Abstract
Let and be two closed linear relations acting between two Banach spaces and , and let be a complex number. We study the stability of the nullity and deficiency of when it is perturbed by . In particular, we show the existence of a constant for which both the nullity and deficiency of remain stable under perturbation by for all inside the disk .
1. Introduction
For purposes of introduction, we shall consider bounded linear operators and with domain and range in . As usual, let and denote the null space and range of respectively. The dimensions of and are called the nullity and the deficiency of respectively and denoted by and . It is well known that and have some kind of stability when is subjected to some kind of perturbation (see for example [1]). More precisely, and are unchanged when is perturbed by some bounded linear operator under certain prescribed conditions. This stability can be described in the form
Another convenient way of describing this stability is to put it in the form
The stability concept described here is very useful in studying eigenvalue problems of the form and , where denotes the adjoint operator.
This paper deals with the stability theory for nullity and deficiency of linear relations, and it can be seen as a generalization of the classical theory for the corresponding quantities for linear operators. The theory and exposition developed here goes along the lines of the classical texts on the perturbation theory for linear operators (see for example [1, 2]), but in a more general setting. Some stability theorems for multivalued linear operators or what we refer to here as linear relations have been considered in [3] and more recently in [4]. In either of these cases, the perturbing multivalued linear operator does not vary with the varying as the case we consider here.
2. Preliminaries
2.1. Relations on Sets
In this section, we introduce some notation and consider some basic concepts concerning relations on sets. Let and be two nonempty sets. By a relation from to , we mean a mapping whose domain is a nonempty subset of and, taking values in , the collection of all nonempty subsets of . Such a mapping is also referred to as a multivalued operator or at times as a set-valued function. If maps the elements of its domain to singletons, then is said to be a single valued mapping or operator. Let be a relation from to , and let denote the image of an element under . If we define for and , then the domain of is given by
Denote by the class of all relations from to . If belongs to , the graph of , which we denote by , is the subset of defined by
A relation is uniquely determined by its graph, and conversely, any nonempty subset of uniquely determines a relation .
For a relation , we define its inverse as the relation from to whose graph is given by
Let . Given a subset of , we define the image of , to be
With this notation we define the range of by
Let be a nonempty subset of . The definition of given in (5) above implies that
If in particular , then
For a detailed study of relations, we refer to [3, 5–8], and [9].
2.2. Linear Relations
Let and be linear spaces over a field (or ), and let . We say that is a linear relation or a multivalued linear operator if for all and any nonzero scalar we have (1)(2)
The equalities in items (1) and (2) above are understood to be set equalities. These two conditions indirectly imply that the domain of a linear relation is a linear subspace. The class of linear relations in will be denoted by . If , then we denote by . We say that is a linear relation in if . We shall use the term operator to refer to a single valued linear operator while a multivalue linear operator will be generally referred to as a linear relation.
If and are normed linear spaces, we say that is closed if its graph is a closed subspace of . The collection of all such will be denoted by .
We conclude this section with the following theorems which are taken from [3].
Theorem 1. Let . The following properties are equivalent. (i) is a linear relation(ii) is a linear subspace of (iii) is a linear relation(iv) is a linear subspace of
Corollary 2. Let . (i)Then, is a linear relation if and only if holds for all and some nonzero scalars and (ii)If is a linear relation, then and are linear subspaces
For a linear relation , the subspace is called the null space (or kernel) of and is denoted by .
Theorem 3. Let be a linear relation in a linear space , and let . Then, if and only if
Theorem 3 shows that is single valued if and only if .
Theorem 4. Let . Then, is a linear relation if and only if for all and all scalars and
Theorem 5. Let . Then, (a) for and (b) for (c) for
2.3. Normed Linear Relations
Let be a normed linear space. By , we shall mean the set
For a closed linear subspace of , we denote by the natural quotient map with domain and null space . For , we shall denote by . It is well known that for , the operator is single valued (see [3]).
For , we set for and . Note that these notions do not define a norm since nonzero relations can have zero norm.
Lemma 6. Let be such that and . If are such that , then .
Proof. Let . Since and are single valued, we see that Hence, .☐
The following lemma is proved in [3].
