Properties of L 1

Theorem 5.4.5 The Dunford-Pettis Theorem)

6.3 Properties of L 1

142 6 TheLp-Spaces for 1≤p <∞

≥ⁿ

i=1

f_i²

p/2

1/2

=ⁿ

i=1

fi²_p1/2

.

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

j=1ε_jf_j^pp)^1/pby (En

j=1ε_jf_j²p)^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

ε_if_i²

p

1/2

≤Cⁿ

i=1

|f_i|²¹₂

p

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

ⁿ

i=1

|f_i|²¹₂

p

= n i=1

|f_i|^21/2

p/2≤ⁿ

i=1

f_i²p/2

1/2

= ⁿ

i=1

f_i²p

1/2

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

The last statement of the theorem follows from Remark 6.2.11 and the fact thatL_p(µ) contains_pas 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).

6.3 Properties ofL1 143

f_N = 2^1−2Nχ_[0,1]+ 2N j=0

2^j+1−2Nh_[0,2−j). Put

g_N = N j=0

2^2j+1−2Nh_[0,2−2j). It is clear that

g_N(t) =−2^2j+1−2N for 2^−2j−1≤t <2^−2j and 0≤j ≤N.

Thus

g_N1≥ N j=0

2^2j+1−2N2^−2j−1= (N+ 1)2^−2N = (N+ 1)f_N1. This shows immediately that the Haar system cannot be unconditional.

In fact we will show thatL₁cannot be embedded in a space with an uncon- ditional basis; this result is due to Pelczy´nski (1961) [170]. In Theorem 4.5.2 we showed, by the technique of testing property (u), thatC(K) embeds in a space with unconditional basis if and only ifC(K)≈c₀. ForL₁this approach does not work becauseL₁ is weakly sequentially complete and therefore has property (u). A more sophisticated argument is therefore required. The ar- gument we use was originally discovered by Milman [151]; first we need a lemma:

Lemma 6.3.2.For everyf ∈L₁ we have lim

n→∞

f(t)r_n(t)dt= 0.

Thus (f r_n)^∞_n=1 is weakly null for every f ∈L₁.

Proof. (r_n)^∞_n=1 is an orthonormal sequence in L₂, which implies (by Bessel’s inequality) that

lim

n→∞

f(t)r_n(t)dt= 0 for all f ∈L₂.

Since (r_n)^∞_n=1 is uniformly bounded inL_∞, and L₂is dense in L₁we deduce lim

n→∞

f(t)r_n(t)dt= 0 for all f ∈L₁. Thus iff ∈L₁ andg∈L_∞, sincef g∈L₁ we obtain

lim

n→∞

g(t)f(t)r_n(t)dt= 0, which gives the latter statement in the lemma.

144 6 TheLp-Spaces for 1≤p <∞

Theorem 6.3.3.L1 cannot be embedded in a Banach space with uncondi- tional basis.

Proof. Let X be a Banach space with K-unconditional basis (e_n)^∞_n=1 and suppose that T : L₁ → X is an embedding. We can assume that for some constantM ≥1,

f1≤ T f ≤Mf1, f ∈L₁.

By exploiting the unconditionality of (e_n)^∞_n=1 we are going to build an unconditional basic sequence inL1 using a gliding-hump type argument.

Take (δk)^∞_k=1 a sequence of positive real numbers with _∞

k=1δk <1.Let f0= 1 =χ_[0,1],n1= 1,s0= 0 and picks1∈Nsuch that

∞ j=s₁+1

e^∗_j(T(f₀r_n1))e_j<1 2δ₁. Put

x₁=

s₁

j=s₀+1

e^∗_j(T(f₀r_n1))e_j.

Next take f₁ = (1 +r_n1)f₀. Since the sequence (f₁r_k)^∞_k=1 is weakly null by Lemma 6.3.2, (T(f₁r_k))^∞_k=1 is also weakly null, hence we can find n₂ ∈ N, n₂> n₁, so that

s₁

j=1

e^∗_j(T(f₁r_n2))e_j<1 2δ₂. Now, picks₂∈N,s₂> s₁, for which

∞ j=s₂+1

e^∗_j(T(f₁r_n2))e_j<1 2δ₂.

