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Averaging in Banach spaces

Dalam dokumen Graduate Texts in Mathematics (Halaman 139-150)

Theorem 5.4.5 The Dunford-Pettis Theorem)

6.2 Averaging in Banach spaces

F(t) =a2k

Ik

2ϕ

2x(ut, vt) +2ϕ

2y(ut, vt) + 2k 2ϕ

∂x∂y(ut, vt)

ds.

By Lemma 6.1.5 (ii),F(t)0. HenceF(1)≥F(0)0 and thus(6.4) holds.

To complete the proof whenp > 2 we plug x=fn and y =gn in (6.2).

Integrating both sides of this inequality and using (6.4) we obtain 1

0

(p−1)p|fn(s)|p− |gn(s)|pds≥0.

The case when 1 < p < 2 now follows by duality: with fn, gn as before choose gn∈Lq(Bn) so thatgnq = 1 and

1 0

gn(s)gn(s)ds=gnp. Thengn =n

j=1bjhj for some (bj)nj=1 and gnp=

n j=1

|Ij|jajbj ≤ fnp n j=1

jbjhj

q (q−1)fnp.

The constant p1 in Burkholder’s theorem is sharp, although we will not prove this here.

132 6 TheLp-Spaces for 1≤p <∞

Definition 6.2.1.TheRademacher functions(rk)k=1 are defined on [0,1] by rk(t) = sgn (sin 2kπt).

Alternatively, the sequence (rk)k=1 can be described as r1(t) =

1 if t∈[0,12)

1 if t∈[12,1) r2(t) =

1 if t∈[0,14)[12,34)

1 if t∈[14,12)[34,1) ...

rk+1(t) =

1 if t∈"2k

s=1[2s2k+12,2s2k+11)

1 if t∈"2k

s=1[2s2k+11,2k+12s ).

That is,

rk+1=

2k

s=1

h2k+s, k= 0,1,2, . . .

Thus (rk)k=1is a block-basic sequence with respect to the Haar basis in every Lp for 1≤p <∞. The key properties we need are the following:

rk(t) =±1 a.e. for allk,

rk1rk2(t). . . rkm(t)dt= 0, wheneverk1< k2<· · ·< km.

The Rademacher functions were first introduced by Rademacher in 1922 [191] with the idea of studying the problem of finding conditions under which a series of real numbers

±an, where the signs were assigned randomly, would converge almost surely. Rademacher showed that if

|an|2<∞then indeed

±an converges almost surely. The converse was proved in 1925 by Khintchine and Kolmogoroff [111].

For our purposes it will be convenient to replace the concrete Rademacher functions by an abstract model. To that end we will use the language and methods of probability theory.

Let us recall that arandom variable is a real-valued measurable function on some probability space (Ω,Σ,P). The expectation(or mean) of a random variablef is defined by

Ef =

f(ω)dP(ω).

A finite set of random variables {fj}nj=1 on the same probability space is independent if

P

$n j=1

fj ∈Bj

= n j=1

P(fj∈Bj)

for all Borel setsBj. Therefore if (fj)nj=1 are independent, E

f1f2· · ·fn

=E(f1)E(f2)· · ·E(fn).

An arbitrary set of random variables is said to be independent if any finite subcollection of the set is independent.

Definition 6.2.2.A Rademacher sequence is a sequence of mutually inde- pendent random variables (εn)n=1 defined on some probability space (Ω,P) such thatP(εn= 1) =P(εn =1) = 12 for everyn.

The terminology is justified by the fact that the Rademacher functions (rn)n=1 are a Rademacher sequence on [0,1]. Thus,

1 0

n

i=1

ri(t)xidt=E n

i=1

εixi=

n i=1

εi(ω)xidP.

Historically, the subject of finding estimates for averages over all choices of signs was initiated in 1923 by the classical Khintchine inequality [110], but the usefulness of a probabilistic viewpoint in studying theLp-spaces seems to have been fully appreciated quite late (around 1970).

Theorem 6.2.3 (Khintchine’s Inequality). There exist constantsAp,Bp (1 p < ∞) such that for any finite sequence of scalars (ai)ni=1 and any n∈N we have

Apn

i=1

|ai|21/2

n i=1

airi

pn

i=1

|ai|21/2

if 1≤p <2,

and n

i=1

|ai|21/2

n i=1

airi

p≤Bpn

i=1

|ai|21/2

if p >2.

