Theorem 5.4.5 The Dunford-Pettis Theorem)
6.1 Conditional expectations and the Haar basis
Let (Ω,Σ, µ) be a probability measure space, and Σ a sub-σ-algebra of Σ.
Givenf ∈L1(Ω,Σ, µ) we define a (signed) measure,ν, on Σ: ν(E) =
E
f dµ, E∈Σ.
ν is absolutely continuous with respect toµ|Σ, hence by the Radon-Nikodym theorem, there is a (unique, up to sets of measure zero) function ψ ∈ L1(Ω,Σ, µ) such that
126 6 TheLp-Spaces for 1≤p <∞ ν(E) =
E
ψ dµ, E∈Σ. Thenψis the (unique) function that satisfies
E
f dµ=
E
ψ dµ, E∈Σ.
ψ is called the conditional expectation of f on the σ-algebra Σ and will be denoted byE(f|Σ).
Let us notice that if Σconsists of countably many disjoint atoms (An)∞n=1, the definition ofE(f|Σ) is specially simple:
E(f|Σ)(t) = ∞ j=1
1
µ(Aj) Ajf dµ χAj(t).
We also observe that iff ∈Lp(µ) where 1≤p <∞andg∈Lq(Ω,Σ, µ) where 1p+1q then
E
g dν =
f g dµ, E∈Σ, and
E(f g|Σ) =gE(f|Σ).
Lemma 6.1.1.Let (Ω,Σ, µ) be a probability measure space and suppose Σ is a sub-σ-algebra of Σ. Then for every1 ≤p≤ ∞, E(· |Σ) is a norm-one linear projection fromLp(Ω,Σ, µ)ontoLp(Ω,Σ, µ).
Proof. We denote E =E(· |Σ). It is immediate to check that E2 =E for all 1≤p≤ ∞.
Fix 1≤p <∞(we leave the casep=∞to the reader). Iff ∈Lp(µ), E(f)p= sup
Ω
E(f)g dµ:g∈Lq(Ω,Σ, µ),gq ≤1
= sup
Ω
E(f g)dµ:g∈Lq(Ω,Σ, µ),gq ≤1
= sup
Ω
f g dµ:g∈Lq(Ω,Σ, µ),gq ≤1
≤ fp.
Definition 6.1.2.The sequence of functions on [0,1], (hn)∞n=1, defined by h1= 1 and forn= 2k+s(wherek= 0,1,2, . . ., ands= 1,2, . . . ,2k),
hn(t) =
⎧⎪
⎨
⎪⎩
1 if t∈[2s2k+1−2,2s2k+1−1)
−1 if t∈[2s2k+1−1,2k+12s ) 0 otherwise
=χ[2s−2 2k+1,2s−1
2k+1)(t)−χ[2s−1 2k+1, 2s
2k+1)(t) is called theHaar system.
Givenk = 0,1,2, . . . and 1≤s≤2k, each interval of the form [s2−k1,2sk) is calleddyadic. It is often useful to label the elements of the Haar system by their supports; thus we write hI to denote hn whenI is the dyadic interval support ofhn.
Proposition 6.1.3.The Haar system is a monotone basis inLp for1≤p <
∞.
Proof. Let us consider an increasing sequence of σ-algebras, (Bn)∞n=1, con- tained in the Borel σ-algebra of [0,1] defined as follows: we let B1 be the trivialσ-algebra,{∅,[0,1]}, and forn= 2k+s(k= 0,1,2, . . .,1≤s≤2k) we letBn be the finite subalgebra of the Borel sets of [0,1] whose atoms are the dyadic intervals of the family
Fn=
[2jk+1−1,2k+1j ) forj = 1, . . . ,2s [j2−k1,2jk) forj =s+ 1, . . . ,2k.
Fix 1 ≤ p < ∞. For each n, En will denote the conditional expectation operator on theσ-algebra Bn. By Lemma 6.1.1,En is a norm-one projection from Lp onto Lp([0,1],Bn, λ), the space of functions which are constant on intervals of the familyFn. We will denote this space byLp(Bn). Clearly, rank En=n. Furthermore,EnEm=EmEn=Emin{m,n}for any two positive integers m, n.
On the other hand, the set
f ∈Lp : En(f)−fp→0
is closed (using the partial converse of the Banach-Steinhaus theorem, see the Appendix) and contains the set∪∞k=1Lp(Bk),which is dense inLp. Therefore En(f)−fp→0 for allf ∈Lp. By Proposition 1.1.7, Lp has a basis whose natural projections are (En)∞n=1. This basis is actually the Haar system because for eachn∈N,Em(hn) =hn form≥nandEmhn = 0 form < n. The basis constant is supnEn= 1.
The Haar system as we have defined it is not normalized inLpfor 1≤p <
∞(it is normalized inL∞). To normalize in Lp one should takehn/hnp=
|In|−1/phn, where In denotes the support of the Haar function hn. Let us observe that iff ∈Lp (1≤p <∞), then
Enf− En−1f = 1
|In|
f(t)hn(t)dt
hn.
We deduce that the dual functionals associated to the Haar system are given by
128 6 TheLp-Spaces for 1≤p <∞ h∗n= 1
|In|hn, n∈N,
and the series expansion off ∈Lp in terms of the Haar basis is f =
∞ n=1
1
|In|
f(t)hn(t)dt
hn.
Notice that ifp= 2 then (hn/hn2)∞n=1 is an orthonormal basis for the Hilbert space L2and is thus unconditional.
