dH(x)

exp Cv J

and denote by Z a r.v. with this characteristic function.

Proposition 12. For every r.i. space E and v > 0 the conditions Z E E and Z E E are equivalent.

Proof: It is enough to show that Z E E implies Z E E. We choose an integer n so that n < v < it + 1 and suppose that Z E E. Let {Yk }k±i be independent r.v.s equidistributed with Z, X be a r.v. with the characteristic and X and Z be independent. Then function

X+Z

n+1

Yk EE. k=1

From here and Proposition 1.10, Z, E E. We continue to prove Lemma 10. It follows from Y E E that E contains a r.v. Z with the distribution function FH(x). According to Proposition 12,

we may assume H(oo) = 1. But at the same time H(x) = FH(x), which follows from (44). Hence, the characteristic function (49) corresponds to the

distribution H(FH). Since E E K, then E contains a r.v. U E ,C(II(FH)). From here and the above X2 E E. So, X E E. (Y E E): From Proposition 1.10 and the The implication (X E E) relation (48) X2 E E. As shown above, X2 a U + b, where U has the characteristic function (49). Hence U E E. Proposition 12 permits us to assume H(oo) = 1. From here and the Kruglov property Z E E, where Z is a r.v. with the distribution function H(x). Taking into account Proposition 11, we get YEE.El 3. Proof of Theorem 7. It follows from Kolmogorov's theorem about three series (see [35]) that under the conditions of Theorem 7 the series ix 1 akXk converges almost surely if {ak}° 1 E 12. The next statement is well known and has a standard proof.

3. Linear combinations of independent random variables

75

Proposition 13. The estimate (40) is fulfilled if and only if when CO

E akXk E E kc1

for each {ak}k 1 E 12.

Suppose at first that the r.v.s Xk are symmetric. Let F be the corresponding distribution and {Yk}k 1 be independent r.v.s with the common distribution II(F). It follows from (1.6) that Yk are also symmetric. Kruglov's Theorem (see the section 1.6) yields that EYk < oo. Hence, the series F'k 1 akXk E E converges almost surely for every sequence {ak}k 1 E 12. Prokhorov's inequality (see the section 1.4) yields that for all x > 0, ak E R and integers n 00

t akXk

E akYk

ko1

k-1

So, the condition FM 1 akYk E E implies Ex 1 akXk E E. Taking into account Proposition 13, we conclude that the problem is reduced to the

proof that To akYk E E for every sequence {ak}' 1 E 12. 1

The outline of the proof is the following. The distribution of the r.v.s Yk is infinite divisible and, therefore, the distribution of the sum CO

S=

(51)

akYk

k-1

is the same. Denote the corresponding Levy-Khinchin spectral function by H(x). Using the A2(E)-property, we show that E contains a r.v. Y with the distribution FH. Lemma 10 yields S E E. Since the r.v.s Yk are symmetric, the distribution of S depends on lak only. Hence, we may assume that ak > 0 (k = 1,2,...)

, E00ak < 1.

(52)

k-1

The characteristic function g(t) of Yk is real. So, g(t) = exp (100"0 (cos(tx) - 1) dF(x))

.

Levy-Khinchin's formula has the form

g(t) = exp (f

'

00

(cos(tx) - 1) 1 +2 2 dG(x)

/

(53)

Rearrangement Invariant Spaces

76

Comparing these formulae, we obtain

G(x) =

J

: 1+y2 dF(x).

(54)

y

Let Ga(x) be the Levy-Khintchin spectral function of the r.v. aYk. Since ak > 0 and the r.v. akYk has the characteristic function g(akt), the related Levy-Khinchin spectral function is determined by the formula Gak (x) = J-oo l + y2 dG

(55)

ak

Proposition 14. The series 00

H(x) = 1: G.,, (x)

(56)

k=1

is uniformly convergent on R and 00

H(oo) = EG.,.(oo) k=1

Proof: Putting u = y/ak, we obtain from (55) n

n

Gak (co) = k=1

"

°O a2(1+u2)

k=1 J_00

1 + aku2 dG(u)

00

<E k=1 a2 1

oo

(1 + u2) dG(u)

for all integer n. Since cQ

l.. x2dF(x) = EX1 < ()0 then (54) implies that the integral in the right-hand side is finite, which with the inequality 0 < G,,, (x) < Gak (oo) gives us that 00

E Ga,, (oo) < oo. k=1

Thus, the series (55) uniformly converges on R.

For every x > 0 and an integer n we have H(x) > k G.,, (x). Letting x -> oo and then n -> oo, we obtain 00

H(oo) ?

Ga,;(oo) k=1

3. Linear combinations of independent random variables

77

The converse estimate is obvious. We put for a bounded non-decreasing function T(x) on R and x > 0

R(T, x) = (T(co) - T(x)) + (T(-x) - T(-oo)). Proposition 15. Under the condition (52)

(57)

R(H,x)RF, xl k-1

ak lJJ

Proof: Proposition 14 gives us the equality

R(H, x) _ E00R(Ga,, , x). k-1

From (52) 0 < ak < 1. Therefore, we have (ak + y2)/(1 + y2) < 1 and (55) yields R(Gak, x) < R(G, x/ak) . From (54) R(G, x) < R(F, x). The last bounds imply the needed relation. Now we may prove Theorem 7 for symmetric Xk. From (57) R(F, x) < P {IX1J > x}, which together with (41) and Proposition 15 gives us the estimate R(H,x) < Q,,(x), where H(x) is determined by the formula (56). According to the A2(E)-property, E contains a r.v. with the distribution function H(x)/H(co). Since H(x) is the Levy-Khintchin spectral function corresponding to the sum (51), Lemma 10 pemits us to conclude that this sum belongs to E. As mentioned above, this implies (40). We turn to the general case. Put Zk = X2k - X2k_1. Since EX1 = 0, the inequality (1.13) holds. Thus, the estimate (40) for the r.v.s {Xk}0 1 is equivalent to the same one for {Zk}°k° 1. Let µX be the median of the r.v. X. From the well known symmetrization inequalities (see [20]) it follows that

2P{IX1I > x} < P{IZiI > x) < 2P{1X1i > 2 for x > 11uX 1. Thus, the conditions X1 E A2(E) and Z1 E A2(E) are equivalent. So, the general case is reduced to the symmetric one and Theorem 7 follows.

3. Stable random variables with different exponents 1. Results. If {Xk}k 1 are i.i.cl.r.v.s with the symmetric q-stable distribution (0 < q < 2), then the relation (1.21) holds. The question arises

as to how (1.21) changes if the Xk have q-stable distributions with different exponents. Here we give the answer.

Rearrangement Invariant Spaces

78

Let 0 < p < qk < 2 (k E N). We put for a real sequence a =

00

{ak}k=1

laklgk

(a)= E

(58)

-p

k=1 qk

and

,$(a) = inf It > 0 : lI(t-la) < 1}

(59)

.

If'I'(a) < oo, then the function'Y(t-la) is non-decreasing and continuous with respect to t. We have 41(t-'a) oo as t -+ 0 and @(t-la) -} 0 as t -> oo. Hence, the infimum in (59) is attained and laklgk

'y

a)

(Q(a)

(60)

1.

- F- Q(a)gk(gk - P) 00

If the exponents qk are identical, then laklgi

Q(a)

L=1q1-p)

We say, as in [53], that a r.v. X has the standard q-stable distribution if in the formula (1.16) ry = 1.

Theorem 8. Let {Xk}i 1 be independent r.v.s with the standard qk-stable distributions and let p be a fixed number such that 0 < p < qk < 2. Then for every finite collection a = {ak }k=1 of real numbers p

1/p

A(p)Q(a) < (E EakXk

B(P)p(a),

(61)

k=1

where

A(p) _

B(p) =

(7-1(1

- e-2)f(1 +p) sin

(27r-'(l +p1)t(1+p)sin (

1/P 2P )

11

Applying this result we show that the estimate (37) cannot be improved.

Theorem 9. Let b > 0

0 and 0 < p < 2. There exist symmetric

,

i.i.d.r.v.s {Yk}k 1 with finite pth absolute moment and such that n

EI1: YkI k=1

P

bn

3. Linear combinations of independent random variables

79

for all integers n.

