An Extension Theorem for a Sequence of Krein Space Contractions
Consider Krein spaces and and let and be regular subspaces of and , respectively, such that and . For each , let be a contraction. We derive necessary and sufficient conditions for the existence of a contraction such that . Some interesting results are proved along the way.
A number of extensions and generalizations of classical function theoretic interpolation problems are driven by factors such as applications in systems and control theory. Many such extensions and generalizations make use of the commutant lifting theorem in one way or another. In particular, the commutant lifting theorem in the Hilbert space case which was obtained by Sz.-Nagy and Foias has been used to solve extension problems like the ones of Nevanlinna-Pick, Nudelman, Nehari, and many others. Extensions of this theorem to an indefinite setting are given in [1–5]. In  (see also [7, 8]), a time-variant version of the commutant lifting theorem is developed. This time-variant version is called the three-chain completion theorem and is used to solve a number of nonstationary norm constrained interpolation problems on Hilbert spaces. Recall (see ) that the given data for the three-chain completion problem are bounded linear operatorswhere and for are Hilbert spaces satisfying the inclusion relationsGiven operators (1) and tolerance the problem is to find an operator such that andAs far as we know, no extension of this theorem to an indefinite setting has been developed so far.
The three-chain theorem mentioned above is the motivation for the extension problem considered in this paper. For , consider a sequence of Krein space contractions , where and are nested regular subspaces of some Krein spaces and , respectively. The problem is to find a contraction such that for all .
In order to keep this paper as self-contained as possible, we briefly outline some definitions and some elementary facts about Krein spaces and bounded linear operators defined on them. This is done in Section 2. Some important results are stated and proved in Section 3. An extension theorem is considered in Section 4 while Section 5 contains some simple application of the extension theorem discussed in Section 4.
Let be a linear space and let be a sesquilinear form on . This sesquilinear form is called an indefinite metric on . If admits a direct orthogonal sum decompositionin which are Hilbert spaces, then (or ) is called a Krein space. Decomposition (5), which is not unique in general, is called a fundamental decomposition and gives rise to orthogonal projections from onto , which we denote by , respectively. The self-adjoint and unitary operator on defined by , is called a fundamental symmetry on .
The space with the inner productis a Hilbert space. This inner product is used to define the norm of an element of the Krein space byTopological notions such as convergence and continuity are understood to be with respect to this norm topology. The inner product in (6) depends on decomposition (5), as does the norm in (7), but the norms generated by different decompositions of are equivalent. We will denote the norm in (7) by where clarity is needed.
An orthogonal projection in a Krein space is a bounded self-adjoint operator in such that . Note that the norm of an orthogonal projection in a Krein space need not be less than 1. The range of an orthogonal projection in a Krein space is a closed subspace of and the space can be decomposed aswhere . On the other hand, given a closed subspace of such that decomposition (8) holds, then is the range of an orthogonal projection in . In this case, is again a Krein space. Unlike in the Hilbert space case, the decomposition (8) need not hold for a given closed subspace. Closed subspaces for which this decomposition holds are referred to as regular subspaces.
Let and be Krein spaces and let be a bounded linear operator. We say that is a contraction if for all , . If both and its Krein space adjoint are contractions, then is called a bicontraction.
Let and be Krein spaces and let , the space of bounded linear operators from into . A column extension of is an operator of the form where is a Krein space and . By a row extension of we shall mean an operator of the form where is a Krein space and . It is shown in  (see also ) that if is a contraction, then there exist contractive row and column extensions of where the extension space is a Hilbert space. Contractive matrix extensions of a contractive operator where the extending space is a Hilbert space are thoroughly discussed in  where Lemma 1, Theorem 2, and Lemma 3 can be found. Results more general than those provided by Lemma 1 and Theorem 2 can be found in  while Lemma 3 can also be found in .
Lemma 1. Let and be Krein spaces and let be a contraction. Let be a contractive column extension of with a Hilbert space. If norms are computed relative to some fixed fundamental decompositions of and and the induced fundamental decomposition of , then
The above norm estimate enables one to fix a bound for the norm of the operator . The following lemma is helpful in finding matrix extensions of a contractive operator . See  for a similar result in a Hilbert space setting.
