© Hindawi Publishing Corp.
ON THE CLOSURE OF THE SUM OF CLOSED SUBSPACES
IRWIN E. SCHOCHETMAN, ROBERT L. SMITH, and SZE-KAI TSUI
(Received 30 May 2000)
Abstract.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.
2000 Mathematics Subject Classification. 46C05, 90C20.
1. Introduction. In this research/expository article, we supposeH is an arbitrary Hilbert space (real or complex) withKandNclosed subspaces ofH. We consider the question of when the (ordinary) sumK+Nis closed inH. It is obviously true if either K⊆NorN⊆K. It is alsotrue if eitherKorNis finite-dimensional. IfKorNis of codimension 1 (i.e., one of the subspaces is a hyperplane through the origin), then K+Nis clearly closed. However, it is not closed in general (seeExample 2.2).
In what follows, we will give necessary and sufficient conditions for K+N tobe closed. Although we have found no such conditions in the published literature, con-ditions were evidently provided without proof in unpublished course notes of Carl Pearcy [3].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 inTheorem 2.1. Our first equivalent condition involves orthog-onal complements and projections. LetL=K⊥ denote the orthogonal complement ofK inH, and letM=N⊥ denote the orthogonal complement ofN inH. Also, let EL:H→L be the corresponding orthogonal projection ontoL and EM :H→M be the corresponding orthogonal projection ontoM. InSection 2, we see thatEL(N)is closed if and only ifEM(K)is closed, and more important for our purposes, thatK+N is closed if and only if each of these subspaces is closed (seeTheorem 2.1(ii)). Thus, the closure ofN+Kis equivalent to the closure of the orthogonal projection ofNinto L(resp.,KintoM). Our next equivalent condition involves the (cosine of the) angle θ(K,N)betweenKandN. Its definition is given inSection 2, where we alsosee that K+N is closed if and only ifθ(K,N) <1, that is, the angle betweenKandNis not equal to0 (seeTheorem 2.1(iii)). Finally, inSection 2we give some sufficient condi-tions forTheorem 2.1to hold in terms of the orthogonal projectionsEK:H→Kand EN:H→N(seeTheorem 2.3).
quadratic objective functionx,Qxsubject tothe linear equality constraintAx=b, forx∈H, whereQandAare bounded linear operators, andQis alsoselfadjoint and positive semidefinite. For such a problem, we letKdenote the kernel ofQ,Nthe kernel ofAandF the feasible region, whereF=N+x, for anyx∈F. Then it is known that this problem admits an optimal solution if the (positive-definite) restriction ofQto L=K⊥is strictly positive definite (i.e., coercive) and the projectionEL(F)ofF intoLis closed [5]. The restrictionQ|Lis 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 projectionEL(F)is closed. Our main results give equivalent conditions for this to happen. In particular, it happens precisely whenK+Nis closed, or ifθ(K,N) <1.
2. Main results. We begin this section by defining the quantityθ(K,N), which is essentially the angle between the subspacesKandN. Let
S(K,N)=(x,y):x∈K∩(K∩N)⊥,y∈N∩(K∩N)⊥,x = y =1. (2.1)
Note thatS(K,N)=S(N,K)andS(K,N)= ∅if and only if eitherN⊆KorK⊆N. Let
θ(K,N)= sup
|x,y|:(x,y)∈S(K,N), forS(K,N)≠∅,
−∞, forS(K,N)= ∅, (2.2)
sothat−∞ ≤θ(K,N)=θ(N,K)≤1. In particular, ifN∩K= {0}, then(K∩N)⊥=H, N=N∩(K∩N)⊥,K=K∩(K∩N)⊥, and
θ(K,N)=sup|x,y|:x∈N,y∈K,x = y =1. (2.3)
IfL=K⊥andM=N⊥are hyperplanes through the origin, thenθ(K,N)is the (cosine of the) conventional angle between the one-dimensional subspacesKandN.
Theorem2.1. The following statements are equivalent: (i) EL(N)is closed inL.
(ii) K+Nis closed inH. (iii) θ(K,N) <1.
(iv) EM(K)is closed inM.
Proof. Given the expository nature of this paper, we find it instructive to show all the implications involved.