2. Proof of Theorem 8. First we prove the right-hand side of (61). We use the following well known formula (see [55]). Let f (t) be the characteristic

function of the r.v. X and E IX IP < oo, where 0 < p < 2. Then E IX IP = C(p) f 0" (1 - Ref (t)) t-P-1dt,

where C(p) = 27r-lr(1 +p)sin(irp/2). The sum S = /3(a)-1 Ek-1 akXk has the characteristic function n

tak

f (t) = exp

qk\ (62)

(_ kk1IQ(a) The last formula implies n

P

E )3(a)-1 > akXk k=1

J00 C1 - exp

= C(p)

Iqk))i_P_'dt.

a C- kn I Q(k)

\

(63)

We break up (0, oo) into the parts (0,1) and [1, oo). The difference in the brackets under the integral sign does not exceed 1. Hence, the integral over [1, oo) has the upper estimate 1/p. For estimating the integral over (0, 1) we

--t

use that 1 - exp(-x) < x for all x > 0. Applying (60), we find that this

j

integral is not greater than

JO

Qtak

(

\k>

I

(a)

Iqk

)tP1dt=kI

qk

Q (a)

7

qk -P

= 1.

So,

n

P

E Q(a)-1 E akXk

< C(p)(1 + p 1),

k=1

which implies the right-hand side of (61).

We turn to the left-hand side. Since 0 < p < qk < 2, it follows that 1/(qk - p) > 1/2. Considering (60), we conclude that n

E

k=1 I Q(a)

qk

ak

tak

< 2 n I

k=1 Q(a) I

I

qk - p

2

Rearrangement Invariant Spaces

80

for 0 < t < 1. It is not hard to verify 1 - exp(-x) > Ax for 0 < x < 2, where A = (1 - exp(-2))/2. Using (60) again, we get n

E #(a) -1

)t_P_[dt

AC(p) Jo

akXk

J,

kcl

I Q(ak) jqk)

_ AC(p) kL,

/I(a) I°k

I

qk

1p =

AC(p)

The last relations lead to the needed inequality.

3. On infinite sums. Here we show that the inequalities (61) are true for infinite sums, a result which will be applied for the proof of Theorem 9.

Lemma 11. Assume the conditions of Theorem 8 and suppose that the sequence a = {ak} 1 is such that *(a) < oo. Then the series Ek l akXk converges almost surely, and the estimate (61) holds for it.

Proof: First of all we show that the series under consideration is weakly convergent. The characteristic function fn(t) of the sum n

a -1 E akXk sn = fl(a) k=1

is determined by the formula (62). Since 0 < p < qk < 2, then Itlak < max 11, t2 j and 1/(qk - p) > 1/2. This, (58) and the assumption 111(a) < 00 imply >k,_1 laktl4k < oo for all t E R. Consequently,

1

f "(t) = exp

-

( kt

9k

to I

k

I

which implies the needed assertion (see [35]). From the assumptions of Theorem 8, ItI°k > ltl' if Itl > 1. Using (60), we obtain k=1 f(t)<exp -itI1'E

a ak I

Na)

9k

= exp (- itlP)

(iii > 1).

So, f (t) is integrable on R and the corresponding distribution is absolutely continuous [36]. According to the well known results of the equivalence of different forms of convergence, the considered series is convergent almost surely [53].

The inequality (61) for the partial sums yields the same estimate for the series.

3. Linear combinations of independent random variables

81

4. Proof of Theorem 9. Let {Xk}k be a sequence of independent 1

standard qk-stable r.v.s and 0 < p < qk < 2. Consider independent r.v.s {X j,k }' k=1 such that

Xk.

Xj,k

(64)

Let a = {ak } 1 , ak > 0 and 'Y(a) < oo. Define 00

Yj =

akXj,k

(65)

k=1

According to Lemma 11 this series is convergent almost surely. The r.v.s {Yj }'1 are symmetric, independent and identically distributed. Lemma 11 and the condition *(a) < oo imply E jYj IP < oo. The numbers ak and qk will be chosen below. From (64) and (1.16)

Xj,k = nllgkXk j=1

Using (65), we get n

n

00

00

Yj = rakEXj,k

j=1

d

j=1

k=1

E ak'nl/gkXk_ k=1

Let

Nn = j3 ({aknllgk J

2 1)

(66)

Lemma 11 leads to the estimate P

n

P

J00

E EYj = E E aknllgkXk j=1

>_ (A(p)(3n)P .

k=1

We write (60) for the sequence {aknllgk}k 1 and get, using (66), 0o

)3n

k=1

Denote

Then h(/8n) = n-1.

_

qk - p

1

n

h(t) = > vkt-gk

ak qk

/

1

00

qk

Vk =

qk

at

F

k=1

(67)

Rearrangement Invariant Spaces

82

The function h(t) is decreasing. So, the estimate (A(p),6,,)P > bn is equivalent to the condition bnlP

1

>

h (A(p)

h(3n)

(69)

n

Thus, the problem is reduced to the following: given a sequence bn, bn/n \ 0 as n -> oo, select numbers ak and qk such that (69) holds. Let En

_

bn (70)

A(p)Pn

According to the conditions of the theorem, En \ 0. Put (71)

Vk = 2(Ek - Ek+1)

p. From (68), (70) and (71)

and let qk

1/p

00

h

A(p)

h I (nen)1/P) = 2 E(ek - Ek+1)(f fn)-gk/P l

k=1

It can be assumed that en < 1 for all n. Since qk \ p, then n-qk/P > .n-q, /P for k > n and En gk1P > C-1 for k > 1. From here and the above, 00

1/p

h

A(p)/ > 2 E(Ek -

Ek+1)(nEn)-qk/P

k-n

00

> 2n-qk/PCn 1 E(Ek - ck+l) =

2n-qk/n.

(72)

k=n

We have used the equality >k n(ek - Ek+1) = En 1, which follows from the

condition en \ 0. We choose qn so that 2n-q+.lp =

1

n

From here

(73)

p(1 + log 2) log(n)

'in

It is clear that qn \,,, p. We define c,, by the formula (70). According to (71) and (68), vk and ak are thereby uniquely determined. By (58) and (71) 00

*(a) = > "0 Vk = 2 E(ek - Ek+1) = E1 1 < 00, k=1

kc1

3. Linear combinations of independent random variables

83

where a = {ak}k 1. The relations (72) and (73) imply (69) and the estimate (A(p)#,,)p > b for all n. Using (67), we get Theorem 9.

4. Equidistributed random variables in exponential Orlicz spaces 1. Results. Let {Uk}k 1 be independent r.v.s. with the symmetric Bernoulli distribution, i.e. P {Uk = 1} = P {Uk = -1} = 1/2. Rodin and

Semenov [48] proved that the norm IIEk= akUkII(p) is equivalent to II{ak}II2

if 1 < p < 2 and to

I I {ak } I Ip, oo

if p > 2 (the definition of the Lorentz

sequence space 1r,, see in the section 1.7). It was mentioned in the section 2 that in the case 1 < p < 2 this relation holds for every sequence of i.i.d.r.v.s {Xk}' 1 C L(p)(S2). But it is not true in the case p > 2 (see Theorem 2.10). Here we investigate the conditions under which such a relation takes place for p > 2, i.e. we study the inequality n

C1 IIaIIp-,,,. <_

< C2 IIaIIp,,.

E akXk k=1

,

(74)

(p)

where a = {ak}k=1 and C1 and C2 are positive and independing of n and a.

Theorem 10. Let p > 2. The lower bound in (74) takes place for every sequence of i.i.d.r.v.s {Xk}k 1 C L(p)(Sl) with mean zero, where C1= B(p) 11X111,

and B(p) is a positive constant.

Let f(t) be the characteristic function of the r.v. X E L(p)(t ). As was mentioned in the section 2.4, f (t) is extended to a entire function. The function

h(t) = log (f (-it)) = log (E exp (tX ))

.

(75)

plays an important role in our considerations.

Theorem 11. Let p > 2, {Xk}k 1 C L(p)(fl) be i.i.d.r.v.s with mean zero and h(t) be the related function determined by the formula (75). The upper estimate in (74) holds if and only if

h(t)t-p'-ldt < oo.

A(h)

(76)

0

In this case C2 = D(p)A(p)1/p, where D(p) is a constant.

We may formulate this result using the distribution of Xk directly. For this we need to recall some definitions.

84

Rearrangement Invariant Spaces

We write f (x) r. g(x) (x -> xo) if there are positive constants a and b and a neighbouhood V of xo such that f (ax) < g(x) < f (bx) for all x E V.

