Reflexive Spaces

Exercises for Chapter 2

3.5 Reflexive Spaces

K₁= &

x∈E

[−x,+x].

Let us recall that (arbitrary)products of compact spaces are compact—a deep theo- rem due to Tychonoff; see, e.g., H. L. Royden [1], G. B. Folland [2], J. R. Munkres [1], A. Knapp [1], or J. Dixmier [1]. ThereforeK₁is compact. On the other hand, K2is closed inY; indeed, for eachfixedλ∈R,x, y∈Ethe sets

Ax,y= {ω∈Y;ωx+y−ωx−ωy=0}, B_λ,x = {ω∈Y;ω_λx−λω_x=0},

are closed inY (since the mapsω→ ω_x+y−ω_x−ω_y andω →ω_λx−λω_xare continuous onY) and we may writeK₂as

K2=' #

x,y∈E

A_x,y

(∩' #

x∈E λ∈R

B_λ,x (

.

Finally,Kis compact since it is the intersection of a compact set (K1) and a closed set (K2).

3.5 Reflexive Spaces

Definition. LetEbe a Banach space and letJ :E→Ebe the canonical injection fromEintoE(see Section 1.3). The spaceEis said to bereflexiveifJis surjective, i.e.,J (E)=E.

WhenEis reflexive,Eis usually identified withE.

Remark13.Many important spaces in analysis are reflexive. Clearly,finite-dimen- sionalspaces are reflexive (since dimE =dimE =dimE). As we shall see in Chapter 4 (see also Chapter 11),L^p(and^p) spaces are reflexive for 1< p <∞. In Chapter 5 we shall see that Hilbert spaces are reflexive. However, equally important spaces in analysis arenotreflexive; for example:

• L¹andL^∞(and¹,^∞) are not reflexive (see Chapters 4 and 11);

• C(K),the space of continuous functions on an infinite compact metric spaceK, is not reflexive (see Exercise 3.25).

Remark14.It is essential to use J in the above definition. R. C. James [1] has constructed a striking example of a nonreflexive space with the property that there exists a surjective isometry fromEontoE.

Our next result describes a basic property of reflexive spaces:

•Theorem 3.17 (Kakutani).LetEbe a Banach space. ThenEis reflexive if and only if

B_E= {x∈E; x ≤1}

is compact in the weak topologyσ (E, E).

Proof. Assume first thatE is reflexive, so thatJ (B_E) = B_E. We already know (by Theorem 3.16) thatB_E is compact in the topologyσ (E, E). Therefore, it suffices to check thatJ⁻¹is continuous fromEequipped withσ (E, E)with values inEequipped withσ (E, E). In view of Proposition 3.2, we have only to prove that for every fixedf ∈ E the mapξ → f, J⁻¹ξ is continuous onE equipped withσ (E, E). Butf, J⁻¹ξ = ξ, f, and the mapξ → ξ, fis indeed continuous onEfor the topologyσ (E, E). Hence we have proved that B_Eis compact inσ (E, E).

The converse is more delicate and relies on the following two lemmas:

Lemma 3.3 (Helly).LetE be a Banach space. Letf₁, f₂, . . . , f_k be given inE and letγ₁, γ₂, . . . , γ_k be given inR. The following properties are equivalent:

(i)∀ε >0∃x_ε∈Esuch thatx_ε ≤1and

|f_i, x_ε −γ_i|< ε ∀i=1,2, . . . , k, (ii)|_k

i=1β_iγ_i| ≤ _k

i=1β_if_i ∀β₁, β₂, . . . , β_k ∈R.

Proof. (i)⇒(ii). Fixβ₁, β₂, . . . , β_kinRand letS=_k

i=1|β_i|. It follows from (i)

that

k

i=1

β_if_i, x_ε − k

i=1

β_iγ_i ≤εS and therefore

k i=1

β_iγ_i ≤

k i=1

β_if_i

x_ε +εS ≤

k i=1

β_if_i +εS.

Since this holds for everyε >0, we obtain (ii).

(ii)⇒(i). Setγ =(γ1, γ2, . . . , γk)∈ R^k and consider the mapϕ : E → R^k defined by

ϕ(x)=(f1, x, . . . ,fk, x).

