Separable Spaces

Exercises for Chapter 2

3.6 Separable Spaces

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.

3.6 Separable Spaces 73 Many important spaces in analysis are separable. Clearly, finite-dimensional spaces are separable. As we shall see in Chapter 4 (see also Chapter 11),L^p(and^p) spaces are separable for 1≤p <∞. AlsoC(K),the space of continuous functions on a compact metric spaceK, is separable (see Problem 24). However,L^∞and^∞ arenotseparable (see Chapters 4 and 11).

Proposition 3.25.LetEbe a separable metric space and letF ⊂Ebe any subset.

ThenF is also separable.

Proof. Let(u_n)be a countable dense subset ofE. Let(r_m)be any sequence of positive numbers such thatr_m→0. Choose any pointa_m,n∈B(u_n, r_m)∩F whenever this set is nonempty. The set(a_m,n)is countable and dense inF.

Theorem 3.26.Let E be a Banach space such that E is separable. Then E is separable.

Remark19.The converse is not true. As we shall see in Chapter 4, E = L¹ is separable but its dual spaceE =L^∞isnotseparable.

Proof. Let(f_n)_n≥1be countable and dense inE. Since f_n = sup

x∈E x≤1

f_n, x,

we can find somex_n∈Esuch that

x_n =1 andf_n, x_n ≥ 1 2f_n.

Let us denote byL0the vector space overQgenerated by the(x_n)_n≥1; i.e.,L0consists of all finitelinear combinations with coefficients in Q of the elements (xn)n≥1. We claim thatL0 is countable. Indeed, for every integer n, let n be the vector space overQgenerated by the(xk)1≤k≤n. Clearly,nis countable and, moreover, L0=

n≥1n.

LetLdenote the vector space overRgenerated by the(x_n)_n≥1. Of course,L₀is a dense subset ofL. We claim thatLis a dense subspace ofE—and this will conclude the proof(L₀will be a dense countable subset ofE). Letf ∈ E be a continuous linear functional that vanishes onL; in view of Corollary 1.8 we have to prove that f =0. Given anyε >0, there is some integerN such thatf−f_N< ε. We have

1

2f_N ≤ f_N, x_N = f_N−f, x_N< ε

(sincef, x_N =0). It follows thatf ≤ f−f_N + f_N<3ε. Thusf =0.

Corollary 3.27.LetEbe a Banach space. Then

[Ereflexive and separable] ⇔ [Ereflexive and separable].

Proof. We already know (Corollary 3.21 and Theorem 3.26) that [Ereflexive and separable]⇒ [Ereflexive and separable].

Conversely, ifEis reflexive and separable, so isE=J (E); thusE is reflexive and separable.

Separabilityproperties are closely related to themetrizabilityof the weak topolo- gies. Let us recall that a topological spaceX is said to bemetrizableif there is a metric onXthat induces the topology ofX.

Theorem 3.28.LetEbe a separable Banach space. ThenB_E is metrizable in the weaktopologyσ (E, E).

Conversely, ifB_Eis metrizable inσ (E, E), thenEis separable.

There is a “dual” statement.

Theorem 3.29.Let E be a Banach space such thatE is separable. ThenBE is metrizable in the weak topologyσ (E, E).

Conversely, ifB_Eis metrizable inσ (E, E), thenEis separable.

Proof of Theorem3.28.Let(x_n)_n≥1be a dense countable subset ofB_E. For every f ∈Eset

[f] = ∞ n=1

1

2ⁿ|f, x_n|.

Clearly, [ ] is a norm on E and[f] ≤ f. Let d(f, g) = [f −g] be the corresponding metric. We shall prove that the topology induced byd onBE is the same as the topologyσ (E, E)restricted toBE.