Suppose q(x) is positive function on (0, oo) such that 0(0) = 0 and 4(x)/x , oc as x , oo. Put for real t 0(t) = sup{Itxl - ¢(x) : x > 0} .

(77)

This function is said to be complemented to 4(x) (see [28], Ch.1). It may be easy verified that O(t)/t -> co as t -+ oo and that 0(t) is even and increasing on (0,oo).

Suppose X E L(p)(0) and P{IXI > x} # 0 for all positive x. Then the function 4(x) = - log(P{IXI > x}) is determine on (0,oo) and 4(x) > Cxp for all x > 0 and some positive constant C. Therefore, if p > 1, there exists the corresponding complemented function b(t). Using the inequality ItsI > 4(x)+V(t), which follows directly from (77), one may easily prove the next statement.

Proposition 16. Let P{IX I > x} i4 0 for all positive x and

q(x) =-log (P{IXI > x})

.

Suppose O(x)lx -> oo as x , oo and 1'(t) is the corresponding complemented function. Then h(t) V'(t) as t -+ oo.

So, condition (76) for h(t) is equivalent to the same one for '(t). Now we describe some cases where (76) holds. If lXI < C, then h(t) sze ItI (t oo) (see [34], Ch.2). and (76) is fulfilled. Suppose q5(x) = x9 and q > p. Then ii(t) = c Itl4 , where c is a positive constant. Since q' < p', then (76) holds. Let O(x) = xp log'Y x for x > 1, where -y > 0 is a constant. Then b(t) _ tp' log-y(p'-t fort sufficiently big (see [28], Ch.l) and (76) is equivalent to

the condition y > l/(p' - 1). Using Kwapien-Richlick's inequality (see the section 1.4), Theorem 2.10 and Theorem 10 we conclude that for every sequence of i.i.d.r.v.s {Xk}k 1 C L(p)(Q) (p > 2) with mean zero the estimate n

< D2 Ilallp,

Dl IIaII,,,. < E akXk k=1

(p)

holds. Here positive D1 and D2 are independent of n and ak.

2. Proof of Theorem 10. First we establish some auxiliary statemants.

3. Linear combinations of independent random variables

85

Proposition 17. Suppose that E exp (tX) < exp (cip") for all t > 0. Then Cp_'

P{X>x}<exp (-b(p

)

for all x > 0, where b(p) = (p' - 1)/(p')p

Proof: We have E exp (tX) > exp(tx)P {X > x} for all x > 0, which yields P {X > x} < exp (cip, - tx) . Taking the infimum of the right-hand side with respect to t we obtain the desired bound. The following statement may easily be proved.

Proposition 18. Suppose P {X > x} < exp (-Bxp) for all x > 0 and

EX =0. Then IIXII(p),0 asB,oo. Let's denote by hX(t) the function determined by the formula (75). The next proposition follows directly from the previous ones.

Proposition 19. Let p > 2. The following relation holds: d(p) = inf {sup hX(t) t#0

:

I

IXI

I

(P)

= 1, EX = 0} > 0.

ItI

Now we may prove Theorem 10. Without loss of generality we may assume akXk II(p) = 1. Lemma 2.5 and Xk to be symmetric, IIXk II(p) = 1 and (75) yield IIEk=1

h(tak) < C(p) min {t2 ,

I tI p } .

(78)

k=1

Fixing j < n and putting t = s/a we obtain

Eh (s*ai)
not less then jh(s), which imlpies that jh(s) (ap)p < C(p)sp'. Proposition 19 gives us IIaIIp,,. <-

C(p)

1/p

Cd(p))

So, the lower estimate in (74) follows with the needed constant.

Rearrangement Invariant Spaces

86

3. Proof of Theorem 11. As above we may assume Xk to be symmetric.

n Put for a ={ak}k=1

n

Ph(a) = sup it I-1 t;60

lip'

\E k=1

h(akt) I

(79)

f

Since p > 2, then p' < 2. Using Lemma 2.9 we conclude that Ph (a) < oo for each finite a. Since Xi is symmetric, then h(t) is even. From here ph({ak}) = P({Iakl})

,

Ph(va) = IVIPh(a),

(80)

where v E R. As was mentioned above h(t) is increasing on (0, oo), which yields that if lad < Ibk I , k = 1,... , n, then ph(a) < ph(b).

Lemma 12. Let p > 2. There are positive constants C = C(p) such that C-1 Ph (a) < EakXk11 k=1


for each sequence {Xk}k i C L(p)(1) of symmetric i.i.d.r.v.s and all a = {ak}k=1

Proof: From (79) Ek=1 h(aki) < (ItI ph(a))p . This and Lemma 2.7 imply the upper bound. Turning to the lower one. Assume that IIEk=1 akXkII(p) = 1. Then (78) takes place, which yields ph(a) < C(p)i/p'. Taking into account (80), we get the needed estimate and complete the proof.

Lemma 13. Let h(t) be an even positive function on R, increasing on (0, oo). Let r > 0 and Ph be determined by the formula (79), where the exponent p' be substituted by r. Then the following conditions are equivalent:

(i) ph(a) < CIIaIlr,oo for all finite sequences a, where C is a constant independent on a;

(ii) B(h) _ fo h(y)y-'-1dy < co. If (ii) holds, then C =

(2rB(h))1"r

Proof: (i) = (ii). Let t > 0 and ak = n-1/r, k = 1, ... , n.

Then IIaIIr,oo = 1 and C' > ph(a) > t-' Ek=1 h(ik-1/r). Since h(t) is increasing for positive t, we get n

n

Eh(tk-11r) > 1 Ik k=1

k=1

n+1

k+1

h(ts-'/r)ds

=

J

h(ts-11r)ds. 1

3. Linear combinations of independent random variables

87

Putting y = ts-11', we get for all positive t and integers n t

Cr >

rh(y)y r-ldy,

J(n+l)-1/r

which yields (ii). (ii) (i). Without loss of generality jjajjr

, = 1, which gives ak <

k-'1'', k = 1, ... , n. If k < s < k + 1, then k-1/'' > s-l/'' > (k + 1)-1/r > (2k)'1/'. From here, as above, n+1

h(tak) < f

S(t)

h(21/rts-l"r)ds

k=1

for all t > 0. Substituting variables by letting y = 21/rts-l/,, we get S(t) < 2rtrB(h), which implies (i), where C = (2rB(h))1/r.

Theorem 11 follows from Lemmas 12 and 13.

CHAPTER IV

COMPLEMENTABILITY OF SUBSPACES GENERATED BY INDEPENDENT RANDOM VARIABLES

1. Subspaces generated by bounded random variables 1. Introduction and result. We recall that a subspace A of a Banach space B is said to be complemented if there is a bounded linear operator T : B -f A such that Tx = x for every x E A. The operator T is called a projector. The closure subspace in B generated by elements {xk}k1 C B, is denoted by span {xk}k=1 In this chapter we suppose that the probability space is the segment [0, 1] with Lebesgue measure, i.e. all considered r.v.s are determined on this segment. Applying the well known theorems about isomorphism of the measure spaces, one may pass to the general case. The main result of this section is the following.

Theorem 1. Let E be a r.i. space and {Xk}°k° 1 be independent bounded r.v.s such that 0 < inf IlXk - EXk I I1 k

SUP IIXk

- EXk I( < oo.

Then the subspace span {Xk }k1 C E is complemented if and only if each of the spaces E' and E" contains a r.v. with the normal distribution.

For the Rademacher system this result has been proved in [33] and [49]. First we show that, without loss of generality, we may assume EXk =

0 (k E N). Put Yk = Xk - EXk and L = span {{Xk}k 1 U {I(o,1)}} C E.

The subspaces span {Xk}k 1 and span {Yk}k1 belong to L and have codimension not greater than 1. So, these subspaces are simultaneously complemented or not.

2. Proof of Theorem 1. Sufficiency. Since EXk = 0, then (1) yields 0 < inf { EXk } < sup { EXk } < c,0' k

which implies the r.v.s

that EXk=1forallkEN.

k (EXk)-112Xk

also satisfy (1). So, we may assume

4. Complementability of subspaces

89

(X, Y) = E(XY).