(i)(ii). SupposeEL(N)is closed. Letxk=νk+ηk, whereνk∈Nandηk∈K, for all k=1,2,...,andxk→x∈H. Toshow thatx∈N+K. SinceL=K⊥, we have that
ELxk →EL(x)∈L, (2.4)
where
ELxk=ELνk+ηk=ELνk+ELηk=ELνk∈EL(N), ∀k=1,2,.... (2.5)
Now supposeN+Kis closed inH. Let{ξk}be a sequence inEL(N)which converges toξ∈L. Toshow thatξ∈EL(N), for eachk, there existsyk∈Nsuch thatEL(yk)=ξk. Letηk=ξk−yk, for allk. Then
ELηk=ELξk−ELyk=ELELyk−ELyk=ELyk−ELyk=0, (2.6)
sothatηk∈K, for allk=1,2,....Sinceξk=yk+ηk∈N+K, for allK, andξk→ξ, it follows, by hypothesis, thatξ∈N+K. Hence, there existν∈Nandη∈Kfor which ξ=ν+η. Therefore,
ELELyk=ELξk →EL(ξ)=EL(ν)+EL(η)=EL(ν),
ELELyk=ELyk=ξk →ξ, (2.7)
sothatξ=EL(ν), fo rν∈N, that is,ξ∈EL(N).
(ii)(iii). First observe that if N⊆K or K ⊆N, thenN+K is closed inH and θ(N,K)= −∞. Thus, we may assume thatNKandKN.
Next we show that, without loss of generality, we may also assume thatN∩K= {0}. If not, then we may write
N=N1⊕(N∩K), K=K1⊕(N∩K), (2.8)
whereN1(resp.,K1) is the orthogonal complement ofN∩KinN(resp.,K). Then N+K=N1⊕(N∩K)+K1⊕(N∩K)
=N1+K1⊕(N∩K)+(N∩K)
=N1+K1⊕(N∩K).
(2.9)
Thus,N+Kis closed if and only ifN1+K1is closed, whereN1∩K1= {0}.
SupposeN+K is closed withN∩K= {0}. Consider the canonical Hilbert space mapping
f:N⊕K →N+K, (2.10)
given by
f (ν,η)=ν+η, ∀ν∈N, ∀η∈K. (2.11) This mapping is clearly linear and onto. It is also one-to-one sinceN∩K= {0}. We next show that it is also bounded.
Letν∈Nandη∈K, then
f (ν,η)2= ν+η2= ν+η,ν+η = ν,ν+ν,η+η,ν+η,η
= ν,ν+2ν,η+η,η = ν2+2ν,η+η2
≤ ν2+2|ν,η|+η2≤ ν2+2νη+η2.
(2.12)
Moreover,
2νη ≤ ν2+η2. (2.13)
Hence,
By the open mapping theorem [2, page 57],fhas a bounded linear inverse, that is, there existsc >0 such that
c(ν,η) ≤ f (ν,η) = ν+η ≤√2(ν,η). (2.15)
Consequently,
c2ν2+η2=c2(ν,η)2≤ ν+η2= ν2+2ν,η+η2. (2.16)
Hence,
c2ν2+η2≤ ν2+2ν,η+η2, (2.17)
that is,
c2−1ν2+η2≤2ν,η, (2.18) for all suchν,η.
Now assume in addition thatν = η =1. Then
2c2−1=c2−1ν2+η2≤2ν,η, (2.19)
that is,
1−c2≥ −ν,η, (2.20)
where−νis an arbitrary element ofNof norm 1, sinceνis such. Ifν,η ≤0, then −ν,η = |ν,η|, sothat 1−c2≥ ν,η, and 0< c≤1. If ν,η ≥0, thenν,η =
|ν,η|, sothat 1−c2≥ −|ν,η|. Lettingν= −ν, we get
1−c2≥ − ν,η= ν,η = |ν,η|. (2.21)
Hence, in either case,
|ν,η| ≤1−c2<1, (2.22)
for allν∈N,η∈Kwithν = η =1. Thus,θ(N,K) <1.
