Subspaces of L p

Theorem 5.4.5 The Dunford-Pettis Theorem)

6.4 Subspaces of L p

148 6 TheLp-Spaces for 1≤p <∞

Corollary 6.3.9.L₁ does not have a boundedly-complete basis.

Proof. We need only recall that, by Theorem 3.2.10, a space with a boundedly- complete basis is (isomorphic to) a separable dual space.

Theorem 6.3.7 is rather classical: it is due to Gelfand [66]. In fact the argument we have given is somewhat ad hoc; to be more precise, one should use the concept of the Radon-Nikodym Property which we discussed earlier in Section 5.4. The main point here is that neitherL₁norc₀ have the Radon- Nikodym Property while separable dual spaces do. Gelfand approaches this through differentiability of Lipschitz maps: a Banach space X has (RNP) if and only if every Lipschitz mapF : [0,1]→X is differentiable a.e. InL₁ the Lipschitz map

F(t) =χ_(0,t), 0≤t≤1 is nowhere differentiable. Inc0we can consider the map

F(t) = 1 nsinnt

_∞

n=1

, 0≤t≤1

which is again nowhere differentiable (note that formally differentiating takes us into the bidual!). These examples are due to Clarkson [30].

Let us conclude this section with the promised result thatL₁ is comple- mented in its bidual.

Proposition 6.3.10.There is a norm-one linear projection P :L^∗∗₁ →L₁. Proof. Let us first define R : L^∗∗₁ → M[0,1] to be the restriction mapϕ→ ϕ|_M[0,1]. Clearly Rϕ ≤ ϕ. Next we define a map S : M[0,1] →L1 by Sµ=f where

dµ=dν+f dt

is the Lebesgue decomposition of µ (i.e., ν is singular with respect to the Lebesgue measure). ThenS= 1.We conclude thatP =SR is a norm-one projection ofL^∗∗₁ ontoL₁.

6.4 Subspaces ofLp 149 Proposition 6.4.1.Let (f_n)^∞_n=1 be a sequence of norm-one, disjointly sup- ported functions in L_p. Then (f_n)^∞_n=1 is a complemented basic sequence iso- metrically equivalent to the canonical basis of_p.

Proof. The case p= 1 was seen in Lemma 5.1.1. Let us fix 1< p < ∞. For any sequence of scalars (a_i)^∞_i=1∈c₀₀, by the disjointness of thef_i’s we have

∞ i=1

a_if_i^p

p

= ^∞

i=1

a_if_i(t)^pdt

= ∞

i=1

|a_if_i(t)|^pdt

= ∞ i=1

|a_i|^p

|f_i(t)|^pdt

= ∞ i=1

|ai|^p.

By the Hahn-Banach theorem, for eachi∈Nthere exists g_i ∈L_q (q the conjugate exponent of p) withg_iq = 1 so that 1 =f_ip =

f_i(t)g_i(t)dt.

Furthermore, without loss of generality, we can assume g_i to have the same support asf_ifor alli. Let us define the linear operator fromL_ponto [f_i] given by

P(f) = ∞ i=1

f(t)g_i(t)dt

f_i, f ∈L_p. Then,

P(f)p= ^∞

i=1

f(t)g_i(t)dt^p1/p

= ^∞

i=1

{|f_i|>0

f(t)g_i(t)dt^p1/p

≤^∞

i=1

{|f_i|>0

|f(t)|^pdt 1/p

≤ |f(t)|^pdt 1/p

.

The following proposition allows us to deduce that L_p is not isomorphic to _p forp= 2, and already hints at the fact that the the L_p-spaces have a more complicated structure than the spacesp.

Proposition 6.4.2.₂ embeds in L_p for all 1 ≤ p < ∞. Furthermore, ₂ embeds complementably in L_p if and only if 1< p <∞.

150 6 TheLp-Spaces for 1≤p <∞

Proof. For each 1 ≤ p < ∞ let R_p be the closed subspace spanned in L_p by the Rademacher functions (r_n)^∞_n=1. By Khintchine’s inequality, (r_n)^∞_n=1 is equivalent to the canonical basis of₂, then R_p is isomorphic to₂.

