Stability Theory for Nullity and Deficiency of Linear Relations

Let $\mathcal A$ and $\mathcal B$ be two closed linear relation acting between two Banach spaces $X$ and $Y$ and let $\lambda$ be a complex number. We study the stability of the nullity and deficiency of $\mathcal A$ when it is perturbed by $\lambda\mathcal B$. In particular, we show the existence of a constant $\rho>0$ for which both the nullity and deficiency of $\mathcal A$ remain stable under perturbation by $\lambda\mathcal B$ for all $\lambda$ inside the disk $\vert \lambda\vert<\rho$.


Introduction
For purposes of introduction we shall consider bounded linear operators A and B with domain X and range in Y . As usual, let N (A) and R(A) denote the null space and range of A respectively. The dimensions of N (A) and Y /R(A) are called the nullity and the deficiency of A respectively and denoted by α(A) and β(A). It is well known that α(A) and β(A) have some kind of stability when A is subjected to some kind of perturbation (see for example [7]). More precisely, α(A) and β(A) are unchanged when A is perturbed by some bounded linear operator B under certain prescribed conditions. This stability can be described in the form The stability concept described here is very useful in studying eigenvalue problems of the form Ax = λBx and A * y = λB * y, where A * 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 [7] and [6]), 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 [5] and more recently in [8]. In either of these cases, the perturbing multivalued linear operator B does not vary with the varying λ as the case we consider here.

Preliminaries
2.1. Relations on sets. In this section we introduce some notation and consider some basic concepts concerning relations on sets. Let U and V be two nonempty sets. By a relation T from U to V we mean a mapping whose domain D(T ) is a nonempty subset of U , and taking values in 2 V \ ∅, the collection of all nonempty subsets of V . Such a mapping T is also referred to as a multi-valued operator or at times as a set valued function. If T maps the elements of its domain to singletons, then T is said to be a single valued mapping or operator. Let T be a relation from U to V and let T (u) denote the image of an element u ∈ U under T . If we define T (u) = ∅ for u ∈ U and u / ∈ D(T ) then the domain D(T ) of T is given by Denote by R(U, V ) the class of all relations from U to V . If T belongs to R(U, V ), the graph of T , which we denote by G(T ) is the subset of U × V defined by A relation T ∈ R(U, V ) is uniquely determined by its graph, and conversely any nonempty subset of U × V uniquely determines a relation T ∈ R(U, V ). For a relation T ∈ R(U, V ) we define its inverse T −1 as the relation from V to U whose graph G(T −1 ) is given by If in particular v ∈ R(T ), then For a detailed study of relations, we refer to [1], [4], [3], [2], [5], and [9].
2.2. Linear Relations. Let X and Y be Linear spaces over a field K = R (or C) and let T ∈ R(X, Y ). We say that T is a linear relation or a multi-valued linear operator if for all x, z ∈ D(T ) and any nonzero scalar α we have (1) T (x) + T (z) = T (x + z), (2) αT (x) = T (αx). The equalities in (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 R(X, Y ) will be denoted by LR(X, Y ). If X = Y then we denote LR(X, X) by LR(X). We say that T is a linear relation in X if T ∈ LR(X). We shall use the term operator to refer to a single valued linear operator while a multi-value linear operator will be generally referred to as a linear relation.
If X and Y are normed linear spaces, we say that T ∈ LR(X, Y ) is closed if its graph G(T ) is a closed subspace of X × Y . The collection of all such T will be denoted by CLR(X, Y ).
We conclude this section with the following theorems which are taken from [5].
The following properties are equivalent.

