E extracta mathematicae Vol. 33, Núm. 2, 191 – 208 (2018) Characterization of some Classes Related to the Class of Browder Linear Relations Attaif Farah, Maher Mnif Department of Mathematics, Faculty of Science of Sfax, University of Sfax Soukra Road, Km 3.5, PO Box 1171, 3000, Sfax, Tunisia farah-lotfi1@hotmail.fr , maher.mnif@gmail.com Presented by Manuel González Received March 3, 2018 Abstract: In this paper, we introduce the sets of left and right invertible linear relations and we give some of their properties. Furthermore, we study the connection between these sets and the classes of Fredholm linear relations. The obtained results are used to give some characterizations of some classes related to the class of Browder linear relations. Key words: Closed linear relation, left- and right-Browder linear relation, left and right invertible linear relation. AMS Subject Class. (2000): Primary 47A06, 47A53, 47A05. 1. Introduction First, let us notice that, throughout this paper, (X, || ||) and (Y, || ||) will represent complex Banach spaces. A linear relation T : X → Y is a mapping from a subspace D(T) = {u ∈ X : T(u) ̸= ∅} ⊆ X, called the domain of T , which takes values in P(Y )\{∅} (the collection of nonempty subsets of Y ) and is such that T(αx1 + βx2) = αT(x1) + βT(x2) for all non-zero scalars α, β ∈ K and x1, x2 ∈ D(T). The class of all linear relations from X to Y will be denoted by LR(X, Y ). We write LR(X) = LR(X, X). If T maps the points of its domain to singletons, then T is said to be an operator, which is equivalent to T(0) = {0}. The class of linear bounded operators defined on all X is denoted by B(X, Y ). A linear relation T is uniquely defined by its graph G(T) = {(u, v) ∈ X ×Y : u ∈ D(T), v ∈ T(u)}. The inverse of T is the relation T −1 given by: G(T −1) = {(v, u) ∈ Y × X : (u, v) ∈ G(T)}. If G(T) is closed, then T is said to be closed. The class of such relations is denoted by CR(X, Y ). We denote by R(T) = T(D(T)) the range of T and 191 192 a. farah, m. mnif by N(T) := {x ∈ X : (x, 0) ∈ G(T)} the kernel of T. If R(T) = Y , then T is called surjective, and if N(T) = {0}, then T is called injective. We may write n(T) = dim N(T) and d(T) = codim R(T) and the index of T , namely i(T), is defined by i(T) = n(T) − d(T), provided that n(T) and d(T) are not both infinite. For S, T ∈ LR(X, Y ) and λ ∈ K, the linear relations S+T , S+̂T , S⊕T and λS are defined by G(S+T) := {(x, y+z) ∈ X×Y : (x, y) ∈ G(S) and (x, z) ∈ G(T)}, G(S+̂T) := {(x+y, z+t) ∈ X×Y : (x, z) ∈ G(S) and (y, t) ∈ G(T)}, this last sum is direct when G(S) ∩ G(T) = {(0, 0)}. In such case, we write S ⊕ T, G(λS) := {(x, λy) ∈ X × Y : (x, y) ∈ G(S)}, and S ⊂ T means that G(S) ⊂ G(T). For T ∈ LR(X, Y ) and S ∈ LR(Y, Z), the product ST is given by G(ST) := {(x, z) ∈ X ×Z : (x, y) ∈ G(T), (y, z) ∈ G(S) for some y ∈ Y }. Let T ∈ LR(X). If α ∈ K, then α − T stands for αI − T, where I is the identity operator in X. The resolvent set of T is the set ρ(T) = {z ∈ C : (z − T)−1 ∈ B(X)}. If M is a subspace of X, then TM is the linear relation whose graph is G(T) ∩ (M × M). Recall that the class of upper semi-Fredholm linear relations is denoted by: ϕ+(X, Y ) = {T ∈ CR(X, Y ) : R(T) is closed and n(T) < ∞}. Moreover, the class of lower semi-Fredholm linear relations is denoted by: ϕ−(X, Y ) = {T ∈ CR(X, Y ) : R(T) is closed and d(T) < ∞}. T is called a Fredholm relation, if T ∈ ϕ+(X, Y ) ∩ ϕ−(X, Y ). The class of all Fredholm relations is denoted by ϕ(X, Y ). Recall that a closed subspace M in a normed space X is said to be comple- mented in X if there exists a closed subspace N of X such that X = M + N and {0} = M ∩ N (in short, X = M ⊕ N)). If a linear relation T ∈ LR(X, Y ) is upper semi-Fredholm and R(T) is complemented in Y , then T is said to be left-Fredholm linear relation. A linear relation T ∈ LR(X, Y ) is right-Fredholm relation if it is lower semi-Fredholm