By Proposition 5.6.1, L₁ has no infinite-dimensional complemented re- flexive subspaces, so R₁ is not complemented in L₁. Let us prove that if 1< p <∞,R_p is complemented inL_p.

Assume first that 2≤p < ∞. Consider the map fromLp ontoRp given by

P(f) = ∞ n=1

f(t)rn(t)dt

rn, f ∈Lp.

P is linear and well defined. Indeed, the series is convergent inL_pbecausef ∈ L_p ⊂ L₂ implies _∞

n=1(

f(t)r_n(t)dt)² < ∞. Now, Khintchine’s inequality and Bessel’s inequality yield

P(f)²p= ∞ n=1

f(t)r_n(t)dt

r_n²

p

≤B_p² ∞ n=1

f(t)r_n(t)dt²

≤B_p²f²2

≤B_p²f²p.

If 1≤p <2 we defineP as before for eachf ∈L_p∩L₂(which is a dense subspace inL_p). Then, using Khintchine’s inequality, we obtain

P(f)p≤^∞

n=1

f(t)r_n(t)dt²1/2

= sup _∞

n=1

α_n

f(t)r_n(t)dt

: ∞ n=1

α²_n= 1

= sup

f(t) ^∞

n=1

αnrn(t)

dt : ∞ n=1

α²_n= 1

≤sup

f_p ∞ n=1

αnrn(t)

q

: ∞ n=1

α²_n= 1

≤sup

f_pBq ∞ n=1

αnrn(t)

2

: ∞ n=1

α²_n= 1

=B_qfp.

By density,P extends continuously toL_p with preservation of norm.

6.4 Subspaces ofLp 151 Proposition 6.4.3.If_q embeds inL_p then eitherp≤q≤2 or2≤q≤p.

Proof. Let us start by noticing that if (e_i)^∞_i=1 is the canonical basis of_q, for eachnwe have

E n i=1

ε_ie_i

q

=n^1/q.

If_qembeds inL_pfor somep <2, by Theorem 6.2.14 there exist constants c₁ andc₂ (given by the embedding and the type and cotype constants) such that

c1n¹² ≤n^1q ≤c2n^1p.

For these inequalities to hold for all n ∈N it is necessary that q ∈[p,2]. If q embeds in Lp for some 2 < p <∞, with the same kind of argument we deduce thatq must belong to the interval [2, p].

Definition 6.4.4.Suppose (Ω,Σ, µ) is a probability measure space and let X be a closed subspace of L_p(µ) for some 1 ≤ p < ∞. X is said to be strongly embeddedin Lp(µ) if, inX, convergence in measure is equivalent to convergence in theLp(µ)-norm; that is, a sequence of functions (fn)^∞_n=1 inX converges to 0 in measure if and only iffn_p→0.

Proposition 6.4.5.Suppose(Ω,Σ, µ)is a probability measure space and let 1 ≤p <∞. Suppose X is an infinite-dimensional closed subspace of L_p(µ).

Then the following are equivalent:

(i)X is strongly embedded inLp(µ);

(ii) For each0< q < pthere exists a constant Cq such that f_q ≤ f_p≤C_q f_q for all f ∈X;

(iii) For some 0< q < p there exists a constant C_q such that fq ≤ fp≤C_q fq for all f ∈X.

Proof. Let us suppose that X is strongly embedded in L_p(µ) but (ii) fails.

Then there would exist a sequence (f_n)^∞_n=1 in X such that f_np = 1 and f_nq → 0 for some 0 < q < p. Obviously, this implies that (f_n)^∞_n=1 con- verges to 0 in measure, which would force (f_np)^∞_n=1 to converge to 0. This contradiction shows that (i)⇒(ii).

Suppose now that (iii) holds and there is a sequence of functions (f_n)^∞_n=1 in X such that (fn)^∞_n=1 converges to 0 in measure but (fn_p)^∞_n=1 does not tend to 0. By passing to a subsequence we can assume that (fn)^∞_n=1converges to 0 almost everywhere andfn_p= 1 for alln.

For eachM >0, sinceq < pwe have

152 6 TheLp-Spaces for 1≤p <∞

Ω

|f_n|^qdµ=

{|f_n|≥M}|f_n|^qdµ+

{|f_n|<M}|f_n|^qdµ

≤

{|f_n|≥M}

M^q−p|f_n|^pdµ+

{|f_n|<M}|f_n|^qdµ

≤ 1 M^p−q +

{|f_n|<M}|fn|^qdµ.