Lemma 7. The following properties are equivalent for a linear relation . (i) is closed(ii) is closed and is closed
Lemma 8. (a)Let be bounded. Then, (b)For with and , we have
Proof. (a)From ([3], II.1.6), we have so thatThe inequality , for all , then follows from ([3], II.1.5). (b)Since , we see that since is a subspace (linear subset). For , let and let . Then, , and so by ([3], II.1.4), we get☐
Let be a normed space. By , we denote the norm dual of , that is, the space of all continuous linear functionals defined on , with norm where denotes the action of on . If and , we write and to mean
Let be a linear relation with and . We define the adjoint of by where
This means that
From (22), we see that for all . Hence,
This means that is an extension of , and therefore, the adjoint can be characterized as follows:
Please note that (see [3], III.1.2).
Lemma 9 ([3], III.1.4). Let be a closed linear relation. Then, (1)(2)(3)(4)
Remark 10. If and are closed linear relations with and , then by Lemma 9(2).
3. Lower Bound of a Closed Linear relation
Consider a closed linear relation on a Banach space , and let denote the null space of which is closed since is closed. Since , a coset which contains a point of consists entirely of points of . To see that this is the case, let and let with . Then, and the linearity of implies that . Let denote the collection of all such cosets . On setting, we define a linear operator , where . To see that (25) is well defined, let . Then, , and therefore,
We see from (26) that . So, let . Then, so that . We have
Remark 11. Since , we also have that a coset that contains a point of consists entirely of element of . To see that this is the case, let be a coset in and let with . Then, . The linearity of implies that .
Lemma 12. The linear operator defined by (25) is closed.
Proof. Let be a sequence in such that , and let be a sequence in such that . Let and . Since , we see that . This means that converges to some element of , say,
From (29), we see that .
Since , that is, , we see that dist as and so for some (where for each ). The closedness of implies that and . Hence, and , showing that is closed.☐
We see that is single valued since . We now introduce the quantity called the lower bound of the linear relation . By definition, with the understanding that if is unbounded and that if . It follows from (30) that
Note that if and only if for all . In order for (31) to hold even for this case, one should stipulate that . Obviously, .
Please note that characterization (31) implies that if then the domain of cannot consist of the zero element alone.
The fact that if and only if for all leads to Lemma 13 (see also [3], Proposition II.2.2).
Lemma 13. For , we have
Remark 14. A bounded linear operator is closed if and only if is closed.
Proof. Suppose that with . The boundedness of implies that is a Cauchy sequence and therefore converges, say . The closedness of implies that and . This shows that is closed.
If is a closed linear relation from to , the graph of , is a closed subset of . Sometimes, it is convenient to regard it as a subset of . More precisely, let be the linear subset of consisting of all pairs of the form , where and . We shall call the inverse graph of . As in the case of the graph , is closed if and only is closed. Clearly, . Thus, is closed if and only is closed.
Lemma 15. If is a closed linear relation in a Banach space , then is closed if and only if .
Proof. By definition, if and only if is bounded (where is the operator defined in (25)), and this is true if and only if is closed (we use the fact that is closed because is closed, and then apply Remark (14)).
Now, assume that and let be a convergent sequence in with
Since is a bounded linear operator, the sequence is a Cauchy sequence in and therefore converges to a point since is closed. We see that dist as so that for some , that is,
Since , a coset that contains a point of consists entirely of element of . To see that this is the case, let be a coset in and let with . Then, . The linearity of implies that .
We see from (33) and (34) that and that since and . This shows that is closed.
On the other hand, assume that is closed. Since is closed (since is closed), it is enough, by the closed graph theorem, to show that is closed. So, assume that is a sequence in such that . Then, dist as . Hence, there exists an element such that . The closedness of implies that so that .
Please see ([3], III.5.3) for another proof of Lemma 15.
For the definition of continuity and openness of a linear relation mentioned in Lemmas 16 and 17, please refer to [3].
Lemma 16 ([3], II.3.2, III.1.3, III.1.5, III.4.6). Let . Then, (a) is continuous if and only if (b)(c) is open if and only if (d)If and then
Lemma 17 ([3], III.4.6). (a) is continuous if and only if (b) is open if and only if (c)If is continuous, then (d)If is open, then
4. The Gap between Closed Linear Manifolds and Their Dimensions
Let be a Banach space, and let be a closed subspaces of . We denote by the unit sphere of , that is, . For any two closed linear manifolds and of with , define the gap between and , denoted by to be and set if . can also be characterized as the smallest number for which
It can be seen from the definition that .
Lemma 18. Let and be linear manifolds in a Banach space . If , then there exists an such that
Lemma 18 can be expressed in the language of the quotient space as follows.
