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Manuel González, J. Pello (2016)
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Let SS and SC be the strictly singular and the strictly cosingular operators acting between Banach spaces, and let P and P be the perturbation classes for the + + upper and the lower semi-Fredholm operators. We study two classes of operators S and C that satisfy SS ⊂ S ⊂ P and SC ⊂ C ⊂ P . We give some + − conditions under which these inclusions become equalities, from which we derive some positive solutions to the perturbation classes problem for semi-Fredholm operators. Keywords Perturbation classes problem · Semi-Fredholm operator · Strictly singular operator Mathematics Subject Classification 47A55 · 47A53 1 Introduction The perturbation classes problem asks whether the perturbation classes for the upper semi-Fredholm operators P and the lower semi-Fredholm operators P coincide + − with the classes of strictly singular operators SS and strictly cosingular operators SC, respectively. This problem was raised in [9] (see also [5, 19]), and it has a positive answer in some cases [11, 13–15, 21], but the general answer is negative in both cases [10], [8, Theorem 4.5]. However, it remains interesting to find positive answers in special cases because the definitions of SS and SC are intrinsic: to check that K is in Communicated by Qingxiang Xu. Manuel González manuel.gonzalez@unican.es Margot Salas-Brown salasbrown@gmail.com Departamento de Matemáticas, Facultad de Ciencias, Universidad de Cantabria, 39071 Santander, Spain Escuela de Ciencias Exactas e Ingeniería, Universidad Sergio Arboleda, Bogotá, Colombia 0123456789().: V,-vol 41 Page 2 of 9 M. González, M. Salas-Brown one of them only involves the action of K , while to check that K is in P or P + − we have to study the properties of T + K for T in a large set of operators. In this paper, we consider two classes S and C introduced in [2] that satisfy SS(X , Y ) ⊂ S(X , Y ) ⊂ P (X , Y ) and SC(X , Y ) ⊂ C(X , Y ) ⊂ P (X , Y ). We study conditions on the Banach spaces X , Y so that some of these four inclu- sions become equalities, and we derive new positive answers to the perturbation classes problem for semi-Fredholm operators. When (X , Y ) =∅, we show that S(X , Y ) = P (X , Y ) when Y × Y is isomorphic to Y , SS(X , Y ) = S(X , Y ) when some quotients of X embed in Y (Theorem 3.1), and adding up we get conditions implying that SS(X , Y ) = P (X , Y ) (Theorem 3.2). When (X , Y ) =∅ we + − prove similar results (Theorems 4.1 and 4.2). We also state some questions concerning the classes S and C. Notation. Along the paper, X , Y and Z denote Banach spaces and L(X , Y ) is the set of bounded operators from X into Y . We write L(X ) when X = Y . Given a closed subspace M of X , we denote J the inclusion of M into X , and Q the quotient map M M from X onto X /M . An operator T ∈ L(X , Y ) is an isomorphism if there exists c > 0 such that Tx≥ cx for every x ∈ X . The operator T ∈ L(X , Y ) is strictly singular, and we write T ∈ SS, when there is no infinite dimensional closed subspace M of X such that the restriction TJ is an isomorphism; and T is strictly cosingular, and we write T ∈ SC, when there is no infinite codimensional closed subspace N of Y such that Q T is surjective. Moreover, T is upper semi-Fredholm, T ∈ , when the range R(T ) is closed and the kernel N (T ) is finite dimensional; T is lower semi-Fredholm, T ∈ , when R(T ) is finite codimensional (hence closed); T is Fredholm, T ∈ , when it is upper and lower semi-Fredholm; and T is inessential, T ∈ In, when I − ST ∈ for each S ∈ L(Y , X ). 