Let > 0. By the Lebesgue Bounded Convergence theorem, there is N0 ∈ N such that

{|f_n|<M}|fn|^qdµ < /2 for all n > N0. So, if we pick M >

(2⁻¹)^p−q1 we get

Ω

|f_n|^qdµ < ,

contradicting (iii). Hence (iii)⇒(i), and so the proof is over because, triv- ially, (ii)⇒(iii).

Example 6.4.6.For each 1 ≤ p < ∞ the closed subspace spanned in Lp

by the Rademacher functions, Rp, is strongly embedded in Lp since, using Khintchine’s inequality, the L_q-norm and the L_p-norm are equivalent in R_p for all 1≤q <∞.

Theorem 6.4.7.Suppose that X is an infinite-dimensional closed subspace of Lp for some 1 ≤ p < ∞. If X is not strongly embedded in Lp then X contains a subspace isomorphic to p and complemented inLp.

Proof. If X is not strongly embedded in L_p, by Proposition 6.4.5 there is a sequence (f_n)^∞_n=1 in X, f_np = 1 for all n, such that f_n → 0 a.e. By Lemma 5.2.1 there is a subsequence (f_nk)^∞_k=1 of (f_n)^∞_n=1 and a sequence of disjoint subsets (A_k)^∞_k=1 of [0,1] such that if B_k = [0,1]\ A_k, then (|f_nk|^pχ_Bk)^∞_k=1 is equi-integrable. Lemma 5.2.7 implies

|f_nk|^pχ_Bkdµ→0.

That is, fn_k−fn_kχA_kp → 0. Now, by standard perturbation arguments we obtain a subsequence (fn_kj)^∞_j=1 of (fn_k)^∞_k=1 such that (fn_kj)^∞_j=1 is equiva- lent to the canonical basis of_p and [f_nkj] is complemented inL_p.

The following theorem was proved in 1962 by Kadets and Pelczy´nski [98]

in a paper which really initiated the study of L_p-spaces by basic sequence techniques. We will see that the case p > 2 is quite different from the case p <2 and this theorem emphasizes this point.

Theorem 6.4.8 (Kadets-Pelczy´nski). Suppose that X is an infinite- dimensional closed subspace of L_p for some 2 < p <∞. Then the following are equivalent:

(i)_p does not embed in X;

(ii)_p does not embed complementably in X;

6.4 Subspaces ofLp 153 (iii)X is strongly embedded inL_p;

(iv)X is isomorphic to a Hilbert space and is complemented inL_p; (v)X is isomorphic to a Hilbert space.

Proof. (i)⇒(ii) and (iv)⇒(v) are obvious, and (ii)⇒(iii) was proved in Theorem 6.4.7. Let us complete the circle by showing that (iii)⇒ (iv) and that (v)⇒(i).

(iii)⇒ (iv) IfX is strongly embedded in L_p, Proposition 6.4.5 yields a constantC₂ such thatf2≤ fp≤C₂f2 for allf ∈X. This shows that X embeds inL₂and hence it is isomorphic to a Hilbert space. Let us see that X is complemented inL₂.

Sincep >2,L_p is contained inL₂ and the inclusionι:L_p→L₂ is norm decreasing. The restriction of ι to X is an isomorphism onto the subspace ι(X) ofL2, and ι(X) is complemented inL2 by an orthogonal projectionP:

L_p ^ι //L₂

P

X?OO

ι|X //ι(X)

Thenι⁻¹P ιis a projection ofL_pontoX (this projection is simply the restric- tion ofP toL_p).

(v) ⇒ (i) If X ≈ ₂ then X cannot contain an isomorphic copy of _p for anyp= 2 because the classical sequence spaces are totally incomparable (Corollary 2.1.6).

The Kadets-Pelczy´nski theorem establishes a dichotomy for subspaces of L_p when 2< p <∞:

Corollary 6.4.9.SupposeX is a closed subspace of L_p for some2< p <∞. Then either

(i)X is isomorphic to ₂, in which caseX is complemented inL_p, or (ii)X contains a subspace that is isomorphic to _p and complemented inL_p.