(2)

Denote We have

(X7, Xk) = bj,k, (3) which yields that the operator Ek=1 (X , Xk) Xk is a projector on the subspace span {Xk}k=1. First we show the uniform boundedness of the progectors {Q.}°0 n=1'

It is easy to verify that from (1) the conditions for Paley-Zygmund's

and Bernstein's inequalities follow (see the section 1.4). As mentioned in the section 1.1, if X is bounded, then IIXIIE = IIXIIEII. Since each of the spaces

E' and E" contains a r.v. with the normal distribution, there are positive constants C and C1 such that

(n

1/2

C-1 E ak I

(a)E

J

k_1

1/2

n

<

ko l (4) 1/2

n

C1 -1 E bk

E akXk k=1

k=1

1/2

k=n

< C1 EJ

(E

bk

1

for all real ak and bk and all integers n. From here and (3)

E ak n

C E

1/2

k=1

= Csup

l

\kakXk' =1

E k=l bkXk/

.

kbk < 1

< C { (E akXk, Y) Y E span {Xk}k=1 , IIYIIE' < C1 J

/

l \k=1

Since

(X,QnY), then Y E span {Xk}k=1

IIQnXIIE < CC1 sup

If Y E span {Xk}k=1, then

,

IIYIIE' < 1}.

Y and we have

IIQnXIIE < CC1 sup {(X,Y) : Y E E', IIYIIE, < 1} < CC1 IIXIIE

Let's put

QX = > (X, Xk) Xk. k=1

(5)

90

Rearrangement Invariant Spaces

We show that the series converges in E for each X E E. According to (4) and (5) 1/2

E (X, Xk)2

C-1

(k="

<_ IIQnXIIE < CC1 IIXIIE

1

Hence EL 1 (X,Xk)2 < C4C2 IIXIIE < co. Using (4) once more, we get for all m < n n

n

E (X,Xk) Xk k=m


((x,xk)2)

1/2

k=m

From here and the above, the convergence of the series follows. So, the operator Q is defined on E. It is obvious that Q is a projector on the subspace span {Xk}k1. According to (5) IIQXIIE = liM IIQnXIIE< CC1 IIXIIE

ri

and sufficiency is proved.

3. Outline proof of the necessity. The proof of the necessity is very bulky and therefore we describe it in outline. First we reduce the proof to the case of separable E and symmetric Xk. In such a space the set of all finitevalued r.v.s is dense, which permits us to assume that Xk is finite-valued. For every collection {Xk }k=1 (n E N) we construct an equidistributed collection {Xk}k=1 such that each set of the type {Xk = cont.} is a finite union of mutually disjoint intervals. Then applying the assumption of complementability of the subspace span {Xk}k 1, we show that there is a sequence of uniformly bounded projectors on the subspaces span {Xk }k=1 (n E N).

The simple structure of the r.v.s Xk permits us to determine for each n a group of measure-preserving transformations 0 on the interval [0, 1] such that each operator TX (t) = X (¢(t)) maps the subspace span {Xk }k=1 into itself. Averaging, we construct, as in [49], the projector Qn : E -* span {Xk}k-1

commuting with the action of the group. The projectors Q,, are uniformly bounded, which permits us to construct a system of uniformly bounded projectors on the subspaces span{rk}k=1 (n E N), where rk(t) = sign (sin (2kirt)) are Rademacher functions. Applying the results of [33] and [49], we complete the proof.

4. Symmetrization. The next statement reduces our consideration to the case of symmetric r.v.s.

4. Complementability of subspaces

91

Lemma 1. Let an r.i. space E be separable or maximal and {Xk}k1 C E be independent r.v.s. There are independent symmetric r.v.s {Yk}k1 such that 1) the subspaces span {Xk}k 1 C E and span {Yk}k 1 are isomorphic; 2) if the subspace span {Xk }k 1 is complemented in E, then the subspace span {Yk}oko 1 has the same property;

3) if the sequence {Xk}k 1 satisfies (1), then {Yk}k1 also satisfies (1); 4) if {Xk}k1 are equidistributed r.v.s, then so are {Yk}k 1.

Proof: Let {Xk}k be an independent copy of the sequence {Xk}k1 and 1

Yk = Xk - Xk. As mentioned above, we may assume EXk = 0. We put L = span {Xk}k 1 , N = span {Yk}k1 and define the operator A from L to N by the formula AXk = Yk (k = 1, 2.... ). According to Proposition 1.11 and the remark following it, IIXIIE C II AX IIE :S 2 IIXIIE

(6)

for every X E L. Hence, A is an isomorphism. Suppose L is complemented in E and denote the corresponding projector

by Q. Let fl be the u-algebra generated by the r.v.s {Xk}k1. Since E is separable or maximal, then the operator Ep of the conditional expectation is bounded in E and IIE16 1IE-.E = 1. Putting Q1 = AQEA we have EOYk = EP(Xk - Xk) = Xk - EXk = Xk, which implies that QIYk = AQ1Xk = AXk = Yk. Taking into account (6), we conclude that Q is a bounded projector on N. The third assertion of the lemma follows from Proposition 1.11 applied to L1(0,1) and Lo,(0,1). The last assertion is obvious.

5. Reduction to separable spaces. Let's recall that Eo is the closure of Lo, (0, 1) in the r.i. space E. If E # Lo, (0, 1), then Eo is a separable r.i. space (see [48]). From the conditions of Theorem 1, Xk E Eo. The complementability of span {Xk }L 1 in E implies the same in E0. Since (Eo)' = E, then, if E # Lo,(0,1), we may assume E to be separable. The next assertion permits to us to eliminate this assumption.

Proposition 1. Each infinite diinentional subspace generated in L".(0, 1) by independent r.v.s is non-complemented.

Proof: Let {Xk}L 1 C Lo,(0,1) be independent r.v.s. According to Lemma 1 we may assume Xk to be symmetric. In addition, without loss of generality, IIXk Iloo = 1

(7)

for all k E N. First we show that the subspace span {Xk}k 1 C Lo,(0,1) is isomorphic to the space 11. Consider a collection {ak }k=1 of real numbers

Rearrangement Invariant Spaces

92

and put ck = sign(ak). Denoting the Lebesgue measure by p, we have

k=1

Hence, the strong inequality n

11 t

E akXk > 2EIakIJ>ta{ekXk> 2 (10 k=1

k=1

J

111

111

holds and we get n

1

E akXk k=1

Besides,

n

n

>> .

Iak I

00

n

n

< EIakI IIXkII. = EIak1.

E akXk k=1

.

k=1

00

k=1

k=1

These estimates imply the needed statement. It was proved in [41], that the spaces L00(0,1) and 100 are isomorphic. According to [42] each complemented infinite dimentional subspace in 100 is not separable. Since the space 11 is separable, the considered subspace span {Xk }k 1 is non-complemented in L00(0,1).

6. The construction of invariant projections. Let n be fixed and {Xk}k_1 be independent symmetric finite-valued r.v.s. Denote the positive values of the r.v. Xk in increasing order by a(k)(vk) (2 < vk < mk). We put

a(k)(±1) = 0

,

a(')(-vk) _ -a(k)(vk) m = {fl,... ,±m}.

Let's denote a(k)(±l) = 2p {Xk = 0} and

a(') (vk) _ {Xk = a(k)(vk)}

(vk E em,, , vk

±1)

(8)

Since Xk is symmetric, then

a(k)(-vk) = a(k)(vk)

(9)

The collection of integral n-dimensional vectors

vkE(1
(10)

4. Complementability of subspaces

93

Proposition 2. There are mutually disjoint intervals 0(v) (v E Un) such that 1) {t(A(v)) = l lk_1 a(') (vk) , v = (v1, ..., vn); 2) UvEU, 0(v) = (0,1).

Proof: From the definition

1: a(k)(vk)

=1

(1 < k < n),

(11)

vkEOmk

which implies n

n

II a(') (vk) _ H E a(k) (vk) vEUn k=1

k=1

= 1.

(vkEe-,

So, there exist intervals with the needed properties. We determine the r.v. Xk by the equality Xk(t) = a(k) (vk)

,

t E 0(v) , v = (vl,... ,vn).

(12)

The next assertion easily follows from the previous.

Proposition 3. The r.v.s {Xk}k_1 are independent and symmetric, and a Xk Xk (1 < k < n). Now we reduce the problem to a study of the subspaces generated by the r.v.s .9k.

