Documenta Math. 597 Characterising Weak-Operator Continuous Linear Functionals on B(H) constructively

Abstract. Let B(H) be the space of bounded operators on a notnecessarily-separable Hilbert space H . Working within Bishop-style constructive analysis, we prove that certain weak-operator continuous linear functionals on B(H) are finite sums of functionals of the form T 〈Tx, y〉. We also prove that the identification of weakand strong-operator continuous linear functionals on B(H) cannot be established constructively.


Introduction
Let H be a complex Hilbert space that is nontrivial (that is, contains a unit vector), B(H) the space of all bounded operators on H, and B 1 (H) the unit ball of B(H).In this paper we carry out, within Bishop-style constructive mathematics (BISH), 1 an investigation of weak-operator continuous linear functionals on B(H).Depending on the context, we use, for example, x to represent either the element (x 1 , . . ., x N ) of the finite direct sum H of N copies of H or else the element (x n ) n≥1 of the direct sum H ∞ ≡ n≥1 H of a sequence of copies of H.We use I to denote the identity projection on H.The following are the topologies of interest to us here.
⊲ The weak operator topology: the weakest topology on B(H) with respect to which the mapping T T x, y is continuous for all x, y ∈ H; sets of the form {T ∈ B(H) : | T x, y | < ε} , with x, y ∈ H and ε > 0, form a sub-base of weak-operator neighbourhoods of 0 in B(H).
⊲ The ultraweak operator topology: the weakest topology on B(H) with respect to which the mapping T ∞ n=1 T x n , y n is continuous for all x, y ∈ H ∞ ; sets of the form with x, y ∈ H ∞ and ε > 0, form a sub-base of ultraweak-operator neighbourhoods of 0 in B(H).
These topologies are induced, respectively, by the seminorms of the form T | T x, y | with x, y ∈ H, and those of the form T | ∞ n=1 T x n , y n | with x, y ∈ H ∞ .An important theorem in classical operator algebra theory states that the weak-operator continuous linear functionals on (any linear subspace of) B(H) all have the form T N n=1 T x n , y n with x, y ∈ H N for some N ; and the ultraweak-operator continuous linear functionals on B(H) have the form T ∞ n=1 T x n , y n , where x, y ∈ H ∞ .However, the classical proofs, such as those found in [10, 11, 14], depend on applications of nonconstructive versions of the Hahn-Banach theorem, the Riesz representation theorem, and polar decomposition.The foregoing characterisation of ultraweak-operator continuous functionals was derived constructively, when H is separable, in [9]. 2 A variant of it was derived in [8] (Proposition 5.4.16)without the requirement of separability, and using not the standard ultraweak operator topology but one that is classically, though not constructively, equivalent to it.Our aim in the present work is to provide a constructive proof of the standard classical characterisation of weakoperator continuous linear functionals (Theorem 10) on B(H), without the requirement of separability but with one hypothesis in addition to the classical ones.In presenting this work, we emphasise that, in contrast to their classical counterparts, our proofs contain extractable, implementable algorithms for the desired representation of weak-operator continuous linear functionals; moreover, the constructive proofs themselves verify that those algorithms meet their specifications.

Preliminary Lemmas
The proof of our main theorem depends on a sequence of (at-times-complicated) lemmas.For the first one, we remind the reader of two elementary definitions in constructive analysis: we say that an inhabited set S-that is, one in which we can construct an element-is finitely enumerable if there exist a positive integer N and a mapping of {1, . . ., N } onto S; if that mapping is one-one, then S is called finite.
Lemma 1 If u is a weak-operator continuous linear functional on B(H), then there exist a finitely enumerable subset F of H × H and a positive number Proof.This is an immediate consequence of Proposition 5.4.1 in [8].
We shall need some information about locally convex spaces.Let (p j ) j∈J be a family of seminorms defining the topology on a locally convex linear space V , and let A be a subset of V .A subset S of A is said to be located exists for each x ∈ A and each finitely enumerable subset F of J.We say that A is totally bounded if for each finitely enumerable subset F of J and each ε > 0, there exists a finitely enumerable subset S of A-called an εapproximation to S relative to (p j ) j∈F -such that for each x ∈ A there exists s ∈ S with j∈F p j (x − s) < ε.The unit ball B 1 (H) is weak-operator totally bounded ([8], Proposition 5.4.15);but, in contrast to the classical situation, it cannot be proved constructively that B 1 (H) is weak-operator complete [5].A mapping f between locally convex spaces X, (p j ) j∈J and Y, (q k ) k∈K is uniformly continuous on a subset S of X if for each ε > 0 and each finitely enumerable subset G of K, there exist δ > 0 and a finitely enumerable subset F of J such that if x, x ′ ∈ S and j∈F p j (x − x ′ ) < δ, then k∈G q k (f (x) − f (x ′ )) < ε.We recall four facts about total boundedness, locatedness, and uniform continuity in a locally convex space V .The proofs are found on pages 129-130 of [8].
⊲ If f is a uniformly continuous mapping of a totally bounded subset A of V into a locally convex space, then f (A) is totally bounded.
⊲ If f is a uniformly continuous, real-valued mapping on a totally bounded subset A of V , then sup x∈A f (x) and inf x∈A f (x) exist.
⊲ A totally bounded subset of V is located in V .

