Bull. Iran. Math. Soc. https://doi.org/10.1007/s41980-018-0021-1 ORIGINAL PAPER

Quasicompact and Riesz Composition Endomorphisms of Lipschitz Algebras of Complex-Valued Bounded Functions and Their Spectra Maliheh Mayghani1 · Davood Alimohammadi2

Received: 29 November 2016 / Accepted: 21 October 2017 © Iranian Mathematical Society 2018

Abstract In this paper, we study quasicompact and Riesz composition endomorphisms of Lipschitz algebras of complex-valued bounded functions on metric spaces, not necessarily compact. We give some necessary and some sufficient conditions that a composition endomorphism of these algebras to be quasicompact or Riesz. We also establish an upper bound and a formula for the essential spectral radius of a composition endomorphism T of these algebras under some conditions which implies that T is quasicompact or Riesz. Finally, we get a relation for the set of eigenvalues and the spectrum of a quasicompact and Riesz endomorphism of these algebras. Keywords Essential spectral radius · Lipschitz algebra · Quasicompact endomorphism · Riesz endomorphism · Spectrum Mathematics Subject Classification Primary 46J10; Secondary 47B48 · 47B38

1 Introduction and Preliminaries Let E be an infinite dimensional Banach space. We denote by B(E) and K(E) the set of all bounded linear operators and compact linear operators on E, respectively. It

Communicated by Hamid Reza Ebrahimi Vishki.

B

Davood Alimohammadi [email protected] Maliheh Mayghani [email protected]

1

Department of Mathematics, Payame Noor University (PNU), P.O. Box 19395-3697, Tehran, Iran

2

Department of Mathematics, Faculty of Science, Arak University, Arak 38156-8-8349, Iran

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is known that B(E) with the operator norm is a unital complex Banach algebra and K(E) is a closed ideal of B(E). The essential norm T e of T ∈ B(E) is the norm of T + K(E) in the Calkin algebra B(E)/K(E), i.e., T e = T + K(E) = inf{T − S : S ∈ K(E)}. The essential spectral radius of T is the spectral radius of T + K(E) in B(E)/K(E), i.e., 1

1

re (T ) = lim (T n e ) n = inf{(T n e ) n : n ∈ N}. n→∞

The linear operator T ∈ B(E) is called power compact if T N ∈ K(E) for some N ∈ N, Riesz if re (T ) = 0 and quasicompact if re (T ) < 1. Clearly, T ∈ K(E) if and only if T e = 0, and T is quasicompact if and only if T n e < 1 for some n ∈ N. Moreover, T is quasicompact if T is a Riesz operator and T is Riesz operator if T is power compact. Let A be a unital complex commutative Banach algebra with unit e A . We denote by M A the character space of A, and for f ∈ A we denote by fˆ the Gelfand transform of f . A linear map T : A → A is an endomorphism of A if T ( f g) = (T f )(T g) for all f, g ∈ A. It is known that every endomorphism of A is continuous if A is semisimple. An endomorphism T of A is unital if T (e A ) = e A . If A is semisimple f ◦ ψ, and T : A → A is a unital endomorphism, then T ∗ (M A ) ⊆ M A and Tf =  for each f ∈ A where ψ = T ∗ |M A . In this case ψ = T ∗ |M A is called the self-map of M A associated with T . Feinstein and Kamowitz studied quasicompact and Riesz unital endomorphisms of commutative semisimple unital complex Banach algebras in [3]. They also obtained some results in [4] for semiprime Banach algebras. We recall that a complex algebra A is semiprime if J = {0} is the only ideal in A with J 2 = {0}. It is known that semisimple algebras are semiprime. Let X be a Hausdorff topological space. We denote by C(X ) the set of all complexvalued continuous functions on X . Clearly, C(X ) is a commutative complex algebra that 1 X , the constant function on X with value 1, is the unit of C(X ). The set of all f ∈ C(X ) for which f is bounded, is denoted by C b (X ). It is known that C b (X ) is a unital commutative complex Banach algebra with unit 1 X where equipped with the uniform norm  f  X = sup{| f (x)| : x ∈ X } ( f ∈ C b (X )). Let A be a unital complex subalgebra of C b (X ) which is a Banach algebra under an algebra norm  · . For each x ∈ X , the map δx : A → C defined by δx ( f ) = f (x) ( f ∈ A), is an element of M A . This fact, implies that A is semisimple. Moreover, the map E X : X → M A defined by E X (x) = δx (x ∈ X ), is continuous. Clearly, E X is injective if A separates the points of X . If ϕ : X → X is a continuous self-map of X such that f ◦ ϕ ∈ A for each f ∈ A, then the map T : A → A defined by T f = f ◦ ϕ ( f ∈ A), is a unital endomorphism of A which is called the composition