(a) Letf0 ∈BE and letV be a neighborhood off0forσ (E, E). We have to find somer >0 such that

U = {f ∈B_E;d(f, f₀) < r} ⊂V . As usual, we may assume thatV has the form

V = {f ∈B_E; |f −f₀, y_i|< ε ∀i=1,2, . . . , k}

withε >0 andy₁, y₂, . . . , y_k ∈E. Without loss of generality we may assume that y_i ≤1 for everyi=1,2, . . . , k. For everyithere is some integern_isuch that

y_i−x_ni< ε/4 (since the set(xn)n≥1is dense inBE).

Chooser >0 small enough that

2ⁿⁱr < ε/2 ∀i=1,2, . . . , k.

We claim that for suchr,U⊂V. Indeed, ifd(f, f₀) < r, we have

3.6 Separable Spaces 75 1

2ⁿⁱ|f −f0, xn_i|< r ∀i=1,2, . . . , k and therefore,∀i=1,2, . . . , k,

|f −f₀, y_i| = |f−f₀, y_i−x_ni + f −f₀, x_ni|< ε 2 +ε

2. It follows thatf ∈V.

(b) Letf₀ ∈ B_E. Given r >0, we have to find some neighborhoodV of f₀for σ (E, E)such that

V ⊂U= {f ∈B_E; d(f, f₀) < r}. We shall chooseV to be

V = {f ∈B_E; f −f₀, x_i|< ε ∀i=1,2, . . . , k} withεandkto be determined in such a way thatV ⊂U. Forf ∈V we have

d(f, f0)= k n=1

1

2ⁿ|f −f0, xn| + ∞ n=k+1

1

2ⁿ|f −f0, xn|

< ε+2 ∞ n=k+1

1

2ⁿ =ε+ 1 2^k−1.

Thus, it suffices to takeε= ^r₂andklarge enough that _2k¹₋₁ < ^r₂.

Conversely, suppose B_E is metrizable inσ (E, E)and let us prove thatEis separable. Set

U_n= {f ∈B_E; d(f,0) <1/n}

and letV_nbe a neighborhood of 0 inσ (E, E)such thatV_n⊂U_n. We may assume thatV_nhas the form

V_n= {f ∈B_E; |f, x|< ε_n ∀x∈_n} withε_n>0 and_nis a finite subset ofE. Set

D= ∞ n=1

_n,

so thatDis countable.

We claim that the vector space generated byDis dense inE(which implies that Eis separable). Indeed, supposef ∈Eis such thatf, x =0 ∀x∈D. It follows thatf ∈V_n ∀nand thereforef ∈U_n ∀n, so thatf =0.

Proof of Theorem3.29.The proof of the implication

[Eseparable]⇒ [B_Eis metrizable inσ (E, E)]

is exactly the same as above—just change the roles ofEandE. The proof of the converse is more delicate (find where the proof above breaks down); we refer to N. Dunford–J. T. Schwartz [1] or Exercise 3.24.

Remark20.One should emphasize again (see Remark 3) that in infinite-dimensional spaces the weak topologyσ (E, E)(resp. weak topologyσ (E, E))onallofE (resp.E) isnotmetrizable; see Exercise 3.8. In particular, the topology induced by the norm [ ] on all ofEdoes not coincide with the weaktopology.

Corollary 3.30.Let E be a separable Banach space and let (f_n) be a bounded sequence inE. Then there exists a subsequence(fn_k)that converges in the weak topologyσ (E, E).

Proof. Without loss of generality we may assume thatf_n ≤1 for alln. The setB_E is compact and metrizable for the topologyσ (E, E)(by Theorems 3.16 and 3.28).

The conclusion follows.

We may now return to the proof of Theorem 3.18:

Proof of Theorem3.18.Let M₀ be the vector space generated by the x_n’s and let M = M₀. Clearly,M is separable (see the proof of Theorem 3.26). Moreover,M is reflexive (by Proposition 3.20). It follows thatB_M is compact and metrizable in the weak topologyσ (M, M), sinceM is separable (we use here Corollary 3.27 and Theorem 3.29). We may thus find a subsequence(xn_k)that converges weakly σ (M, M), and hence (xn_k)converges also weaklyσ (E, E)(as in the proof of Proposition 3.20).