and N(T) is complemented in D(T). The set of left-Fredholm linear relations (right-Fredholm linear relations) is denoted by ϕl(X, Y ) (ϕr(X, Y )). Let T ∈ LR(X). We define T n ∈ LR(X), n ∈ N by T 0 = I, T 1 = T and T n = TT n−1. We define N∞(T) = ∪nN(T n) and R∞(T) = ∩nR(T n). The singular chain manifold of T ∈ LR(X), Rc(T) is defined by Rc(T) := ( ∞∪ n=1 N(T) )∩( ∞∪ n=1 T n(0) ) . class of browder linear relations 193 The ascent and the descent of T ∈ LR(X) are defined as follows: asc(T) := min{p ∈ N : N(T p) = N(T p+1)}, des(T) := min{p ∈ N : R(T p) = R(T p+1)}, respectively, whenever these minima exists. If no such numbers exist, the ascent and descent of T are defined to be ∞. A relation T ∈ CR(X) is upper semi-Browder if it is upper semi-Fredholm with finite ascent. If T ∈ CR(X) is lower semi-Fredholm with finite descent, then T is lower semi- Browder. Let B+(X) (B−(X)) denotes the set of all upper (lower) semi- Browder linear relations. The set of Browder linear relations is defined by B(X) = B+(X) ∩ B−(X). T ∈ CR(X) is said to be left-Browder relation if it is left-Fredholm with finite ascent. If T ∈ CR(X) is right-Fredholm with finite descent, then T is right-Browder relation. Let Bl(X) (Br(X)) denotes the set of all left-(right-) Browder linear relations. A study of left and right Browder linear relations has been carried by a number of authors in the recent past (see [5], [7], [9]). In a recent paper of (2016) [7], the authors prove that a left (right) Browder linear relation T in a Banach space can be expressed in the form T = A+B where A is an injective (onto) left (right) Fredholm linear relation and B is a bounded finite rank operator with BT ⊂ TB. The purpose of the present paper is to consider the notion of left and right invertible linear relations and we give some characterizations of left- and right-Browder closed linear relations. To make the paper easily accessible, some results from the theory of linear relations due to Cross [8] are recalled in Section 2. In Section 3, we extend to the general case of closed linear relations in Banach spaces, some results con- cerning upper and the lower semi-Browder closed operators proved by Snez̆ana C̆ in [11, Theorem 3 and Theorem 4]. In particular, we prove that the upper (lower) semi-Browder linear relation T is a upper (lower) semi-Fredholm and almost bounded below (onto) linear relation. Finally, in Section 4, the defi- nition of left (right) invertible linear relation is given, and some properties of these relations are shown wich have been used to characterize the left (right) Browder linear relations. In particular, we prove that the linear relation T is left (right) Browder if and only if there exists a bounded operator projector P , such that TP −PT = T −T, dim R(P) < ∞, (TP)d = T(0) for some d ∈ N and T + P is left (right) invertible linear relation if and only if there exists a compact operator B satisfying TB − BT = T − T and T − B is a left (right) invertible linear relation. These results are generalizations of the results in the 194 a. farah, m. mnif case of linear operators shown by Snez̆ana C̆. Z̆ivković-Zlatanović, Dragan S. Djordjević and Robin E. Harte [11, Theorem 5 and Theorem 6]. 2. Auxiliary results In this section, we recall some auxiliary results from the theory of linear relations in Banach spaces. Let T be a linear relation in a Banach space X. Recall that T is said to be continuous if for each neighborhood V in R(T), the inverse image T −1(V ) is a neighborhood in D(T), bounded if it is continuous and its domain is whole X, open if its inverse is continuous. In order to give some characterizations of these classes of linear relations, one introduces the following notations. Let QT denotes the quotient map from X onto X/T(0). We note that QT T is single-valued and so we can define ∥Tx∥ := ∥QT Tx∥, x ∈ D(T) and ∥T∥ := ∥QT T∥ called