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endomorphism of A induced by ϕ. Recall that if T : A → A is the composition endomorphism of A induced by a continuous self-map ϕ : X → X , then T n : A → A is the composition endomorphism of A induced by the self-map ϕn : X → X , where ϕn is the nth iterate of ϕ. We also consider ϕ0 as the identity map on X . Let (X, d) and (Y, ρ) be metric spaces such that X and Y have infinitely many points. A map ϕ : X → Y is called a Lipschitz mapping from (X, d) into (Y, ρ) if there exists a constant C such that ρ(ϕ(x), ϕ(y)) ≤ Cd(x, y) for all x, y ∈ X . The constant Lipschitz of ϕ is denoted by p(ϕ) and defined by p(ϕ) = sup{

ρ(ϕ(x), ϕ(y)) : x, y ∈ X, x = y}. d(x, y)

A map ϕ : X → Y is called a supercontractive mapping from (X, d) into (Y, ρ) if for each ε > 0, there exists δ > 0 such that ρ(ϕ(x), ϕ(y))/d(x, y) < ε whenever x, y ∈ X with 0 < d(x, y) < δ. Let (X, d) be a metric space such that X has infinitely many points. A self-map ϕ : X → X is called a Lipschitz mapping on (X, d) if ϕ is a Lipschitz mapping from (X, d) into (X, d). In this case, p(ϕn ) ≤ p(ϕ)n for each n ∈ N and p(ϕm+n ) ≤ 1 1 p(ϕm ) p(ϕn ) for all m, n ∈ N. Hence, limn→∞ p(ϕn ) n exists and limn→∞ p(ϕn ) n = 1 inf{ p(ϕn ) n : n ∈ N} by [2, Proposition A.1.26(iii)]. A function f : X → C is called a complex-valued Lipschitz function on (X, d) if f is a Lipschitz mapping from (X, d) to the Euclidean metric space C. We denote by p(X,d) ( f ) the constant f (y)| : x, y ∈ X, x = y}. For α ∈ Lipschitz of f , i.e., p(X,d) ( f ) = sup{ | f (x)− d(x,y) (0, 1], the formula d α (x, y) = (d(x, y))α defines a new metric d α on X such that the generated topology on X by d α coincides with the generated topology on X by d. We denote by Lip(X, d α ) the set of all complex-valued bounded Lipschitz functions on (X, d α ). Then Lip(X, d α ) is a complex subalgebra of C b (X ) containing 1 X . Moreover, Lip(X, d α ) is a commutative unital complex Banach algebra under the d α -Lipschitz norm  f Lip(X,d α ) =  f  X + p(X,d α ) ( f ) ( f ∈ Lip(X, d α )). Note that the infiniteness of X implies that Lip(X, d α ) is infinite dimensional. A function f : X → C is called a supercontractive function on (X, d) if f is a supercontractive mapping from (X, d) into the Euclidean space C. For α ∈ (0, 1), the set of all bounded supercontractive functions on (X, d α ) is denoted by lip(X, d α ). Clearly, lip(X, d α ) is a complex subalgebra of Lip(X, d α ) containing 1 X and closed in (Lip(X, d α ), ·Lip(X,d α ) ). So (lip(X, d α ), ·Lip(X,d α ) ) is a unital commutative complex Banach algebra. It is clear that Lip(X, d β ) ⊆ lip(X, d α ) where 0 < α < β ≤ 1. The algebras Lip(X, d α ) for α ∈ (0, 1] and lip(X, d α ) for α ∈ (0, 1) are called Lipschitz algebra and little Lipschitz algebra of order α on (X, d), respectively. These algebras were first introduced by Sherbert in [12,13]. Note that Lipschitz algebras and little Lipschitz algebras are semisimple and so semiprime. It is clear that if ϕ : X → X is a Lipschitz mapping on (X, d), then f ◦ϕ ∈ Lip(X, d α ) ( f ◦ϕ ∈ lip(X, d α ), respectively) for each f ∈ Lip(X, d α ) ( f ∈ lip(X, d α ), respectively). If x ∈ X and δ > 0,

