Kenneth Kuttler

2354 CHAPTER 69. GELFAND TRIPLES

Consider a closed ball B(v0,r) in X . This equals{v ∈ X : sup

n|φ n (v)−φ n (v0)| ≤ r

}= ∩∞

n=1φ−1n

(B(φ n (v0) ,r)

)and this last set is in σ (D′). Therefore, every closed ball is in σ (D′) which implies everyopen ball is also in σ (D′) since open balls are the countable union of closed balls. SinceX is separable, it follows every open set is the countable union of balls and so every openset is in σ (D′). It follows B (X) ⊆ σ (D′) ⊆ σ (X ′). On the other hand, every φ ∈ X ′ iscontinuous and so it is Borel measurable. Hence σ (X ′)⊆B (X).

Now consider the last claim. From Lemma 21.1.6 and density of H ′ = H in V ′, it canbe assumed D′ ⊆ H = H ′. Therefore, from the first part of the argument

B (V )⊆ σ(D′)⊆ σ

(i∗H ′

)Also each i∗h is continuous on V so in fact, equality holds in the above because σ (i∗H ′)⊆B (V ). This proves the proposition.

Next I want to verify that V is in B (H). This will be true if V is reflexive. Moregenerally, here is an interesting result.

Proposition 69.0.4 Let X ⊆Y, X dense in Y and suppose X, Y are Banach spaces and thatX is reflexive. Then X ∈B (Y ).

Proof: Define the functional

φ (x)≡{||x||X if x ∈ X∞ if x ∈ Y \X

Then φ is lower semicontinuous on Y . Here is why. Suppose (x,a) /∈ epi(φ) so that a <φ (x) . I need to verify this situation persists for (x,b) near (x,a). If this is not so, thereexists xn→ x and an→ a such that an ≥ φ (xn) . If liminfn→∞ φ (xn)< ∞, then there exists asubsequence still denoted by n such that ||xn||X is bounded. Then by the Eberlein Smuliantheorem, there exists a further subsequence such that xn converges weakly in X to somez. Now since X is dense in Y it follows Y ′ can be considered a subspace of X ′ and so forf ∈ Y ′

f (xn)→ f (z) , f (xn)→ f (x)

and so f (z− x) = 0 for all f ∈Y ′ which requires z = x. Now x→ ||x||X is convex and lowersemicontinuous on X so it follows from Corollary 18.2.12

a = lim infn→∞

an ≥ lim infn→∞

φ (xn)≥ φ (x)> a

which is a contradiction. If liminfn→∞ φ (xn) = ∞, then

∞ > a = lim infn→∞

an = ∞

another contradiction. Therefore, epi(φ) is closed and so φ is lower semicontinuous asclaimed. Therefore,

X = Y \(∩∞

n=1φ−1 ((n,∞))

)and since φ is lower semicontinuous, each φ

−1 ((n,∞)) is open. Hence X is a Borel subsetof Y . This proves the proposition.

2354 CHAPTER 69. GELFAND TRIPLESConsider a closed ball B(vo,r) in X. This equals{ve x:suplo, (0) ~ey(0)] <r} = 105! (BG, 001-7)and this last set is in o (D’). Therefore, every closed ball is in o (D’) which implies everyopen ball is also in o (D’) since open balls are the countable union of closed balls. SinceX is separable, it follows every open set is the countable union of balls and so every openset is in o(D’). It follows 4(X) C o(D’) C o (X'). On the other hand, every @ € X’ iscontinuous and so it is Borel measurable. Hence o (X’) C #(X).Now consider the last claim. From Lemma 21.1.6 and density of H’ = H in V’, it canbe assumed D’ C H = H’. Therefore, from the first part of the argumentBV) Co(D') Co(i*H’')Also each i*h is continuous on V so in fact, equality holds in the above because o (i*H’) C£(V). This proves the proposition.Next I want to verify that V is in @(H). This will be true if V is reflexive. Moregenerally, here is an interesting result.Proposition 69.0.4 Let X CY, X dense in Y and suppose X, Y are Banach spaces and thatX is reflexive. Then X € B(Y).Proof: Define the functional_ |x||y ifx EXoi={ oo if x EY \XThen @ is lower semicontinuous on Y. Here is why. Suppose (x,a) ¢ epi(@) so that a <@ (x). I need to verify this situation persists for (x,b) near (x,a). If this is not so, thereexists x, — x and a, — a such that a, > @ (x,) . If liminf,_,.0 @ (x,) <9, then there exists asubsequence still denoted by n such that ||x,||y is bounded. Then by the Eberlein Smuliantheorem, there exists a further subsequence such that x, converges weakly in X to somez. Now since X is dense in Y it follows Y’ can be considered a subspace of X’ and so forfey’Pn) f(s Fan) > £09)and so f (z—x) =0 for all f € Y’ which requires z = x. Now x — ||x||y is convex and lowersemicontinuous on X so it follows from Corollary 18.2.12a=lim inf a, > lim inf @ (x) > (x) >an—-yoo n—oowhich is a contradiction. If liminf;_,.. @ (x) = °°, thencoo > a=lim inf a, =n-ooanother contradiction. Therefore, epi(@) is closed and so @ is lower semicontinuous asclaimed. Therefore,X=¥\ (19! ((2,~)))and since @ is lower semicontinuous, each @~! ((n,co)) is open. Hence X is a Borel subsetof Y. This proves the proposition.