Integr. equ. oper. theory 48 (2004), 147–157 0378-620X/020147-11, DOI 10.1007/s00020-002-1183-y c 2004 Birkh¨
auser Verlag Basel/Switzerland

Integral Equations and Operator Theory

Stability of P(S) under Finite Perturbation Dominique Gagnage Abstract. T. Kato [9] found an important property of semi-Fredholm pencils, now called the Kato decomposition. Few years later, M.A. Kaashoek [7] introduced operators having the property P(S) as a generalization of semiFredholm operators. In [4], it is proved that these two notions are linked. The aim of this work is to study the stability of the property P(S) under finite perturbation. Keywords. Kato decomposition of finite type, P (S : k) property.

1. Introduction Let X and Y be two Banach spaces. Denote by B(X, Y ) the set of all bounded operators from X to Y . For an operator A in B(X, Y ), denote by N (A) and R(A) its kernel and its range, respectively. In this article, we will consider T and S, two operators of B(X, Y ) not equal to zero. Before explaining the notion of having the property P(S), let us recall the definition of the subspaces relative to (T, S), introduced by T. Kato and M. A. Kaashoek ([8], [7]) :  D0 (T : S) = X, R0 (T : S) = Y,    Rn+1 (T : S) = T Dn (T : S), Dn+1 (T : S) = S −1 Rn+1 (T : S) for n ≥ 0, N0 (T : S) = {0}, M0 (T : S) = {0},    Nn+1 (T : S) = T −1 Mn (T : S), Mn+1 (T : S) = SNn+1 (T : S) for n ≥ 0. If it is not ambiguous, we will write Dn , Nn , Rn and Mn for the corresponding subspaces. Notice that the sequences (Dn (T : S))n≥0 , (Rn (T : S))n≥0 are decreasing and the sequences (Nn (T : S))n≥0 , (Mn (T : S))n≥0 are increasing.

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    D(T : S) = Dn (T : S), R(T : S) = Rn (T : S),   n≥0 n≥0    N (T : S) and M (T : S) = Mn (T : S). N (T : S) =  n  n≥0

n≥0

We can easily see that   T D(T : S) ⊂ R(T : S), S −1 R(T : S) = D(T : S),  for all λ = 0, N (T + λS) ⊂ D(T : S). We notice that in the particular case that X = Y and S = I, we have Dn = Rn = R(T n ) and Nn = Mn = N (T n ). Before giving the definition of the operators which we will study in this paper, let us recall a notation. For M and N two subspaces of X, we write M ⊂e N if there exists a finite dimensional subspace F of X such that M ⊂ N + F , i.e. dim [M/(M ∩N ) ] < +∞. Now we can introduce the notion we are interested in. According to M. A. Kaashoek, [7], T is said to have the property P (S : k) if R(T ) is closed and dim N (T )/[D(T :S)∩N (T )] = k. We will write T ∈ P(S) if there exists k ∈ N such that T has the property P (S : k), i.e. if R(T ) is closed and N (T ) ⊂e D(T : S). This class of operators has been studied in several articles ([1], [2], [3], [4], [5], [7]). Let us recall a definition: the pair (T, S) has a Kato decomposition (DK) if there exist M , N , closed subspaces of X, M  , N  , closed subspaces of Y , such that i) X = M ⊕ N , Y = M  ⊕ N  , ii) T M ⊂ M  , T N ⊂ N  , SM ⊂ M  , SN ⊂ N  , iii) T |M has the property P (S : 0), iv) S : N → N  is bijective and S −1 T |N is nilpotent. We will say that (T, S) has a Kato decomposition of finite type (DKF) if the previous assertions hold and dim N < +∞. In [4], we have proved the following theorem which links the two notions : Theorem 1.1. (Theorem 2.3, [4]) : If R(T ) is closed, the following are equivalent : 1) T ∈ P(S), i.e. N (T ) ⊂e D(T : S). 2) N (T : S) ⊂e S −1 R(T ). 3) N (T : S) ⊂e D(T : S). 4) The pair (T, S) has a Kato decomposition of finite type. Example. i) The Kato decomposition has been studied for the first time for semiFredholm operators by T. Kato, Theorem 4, [9]. He proved that if T is semiFredholm, i.e. R(T ) is closed and min{dim N (T ), codim R(T )} < +∞, then the pair (T, S) has a DKF. ii) In the case that X = Y = H, a Hilbert space, and S = I, the Kato decomposition caracterizes the quasi-Fredholm operators, Theorem 3.2.2, [11]. iii) In Theorem 2.3, [4], it is shown that (T, S) has a DKF if and only if T has