the norm of Tx and T respectively, and the minimum modulus of T is the quantity γ(T) := sup{λ ≥ 0 : λ dist(x, N(T)) ≤ ∥Tx∥, x ∈ D(T)}. In [8, II.3.2 and II.5.3] Ronald Cross proves that: (i) T is continuous if and only if ∥T∥ < ∞; (ii) T is open if and only if γ(T) > 0; (iii) T is closed if and only if QT T is a closed operator and T(0) is a closed subspace. Recall that T is said to be regular linear relation if R(T) is closed and T verifies one of the equivalent conditions: (i) N(T) ⊆ R(T m), for all nonnegative integer m; (ii) N(T n) ⊆ R(T), for all nonnegative integer n; (iii) N(T n) ⊆ R(T m), for all nonnegative integers n and m. The Kato decomposition of left and right-Fredholm linear relations are collected in the following lemma. Lemma 2.1. ([7, Theorem 5.1 and Theorem 6.1]) Let T ∈ CR(X, Y ). Then: class of browder linear relations 195 (i) If T ∈ ϕl(X, Y ), then there exist two closed subspaces M and N of X such that X = M ⊕ N with N ⊂ D(T) and dim N < ∞; T = TM ⊕ TN, such that TM is a regular left-Fredholm linear relation in M and TN is a bounded nilpotent operator in N. (ii) If T ∈ ϕr(X, Y ), be such that D(T) = X and ρ(T) ̸= ∅, then there exist two closed subspaces M and N of X such that X = M ⊕ N with N ⊂ D(T) and dim N < ∞; T = TM ⊕ TN, such that TM is a regular right-Fredholm linear relation in M and TN is a bounded nilpotent operator in N. Let M, L be two subspaces of a Banach space X and let δ(M, L) = sup x∈M ∥x∥≤1 dis(x, L). The gap between M and L is defined by δ̂(M, L) = max{δ(M, L), δ(L, M)}. Finally, we give the main result of this section. Theorem 2.1. Let T be a bounded regular linear relation with ρ(T) ̸= ∅. Then (i) If T is almost bounded below, then T is bounded below. (ii) If T is almost onto, then T is onto. Proof. (i) We have T is almost bounded below, then there exists δ > 0 such that T −λ is injective and open for all 0 < |λ| < δ. Hence N(T −λ) = {0} and γ(T − λ) > 0. On the other hand, we have T is regular, then γ(T) > 0. By using [2, Lemma 2.10 (ii)] and [1, Theorem 23 (5)], we deduce that δ̂ ( N(T − λ), N(T) ) ≤ | λ | min { γ(T − λ), γ(T) } ≤ | λ | γ(T) − 3 | λ | . Hence limλ→0 δ̂ ( N(T − λ), N(T) ) = 0. Therefore there exists λ > 0 such that δ̂ ( N(T − λ), N(T) ) < 1. Thus by [10, Corollary 10] we have dim N(T) = dim N(T − λ). Then N(T) = {0}. Therefore T is bounded below. 196 a. farah, m. mnif (ii) Suppose that T is almost onto, then there exists δ > 0 such that for all 0 < |λ| < δ we have R(T −λI) = X. Hence N((T −λI)′) = R(T −λI)⊥ = {0}. By [12, Proposition III.1.5] we have N(T ′ −λI) = {0}. Therefore T ′ is almost bounded below and regular relation. Then, by (i), we have T ′ is bounded below. Hence N(T ′) = {0}, then R(T) = X. Thus, T is onto. 3. Some properties of upper and lower semi-Browder linear relations The goal of this section is to discuss some properties of upper and lower semi-Browder linear relations that will be used in the last section. Definition 3.1. Let T ∈ LR(X) and S ∈ B(X). We say that S com- mutes with T if S(D(T)) = D(T) and for all x ∈ D(T), we have, STx = TSx. We shall write comm−1ϵ (T) = { S ∈ B(X) : S invertible, commutes with T and ∥S∥ < ϵ } . Proposition 3.1. Let X be a Banach space and T ∈ CR(X) be such that D(T) = X and ρ(T) ̸= ∅. Let S ∈ comm−1ϵ (T). Then (i) S−1 commutes with T. (ii) S′ commutes with T ′. (iii) For all n ∈ N∗, Sn commutes with T n. Proof. (i) Let x ∈ D(T). Then there exists u ∈ D(T) such that Su = x. We have TSu = STu, hence Tx = STS−1x, and S−1Tx = TS−1x. Therefore S−1 commutes with T. (ii) First we claim that S′(D(T ′)) = D(T ′). Indeed, for y′ ∈ D(T ′) we have ∥S′y′(Tx)∥ = ∥(y′S)(Tx)∥ = ∥(y′T)(Sx)∥ ≤ ∥y′T∥∥S∥∥x∥ for every x ∈ D(T). Hence, S′y′T is continuous. We have S′y′T(0) = S′(T ′y′(0)) = S′(0) = 0, then, by [8, Proposition III.1.2] we deduce that S′y′ ∈ D(T ′). Therefore, S′(D(T ′)) ⊂ D(T ′). We have S ∈ comm−1ϵ (T) then, S′ is bijective and ∥S′∥ = ∥S∥. Let y′ ∈ D(T ′). Then there exists a unique functional z′ ∈ X′ such that y′ = S′z′ = z′S. It follows that z′ = y′S−1 and by (i) we get: ∥(z′T)x∥ = ∥y′(S−1(Tx))∥ = ∥y′(T(S−1x))∥ = ∥(y′T)(S−1x)∥ ≤ ∥y′T∥∥S−1∥∥x∥. class of browder linear relations 197 Therefore, z′T is continuous. We have z′T(0) = y′S−1T(0) = y′T(0) = 0, then, by [8, Proposition III.1.2] we deduce that z′ ∈ D(T ′). Now, we show that T ′S′ = S′T ′. We have S and S′ are bijective, then R(S′) = X′, D(S′) = X′, D(T) ⊂ R(S) = X, and R(T) ⊂ D(S) = X. Hence by [9, Theorem III.1.6], we have (ST)′ = T ′S′ and (TS)′ = S′T ′. Therefore T ′S′ = (ST)′ = (TS)′ = S′T ′. Then S′ commutes with T ′. (iii) For n = 2 we will show that S2(D(T 2)) = D(T 2) and for all x ∈ D(T 2), we have, S2T 2x = T 2S2x. Indeed, let x ∈ D(T 2). Then x ∈ D(T) and Tx ∩ D(T) ̸= ∅. Using that x ∈ D(T) and S(D(T)) = D(T), we get S2x ∈ D(T). On another hand, we have TS2x = S2Tx, Tx ∩ D(T) ̸= ∅ and S(D(T)) = D(T) then TS2x ∩ D(T) ̸= ∅. So, S2x ∈ D(T 2). Therefore D(T 2) ⊂ S2(D(T 2)). Let x ∈ D(T 2). Then x ∈ D(T) and Tx ∩ D(T) ̸= ∅. We have S(D(T)) = D(T), then there exists y ∈ D(T) such that x = S2y. It remains to prove that Ty ∩ D(T) ̸= ∅. We have, Tx ∩ D(T) = TS2y ∩ D(T) = S2Ty ∩ D(T) ̸= ∅ . Then Ty ∩ D(T) ̸= ∅. Hence S2(D(T 2)) ⊂ D(T 2). Therefore S2(D(T 2)) = D(T 2). Let x ∈ D(T 2). Then S2T 2x = SSTTx = STSTx = TSTSx = TTSSx = T 2S2x. The case n > 2, is deduced by using an induction argument. Now, we prove the following result useful for the proof of the first main result of this section. Proposition 3.2. Let T ∈ LR(X) and S ∈ comm−1ϵ (T). Then N(T − S) ⊂ R∞(T). Proof. First we show by induction that, if x ∈ N(T − S) then for all n ≥ 1, we have T nx = Snx + T n(0). The case n = 1 is obvious. Assume that T nx = Snx + T n(0) and we shall prove that T n+1x = Sn+1x + T n+1(0). Indeed, T n+1x = TSnx + TT n(0) = SnTx + T n+1(0) = Sn(Sx + T(0)) + T n+1(0) = Sn+1x + T n+1(0). Now, let x ∈ N(T −S). We have S−1 commutes with T , then for all n ≥ 1, T n(S−1)nx = (S−1)nT nx = x + T n(0). Hence N(T − S) ⊂ R(T n). Therefore N(T − S) ⊂ R∞(T). 198 a. farah, m. mnif Now, we are ready to state the first main result of this section. Theorem 3.1. Let T ∈ B+(X) be such that Rc(T) = {0}. Then, the following statements holds: (i) T ∈ ϕ+(X), and there exists ϵ > 0 such that for all S ∈ comm−1ϵ (T), we have T − S is bounded below. (ii) T ∈ ϕ+(X), and almost bounded below. Proof. (i) We have T ∈ ϕ+(X), then, by [5, Lemma 2.5], T n ∈ ϕ+(X) for all n ∈ N. Let X1 = R∞(T). Then X1 is a closed subspace. Let T1 : X1 → X1; the restriction of T to X1. Then, by using [1, Lemma 20] we deduce that β(T1) = 0 and α(T1) < ∞. Then, T1 ∈ ϕ(X). Clearly we have S(X1) ⊂ X1. Writing S1 : X1 → X1, the restriction of S to X1. Then, by using [12, Proposition 2.4 and Proposition 2.6], we deduce that α(T1 − S1) ≤ α(T1); β(T1 − S1) ≤ β(T1), and i(T1 − S1) = i(T1). From Proposition 3.2 we deduce that α(T − S) = α(T1 − S1). Therefore α(T − S) = i(T1 − S1) = i(T1) = α(T1). Now, by using Rc(T) = {0} and asc(T) ≤ p for some p ∈ N, we get N(T k) ∩ R(T p) = {0} for all k ∈ N. Hence α(T1) = 0 and therefore α(T − S) = 0. Furthermore by [4, Proposition 3], T − S has a closed range, then T − S is bounded below. (ii) is obvious. Now, we are in position to give the second main result of this section. Theorem 3.2. Let X be a Banach space and T ∈ B−(X) be such that D(T) = X and ρ(T) ̸= ∅. Then (i) T ∈ ϕ−(X) and there exists ϵ > 0 such that for all S ∈ comm−1ϵ (T), we have T − S is onto. (ii) T ∈ ϕ−(X) and almost onto. Proof. (i) Let T ∈ B−(X) be such that D(T) = X and ρ(T) ̸= ∅. Then, by using [6, Theorem 2.1], [5, Lemma 2.3] and [8, V.1.1] we deduce that T ′ ∈ B+(X) and Rc(T ′) = 0. Hence by Theorem 3.1, there exists ϵ > 0 such that for all A ∈ comm−1ϵ (T ′), T ′ − A is bounded below. Let S ∈ comm−1ϵ (T). Then by Proposition 3.1, we have S′ ∈ comm−1ϵ (T ′). Therefore β(T − S) = α((T − S)′) = α(T ′ − S′) = 0. Then T − S is onto. (ii) is obvious. class of browder linear relations 199 4. Characterization of left- and right-Browder linear relations This section concerns the characterization of left- and right-Browder linear relations in Banach spaces. 