We will not prove this here but it will be derived as a consequence of a more general result below. Theorem 6.2.3 was first given in the stated form by Littlewood in 1930 [141] but Khintchine’s earlier work (of which Littlewood was unaware) implied these inequalities as a consequence.

Remark 6.2.4.(a) Khintchine’s inequality says that (ri)i=1 is a basic se- quence equivalent to the2-basis in everyLp for 1≤p <∞. InL, though, one readily checks that (ri)i=1 is isometrically equivalent to the canonical 1-basis.

(b) (ri)i=1 is an orthonormal sequence inL2, which yields the identity

n i=1

airi

2

= n

i=1

|ai|21/2

,

134 6 TheLp-Spaces for 1≤p <∞

for any choice of scalars (ai). But (ri)i=1is not a complete system inL2, that is, [ri]=L2 (for instance, notice that the functionr1r2 is orthogonal to the subspace [ri]). However, one can obtain a complete orthonormal system for L2 using the Rademacher functions by adding to (rn) the constant function r0= 1 and the functions of the formrk1rk2. . . rknfor anyk1< k2<· · ·< kn. This collection of functions are theWalsh functions.

Thus we can also interpret Khintchine’s inequality as stating that all the normsp: 1≤p <∞}are equivalent on the linear span of the Rademacher functions in Lp. It turns out that in this form the statement can be general- ized to an arbitrary Banach space. This generalization was first obtained by Kahane in 1964 [101].

Theorem 6.2.5 (Kahane-Khintchine Inequality). For each 1≤p <∞ there exists a constant Cp such that, for every Banach spaceX and for any finite sequence(xi)ni=1 inX, the following inequality holds:

E n i=1

εixi E

n i=1

εixip1/p

≤CpE n i=1

εixi.

We will prove the Kahane-Khintchine inequality (and this will imply the Khintchine inequality by takingX =RorX=C) but first we shall establish three lemmas on our way to the proof. To avoid repetitions, in all three lemmas (Ω,Σ,P) will be a probability space and X will be a Banach space. Let us recall that an X-valued random variable on Ω is a functionf : Ω→X such that f1(B) Σ for every Borel setB ⊂X. f issymmetric if P(f ∈B) = P(−f ∈B) for all Borel subsetsB ofX.

Lemma 6.2.6.Letf : Ω→X be a symmetric random variable. Then for all x∈X we have

P

f+x ≥ x

1/2.

Proof. Let us take any x X. For every ω Ω, using the convexity of the norm of X, clearly f(ω) +x+x−f(ω) 2x. Then, either f(ω) +x ≥ x orx−f(ω) ≥ x. Hence

1P

f+x ≥ x +P

x−f ≥ x .

Sincef is symmetric,x+f andx−f have the same distribution and so the lemma follows.

Let (εi)i=1 be a Rademacher sequence on Ω. Given n N and vectors x1, . . . , xn in X, we shall consider Λm: Ω−→X (1≤m≤n) defined by

Λm(ω) = m i=1

εi(ω)xi.

Lemma 6.2.7.For allλ >0, P

max

mnΛm> λ

2P

Λn> λ . Proof. Givenλ >0, form= 1, . . . , nput

(λ)m =

ω∈Ω :Λm(ω)> λ and Λj(ω) ≤λ for all j= 1, . . . , m−1 . Since{ω∈Ω : maxmnΛm(ω)> λ}=nm=1(λ)m , by the disjointedness of the sets Ω(λ)m it follows that

P max

mnΛm> λ

= n m=1

P(Ω(λ)m ). (6.5) Therefore,

P

Λn> λ

= n m=1

P

(λ)m (Λn> λ)

. (6.6)

Notice that every Ω(λ)m can be written as the union of sets of the type {ω∈Ω :εj(ω) =δj for 1≤j≤m}

for some choices of signsδj =±1.For each of these choices of signsδ1, . . . , δm we observe that by Lemma 6.2.6,

P m j=1

δjxj+ n j=m+1

εjxj>

m j=1

δjxj

1 2. Summing over the appropriate signs (δ1, . . . , δm) it follows that

P

(λ)m (ΛnΛm)

1

2P(Ω(λ)m ).