Documenta Mathematica 16 (2011) 597-617
⊲ If A ⊂ V is totally bounded and S ⊂ A is located in A, then S is totally bounded.
We remind the reader that a bounded linear mapping T : X → Y between normed linear spaces is normed if its norm, exists.If X is finite-dimensional, then T exists; but the statement 'Every bounded linear functional on an infinite-dimensional Hilbert space is normed'3 is essentially nonconstructive.
Lemma 2 Every weak-operator continuous linear functional on B(H) is normed.
Proof.This follows from observations made above, since, in view of Lemma 1, the linear functional is weak-operator uniformly continuous on the weakoperator totally bounded set B 1 (H).
We note the following stronger form of Lemma 1.
Lemma 3 Let u be a weak-operator continuous linear functional on B(H).Then there exist δ > 0, and finitely many nonzero4 elements ξ 1 , . . ., ξ N and Proof.By Lemma 1, there exist a positive integer ν,C > 0, and vectors x, y ∈ H ν such that |u(T )| ≤ C ν n=1 | T x n , y n | for all T ∈ B(H). 5 For each n ≤ ν, construct nonzero vectors x ′ n , y ′ n such that x ′ n = x n and y ′ n = y n .The desired result follows from the inequality the fact that each of the vectors x ′ n , x n − x ′ n , y ′ n , and y n − y ′ n is nonzero, and scaling to get the desired norm sums equal to 1 and then the positive δ.
The next lemma will be used in an application of the separation theorem in the proof of Lemma 6.
Let K be a finite-dimensional subspace of H N , and let * be the standard norm on the dual space K * of K: Then F is weak-operator uniformly continuous on B 1 (H).
Proof.Given ε > 0, let {x 1 , . . ., x m } be an ε-approximation to the (compact) unit ball of K. Writing For each x in the unit ball of K, there exists i such that x − x i < ε.We compute In order to ensure that the unit kernel B 1 (H) ∩ ker u of a weak-operator continuous linear functional u on B(H) is weak-operator totally bounded, and hence weak-operator located, we derive a generalisation of Lemma 5.4.9 of [8].
Lemma 5 Let V, (p j ) j∈J be a locally convex space.Let V 1 be a balanced, convex, and totally bounded subset of V .Let u be a linear functional on V that, on V 1 , is both uniformly continuous and nonzero.Then V 1 ∩ ker u is totally bounded.
Proof.Since u is nonzero and uniformly continuous on the totally bounded set y 1 belongs to the balanced set V 1 , and u(y 0 ) = C/2.Let ε > 0, and let F be a finitely enumerable subset of J. Since each p j is uniformly continuous on V , it maps the totally bounded set V 1 onto a totally bounded subset of R. 6 Hence there exists b > 0 such that p j (x) ≤ b for each j ∈ F and each is totally bounded.Pick a t-approximation {s 1 , . . ., s n } of S t relative to (p j ) j∈F , and set Then Thus Now consider any element y of V 1 ∩ ker(u).Since y ∈ S t , there exists k such that j∈F p j (y − s k ) < t and therefore Thus {y 1 , . . ., y n } is a finitely enumerable ε-approximation to V 1 ∩ ker(u) relative to the family (p j ) j∈F of seminorms.
The next lemma, the most complicated in the paper, extracts much of the sting from the proof of our main theorem by showing how to find finitely many mappings of the form T T x, ζ whose sum is small on the unit kernel of u. and Then for each ε > 0, there exists a unit vector x in the subspace Proof.First note that since each ξ n is nonzero, K is an N -dimensional subspace of H N .Now, an application of Lemma 5 tells us that the unit kernel B 1 (H) ∩ ker u of u is weak-operator totally bounded.For each x ∈ H N , since the mapping T N n=1 T x n , ζ n is weak-operator uniformly continuous on the unit kernel, we see that x 0 is a seminorm on H N satisfying x 0 ≤ ζ x = x ; whence the identity mapping from (H N , ) to (H N , 0 ) is uniformly continuous.Since the subset ) is totally bounded, it follows that exists.It will suffice to prove that β = 0.For then, given ε with 0 < ε < 1, we can construct a unit vector x ′ ∈ K such that To prove that β = 0, we suppose that β > 0. Then 0 is a norm equivalent to the original norm on K, so (K, 0 ) is an N -dimensional Banach space.Define norms * and * 0 on the dual K * of K by For each T ∈ B(H) and each x ∈ K let Then, by Lemma 4, F is weak-operator uniformly continuous as a mapping of B 1 (H) into K * , * ; since the norms * and * 0 are equivalent on the finite-dimensional dual space K * , F is therefore weak-operator uniformly continuous as a mapping of B 1 (H) into K * , * 0 .Hence is a totally bounded, and therefore located, subset of K * , * 0 .Moreover, for each T ∈ B 1 (H) ∩ ker u and each x ∈ K, |F (T ) (x)| ≤ x 0 ; so D is a subset of the unit ball S * 0 of K * , * 0 .We shall use the separation theorem from functional analysis to prove that D is * 0 -dense in S * 0 .Consider any φ in S * 0 , and suppose that Now, D is bounded, convex, balanced, and located; so, by Corollary 5.2.10 of [8], there exists a linear functional v on K * , * 0 with norm 1 such that Documenta Mathematica 16 (2011) 597-617