It is an important fact that, actually, the Haar basis is an unconditional basis inLp for 1< p <∞.This was first proved by Paley [165] in 1932. Much more recently, Burkholder [20] established the best constant.
We are going to present another proof of Burkholder from 1988 [21]. We will only treat the real case here, although, remarkably, the same proof works for complex scalars with the same constant; however, the calculations needed for the complex case are a little harder to follow. For our purposes the constant is not so important, and we simply note that if the Haar basis is unconditional for real scalars, one readily checks it is also unconditional for complex scalars.
There is one drawback to Burkholder’s argument: it is simply too clever in the sense that the proof looks very like magic.
We start with some elementary calculus.
Lemma 6.1.4.Supposep >2 and 1p+1q = 1. Then for 0≤t≤1 we have tp−ppq−p(1−t)p≤p2q1−p(t−1
q). (6.1)
Proof. For 0≤t≤1 put
f(t) =tp−ppq−p(1−t)p−p2q1−p(t−1 q).
Then
f(t) =ptp−1+pp+1q−p(1−t)p−1−p2q1−p and
f(t) =p(p−1)tp−2−pp+1(p−1)q−p(1−t)p−2.
Observe that f(0) = −ppq−p+p2q−p < 0 and f(1) = 1−pq1−p. Since p >2 we have (1−1p)p−1> 1p, i.e., pq1−p>1; thusf(1)<0.
Next note thatf(1q) = 0 and f(1
q) =pq1−p+p2q−p−p2q1−p= 0.
We also have
f(1
q) = (p−1)(pq2−p−p3q−p) = (p−1)pq−p(q2−p2)<0.
Assume that there exists some 0 < s < 1 with f(s) > 0. Then there must exist at least three solutions of f(t) = 0 in the open interval (0,1), including 1/q. By Rolle’s theorem this means there are at least two solutions off(t) = 0, which is clearly false.
In the next lemma we introduce a mysterious function which will enable us to prove Burkholder’s theorem. This function appears to be plucked out of the air although there are sound reasons behind its selection. The use of such functions to prove sharp inequalities has been developed extensively by Nazarov, Treil, and Volberg who term them Bellman functions. We refer to [156] for a discussion of this technique.
Lemma 6.1.5.Supposep >2 and defineϕ:R2→R by ϕ(x, y) = (|x|+|y|)p−1
(p−1)|x| − |y| .
(i)If 1/p+ 1/q= 1, the following inequality holds for all(x, y)∈R2
(p−1)p|x|p− |y|p≥pq1−pϕ(x, y). (6.2) (ii)ϕ is twice continuous differentiable and satisfies the condition
∂2ϕ
∂y2 +∂2ϕ
∂x2 = 2 ∂2ϕ
∂x∂y
≥0. (6.3)
Proof. (i) If we substitutet=|y|(|x|+|y|)−1(for (x, y)= (0,0)) in equation (6.1) we have
|y|p−ppq−p|x|p≤pq1−p(|y| −(p−1)|x|)(|x|+|y|)p−1. Thus
ppq−p|x|p− |y|p≥pq1−pϕ(x, y).
Note thatppq−p= (p−1)p.
(ii) The fact thatϕis twice continuously differentiable is immediate since p >2.
Clearly, it suffices to prove (6.3) in the first quadrant, wherex >0,y >0.
Letu=x+y andv= (p−1)x−y.Then ϕ(x, y) =up−1v. Hence
∂2ϕ
∂y2 = (p−1)(p−2)up−3v−2(p−1)up−2 while
∂2ϕ
∂x2 = (p−1)(p−2)up−3v+ 2(p−1)2up−2.
130 6 TheLp-Spaces for 1≤p <∞ Hence
∂2ϕ
∂x2 +∂2ϕ
∂y2 = 2(p−1)(p−2)up−3(u+v)≥0.
On the other hand, since ϕis linear on any line of slope one (or by routine calculation) we must also have
∂2ϕ
∂x∂y = (p−1)(p−2)up−3(u+v).
Theorem 6.1.6.Suppose 1 < p < ∞ and 1p + 1q = 1. Let p∗ = max(p, q).
The Haar basis(hk)∞k=1 inLp is unconditional with unconditional constant at mostp∗−1. That is,
n j=1
jajhj
p≤(p∗−1) n j=1
ajhj
p
,
whenevern∈N, for any real scalarsa1, . . . , an and any signs 1, . . . , n. Proof. Suppose p > 2, in which case p∗ = p. For each fixed n ∈ N, let f0=g0= 0 and for 1≤k≤nput
fk= k j=1
ajhj and gk = k j=1
jajhj. We will prove by induction onk that
1 0
ϕ(fk(s), gk(s))ds≥0, 1≤k≤n, (6.4) whereϕis the function defined in Lemma 6.1.5. This is trivial whenk= 0.In order to establish the inductive step, for a givenklet us consider the function F : [0,1]→Rdefined by
F(t) = 1
0
ϕ
(1−t)fk−1(s) +tfk(s),(1−t)gk−1(s) +tgk(s) ds, and show thatF(1)≥0 assuming thatF(0)≥0.
Letut= (1−t)fk−1+tfk andvt= (1−t)gk−1+tgk.Then F(t) =ak
1 0
∂ϕ
∂x(ut, vt)hkds+kak 1
0
∂ϕ
∂y(ut, vt)hkds.
Observe that F(0) = 0 since ∂ϕ∂x(u0, v0) and ∂ϕ∂y(u0, v0) are constant on the support interval of hk.
Differentiating again gives
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 p∗−1 in Burkholder’s theorem is sharp, although we will not prove this here.