LOCATING Ax, WHERE A IS A SUBSPACE OF B(H)

Logical Methods in Computer Science Vol. 10:9)014, pp. 1 10 www.lmcs-online.org Submitted Jan. 4, 013 Published Jun. 16, 014 LOCATING Ax, WHERE A IS A SUBSPACE OF BH) DOUGLAS S. BRIDGES University of Canterbury, Christchurch, New Zealand e-mail address: douglas.bridges@canterbury.ac.nz Abstract. Given a linear space of operators on a Hilbert space, any vector in the latter determines a subspace of its images under all operators. We discuss, within a Bishop-style constructive framework, conditions under which the projection of the original Hilbert space onto the closure of the image space exists. We derive a general result that leads directly to both the open mapping theorem and our main theorem on the existence of the projection. 1. Introduction Let H be a real or complex Hilbert space, BH) the space of bounded operators on H, and A a linear subspace of BH). For each x H write Ax {Ax : A A}, and, if it exists, denotetheprojection of H onto theclosure Ax of Axby [Ax]. Projections of this type play a very big part in the classical theory of operator algebras, in which context A is normally a subalgebra of BH); see, for example, [10, 11, 13, 15]. However, in the constructive 1 setting the one of this paper we cannot even guarantee that [Ax] exists. Our aim is to give sufficient conditions on A and x under which [Ax] exists, or, equivalently, the set Ax is located, in the sense that ρv,ax) inf{ v Ax : A A} exists for each v H. We require some background on operator topologies. Specifically, in addition to the standard uniform topology on BH), we need the strong operator topology: the weakest topology on BH) with respect to which the mapping T Tx is continuous for all x H; the weak operator topology: the weakest topology on BH) with respect to which the mapping T Tx,y is continuous for all x,y H. 01 ACM CCS: [Theory of computation]: Constructive Mathematics. 010 Mathematics Subject Classification: 03F60,46S30,47S30. Key words and phrases: constructive, Hilbert space, space of operators, located. 1 Our constructive setting is that of Bishop [, 3, 6], in which the mathematics is developed with intuitionistic, not classical, logic, in a suitable set- or type-theoretic framework [1, 1] and with dependent choice permitted. LOGICAL METHODS c Douglas S. Bridges Ð IN COMPUTER SCIENCE DOI:10.168/LMCS-10:9)014 CC Creative Commons

DOUGLAS S. BRIDGES These topologies are induced, respectively, by the seminorms of the form T Tx with x H, and T Tx,y with x,y H. The unit ball B 1 H) {T BH) : T 1} of BH) is classically weak-operator compact, but constructively the most we can say is that it is weak-operator totally bounded see [4]). The evidence so far suggests that in order to make progress when dealing constructively with a subspace or subalgebra A of BH), it makes sense to add the weak-operator total boundedness of A 1 A B 1 H) to whatever other hypothesis we are making; in particular, it is known that A 1 is located in the strong operator topology and hence A 1 x is located for each x H if and only if it is weak-operator totally bounded [7, 14]. Recall that the metric complement of a subset S of a metric space X is the set S of those elements of X that are bounded away from X. When Y is a subspace of X, y Y, and S Y, we define if that infimum exists. We now state our main result. ρ Y y, S) inf{ρy,z) : z Y S} Theorem 1.1. Let A be a uniformly closed subspace of BH) such that A 1 is weak-operator totally bounded, and let x be a point of H such that Ax is closed andρ Ax 0, A 1 x) exists. Then the projection [Ax] exists. Before proving this theorem, we discuss, in Section, some general results about the locatedness of sets like Ax, and we derive, in Section 3, a generalisation of the open mapping theorem that leads to the proof of Theorem 1.1. Finally, we show, by means of a Brouwerian example, that the existence of ρ Ax 0, A 1 x) cannot be dropped from the hypotheses of our main theorem.. Some general locatedness results for Ax We now prove an elementary, but helpful, result on locatedness in a Hilbert space. Proposition.1. Let S n ) n 1 be a sequence of located, convex subsets of a Hilbert space H such that S 1 S, let S = n 1S n, and let x H. For each n, let x n S n satisfy x x n < ρx,s n )+ n. Then ρx,s ) = inf n 1 ρx,s n) = lim n ρx,s n),.1) in the sense that if any of these three numbers exists, then all three do and they are equal. Moreover, ρx,s ) exists if and only if x n ) n 1 converges to a limit x H; in that case, ρx,s ) = x x, and x y > x x for all y S with y x. Note that it is not constructively provable that every element T of BH) is normed, in the sense that the usual operator norm of T exists. Nevertheless, when we write T 1, we are using a shorthand for Tx x for each x H. Likewise, T < 1 means that there exists c < 1 such that Tx c x for each x H; and T > 1 means that there exists x H such that Tx > x.