Now suppose−∞ ≤θ=θ(N,K) <1. Observe thatN,Kare weakly closed inH, as well as closed. Also, from the definition ofθwe have that
|ν,η| ≤θνη, (2.23)
sothat
−2θνη ≤2ν,η ≤2θνη, ∀ν∈N, ∀η∈K. (2.24) Suppose{νk}is a sequence inNand{ηk}is a sequence inKsuch thatxk=νk+ηk→ xinH. Toshow thatx∈N+K. First, we show that
sup k
νk<∞, sup k
ηk<∞. (2.25)
We have
xk2=νk+ηk2=νk+ηk2+2 νk,ηk
≥νk2+ηk2−2θνkηk
=θνk−ηk2+1−θ2νk2,
that is,
νk+ηk2≥θνk−ηk2+1−θ2νk2. (2.27)
Interchangingνkandηk, we alsoobtain
νk+ηk2≥θηk−νk2+1−θ2ηk2, (2.28)
where 1−θ2>0 by hypothesis. We see from these inequalities that if either{νk}or
{ηk}is unbounded, we obtain a contradiction, since the convergent sequence{νk+ηk} must be bounded.
By the Banach-Alaoglu theorem [2], passing tosubsequences if necessary, we may assume that there existsν∈N andη∈Ksuch thatνk νand ηk η(weak con-vergence), ask→ ∞. Thus,νk+ηk ν+ηandνk+ηk→x, ask→ ∞. Necessarily, x=ν+η∈N+K, since the weak topology onHseparates points.
(iii)(iv). InterchangeKandNin (i), (ii), and (iii).
Example2.2. For eachj=1,2,...,letψj=arcsin(1/j)sothat cosψj=j2−1/j. Define
H=
yiui∞i=1:yi, ui∈R, ∞
i=1
y2 i+u2i
<∞ , K=
yi0∞i=1:yi∈R, ∞
i=1 y2
i<∞
, N=yiui∞i=1∈H:yi=uicotψi,∀i. (2.29)
Clearly,KandNare closed subspaces of the real Hilbert spaceHwithK∩N= {0}and
L=K⊥=
0ui∞i=1:ui∈R, ∞
i=1 u2
i<∞
. (2.30)
Now, for eachj=1,2,...,definexj=(xj i)∞i=1by
xji=
[[0 01 0],], forforii≠=j,j, (2.31)
sothatxj∈Kandxj =1. Similarly, defineνj=(νj i)∞i=1by
νij=
[0 0cos],ψj sinψj, forforii≠=j,j, (2.32)
sothatνj∈Nandνj =1. Note that
νjj=
cosψj sinψj=sinψjcotψj1, ∀j. (2.33)
We then have
xj,νj=
∞
i=1
Consequently, we see that
θ(N,K)≥sup
j cosψj=1, (2.35)
that is,θ(N,K)=1, sothatEL(N)is not closed byTheorem 2.1.
We next show thatEL(N)is not closed. Suppose it is. Observe thatEL(N)is the set of([0uj])∞
j=1∈Lfor which there existsyj∈Rsuch thatyj=ujcotψj,∀j, and(yj) is square summable. For eachj=1,2,...,letuj=1/j, sothatu=([0uj])∈L. Set yj=ujcotψj, and define
xij=
yiui, fori≤j,
[0 0], fori > j, (2.36)
sothatxj∈N, anduj=EL(xj)∈EL(N), where
uji=
0ui, fori≤j,
[0 0], fori > j, (2.37)
for allj=1,2,....Clearly,{uj}is a Cauchy sequence inEL(N). SinceEL(N)is closed, then there existsu∈EL(N)such thatuj→u, asj→ ∞. Therefore, there existsy= (yj)such thatyj=ujcotψj, for allj, andy2
j <∞, that is,([yjuj])∈H. But
yjuj2=∞ j=1
y2 j+u2j
= ∞ j=1
cot2ψj+1u2 j= ∞ j=1 u2 j
sin2ψj = ∞. (2.38)
This is a contradiction. Hence,EL(N)is not closed in this example.
We next show thatN+Kis not closed. Let[yi,ui]be an arbitrary element ofN, so that∞i=1u2i ≤ ∞, in particular. Once again let
xij=
yiui, fori≤j,
[0 0], fori > j, (2.39)
sothatxj∈N, and
zj=
yi0, fori≤j,
[0 0], fori > j, (2.40)
sothat−zj∈K, for allj. Consequently,uj∈N+K, for allj, where
uj=
0ui, fori≤j,
[0 0], fori > j. (2.41)
It follows that the sequence{uj}is Cauchy since∞
j=1u2i <∞. By hypothesis, there existsu∈Hsuch thatuj→uasj→ ∞. Necessarily,uj
i→ui, for alli. Hence,
If u∈ K+N, then there exists z= ([z1,0],[z2,0],...) ∈K and x ∈ N such that z+x=u. Thus,
x=u−z= −z1,u1,z2,u2,.... (2.43) Since this belongs toN, we must have
zi=uicotψi, ∀i. (2.44)
Consequently,
x2=∞ j=1
cotψ2
i+1
ui= ∞, (2.45)
which is a contradiction. Therefore,K+Nis not closed.