2 Preliminaries The perturbation class PS of a class of operators S is defined in terms of its components: Definition 2.1 Let S denote one of the classes , or . For spaces X , Y such + − that S(X , Y ) =∅, PS(X , Y ) ={K ∈ L(X , Y ) : for each T ∈ S(X , Y ), T + K ∈ S}. We could define PS(X , Y ) = L(X , Y ) when S(X , Y ) is empty, but this is not useful. The components of P coincide with those of the operator ideal of inessential operators In when they are defined [1], but given S ∈ L(Y , Z ) and T ∈ L(X , Y ), S or T in P does not imply ST ∈ P , and similarly for P [10]. However, the + + − following result holds, and it will be useful for us. Two classes of operators related to the perturbation… Page 3 of 9 41 Proposition 2.1 Suppose that K ∈ P (X , Y ), L ∈ P (X , Y ), S ∈ L(Y ) and + − T ∈ L(X ). Then, SK , KT ∈ P (X , Y ) and SL, LT ∈ P (X , Y ). + − Proof Suppose that K ∈ P (X , Y ) and let S ∈ L(Y ) and U ∈ (X , Y ). If S is + + −1 −1 bijective, then S U ∈ (X , Y ), hence U + SK = S(S U + K ) ∈ ; thus + + SK ∈ P (X , Y ). In the general case, S = S + S with S , S bijective; thus, + 1 2 1 2 SK = S K + S K ∈ P (X , Y ). 1 2 + The proof of the other three results is similar. 3 The perturbation class for 8 Given two operators S, T ∈ L(X , Y ), we denote by (S, T ) the operator from X into Y × Y defined by (S, T )x = (Sx , Tx ), where Y × Y is endowed with the product norm (y , y ) =y +y . 1 2 1 1 2 Inspired by the results of Friedman [6], the authors of [2] defined the following class of operators. Definition 3.1 Suppose that (X , Y ) =∅ and let K ∈ L(X , Y ). We say that K is -singular, and write K ∈ S, when for each S ∈ L(X , Y ), (S, K ) ∈ implies S ∈ . The definition of S(X , Y ) is similar to that of P (X , Y ), but the former one is easier to handle because the action of S and K is decoupled when we consider (S, K ) instead of S + K . With our notation, [6, Theorems 3 and 4] can be stated as follows: Proposition 3.1 [2, Proposition 2.2] Suppose that (X , Y ) =∅. Then, SS(X , Y ) ⊂ S(X , Y ) ⊂ P (X , Y ). Note that SS is an operator ideal but P is not; P (X , Y ) is a closed subspace + + of L(X , Y ), and P (X ) is an ideal of L(X ). Proposition 3.2 S(X , Y ) is closed in L(X , Y ). Proof Let {T } be a sequence in S(X , Y ) converging to T ∈ L(X , Y ). Suppose that S ∈ L(X , Y ) and (S, T ) ∈ (X , Y × Y ). Note that the sequence (S, T ) converge + n to (S, T ), because (S, T ) − (S, T )=T − T . n n Since (X , Y × Y ) is an open set, there exists a positive integer N such that (S, T ) ∈ (X , Y × Y ). Then, T ∈ S(X , Y ) implies S ∈ (X , Y ). Thus N + N + T ∈ S(X , Y ). We state some basic questions on the class S. Question 3.1 Suppose that (X , Y ) =∅. (a) Is S(X , Y ) a subspace of L(X , Y )? (b) Is S an operator ideal? 