LOCATING Ax, WHERE A IS A SUBSPACE OF BH) 3 Proof. Suppose that ρx,s ) exists. Then ρx,s ) ρx,s n ) for each n. On the other hand, given ε > 0 we can find z S such that x z < ρx,s )+ε. Pick N such that z S N. Then for all n N, ρx,s ) ρx,s n ) ρx,s N ) x z < ρx,s )+ε. The desired conclusion.1) now follows. Next, observe that by the parallelogram law in H) if m n, then x m x n x x m ) x x n ) = x x m + x x n 4 x 1 x m +x n ) ρx,s m )+ m) + ρx,sn )+ n) 4ρx,Sm ), since 1 x m +x n ) S m. Thus x m x n ρx,sm )+ m) ρx,sm ) ) + ρx,sn )+ n) ρx,sm ) )..) If ρx,s ) exists, then, by the first part of the proof, ρx,s n ) ρx,s ) as n. It follows from this and.) that x m x n 0 as m,n ; whence x n ) n 1 is a Cauchy sequence in H and therefore converges to a limit x S. Then ρx,s ) = ρ x,s ) x x = lim x x n n lim ρx,sn )+ n) = ρx,s ). n Thus ρx,s ) = x x. Conversely, suppose that x = lim n x n exists. Let 0 < α < β and ε = 1 3 β α). Pick N such that N < ε and x x n < ε for all n N. Either x x > α+ε or x x < β. In the first case, for all n N, ρx,s n ) > x x n n x x x x n ε > α+ε) ε ε = α. In the other case, there exists ν > N such that x x ν < β; we then have ρx,s ν ) x x ν < β. It follows from this and the constructive least-upper-bound principle [6], Theorem.1.18) that inf{ρx,s n ) : n 1} exists; whence, by.1), d ρx,s ) exists.

4 DOUGLAS S. BRIDGES Finally, suppose that x exists, and consider any y S with y x. We have so x y > d. 0 < y x = y x x x) = y x + x x 4 = y x d ) + For each positive integer n we write y +x x x x d ) = A n na 1 = {na : A A 1 }. y x d ), If A 1 is weak-operator totally bounded and hence strong-operator located, then A n has those two properties as well. Our interest in Proposition.1 stems from this: Corollary.. Let A be a linear subspace of BH) with A 1 weak-operator totally bounded, and let x,y H. For each n, let y n A n satisfy y y n < ρx,a n x)+ n. Then ρy,ax) = inf n 1 ρy,a nx) = lim n ρy,a nx). Moreover, ρy,ax) exists if and only if y n ) n 1 converges to a limit y H; in which case, ρy,ax) = y y, and y Ax > y y for each A A such that Ax y. One case of this corollary arises when the sequence ρy,a n x)) n 1 stabilises: Proposition.3. Let A be a linear subspace of BH) such that A 1 is weak-operator totally bounded. Let x,y H, and suppose that for some positive integer N, ρy,a N x) = ρy,a N+1 x). Then ρy,ax) exists and equals ρy,a N x). Proof. By Theorem 4.3.1 of [6], there exists a unique z A N x such that ρy,a N x) = y z. We prove that y z is orthogonal to Ax. Let A A, and consider λ C so small that λa A 1. Since, z λax A N+1 x, we have y z λax,y z λax ρy,a N+1 x) = ρy,a N x) = y z,y z. This yields λ Ax +Reλ y z,ax ) 0. Suppose that Re y z,ax 0. Then by taking a sufficiently small real λ with λre y z,ax < 0, we obtain a contradiction. Hence Re y z,ax = 0. Likewise, Im y z,ax = 0. Thus y z,ax = 0. Since A A is arbitrary, we conclude that y z is orthogonal to Ax and hence to Ax. It is well known that this implies that z is the unique closest point to y in the closed linear subspace Ax. Since Ax is dense in Ax, it readily follows that ρy,ax) = ρ y,ax ) = y z.