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then the function f x,δ : X → C, defined by f x,δ (t) = max{0, (δ − d(x, t))δ} (t ∈ X ), belongs to Lip(X, d 1 ). This fact implies that A separates the points of X and so the map E X : X → M(A), defined by E X (x) = δx (x ∈ X ), is injective, where A = Lip(X, d α ) for α ∈ (0, 1] or A = lip(X, d α ) for α ∈ (0, 1). Kamowitz and Scheinberg in [9] characterized compact unital endomorphisms of Lip(X, d α ) for α ∈ (0, 1] and lip(X, d α ) for α ∈ (0, 1), whenever (X, d) is compact. Golbaharan and Mahyar in [6] provided a complete description of linear operators of Banach spaces with range in Lipschitz algebras on compact metric spaces and obtained necessary and sufficient conditions to ensure boundedness and (weak) compactness of these operators. Jiménez-Vargas and Villegas-Vallecillos in [7] studied compact composition operators of Banach spaces of scalar-valued bounded Lipschitz functions on metric spaces, not necessarily compact. Behrouzi in [1] studied quasicompact and Riesz unital endomorphisms of Lip(X, d α ) for α ∈ (0, 1] and lip(X, d α ) for α ∈ (0, 1) whenever (X, d) is compact. Mahyar and Sanatpour studied quasicompact unital endomorphisms of analytic Lipschitz algebras on compact plane sets in [10]. Jiménez-Vargas, Lacruz and Villegas-Vallecillos in [8] studied the essential norm of composition operators of Banach spaces of Hölder functions on pointed compact metric spaces and gave a formula for the essential norm. Golbaharan and Mahyar in [5] generalized some results of [1], studied essential spectral radius of a unital endomorphism of Lip(X, d α ) for α ∈ (0, 1] and lip(X, d α ) for α ∈ (0, 1) and obtained a formula for the essential spectral radius whenever (X, d) is a compact metric space. Sanatpour investigated quasicompact composition operators on certain classes of Lipschitz algebras and obtained certain properties of power-contractive self-maps of compact plane sets in [11]. In this paper, many results of [5] have been generalized by omitting the compactness condition of considered metric spaces. The main idea of this work is based on the corresponding results in [5] and [7].

2 Quasicompact and Riesz Composition Endomorphisms We first give a necessary condition for a composition endomorphism of A to be quasicompact, where A is either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1), and (X, d) is a connected metric space, not necessarily compact. For this purpose, we need the interesting result of Golbaharan and Mahyar [6, Lemma 2.7] which is also valid for Lip(X, d α ), α ∈ (0, 1], whenever (X, d) is a metric space, not necessarily compact, as follows. Lemma 2.1 Let (X, d) be a metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Suppose that x, y ∈ X . Then δx − δ y  =

2d α (x, y) , 2 + d α (x, y)

where δx and δ y are regarded as elements of dual space A∗ . In particular, δx −δ y  ≤ 1 if and only if d α (x, y) ≤ 2. Moreover, if δx − δ y  ≤ 1 then d α (x, y) ≤ 2δx − δ y .

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Theorem 2.2 Let (X, d) be a connected metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and the operator T : A → A be the composition endomorphism of A induced by ϕ. If T is quasicompact, then limn→∞ diam(ϕn (X )) = limn→∞ p(ϕn ) = 0, where ϕn is the nth iterate of ϕ. Proof Let T be quasicompact. Since (X, d) is a connected metric space, we deduce that the only idempotents in A are 0 and 1 X . So M A with the Gelfand topology is a connected topological space by the Shilov idempotent theorem [2, Theorem 2.4.33]. Let ψ : M A → M A be the associated self-map of M A with T . By [3, Theorem 1.2], ψ has a unique fixed point θ0 in M A and the sequence {T n }∞ n=1 converges in operator norm to the rank-one unital endomorphism S0 of A defined by f (θ0 )1 X ( f ∈ A). S0 f =  Moreover, for each ε (0 < ε ≤ 1), there exists a positive integer m such that ψm satisfies the inclusion ψm (M A ) ⊆ {θ ∈ M A : θ − θ0  <

ε }. 4

(2.1)