Proposition 4. Let {Xk}k=1 and {Xk}k=1 be equidistributed collections of finite-valued r.v.s. Then there is an inverse measure-preserving transformation Y'n [0,1] -+ [0,1] such that Xk (n (t)) = Xk (t) (1 < k < n) . :

Proof: Without loss of generality we may assume the r.v.s Xk and Xk to be Borel measurable. We denote the values taken by Xk by b(k) (jk) (1 < jk < sk). Let Vn be the collection of all integral vectors of the type j =

(j',...,jn)(1<jk<Sk,1
el={Xk=b(k)(ik),(1
Rearrangement Invariant Spaces

94 a

Xk, then peg = peg. From here and the well known theorem about isomorphism of o-algebras the existence of an isomorphism h3 13j -+ /3j Xk

:

follows for each j E Vn

.

The sets ej are mutually disjoint and their union is a set of full measure and the sets ej are the same. So, the isomorphisms h, generate the isomorphism hn : (3 -> ,Q. It is well known that such an isomorphism is induced by a measure-preserving transformation Ybn : [0, 1] -> [0, 1].

According to the construction, iin(ej) = ej for all j E Vn, which implies the needed relations.

Proposition 5. Let the condition of Proposition 4 hold, E be a r.i. space and Qn E -+ span {Xk}ko1 be a projector. Then there exist a projector Qn : E -+ span {Xk}k_1 such that IIQ-IIE-E <

IIQnhIE-E'

Proof: We put (T, X) (t) = X (On (t)) and Qn

It is not hard to verify that Qn is the needed projector. Denote by Vn the group of all linear operators in Rn changing the signs of some coordinates of vectors. More precisely, if g E Vn, then there is a collection of numbers ek(g) = ±1 such that g(v) = (e (9)vl,...,en(9)vn),

where v = (vl, ..., vn). It follows from (10) and the definition of On, that g(Un) = Un for all g E Vn. According to (9) and Proposition 2, pO(g(v)) _

(v) for all v E U, andgEVn. Let O(v) = and 0(g(v)) = (-y, b). Put

t _ t+(y-a) iftE(a,Q)' t + (a - -) 'iftE(^y,a)'

() 13

This equality and Proposition 2 imply that ¢g(t) is an inverse measurepreserving transformation on [0, 1]. Moreover,

0(9(v))

(14)

which yields

We put for9EVn T(g)X(t) = X (¢g(t)) a

.

Since T(g)X X, then T(g) acts isometrically in every r.i. space. The mapping g -+ T(g) is the representation of the group Vn in such a space. We show that the subspace span {Xk }k-1 is invariant with respect to the representation T(g).

4. Complementability of subspaces

95

Proposition 6. The equalities T(g)Xk = Ek(g)Xk are valid.

Proof: Let v E U,,. According to (14), T(g)Xk

= Xkl

o(B(v))

From the definition, a(k)(-vk) = -a(k)(vk). This and (12) imply XkI

= a(k)(ek(g)vk) = fk(g)a(k)(vk) = ck(g)Xk

0(v) Hence,

T(g)Xkln(v)

=ek(g)XklA(v).

Using Proposition 2 we get the needed equality.

The next step is to construct an invariant projector on the subspace span {Xk}k=1 C E.

Lemma 2. Let P . E --> span {Xk }k=1 be a projector. There is a collection { Yk } C E' such that 1) (Xj , Rk) = bj,k; 2) the r.v.s Yk are constant on each interval A(v), v E Un; 3) the projector k=1

n

QnX =E(X ,Yk) Xk

(15)

k=1

commutes with every operator T(9), g E Vn; 4) 11QnhIE-.E

11PfhIE-E'

Proof: Since E is separable, then E' = E*, where E* is the conjugate space. Using Proposition 1.14, we conclude that there is a collection {Zk}k=1 C E' such that

(X, Zk)Xk. k=1

Let f3 be the a-algebra generated by all the intervals 0(v) , v E U. According to the definition, the r.v.s Xk (1 < k < n) are ft-measurable. Hence, is a projector on the subspace EpXk = Xk and the operator P,, = It is easy to verify that (EI3X, Y) = (X, EQY). From here span {Xk }k=1.

n

(X,EQZk)Xk. k=1

96

Rearrangement Invariant Spaces

It is clear that each r.v. Zk = DOZk is a constant on 0(v), v E U. Since E is separable, the operator Ep is bounded on E and has the norm 1 (see Proposition 1.4). Hence, V-11._. < 11P-11E-E- We Put Q"X = 2-n

(T(9)P"T(9-1)) X. 9EVa

As mentioned above, T(g) acts isometrically in E. Since the group V" consists of 2" elements, then the estimate IIQ"IIE.E < I1PnIIE-E holds. It is

not hard to verify that T(g)Q" = Q"T(g) for each g E V". Proposition 6 permits us to conclude that Q, is a projector on span {Xk)k=1 Since

(T (g-1) X, Y) = (X, T(g)Y)

(16)

for every g E V", then using Proposition 6, we get

1`

Q"X = k=1

\2-" gEV°

(X, T(g)Zk) Ek(9)) Xk.

/

The r.v.s T(g)2k are constants on each of the segments .(v). Put

Yk = 2-" E fk(g)T(g)Zk. gEV"

Then we get (15) and the statements 2) and 4) of the lemma. Since Q" is a projector, then 1) holds.

It is obvious that g-1 = g for all g E V", which yields that T(g-1) _ T(g) = (T(g))-1. This, (16), Proposition 6 and the permutability of Q" and the operators T(g) imply that T(g)Yk = Ek(9)Yk

(1 < k < n, g E V").

(17)

7. Reduction to the Rademacher system. For v = (vi, ... v") and ,

t E A(v) we put if

00,

Yk

0(v)

rk(t) = sign (vk)

(18)

= 0.

if Yk IA(V)

This definition is valid because Yk is constant on each interval 0(v).

4. Complementability of subspaces

97

Proposition 7. Forevery gEV' andk=1,...,n T(9)T'k = ck(9)r'k

Proof: We have = rk

(T(9)''k) IA(V)

According to (17)

= (T(9)kk)

Ykl

I

o(v)

o(s(v))

=

ck(9)Yklo(v).

From here and (18) the needed equality follows.

Proposition 8. The r.v.s {Tk}kc1 are symmetric and independent.

Proof: Let k, 1 < k < n, be fixed and g E V' be such that ck (g) _ -1. From Proposition 7, T(g)Tk = -Tk. Since 09 is a measure-preserving transformation, it follows from the definition of T(g) that d

rk= - rk i.e. Tk is symmetric.

Let e(g) = {rk = ck(g) , 1 < k < n}, where g E V. The sets e(g), g E V", are mutually disjoint and their union is (0, 1). From Proposition 7 C(991) = {'rk = fk(9)ek(91) , 1 < k < n}

= {T(9)1'k = ck(91), 1 < k < n} _ 91(e(gi))

This relation implies pe(g) = pe(g') for all 9, g' E V". Since the group V" consists of 2' elements, then pe(g) = 2-'". By virtue of symmetry, p {Tk = fk(g)} = 1/2. Each collection of numbers ck = ±1 (1 < k < n) determines an element g E Vn. So, n

p {Tk = ck , 1 < k < n} = pe(g) = 2-n = II P {Tk = ck) . k=1

Hence, the r.v.s {Tk}k-1 are independent.

Proposition 9. The following equalities are valid:

(i;k,ii)=Si,kEIYkI

(1 <j,k
Rearrangement Invariant Spaces

98

Proof: Let j # k and choose g E Vn so that cg(g) = 1 and ck(g) = -1. As mentioned above, T(g-1) = T(g). Using (16), (17) and Proposition 7, we obtain

(rk, V, = (Ek (9)T (9)rk , k.1) = Ek (9) (rk, T (9)Yj ) = Ek (9) El (9) (rk , Yj) = - (rk , Y, )

Thus, for j # k the equality is proved. According to Lemma 2, the r.v. Yk is a constant on each interval 0(v). Using (18), we get

(rk,YJ) =

µA(v) = E IYk I . O

Yk

vE U

We put

QnX=E(X,Yk)pk

(19)

k=1 (E IYk ()

Proposition 9 implies On is a projector on the subspace span {7'k}k=1 C E. We estimate now the norm On .

We recall X Y denotes that t

X*(s)ds < 10

rt Y*(s)ds 0

for each t E (0,1).