LOCATING Ax, WHERE A IS A SUBSPACE OF BH) 5 The final result in this section will be used in the proof of our main theorem. Proposition.4. Let A be a linear subspace of BH) with weak-operator totally bounded unit ball, and let x H. Suppose that there exists r > 0 such that A 1 x B Ax 0,r) Ax B0,r). Then Ax is located in H; in fact, for each y H, there exists a positive integer N such that ρy,ax) = ρy,a N x). Proof. Fixing y H, compute a positive integer N > y /r. Let A A, and suppose that y Ax < ρy,a N x). We have either Ax < Nr or Ax > y. In the first case, N 1 Ax B Ax 0,r), so there exists B A 1 with N 1 Ax = Bx and therefore Ax = NBx. But NB A N, so y Ax = y NBx ρy,a N x), a contradiction. In the case Ax Nr > y, we have y Ax Ax y > y ρy,a N x), another contradiction. We conclude that y Ax ρy,a N x) for each A A. On the other hand, given ε > 0, we can find A A N such that y Ax < ρy,a N x)+ε. It now follows that ρy,ax) exists and equals ρy,a N x). 3. Generalising the open mapping theorem Thekey toourmain resultontheexistence of projections of theform[ax] is ageneralisation of the open mapping theorem from functional analysis [6], Theorem 6.6.4). Before giving that generalisation, we note a proposition and a lemma. Proposition 3.1. If C is a balanced, convex subset of a normed space X, then V n 1nC is a linear subspace of X. Proof. Let x V and α C. Pick a positive integer n and an element c of C such that x = nc. If α 0, then since C is balanced, α 1 αc C, so αx = αnc = α n α 1 αc α nc 1+ α )nc. In the general case, we can apply what we have just proved to show that Now, since C is balanced, Hence, by the convexity of + α )nc, 1+α)x 1+ 1+α )nc + α )nc. x = n c) nc + α )nc. αx = 1+α)x x + α )nc. Taking N as any integer > + α )n, we now see that αx NC V. In view of the foregoing and the fact that nc) n 1 is an ascending sequence of sets, if x also belongs to V

6 DOUGLAS S. BRIDGES we can take N large enough to ensure that αx and x both belong to NC. Picking c,c C such that αx = Nc and x = Nc, we obtain ) c+c αx+x = N NC, so αx+x V. We call a bounded subset C of a Banach space X superconvex if for each sequence x n ) n 1 in C and each sequence λ n ) n 1 of nonnegative numbers such that n=1 λ n converges to 1 and the series n=1 λ nx n converges, we have n=1 λ nx n C. In that case, C is clearly convex. Lemma 3.. Let C be a located, bounded, balanced, and superconvex subset of a Banach space X, such that X = n 1nC. Let y X and r > y. Then there exists ξ C such that if y ξ, then ρz,c) > 0 for some z with z < r. Proof. Either ρy,c) > 0 and we take z = y, or else, as we suppose, ρy,c) < r/. Choosing x 1 C such that y 1 x 1 < r/ and therefore y x1 < r, set λ 1 = 0. Then either ρy x 1,C) > 0 or ρy x 1,C) < r/. In the first case, set λ k = 1 and x k = 0 for all k. In the second case, pick x C such that y x1 1 x < r/ and therefore y x 1 x < r, and set λ = 0. Carrying on in this way, we construct a sequence x n ) n 1 in C, and an increasing binary sequence λ n ) n 1 with the following properties. If λ n = 0, then ) n ρ n 1 y n i 1 x i,c < r and If λ n = 1 λ n 1, then ρ n y n 1 y n n i x i < r. ) n n i 1 x i,c > 0 and x k = 0 for all k n. Compute α > 0 such that x < α for all x C. Then the series i x i converges, by comparison with α i, to a sum ξ in the Banach space X. Since i = 1 and C is superconvex, we see that ) 1 i x i = i x i C. If y ξ, then there exists N such that N y i x i > N r

