On the closure of the sum of closed subspaces

We give necessary and sufficient conditions for the sum of closed subspaces of a Hilbert space to be closed. Specifically, we show that the sum will be closed if and only if the angle between the subspaces is not zero, or if and only if the projection of either space into the orthogonal complement of the other is closed. We also give sufficient conditions for the sum to be closed in terms of the relevant orthogonal projections. As a consequence, we obtain sufficient conditions for the existence of an optimal solution to an abstract quadratic programming problem in terms of the kernels of the cost and constraint operators.


Introduction.
In this research/expository article, we suppose H is an arbitrary Hilbert space (real or complex) with K and N closed subspaces of H.We consider the question of when the (ordinary) sum K +N is closed in H.It is obviously true if either K ⊆ N or N ⊆ K.It is also true if either K or N is finite-dimensional.If K or N is of codimension 1 (i.e., one of the subspaces is a hyperplane through the origin), then K + N is clearly closed.However, it is not closed in general (see Example 2.2).
In what follows, we will give necessary and sufficient conditions for K + N to be closed.Although we have found no such conditions in the published literature, conditions were evidently provided without proof in unpublished course notes of Carl Pearcy [3].In this paper, we state and prove a collection of similar necessary and sufficient conditions in Theorem 2.1.Our first equivalent condition involves orthogonal complements and projections.Let L = K ⊥ denote the orthogonal complement of K in H, and let M = N ⊥ denote the orthogonal complement of N in H. Also, let E L : H → L be the corresponding orthogonal projection onto L and E M : H → M be the corresponding orthogonal projection onto M. In Section 2, we see that E L (N) is closed if and only if E M (K) is closed, and more important for our purposes, that K +N is closed if and only if each of these subspaces is closed (see Theorem 2.1(ii)).Thus, the closure of N +K is equivalent to the closure of the orthogonal projection of N into L (resp., K into M).Our next equivalent condition involves the (cosine of the) angle θ(K, N) between K and N. Its definition is given in Section 2, where we also see that K + N is closed if and only if θ(K, N) < 1, that is, the angle between K and N is not equal to 0 (see Theorem 2.1(iii)).Finally, in Section 2 we give some sufficient conditions for Theorem 2.1 to hold in terms of the orthogonal projections E K : H → K and E N : H → N (see Theorem 2.3).
In Section 3, we give an application of our main results.This involves an abstract positive semidefinite quadratic programming problem given by minimizing the quadratic objective function x, Qx subject to the linear equality constraint Ax = b, for x ∈ H, where Q and A are bounded linear operators, and Q is also selfadjoint and positive semidefinite.For such a problem, we let K denote the kernel of Q, N the kernel of A and F the feasible region, where F = N + x, for any x ∈ F .Then it is known that this problem admits an optimal solution if the (positive-definite) restriction of Q to L = K ⊥ is strictly positive definite (i.e., coercive) and the projection E L (F ) of F into L is closed [5].The restriction Q|L is well known to be strictly positive definite if and only if its (positive) spectrum is bounded away from zero.That leaves the question of when the projection E L (F ) is closed.Our main results give equivalent conditions for this to happen.In particular, it happens precisely when K + N is closed, or if θ(K, N) < 1.

