30 Ns 1 Maig 1986 DUALITY OF INFINITE-DIMENSIONAL SUBSPACES IN AN INDEFINITE INNER PRODUCT SPACE

A characterization of the subspaces of an inner product space which admit a dual companion and the relation betWeen duality and orthogonal projections are given . Following Bognár [1] we mean, by "inner product space", a complex vector space E, endowed with a sesquilinear form ( 1 ) -the "inner product on E"not necessarily positive defined . If A is a subset of E, we symbolize by A1 the set of all vectors x E.E such that (ájx)=0 for every a E A . Two subspaces L and M are "dual companions" (L#M) if L (1 M 1 = L1 (1 M = 0 . A locally convex topology L on E is "admissible" if the inner product is separately Z-continous and, for every linear form c-continous y there exists a vector xo with lf(y) = (ylxo ) for every y . In [1], [2] and [3] duality between finite-dimensional subspaces is studied . The purpose of this note is to give a characterisation of the subspaces L of E which admit a dual companion (Corollary 1) that generalizes the results of the quoted works . Theorem 2 gives a construc tive method of dual companions in a particular case, which extends a result of [1] . Finally, if the subspaces considered are orthocomplemen ted (L+L.l =E), we express the duality by means of orthogonal projections . THEOREM 1 : Let L and M be subspaces of the inner product space E with Ll1M 1 = 0 . Then, M contains a dual companion of L . Proof : The result is obvious when L = 0 . In the other case, let 91 be the family of the pairs (L' .,M') of subspaces of E such that L'#M', L'C L and M' C M . Because 0#0,11 is not void . determines a partial ordering of S1 . If ~1 = {(L,,Mi ) : i6I} is a totally ordered subset of S1, let be i .e ., z=0, , because Li # Mi . o 0 Besides, the relation (L',M') é (L",M") <==:P L'G L" and M'C M" L' _ +L M' = +M . ~cl i ¡CI i Since n1 is totally ordered, we get the identities L' = UL . M' = UM . . i.eI 1 -,c1 1 Consequently, if z is a vector of L' (1 M' l. . there must exist an index i in I such that 0 z£Li l1M'1'=Li (1((1Mi )C-Li nMi1 , 0 o ¡El 0 0 Similarly, L'1n M' = 0, so (L',M') is an upper bound of S11 , Thus ; from the Zorn Lemma, we infer the existente of a maximal element (L1 ,M1 ) in n . The proof will conclude if we get the identity L = L1 . Let la assume now LéL 1 ; if so, we can consider a vector x in L-L 1 and the subspace L2 = Ll +<x> G L «x> symbolizes the linear span of x) . By virtue of maximality of (Li , M1 ), there must be uj0 in L2l1 Mi since L2n M1 G Li n M1 = 0 . Thus, u = z+ax, with z E L1 and a~0, which implies L2=L1 +<u> . On the other hand, since L^ M I = 0, it is possible to get y E M with (uly) = 1 (1) and consider the subspace of M, M2 = M1+<y> . Then immediately follows M L f) L2 = 0, so, M2 (1 L2 1 0 and there exists v~0 in M2(1 L2 . So, M2 = M1+<v>, since v = m+dy, m E M1 , dé0 ( 2) 1 By considering a not vanishing vector ml+bv in 2 n L 2 (consequently, bIO), for every t E L1 . (ml +bvit) = 0, since L2 C Li. From it, (ml lt) = 0, and so, ml 0 . i Finally, since uE L2 , ve L2 , (ml+bviu) = b(viu) = 0 . But, taking into account (1) and (2), (VIU) = (ml+bylu) = b 1 O. COROLLARY 1 : A subspace L of the inner product space E admits a dual com panion if and only if LOE -J " = 0 . 1 COROLLARY 2: Let E be an inner product space and let e be an admissible topology on E . The following propositions are equivalent : i) E is non degenerate (i .e ., E1 = 0) ii) Every subspace of E admits a dual companion iii) There exists a subspace L, ¿-closed in E, which admits dual companion . Proof: By using Corollary 1 it is enough to prove that i) follows from iii) . If M is a dual companion of the Z-closed subspace L, we obtain, El ( (L£ +M)1 = Lll nM l = Lr%M1 =0 since L11 coincides with the Z-closure of L given that Z is admissible ./ In Corollary 2, iii) the hypothesis of L be closed is necessary as the following example proves : Let E = <e,f>, (eje) = (elf) = (fle) = 0, (flf) = 1 . Then <f> # <f>, but E is degenerate . DEFINITION : Two families .of vectors te, : i E Ij and t fi. : i E Ij in the inner product space E form a "dual pair" if, for every i,j E I, with iéj, the relations