LOCATING Ax, WHERE A IS A SUBSPACE OF BH) 7 and therefore N y N N i x i > r. It follows that we cannot have λ N = 0, so λ N = 1 and therefore there exists ν N such that λ ν = 1 λ ν 1. Setting ν 1 z ν 1 y ν i 1 x i, we see that ρz,c) > 0 and z < r, as required. We now prove our generalisation of the open mapping theorem. Theorem 3.3. Let X be a Banach space,and C a located, bounded, balanced, and superconvex subset of X such that ρ0, C) exists and X = n 1nC. Then there exists r > 0 such that B0,r) C. Proof. Consider the identity X = n 1nC. By Theorem 6.6.1 of [6] see also [8]), there exists N such that the interior of NC is inhabited. Thus there exist y 0 NC and R > 0 such that By 0,R) NC. Writing y 1 = N 1 y 0 and r = N) 1 R, we obtain By 1,r) C.It follows from Lemma 6.6.3 of [6] that B0,r) C. Now consider any y B0,r). By Lemma 3., there exists ξ C such that if y ξ, then there exists z B0,r) with ρz,c) > 0. Since B0,r) C, this is absurd. Hence y = ξ C. It follows that B0,r) C and hence that B0,r) C. Note that in Lemma 3. and Theorem 3.3 we can replace the superconvexity of C by these two properties: C is convex, and for each sequence x n ) n 1 in C, if n=1 n x n converges in H, then its sum belongs to C. We now derive two corollaries of Theorem 3.3. Corollary 3.4 The open mapping theorem[6], Theorem6.6.4) 3 ). Let X,Y be Banach ) spaces, and T a sequentially continuous linear mapping of X onto Y such that T B0, 1) )) is located and ρ 0, T B0, 1) exists. Then there exists r > 0 such that B0, r) ) T B0,1). ) Proof. In view of Theorem 3.3, it will suffice to prove that C T B0,1) is superconvex. But if x n ) n 1 is a sequence in B0,1) and λ n ) n 1 is a sequence of nonnegative numbers such that n=1 λ n = 1, then λ n x n λ n for each n, so n=1 λ nx n converges in X; moreover, λ n x n λ n = 1, n=1 n=1 3 This is but one version of the open mapping theorem; for another, see [5].

8 DOUGLAS S. BRIDGES so, by the sequential continuity of T, ) T λ n x n C. Thus C is superconvex. n=1 Theorem 3.3 also leads to the proof of Theorem 1.1: Proof. Taking C A 1 x, we know that C is located since A 1 is weak-operator totally bounded and hence, by [7, 14], strong-operator located), as well as bounded and balanced. To prove that C is superconvex, consider a sequence A n ) n 1 in A 1, and a sequence λ n ) n 1 of nonnegative numbers such that n=1 λ n converges to 1. For k j we have k k λ n A n λ n, n=j so n=1 λ na n converges uniformly to an element A of B 1 H). Since A is uniformly closed, A A 1, so n=1 λ na n x = Ax A 1 x. Thus C is superconvex. We can now apply Theorem 3.3, to produce r > 0 such that B Ax 0,r) C. The locatedness of Ax, and the consequent existence of the projection [Ax], now follow from Proposition.4. We now discuss further the requirement, in Theorem 1.1, that ρ Ax 0, A 1 x) exist, where A 1 is weak-operator totally bounded. We begin by giving conditions under which that requirement is satisfied. If Ax has positive, finite dimension in which case it is both closed and located in H then Ax A 1 x is inhabited, so Proposition 1.5) of [9] can be applied to show that Ax A 1 x is located in Ax. In particular, ρ Ax 0, A 1 x) exists. On the other hand, if P is a projection in BH) and A {PTP : T BH)}, then A can be identified with BPH)), so A 1 is weak-operator totally bounded. Moreover, if x 0, then Ax = PH) and so is both closed and located, A 1 x = B0, Px ) PH), and ρ Ax 0, A 1 x) = Px. We end with a Brouwerian example showing that we cannot drop the existence of ρ Ax 0, A 1 x) from the hypotheses of Theorem 1.1. Consider the case where H = R R, and let A be the linear subspace actually an algebra) of BH) comprising all matrices of the form ) a 0 T a,b 0 b with a,b R. It is easy to show that A is uniformly closed: if a n ),b n ) are ) sequences in a p R such that T an,b n ) n 1 converges uniformly to an element T, then q b ) ) 1 1 a n = T an,b n T = a 0 0, n=j Likewise, b n b, p = 0, and q = 0. Hence T = T a,b A.