Since Tf =  f ◦ ψ and T f = f ◦ ϕ for each f ∈ A, it follows that ψ(δx ) = δϕ(x) . By repeating this process, we conclude that ψn (δx ) = δϕn (x) for all n ∈ N. Since δϕm (x) − δϕm (y)  = ψm (δx ) − ψm (δ y ) ≤ ψm (δx ) − θ0  + ψm (δ y ) − θ0  ε < < 1, 2 we deduce that d α (ϕm (x), ϕm (y)) ≤ 2δϕm (x) − δϕm (y)  < ε, for all x, y ∈ X by Lemma 2.1. Therefore, (diam(ϕm (X )))α < ε. Since ϕn (X ) ⊆ ϕm (X ) for all n ≥ m, it follows that limn→∞ diam(ϕn (X )) = 0. To prove the next assertion of the theorem we take m large enough so that diam(ϕm (X )) < 1. Now let x ∈ X and n > m. (i) Let A = Lip(X, d α ), where α ∈ (0, 1]. We define f : ϕm (X ) → R by f (t) = d α (t, ϕn (x)). It is clear that f ∈ Lip(ϕm (X ), d α ) and  f Lip(ϕm (X ),d α ) ≤ 2. Let F be the Sherbert extension of f to all of X by [13, Proposition 1.4]. Clearly, F ∈ A and FLip(X,d α ) ≤ 2. (ii) Let A = lip(X, d α ), where α ∈ (0, 1). We take β > α and define f : ϕm (X ) → R by f (t) = d β (t, ϕn (x)). Clearly  f ϕm (X ) ≤ (diam(ϕm (X )))β ≤ 1 and |d β (t, ϕn (x)) − d β (t , ϕn (x))| d β (t, t ) | f (t) − f (t )| = ≤ = 1, d β (t, t ) d β (t, t ) d β (t, t ) for all t, t ∈ ϕm (X ), with t = t . Hence, f ∈ Lip(ϕm (X ), d β ) and p(ϕm (X ),d β ) ≤ 1. By Sherbert extension theorem, there exists F ∈ Lip(X, d β ) such that F|ϕm (X ) = f ,

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F X ≤ 1 and p(X,d β ) (F) ≤ 1. Therefore, FLip(X,d β ) ≤ 1 + 1 = 2. Moreover, F ∈ lip(X, d α ), since Lip(X, d β ) ⊆ lip(X, d α ). On the other hand, 

 |F(t) − F(t )| ∈ X, t

= t : t, t d α (t, t )  α α |F(t) − F(t )| β 1− β = 1 + sup |F(t) − F(t )| : t, t ∈ X, t = t d β (t, t )

FLip(X,d α ) ≤ 1 + sup

α

≤ 1 + ( p(X,d β ) (F)) β (2F X ) ≤1+2 ≤ 3.

1− βα

1− βα

Consequently the following inequalities hold for both cases (i) and (ii), whenever x, y ∈ X with x = y: 3T n − S0  ≥ FLip(X,d α ) T n − S0  ≥ T n (F) − S0 (F)Lip(X,d α )  0 )1 X Lip(X,d α ) ≥ p(X,d α ) (F ◦ ϕn ) = F ◦ ϕn − F(θ ≥

|F(ϕn (x)) − F(ϕn (y))| | f (ϕn (x)) − f (ϕn (y))| = . α d (x, y) d α (x, y)

Therefore, in the case (i) we have

d α (ϕn (x), ϕn (y)) ≤ 3T n − S0  and in the case d α (x, y)

d β (ϕn (x), ϕn (y)) ≤ 3T n − S0 . Since β (β > α) is arbitrary, we may d α (x, y) assume that β → α. Hence for both cases we conclude that ( p(ϕn ))α ≤ 3T n − S0   for all n > m. Since limn→∞ T n − S0  = 0, it follows that limn→∞ p(ϕn ) = 0. (ii) we have

We now give a sufficient condition for the quasicompactness of a composition endomorphism of A, where A is either Lip(X, d α ) for α ∈ (0, 1] or A = lip(X, d α ) for α ∈ (0, 1), and (X, d) is a metric space, not necessarily compact. Theorem 2.3 Let (X, d) be a metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and the operator T : A → A be the composition endomorphism of A induced by ϕ. If diam(ϕn 0 (X )) < ∞, p(ϕn 0 ) < 1 and ϕn 0 has a fixed point in X for some α n 0 ∈ N, then re (T ) ≤ limn→∞ ( p(ϕn )) n and T is quasicompact. Proof Let diam(ϕn 0 (X )) < ∞, p(ϕn 0 ) < 1 and ϕn 0 has a fixed point in X . Since ϕn 0 is a Lipschitz mapping on (X, d) and p(ϕn 0 ) < 1, we deduce that {x ∈ X : ϕn 0 (x) = x} is a singleton. Assume that {x ∈ X : ϕn 0 (x) = x} = {x0 }.