Proposition 10. Let {Xk}ki 1 be independent symmetric r.v.s such that E IXk I = a > 0 for all k E N and {rk } n

1 be the Rademacher system. Then n

Eakrk

EakXk

k=1

k=1

for all ak E R and n E N. Proof: Suppose first that p {Xk = 0} = 0 for all k. We put hk = {Xk > 0} and hk = {Xk < 0}. Let /3k be the o-algebra generated by {hk , hk } and let

rk = a-1EAkXk. a

Then the r.v.s {rk}k=1 are independent and rk rk. Let /3(n) be the o-algebra generated by Uk=1,(3k . By virtue of independence (see [35])

4. Complementability of subspaces

EokXk

99

(1 < k < n).

Therefore n

n

n

E ak rk = a-1 E ak EPk Xk = - ' E # " ' 1: akXk . 1

k=1

k=1

k=1

The operator EO(") acts in L1(0, 1) and L.(0,1) with the norm 1. Hence E' ' X -.< X (see [29]), which implies the desired relation. Consider the general case. Each symmetric r.v. Xk may be uniformly approximated by symmetric r.v.s Xk,n such that p {Xk,n = 0} = 0. We may choose {Xk n}i 1 to be independent for each integer n. From here and the above, Proposition 10 follows.

Proposition 11. Let E be a separable r.i. space. Then

IIQnIIE-+E -

minnl
Proof: The collections {rk}k=1 and {rk}k=1 are equidistributed. From Propositions 8, 10 and 1.3 n

n

E

E akrk kol

E

IIXkII1Xk

for every ak E R. This and (19) give us

IIQnXIIE:

(X Yk ) k=1

(E IYkI) IIXkI1 Xk

Applying Propositions 1.14 and 1.15, we obtain

IIQnXIIE

`

(mini
IIlIIXkII)

( kmnl k<

We have 1 = (Xk, Yk) G E IYkI IIXk II0. Hence, 1

C1mm

E IYkI I

G 1max IIXkIIoo

The last relations and (15) imply the needed estimate. The next statement was proved in [33] and [49].

EIYkI)

Rearrangement Invariant Spaces

100

Lemma 3. Let E be a separable r.i. space. Suppose there are uniformly bounded projectors P : E -> span {rk}k=1 (n E N). Then each of the spaces E' and E" contains a r.v. with the normal distribution. First we establish two auxiliary statements. Let 00

CO

t=E

ak2-k

S = L:

,

k=1

bk2-k

(ak, bk = 0; 1)

k=1

be the dyadic decomposition oft , s E (0, 1). Let's put eo // t 'F s = E 2-k ((ak + bk) mod 2)

k=1

and consider the transformation Ot(s) = t -F s, where s, t E [0, 1]. It is easy to verify that 4t is an inverse measure-preserving transformation on [0, 1].

We putA+ ={rk=1} andA-={rk=-1}. Proposition 12. For each t E Ak

qt(A+)=Ak

,

Ot(A;)=Ak.

Ot(Ak) = Ax

,

Ot(Ak) = A+

If t E Ak, then Proof: Let t E [0, 1] and ak be the coefficients of the dyadic decomposition oft. It is easy to verify that the conditions t E Ak and ak = 0 (respectively, t E Ak and ak = 1) are equivalent. This and the definition of the operation + yield the needed equalities. p

We put (S(t)X)(s) = X(Ot(s)) (0 < t < 1) and have S(t1 -j- t2) _ S(t1)S(t2). The equalities

S(t)rk = rk (t E A,)

,

S(t)rk = -rk (t E Ak)

(20)

follows from Proposition 12.

Proposition 13. Let Y be a r.v. such that S(t)Y = Y (t E Ak) and S(t)y = -Y (t E Ak ). Then Y = crk for some constant c.

Proof: Suppose t E Ak . We have Y(t -F v) = Y(v) for every v E [0,1]. Proposition 12 implies {t + v : t E Ak } (Ak) = At for fixed v E Ak . Hence, Y is a constant on At and, similarly, on Ak . Let t E Ak . Then Y(t + v) = -Y(v) (0 < v < 1). According to Proposition 12, t -j- v E Ak if v E Ak . Hence Y

I

= -Y Ak I

Ak

b. Complementability of subspaces

101

From here Y = crk, where c is a constant.

Proof of Lemma 3.: We prove that for the Rademacher system and the spaces E and E' the estimates (4) hold. From here Lemma 3 follows (see the section 3.2). For all t E [0, 1] and n E N the operator S(t) maps the subspace {rk}k=1 into itself, which follows from (20). This operator acts isometrically on E. We put r

J0

Xdt.

1

Then Qn is a projector on the subspace span {rk}k=1 commuting with each operator S(t) (0 < t < 1). In addition IIQnIIE-+E < IIPIIE_E. Since E is separable, there are r.v.s {Yk,n}k=1 C E' such that n

QnX = E (X

,

Yk,n) rk

k=1

(see the proof of Lemma 2). By virtue of the permutability of Qn and S(t) and the formulae (20), the r.v.s Yk n satisfy the conditions of Proposition 13. Hence Yk,n = ck,nrk and ck,n is a constant. As Qn is a projector on the subspace span {rk}k=1, then (rk , Yk,n) = 1 and ck,n = 1 (1 < k < n; n E N). So, QnX = Ek=1 (X, rk) rk. From the assumption of the lemma C = sup IIQnIIE-E < IIPnIIE-.E < oo .

(21)

n

Y) = (X,

It is not hard to verify that

Since E' = E*, we

have

IIQfXIIE = sup {MIX , Y) : Y E E', IIYIIE' < 1} . These relations give us I IQn I IE'-.E' = IIQn IIE_E . Using (21), we get n

Q. E akrk E

k=1

E

=sup((Eakrk,QnY) :YEE',IIYIIE,<11 Illlll \k=1 /

n 1(

< suP C

Il

(

akrk ,

Z)

: Z E span {rk}k=1

,

IIZIIE- < C

k=1

According to Lemma 3.1, there are positive constants a and b such that for every ak, bk E R and n E N

/n

n

a k=1

E

\1/2

(Ea2) k=1

(22)

102

Rearrangement Invariant Spaces n

1/2

>b E b2 k=1

(23)

k-1

E

From here and the above G slip

l

\kakrk k>bkrk) Otirk n (/n 11/2

< sup E akbk

:

k=1

< b-1C

1/2

n

= b-1C

E b2 I k=1 J

E'

E

a2

k=1

Similarly

a- 1C k=1

El

(

n

1/2

k=1

Therefore, for the sequence Irk l',1 and the spaces E and E' the estimates (4) are fulfilled. Reasoning as in the section 3.2 and using the relation (E')" = E' we get that the spaces E' and E" contain a r.v. with the normal distribution.

8. Proof of Theorem 1. Necessity. We may suppose the r.v.s Xk to be symmetric and the r.i. space E to be separable. We denote by P : E -* span {Xk }' 1 a bounded projector and put RnXk = Xk for 1 < k < n and RnXk = 0 for k > n. Propositions 1.14 and 1.15 yield that the operator Rn acts in the subspace span {Xk}k 1 C E with the norm 1. Hence, for the operator Pn = Rn P and all n E N we have IIPfhIE_E < IIPIIE.E

(24)

We reduce the proof to the case of finite-valued random variables.

Proposition 14. Let {Xk }k=1 be independent bounded symmetric r.v.s and P : E -> span {Xk }k=1 be a projector. Then for every e > 0 there exists a collection of independent symmetric finite-valued r.v.s {Xk,e}k_1 with the following properties:

1)IIXk-Xk,eIIE<e(1 span {Xk e}k=1 such that

(1 - e) IIPIIE.E

IIP°IIE-YE

(1 + e) IIPIIE-.E

4. Complementability of subspaces

103

Proof: Since the r.v.s Xk are bounded, we may construct for every integer m a collection of independent finite-valued symmetric r.v.s {Xk,m}k_1 such that

IIXk-Xk,mll,,. <m-1 (1 oo. We define the operator Jm : span {Xk m}k .1 -> span {Xk}k=1 by the equality JmXk,,,, = PXk,,,,. For large enough m the operator Jm is invertible

andIiJ,,,Il,1and IIJ,;,111-p1as m-+oo. We putPm=J,;,1P. Then P.. is a projector on span {Xk,,,,}k .1 and IIPIIE_E IIJmII-1 _< IIPmIIE_E < IIPIIE.E IIJm1I1

'

Choosing m sufficiently large and putting Xk,, = Xk,m, we get the needed assertion. We continue to prove Theorem 1. According to Proposition 14 there exist collections of finite-valued symmetric independent r.v.s {Xk n}k_1 (n E N) with the following properties: 1)IIXk,nll. < 211XkIIO. (1 < k < n)

(1
II

PIIE-E <-

2IIPIIE-.E

Let the r.v.s {Xk,n}k_1 be constructed by the formula (12) and the r.v.s Consider the r.v.s {rk}k=1 determined by the formula (18). Ack cording to Lemma 2 and Proposition 10 there is a projector span {7k}k_1 such that {Xk

IIQn IIE-.E

C m n1
VILE

Xk,n, then the assumptions of Theorem 1, the estimate (24) and the properties 1)-3) imply Since Xk,n

IIQn

IIE-+E

< 8maxl
According to (18) and Proposition 8, the collections {rk}k.1 and {rk}k-1 are equidistributed. Proposition 5 yields that there is a sequence of uniformly bounded projectors from E on the subspaces span {rk}k_1 (n E N). Applying Lemma 3 we obtain Theorem 1.