Lemma 19. Let and be linear manifolds in a Banach space . If , then there exists an such that
Lemma 20 is a direct consequence of the preceding one.
Lemma 20. If for every none zero , where , then .
See ([1], Page 200) and [2] for Lemmas 21 and 22 respectively.
Lemma 21. Let and be closed linear manifolds of a Banach space . If , then .
Lemma 22. Let be an element of a normed linear space , and let and be closed linear subspaces of . Consider the quotient space , and let denote the quotient class of . For any , there exists such that
5. The Quantity
Let and be two linear spaces and let with . For , let and be the linear manifolds of and and be the linear manifolds of defined inductively as follows:
If then , and therefore,
Since , we conclude by induction that
Similarly,
Note that
Lemma 23. Let be a positive integer. The following first conditions are equivalent to one another, and they in turn imply that condition holds.
,
.
Proof. First, we prove the equivalence of the conditions to . For each , implies . In fact if , then (44), (45), and (47) imply that
Conversely, implies . In fact, if , then
so that each has the property that there exists a such that for some . Then, and so This proves that .
Next, we prove that implies . So, suppose that is satisfied. Then, for , so that for each , there exists a such that for some . Then, and so .☐
If , then , for all since is a nonincreasing sequence. We denote by the smallest number for which the condition (or any one of the other equivalent conditions) is not satisfied. We set if there is no such . This is the case if for example .
Lemma 24. Let and be Banach spaces and let with . Then,
Proof. First, we show that (49) holds for . To begin with, let and let . Then, by definition, and . Hence, there exists an element such that . Since , there exists an element such that . Since , (22) implies that so that . So, for , , showing the .
The second inclusion follows from (see Lemma 9(1)).
We shall therefore assume that (49) has been proved for and prove it for . So, let and let . Then, and . Hence, there exists an element such that . Since , it follows that there exists an element such that . The fact that means that there is an element such that . This means that and since . So, for , meaning that and that . This proves the first inclusion in (49). The second inclusion can be proved in a similar way.☐
Lemma 25. Let . For every , there exists such that , for all , and all .
Proof. Define a linear functional on by setting for all and all . Then, is defined on and is bounded. To show that is indeed bounded, we first note that for , and consider the quotient space . Let . Then, for some so that . This equality means that in (51) can be replaced with , for any without changing the inequality. This therefore means that that is, is bounded on . The Hahn-Banach extension theorem implies that can be extended to the whole of without changing its bound.☐
Remark 26. Lemma 25 above implies that and that by Lemma 9(3).
Lemma 27. Let with , closed and bounded. If , then
Proof. Let . Since , Lemma 16 together with Lemma 17 (a) imply that . So, let , that is, . This means that for and , , which shows that and therefore and so by Remark 26. It follows that . This shows that . Equality (53) then is followed by (49). To prove the second equality, let . Then, for all . Since , we see that
where the last equality follows from the fact that and is a linear space. We see from (55) that . It then follows from (49) and (53) that . This means that and that .
To prove the opposite inequality, let . Then, we have . If follows from Lemmas 9(1), (47), and (49) that . Since is closed, this implies that . Since , we see that . This shows that and therefore .☐
6. Nullity and Deficiency
In this section, we study the behaviour of the nullity and deficiency for linear relations under some perturbations. For , the nullity and the deficiency are defined by
Lemma 28 ([3], III.7.2). Let be a closed linear relation with . Then .
Let and be Banach spaces, and let be a closed linear relation with and . Let be such that for any there exists an -dimensional closed linear subset of such that while this is not true if is replaced by a larger number. In such a case, we set and define to be
Lemmas 29 and 30 show that is defined for every closed linear relation .
Lemma 29. Assume that for every and any closed linear subset of of finite codimension; there is an such that and , then .
Proof. We have to show that for each , there exists an infinite dimensional closed linear subset with the property (57). First, we construct two sequences and such that For , the result holds by ([1], III-Corollary 1.24). Suppose that have been constructed for . Then, and can be constructed in he following way. Let be the collection of all such that . Since is a closed linear subset of with finite codimension (dim and use codim =dim ), there is an such that and . For this , there exists an such that and (see [1], III-Corollary 1.24). It follows from (59) that the are linearly independent so that is infinite dimensional. Each has the form for some positive integer . Hence, for , We show that the coefficients satisfy the inequality For , this is clear from (59) and (61). If we assume that (62) has been proved for , we see from (61) that It follows from (59), (61), and (62) that Let , and let be a sequence in such that . The boundedness of on implies that is a Cauchy sequence in and therefore converges, say . This means that dist as , that is, for some . In other words, . The closedeness of implies that and . Hence, is defined and bounded on the closure of with the same bound.☐
Lemma 30. If is a closed linear relation with closed range(that is, , then and .