a linear relation if and only if
holds for all x 1 , x 2 ∈ D(T ) and some nonzero scalars α and β. (ii) If T is a linear relation then T (0) and T −1 (0) are linear subspaces.
For a linear relation T , the subspace T −1 (0) is called the null space (or kernel ) of T and is denoted by N (T ).
2.3. Normed linear relations. Let X be a normed linear space. By B X we shall mean the set B X := {x ∈ X : |x| ≤ 1}. For a closed linear subspace E of X, we denote by Q E the natural quotient map with domain X and null space E. For T ∈ LR(X, Y ), we shall denote Q T (0) by Q T . It is well known that for T ∈ LR(X, Y ), the operator Q T T is single valued (see [5]).
For T ∈ LR(X, Y ), we set T x = Q T T x for x ∈ D(T and T = Q T T . Note that these notions do not define a norm since nonzero relations can have zero norm. Proof. Let z ∈ A(x 1 ) ∩ B(x 2 ). Since Q A and Q B are single valued, we see that The following lemma is proved in [5].
Lemma 2.7. The following properties are equivalent for a linear relation A.
(i) A is closed.
(ii) Q A A is closed and A(0) is closed. Proof.
(a) From [5, II.1.6] we have T = sup The inequality T x ≥ T x for all x ∈ D(T ) then follows from [5, is a subspace (linear subset). For x ∈ D(S), let s ∈ S(x) and let t ∈ T (x). Then s + t ∈ (S + T )(x) = S(x) + T (x) and so by [5, II.1.4] we get Let X be a normed space. By X ′ we denote the norm dual of X, that is, the space of all continuous linear functionals x ′ defined on X, with norm where [x, x ′ ] := x ′ (x) denotes the action of x ′ ∈ X ′ on x ∈ X. If M ⊂ X and N ⊂ X ′ , we write M ⊥ and N ⊤ to mean Let T be a linear relation with D(T ) ⊂ X and R(T ) ⊂ Y . We define the adjoint From (2.3) we see that y ′ (y) = x ′ (x) for all y ∈ T (x), x ∈ D(T ). Hence This means that x ′ is an extension of y ′ T (x) and therefore the adjoint T ′ can be characterized as follows: Please note that T ′ ∈ CLR(Y ′ , X ′ ) (see [ we define a linear operator A : X → X where X := X/A(0). To see that (3.1) is well defined, let x, y ∈ x. Then x − y ∈ N (A) and therefore We see from (3.2) that A(x) ∩ A(y) = ∅. So, let u ∈ A(x) ∩ A(y). Then Remark 3.1. Since A(0) ⊂ R(A), we also have that a coset x ∈ X that contains a point of R(A) consists entirely of element of R(A). To see that this is the case, let x be a coset in X and let u, v ∈ x with u ∈ R(A). Proof. Let { x n } be a sequence in D such that x n → x ∈ X and let {Q A Ax n } be a sequence in R(A) such that Q A Ax n → y ∈ X. Let x n ∈ x n and x ∈ x. Since x n → x, we see that dist(x n − x, N (A)) → 0. This means that x n − x converges to some element of N (A), say, From (3.4) we see that x n → x + u = w ∈ x.
Since Q A Ax n → y ∈ X, that is, z n → y, we see that dist(z n − y, A(0)) → 0 as n → ∞ and so z n → y + v = z ∈ y for some v ∈ A(0) (where z n ∈ A(x n ) for each n ∈ N). The closedness of A implies that w ∈ D(A) and z ∈ A(w). Hence x ∈ D and A( x) = Q A A(x) = y, showing that A is closed.
We see that A −1 is single valued since A −1 0 = 0 . We now introduce the quantity γ(A) called the lower bound of the linear relation A. By definition,

Remark 3.4. A bounded linear operator T is closed if and only if D(T ) is closed.
Proof. Suppose that u n → u with u n ∈ D(T ). The boundedness of T implies that T (u n ) is a Cauchy sequence and therefore converges, say T (u n ) → v. The closedness of T implies that u ∈ D(T ) and T (u) = v. This shows that D(T ) is closed.
If S is a closed linear relation from X to Y , the graph of S, G(S) is a closed subset of X × Y . Sometimes it is convenient to regard it as a subset of Y × X. More precisely, let G ′ (S) be the linear subset of Y × X consisting of all pairs of the form (v, u) where u ∈ D(S) and v ∈ S(u). We shall call G ′ (S) the inverse graph of S. As in the case of the graph G(S), G ′ (S) is closed if and only S −1 is closed.

)), and this is true if and only if
is closed (we use the fact that A −1 is closed because A is closed, and then apply Remark (3.4)). Now assume that γ(A) > 0 and set {y n } be a convergent sequence in R(A) with (3.7) y n → y.
Since Q A is a bounded linear operator, the sequence {Q A y n } is a Cauchy sequence in R(Q A A) and therefore converges to a point z ∈ R( Since A(0) ⊂ R(A), a coset x ∈ H that contains a point of R(A) consists entirely of element of R(A). To see that this is the case, let x be a coset in H and let We see from (3.7) and (3.8) that y ∈ z and that y ∈ R(A) since z ∈ R(A) and y ∈ z. This shows that R(A) is closed.
On the other hand, assume that Please see [5,III.5.3] for another proof of Lemma 3.5. For the definition of continuity and openness of a linear relation T mentioned in the following two lammas, please refer to [5].