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the property P (S : k), for some integer k. In the particular case that X = Y and S = I, T has a DKF if and only if T is essentially s-regular, a class of operators studied in [10], [12], [13], for example.

2. Preliminaries First, let us recall a lemma. Lemma 2.1. (Lemma 2.1, [6]) : Let U, V, W be subspaces of X. If U ⊂ W , then (U + V ) ∩ W = U + (V ∩ W ). Let us underline some relations between the subspaces relative to the pair (T, S). Lemma 2.2. The following are equivalent : 1) For all m ∈ N, N (T ) ⊂ Dm (T : S). 2) For all n ∈ N, Nn (T : S) ⊂ D1 (T : S). 3) For all m, n ∈ N, Nn (T : S) ⊂ Dm (T : S). 4) For all m, n ∈ N, Nn (T : S) ⊂ (S −1 T )m Nn+m (T : S). Proof. 1)⇒2) : As the cases n = 0, 1 are clear, let n ≥ 2. We prove that for k ∈ {0, . . . , n}, we have Nk (T : S) ⊂ Dn+1−k (T : S). Then, for k = n, we will obtain Nn (T : S) ⊂ D1 (T : S). As the cases k = 0, 1 are true, assume that the inclusion is verified for some k ∈ {0, . . . , n − 1} : Nk (T : S) ⊂ Dn+1−k (T : S) = S −1 T Dn−k (T : S). Then SNk (T : S) ⊂ T Dn−k (T : S) ∩ R(S) ⊂ T Dn−k (T : S). So Nk+1 (T : S) = T −1 SNk (T : S) ⊂ Dn−k (T : S) + N (T ) = Dn+1−(k+1) (T : S), as N (T ) ⊂ Dn−k (T : S) by 1). Thus the result is proved at the rank k + 1. So, for k = n, we obtain Nn (T : S) ⊂ D1 (T : S). 2)⇒3) : We prove the result by induction on m. As the cases m = 0, 1 are true, assume that for some m ≥ 1, Nn (T : S) ⊂ Dm (T : S) for all integers n. Let n ∈ N and x ∈ Nn (T : S) ⊂ S −1 R(T ), by hypothesis. So there exists y ∈ X such that Sx = T y. Then y ∈ T −1 Sx ⊂ T −1 SNn (T : S) = Nn+1 (T : S) ⊂ Dm (T : S), by the induction assumption. Thus x ∈ S −1 T y ⊂ S −1 T Dm (T : S) = Dm+1 (T : S). 3)⇒4) : We prove the result by induction on m. As the case m = 0 is clear, assume that the property is verified for some m ≥ 0 and let n ∈ N. As Nn+m+1 (T : S) = T −1 SNn+m (T : S), we have : S −1 T Nn+m+1 (T : S) = Nn+m (T : S) ∩ S −1 R(T ) + N (S), by Lemma 2.1 = Nn+m (T : S) + N (S), by 3) ⊃ Nn+m (T : S). Thus (S −1 T )m+1 Nn+m+1 (T : S) ⊃ (S −1 T )m Nn+m (T : S) ⊃ Nn (T : S), by the

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induction assumption, which proves 4). 4)⇒1) : With n = 1, we obtain, for all integers m : N (T ) = N1 (T : S) ⊂ (S −1 T )m Nm+1 (T : S) ⊂ Dm (T : S).