4.1. Characterization of left-Browder linear relations. We begin by introducing the new concept of left invertible linear relation and give some of its properties. Definition 4.1. Let T ∈ LR(X). We say that T is left invertible, if there exists a bounded operator A such that for all x ∈ D(T), ATx = x. In this case we say that A is a left inverse of T. The following lemmas give the relationship between the notion of bounded below linear relations and the notion of left invertible linear relations. Lemma 4.1. Let T be an everywhere defined closed bounded below linear relation. If R(T) is complemented in X, then T is left invertible. Proof. Since T is injective and open then T −1 is a continuous operator. We have R(T) is complemented, then there exists a closed subspace F of X such that X = R(T) ⊕ F and there exists a continuous projector P such that R(P) = F and N(P) = R(T). Take A = T −1(I − P) + P . Then A is a bounded operator and we have for all x ∈ D(T), ATx = x. Therefore T is left invertible. Lemma 4.2. Let T be a closed left invertible linear relation. Then T is bounded below. Proof. Let A be a bounded operator such that for all x ∈ D(T), ATx = x. Let x ∈ N(T). Then T(x) = T(0). Hence x = AT(x) = AT(0) = 0. Therefore N(T) = {0}. On the other hand, for all x ∈ D(T), T −1T(x) = x = AT(x). Then for all y ∈ R(T) we have T −1(y) = A(y). Hence ∥T −1y∥ = ∥Ay∥ ≤ ∥A∥∥y∥, for all y ∈ D(T −1). Hence T −1 is a continuous relation. Thus T is bounded below. Proposition 4.1. Let T be an everywhere defined closed left invertible linear relation on a Banach space X such that T(0) is complemented. Then T ∈ ϕl(X). 200 a. farah, m. mnif Proof. Using Lemma 4.2 and [8, V.18] we deduce that T ∈ ϕ+(X). Let A be a left inverse of T, we have (TA)2x = TATAx = TAx. Then TA is a multivalued projector. Let N = R(T) = R(TA) = N(I − TA) and M = N(TA) = R(I − TA). We claim that M ∩ N = T(0). Indeed, let x ∈ M ∩ N. Then there exists y ∈ X such that x ∈ TAy and 0 ∈ TAx. Hence 0 ∈ TATAy = TAy. Therefore TAy = TA(0) = T(0) and, as a result, M ∩ N ⊂ T(0). Conversely, let x ∈ T(0). Then x ∈ TA(0) and so x ∈ R(TA) and TAx = TA(0). Hence x ∈ N(TA). Thus M ∩ N = R(TA) ∩ N(TA) = T(0). We have M, N and M + N = X are closed, then by [5, Lemma 3.1 (i)], P = TA is a continuous multivalued projector. On the other hand, we have M ∩ N = T(0) is complemented in X. Then by using [5, Lemma 3.1 (ii)], we deduce that R(T) is complemented in X. Therefore T ∈ ϕl(X). Lemma 4.3. If T is an injective everywhere defined linear relation and S be a bounded operator such that ST ⊂ TS, then, for all n ∈ N, T −1(T + S)−n(0) = (T + S)−n(0) ⊆ T(X). Proof. We have (T +S)−n(0) = (T +S)−nT −1(0) = (T(T +S)n)−1(0). By using [9, Proposition 3 (iii)] we deduce that (T +S)−n(0) = ((T +S)nT)−1(0) = T −1(T + S)−n(0). Then T(T + S)−n(0) = (T + S)−n(0) + T(0). Hence (T + S)−n(0) ⊂ T(T + S)−n(0) ⊂ T(X). Proposition 4.2. Let T be a bounded closed bounded below linear rela- tion and S be a compact operator such that ST ⊂ TS. Then asc(T +S) < ∞. Proof. We have T is injective then by Lemma 4.3, we deduce that T −1(T + S)−n(0) = (T + S)−n(0) ⊆ T(X). (4.1) If k > 0 be such that ∥x∥ ≤ k∥Tx∥ for each x ∈ X, then dis(x, (T + S)−n(0)) ≤ k dis(QT+STx, QT+S(T + S)−n(0)). Indeed; if x ∈ X and yn ∈ (T + S)−n(0) are arbitrary, then by (4.1), there exists zn ∈ (T + S)−n(0) for which yn ∈ Tzn, (Tzn = yn + T(0) = yn + T(0)); dis(x, (T + S)−n(0)) ≤ ∥x − zn∥ ≤ k∥T(x − zn)∥ ≤ k∥QT T(x − zn)∥ ≤ k∥QT+STx − QT+Syn∥. class of browder linear relations 201 We deduce that dis(x, (T + S)−n(0)) ≤ k dis(QT+STx, QT+S(T + S)−n(0)). Assume that asc(T + S) = ∞, then there exists (xn) ⊂ X such that ∥xn∥ = 1; xn ∈ (T + S)−n−1(0) and dis(xn, (T + S)−n(0)) ≥ 12. It follows that if n and m ≥ n + 1 are arbitrary, then k∥QT+SSxm − QT+SSxn∥ ≥ k∥QT+SSxn − QT+S(T + S)xm + QT+STxm∥. We have xn ∈ (T + S)−n−1(0), then 0 ∈ (T + S)n+1(xn). Hence, using [9, Proposition 3], we get 0 = S(0) ∈ S(T + S)n+1(xn) ⊂ (T + S)n+1(S(xn)). Therefore S(xn) ∈ (T + S)−n−1(0). Thus, QT+SS(xn) ∈ QT+S(T + S)−n−1(0) ⊂ QT+S(T + S)−m(0). Now, we have xm ∈ (T + S)−m−1(0), then (T + S)(xm) ⊂ (T + S)(T + S)−m−1(0) = (T + S)−m(0) + (T + S)(0). Hence QT+S(T + S)(xm) ∈ QT+S(T + S)−m(0). Therefore k∥QT+SSxm − QT+SSxn∥ ≥ k dis(QT+STxm, QT+S(T + S)−m(0)) ≥ dis(xm, (T + S)−m(0)) ≥ 1 2 . Which contradicts the compactness of the operator QT+SS. Definition 4.2. We say that a relation T ∈ CR(X) is almost left invert- ible if there exists δ > 0 such that for all 0 < |λ| < δ we have T − λI is left invertible. The following theorem is our first main result of this section where we give several sufficient and necessary conditions for a closed bounded linear relation to be left-Browder. Theorem 4.1. Let T be a bounded closed linear relation such that ρ(T) ̸= ∅ and T(0) is complemented in X. Then the following properties are equiva- lent: (i) T ∈ Bl(X). (ii) T ∈ ϕl(X) and there exists ϵ > 0 such that for all S ∈ comm−1ϵ (T), we have T − S is bounded below. (iii) T ∈ ϕl(X) and almost bounded below. 202 a. farah, m. mnif (iv) There exists a bounded operator projector P , such that TP − PT = T −T, dim R(P) < ∞, T is completely reduced by the pair (N(P), R(P)) with TN(P) is regular left invertible linear relation in N(P) and TR(P) is a bounded nilpotent operator in R(P). (v) There exists a bounded operator projector P , such that TP − PT = T − T, dim R(P) < ∞, (TP)d = T(0) for some d ∈ N and T + P is left invertible linear relation. (vi) There exists a compact operator B satisfying TB − BT = T − T and T − B is left invertible. Proof. (i) ⇒ (ii) : An immediate consequence of Theorem 3.1. (ii) ⇒ (iii) : obvious. (iii) ⇒ (iv) : From Lemma 2.2 it follows that there exist two closed sub- spaces M and N of X such that X = M ⊕N with dim N < ∞; T = TM ⊕TN, such that TM is regular left-Fredholm linear relation in M and TN is bounded nilpotent operator in N. Let P be the projector such that R(P) = N and N(P) = M. We claim that TP −PT = T −T. Indeed, let x ∈ X. Then there exist x1 ∈ M and x2 ∈ N such that x = x1 + x2. Hence (TP − PT)x = TPx − PTx = T(x2) − P(TM(x1) + TN(x2)) = TM(0) + TN(x2) − TN(x2) = TM(0) = T(0). Therefore TP − PT = T − T . We have TR(P) = TN and TN(P) = TM. Evidently TR(P) is a bounded nilpotent operator. We claim now that TN(P) is a regular left invertible linear relation. Indeed, we have TM = TN(P) is regular and left Fredholm linear relation in N(P). So R(TN(P)) is complemented in N(P). On another hand T(0) = TN(P)(0) is complemented in X. So there exists a closed subspace F of X such that T(0) ⊕ F = X. Hence (T(0) ⊕ F) ∩ N(P) = N(P). Since T(0) ⊂ N(P), then T(0) ⊕ F ∩ N(P) = N(P). Therefore T(0) = TN(P)(0) is complemented in N(P). We have T is almost bounded below, then there exists δ > 0 such that for all 0 < |λ| < δ there exists kλ > 0 such that for all x ∈ X; ∥x∥ ≤ kλ∥(T − λI)x∥. So for all x ∈ N(P); ∥x∥ ≤ kλ∥(T − λI)x∥ = kλ∥TN(P)x − λx + TR(P)(0) − λ0∥ = ∥(TN(P) − λIN(P))x∥. class of browder linear relations 203 Therefore TN(P) is almost bounded below in N(P). By using Theorem 2.1, we deduce that TN(P) is bounded below. Using Lemma 4.1, we get TN(P) is left invertible. (iv) ⇒ (v) : Let P be the projector in (iv) and x = u + v such that u ∈ N(P) and v ∈ R(P). We have TR(P) is a nilpotent operator. Then there exists d ∈ N such that T d R(P) = 0. Hence: (TP)dx = (TP)(TP) . . . (TP)︸ ︷︷ ︸ d−1 T(v) = (TP)(TP) . . . (TP)︸ ︷︷ ︸ d−1 (TN(P)(0) + TR(P)(v)). And by iteration, we get (TP)dx = T dR(P)(v) + T(0) = T(0). Now, if we show that T + P is bounded below and R(T + P) is comple- mented, then we can use Lemma 4.1 to deduce that T + P is left invertible. For that, since T is closed and P is a bounded linear operator, then T + P is closed.On another hand we have TR(P) is a bounded nilpotent operator, so TR(P) +I is invertible, and hence N(T +P) = N(TN(P))⊕N(TR(P) +I) = {0} and R(T +P) = R(TN(P))⊕R(TR(P) +I) = R(TN(P))⊕R(P) which is closed. Therefore, T + P is injective with closed range. Then by the closed graph theorem and Lemma 2.7 we get T + P is bounded below. Now, by Proposi- tion 4.1, TN(P) is ϕl(N(P)). Then R(TN(P)) is complemented in N(P). So there exists a closed subspace F1 such that R(TN(P))⊕F1 = N(P). Therefore R(T + P) + F1 = X. Let x ∈ R(T + P) ∩ F1. Then x = xR(TN(P )) + xR(P) and x = xF1. So, xR(P) = xF1 − xR(TN(P )). By according to xR(P) ∈ R(P), xF1 − xR(TN(P )) ∈ N(P) and R(TN(P)) ∩ F1 = {0}, we may deduce that xR(P) = xF1 = xR(TN(P )) = 0. Therefore x = 0. So R(T + P) ⊕ F1 = X. Hence R(T + P) is complemented in X. (v) ⇒ (vi) : As, P is a bounded operator with finite rank, then P is compact. So, just take B = −P , we deduce the desired result. (vi) ⇒ (i) : Let B be a compact operator satisfying TB −BT = T −T and T − B is left invertible. By Proposition 4.1, we deduce that T − B ∈ ϕl(X). By using [2, Theorem 11] we infer that T ∈ ϕl(X). We claim now that asc(T) < +∞. Indeed, let y ∈ BTx. We have TBx−BTx = T(0), then there exist z ∈ TBx and α ∈ T(0) such that y−z = α. Therefore y ∈ TBx+T(0) = 204 a. farah, m. mnif TBx. So G(BT) ⊂ G(TB). Then B(T − B) ⊂ BT − BB ⊂ TB − BB = (T − B)B. By using Lemma 4.2 and Proposition 4.2 we deduce that asc(T) < ∞ and, as a result, T ∈ Bl(X). 4.2. Characterization of right-Browder linear relation. We begin by introducing the new concept of right invertible linear relation and giving some of its properties. Definition 4.3. Let T ∈ LR(X). We say that T is right invertible, if there exists a bounded operator B such that TB = I+T(0) and R(B) ⊂ D(T). In this case we say that B is a right inverse of T. Proposition 4.3. Let T ∈ ϕr(X) be such that T is bounded and onto. Then T is right invertible. Proof. We have T is onto and closed, then T is open. Hence T −1 is con- tinuous. Since T ∈ ϕr(X), then N(T) is complemented. Hence there exists a continuous projector P such that R(P) = N(T). Let P1 = I − P and A = P1T −1. Then, A is a continuous selection of T −1. Hence TAx = x+T(0) for all x ∈ X and R(A) ⊂ R(T −1) = D(T). Therefore T is right invertible. Remark 4.1. Let T ∈ CR(X) be everywhere defined. If T is right invert- ible and T(0) is complemented, then there exists a bounded operator B such that TB = I + T(0) and T(0) = N(B). Proof. We have T is right invertible, then there exists a bounded operator A such that TA = I +T(0). Using that T(0) is complemented we deduce that there exists a closed subspace G ⊂ X such that T(0)⊕G = X. Let B = APG, where PG is the projector onto G with kernel T(0). Hence for all x ∈ X, we have TBx = TAPGx = PGx + T(0) = xG + T(0) = xG + xT(0) + T(0) = x + T(0). Therefore TB = I + T(0) and N(B) ⊂ T(0). Now, let x ∈ T(0). Then Bx = APGx = A(0) = 0. Hence T(0) ⊂ N(B). Therefore T(0) = N(B). class of browder linear relations 205 Proposition 4.4. Let T ∈ CR(X) be everywhere defined. If T is right invertible with T(0) is complemented, then T ∈ ϕr(X). Proof. Let A be a right inverse of T and x ∈ X. Then TAx = x + T(0). Hence x ∈ TA(x) ⊂ R(TA) ⊂ R(T). Therefore T is onto and, as a result, T ∈ ϕ−(X). Now, by Remark 4.1 we deduce that there exists a bounded operator B such that TB = I + T(0) and T(0) = N(B). Let x ∈ X = D(T). Then there exists y ∈ X such that x ∈ T −1(y). We have TBy = y + T(0), then T −1y = By + T −1(0). Hence x ∈ R(B) + N(T). Let x ∈ R(B) ∩ N(T). Then 0 ∈ Tx and there exists y ∈ X such that x ∈ By. Hence 0 ∈ TBy = y + T(0). Therefore y ∈ T(0). Since N(B) = T(0) we deduce that x = 0 and so, X = R(B) ⊕ N(T). Let S : (X/N(B)) ⊕ N(T) → X defined by S(x, y) = Bx + y. We have R(S) = R(B) + N(T) = X, then S is onto. Let x ∈ X/N(B) and y ∈ N(T) be such that S(x, y) = Bx + y = 0. Then Bx = 0 and y = 0, hence x = 0 and y = 0. Therefore S is injective. We have S is bijective and continuous, then S−1 is continuous. Since (X/N(B))⊕{0} is closed and S((X/N(B))⊕{0}) = R(B), then we deduce that R(B) is closed. Hence N(T) is complemented and T ∈ ϕr(X). Definition 4.4. We say that a relation T ∈ LR(X) is almost right in- vertible if there exist δ > 0 such that for all 0 < |λ| < δ we have T − λI is right invertible. We finish this section by giving a characterization of right Browder linear relations. Theorem 4.2. Let X be a Banach space and T ∈ CR(X) be such that D(T) = X, T(0) is complemented and ρ(T) ̸= ∅. Then the following proper- ties are equivalent: (i) T ∈ Br(X). (ii) T ∈ ϕr(X) and there exists ϵ > 0 such that for all S ∈ comm−1ϵ (T), we have T − S is onto. (iii) T ∈ ϕr(X), and almost onto. (iv) There exists a bounded projector operator P , such that TP −PT = T − T, dim R(P) < ∞, T is completely reduced by the pair (N(P), R(P)), with TN(P) is a regular right invertible linear relation in N(P) and TR(P) is a bounded nilpotent operator in R(P). 206 a. farah, m. mnif (v) There exists a bounded projector operator P , such that TP − PT = T −T, dim R(P) < ∞, (TP)d = T(0) for some d ∈ N, and T +P is right invertible. (vi) There exists a compact operator B satisfying TB − BT = T − T and T − B is right invertible. Proof. (i) ⇒ (ii) : An immediate consequence of Theorem 3.2. (ii) ⇒ (iii) : obvious. (iii) ⇒ (iv) : If T ∈ ϕr(X), then by Lemma 2.2 and as in the proof of Theorem 4.1 there exists a projector P = P 2, such that TP − PT = T − T, TR(P) is a bounded nilpotent operator and TN(P) is a regular right-Fredholm linear relation. We claim now that TN(P) is a bounded regular right invertible linear relation. Indeed, we have T is almost onto, then there exists δ > 0 such that for all 0 < |λ| < δ, T − λ is onto. Then, N(P) ∩ R(T − λI) = N(P). Let y ∈ N(P) ∩ R(T − λI) = N(P). Then, there exists x ∈ X such that y ∈ Tx − λx. Therefore there exist xR(P) ∈ R(P) and xN(P) ∈ N(P) such that y ∈ TR(P)xR(P) − λxR(P) + TN(P)xN(P) − λxN(P). Then, −TR(P)xR(P) + λxR(P) ∈ −y + TN(P)xN(P) − λxN(P). By using −TR(P)xR(P) + λxR(P) ∈ R(P) and −y + TN(P)xN(P) − λxN(P) ∈ N(P) we deduce that −TR(P)xR(P) + λxR(P) = 0. Then, y ∈ R(TN(P) − λIN(P)). Therefore, TN(P) − λIN(P) is onto. So TN(P) is almost onto. By using TN(P) is regular and by Theorem 2.2, we deduce that TN(P) is onto. Finally, according to TN(P) ∈ ϕr(N(P)) and Proposition 4.3, we infer that TN(P) is right invertible. (iv) ⇒ (v) : Suppose that there exists a projector P , such that TP −PT = T − T, TR(P) is a nilpotent operator of degree d and TN(P) is onto. As in the proof of Theorem 4.1 we get that (TP)d = T(0). From R(T + P) = R(TN(P)) ⊕ R(TR(P) + I) = N(P) ⊕ R(P) = X, we see that T +P is onto. Now, we claim that N(T +P) is complemented. In- deed, first we note that N(T + P) = N(TN(P)). On another hand, N(TN(P)) is complemented in N(P), then there exists a closed subspace F such that N(TN(P)) ⊕ F = N(P). Hence N(TN(P)) + F + R(P) = X. Let x ∈ (N(TN(P)) + R(P)) ∩ F. Then x = xN(TN(P )) + xR(P) and x = xF . Hence class of browder linear relations 207 xN(TN(P )) + xR(P) = xF . So, xR(P) = xF − xN(TN(P )). By according to xR(P) ∈ R(P) and xF − xN(TN(P )) ∈ N(P), we may deduce that xR(P) = 0 and xF = xN(TN(P )). Therefore xF = 0 and xN(TN(P )) = 0. Thus x = 0. So N(T + P) is complemented in X. Then T + P ∈ ϕr(X) such that T + P is onto. By using Proposition 4.3 we deduce that T + P is right invertible. (v) ⇒ (vi) : As, P is a bounded operator with finite rank, then P is compact. 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