Notice that, in particular, this settles the question of whichLq-spaces for 1≤q <∞embed inL_pforp >2:

Corollary 6.4.10.For2< p <∞and1≤q <∞withq=p,2,L_p does not have any subspace isomorphic to L_q or_q.

We are now ready to find a more efficient embedding of₂ into the L_p- spaces, replacing the Rademacher sequences by sequences of independent Gaussians. We consider only the real case, although modifications can be made to handle complex functions. In order to introduce these ideas, we will require some more basic notions from probability theory.

154 6 TheLp-Spaces for 1≤p <∞

Iff is a real random variable, itsdistribution is the probability measure µ_f onRgiven by

µ_f(B) =P(f⁻¹B)

for any Borel setB. f is calledsymmetriciff and−f have the same distri- bution.

Conversely, for each probability measureµ onR there exist real random variablesf withµf =µ, and the formula

Ω

F(f(ω))dP(ω) = _∞

−∞

F(x)dµf(x) (6.12) holds for any positive Borel functionF :R→R.

Thecharacteristic functionφ_f of a random variablef is the functionφ_f : R→Cdefined by

φ_f(t) =E(e^itf).

This is related toµ_f via the Fourier transform:

ˆ

µ_f(−t) =

R

e^itxdµ_f(x) =φ_f(t).

In particular φ_f determines µ_f, i.e., if f and g are two random variables (possibly on different probability spaces) withφ_f =φ_g then µ_f =µ_g.Other basic useful properties of characteristic functions are:

• φ_f(−t) =φ_f(t);

• φ_cf+d(−t) =e^idtφ_f(ct), forc, dconstants;

• φ_f+g=φ_fφ_g iff andgare independent.

Remark 6.4.11.If f₁, . . . , f_n are independent random variables (not nec- essarily equally distributed) on some probability space, then we can exploit independence to compute the characteristic function of any linear combination n

j=1a_jf_j: E

e^itPnj=1ajfj

= n j=1

E e^itajfj

= n j=1

φ_fj(a_jt). (6.13) Suppose we are given a probability measureµonR.The random variable f(x) = x has distribution µ with respect to the probability space (R, µ).

Next consider the countable product space R^N with the product measure P=µ×µ× · · ·.(R^N,P) is also a probability space and the coordinate maps fj :R^N→R,

f_j(x₁, . . . , x_n, . . .) =x_j,

are identically distributed random variables onR^Nwith distributionµ. More- over, the random variables (f_j)^∞_j=1are independent.

6.4 Subspaces ofLp 155 Although we created the sequence of functions (f_j)^∞_j=1on (R^N,P) we might just as well have worked on ([0,1],B, λ). As we discussed in Section 5.1 there is a Borel isomorphismσ:R^N→[0,1] which preserves measure, that is,

λ(B) =P(σ⁻¹B), B ∈ B

and the functions (fj◦σ⁻¹)^∞_j=1 have exactly the same properties on [0,1].

This remark, in particular, allows us to pick an infinite sequence of inde- pendent identically distributed random variables on [0,1] with a given distri- bution.

Thestandard normal distributionis given by the measure onR dµ_G = 1

√2πe^−x2/2dx.

We will call any random variable with this distribution a(normalized) Gaus- sian. In this case we have

ˆ

µ_G(−t) = 1

√2π _∞

−∞

e^itx−x2/2dx=e^−t2/2, so the characteristic function of a Gaussian ise^−t2/2.

Proposition 6.4.12.If g is a Gaussian on some probability measure space (Ω,Σ, µ)theng∈L_p(µ)for every 1≤p <∞.

Proof. This is because

Ω

|g(ω)|^pdω= 1

√2π _∞

−∞|x|^pe^−12x2dx,

and the last integral is finite and indeed computable in terms of the Γ function as

2^p/2

√πΓ(p+ 1 2 ).

Proposition 6.4.13.₂ embeds isometrically inL_p for all1≤p <∞. Proof. Take (gj)^∞_j=1, a sequence of independent Gaussians on [0,1]. By Propo- sition 6.4.12, (gj)^∞_j=1⊂Lp. We will show that [gj] is isometrically isomorphic to2.