Theorem2.3. The following are sufficient forK+Nto be closed: (i) ENEK=0orEKEN=0.
(ii) ENEK−EKEN=0.
(iiia) sup{ENη:η∈K∩(N∩K)⊥,η =1}<1. (iiib) sup{EKν:ν∈N∩(N∩K)⊥,ν =1}<1. Proof. (i) IfENEK=0, then
EKEN=ENEK∗=E∗KE∗
N=0, (2.46)
sothatL+M=L⊕M, and is therefore closed inH. Similarly forEKEN=0. (ii) We have
ENEK=EKEN=EN∩K, (2.47)
(where, in general,EXdenotes the orthogonal projection ofHonto the closed subspace X) and
N+K=(N∩K)⊕N∩(N∩K)⊥+K∩(N∩K)⊥. (2.48) We alsohave
ENI−ENEK=EN∩(N∩K)⊥, (2.49)
whereIis the identity operator onHand
I−ENEK=E(N∩K)⊥. (2.50)
Thus, by hypothesis,
ENI−ENEK=EN−E2
NEK=EN−ENEKEN=
I−ENEKEN. (2.51)
Similarly,
EKI−ENEK=EK∩(N∩K)⊥. (2.52)
But
ENI−ENEKI−ENEKEK=2ENEK−EKEN=0, (2.53) by hypothesis. (Note that part (i) is valid for any closed subspaceNandKand their corresponding projections EN and EK.) Hence, by (i) applied to EN and E(N∩K)⊥, we
obtain that
N∩(N∩K)⊥+K∩(N∩K)⊥ (2.54)
(iiia) For convenience, let
β=supENη:η∈K∩(N∩K)⊥,η =1. (2.55)
(We may exclude the case where the defining set is empty. This happens only ifK⊆N, in which caseK+Nis trivially closed.) Thenβ <1 andβ2<1, that is,δ=1−β2>0. Fix
η∈K∩(N∩K)⊥, ν∈N∩(N∩K)⊥, (2.56)
such thatη = ν =1. Then
|η,ν| = ηνcosψ, (2.57)
whereψis the angle betweenηandν, if this angle is at mostπ/2, or the supplement of this angle, if it is greater than π/2. (Alternately, we can replaceηby−ηwhere necessary.) Thus, 0≤ψ≤π/2. By the law of cosines, we have
η+ν2= η+ν2−2ηνcosψ=2−2cosψ. (2.58)
On the other hand, since ν∈N, and ENηis the best approximation inN toη, we have that
η−ν2≥ η−ENη2= η2−ENη2=1−ENη2, (2.59)
by the Pythagorean identity, whereη−ENηis orthogonal toENη. Thus,
1−ENη2≤2−2cosψ. (2.60)
However, by definition ofβ, we have
ENη2≤β2, (2.61)
sothat
1−ENη2≥1−β2=δ >0. (2.62)
Consequently,
0< δ≤2−2cosψ, (2.63)
that is,
0≤cosψ≤1−δ2<1, (2.64)
sothat
|η,ν| ≤1−δ2<1. (2.65)
This completes the proof of (iiia), sinceηandνare arbitrary.
3. An application to quadratic programming. We consider the general infinite qua-dratic programming problem given by
minx,Qx (3.1)
subject to
Ax=b, x∈H, (3.2)
whereHandMare real Hilbert spaces,b∈M, the constraint operatorA:H→Mis a bounded linear operator and the cost operatorQ:H→H is a nonzero, (selfadjoint) positive semidefinite, bounded linear operator. The feasible region
F= {x∈H:Ax=b} (3.3)
is a closed, affine subset ofH(which we assume tobe nonempty), and the kernel
N= {x∈H:Ax=0} (3.4)
ofAinHis a closed subspace ofH. Of course,F=N+x, for anyx∈F.