Douglas S. Bridges
It is a simple exercise8 to show that since K * , * 0 is N -dimensional, there exists y ∈ K such that y 0 = 1 and v(f ) = f (y) for each f ∈ K * .Hence which contradicts the fact that φ ∈ S * 0 .We conclude that d = 0 and therefore that D is * 0 -dense in S * 0 .Continuing our proof that β = 0, pick T 0 ∈ B 1 (H) with u(T 0 ) > 0. Replacing u by u(T 0 ) −1 u if necessary, we may assume that u(T 0 ) = 1.Define a linear functional Ψ on (K, 0 ) by setting Note that for x ∈ K we have Hence Ψ ∈ S * 0 .By the work of the previous paragraph, we can find T ∈ B 1 (H)∩ ker u such that Ψ − F (T ) * 0 < β/2δ.In particular, since In order to apply the defining property of δ and thereby obtain a contradiction, we need to estimate not the sum on the left hand side of (2), but To do so, we write where P, Q are disjoint sets, .
We shall apply Lemma 6 shortly; but its application requires another construction.
Lemma 7 Let N be a positive integer, let ξ 1 , . . ., ξ N be linearly independent vectors in H, and let ζ 1 , . . ., ζ N be nonzero elements of H, such that Then there exists a positive number c with the following property: for each unit vector z in the subspace Define a norm on the N -dimensional span V of {ξ 1 , . . ., ξ N } by which is absurd.Hence we can pick ν such that and T x = 0 whenever x is orthogonal to V .Then The next lemma takes the information arising from the preceding two, and shows that when the vectors ξ n in (1) are linearly independent, we can approximate u by a finite sum of mappings of the form T T x, y , not just on its unit kernel but on the entire unit ball of B(H).At the same time, we produce a priori bounds on the sums of squares of the norms of the components of the vectors x, y that appear in the terms T x, y whose sum approximates u(T ).Those bounds will be needed in the proof of our characterisation theorem.
Lemma 8 Let H be a Hilbert space, and u a nonzero weak-operator continuous linear functional on B(H).Let δ be a positive number, ξ 1 , . . ., ξ N linearly independent vectors in H, and and (1) holds.Let c > 0 be as in Lemma 7.Then for each ε > 0, there exists Proof.Pick T 0 ∈ B 1 (H) with u(T 0 ) > 0. To begin with, take the case where u(T 0 ) = 1 and therefore u ≥ 1.Given ε > 0, set Applying Lemma 6, we obtain nonzero vectors By Lemma 7, there exists we have 0 = x n ∈ Cξ n for each n, and Moreover, for each T ∈ B 1 (H), We now remove the restriction that u(T 0 ) = 1.Applying the first part of the theorem to v ≡ u(T 0 ) −1 u, we construct y ∈ K such that each component and we obtain the desired conclusion by taking x ≡ u(T 0 )y.
Lemma 9 Under the hypotheses of Lemma 8, but without the assumption that u is nonzero, for all ε, ε ′ > 0, there exists Proof.Either u > 0 and we can apply Lemma 8, or else u < ε/2.In the latter event, pick Then for each T ∈ B 1 (H) we have and therefore