Now let us prove some elementary results relative to the Kato decomposition. Lemma 2.3. Assume that (T, S) has a DK (M, N, M  , N  ). Then, we have : 1) N ⊂ N (T : S) and D(T : S) ⊂ M . 2) D(T : S) is closed and, for all n ∈ N, Nn (T : S) ∩ D(T : S) = Nn (T : S) ∩ M . 3) D(T : S) + N (T : S) = D(T : S) ⊕ N is closed. Proof. 1) Let d be the index of nilpotency of (S −1 T )|N . Then N ⊂ Nd (T : S) = (T −1 S)d {0} and Dd (T |N : S|N ) = (S −1 T )d N = {0}. So Dd (T : S) = Dd (T |N : S|N ) + Dd (T |M : S|M ) = Dd (T |M : S|M ) ⊂ M . As the sequence (Nn (T : S))n≥0 is increasing and the sequence (Dn (T : S))n≥0 is decreasing, we have the desired inclusions. 2) By 1), D(T : S) = D(T : S) ∩ M = D(T |M : S|M ). By Theorem 3.1, [7], D(T |M : S|M ) is closed, as T |M has the property P (S|M : 0). So D(T : S) is closed. Moreover, for n ∈ N, Nn (T |M : S|M ) ⊂ D(T |M : S|M ) = D(T : S). So we have : Nn (T : S) ∩ D(T : S) = Nn (T : S) ∩ D(T : S) ∩ M = Nn (T |M : S|M ) ∩ D(T |M : S|M ) = Nn (T |M : S|M ) = Nn (T : S) ∩ M. In the same manner, we prove that N (T : S) ∩ D(T : S) = N (T : S) ∩ M . 3) By 1), N +D(T : S) ⊂ N (T : S)+D(T : S). Let us prove the converse inclusion. By Lemma 2.2, N (T |M : S|M ) ⊂ D(T |M : S|M ), as T |M has the property P (S|M : 0). Thus we have N (T : S) = N (T : S) ∩ M + N , as N ⊂ N (T : S) ⊂ D(T : S) ∩ M + N = D(T : S) + N , as D(T : S) ⊂ M. So N (T : S) + D(T : S) ⊂ N + D(T : S), and we obtain the equality. Moreover, as D(T : S) ⊂ M , D(T : S) ∩ N = {0}. So D(T : S) + N (T : S) = D(T : S) ⊕ N , which is closed as N and D(T : S) are.
Denote by Φ+ (X, Y ) the set of upper semi-Fredholm operators, i.e. operators with closed range and finite dimensional kernel. Lemma 2.4. Assume that T ∈ P(S). Then dim [N (T ) ∩ D(T : S)] < +∞ if and only if T ∈ Φ+ (X, Y ). Proof. Assume that dim [N (T ) ∩ D(T : S)] < +∞. By Theorem 1.1, as T ∈ P(S), there exists (M, N, M  , N  ), a Kato decomposition of finite type associated to the pair (T, S). By Lemma 2.3, N (T ) ∩ D(T : S) = N (T ) ∩ M .

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Thus N (T ) = N (T ) ∩ N ⊕ N (T ) ∩ D(T : S). As dim N < +∞ and R(T ) is closed, T ∈ Φ+ (X, Y ). As the converse implication is always true, we have the equivalence.
Lemma 2.5. Let A, B ∈ B(X, Y ) be such that R(A) is closed. Assume that for all integers k, N (A) ⊂e Dk (A : B). Then Dk (A : B) is closed for all k ∈ N. Thus if R(A) is closed and N (A) ⊂e D(A : B), then A ∈ P(B). Proof. For every integer k, let Fk , a finite dimensional subspace of N (A), be such that N (A) ⊂ Dk (A : B) + Fk . We prove the desired result by induction on k. As the cases k = 0, 1 are true, assume that Dk (A : B) = Dk (A : B) for some k ≥ 1 and let u ∈ Dk+1 (A : B). As the sequence (Dn (A : B))n≥0 is decreasing, u ∈ Dk (A : B) = Dk (A : B), i.e. there exist u0 , . . . , uk such that uk = u and Bui+1 = Aui for i = 0, . . . , k − 1. Moreover, there exists a sequence (v n )n≥0 of Dk+1 (A : B) such that lim v n = u. n→+∞

n n n such that vk+1 = v n and Bvi+1 = Avin for For all n ∈ N, there exist v0n , . . . , vk+1 i = 0, . . . , k. As Bv n = Avkn and Bu = Auk−1 , we have lim A(vkn − uk−1 ) = 0. n→+∞