Recall thatQis positive semidefinite ifx,Qx ≥0, for allx∈H, and thatQis positive definite ifx,Qx>0, for allx∈H,x≠0. We say that the operatorQis strictly positive definite if it is coercive, that is, if there existsσQ>0 satisfying
σQx2≤ x,Qx, ∀x∈H. (3.5)
This condition is known (see [1, page 73]) tobe necessary and sufficient for (3.1) to admit a (unique) optimal solution for any (nonempty) closed, convex subsetF ofH. Thus, even ifF is a closed, convex set inH, andQis 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) toadmit an optimal solution.
SinceQis positive semidefinite, its kernelKis given by
K=x∈H:x,Qx =0. (3.6)
Moreover, sinceQis selfadjoint, it follows thatKandL=K⊥are invariant underQ. Hence,Qalso decomposes into 0⊕P, where 0 is the zero operator onKandP:L→Lis the restriction operatorQ|L. Note thatPis a positive-definite, bounded linear operator onL. It need not be strictly positive definite.
Also, sinceF⊆H, we have that the image ofFunderEKis
EK(F)= {η∈K:η+ξ∈F,for someξ∈L}. (3.7)
It is nonempty and convex inK, since this is the case forF inH. It is alsotrue thatF is closed inH; however,EK(F)need not be closed inK.
Analogously, the image ofF underELis
As withEK(F), the setEL(F)is nonempty and convex, but not necessarily closed inL. Moreover,F⊆EK(F)⊕EL(F). The same is true ofN,EK(N), andEL(N).
We may now consider the following related problem:
min
ξ∈EL(F)ξ,Pξ, (3.9)
where, as we have seen,P is positive definite onLandEL(F)is a nonempty, convex subset ofL, which may not be closed. Moreover,
ξ,Pξ = x,Qx, (3.10)
for allξ∈L,η∈K, andx=η+ξ.
Note that solving (3.9) is equivalent tosolving (3.1) in the following sense. Ifξ∗∈ EL(F)is an optimal solution to (3.9), then there existsη∗∈EK(F)such that x∗= η∗+ξ∗∈Fandx∗is optimal for (3.1). Conversely, ifx∗∈F is optimal for (3.1), then x∗=η∗+ξ∗, fo rη∗∈EK(F)andξ∗∈EL(F), whereξ∗is optimal for (3.9).
We are interested in when the feasible regionEL(F)for (3.9) is closed. Lemma3.1. The following statements are equivalent forF,K, andN:
(i) EL(F)orEL(N)is closed inL. (ii) EM(K)is closed inM. (iii) N+Kis closed inH. (iv) F+Kis closed inH.
(v) θ(N,K) <1.
Proof. ApplyTheorem 2.1 together with the fact that F =N+x, for any fixed x∈F.
Theorem 3.2. If Q|L is strictly positive definite and any one of the conditions in Lemma 3.1holds, then (3.1) admits an optimal solution.
Proof. Observe that (3.9) admits an optimal solution ifP=Q|Lis strictly positive definite andEL(F)is closed inL[1, page 73].
Acknowledgements. We wish to thank Peter Duren of the Department of Math-ematics at the University of Michigan at Ann Arbor, and Stephan Richter of the De-partment of Mathematics at the University of Tennessee for bringing the information o f [3] to our attention.
The second author was supported in part by the National Science Foundation under grant DMI-9713723.
References
[1] A. L. Dontchev and T. Zolezzi,Well-Posed Optimization Problems, Lecture Notes in Mathe-matics, no. 1543, Springer-Verlag, Berlin, 1993.MR 95a:49002. Zbl 797.49001. [2] N. Dunford and J. T. Schwartz,Linear Operators. Part I: General Theory, Wiley Classics
Library, John Wiley and Sons, New York, 1988.MR 90g:47001a. Zbl 635.47001. [3] P. Duren,Private Communication, University of Michigan, Michigan, 1997.
[5] ,Solution existence for infinite quadratic programming, Mathematical Programming with Data Perturbations, Lecture Notes in Pure and Applied Mathematics, no. 195, Dekker, New York, 1998, pp. 363–385.MR 98i:90064.
Irwin E. Schochetman: Mathematics and Statistics, Oakland University, Rochester, MI48309, USA
E-mail address:schochet@oakland.edu
Robert L. Smith: Industrial and Operations Engineering, The University of Michigan, Ann Arbor, MI48109, USA
E-mail address:rlsmith@umich.edu
Sze-Kai Tsui: Mathematics and Statistics, Oakland University, Rochester, MI48309, USA