Theorem 2. Let , , and be Krein spaces and let be a Hilbert space. Assume that are contractions. Then there exists an operator such that is a contraction. If is a bicontraction, may be chosen such that is a bicontraction.
We conclude this section by stating the following lemma.
Lemma 3. Let be regular subspaces of a Krein space such that . Suppose that the projections of onto the subspaces are uniformly bounded. Then are regular subspaces of with = . If is the projection of onto , then with convergence in the strong operator topology.
3. Some Preliminary Results
In this section we prove some useful results regarding sequences of regular subspaces. See [4, Corollary 3.3 and Lemma 3.4] for closely related results.
Theorem 4. For , let be a sequence of regular subspaces of a Krein space such that , . For each , let be the projection of onto . If the projections are uniformly bounded then is a regular subspace of .
Proof. Set , . Then is a sequence of regular subspaces of such that . Let be the projections of onto the subspaces . Then and so . Since the projections are uniformly bounded we see that the projections are also uniformly bounded. Hence is a regular subspace of by Lemma 3.
Theorem 5. For , let be regular subspaces of a Krein space such that . Let be the projection of onto . If is a fundamental decomposition of , we denote by the projection of onto . Let Then the following are equivalent: (1) is regular and there exists a fundamental symmetry on such that is a fundamental symmetry on .(2)There exist fundamental decompositions such that(a),(b), where is any fundamental symmetry on .
Proof. Suppose that (1) holds and let be the fundamental decomposition of which gives rise to the fundamental symmetry with the stated property, where . Then where . To show that (a) holds, we let . Then Hence and so . This means that and that . Therefore . Similarly, we have that . Hence (a) holds. To show that (b) also holds, we let Let , and let be a fundamental symmetry on . Then is a fundamental symmetry on . Now, where with , and . Therefore Hence . Similarly .
Conversely, assume that (2) holds. We start by showing that is regular. To do this we note that . So for some constant . Hence is regular by Theorem 4.
Next set and let be the projection of onto . Then . Hence for some constant since (2) holds. Hence and are regular subspaces by Lemma 3. Let . Then there exists a sequence such that . Hence . This means that since . Hence is regular and nonnegative and so it is a uniformly positive subspace of . Similarly is a uniformly negative subspace of . We now show that Let . Then there exists a sequence such that . If then there exists a sequence such that . Hence = = 0. This implies that . Next, assume that . Then and . This in turn means that and and so . Let . Then there exists a sequence such that . Hence for all and so . Hence . This implies that is maximal positive and is maximal negative. Hence is a maximal uniformly positive space and is a maximal uniformly negative space. Since , is uniformly positive and is uniformly negative, there exists a fundamental decomposition such that . But maximal implies . Hence is a fundamental decomposition of . This decomposition gives rise to a fundamental symmetry where .
We show that . Let . Then , . Hence . Similarly, and so .
Corollary 6. For , let , , , and be as in Theorem 5. Then there exists a fundamental symmetry on which commutes with the projections onto if and only if is regular and there exists a fundamental symmetry on such that is a fundamental symmetry on .
Proof. Suppose that is regular and exists which has the stated property. Let be the fundamental decomposition which gives rise to . Since is regular there exists a fundamental decomposition such that . This gives rise to on such that is a fundamental symmetry on . Let . Then , where and . Now, (where ). On the other hand, Hence .
Conversely, assume that there exists a fundamental symmetry on that commutes with the projections . To show that ’s are uniformly bounded we consider the fundamental decomposition which gives rise to the fundamental symmetry . Let be the matrix representation of with respect to these decompositions. Then the commutativity condition implies that and so the matrix representation of is diagonal; that is to say, Since and for all , we see that ’s are uniformly bounded. The uniform boundedness of ’s implies that is regular. From the matrix representation of above we see that and and so . Hence condition (b) in Theorem 5 holds. Since, for each , is a regular subspace and , we have that, for any fundamental decomposition , there exist a fundamental decomposition such that . Hence condition (a) in Theorem 5 also holds. This shows that part of Theorem 5 holds and this completes the proof.