The gap between closed linear manifolds and their dimensions
Let Z be a Banach space and let L be a closed subspaces of Z. We denote by S L the unit sphere of L, that is, It can be seen from the definition that 0 ≤ δ(M, N ) ≤ 1. See [7] for the following lemma.
The above lemma can be expressed in the language of the quotient space as follows.
The following lemma is a direct consequence of the preceding one.

The quantity ν(A : B)
Let X and Y be two linear spaces and let A, B ∈ LR(X, Y ) with B(0) ⊂ A(0). For n ∈ N, let M n and N n be the linear manifolds of X and M ′ n and N ′ n be the linear manifolds of Y ′ defined inductively as follows: Similarly, Note that Lemma 5.1. Let n be a positive integer. The following first n conditions are equivalent to one another and they in turn imply that condition (κ) holds.
If N 1 ⊂ M n then N 1 ⊂ M n ′ , for all n ′ < n since M n is a non increasing sequence. We denote by ν(A : B) the smallest number n for which the condition N 1 ⊂ M n (or any one of the other equivalent conditions) is not satisfied. We set ν(A : B) = ∞ if there is no such n. This is the case if for example A −1 (0) ⊂ B −1 (0).
The second inclusion follows from . We shall therefore assume that (5.9) has been proved for n = k and prove it for n = k + 1. So, let g ′ ∈ M ′ k+1 and let z ∈ D(B) ∩ N k+1 . Then The fact that z ∈ N k+1 means that there is an element w ∈ B(N k ) such that (z, w) ∈ G(A). This means that l ′ (w) = h ′ (z) and This proves the first inclusion in (5.9). The second inclusion can be proved in a similar way.
for all y ∈ Ax, and all x ∈ D(A).
Proof. Define a linear functional g ′ on Y ′ by setting g ′ (y) = f ′ (x) for all y ∈ A(x) and all x ∈ D(A). Then g ′ is defined on R(A) and is bounded. To show that g ′ is indeed bounded, we first note that for y ∈ A(x), . This equality means that x in (5.10) can be replaced with x 1 for any x 1 ∈ x without changing the inequality. This therefore means that that is, g ′ is bounded on R(A). The Hahn-Banach extension theorem implies that g ′ can be extended to the whole of Y ′ without changing its bound. Proof . This means that for x ∈ N 1 and y ∈ B(x), g ′ (x) = f ′ (y) = 0, which shows that g ′ ∈ N ⊥ 1 = N (A) ⊥ and therefore g ′ ∈ R(A ′ ) and so To prove the opposite inequality, let n < v ′ . Then we have N ′ 1 ⊂ M ′ n . If follows from Lemma 2.9 (a), (5

.8), and (5.9) that [A(X)]
This shows that v > n and therefore v ≥ v ′ .