Rearrangement Invariant Spaces

104

2. Subspaces generated by independent equidistributed random variables 1. Result. In this section we consider i.i.d.r.v.s {Xk}k 1 C E and study the conditions of complementability of the corresponding subspace.

Theorem 2. Let E be a separable r.i. space and {Xk }k 1 C E be i.i.d.r.v.s. The subspace span {Xk }°k° 1 C E is complemented if and only if the following hold:

1) each of the spaces E' and E" contains a r.v. with the normal distribution; 2) the considered subspace is isomorphic to 12

We examine more explicitly the condition 2). We may assume that EX1 = 0. According to Proposition 1.14 the sequence {Xk}k 1 is the unconditional

basis in the subspace span {Xk} ° 1 C E. From Theorem 1.4 {Xk}k 1 is equivalent to the orthogonal basis in Hilbert space. Hence, if EX1 = 0, then 2) is equivalent to the estimate which has been considered in the section 3.2. 2.

Proof of Theorem 2. Sufficiency. Without loss of generality

EX1 = 0. We denote hk = {Xk > 0} and hk = {Xk < 0}. Each of these sets has a positive measure. We put

Yk = Ihk - alh- ,

(25)

where a > 0 is chosen under the condition EYk = 0. Then the r.v.s {Yk}k_1 are independent and equidistributed,

E(X,Yk) = 0 (j f k)

,

E(XkYk) = b > 0,

(26)

where b is independent of k. The sequence {Yk }k 1 and the space E' satisfy the conditions of Theorem 1. Hence the estimates (4) holds, i.e. there is a positive constant B such that for every bk E E and n E N

E<

/n

B-t(b2) k-1

1/2

n


EbkYk k=1

1/2

n

E'

bk I k=1

(27)

1

We pu t 00

QX = E(X,Yk)Xk kc1

Reasoning as in the proof of Theorem 1 and using (26) and (27), we get that Q is a bounded projector on the considered subspace.

4. Complementability of subspaces

105

3. The construction of the special projectors. We turn to the proof of necessity. According to Lemma 1, we may assume the r.v.s Xk to be symmetric.

The scheme of the proof is the following. First we construct bounded symmetric i.i.d.r.v.s {Vk}' 1 such that there exists a sequence of uniformly

bounded projectors from E on the subspaces span{Vk}k_1 (n E N). As above, this yields the condition 1). Applying 1), we obtain the condition 2). Consider first a collection {Xk}kc1 of symmetric finite-valued i.i.d.r.v.s. Denote the group of all linear operators on R, which rearrange the coordinates of vectors and change their signs, by W'. More explicitly, let S" be the group of all permutations of the set 11,... , n} (the symmetric group). For each g E W' there is a permutation o9 E Sn and numbers ek(g) = ±1 such that 9(v) = (E1(9)va9(1),...)En(9)vo9(n))

,

(28)

where v = (v1i... ,vn). Let Un be defined by (10). Since the r.v.s {Xk}kc1 are equidistributed, the sets 0,,,,, are identical. From here g(Un) = Un for every g E Wn. As in the section 1, we construct the collection of segments O(v) , v E Un, with the properties described in Proposition 2. Using the formulae (12) and (13) we define the r.v.s { k } 1 and the invertible measure-preserving transformations ¢9 (g E W'). As above, T(g) are the corresponding operators. Proposition 15. The following relations are true: (1 < k < n, g E Wn). The proof is similar to the proof of Proposition 6 . One has to take into account that the r.v.s {Xk}k=1 are equidistributed, which yields that T(9)Xk = Ek(9)Xa9(k)

a(k)(va9(k)) = a(a9(k))(vae(k))

for each v = (v1i... vn) E U. Hence, XkL(A-D = Ek(9)Xa9(k)I

.

A(v)

Let the r.v.s {Yk}k=1 C E' define the projector from E on span {Xk}k.1 by the formula n

(X , Yk) Xk

QX =

(29)

k=1

and suppose that Q commutes with all operators T(9), g E Wn. The equality (16) holds for W'. Applying Proposition 15, we easily verify that (30) T(g)Yk = ek(g)Ya9(k) (1 < k < n, g E Wn) . The next step is to construct a projector on the subspace span {Xk}ko1 with some special properties.

Rearrangement Invariant Spaces

106

Proposition 16. Let {Xk}k=1 be finite-valued symmetric i.i.d.r.v.s, E be a separable r.i. space and Pn : E -> span {Xk }kc1 be a projector. Then there are r. v.s {Yk}k=1 C E' with the following properties: 1) (X1 , Yk) = bi,k;

2)E(YkI{a<xj
3) the variable E (YkI{a<xk
QnX =E(X ,Yk)Xk k=1

the estimate IIQnIIE-E < 11PfIIE-.E holds .

Proof: According to Proposition 5 there is a projector Pn : E , span {Xk }k-1 such that IIPnIIE-E' where the r.v.s Xk are determined by the 11PnIIE-.E C formula (12). We put

QnX = 2nn(

(T(g)PPT (g-1)) X. gEWn

It follows from Proposition 15 that every operator T(g) (g E Wn) maps the span {Xk}k=1 into itself. Since the group Wn has 2nn! elements, On is a projector on this subspace and 111QnIIE-.E

IIPnIIE_E

11PfhIE-E

In addition, On commutes with every T(g) (g E W"). As E is separable, Qn is represented in the form (29) and 1) is fulfilled. Let k # j. Using (28) we may choose g E Wn so that fk(g) = 1, ei(g) _ -1 and vg(m) = m (1 < m,, < n). From (3) and Proposition 15 T(g)Yk = -Yk

,

T(g)X., = Xj .

So for alla
Since g = g-1, then from here and (16) E (YkI{a<X; CI{a<X;

_ -E (YkI{a<X;
4. Complementability of subspaces

107

This implies 2).

Now we choose g E W' such that e,n(g) = 1 (1 <m < n) and o9(k) = j for fixed j # k. From (30) and Proposition 15, T(g)Xk = X j and T(g)Yk = Y. Since ¢9 is measure-preserving, the distributions of the pairs (Xk , Yk) and (Xj , Y1) are identical and 3) follows. According to Proposition 4, Xk = Xk (bn(t)) (1 < k < n), where ,b [0, 1] is a measure-preserving transformation. Putting Yk(t) _ [0, 1] Yk (fin 1(t)) we get the r.v.s with the needed properties. El Now we eliminate the assumption that Xk is finite-valued.

Proposition 17. Let r.i. the space E be separable and {Xk}k_1 C E be symmetric i.i.d.r.v.s. Let Pn be a projector on the subspace span {Xk} Then there are r.v.s {Yk}k_1 C E' such that the relations 1)-4) of Proposition 16 are valid.

Proof: Since E is separable, there exist finite-valued r.v.s Xk,,n (1 < k < n ; m E N) with the following properties: (i) for every m the r.v.s {Xk,m}k_1 are symmetric, independent and equidistributed;

(ii)Xk,,,,- Xk inEasm -goo (1
,

lim IIPn,mIIE_E = IIPnIIE-.E .

(31)

Proposition 16 yields that for each m there are r.v.s {Yk,n}k_1 C E' having the properties 1)-4). Specifically, for the projectors n

Qn,mX = E (X ,

Yk,m) Xk,m

k=1

the estimate IIQr,nIIE_E : IIPm,nllE_.E holds. From Propositions 1.14 and 1.15 I(X , Yk,m)I IIXk,tIIE

IIQn,mXIIE

(1 < k < n).