The purpose of this note is to give a characterisation of the subspaces L of E which admit a dual companion (Corollary 1) that generalizes the results of the quoted works .Theorem 2 gives a construc tive method of dual companions in a particular case, which extends a result of [1] .Finally, if the subspaces considered are orthocomplemented (L+L.l=E), we express the duality by means of orthogonal projections .Consequently, if z is a vector of L' (1 M' l. .there must exist an index i in I such that 0 Similarly, L' 1n M' = 0, so (L',M') is an upper bound of S1 1 , Thus ; from the Zorn Lemma, we infer the existente of a maximal element (L 1 ,M 1 ) in n .The proof will conclude if we get the identity L = L1 .Let la assume now LéL 1 ; if so, we can consider a vector x in L-L 1 and the subspace L 2 = L l +<x> G L «x> symbolizes the linear span of x) .By virtue of maximality of (L i , M1 ), there must be uj0 in L 2l1 Mi since L2 n M1 G Li n M1 = 0 .Thus, u = z+ax, with z E L 1 and a~0, which implies L2=L1 +<u> .
On the other hand, since L^M I = 0, it is possible to get y E M with and consider the subspace of M, M 2 = M1+<y> .
The countable families obtained by means of this process form dual pair and, obviously, the former one is a Hamel basis for LI If the subspace L is orthocomplemented, every vector of E can be expressed (in a not necessarily only way) as the sum of one of L and another of LL .Thus, in a natural way, the "orthogonal projection" of a subspace M on L, P L M, can be defined as the set of the vectors x a L such that x-z E L1 for some z in M .
For a Hilbert space it is well known the fact that, given two closed subspaces L and M, L (1 M 1 = 0 if and only if PL M is dense in L .
The following lemma expresses the best possible generalization of this result for an inner product space .
LEMMA 1 : Let L be an orthocomplemented subspace of the inner product space E such that Ll1Ll = 0 (i .e., L is nondegenerate) .Then, for every

THEOREM 1 :
Let L and M be subspaces of the inner product space E with Ll1M 1 = 0 .Then, M contains a dual companion of L .Proof : The result is obvious when L = 0 .In the other case, let 91 be the family of the pairs (L' .,M') of subspaces of E such that L'#M', L'C L and M' C M .Because 0#0,11 is not void .determines a partial ordering of S1 .If ~1 = {(L,,M i ) : i6I} is a totally ordered subset of S1, let be i .e ., z=0,, because L i # Mi .o 0 Besides, the relation (L',M') é (L",M") <== :P L'G L" and M'C M"

THEOREM 2 :
By considering a not vanishing vector ml+bv in M 2 n L 2 (consequently, bIO), for every t E L1 .(m l +bvit) = 0, since L2 C Li. From it, (m l lt) = 0, ) = b(viu) = 0 .But, taking into account (1) and (2), (VIU) = (ml+bylu) = b 1 O. COROLLARY 1 : A subspace L of the inner product space E admits a dual com panion if and only if LOE -J " = 0 . 1 COROLLARY 2: Let E be an inner product space and let e be an admissible topology on E .The following propositions are equivalent : i) E is non degenerate (i .e., E 1 = 0) ii) Every subspace of E admits a dual companion iii) There exists a subspace L, ¿-closed in E, which admits dual companion .Proof: By using Corollary 1 it is enough to prove that i) follows from iii) .If M is a dual companion of the Z-closed subspace L, we obtain, El ( (L £ +M)1 = Lll nM l = Lr%M 1 =0 since L11 coincides with the Z-closure of L given that Z is admissible ./In Corollary 2, iii) the hypothesis of L be closed is necessary as the following example proves : Let E = <e,f>, (eje) = (elf) = (fle) = 0, (flf) = 1 .Then <f> # <f>, but E is degenerate .DEFINITION : Two families .ofvectors te, : i E Ij and t f i .: i E Ij in the inner product space E form a "dual pair" if, for every i,j E I, with iéj, the relations are verified .,As is easily checked, if two families of vectors form dual pair each of them is linearly independent and their linear envelopes are dual companions .Let L and M be subspaces of the inner product space E such that L0 M i = 0.If L admits a countable Hamel basis then it exists a dual pair of families of vectors, their linear envelopes being L and a subspace of M .Proof: We will construct recurrently the dual pair .Let `gn : n=1,2, . . .j a Hamel basis of L. Since gl `F M' , it is possible to find fl in M with (g 1 If1 ) = 1 .Let el = 91* in M ..re given . . .e2, . . ..en> = <gl,g2, . . . .If j ) = 0, Assuming that the vectors e l , e2 , . ., en in L and fl , f2 , . ., fn (gn+1 Ifk )e k' K=1 and, because en+lE M , it exists h E M such that (e n+l lh)

COROLLARY 3 :
subspace M in E and for every admissible topology L on E, M Ln L = 0 F==> P LM is e-dense in L .Proof : Since the closures of the subspaces are the same for every admissible topologies it is enough to work with one of them, in particular with the weak topology 6(E) .Following a result of Scheibe (see (3)) if L is orthocomplemented then the weak topology of L, <3r(L) coincides with the relative one of v(E) .Besides, in [1] it is proved that (P LM) 1 0 L = 0 if and only if P L M is o-(L)-dense in L .Finally, it is easily checked that (for L orthocomplemented) M. L /1 L = (P LM ~(1 L, which concludes the proof .eLemma 1 is a particular case of the following fact : if L is an orthocomplemented subspace of E, then for every admissible topology and for every subspace M, P LM is dense in L if and only if L(IM l = LA L1 , result established in[4] .From Lemma 1 the proof of the next theorem follows straigthforwardly THEOREM 3 : Let L and M be subapaces of the inner próduct apace E and assúme that L is orthocomplemented and nondegenerate .Then, if t is an admissible topology on E, L#M P LM is L-dense in L and MAL 1 = 0.1 If L and M are orthocomplemented nondegenerate subspaces of the inner produc't apace E, then L#M if and onlyy if PL M is weakly dense in L and PML is weakly dense in M .N