Nullity and Deficiency
In this section we study the behaviour of the nullity and deficiency for linear relations under some perturbations.  The following two lemmas show that α ′ (A) is defined for every closed linear relation A. Proof. We have to show that for each ε > 0, there exists an infinite dimensional closed linear subset N ε ⊂ D(A) with the property (6.1). First we construct two sequences x n ∈ D(A) and f n ∈ X ′ such that x n = 1, f n = 1, f n (x n ) = 1, f k (x n ) = 0, k = 1, 2, · · · , n − 1, (6.3) A(x n ) ≤ 3 −n ε, n ∈ N.
For n = 1, the result holds by [7, III-Corollary 1.24]. Suppose that x n , f k have been constructed for k = 1, 2, · · · , n − 1. Then x n and f n can be constructed in he following way. Let M ⊂ X be the collection of all x ∈ X such that f k (x) = 0, k = 1, 2, · · · , n − 1. Since M is a closed linear subset of X with finite codimension (dim M ⊥ ≤ n − 1 and use codim M =dim M ⊥ ), there is an x n ∈ M ∩ D(A) such that x n = 1 and A(x n ) ≤ 3 −n ε. For this x n , there exists an f n ∈ X ′ such that f n = 1 and f n (x n ) = 1 (see [7, III-Corollary 1.24]. It follows from (6.3) that the x n are linearly independent so that M ′ ε := span {x 1 , x 2 , · · · } is infinite dimensional. Each x ∈ M ′ ε has the form (6.4) x = ξ 1 x 1 + ξ 2 x 2 + · · · + ξ n x n for some positive integer n. Hence for k = 1, 2, · · · , n, We show that the coefficients ξ k satisfy the inequality (6.6) |ξ k | ≤ 2 k−1 x , k = 1, 2, · · · , n.
For k = 1, this is clear from (6.3) and (6.5). If we assume that (6.6) has been proved for k < j, we see from (6.5) that It follows from (6.3), (6.1) and (6.1) that The closedeness of A implies that x ∈ D(A) and w ∈ A(x). Hence Q A A is defined and bounded on the closure of M ′ ε with the same bound. that T (X) is a closed subset of Y . To see why this is true, let {y n } be a convergent sequence in T (X) with y n → y ∈ Y . Then {Qy n } is a Cauchy sequence in Y and therefore converges to some point z ∈ QT (X). In other words, y n − z → w ∈ T (0), so that y n → z + w ∈ z. The uniqueness of the limit implies that y = z + w ∈ z and that y ∈ R(T ) since z ∈ R(T ) and every coset that contains and element of R ( Then the linear relation A + B is closed and has closed range. Moreover, Proof. Let {x n } be a sequence in D(A) such that x n → x ∈ X and let {y n } be a sequence in R(A + B) such that y n → y ∈ Y , where y n = u n + v n with u n ∈ A(x n ) and v n ∈ B(x n ) for each n ∈ N. In other words, (6.10) u n + v n → y.
Note that (6.8) implies that {Q B B(x n )} is a Cauchy sequence in Y := Y /B(0) and therefore converges to a point of Y , say The closedness of B implies that x ∈ D(B) and v + z ∈ B(x). Hence y = y − v − z + (v + z) ∈ A(x) + B(x) and so A + B is closed.
To complete the proof, it is enough to show that It then follows form (6.12) and Lemma 2.8 that where x ∈ X := X/N (A). If we pick ε such that 0 < ε < γ(A) − B we see from (6.13) that x < x for all non-zero x ∈ D(A). It therefore follows from To prove the second inequality, we note that Lemma 3.6 together with Lemma 3.7 imply that B ′ = B , γ(A ′ ) = γ(A), and (A + B) ′ = A ′ + B ′ . It therefore follows that B ′ ≤ γ(A ′ ). Applying what has been proved above to the pair A ′ , B ′ , we see that where the last equality follows from Lemma 6.1. Lemma 6.6. Let X and Y be Banach spaces and let T be a closed linear relation with D(T ) ⊂ X and R(T ) ⊂ Y . Set (6.14) x Then D(T ) becomes a Banach space if · D(T ) is chosen as the norm.
Proof. That · D(T ) defines a norm on D(T ) is clear. To prove completeness, assume that {x n } is a Cauchy sequence in D(T ). Then {x n } and {Q T T (x n )} are Cauchy sequences in X and Y = Y /T (0) respectively and therefore converge, say, x n → x ∈ X and Q T T (x n ) → u ∈ Y . Let u n ∈ T (x n ) for each n ∈ N. Then u n → u and so dist (u n − u, T (0)) → 0 as n → ∞, that is, u n − u → v ∈ T (0). We therefore see that u n → u + v = s ∈ u. The closedness of T implies that x ∈ D(T ) and that s ∈ T (x). Now, This shows that D(T ) is complete.
Let X and Y be Banach spaces and let A, B ∈ CLR(X, Y ) be such that D(A) ⊂ D(B) and B(0) ⊂ A(0). In the following theorem we write B(x) A to mean the quantity Q A B(x) . The quantities A(x) A and B(x) B are defined in a similar way where σ and τ are non-negative constants such that Then the linear relation A + B is closed and has closed range. If α(A) < ∞ then Proof. Let {x n } be a sequence in D(A) such that x n → x ∈ X and let {y n } be a sequence in R(A + B) such that y n → y ∈ Y , where y n = u n + v n with u n ∈ A(x n ) and v n ∈ B(x n ) for each n ∈ N. Note that (6.15) implies that and that (6.20) Inequality (6.20) and the linearity of Q A implies that It therefore follows that for m, n ∈ N, Since 1 − τ > 0 by (6.16) and both {x n } and {y n } are Cauchy sequences, it follows by (6.23) that {Q A u n } is a Cauchy sequence and therefore converges, say, where we denote Q A u n by u n in Y /A(0). The convergence in (6.24) implies that dist (u n − u, A(0)) → 0 as n → ∞. This means that u n −u converges to an element of A(0) = A(0), say u n − u → z ∈ A(0). This means that u n → z − u = s. The closedness of A implies that x ∈ D(A) and s ∈ A(x). Since u n → s, we see that where we have used the linearily of the natural quotient map and the fact that A(z) = A(0). Hence where we have used the fact that f (t) = t α+t is an increasing function for any constant α.
In view of (6.16), we can make γ(Ȃ) > 1 by choosing ε small enough. Since B ≤ 1, we can apply Theorem 6.5 to the pairÀ,B with the result that R(Ȃ+B) = R(A + B) is closed and (6.9) holds with A, B replaced withȂ,B. The result then follows by (6.26).