 : X/N (A) → R(A) be the operator induced by A. Let A  n − uk−1 + N (A)) = 0. The operator A  is bounded from below, Then lim A(v k n→+∞

so

lim [vkn − uk−1 + N (A)] = 0 in the quotient space X/N (A) . Thus there exist

n→+∞

wn , elements of N (A), such that lim [vkn + wn ] = uk−1 . n→+∞

As vkn + wn ∈ Dk (A : B) + N (A) ⊂ Dk (A : B) + Fk which is closed, we have uk−1 = ak + bk with ak ∈ Dk (A : B) and bk ∈ Fk ⊂ N (A). Hence Bu = Auk−1 = Aak ∈ ADk (A : B).
Thus u ∈ B −1 ADk (A : B) = Dk+1 (A : B) and Dk+1 (A : B) is closed. Definition 2.6. Let n ∈ N∗ . The n-tuple (x1 , . . . , xn ) is a chain for the pair (T, S) if T xi+1 = Sxi for i = 1, . . . , n − 1. Let γn (T : S) be the supremum of all the c ≥ 0 with the property T x1  ≥ c d(xn , Nn (T : S)) for all chains (x1 , . . . , xn ). Notice that γ1 (T : S) is the reduced modulus of T , γ(T ). Lemma 2.7. Let A, B ∈ B(X, Y ) be such that A has the property P (B : 0) and m (A:B) . B = 0. Then for all (m, n) ∈ N∗2 , γn+m (A : B) ≥ γn (A:B)γ B n In particular, γn+1 (A : B) ≥ ( γ(A) B ) γ(A) for every integer n. Thus for all n ≥ 1, γn (A : B) = 0.

Proof. Let n, m ∈ N∗ and x1 , . . . , xn+m ∈ X be such that for all i = 1, . . . , n + m − 1, Axi+1 = Bxi . By definition, we have : Ax1  ≥ γn (A : B)d(xn , Nn (A : B)) = γn (A : B)inf {xn − u; u ∈ Nn (A : B)}. Let u0 ∈ Nn (A : B). As A has the property P (B : 0), by Lemma 2.2, we have Nn (A : B) ⊂ (B −1 A)m Nn+m (A : B), so there exist u1 , . . . , um such that

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um ∈ Nn+m (A : B) and Bui = Aui+1 for i = 0, . . . , m − 1. −Au1  −Bu0  Then xn − u0  ≥ BxnB = Axn+1 B

γm (A:B) B d(xn+m − um , Nm (A : B)). (A:B) Thus d(xn , Nn (A : B)) ≥ γmB d(xn+m , Nn+m (A : B)). γn (A:B)γm (A:B) d(xn+m , Nn+m (A : B)). Hence Ax1  ≥ B

≥

So we have γn+m (A : B) ≥

γn (A : B)γm (A : B) . B

n Thus γn+1 (A : B) ≥ ( γ(A) B ) γ(A) for all integers n. As R(A) is closed, we obtain