Proof. By Lemma 16, implies while Lemma 28 implies that . In view of (58), it is enough to show that . It is clear that . Now suppose that there exists a closed linear manifold with and with property (57). Pick such that where (this is possible by ([2], Lemma 241). For this , on the one hand and on the other hand, leading to the inequality . In other words, there is no with for . This proves that and that . The second equality follows from (58) and Lemma 28.☐
Lemma 31. Let with nonclosed range (that is, ), then
Proof. Let be any closed linear manifold of with finite codimension, and let be denoted by . Consider the mapping defined by setting . Then, is clearly well defined and linear. It is well defined since
for any . It follows that is a finite dimensional space since has finite codimension. ([1], III-Lemma 1.9) implies that is a closed subset of if is a closed subspace of the same space. This would mean that is a closed subset of . To see why this is true, let be a convergent sequence in with . Then, is a Cauchy sequence in and therefore converges to some point . In other words, , so that . The uniqueness of the limit implies that and that since and every coset that contains and element of consists entirely of elements of . Next, we show that if is closed then is closed. So, assume that is closed and let be a sequence in that converges to an element . Then, and so . The closedness of implies that and that .
The contradiction that is both open and closed means that is not closed and that is not closed and therefore . Hence, there exists, for any , an such that and , where . This shows that the conditions of Lemma 29 are satisfied and therefore .☐
Theorem 32. Let and be Banach spaces, and let be a closed linear relation with , having closed range , and with finite. Let be a closed bounded linear relation such that , , and Then, the linear relation is closed and has closed range. Moreover,
Proof. Let be a sequence in such that , and let be a sequence in such that , where with and for each . In other words,
Note that (67) implies that is a Cauchy sequence in and therefore converges to a point of , say . Hence, as , that is, for some . Hence, . The closedness of implies that and . Hence, and so is closed.
To complete the proof, it is enough to show that
and then apply Lemma 31 to conclude that has closed range and Lemma 30 to establish the inequalities in the theorem since by definition and by (58) and Lemma 28.
To prove (70), suppose that for a given there exists a closed linear manifold such that
It then follows from (71) and Lemma 8 that
where . If we pick such that , we see from (72) that for all nonzero . It therefore follows from Lemma 20 that
which means that .
To prove the second inequality, we note that Lemma 16 together with Lemma 17 implies that , , and . It therefore follows that . Applying what has been proved above to the pair , we see that
where the last equality follows from Lemma 28.☐
Lemma 33. Let and be Banach spaces and let be a closed linear relation with and . Set Then, becomes a Banach space if is chosen as the norm.
Proof. That defines a norm on is clear. To prove completeness, assume that is a Cauchy sequence in . Then, and are Cauchy sequences in and , respectively, and therefore converge, say and . Let for each . Then, and so as , that is, . We therefore see that . The closedness of implies that and that . Now, This shows that is complete.☐
Let and be Banach spaces, and let be such that and . In Theorem 34, we write to mean the quantity . The quantities and are defined in a similar way
Theorem 34. Let and be Banach spaces, and let be a closed linear relation with and with closed range . Let be a closed linear relation such that , , , and where and are nonnegative constants such that Then, the linear relation is closed and has closed range. If , then
Proof. Let be a sequence in such that , and let be a sequence in such that , where with and for each . Note that (77) implies that
Since , we see that
and that
Inequality (82) and the linearity of imply that
so that
It therefore follows that for ,
Since by (78) and both and are Cauchy sequences, it follows by (85) that is a Cauchy sequence and therefore converges, say
where we denote by in . The convergence in (86) implies that as . This means that converges to an element of , say . This means that . The closedness of implies that and . Since , we see that . Applying (77) to , we see that , that is, as , . This shows that converges to an element say of , that is, since . Hence, , showing that is closed.
We introduce a norm on by
for some arbitrary but fixed positive constant . Note that the space becomes a Banach space by Lemma 33, which we denote by . We now regard and as linear relations with and denote them by and respectively. Since for every and , we see that . From and the definition of , we also see that .