41 Page 4 of 9 M. González, M. Salas-Brown (c) Is Proposition 2.1 valid for S? Answering a question in [6], an example of an operator K ∈ P \ S was given in [2, Example 2.3], but we do not know if the other inclusion can be strict. Question 3.2 Suppose that (X , Y ) =∅. Is SS(X , Y ) = S(X , Y )? A negative answer to Question 3.2 would provide a new counterexample to the perturbation classes problem for . Let us see that the inclusions in Proposition 3.1 become equalities in some cases. An infinite dimensional Banach space Y is isomorphic to its square, denoted Y ×Y Y , in many cases: L (μ) and (1 ≤ p ≤∞), c , and C [0, 1]. On the other hand, p p 0 James’ space J and some spaces of continuous functions on a compact like C [0,ω ] are not isomorphic to their square, where ω is the first uncountable ordinal. See [4, 20]. Theorem 3.1 Suppose that the spaces X and Y satisfy (X , Y ) =∅. 1. If Y × Y Y , then S(X , Y ) = P (X , Y ). 2. If every infinite dimensional subspace of X has an infinite dimensional subspace N such that X /N embeds in Y , then SS(X , Y ) = S(X , Y ). Proof (1) Let U : Y × Y → Y be a bijective isomorphism and let V , W ∈ L(Y ) such that U (y , y ) = Vy + Wy . If K ∈ P (X , Y ), for each S ∈ L(X , Y ) 1 2 1 2 + such that (S, K ) ∈ we have U (S, K ) = VS + WK ∈ . By Proposition 2.1, + + WK ∈ P (X , Y ). Then, VS ∈ , hence S ∈ . Thus we conclude that + + + K ∈ S(X , Y ). (2) Let K ∈ L(X , Y ), K ∈ / SS. By the hypothesis, there exists an infinite dimen- sional subspace N of X such that KJ is an isomorphism, and there is an isomorphism U : X /N → Y . Then, S = UQ ∈ L(X , Y ) is not upper semi-Fredholm. We will prove that K ∈ / S by showing that (S, K ) ∈ . Recall that Q x= dist(x , N ). We can choose the isomorphism U so that Sx=UQ x≥ dist(x , N ) for each x ∈ X . Moreover, there is a constant c > 0 such that Kn≥ cn for each n ∈ N . Let x ∈ X with x= 1, and let 0 <α < 1 such that c(1 − α) = 2K α. If dist(x , N ) ≥ α, then Sx≥ α. Otherwise, there exists y ∈ N such that x − y <α; hence y > 1 − α. Therefore, Kx≥Ky−K (x − y)≥ c(1 − α) −K α =K α. Then, (S, K )x =Sx+Kx≥ min{K α, α}, hence (S, K ) is an isomorphism; in particular (S, K ) ∈ , as we wanted to show. In the known examples in which SS(X , Y ) = P (X , Y ) in [8, 10], the space Y has a complemented subspace which is hereditarily indecomposable in the sense of [3, 16, 17]. Therefore, the question arises. Question 3.3 Suppose that X and Y satisfy (X , Y ) =∅ and Y × Y Y . Is SS(X , Y ) = P (X , Y )? + Two classes of operators related to the perturbation… Page 5 of 9 41 A Banach space X is subprojective if every closed infinite dimensional subspace of X contains an infinite dimensional subspace complemented in X . The spaces c , 0 p (1 ≤ p < ∞) and L (μ) (2 ≤ q < ∞) are subprojective [22]. See [7, 18] for further examples. Corollary 3.1 Suppose that (X , Y ) =∅ and the space X is subprojective. Then, SS(X , Y ) = S(X , Y ). Proof Every closed infinite dimensional subspace of X contains an infinite dimen- sional subspace N complemented subspace in X ; thus X /N is isomorphic to the complement of N . Since (X , Y ) =∅, the quotient X /N is isomorphic to a subspace of Y and we can apply Theorem 3.1. The next result is a refinement of Theorem 3.1 that is proved using the