Continuing in this way we will inductively select two strictly increasing se- quences of natural numbers (n_k)^∞_k=1 and (s_k)^∞_k=0, a sequence of functions (f_k)^∞_k=0 inL₁given by

f_k= (1 +r_nk)f_k−1 for k≥1, and a block basic sequence (xk)^∞_k=1of (en)^∞_n=1defined by

x_k =

s_k

j=s_k−1+1

e^∗_j(T(f_k−1r_nk))e_j, k= 1,2, . . . .

This is how the inductive step goes: supposen₁, n₂, . . . , n_l−1,s₀, s₁, . . . , s_l−1, and therefore f₁, . . . , f_l−1 have been determined. Since (T(f_l−1r_k))^∞_k=1 is weakly null we can findn_l> n_l−1 so that

6.3 Properties ofL1 145

s_l−1

j=1

e^∗_j(T(f_l−1r_nl))e_j< 1 2δ_l, and then we chooses_l> s_l−1 so that

∞ j=s_l+1

e^∗_j(T(f_l−1r_nl))e_j<1 2δ_l. Note that for k≥1 we have

f_k= k j=1

(1 +r_nj), (6.11)

which yields f_k ≥0 for allk. Expanding out (6.11), it is also clear that for eachk,

f_k1= 1

0

f_k(t)dt= 1.

On the other hand, fork≥1 we have

T f_k−T f_k−1−x_k< δ_k, and hence the estimate

n j=1

x_j< M+ n j=1

δ_j< M+ 1 holds for alln.

Since it is a block basic sequence with respect to (e_n)^∞_n=1, (x_k)^∞_n=1 is an unconditional basic sequence in X with unconditional constant ≤ K (see Problem 3.1). Therefore for all choices of signs _j =±1 and alln= 1,2, . . . we have a bound:

n j=1

_jx_j≤K(M + 1), which implies

n j=1

_j(T f_j−T f_j−1)≤K(M+ 1) + 1, and thus

n j=1

_j(f_j−f_j−1)

1≤K(M + 1) + 1.

This shows that_∞

j=1(f_j−f_j−1) inL₁is a WUC series inL₁(see Lemma 2.4.6).

SinceL₁is weakly sequentially complete (Theorem 5.2.10), by Corollary 2.4.15 the series _∞

j=1(f_j −f_j−1) must converge (unconditionally) in norm in L₁ and, in particular, lim_j→∞f_j − f_j−11 = 0. But for j ≥ 1 we have f_j−f_j−11=r_njf_j−11= 1,a contradiction.

146 6 TheLp-Spaces for 1≤p <∞

In Corollary 2.5.6 we saw thatc₀is not a dual space. We will show thatL₁ is also not a dual space and, even more generally, that it cannot be embedded in a separable dual space. We know thatc₀ is not isomorphic to a dual space becausec₀is uncomplemented in its bidual. This is not the case forL₁as we shall see below. Thus to showL₁ is not a dual space requires another type of argument and we will use some rather more delicate geometrical properties of separable dual spaces.

Lemma 6.3.4.Let X be a Banach space such thatX^∗ is separable. Assume that K is a weak^∗ compact set in X^∗. ThenK has a point of weak^∗-to-norm continuity. That is, there isx^∗∈Ksuch that whenever a sequence(x^∗_n)^∞_n=1⊂ K converges to x^∗ with respect to the weak^∗ topology of X^∗, then (x^∗_n)^∞_n=1 converges to x^∗ in the norm topology ofX^∗.

Proof. Let (_n)^∞_n=1be a sequence of scalars converging to zero. Using thatX^∗ is separable for the norm topology, for each _n there is a sequence of points (x⁽ⁿ⁾_k )^∞_k=1⊂X^∗ such that

K⊂ !^∞

k=1

B(x⁽ⁿ⁾_k , n)∩K .