It is clear that is closed and that
Please note that if . In order to relate to in the other case, we recall that in this case,
where .
But
where we have used the linearity of the natural quotient map and the fact that .
Hence,
where we have used the fact that is an increasing function for any constant .
In view of (78), we can make by choosing small enough. Since , we can apply Theorem 32 to the pair with the result that is closed and (68) holds with replaced with . The result then follows (88).☐
7. Stability Theorems
Consider an eigenvalue problem of the form where and are linear operators from to and the associated problem where the adjoints and exist. The null space of the linear operator is the solution set of the eigenvalue problem (92). Similarly, is the solution set of the eigenvalue problem (93). In studying the above eigenvalue problems, one therefore gets interested in the behaviour of and .
In the setting of linear relations, the eigenvalue problems (92) and (93) can be formulated as where . Conditions (94) and (93) are equivalent to respectively.
As before, the solution sets of (96) and (97) are and , respectively. In this last section, we study the stability of the dimensions of the null spaces of and as varies in some specified subset of the complex plane.
Theorem 35. Let and be Banach spaces and let be such that has closed range, , , and where and are nonnegative constants. Then is closed for , and if , then for .
Proof. It follows from Theorem 34 that is closed if .
If , then . The fact that for every implies that for every . Since , it follows that for every and therefore
so that
Since , we see that there exists at least one in with . Inequality (100) implies that
Since can vary freely in , we conclude that and that .☐
Theorem 36. Let and be Banach spaces, and let be such that has closed range, , , and where and are nonnegative constants. If , then
Proof. Let be as defined in (41) and consider a sequence with the following properties:
where is a positive constant. We show that for each and , there is a sequence that satisfies (104) such that . We set and construct by induction. Suppose have been constructed with properties (104). Since , there exists a such that . Since and can be replaced by any other element of , we can choose such that . Since , we see that . This completes the induction process.
Since and , we see that
For , (105) gives since . For , (104) implies that
We also see from (104) and (106) that
The bounds in (106) and (107) imply that the series
are absolutely convergent for . The convergence of the last series follows from the fact that since .
Let , , , and denote the sequences of the partial sums of the above series in that order. Then, for each , and . Furthermore, and . Since is closed by Lemma 7, we see that and that
Since , a similar argument shows that
One also obtains the equality using the closedness of .
From (109) and (110), we see that
and so .
Furthermore,
Since there is such a for every , we conclude that
☐
We observe that if then Theorem 34 can be used to conclude that has closed range if . However, this conclusion is not possible if no restriction is imposed on . This case is considered in Lemma 37.
Lemma 37. Let and be as in Theorem 36 with . Then, has closed range for .
Proof. In the present case, let and set for any . Lemma 22 implies that for any , Suppose that , and let . Since , we see that, Let denote the quotient space . Since , we see that , and therefore, (115) implies that Letting in (116) leads to the inequality from which we conclude that It therefore follows that , and therefore, is closed if .☐
Finally, we establish the stability of both the nullity and deficiency of for inside the disk for some constant .
Theorem 38. Let and be Banach spaces, and let be such that has closed range, , , and where and are nonnegative constants. If , then and are constants for all for which .
Proof. Let . Then, and we see from (101) that Since , we see from characterization (36) that Since if , Lemma 21 implies that The reverse inequality follows from Theorem 36 by noting that the right-hand side of (103) is less than one if . We therefore conclude by Lemma 21 that if . Combined with (122), we conclude that To show that , we make use of the linear relations and as defined in the proof of Theorem 34. Since is bounded, Lemmas 16 (c), 17 (d), and 15 imply that has closed range. Since by Remark 10 and by Lemma 27, all the assumptions of Theorem 38 are satisfied by the pair and . Since by Lemmas 16 (a) and 17 (c), it follows from (123) that Since by Lemma 16 (b) and (d) and has closed range (since has closed range), it follows from (88), Lemma 28, and (124) that ☐
Theorem 38 remains true if we replace the requirement with .
Data Availability
The data in terms of references used in this manuscript can be publicly accessed.
Conflicts of Interest
The authors declare that there is no conflict of interest.
Acknowledgments
This work was financially supported by the Swedish Sida Phase IV Bilateral Program with Makerere University, 2015 - 2020(project 316) capacity building in mathematics and its applications.