Thus,

P

(λ)m (Λn> λ)

1

2P(Ω(λ)m ).

Summing inmand combining (6.5) and (6.6) we finish the proof.

Lemma 6.2.8.For allλ >0,

P

Λn>2λ

4 P

Λn> λ2

.

Proof. We will keep the notation that we introduced in the previous lemma.

Notice that for each 1 m n, the random variable n

i=mεixi is in- dependent of each of ε1, . . . , εm and hence for all λ > 0 the events : n

i=mεi(ω)xi> λ}and Ω(λ)m are independent. Observe as well that if some

136 6 TheLp-Spaces for 1≤p <∞

ω (λ)m further satisfies Λn(w) >2λ, thenΛn(ω)Λm1(ω) > λ (for m= 1, take Λ0 = 0). Therefore, sinceP(n

i=mεixi> λ)2P(Λn> λ) for eachm= 1, . . . , nby Lemma 6.2.7, we have

P

(λ)m (Λn>2λ)

P(Ω(λ)m )Pn

i=m

εixi> λ

2P(Ω(λ)m )P

Λn> λ . Summing inmand using again Lemma 6.2.7 we obtain

P

Λn>2λ

P max

mnΛm> λ P

Λn> λ

4 P

Λn> λ2

. Proof of Theorem 6.2.5. Fix 1≤p < and let {xi}ni=1 be any finite set of vectors inX. Without loss of generality we will suppose thatEn

i=1εixi= 1. Then, by Chebyshev’s inequality,

P

Λn>8

1

8. (6.7)

Using Lemma 6.2.8 repeatedly we obtain P

Λn>2·8

4(1/8)2, P

Λn>22·8

43(1/8)4, P

Λn>23·8

47(1/8)8, and so on. Hence, by induction, we deduce that

P

Λn>2n·8

42n1(1/8)2n42n(1/8)2n = (1/2)2n. Therefore,

E n i=1

εixip =

0

P

Λnp> t dt

=

0

p tp1P

Λn> t dt

= 8

0

p tp1P

Λn> t dt+

n=1

2n·8 2n−1·8

p tp1P

Λn> t dt

8

0

p tp1dt+ n=1

(1/2)2n1 2n·8

2n−1·8

p tp1dt

8p

1 + n=1

(1/2)2n12np

=Cpp.

Suppose that H is a Hilbert space. The well-known Parallelogram Law states that for any two vectorsx, yin H we have

x+y2+x−y2

2 =x2+y2.

This identity is a simple example of the power of averaging over signs and has an elementary generalization:

Proposition 6.2.9 (Generalized Parallelogram Law). Suppose that H is a Hilbert space. Then for every finite sequence(xi)ni inH,

E n i=1

εixi2= n i=1

xi2.

Proof. For any vectors (xi)ni=1 inH we have E

n i=1

εixi2=E%n

i=1

εixi, n i=1

εixi&

= n i,j=1

xi, xjE(εiεj)

= n i=1

xi2.

Next we are going to study how the averages (En

i=1εixip)1/p are situ- ated with respect to the sums (n

i=1xip)1/p using the concepts oftypeand cotype of a Banach space. These were introduced into Banach space theory by Hoffmann-Jørgensen [79] and their basic theory was developed in the early 1970s by Maurey and Pisier [147]; see [146] for historical comments. However, it should be said that the origin of these ideas was in two very early papers of Orlicz in 1933, [163] and [164]. Orlicz essentially introduced the notion of co- type for the spacesLp although he did not use the more modern terminology.

Definition 6.2.10.A Banach spaceX is said to haveRademacher typep(in short, typep) for some 1≤p≤2 if there is a constantCsuch that for every finite set of vectors{xi}ni=1 inX,

E n i=1

εixip1/p

≤C n

i=1

xip1/p

. (6.8)

The smallest constant for which (6.8) holds is called thetype-pconstant of X and is denotedTp(X).