Observe that for all integers n and k, B(x⁽ⁿ⁾_k , _n) (the closed ball centered in x⁽ⁿ⁾_k of radius_n) is weak^∗ compact by Banach-Alaoglu’s theorem, so the sets B(x⁽ⁿ⁾_k , _n)∩K are weak^∗ closed. Let us call B_k⁽ⁿ⁾ the weak^∗ interior of B(x⁽ⁿ⁾_k , _n)∩K. Hence

Vn=

!∞ k=1

B_k⁽ⁿ⁾ is dense and open.

SinceX^∗ is separable, the weak^∗ topology ofX^∗ relative toK is metriz- able. Then K is compact metric, therefore complete. By the Baire Category theorem, the set V = #_∞

n=1V_n is a dense G_δ-set. We are going to see that all of the elements in V are points of weak^∗-to-norm continuity. Indeed, take v^∗ ∈ V. Then for each n there exists a weak^∗ neighborhood of v^∗ relative to K of diameter at most 2n. Since (n)^∞_n=1 converges to zero, the identity operator

I: (K, w^∗)−→(K,·) is continuous atv^∗.

Lemma 6.3.5.SupposeX is a Banach space which embeds in a separable dual space. Then every closed bounded subsetF ofX has a point of weak-to-norm continuity.

6.3 Properties ofL1 147 Proof. Let F be a closed bounded subset of X. Suppose T : X → Y^∗ is an embedding in Y^∗, where Y is a Banach space with separable dual. We can assume that x ≤ T x ≤ Mx for x ∈ X where M is a constant independent ofx.LetW be the weak^∗closure ofT(F).Then by Lemma 6.3.4 there is y^∗ ∈W which is a point of weak^∗-to-norm continuity. In particular there is a sequence (y_n^∗) inT(F) with y_n^∗−y^∗ → 0. If we let y_n^∗ = T x_n with xn ∈ F for each n, then (xn)^∞_n=1 is Cauchy in X and so converges to some x ∈ F, hence T x = y^∗. Now for any > 0 we can find a weak^∗ neighborhood U of y^∗ so that w^∗ ∈ U ∩W implies w^∗ −y^∗ < . In particular if v ∈ T⁻¹(U)∩C then v−x < . Clearly T⁻¹(U) is weakly open since the map T :X →Y^∗ is weak-to-weak^∗ continuous. This shows x is a point of weak-to-norm continuity.

Lemma 6.3.6.SupposeX is a Banach space which embeds in a separable dual space and letx∈B_Xbe a point of weak-to-norm continuity. If(x_n)is a weakly null sequence inX such that lim supx+x_n ≤1 thenlim_n→∞x_n= 0.

Proof. Put

u_n=

x+x_n if x+x_n ≤1

x+x_n

x+x_n if x+xn>1 and observe that

u_n−x=x_n+ (1−α_n)(x+x_n),

where α_n = (x+x_n −1)₊ → 0. Thus lim_n→∞u_n = x weakly and so lim_n→∞u_n−x= 0.This implies that lim_n→∞x_n= 0.

Theorem 6.3.7.Neither of the Banach spacesL₁andc₀ can be embedded in a separable dual space.

Proof. IfL₁embeds in a separable dual space, Lemma 6.3.5 yields a function f ∈ B_L1 that is a point of weak-to-norm continuity. By Lemma 6.3.2 the sequence (r_nf)^∞_n=1 is weakly null inL₁ and satisfies

f +f r_n1=

(1 +r_n(t))|f(t)|dt−→1.

Therefore by Lemma 6.3.6 it must be lim_n→∞r_nf1 = 0, which implies f = 0. This is absurd since (r_n)^∞_n=1 is a weakly null sequence and r_n1= 1.

Forc₀the argument is similar. Letξbe a point of weak-to-norm continuity in B_c0. Then if (e_n)^∞_n=1 is the canonical basis we have lim_n→∞ξ+e_n= 1 and so lim_n→∞e_n= 0, which is again absurd.

Remark 6.3.8.The fact that c₀ cannot be embedded in a separable dual space can be proved in many ways, and we have already seen this in Prob- lems 2.6 and 2.9.

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.