As IIXk,mIIE - IIXkIIE (m -> oo), then it follows that supIIYk,mIIE' <00 M

(1
Using the well known results about the weak compactness [16] and the re-

lation E' = E*, we obtain that there are r.v.s {Yk}k_1 C E' and integers

m(j) / oo such that ilim (X, Yk,m(j)) = (X, Yk) 00

(1 < k < n)

(33)

Rearrangement Invariant Spaces

108

for every X E E. We show that the r.v.s {Yk}k-1 have the desired properties. It follows from (33) and the condition (ii) that

(Xi m(j)' Yk,m(j)) = bi,k. (Xi, Yk) = jliM 00 To prove the properties 2) and 3) we show that j

lira E (Yk,m(j)I{ a<x,,-c>)
(34)

for every 1 < i, k < n and a < b. Proposition 1.1 implies Xk,m ' Xk in the space L1(0, 1). Therefore, passing to a subsequence, we may suppose that Xk,m -. Xk almost surely [35] . From here for every a < b I{a<x,,m(;)
(35)

almost surely as j --i oo. According to (33),

jlim E (Yk,m(j)Ih) = E (YkIh) 00 for each measurable h C [0, 1]. Using the Vitali-Hahn-Saks theorem (see [16]) we conclude that E (Yk m(j)Ih) , 0 uniformly with respect to j and k as ph -> 0. From here and (35) the relation (34) follows. Using Proposition 16 we get the relations 2) and 3). We Put QnX = Ek-1 (X , Yk) Xk. From (33) Qn,m(j)X -> in E. Taking into account (31) and (32) we obtain IIQnIIE_E < IIPnIIE-.E .0 Lastly we consider an infinite sequence of i.i.d. random variables. Lemma 4. Suppose E is a separable r.i. space, {Xk }k 1 C E are symmetric i.i.d.r.v.s and there exists a family of uniformly bounded projectors Pn : E--+

span {Xk}k-1 (n E N). Then there exist r.v.s {Yk}kl C E' such that the relations 1)-4) of Proposition 16 are valid for each n. Proof: Fix n and let {Yk,n}k-1 be the r.v.s constructed in Proposition 17 for the collection {Xk}k-1. Then for the projectors EL, (X , Yk,n) Xk the relation sup IIQnI n

E-+E -

IIPfhIE_E =

C < Co

holds. Propositions 1.14 and 1.15 imply for every X E E I(X, Yk,n)I IIX1IIE = I(X, Yk,n)I IIXkIIE : IIQnXIIE
,

4. Complementability of subspaces

109

which gives us

C

iiyk,nIIE, <

(1 < k < n, n E N).

IIX I1E

Reasoning as in the proof of Proposition 17 and using the diagonal process,

we may find integers n(j) / oo and r.v.s {Yk}' 1 C E' such that for every X E E and k E N

jlim V, Yk,n(j)) = (X, Yk). The r.v.s {Yk}k 1 have the desired properties. 4. Reduction to Theorem 1. Now we construct bounded symmetric i.i.d.r.v.s {Vk}k1 for which there is a sequence of uniformly bounded projectors from E to the subspaces span {Vk}k=1. We put Vk,c = XkI{IXkI
(36)

where c > 0 is a constant.

Lemma 5. Let the conditions of Lemma 4 hold. Then for large enough c there is a collection of uniformly bounded projectors on the subspaces span {Vk,,}k=1.

Proof: Let {Yk }k 1 C E' be the r.v.s constructed in Lemma 4. The property 2) implies (Vk,,, Yj) = 0 for j # k. According to the property 3) the quantity h(c) = (Vk,,,, Yk)

(37)

is independent of k. We have h(c) -> (Xk , Yk) = 1 as c -> oo. Hence, h(c) > 1/2 for large enough c. We put

Qn,cX = h( E (X

,

Yk) Vk

k_1

and show the projectors Qn,, are uniformly bounded. Let's denote Xk,c = XkI{JXkIc}

By virtue of symmetry Xk,,

a

X. Hence, the sums n

n

QnX =

(X, Yk) Xk k=1

,

Rn,c.X = E (X, Yk) Xk,c k=1

(38)

Rearrangement Invariant Spaces

110

are equidistributed and IIQnXIIE = II Rn,CX IIE for each X E E. From (36) and (38) Qn,CX =

2h(c)

(QnX + Rn,cX),

which yields IIQh()IIE

IIQn,cX IlE <

< 2IIQnIIE-.E IIXIIE

Lemma 5 follows from here and Lemma 4.

5. Proof of Theorem 2. Necessity. We may assume as mentioned above, the r.v.s Xk to be symmetric. It follows from the complementability of span {Xk }k that there exists a sequence of uniformly bounded projectors on the subspaces span {Xk}k=1. Using Lemma 5 and applying to the r.v.s {Vk,c}k 1.the arguments of subsection 8 of the section 1, we get the assertion 1) of Theorem 2. Now we show that span {Xk } 1 C E is isomorphic to the space 12. According to Lemma 3.1

i

E2 n

akXk k=1

1/2 ,

k=1

E

where a > 0 is independent of n and ak. So, we have to prove the corresponding upper estimate. For the r.v.s {Vk,c}0 1 the inequality (4) holds, i.e. there is a positive constant D(c) such that n

D(c)-1

(a2)

1/2

n

<

n

D(c) E a2

E akVk,c k=1

1/2

E

k=1

for all integers n and ak E R. Let {Yk }' 1 C E' be the r.v.s constructed in Lemma 4. We choose c > 0 under the condition h(c) > 1/2, where h(c) is determined by the formula. (37). The last bound gives us

>sup

X Espan {vkc}k=1 ,IIXIIE<1}

\k-1

k=1

/ n

tt

> sup

akbk

h(c)

1/2

n 2

k=1

< D(c)

\k=1

k=1

2D(c)

(a2

b

(39)

4. Complementability of subspaces

111

Now we can prove the desired upper estimate. Let Qn be the projectors constructed in Lemma 4. We have C = sup IIQnIIE-E < CO . n

Since E is separable, then IIXIIE = sup {(X, Y) : Y E E', IIYIIE, < 1} .

It is not difficult to verify that the conjugate operator Qn is the projector from E' on the subspace span{Yk}k=1 and the equality IIQnIIE'-E' = 11QnIIE-.E

holds. Using the relation (39), we get n

n

Qn E akXk

E akXk k=1

E

k=1

E

/

k=1 n :

E bkYk

I

k=1

k=1

k=1

(En k=1

b2 I

CE k=1 f 1/2

a2)

IIQn IIE'-.E' } lIE' 1/2

n

< sup E akbk

which completes the proof.

J

n

< sup {Eakbk

= 2CD(c)

YEE',IIYIIE'_1}

,

< 2D(c) IIQn IIE'-.E'

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Index

Arzela-Ascoli's theorem 63 Banach lattice 19, 25

Kwapien and Richlik's inequality 8,

Banach spaces 17-19 Bernstein's inequality 7, 23, 89 Borel measue 93 Boyd index 5, 38 Calderon-Mityagin's theorem 2 central limit theorem 37, 70 characteristic function 1, 44, 53, 61 closure subspace 88 complemented function 84, 88 decreasing rearrangement 2 disjoint r.v.s 8 disjointification 13 distribution 1 dual space 2 equivalent functions 9 exponential Orlicz spaces 42, 83 function of dilatation 4 fundamental function 4, 30 Holder's inequality 67 indicator 2 infinite sums 80 Kolmogorov's theorem 74 Kruglov's theorem 10, 42, 75 property 11, 16, 17, 42

Lebesgue measure 92

84

Levy-Khintchin spectral function 72, 75-77

mean value 1 median 5 Paley and Zygmund's inequality 7, 23, 34, 52, 89 projector 88, 111

special 105

Prokhorov's 'arcsinh' inequality 7, 35

Prokhorov's inequality 8, 75 random variable 1 Rademacher systems 88 reduction to 96 Rosenthal's inequality 21, 70 standard q-stable distribution 78 symmetrization 90 symmetric group 105

Vitali-Hahn-Saks theorem 108 von-Bahr and Esseen p-property 34 weak Rosenthal p-property 38 weakly convergent 3