γn (A : B) ≥

γ(A)n > 0 for all integers n ≥ 1. Bn−1




Lemma 2.8. Let Z be a subspace of finite codimension in X. Then, for all subspaces V of X, there exists N ⊂ V of finite dimension such that V = V ∩ Z ⊕ N . Lemma 2.9. Assume that T has the property P (S : k). Denote M = D(T : S) and let F be the operator defined by F (x) = f (x)z0 for all x ∈ X, with f ∈ X ∗ and z0 ∈ Y . Then N (T + F ) ⊂e N ((T + F )|M ). Proof. By Proposition 2.1, [4], as T has the property P (S : k), M = D(T : S) is a closed subspace of X such that S −1 T M = M , and the application T˘ : X/M → Y /T M defined by T˘(˘ x) = T x + T M (where x ˘ = x + M ) has closed range and n(T˘) = k. Let x ∈ N (T + F ). As T˘(˘ x) = −f (x)z˘0 , x ˘ ∈ T˘−1 < z˘0 >. Let Π : X → X/M be the canonical projection. Then ΠN (T + F ) ⊂ T˘−1 < z˘0 >. Thus dim ΠN (T + F ) ≤ dim N (T˘) +1. So dim ΠN (T + F ) < +∞. Let {g1 + M, . . . , gm + M } be a basis of ΠN (T + F ), with gi ∈ N (T + F ). Denote by G1 the subspace generated by g1 , . . . , gm . Then G1 ⊂ N (T + F ). For x ∈ N (T +F ), x+M ∈ G1 +M , i.e. ∃g ∈ G1 , ∃m ∈ M ∩N (T +F ), x = g +m. Thus N (T + F ) ⊂ G1 + N ((T + F ) |M ), with dim G1 < +∞.
Hence N (T + F ) ⊂e N ((T + F )|M ). Lemma 2.10. Let U, V be two subspaces of X and k ∈ N. Assume that S −1 R(T ) = X and (T −1 S)k U ⊂ V . Then U ⊂ (S −1 T )k V . Proof. As the case k = 0 is clear, assume that k ≥ 1. We have T T −1 S(T −1 S)k−1 U = S(T −1 S)k−1 U ∩ R(T ) ⊂ T V . Thus [(T −1 S)k−1 U + N (S)] ∩ S −1 R(T ) ⊂ S −1 T V .

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As S −1 R(T ) = X, (T −1 S)k−1 U ⊂ (T −1 S)k−1 U + N (S) ⊂ S −1 T V . Step by step, we prove that U ⊂ (S −1 T )k V .

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3. Stability under finite perturbation Theorem 3.1. Assume that T ∈ P(S) and S is bounded from below. For every finite rank operator F from X to Y , T + F ∈ P(S). Proof. Without loss of generality, we can assume that the dimension of F is equal to 1, i.e. there exist f ∈ X ∗ , z0 ∈ Y such that for all x ∈ X, F (x) = f (x)z0 . As R(T + F ) is closed, we only have to prove that N (T + F ) ⊂e D(T + F : S). Denote M = D(T : S). By Lemma 2.9, N (T + F ) ⊂e N ((T + F )|M ), i.e there exists G1 , finite dimensional subspace, such that N (T + F ) ⊂ N ((T + F )|M ) + G1 . Moreover, it is clear that D((T + F )|M : S|M ) ⊂ D(T + F : S). Thus we only have to prove that N ((T + F )|M ) ⊂e D((T + F )|M : S|M ). If M = {0}, by Lemma 2.4, T ∈ Φ+ (X, Y ), so T + F ∈ Φ+ (X, Y ), and T + F ∈ P(S). If T M = {0}, then M = S −1 T M = S −1 {0} = {0}, and we recover the previous case. Assume that T M = {0}. Denote T1 = T |M , F1 = F |M , S1 = S |M and f1 = f |M . By Theorem 3.1, [7], R(T1 ) = T M = R(T : S) is closed. Notice that R(S1 ) = SM = T M ∩ R(S) and N (S1 ) = N (S) ∩ M = {0}, so S1 is bounded from below. We will distinguish three cases. a) 1st case : Assume that z0 ∈ T M . Thus we can consider T1 , S1 and F1 as operators from M to T M . Then R(T1 + F1 ) ⊂ R(T1 ) ⊂ R(T1 + F1 ) + R(F1 ), so dim [R(T1 )/R(T1 +F1 ) ] < +∞. Let ψ : M/S −1 R(T1 +F1 ) → R(T1 )/R(T1 +F1 ) be the application definied by : 1