Property (i) says precisely thatγ ∈ ϕ(B_E). Suppose, by contradiction, that (i) fails, so thatγ /∈ϕ(B_E). Hence{γ}andϕ(B_E)may be strictly separated inR^k by some hyperplane; i.e., there exists someβ=(β₁, β₂, . . . , β_k)∈R^kand someα∈R such that

β·ϕ(x) < α < β·γ ∀x ∈B_E. It follows that _k

i=1

βifi, x < α <

k i=1

βiγi ∀x∈BE,

3.5 Reflexive Spaces 69

and therefore

k i=1

β_if_i ≤α <

k i=1

β_iγ_i, which contradicts (ii).

Lemma 3.4 (Goldstine).LetEbe any Banach space. ThenJ (B_E)is dense inB_E with respect to the topologyσ (E, E), and consequentlyJ (E)is dense inEin the topologyσ (E, E).

Proof. Letξ ∈B_E and letV be a neighborhood ofξ for the topologyσ (E, E).

We must prove thatV∩J (B_E) = ∅. As usual, we may assume thatV is of the form V =$

η∈E; |η−ξ, f_i|< ε ∀i=1,2, . . . , k%

for some given elementsf₁, f₂, . . . , f_kinEand someε >0. We have to find some x∈B_Esuch thatJ (x)∈V, i.e.,

|f_i, x − ξ, f_i|< ε ∀i=1,2, . . . , k.

Setγi = ξ, fi. In view of Lemma 3.3 it suffices to check that

k i=1

β_iγ_i ≤

k

i=1

β_if_i ,

which is clear since_k

i=1β_iγ_i =) ξ,_k

i=1β_if_i

*

andξ ≤1.

Remark15.Note that J (B_E)is closedin B_E in the strong topology. Indeed, if ξ_n = J (x_n) → ξ we see that (x_n)is a Cauchy sequence in B_E (sinceJ is an isometry) and thereforex_n→x, so thatξ =J x. It follows thatJ (B_E)is not dense inB_E in the strong topology, unlessJ (B_E)=B_E, i.e.,Eis reflexive.

Remark16.See Problem 9 for an alternative proof of Lemma 3.4 (based on a variant of Hahn–Banach inE).

Proof of Theorem3.17,concluded.The canonical injectionJ :E→Eis always continuous fromσ (E, E)intoσ (E, E), since for every fixedf ∈Ethe map x → J x, f = f, xis continuous with respect toσ (E, E). Assuming thatB_E is compact in the topologyσ (E, E), we deduce thatJ (B_E)is compact—and thus closed—in E with respect to the topology σ (E, E). On the other hand, by Lemma 3.4,J (BE)is dense inBE for the same topology. It follows thatJ (BE)= BE and thusJ (E)=E.

In connection with the compactness properties of reflexive spaces we also have the following two results:

•Theorem 3.18.Assume thatEis a reflexive Banach space and let(xn)be a bounded sequence in E. Then there exists a subsequence(x_nk)that converges in the weak topologyσ (E, E).

The converse is also true, namely the following.

Theorem 3.19 (Eberlein–Smulian).ˇ Assume that E is a Banach space such that every bounded sequence in E admits a weakly convergent subsequence (in σ (E, E)). ThenEis reflexive.

The proof of Theorem 3.18 requires a little excursion through separable spaces and will be given in Section 3.6. The proof of Theorem 3.19 is rather delicate and is omitted; see, e.g., R. Holmes [1], K. Yosida [1], N. Dunford–J. T. Schwartz [1], J. Diestel [2], or Problem 10.

Remark17.In order to clarify the connection between Theorems 3.17, 3.18, and 3.19 it is useful to recall the following facts:

(i) IfXis ametricspace, then

[Xis compact] ⇔ [every sequence inXadmits a convergent subsequence]. (ii) There existcompact topologicalspacesXand some sequences inXwithoutany

convergent subsequence. A typical example isX =B_E, which is compact in the topologyσ (E, E); whenE=^∞it is easy to construct a sequence inX withoutany convergent subsequence (see Exercise 3.18).

(iii) If X is a topological space with the property that every sequence admits a convergent subsequence, thenXneednotbe compact.

Here are some further properties of reflexive spaces.

•Proposition 3.20.Assume thatEis a reflexive Banach space and letM⊂Ebe a closed linear subspace ofE. ThenMis reflexive.

Proof. The spaceM—equipped with the norm ofE—has a priori two distinct weak topologies:

(a) the topology induced byσ (E, E), (b) its own weak topologyσ (M, M).

In fact, these two topologies are the same (since, by Hahn–Banach, every continu- ous linear functional onMis the restriction toMof a continuous linear functional on E). In view of Theorem 3.17, we have to check thatBM is compact in the topology σ (M, M)or equivalently in the topologyσ (E, E). However,BE is compact in the topologyσ (E, E)andMis closed in the topologyσ (E, E)(by Theorem 3.7).