For everyn∈Nand scalars (a_j)ⁿ_j=1 such thatn

j=1a²_j = 1, put h_n=

n j=1

a_jg_j.

156 6 TheLp-Spaces for 1≤p <∞

By (6.13) we have

φ_hn(t) =e^{−(a21+···+a2n)t2/2}=e^−t2/2. This means thatµh_n =µg₁ and so by (6.12)

hnp=g1p. It follows that for anya₁, . . . , a_n inR,

n j=1

a_jg_j

p

=g₁p

ⁿ

j=1

|a_j|²1/2

.

Thus the mapping e_n → g₁⁻p¹g_n linearly extends to an isometry from ₂ onto the subspace [g_n] of L_p.

The connection between the Gaussians and₂is encoded in the character- istic function. We are now going to dig a little deeper to try to make copies of _q for other values ofq in theL_p-spaces. A moment’s thought shows that we need a random variablef with characteristic function

φf(t) =e^−c|t|q

for some constantc=c(q). It turns out that if (and only if) 0< q <2 we can construct such a random variable. This has long been known to Probabilists;

here we give a treatment based on some unpublished notes of Ben Garling.

We will need the following classical lemma due to Paul L´evy (see, for instance, [57]).

Lemma 6.4.14.Suppose(µn)^∞_n=1 is a sequence of probability measures onR such that

lim

n→∞µˆ_n(−t) =F(t)

exists for all t∈R. If F is continuous then there is a probability measure µ on Rsuch thatµ(ˆ −t) =F(t).

Proof. It is convenient to compactify the real line by adding one point at∞ to make the one-point compactificationK=R∪ ∞. We can then regard each µ_n as a Borel measure onK which assigns zero mass to {∞}.Let µbe any weak^∗ cluster point of this sequence (viewed as elements of C(K)^∗; such a measure then exists by Banach-Alaoglu’s theorem). The functions x→ e^itx cannot be extended continuously toK. However, fort= 0 the functions

h_t(x) =

⎧⎪

⎨

⎪⎩

t if x= 0

e^itx−1

ix if x∈R\ {0}

0 if x=∞

6.4 Subspaces ofLp 157 are continuous onK.

Ift >0,

K

h_t(x)dµ_n(x) =

R t

0

e^isxds

dµ_n(x)

= t

0 R

e^isxdµ_n(x)

ds

= t

0

ˆ

µn(−s)ds.

Thus

K

h_t(x)dµ(x) = t

0

F(s)ds.

Ift <0 the same calculation works to give

K

h_t(x)dµ(x) =− 0

t

F(s)ds.

Note that fort >0,|h_t(x)| ≤tfor allxand vanishes at∞.Thus

K

h_t(x)dµ(x)≤tµ(R).

Hence, for t >0

1 t

t 0

F(s)ds≤µ(R).

F is continuousand, obviously, F(1) = 1. Thus the left-hand side converges to 1. We conclude that µ(R) = 1, i.e., µ is actually a Borel measure onR. Now ˆµ(−t) is a continuous function oftand ift >0,

t 0

ˆ

µ(−s)ds=

R

h_t(x)dµ(x) = t

0

F(s)ds.

By the Fundamental Theorem of Calculus, since both ˆµ(−t) and F(t) are continuous, ˆµ(−t) =F(t) fort >0.A similar calculation works ift <0.

Theorem 6.4.15.For every 0< p≤2 there is a probability measure µ_p on (R, dx)such that _∞

−∞

e^itxdµ_p(x) =e^−|t|p, t∈R. Proof. It obviously suffices to show the existence ofµ_p with

_∞

−∞

e^itxdµ_p(x) =e^−cp|t|p, t∈R,

158 6 TheLp-Spaces for 1≤p <∞

where c_p is some positive constant. For the case p = 2 this is achieved by using a Gaussian.

Now suppose 0< p <2. Letf be a random variable on some probability space with probability distribution

dµ_f = p 2|x|^p+1

'χ_{(−∞,−1)}(x) +χ_(1,+∞)(x)( dx.

The characteristic function off is the following:

E e^itf

= _∞

−∞

e^itxdµ_f(x)

=p 2

₋1

−∞

e^itx

(−x)^p+1 dx+p 2

_∞

1

e^itx x^p+1 dx

=p _∞

1

e^itx+e^−itx 2

dx x^p+1

=p _∞

1

cos(tx) x^p+1 dx.