Stability Theorms
Consider an eigenvalue problem of the form In studying the above eigenvalue problems, one therefore gets interested in the behaviour of N (A − λB) and N (A * − λB * ).
In the setting of linear relations, the eigenvalue problems (7.1) and (7.2) can be formulated as where A, B ∈ LR(X, Y ). Conditions (7.3) and (7.2) are equivalent to respectively. As before, the solution sets of (7.5) and (7.6) are N (A − λB) and N (A ′ − λB ′ ) = R(A−λB) ⊥ respectively. In this last section we study the stability of the dimensions of the null spaces of A − λB and A ′ − λB ′ as λ varies in some specified subset of the complex plane. This is considered in the following theorems. Proof. If follows from Theorem 6.
Since R(A) = A(0), we see that there exists at least one x in X = X/N (A) with x = 0. Inequality (7.8) implies that Since x can vary freely in x, we conclude that γ(A) ≤ σ|λ|/(1 − |λ|τ ) and that |λ| ≥ γ(A) σ+τ γ(A) . Proof. Let N k be as defined in (5.2) and consider a sequence z k with the following properties: where ξ is a positive constant. We show that for each z ∈ N (A) and ξ < γ(A), there is a sequence z k that satisfies (7.12) such that z = z 1 . We set z = z 1 and construct z k by induction. Suppose z 1 , z 2 , . . . x k have been constructed with properties (7.12). Since z k ∈ N k ⊂ M 1 = B −1 (A(X)), there exists a z k+1 ∈ D(A) such that A(z k+1 ) ∩ B(z k ) = ∅. Since γ(A) z k+1 ≤ A(z k+1 ) and z k+1 can be replaced by any other element of z k+1 we can choose z k+1 such that ξ z k+1 ≤ A(z k+1 ) . Since A(z k+1 ) ∩ B(z k ) = ∅, we see that z k+1 ∈ A −1 (B(N n )) = N k+1 . This completes the induction process.
Since A(z k+1 ) ∩ B(z k ) = ∅ and A(0) ⊃ B(0), we see that For k = 1, (7.13) gives A(z 2 ) ≤ B(z 1 ) ≤ σ z 1 since z 1 ∈ N (A). For k ≥ 2, (7.12) implies that We also see from (7.12) and (7.14) that The bounds in (7.14) and (7.15) imply that the series are absolutely convergent for |λ| < ξ σ+ξτ . The convergence of the last series follows from the fact that Let u n (λ), λ n (A), λ n (B) and λ n (B A ) denote the sequences of the partial sums of the above series in that order. Then for each n, u n (λ) ∈ D() and λ n (A) ∈ Y := Y / A(0). Furthermore, u n (λ) → u(λ) and λ n (A) → λ(A). Since Q A A is closed by Lemma 2.7 we see that u(λ) ∈ D(Q A A) = D(A) and that Since A(z k+1 ) ∩ B(z k ) = ∅, a similar argument shows that One also obtains the equality Q B B(u(λ)) = λ(B) = ∞ k=1 λ k Q B B(z k ) using the closedness of B. From (7.16) and (7.17) we see that Since there is such a u(λ) ∈ N (A − λB) for every z − z 1 ∈ N (A), we conclude that .
We observe that if α(A) < ∞ then Theorem 6.7 can be used to conclude that A − λB has closed range if |λ| < γ(A) σ+τ γ(A) . However, this conclusion is not possible if no restriction is imposed on α(A). This case is considered in the next lemma. 3σ+τ γ(A) . Proof. In the present case, let x ∈ X and set y = x − u for any u ∈ N (A − λB). Lemma 4.5 implies that for any ε > 0, Let X denote the quotient space X/N (A − λB). Since x − y = u ∈ N (A − λB), we see that y ≥ y = x and therefore (7.20) implies that Letting ε → 0 in (7.21) leads to the inequality It therefore follows that γ(A − λB) > 0 and therefore R(A − λB) is closed if |λ| < γ(A) 3σ+τ γ(A) .