ψ(x + S1−1 R(T1 + F1 )) = S1 x + R(T1 + F1 ). Clearly, ψ is injective. So dim [M/S −1 R(T1 +F1 ) ] ≤ dim [R(T1 )/R(T1 +F1 ) ] < +∞. 1 Thus D1 (T1 + F1 : S1 ) = S −1 R(T1 + F1 ) has finite codimension in M . Hence N (T1 + F1 : S1 ) ⊂e D1 (T1 + F1 : S1 ). By Theorem 1.1, we obtain N (T1 + F1 ) ⊂e D(T1 + F1 : S1 ). So N (T +F ) ⊂e N (T1 +F1 ) ⊂e D(T1 +F1 : S1 ) ⊂ D(T +F : S). Thus T +F ∈ P(S). / T M . We first prove that for all In the two following cases, we assume that z0 ∈ integers n, (S −1 T )n N (f1 ) is closed. Let n ∈ N and ϕn : M/(S −1 T )n N (f1 ) → M/(S −1 T )n+1 N (f1 ) be the application defined by :  if there exists y ∈ M such  y + (S −1 T )n+1 N (f1 ) −1 n that T x = Sy, ϕn (x+(S T ) N (f1 )) =  (S −1 T )n+1 N (f1 ) if not.

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This application is well defined. For y ∈ M = S −1 T M , there exists x ∈ M such that Sy = T x. Then ϕn (x + (S −1 T )n N (f1 )) = y + (S −1 T )n+1 N (f1 ) and the application ϕn is surjective. Thus dim M/(S −1 T )n+1 N (f1 ) ≤ dim M/(S −1 T )n N (f1 ) for all n ∈ N. Hence dim M/(S −1 T )n+1 N (f1 ) ≤ dim M/N (f1 ) = 1, using f1 ∈ M ∗ . Then, for all n ∈ N, (S −1 T )n N (f1 ) is finite codimensional in M , and so is closed. b) 2nd case : Assume that z0 ∈ / T M and for all k ≥ 0, f1 (Nk (T1 : S1 )) = {0}. As M = S −1 T M and R(T1 ) is closed, there exists a constant K > 0 such that ∀y ∈ M, ∃x ∈ M, S1 y = T1 x and x ≤ Ky. Let m0 ∈ N (T1 + F1 ). Then T1 m0 + f1 (m0 )z0 = 0 and, as z0 ∈ / T M , T1 m0 = 0. We construct by induction elements mi of M such that T1 mi+1 = S1 mi and mi  ≤ K i m0 . Thus for all i ∈ N, mi ∈ (T −1 S)i m0 ⊂ (T −1 S)i N (T1 ) = Ni+1 (T1 : S1 ) ⊂ N (f1 ). +∞  mi λi , for | λ |< K −1 . As mi ∈ N (f1 ) for all i ∈ N, F1 (g(λ)) = 0. Let g(λ) = i=0

Moreover, (T1 − λS1 )(g(λ)) =

+∞  i=1

λi S1 mi−1 −

Thus, for 0 <| λ |< K −1 , we have :

+∞ 

λi+1 S1 mi = 0.