Main results.
We begin this section by defining the quantity θ(K, N), which is essentially the angle between the subspaces K and N. Let (2.1) Note that S(K, N) = S(N, K) and S(K, N) = ∅ if and only if either If L = K ⊥ and M = N ⊥ are hyperplanes through the origin, then θ(K, N) is the (cosine of the) conventional angle between the one-dimensional subspaces K and N.
Theorem 2.1.The following statements are equivalent: Proof.Given the expository nature of this paper, we find it instructive to show all the implications involved.
(i) (ii).Suppose E L (N) is closed.Let x k = ν k +η k , where ν k ∈ N and η k ∈ K, for all k = 1, 2,..., and where By hypothesis, E L (x) ∈ E L (N).Thus, there exists y ∈ N such that for all K, and ξ k → ξ, it follows, by hypothesis, that ξ ∈ N + K. Hence, there exist ν ∈ N and η ∈ K for which ξ = ν + η.Therefore, Thus, we may assume that N K and K N.
Next we show that, without loss of generality, we may also assume that N ∩K = {0}.If not, then we may write where N 1 (resp., K 1 ) is the orthogonal complement of N ∩ K in N (resp., K).Then (2.9) This mapping is clearly linear and onto.It is also one-to-one since (2.12) By the open mapping theorem [2, page 57], f has a bounded linear inverse, that is, there exists c > 0 such that Consequently, for all such ν, η.Now assume in addition that ν = η = 1.Then where −ν is an arbitrary element of Observe that N, K are weakly closed in H, as well as closed.Also, from the definition of θ we have that We have that is, (2.27) Interchanging ν k and η k , we also obtain where 1 − θ 2 > 0 by hypothesis.We see from these inequalities that if either {ν k } or {η k } is unbounded, we obtain a contradiction, since the convergent sequence {ν k +η k } must be bounded.By the Banach-Alaoglu theorem [2], passing to subsequences if necessary, we may assume that there exists ν ∈ N and η ∈ K such that ν k ν and η k η (weak convergence), as k → ∞.Thus, since the weak topology on H separates points.
Example 2.2.For each j = 1, 2,..., let ψ j = arcsin(1/j) so that cos ψ j = j 2 − 1/j.Define (2.29) Clearly, K and N are closed subspaces of the real Hilbert space H with K ∩ N = {0} and Now, for each j = 1, 2,..., define x j = (x j i ) ∞ i=1 by so that x j ∈ K and x j = 1.Similarly, define for i ≠ j, cos ψ j sin ψ j , for i = j, (2.32) so that ν j ∈ N and ν j = 1.Note that ν j j = cos ψ j sin ψ j = sin ψ j cot ψ j 1 , ∀j. (2.33) We then have Consequently, we see that that is, θ(N, K) = 1, so that E L (N) is not closed by Theorem 2.1.We next show that E L (N) is not closed.Suppose it is.Observe that E L (N) is the set of ([0 u j ]) ∞ j=1 ∈ L for which there exists y j ∈ R such that y j = u j cot ψ j , ∀j, and (y j ) is square summable.For each j = 1, 2,..., let u j = 1/j, so that u = ([0 u j ]) ∈ L. Set y j = u j cot ψ j , and define for i > j, (2.36) so that x j ∈ N, and u j = E L (x j ) ∈ E L (N), where for all j = 1, 2,.... Clearly, {u j } is a Cauchy sequence in E L (N).Since E L (N) is closed, then there exists u ∈ E L (N) such that u j → u, as j → ∞.Therefore, there exists y = (y j ) such that y j = u j cot ψ j , for all j, and y 2 j < ∞, that is, ([y j u j ]) ∈ H.But This is a contradiction.Hence, E L (N) is not closed in this example.We next show that N + K is not closed.Let [y i ,u i ] be an arbitrary element of N, so that ∞ i=1 u 2 i ≤ ∞, in particular.Once again let for i > j, (2.39) so that x j ∈ N, and .40) so that −z j ∈ K, for all j.Consequently, u j ∈ N + K, for all j, where (2.41) It follows that the sequence {u j } is Cauchy since By hypothesis, there exists u ∈ H such that u j → u as j → ∞.Necessarily, u j i → u i , for all i.Hence, u = 0,u 1 , 0,u 2 ,... . (2.42) Since this belongs to N, we must have Consequently, which is a contradiction.Therefore, K + N is not closed.
Theorem 2.3.The following are sufficient for K + N to be closed: so that L + M = L ⊕ M, and is therefore closed in H. Similarly for (where, in general, E X denotes the orthogonal projection of H onto the closed subspace X) and (2.48) We also have where I is the identity operator on H and Thus, by hypothesis, Similarly, by hypothesis.(Note that part (i) is valid for any closed subspace N and K and their corresponding projections E N and E K .)Hence, by (i) applied to E N and E (N∩K) ⊥ , we obtain that (iiia) For convenience, let (2.55) (We may exclude the case where the defining set is empty.This happens only if K ⊆ N, in which case K +N is trivially closed.)Then β < 1 and where ψ is the angle between η and ν, if this angle is at most π/2, or the supplement of this angle, if it is greater than π/2.(Alternately, we can replace η by −η where necessary.)Thus, 0 ≤ ψ ≤ π/2.By the law of cosines, we have On the other hand, since ν ∈ N, and E N η is the best approximation in N to η, we have that by the Pythagorean identity, where η − E N η is orthogonal to E N η.Thus, However, by definition of β, we have This completes the proof of (iiia), since η and ν are arbitrary.
(iiib) The proof in part (iiia) depends only on the fact that N and K are closed subspaces of H. Thus, simply interchange these spaces in the proof.
3. An application to quadratic programming.We consider the general infinite quadratic programming problem given by min x, Qx (3.1) subject to Ax = b, x ∈ H, ( where H and M are real Hilbert spaces, b ∈ M, the constraint operator A : H → M is a bounded linear operator and the cost operator Q : H → H is a nonzero, (selfadjoint) positive semidefinite, bounded linear operator.The feasible region is a closed, affine subset of H (which we assume to be nonempty), and the kernel This condition is known (see [1, page 73]) to be necessary and sufficient for (3.1) to admit a (unique) optimal solution for any (nonempty) closed, convex subset F of H. Thus, even if F is a closed, convex set in H, and Q is only positive definite, problem (3.1) may not admit an optimal solution.See [4] for an example.
In this section, we establish sufficient conditions for (3.1) to admit an optimal solution.
Since Q is positive semidefinite, its kernel K is given by Hence, Q also decomposes into 0⊕P , where 0 is the zero operator on K and P : L → L is the restriction operator Q|L.Note that P is a positive-definite, bounded linear operator on L. It need not be strictly positive definite.Also, since F ⊆ H, we have that the image of F under E K is It is nonempty and convex in K, since this is the case for F in H.It is also true that F is closed in H; however, E K (F ) need not be closed in K.
Analogously, the image of F under E L is As with E K (F ), the set E L (F ) is nonempty and convex, but not necessarily closed in L.Moreover, F ⊆ E K (F ) ⊕ E L (F ).The same is true of N, E K (N), and E L (N).
We are interested in when the feasible region E L (F ) for (3.9) is closed.
Lemma 3.1.The following statements are equivalent for F , K, and N: Proof.Apply Theorem 2.1 together with the fact that F = N + x, for any fixed x ∈ F .Proof.Observe that (3.9) admits an optimal solution if P = Q|L is strictly positive definite and E L (F ) is closed in L [1, page 73].

Theorem 3 . 2 .
If Q|L is strictly positive definite and any one of the conditions in Lemma 3.1 holds, then (3.1) admits an optimal solution.