previous arguments. Theorem 3.2 Suppose that (X , Y ) =∅, Y × Y embeds in Y and every infinite dimensional subspace of X has an infinite dimensional subspace N such that X /N embeds in Y . Then, SS(X , Y ) = P (X , Y ). Proof Since Y × Y embeds in Y , there exist isomorphisms V , W ∈ L(Y ) such that R(V ) ∩ R(W ) ={0} and R(V ) + R(W ) is closed. Hence, there exists r > 0 such that y + y ≥ r (y +y ) for y ∈ R(V ) and y ∈ R(W ), and clearly we can 1 2 1 2 1 2 choose V , W so that r = 1. Let K ∈ L(X , Y ) with K ∈ / SS. Select an infinite dimensional subspace M of X such that KJ is an isomorphism, and let N be an infinite dimensional subspace of M such that there exists an isomorphism U : X /N → Y . We can assume that Uz≥ z for each z ∈ X /N . The operator S = VU Q ∈ / , and proceeding like in the proof of (2) in N + Theorem 3.1, we can show that S +WK ∈ . Then, WK ∈ / P , hence K ∈ / P , + + + by Proposition 2.1. Corollary 3.2 If X is separable, Y × Y embeds in Y , and Y contains a copy of C [0, 1], then SS(X , Y ) = (X , Y ). Proof It is well known that the space C [0, 1] contains a copy of each separable Banach space. The class S is injective in the following sense: Proposition 3.3 Given an operator K ∈ L(X , Y ) and an (into) isomorphism L ∈ L(Y , Y ), if L K ∈ S(X , Y ), then K ∈ S(X , Y ). 0 0 Proof Let K ∈ L(X , Y ) and let L ∈ L(X , Y ) be an isomorphism into Y such that 0 0 LK ∈ S(X , Y ). Take S ∈ L(X , Y ) and suppose that (S, K ) ∈ (X , Y × Y ). 0 + Then, (LS, LK ) = (L × L)(S, K ) ∈ (X , Y × Y ), where (L × L) ∈ L(X × + 0 0 X , Y × Y ) is defined by (L × L)(x , x ) = (Lx , Lx ). 0 0 1 2 1 2 Since LK ∈ S(X , Y ) we obtain LS ∈ (X , Y ). Therefore S ∈ (X , Y ), 0 + 0 + hence K ∈ S(X , Y ). 41 Page 6 of 9 M. González, M. Salas-Brown 4 The perturbation class for 8 Given two operators S, T ∈ L(X , Y ), we denote by [S, T ] the operator from X × X into Y defined by [S, T ](x , x ) = Sx + Tx , where X × X is endowed with the 1 2 1 2 maximum norm (x , x ) = max{y , y }. 1 2 ∞ 1 2 Definition 4.1 Suppose that (X , Y ) =∅ and let K ∈ L(X , Y ). We say that K is -cosingular, and write K ∈ C, when for each S ∈ L(X , Y ), [S, K]∈ implies S ∈ . Like in the case of S, the definition of C(X , Y ) is similar to that of P (X , Y ), but the former one is easier to handle because the action of S and K is decoupled when we consider [S, K ] instead of S + K . Proposition 4.1 [2, Proposition 2.5] Suppose that (X , Y ) =∅. Then, SC(X , Y ) ⊂ C(X , Y ) ⊂ P (X , Y ). Note that SC is an operator ideal but P is not; P (X , Y ) is a closed subspace − − of L(X , Y ), and P (X ) is an ideal of L(X ). Proposition 4.2 C(X , Y ) is closed in L(X , Y ). Proof Let {T } be a sequence in C(X , Y ) converging to T ∈ L(X , Y ). Suppose that S ∈ L(X , Y ) and [S, T]∈ (X × X , Y ). Note that the sequence [S, T ] converge − n to [S, T ]. Since (X × X , Y ) is an open set there exists a positive integer N such that [S, T ]∈ (X × X , Y ). Hence, T ∈ C(X , Y ) implies S ∈ (X , Y ). N − N − Question 4.1 Suppose that (X , Y ) =∅. (a) Is C(X , Y ) a subspace of L(X , Y )? (b) Is C an operator ideal? (c) Is Proposition 2.1 valid for C? Answering a question in [6], an example of an operator K ∈ P \ C was given in [2], but we do not know if the other inclusion can be strict. Question 4.2 Suppose that (X , Y ) =∅. Is SC(X , Y ) = C(X , Y )? A negative answer to Question 4.2 would provide a new counterexample to the perturbation classes problem for . Next we will show that the inclusions in Proposition 4.1 become equalities in some cases. Theorem 4.1 Suppose that the spaces X and Y satisfy (X , Y ) =∅. 