138 6 TheLp-Spaces for 1≤p <∞

Similarly, a Banach spaceXis said to haveRademacher cotypeq(in short, cotype q) for some 2 q ≤ ∞ if there is a constant C such that for every finite sequencex1, x2, . . . , xn in X,

n

i=1

xiq1/q

≤C E

n i=1

εixiq1/q

, (6.9)

with the usual modification of max1inxi replacing ni=1xiq1/q

whenq=. The smallest constant for which (6.9) holds is called thecotype-q constant ofX and is denotedCq(X).

Remark 6.2.11.(a) The restrictions on p and q in the definitions of type and cotype respectively are natural since it is impossible to have typep >2 or cotypeq <2 even in a one-dimensional space. To see this, for eachntake vectors {xi}ni=1 all equal to some x X with x = 1. The combination of Khintchine’s inequality with (6.8) and (6.9) gives us the range of eligible values for pandq.

(b) Every Banach spaceX has type 1 withT1(X) = 1 and cotype with C(X) = 1 by the triangle law. ThusX is said to have nontrivial type if it has typepfor some 1< p≤2; similarlyX is said to havenontrivial cotypeif it has cotypeq for some 2≤q <∞.

(c) The generalized Parallelogram Law (Proposition 6.2.9) says that a Hilbert spaceH has type 2 and cotype 2 withT2(H) =C2(H) = 1. In partic- ular a one-dimensional space has type 2 and cotype 2. But the Parallelogram Law is also a characterization of Banach spaces which are linearly isometric to Hilbert spaces, hence we deduce that a Banach space X is isometric to a Hilbert space if and only ifT2(X) =C2(X) = 1 (see Problem 7.6).

(d) By Theorem 6.2.5, theLp-average (En

i=1εixip)1/pin the definition of type can be replaced by any otherLr-average (En

i=1εixir)1/r(1≤r <

) and this has the effect only of changing the constant. The same comment applies to theLq-average in the definition of cotype.

(e) If X has type p then X has type r for r < p and ifX has cotype q thenX has cotypesfors > q.

(f) The type and cotype of a Banach space are isomorphic invariants and are inherited by subspaces.

(g) Consider the unit vector basis (en)n=1 in p (1≤p <∞) or c0.Then for any signs (k) we have

1e1+· · ·+nenp=np1 and

1e1+· · ·+nen= 1.

Thusp cannot have type greater thanpif 1≤p≤2 or cotype less thanpif 2≤p≤ ∞.

Proposition 6.2.12.If a Banach space X has typepthenX has cotypeq, where 1p+1q = 1 andCq(X)≤Tp(X).

Proof. Let us pick an arbitrary finite set{xi}ni=1 in X. Given >0 we can find x1, . . . , xn in X such that xi = 1 and |xi(xi)| ≥ (1)xi for all i= 1, . . . , n. Thus

n

i=1

|xi(xi)|q1/q

(1)n

i=1

xiq1/q

. On the other hand,

n

i=1

|xi(xi)|q1q

= sup n i=1

aixi(xi) : n i=1

|ai|p 1

. For any scalars (ai)ni=1 withn

i=1|ai|p1 we have n

i=1

aixi(xi) =

n

i=1

εixin

i=1

εiaixi dP

n i=1

εixi n i=1

εiaixidP

n i=1

εixiqdP1q

n i=1

εiaixipdP1p

n i=1

εixiqdP1q Tp(X)

n

i=1

|ai|p1p . Therefore,

n

i=1

xiq1q

(1)1Tp(X) E

n i=1

εixiq1q . Sincewas arbitrary, this shows Cq(X)≤Tp(X).

Curiously, Proposition 6.2.12 does not have a converse statement. At the end of the section we shall give an example showing that if X has cotype q forq <∞thenX may not have typepwhere 1p+1q.

Next we want to investigate the type and cotype ofLp for 1 ≤p < . To do so we will estimate(n

i=1|fi|2)1/2pin relation with the Rademacher averages (En

j=1εjfjpp)1/p on a genericLp(µ)-space.

Theorem 6.2.13.For every finite set of functions{fi}ni=1 inLp(µ)(1≤p <

∞),

140 6 TheLp-Spaces for 1≤p <∞

Apn

i=1

|fi|212

p E

n i=1

εifip

p

1/p

≤Bpn

i=1

|fi|212

p

, whereAp,Bp are the constants in Khintchine’s inequality (in particularAp= 1 for2≤p <∞andBp= 1 for1≤p≤2).