LOCATING Ax, WHERE A IS A SUBSPACE OF BH) 9 Now, if x,y) is in the unit ball of H, then ) x ) a,b T = ax y = a x +b y by = a x +y ) + b a ) y = a + b a ) y. We see from this that if a b, then T a,b a ; moreover, T a,b 1,0) = a, so T a,b = a. If a < b, then a similar argument shows that T a,b = b. It now follows that T a,b exists and equals max{ a, b }. Also, since, relative to the uniform topology on BH), A 1 is homeomorphic to the totally bounded subset {a,b) : max{ a, b } 1} of R, it is uniformly, and hence weak-operator, totally bounded. Consider the vector ξ 1,c), where c R. If c = 0, then Aξ = R {0}, the projection of H on Aξ is just the projection on the x-axis, and ρ0,1),aξ) = 1. If c 0, then Aξ = {a,cb) : a,b R} = R R, the projection of H on Aξ is just the identity projection I, and ρ0,1),aξ) = 0. Suppose, then, that the projection P of H on Aξ exists. Then either ρ0,1),aξ) > 0 or ρ0,1),aξ) < 1. In the first case, c = 0; in the second, c 0. Thus if [Ax] exists for each x H, then we can prove that x R x = 0 x 0), a statement constructively equivalent to the essentially nonconstructive omniscience 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. It follows from this and our Theorem 1.1 that if ρ Ax 0, A 1 x) exists for each x H, then we can derive LPO. Acknowledgement This research was partially done when the author was a visiting fellow at the Isaac Newton Institute for the Mathematical Sciences, in the programme Semantics & Syntax: A Legacy of Alan Turing. The author thanks the referees for helpful comments that improved the presentation of the paper. References [1] P. Aczel and M. Rathjen: Notes on Constructive Set Theory, Report No. 40, Institut Mittag-Leffler, Royal Swedish Academy of Sciences, 001. [] E. Bishop: Foundations of Constructive Analysis, McGraw-Hill, New York, 1967. [3] E. Bishop and D.S. Bridges: Constructive Analysis, Grundlehren der Math. Wiss. 79, Springer Verlag, Heidelberg, 1985. [4] D.S. Bridges: On weak operator compactness of the unit ball of LH), Zeit. math. Logik Grundlagen Math. 4, 493 494, 1978. [5] D.S. Bridges, H. Ishihara: A definitive constructive open mapping theorem?, Math. Logic Quarterly 44, 545 55, 1998.

10 DOUGLAS S. BRIDGES [6] D.S. Bridges and L.S. Vîţă: Techniques of Constructive Analysis, Universitext, Springer Verlag, Heidelberg, 006. [7] D.S. Bridges, H. Ishihara, L.S. Vîţă: Computing infima on convex sets, with applications in Hilbert space, Proc. Amer. Math. Soc. 139), 73 73, 004. [8] D.S. Bridges, H. Ishihara, L.S. Vîţă: A new constructive version of Baire s Theorem, Hokkaido Math. Journal 351), 107 118, 006. [9] D.S. Bridges, A. Calder, W. Julian, R. Mines, and F. Richman: Locating metric complements in R n, in Constructive Mathematics F. Richman, ed.), Springer Lecture Notes in Math. 873, 41 49, 1981. [10] J. Dixmier: Les algèbres d opérateurs dans l espace hilbertien: algèbres de von Neumann, Gauthier- Villars, Paris, 1981. [11] R.V. Kadison and J.R. Ringrose: Fundamentals of the Theory of Operator Algebras, Academic Press, New York, 1983 Vol 1) and 1988 Vol ). [1] P. Martin-Löf: An intuitionistic theory of types, in Twenty-five Years of Constructive Type Theory G. Sambin, J. Smith, eds), 17 17, Oxford Logic Guides 36, Clarendon Press, Oxford, 1998. [13] S. Sakai: C*-algebras and W*-algebras, Springer Verlag, Heidelberg, 1971. [14] B. Spitters: Constructive results on operator algebras, J. Univ. Comp. Sci. 111), 096 113, 005. [15] D.M. Topping: Lectures on von Neumann Algebras, van Nostrand Reinhold, London 1971. This work is licensed under the Creative Commons Attribution-NoDerivs License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nd/.0/ or send a letter to Creative Commons, 171 Second St, Suite 300, San Francisco, CA 94105, USA, or Eisenacher Strasse, 10777 Berlin, Germany