1. If X × X X, then C(X , Y ) = P (X , Y ). 2. If every infinite codimensional closed subspace of Y is contained in an infinite codimensional closed subspace N which is isomorphic to a quotient of X, then C(X , Y ) = SC(X , Y ). Two classes of operators related to the perturbation… Page 7 of 9 41 Proof (1) Let U : X → X × X be a bijective isomorphism and let U , U ∈ L(X ) 1 2 such that U (x ) = (U x , U x ). If K ∈ P (X , Y ), for each S ∈ L(X , Y ) such that 1 2 − [S, K]∈ (X × X , Y ) we have [S, K ]U = SU + KU ∈ (X , Y × Y ). By − 1 2 − Proposition 2.1, KU ∈ P (X , Y ). Then, SU ∈ (X , Y ), hence S ∈ (X , Y ). 2 − 1 − − Thus we conclude that K ∈ C(X , Y ). (2) Let K ∈ L(X , Y ) K ∈ / SC. Then, there exists an infinite codimensional closed subspace M of Y such that Q K is surjective. By the hypothesis, there exist an infinite codimensional closed subspace N such that M ⊂ N and a surjective operator V ∈ L(X , M ). Observe that S = J V ∈ L(X , Y ) is not in . We prove that K ∈ / C by showing N − that [S, K]∈ (X × X , Y ). Indeed, note that R(S) = N . Moreover Q K surjective implies R(K ) + N = Y . Since R([S, K ]) = R(S) + R(K ), [S, K ] is surjective. In the known examples in which SC(X , Y ) = P (X , Y ) in [8, 10], the space X has a complemented subspace which is hereditarily indecomposable. Therefore, the question arises. Question 4.3 Suppose that (X , Y ) =∅ and X × X X . Is SC(X , Y ) = P (X , Y )? A Banach space X is superprojective if each of its infinite codimensional closed subspaces is contained in some complemented infinite codimensional subspace. The spaces c , (1 < p < ∞) and L (μ) (1 < q ≤ 2) are superprojective. See [7, 12] 0 p q for further examples. Corollary 4.1 Suppose that (X , Y ) =∅ and the space Y is superprojective. Then, C(X , Y ) = SC(X , Y ). Proof Every closed infinite codimensional subspace of Y is contained in an infinite codimensional complemented subspace N . Since (X , Y ) =∅, there exists T ∈ L(X , Y ) with R(T ) ⊃ N , and composing with the projection P on Y onto N we get R(PT ) = N , and we can apply Theorem 4.1. The following result is a refinement of the previous results in this section. Theorem 4.2 Suppose that (X , Y ) =∅, there exists a surjection from X × X onto X , and every closed infinite codimensional subspace of Y is contained in a closed infinite codimensional subspace N which is isomorphic to a quotient of X . Then, P (X , Y ) = SC(X , Y ). Proof K ∈ L(X , Y ), K ∈ / SC. Then, there exists an infinite codimensional subspace N of Y such that Q K is surjective. By hypothesis, we can assume that there exists a surjective operator U : X → N . Then, S = J U ∈ L(X , Y ) is not in . N − Moreover, [S, K ] is surjective: R([S, K ]) = R(S) + R(K ) = N + R(K ) = Y , hence [S, K]∈ (X × X , Y ). Let V : X → X × X be a surjection and let V , V ∈ L(X ) such that V (x ) = 1 2 (V x , V x ) for each x ∈ X . Then, [S, K ]V = SV + KV ∈ (X , Y ). Since 1 2 1 2 − SV ∈ / , we get KV ∈ / P ; hence K ∈ / P by Proposition 2.1. 