i=0

g(λ) ∈ N (T1 + F1 − λS1 ) ⊂ N (T + F − λS) ⊂ D(T + F : S). As g is continuous, m = g(0) ∈ D(T + F : S). Hence N (T + F ) ⊂ G1 + N (T1 + F1 ) ⊂ G1 + D(T + F : S). By Lemma 2.5, T + F ∈ P(S). c) 3rd case : Assume that z0 ∈ / T M and f1 (Nk (T1 : S1 )) = {0} for some k > 0. Let k be the smallest integer such that the following property holds : f1 (Nk−1 (T1 : S1 )) = {0}. For x0 ∈ M , we construct a sequence (xn )n≥1 in M by the following manner : i) If T xi ∈ R(S1 ), let xi+1 be the unique element of M such that T xi = Sxi+1 , / R(S1 ), let xn = 0 for n ≥ i + 1. ii) If T xi ∈ Let Ai (x0 ) = xi and A : M → M ⊕ C be the application defined by A(x) = (Ak (x), f1 (x)). As Nk (T1 : S1 ) is not a subset of N (f1 ), there exists s ∈ Nk (T1 : S1 ) such that f1 (s) = 1 (notice that Ak (s) = 0). We prove that A is surjective : let m ∈ M and α ∈ C. Then m ∈ M = S −1 T M and we can construct m0 , . . . , mk in M such that mk = m and Smi+1 = T mi for i = 0, . . . , k − 1. Notice that Ak (m0 ) = m. Hence A[m0 + s(α − f1 (m0 ))] = (Ak (m0 ) + Ak (s)(α − f1 (m0 )), α) = (m, α). So A is surjective. Thus there exists K > 0 such that ∀m ∈ M, ∃x ∈ N (f1 ), Ak (x) = m and x ≤ Km.

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We proved that for all n ∈ N, (S −1 T )n N (f1 ) is a closed finite codimensional subspace of M . k−1  Let Z = (S −1 T )i N (f1 ), Z ⊂ N (f1 ). Z is a closed finite codimensional subspace i=0

of M . By Lemma 2.8, as N (T1 + F1 ) is closed, there exists G2 ⊂ N (T1 + F1 ) of finite dimension such that N (T1 + F1 ) = [N (T1 + F1 ) ∩ Z] ⊕ G2 . We now want to prove that N (T1 + F1 ) ∩ Z ⊂ D(T + F : S). / TM. Let m0 ∈ N (T1 + F1 ) ∩ Z. Then T1 m0 = 0, using z0 ∈ By hypothesis, S1 is bounded from below (and so S1 has the property P (T1 : 0)), hence, by Lemma 2.7, γk (S1 : T1 ) = 0. We construct by induction elements mi of Z such that T1 mi+1 = S1 mi and mi  ≤ (

KS1  i ) m0 . γk (S1 : T1 )

Assume that we have constructed m0 , . . . , mj verifying these properties. There exists u ∈ M such that Ak (u) = mj , u ≤ Kmj  and f1 (u) = 0. As Ak (u) = mj , there exist u0 , . . . , uk in M such that u0 = u, uk = mj and Sui+1 = T ui for i = 0, . . . , k − 1. Thus uk−1 γk (S1 : T1 ) ≤ S1 u0  ≤ S1 u ≤ S1 Kmj . Thanks to the hypothesis, we obtain uk−1  ≤ (

S1 K j+1 ) m0 . γk (S1 : T1 )

Let mj+1 = uk−1 . We have to see whether mj+1 ∈ Z =

k−1 

(S −1 T )i N (f1 ).

i=0

We have mj+1 = uk−1 ∈ (S −1 T )k−1 u0 ⊂ (S −1 T )k−1 N (f1 ). Assume that k ≥ 2. We prove that mj+1 ∈ (S −1 T )i N (f1 ) for i = 0, . . . , k − 2. We have T1 mj+1 = S1 mj ∈ S1 (S −1 T )i N (f1 ) for i = 0, . . . , k − 1. Thus T mj+1 ∈ SS −1 T (S −1 T )i−1 N (f1 ) = T (S −1 T )i−1 N (f1 ) ∩ R(S1 ) for i = 1, . . . , k − 1. So, for all i ∈ {0, . . . , k − 2}, we have : mj+1

∈ (S −1 T )i N (f1 ) + N (T1 ).

For i ∈ {0, . . . , k − 2}, Ni+1 (T1 : S1 ) = (T −1 S)i N (T1 ) ⊂ N (f1 ). So, by Lemma 2.10, as S −1 T M = M , N (T1 ) ⊂ (S −1 T )i N (f1 ), for i = 0, . . . , k − 2. Hence mj+1 ∈ Z. ∞  1 −1 mi λi , for | λ |< K1 = ( γkKS . Let g(λ) = (S1 :T1 ) ) i=0

Then (T1 − λS1 )(g(λ)) =

∞  i=1

λi S1 mi−1 −

∞  i=0

λi+1 S1 mi = 0.