The Characterisation Theorem
We are finally able to prove our main result, by inductively applying Lemma 9.
Theorem 10 Let H be a nontrivial Hilbert space, and u a nonzero weakoperator continuous linear functional on B(H).Let δ be a positive number, ξ 1 , . . ., ξ N linearly independent vectors in H, 9 and for all T ∈ B(H).
Suppose that for some positive integer k we have constructed vectors 9 The requirement that the vectors ξn be linearly independent is the one place where we have a stronger hypothesis than is needed in the classical theorem.It is worth noting here that if u(T ) has the desired form N n=1 T ξn, ζn , then classically we can find a set F of indices n ≤ N such that (i) the set S of those ξn with n ∈ F is linearly independent and (ii) if ξ k / ∈ S, then ξ k is linearly dependent on S. We can then write with each λn ∈ C. Constructively, this is not possible, since we cannot necessarily determine whether or not ξn is linearly dependent on the vectors ξ 1 , . . ., ξ n−1 .
For nonzero u, the proof of our theorem can be simplified at each stage of the induction, since we can use Lemma 8 directly.If H has dimension > N , we can then construct the classical representation of u in the general case as follows.
Either u > 0 and there is nothing to prove, or else u < δ (the same δ as in the statement of the theorem).In the latter case, we construct a unit vector ξ N +1 orthogonal to each of the vectors , and consider the weak-operator continuous linear functional We have where I is the identity operator on H; so v is nonzero.We can therefore apply the nonzero case to v, to produce a vector y ∈ Cξ Next we turn briefly to the strong operator topology on B(H): the locally convex topology generated by the seminorms T T x with x ∈ H. (That is, the weakest topology with respect to which the mapping T T x is continuous for each x ∈ H.) Clearly, a weak-operator continuous linear functional on B(H) is strong-operator continuous.The converse holds classically, but, as we now show by a Brouwerian example, is essentially nonconstructive.Let (e n ) n≥1 be an orthonormal basis of unit vectors in an infinite-dimensional Hilbert space, and let (a n ) n≥1 be a binary sequence with at most one term equal to 1. Then for k ≥ j we have 2 converges to T e 1 , we see that a n T e 1 , e n defines a linear functional u on B(H); moreover, |u(T )| ≤ T e 1 for each T , so (by Proposition 5.4.1 of [8]) u is strong-operator continuous.Suppose it is also weak-operator continuous.Then, by Lemma 2, it is normed.Either u < 1 or u > 0. In the first case, if there exists (a unique) ν with a ν = 1, then u(T ) = T e 1 , e ν for each T ∈ B(H).Defining T such that T e 1 = e ν and T e n = 0 for all n = ν, we see that T ∈ B 1 (H) and u(T ) = 1; whence u = 1, a contradiction.Thus in this case, a n = 0 for all n.On the other hand, in the case u > 0 we can find T such that u(T ) > 0; whence there exists n such that a n = 1.It now follows that the statement If H is an infinite-dimensional Hilbert space, then every strongoperator continuous linear functional on B(H) is weak-operator continuous implies the essentially nonconstructive principle LPO: For each binary sequence (a n ) n≥1 , either a n = 0 for all n or else there exists n such that a n = 1 and so is itself essentially nonconstructive.

Concluding Observations
The ideal constructive form of Theorem 10 would have two improvements over the current one.First, the requirement that the vectors ξ n be linearly independent would be relaxed to have them only nonzero in Lemma 8, Lemma 9, and Theorem 10.Second, B(H) would be replaced by a suitable linear subspace R of itself, and our theorem would apply to linear functionals that are weak-operator continuous on R, where "suitable" probably means "having weak-operator totally bounded unit ball R 1 ≡ R ∩ B 1 (H)".With that notion of suitability and with minor adaptations, Lemma 6 holds and the proof of Lemma 8 goes through as far as the construction of the vector z ∈ K.In fact, Theorem 10 goes through with B(H) replaced by any linear subspace R of B(H) that has weak-operator totally bounded unit ball and satisfies the following condition (cf.This condition holds in the special case where N = 1, in which case, if also R 1 is weak-operator totally bounded, we obtain Theorem 1 of [6]. 11However, there seems to be no means of establishing (*) for N > 1 and a general R.So the ideal form of our theorem remains an ideal and a challenge.

ζ n 2 = 1 .
Lemma 7):(*) Let N be a positive integer, let ξ 1 , . . ., ξ N be linearly independent vectors in H, and let ζ 1 , . . ., ζ N be nonzero elements of H, Then there exists a positive number c with the following property: for each unit vector z in the subspaceK ≡ Cξ 1 × • • • × Cξ N there exists T ∈ R 1 such that N n=1 T z n , ζ n > c.