4. An Extension Theorem for a Sequence of Contractions
In this section, we formulate and give a proof of an extension theorem for a sequence of contractions defined on a nested sequence of regular subspaces. Please refer to [14, Lemma 3.1] and  for closely related results.
Theorem 7. Let and be Krein spaces and let and , be sequences of regular subspaces satisfyingLet and be the orthogonal projections of onto and onto , respectively, and let and . Assume that there exist fundamental symmetries on and on such that commutes with and commutes with and that , , , and are all Hilbert space.
For each , letbe a contraction. Then there exists a contraction such thatif and only if
Proof. First, let us assume that such an extension exists and take . Since and (30) holds, we have that and therefore . Hence the first condition in (31) holds. To show that the second condition also holds, we first note that since and commute with and , respectively, Corollary 6 ensures that the subspaces and are regular and that there exist fundamental symmetries on and on such that is a fundamental symmetry on and is a fundamental symmetry on . Let . Then we have that . Hence, Hence, for all and so the norms of ’s are uniformly bounded.
Conversely, let , , be contractions satisfying both conditions in (31). Since , we can decompose the operator as . The fact that is a contraction and is a Hilbert space implies that the operator is a contraction. Since is a contraction, Theorem 2 implies that there exists a bounded operator such thatis a contraction. For each , the operator is clearly a contractive extension of .
We now show that, for , the contractive liftings ’s of ’s are uniformly bounded. Let be the bound for the norm of ’s. Then and so Lemma 1 implies that . Hence ’s are uniformly bounded.
Define an operator by for . The operator is well defined. To see this, assume that and . Then and . For , . The operator is bounded since the contractions are uniformly bounded. To see that this is the case, consider . Then Hence for , we see that and so is a bounded operator. Since it is defined on a dense set we can extend it by continuity to a bounded operator . The operator is clearly an extension of for each since is an extension of for each . To show that is a contraction, let . Then there exists a sequence such that . Since the inequality holds for each we have Hence is a contraction.
Define a new operator by the matrix where is any contractive row extension of and . Since is a Hilbert space, Theorem 2 guarantees the existence of such that is a contraction. Note that since is also a Hilbert space we may set , the zero operator, in the matrix representation of and the above result still holds.
Clearly, and so is the required extension.
In this section we use Theorem 7 to solve a nonstationary extension problem in a Krein space setting. Let and be Krein spaces with fixed fundamental decompositionsand letbe bounded operators with matrix representationswith for and for . Define the Krein spaces and by and let to be a bounded operator such thatWe use Theorem 7 to establish conditions under which such an operator is a contraction. Equality (47) implies that the operator is of the formConsider the decompositions With the above decompositions, the operator takes the form
For each , let denote the operatorWith these notations, we are in a position to state the following theorem.
Theorem 8. Let and be Krein spaces with decompositions (43) with operators in (44). Then there exists a contraction satisfying (47) if and only if, for each , the operator in (51) defines a contraction whose norm has an upper bound independent of .
Proof. Assume that a contraction satisfying (47) exists and consider the operator matrices (50) and (51). We see that, for ,whereSince is a Hilbert space, we see that and so This shows that the operators are uniformly bounded. Since is a contraction and is a Hilbert space, (52) implies that is a contraction for each .
To prove the reverse implication, we assume that for each the operator is a contraction of norm at most where . Since and , it follows that and are regular subspaces. By construction, the subspaces and satisfy the conditions of Corollary 6 and hence do satisfy all the conditions specified in Theorem 7. Since, for each , the operators are defined by (51) and are assumed to be uniformly bounded, condition (31) in Theorem 7 is also fulfilled (it can be easily seen that the first condition in (31) is satisfied by the operators ). Since (31) is satisfied for all , it follows that there exists a contraction satisfying (30) and therefore must be of the form (50). This concludes the proof.
Conflicts of Interest
The author declares that there are no conflicts of interest regarding the publication of this paper.
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