156

Gagnage

IEOT

So g(λ) ∈ N (T1 − λS1 ), for | λ |< K1 . As mi ∈ Z ⊂ N (f1 ) for all i ∈ N, we have g(λ) ∈ N (f1 ), for | λ |< K1 . Hence g(λ) ∈ [N (f1 ) ∩ N (T1 − λS1 )]. Then, for 0 <| λ |< K1 , g(λ) ∈ N (T1 +F1 −λS1 ) ⊂ N (T +F −λS) ⊂ D(T +F : S). As the application g is continue, we obtain g(0) = m0 ∈ D(T + F : S). Hence N (T1 + F1 ) ∩ Z ⊂ D(T + F : S). Denote G = G1 + G2 . Notice that G is finite dimensional. Thus we obtain N (T + F ) ⊂ N (T1 + F1 ) + G1 = N (T1 + F1 ) ∩ Z + G2 + G1 ⊂ D(T + F : S) + G. By Lemma 2.5, T + F ∈ P(S).




Theorem 3.2. Assume that T ∈ P(S) and S is surjective. Then, for every finite rank operator F from X to Y , T + F ∈ P(S). Proof. By Theorem 3.6, [7], T ∗ ∈ P(S ∗ ). Moreover, S ∗ is bounded from below and F ∗ is an operator of finite rank. So, by Theorem 3.1, T ∗ + F ∗ ∈ P(S ∗ ). Then, by Theorem 3.6, [7], T + F ∈ P(S).
Remark 3.3. These two theorems generalize the result proved by V. Kordula, [10].

References [1] Bart, M. A. Kaashoek and D. C. Lay, Stability properties of finite meromorphic operator functions, Nederl. Akad. Wetensh. Proc. Ser. A77 = Indag. Math. 36 (1974), 217–259. [2] H. Bart and D. C. Lay, Poles of a generalized resolvent operator, Proc. Royal Irish Academy (1974), 147–168. [3] H. Bart and D. C. Lay, The stability radius of a bundle of closed linear operators, Studia Math. 66 (1980), 307–320. [4] D. Gagnage, Kato decomposition of linear pencils, to appear. [5] D. Gagnage, On the unicity of the Kato decomposition, submitted. [6] S. Grabiner, Uniform ascent and descent of bounded operators, J. Math. Soc. Japan 34 (1982), 317–337. [7] M. A. Kaashoek, Stability theorems for closed linear operators, Nederl. Akad. Wetensh. Proc. Ser. A68 = Indag. Math. 27 (1965), 452–466. [8] T. Kato, Perturbation theory for linear operators, Berlin, Springer (1966). [9] T. Kato, Perturbation theory for nullity, deficiency and other quantities of linear operators, J. Analyse Math. 6 (1958), 261–322. [10] V. Kordula, The essential Apostol spectrum and finite-dimensional perturbations, Proc. Roy. Irish Acad. 96A (1996), 105–109.

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[11] J. P. Labrousse, Les op´ erateurs quasi-Fredholm : une g´en´eralisation des op´eratuers semi-Fredholm, Rend. Circ. Mat. Palermo 29 (1980), 161–258. [12] M. Mbekhta and A. Ouahab, Op´erateurs s-r´eguliers dans un espace de Banach et th´eorie spectrale, Acta Sci. Math. (Szeged) 59 (1994), 525–543. [13] V. M¨ uller, On the regular spectrum, J. Operator Theory 31 (1994), 363–380.

Acknowledgment I wish to thank my supervisor M. Mbekhta for discussions about the topic of this paper. Dominique Gagnage D´epartement de Math´ematiques Universit´e de Lille I U.M.R., C.N.R.S. 8524 59655 Villeneuve d’Ascq C´edex France e-mail: [email protected] Submitted: June 17, 2002

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