1 − 2 − − 41 Page 8 of 9 M. González, M. Salas-Brown Corollary 4.2 If Y is separable, there exists a surjection from X × X onto X , and X has a quotient isomorphic to , then SC(X , Y ) = (X , Y ). 1 − Proof It is well known that every separable Banach space is isomorphic to a quotient of . The class C(X , Y ) is surjective in the following sense: Proposition 4.3 Given K ∈ L(X , Y ) and a surjective operator Q ∈ L(Z , X ), if KQ ∈ C(Z , Y ), then K ∈ C(X , Y ). Proof Let S ∈ L(X , Y ) such that [S, K]∈ (X × X , Y ), and let Q : Z → X be a surjective operator. Then, the operator Q × Q ∈ L(Z × Z , X × X ) defined by (Q × Q)(a, b) = (Qa, Qb) is surjective. Thus, [S, K ](Q × Q) =[SQ, KQ] is in (Z × Z , Y ). Since KQ ∈ C(Z , Y ), we obtain SQ ∈ (Z , Y ), hence − − S ∈ (X , Y ), and we conclude K ∈ C(X , Y ). ∗ ∗ ∗ The dual space (X × X , · ) can be identified with (X × X , · ) in the ∞ 1 ∗ ∗ ∗ obvious way. Hence, the conjugate operator [S, T ] can be identified with (S , T ). ∗ ∗ Indeed, for x ∈ X and x ∈ X,wehave ∗ ∗ ∗ ∗ [S, T ] x , x=x , [S, T ]x=x , Sx + Tx ∗ ∗ ∗ ∗ ∗ ∗ ∗ =S x + T x , x=(S , T )x , x . ∗ ∗ ∗ As a consequence, [S, T]∈ if and only if (S , T ) ∈ . Similarly, (S, T ) − + ∗ ∗ can be identified with [S , T ]. The following result describes the behavior of the classes of -singular and - cosingular operators under duality. Proposition 4.4 Let K ∈ L(X , Y ). ∗ ∗ ∗ 1. If K ∈ S(Y , X ), then K ∈ C(X , Y ). ∗ ∗ ∗ 2. If K ∈ C(Y , X ), then K ∈ S(X , Y ). Proof (1) Let S ∈ L(X , Y ) such that [S, K]∈ (X × X , Y ). Then, [S, K ] ∈ . − + ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Since [S, K ] ≡ (S , K ), we have (S , K ) ∈ (Y , X × X ), and from K ∈ ∗ ∗ ∗ ∗ ∗ S(Y , X ) we obtain S ∈ (Y , X ); therefore, S ∈ (X , Y ), and hence + − K ∈ C(X , Y ). The proof of (2) is similar. Acknowledgements Supported in part by MINCIN Project PID2019-103961. Funding Open Access funding provided thanks to the CRUE-CSIC agreement with Springer Nature. Declarations Conflict of interest The authors have no competing interests to declare that are relevant to the content of this article. Two classes of operators related to the perturbation… Page 9 of 9 41 Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. References 1. Aiena, P., González, M.: On inessential and improjective operators. Stud. Math. 131, 271–287 (1998) 2. Aiena, P., González, M., Martínez-Abejón, A.: Characterizations of strictly singular and strictly cosingular operators by perturbation classes. Glasg. Math. J. 54, 87–96 (2011) 3. Argyros, S.A., Felouzis, V.: Interpolating hereditarily indecomposable Banach spaces. J. Am. Math. Soc. 13, 243–294 (2000) 4. Bessaga, C., Pełczynski, ´ A.: Banach spaces non-isomorphic to their Cartesian squares. I. Bull. Acad. Pol. Sci. Sér. Sci. Math. Astron. Phys. 8, 77–80 (1960) 5. 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Advances in Operator Theory – Springer Journals
Published: Jul 1, 2023
Keywords: Perturbation classes problem; Semi-Fredholm operator; Strictly singular operator; 47A55; 47A53
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