ThereforeB_Mis compact in the topologyσ (E, E).

Corollary 3.21.A Banach spaceEis reflexive if and only if its dual spaceE is reflexive.

Proof. Ereflexive⇒E reflexive. The idea of the proof is simple, since, roughly speaking, we have thatE = E ⇒ E = E. More precisely, let J be the canonical isomorphism fromEinto E. Letϕ ∈ E be given. The map x → ϕ, J xis a continuous linear functional onE. Call itf ∈E, so that

3.5 Reflexive Spaces 71

ϕ, J x = f, x ∀x∈E.

But we also have

ϕ, J x = J x, f ∀x∈E.

SinceJ is surjective, we infer that

ϕ, ξ = ξ, f ∀ξ ∈E,

which means precisely that the canonical injection fromEintoEis surjective.

E reflexive⇒ E reflexive. From the step above we already know thatEis reflexive. SinceJ (E)is a closed subspace ofEin the strong topology, we conclude (by Proposition 3.20) thatJ (E)is reflexive. Therefore,Eis reflexive.³

•Corollary 3.22.LetE be a reflexive Banach space. LetK ⊂ Ebe a bounded, closed, and convex subset ofE. ThenKis compact in the topologyσ (E, E).

Proof. Kis closed for the topologyσ (E, E)(by Theorem 3.7). On the other hand, there exists a constantmsuch thatK⊂mB_E, andmB_Eis compact inσ (E, E)(by Theorem 3.17).

•Corollary 3.23.LetEbe a reflexive Banach space and letA⊂Ebe a nonempty, closed, convex subset ofE. Letϕ:A→(−∞,+∞]be a convexl.s.c.function such thatϕ ≡ +∞and

(5) lim

x∈A x→∞

ϕ(x)= +∞ (no assumption ifAis bounded).

Thenϕachieves its minimum onA, i.e., there exists somex0∈Asuch that ϕ(x₀)=min

A ϕ.

Proof. Fix anya∈Asuch thatϕ(a) <+∞and consider the set A˜= {x ∈A; ϕ(x)≤ϕ(a)}.

ThenA˜is closed, convex, and bounded (by (5)) and thus it iscompactin the topology σ (E, E)(by Corollary 3.22). On the other hand,ϕ is also l.s.c. in the topology σ (E, E) (by Corollary 3.9). It follows that ϕ achieves its minimum on A˜ (see property 5 following the definition of l.s.c. in Chapter 1), i.e., there existsx0 ∈ ˜A such that

ϕ(x₀)≤ϕ(x) ∀x ∈ ˜A.

Ifx∈A\ ˜A, we haveϕ(x₀)≤ϕ(a) < ϕ(x); therefore ϕ(x₀)≤ϕ(x) ∀x∈A.

3It is clear that ifEandFare Banach spaces, andTis a linear surjective isometry fromEontoF, thenEis reflexive iffFis reflexive. Of course, there is no contradiction with Remark 14!

Remark18.Corollary 3.23 is the main reason whyreflexive spacesandconvex func- tionsare so important in many problems occurring in thecalculus of variationsand inoptimization.

Theorem 3.24.LetEandF be two reflexive Banach spaces. LetA:D(A)⊂E→ F be an unbounded linear operator that is densely defined and closed. ThenD(A) is dense inF. ThusAis well defined(A:D(A)⊂E →F)and it may also be viewed as an unbounded operator fromEintoF. Then we have

A=A.

Proof.

1. D(A)is dense inF. Let ϕ be a continuous linear functional on F that vanishes onD(A). In view of Corollary 1.8 it suffices to prove thatϕ ≡0 onF. SinceF is reflexive,ϕ∈F and we have

(6) w, ϕ =0 ∀w∈D(A).

Ifϕ =0 then[0, ϕ]∈/ G(A)inE×F. Thus, one may strictly separate[0, ϕ]and G(A)by a closed hyperplane inE×F; i.e., there exist some[f, v] ∈E×Fand someα∈Rsuch that

f, u + v, Au< α <v, ϕ ∀u∈D(A).

It follows that

f, u + v, Au =0 ∀u∈D(A) and

v, ϕ =0.

Thusv∈D(A), and we are led to a contradiction by choosingw=vin (6).

2.A=A. We recall (see Section 2.6) that I[G(A)] =G(A)^⊥ and

I[G(A)] =G(A)^⊥. It follows that

G(A)=G(A)^⊥⊥=G(A), sinceAis closed.