Then, ift >0 the substitution u=txin the last integral yields 1−E

e^itf

=p _∞

1

dx x^p+1 −p

_∞

1

cos(tx) x^p+1 dx

=p _∞

1

1−cos(tx) x^p+1 dx

=pt^p _∞

t

1−cosu u^p+1 du.

Let

ω_p(t) =p _∞

t

1−cosu u^p+1 du and

c_p= lim

t→0⁺

ω_p(t) =p _∞

0

1−cosu u^p+1 du.

Note that_∞

0

1−cosu

u^p+1 du is finite and positive for every 0< p <2.

Sincef is symmetric, its characteristic function is even and therefore the equality

E e^itf

= 1− |t|^pω_p(t) holds for allt∈R.

Let (f_j)^∞_j=1be a sequence of independent random variables with the same distribution asf. Then, for everynthe characteristic function of the random variable ^f1+_n^···_1/p^+fn is

6.4 Subspaces ofLp 159 E

e^itf1+n^···+fn1/p

= n i=1

E

e^itn^fi1/p

= E

e^itn1/p^{f n}

=

1−|t|^p

n ω_p |t| n^1/p

ⁿ . Since

nlim→∞

1−|t|^p

n ωp

|t| n^1/p

ⁿ

=e^−cp|t|p,

we can apply the preceding lemma to obtain the required measureµ_p. Definition 6.4.16.A random variablef on a probability space is calledp- stable(0< p <2) if

ˆ

µ_f(−t) =e^−c|t|p, t∈R

for some positive constantc=c(p).f is callednormalized p-stableifc= 1.

Note that the normalization for Gaussians is somewhat different, i.e., the characteristic function of a normalized Gaussian would correspond to the case c= 1/2 in the previous definition.

Theorem 6.4.17.Let f be ap-stable random variable on a probability mea- sure space(Ω,Σ, µ)for some0< p <2. Then

(i)f ∈L_q(µ)for all 0< q < p;

(ii)f ∈Lp(µ).

Proof. Suppose thatf is normalizedp-stable for some 0< p <2 with distri- bution of probabilityµ_p. Then

Ω

|f(ω)|^qdω= _∞

−∞|x|^qdµ_p(x).

For everyx∈Rthe substitutionu=|x|t in the integral_∞

0

1−costx

t^1+q dtyields _∞

0

1−costx

t^1+q dt=|x|^qα_q, whereα_q =_∞

0

1−cosu

u^1+q duis a positive constant for 0< q <2. Hence, _∞

−∞|x|^qdµ_p(x) =α⁻_q¹ _∞

−∞

∞ 0

1−costx t^1+q dt

dµ_p(x)

=α⁻_q¹ _∞

0

1 t^q+1

∞

−∞

(1−costx)dµ_p(x)

dt

=α⁻_q¹ _∞

0

1 t^q+1

∞

−∞

(1− e^ixt)dµ_p(x)

dt

=α⁻_q¹ _∞

0

1

t^q+1(1−e^−tp)dt.

The last integral is finite for 0< q < p and fails to converge whenq=p.

160 6 TheLp-Spaces for 1≤p <∞

Theorem 6.4.18.If 1 ≤p <2 andp≤q≤2, then_q embeds isometrically inL_p.

Proof. We have already seen the cases when q =pand q = 2. For 1≤p <

q <2, let (f_j)^∞_j=1 be a sequence of independent normalized q-stable random variables on [0,1]. Then we can repeat the argument we used in Proposi- tion 6.4.13 to prove that [f_j] is isometric to _q in L_p. The only constraint is that the sequence (f_j) must belong toL_p, which requires thatp < q.

We can summarize our discussion by stating:

Theorem 6.4.19 (_q-subspaces of L_p).

(i) For1≤p≤2,_q embeds inL_p if and only ifp≤q≤2;

(ii) For 2< p <∞,_q embeds inL_p if and only ifq= 2or q=p.

Moreover, if_q embeds in L_p then it embeds isometrically.