Proof. For eachω∈Ω, from Khintchine’s inequality Ap

n

i=1

|fi(ω)|21/2

E

n i=1

εifi(ω)p1/p

, whereAp= 1 for 2≤p <∞. Now, using Fubini’s theorem

Appn

i=1

|fi|21/2p

p

E n i=1

εifi(ω)p

=E

n i=1

εifi(ω)p

=E n i=1

εifip

p

. The converse estimate is obtained similarly.

The next theorem is due to Orlicz for cotype [163, 164] and Nordlander for type [159]. Obviously, the language of type and cotype did not exist before the 1970s and their results were stated differently. Note the difference in behavior of theLp-spaces whenp >2 orp <2.This is the first example where we meet some fundamental change around the indexp= 2 and, as the reader will see, it is really because when p/2 <1 the triangle law for positive functions in Lp/2reverses.

Theorem 6.2.14.

(a)If1≤p≤2,Lp(µ) has typepand cotype 2.

(b)If2< p <∞,Lp(µ)has type2 and cotypep.

Moreover,(a)and(b)are optimal.

Proof. (a) Let us prove first that if 1 ≤p 2, thenLp(µ) has type p. We recall this elementary inequality:

Lemma 6.2.15.Let 0< r≤1. Then for any nonnegative scalars(αi)ni=1 we have

(α1+· · ·+αn)r≤αr1+· · ·+αrn. (6.10)

This way, combining Theorem 6.2.13 with (6.10) we obtain E

n i=1

εifip

p

p1

n

i=1

|fi|21/2

p

= n i=1

|fi|21/2

p/2

n

i=1

|fi|p2/p1/2

p/2

=

n i=1

|fi|p 1/p

=n

i=1

fpp

1/p

.

To show thatLp(µ) has cotype 2 when 1≤p≤2 we need the reverse of Minkowski’s inequality:

Lemma 6.2.16.Let 0< r <1. Then

f+gr≥ fr+gr, wheneverf andg are nonnegative functions inLr(µ).

Proof. Without loss of generality we can assume that f+gr = 1 and so = (f+g)ris a probability measure. This implies

fr=

fr 1/r

=

{f+g>0}

fr

(f+g)r(f +g)r 1/r

{f+g>0}

f

f+g(f+g)rdµ.

Analogously,

gr

{f+g>0}

g

f+g(f+g)rdµ.

Thereforefr+gr1 =f+gr.

Now, combining Theorem 6.2.13 with Lemma 6.2.16,

Ap1 E

n i=1

εifip

p

1/p

n

i=1

|fi|212

p

= n i=1

|fi|2

12

p/2

142 6 TheLp-Spaces for 1≤p <∞

n

i=1

fi2

p/2

1/2

=n

i=1

fi2p1/2

.

To obtain the cotype-2 estimate we just have to replace the Lp-average (En

j=1εjfjpp)1/pby (En

j=1εjfj2p)1/2using Kahane’s inequality (at the small cost of a constant) .

(b) For each 2 < p < , from Theorem 6.2.13 in combination with Ka- hane’s inequality there exists a constantC=C(p) so that

E n i=1

εifi2

p

1/2

≤Cn

i=1

|fi|212

p

. Sincep/2>1, the triangle law now holds inLp/2(µ) and hence

n

i=1

|fi|212

p

= n i=1

|fi|21/2

p/2n

i=1

fi2p/2

1/2

= n

i=1

fi2p

1/2

. This shows that Lp(µ) has type 2. Therefore, from part (a) and Proposi- tion 6.2.12 it follows thatLp(µ) has cotypep.

The last statement of the theorem follows from Remark 6.2.11 and the fact thatLp(µ) containspas a subspace.

Example 6.2.17.To finish the section let us give an example showing that the concepts of type and cotype are not in duality, in the sense that the converse of Proposition 6.2.12 need not hold. The space C[0,1] fails to have nontrivial type because it contains a copy ofL1, whereas its dual,M(K), has cotype 2 (we leave the verification of this fact to the reader).

Dalam dokumen Graduate Texts in Mathematics (Halaman 139-150)