Remark 6.4.20.The alert reader will wonder for which values ofq, the func- tion spaceL_q can be embedded inL_p. In fact, the answer is exactly the same as for the sequence space_q, but we will postpone the proof of this until Chap- ter 11. A direct proof of this facts can be based on a discussion of stochastic integrals (see [106]).

Theorem 6.4.21.Let 1< p, q <∞. Then _q embeds complementably in L_p if and onlyq=porq= 2.

Proof. We know (Proposition 6.4.2 and Proposition 6.4.1) that both q = p andq= 2 allow complemented embeddings. Suppose_q embeds inL_pcomple- mentably andq /∈ {2, p}.By Theorem 6.4.19 we must havep < q <2. Taking duals it follows that _q embeds complementably inL_p, where q, p are the conjugate indices ofqandp. This is impossible.

The L_p-spaces (1 ≤ p < ∞) are primary. Alspach, Enflo, and Odell [3]

proved the result for 1< p <∞in 1977. The case p= 1 was established in 1979 by Enflo and Starbird [55] as we already mentioned in Chapter 5.

The problem of classifying the complemented subspaces ofLp when 1 <

p < ∞ received a great deal of attention during the 1970s. At this stage we know of three isomorphism classes that we can find as complemented subspaces inside anyLp:2,p, andLp, and it is easily seen that we can add _p⊕₂and_p(₂) to that list. In fact, it turns out thatL_phas a very rich class of complemented subspaces and the classification of them seems beyond reach.

In 1981, Bourgain, Rosenthal, and Schechtman [17] showed the existence of uncountably many mutually nonisomorphic complemented subspaces of L_p; curiously it seems unknown (unless we assume the Continuum Hypothesis) whether there is a continuum of such spaces!

6.4 Subspaces ofLp 161

Problems

6.1.This exercise can be considered as a continuation of Problem 5.6.

(a) A closed subspace X of L_p(T) is called translation-invariant if f ∈ X implies τ_φ(f) ∈ X, where τ_φ(f) = f(θ−φ). Show that if X is translation- invariant and E = {n ∈ Z: e^inθ ∈ X}, then X is the closed linear span of {e^inθ:n∈E}. In this case we putX =Lp,E(T).

(b)Eis called a Λ(p)-set ifL_p,E(T) is strongly embedded inL_p(T). Show that ifE is a Λ(p)-set then it is a Λ(q)-set forq < p.

(c) Show that if p > 2, E is a Λ(p)-set if and only if {e^inθ : n ∈ E} is an unconditional basis ofL_p,E(T).

(d) Prove thatE={4ⁿ:n∈N}is a Λ(4)-set. [Hint: Expand|

n∈Ea_ne^inθ|⁴.]

(e)E is called aSidon setif for any (a_n)_n∈E∈_∞(E) there existsµ∈ M(T) with ˆµ(n) =a_n. Show that the following are equivalent:

(i)Eis a Sidon set;

(ii) (e^inθ)_n∈E is an unconditional basic sequence inC(T);

(iii) (e^inθ)_n∈E is a basic sequence equivalent to the canonical₁-basis inC(T).

(f) Show that a Sidon set is a Λ(p)-set for every 1≤p <∞.

(g) Show that E = {4ⁿ : n ∈ N} is a Sidon set. [Hint: For −1 ≤ a_n ≤ 1, consider the functions f_n(θ) = n

k=1(1 +a_kcos 4^kθ), and let µ be a weak^∗ cluster point of the measuresf_n^dθ_2π.]

6.2.In this problem we aim to obtainKhintchine’s inequalitydirectly, not as a consequence of Kahane’s inequality.

(a) Prove that cosht≤e^t2/2 for allt∈R. (b) Show that ifp≥1 thent^p≤p^pe^−pe^t.

(c) Let (n)^∞_n=1 be a sequence of Rademachers and supposef =n k=1akεk

wheren

k=1a²_k = 1.Show that

E(e^f)≤e and deduce that

E(e^|f|)≤2e.

Hence show that

(E(|f|^p))^1/p≤2^1/pe^1/pp e. Finally obtain Khintchine’s inequality forp >2.

(d) Show by using H¨older’s inequality that (c) implies Khintchine’s inequality forp <2.

162 6 TheLp-Spaces for 1≤p <∞