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We define the map S0 : A → A by S0 f = f (x0 )1 X ( f ∈ A). Clearly, S0 ∈ K(A). Let n ∈ N and f ∈ A. Choose k, l ∈ N such that kn 0 = n +n 0 +l. Then for each x ∈ X we have |(T n+n 0 f )(x) − (S0 f )(x)| = | f (ϕn+n 0 (x)) − f (x0 )| = | f (ϕn+n 0 (x)) − f (ϕkn 0 (x0 ))| = | f (ϕn (ϕn 0 (x))) − f (ϕn (ϕn 0 +l (x0 )))|

≤ p(X,d α ) ( f )d α (ϕn (ϕn 0 (x)), ϕn (ϕn 0 +l (x0 ))) ≤ p(X,d α ) ( f )( p(ϕn ))α d α (ϕn 0 (x), ϕn 0 (ϕl (x0 ))) ≤ p(X,d α ) ( f )( p(ϕn ))α (diam(ϕn 0 (X )))α .

Hence, T n+n 0 f − S0 f  X ≤ p(X,d α ) ( f )( p(ϕn ))α (diam(ϕn 0 (X )))α .

(2.2)

Since | f (ϕn+n 0 (x)) − f (ϕn+n 0 (y))| ≤ p(X,d α ) ( f )d α (ϕn+n 0 (x), ϕn+n 0 (y)) ≤ p(X,d α ) ( f )( p(ϕn ))α ( p(ϕn 0 ))α d α (x, y), for all x, y ∈ X with x = y, we deduce that p(X,d α ) (T n+n 0 f − S0 f ) ≤ p(X,d α ) ( f )( p(ϕn ))α ( p(ϕn 0 ))α .

(2.3)

From (2.2) and (2.3) we have (T n+n 0 − S0 ) f Lip(X,d α ) ≤ M f Lip(X,d α ) , where M = ( p(ϕn ))α [(diam(ϕn 0 (X )))α + ( p(ϕn 0 ))α ]. This implies that T n+n 0 − S0  ≤ ( p(ϕn ))α [(diam(ϕn 0 (X )))α + ( p(ϕn 0 ))α ]. Therefore, T n+n 0 e ≤ ( p(ϕn ))α [(diam(ϕn 0 (X )))α + ( p(ϕn 0 ))α ], and then 1

α

1

(T n+n 0 e ) n+n0 ≤ ( p(ϕn )) n+n0 [(diam(ϕn 0 (X )))α + ( p(ϕn 0 ))α ] n+n0 .

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This implies that α

1

lim (T n+n 0 e ) n+n0 ≤ lim ( p(ϕn )) n ,

n→∞

n→∞

and so, α

re (T ) ≤ lim ( p(ϕn )) n . n→∞

1

1

1

Since limn→∞ p(ϕn ) n = inf{ p(ϕn ) n : n ∈ N} ≤ p(ϕn 0 ) n0 < 1, we deduce that 

re (T ) < 1 and hence T is quasicompact. Corollary 2.4 Let (X, d) be a metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and the operator T : A → A be the composition endomorphism of A induced by ϕ. If diam(ϕn 0 (X )) < ∞, ϕn 0 has a fixed point in X for some n 0 ∈ N and 1 limn→∞ ( p(ϕn )) n = 0, then T is Riesz. In the following examples we give some noncompact metric spaces and Lipschitz mappings satisfying the conditions of Theorem 2.3 and Corollary 2.4. Example 2.5 Let (X, ·) be a nontrivial complex normed space and d be the induced metric on X by the norm  · . Suppose that x0 ∈ X with x0  = 1 and define the map ϕ : X → X by ϕ(x) =

1

2 (1 − x)x 0

0

x ∈ X, x < 1, x ∈ X, x ≥ 1.

Then (X, d) is a noncompact connected metric space, ϕ is a Lipschitz mapping on (X, d), p(ϕ) < 1, diam(ϕ(X )) ≤ 1 and Fix(ϕ) = { 13 x0 }. Example 2.6 Let X = R and d be the Euclidean metric on X . Define the map ϕ : X → X by ⎧ x ⎨−2 ϕ(x) = −x ⎩ −1

x < 0, 0 ≤ x < 1, x ≥ 1.

Then ϕ is a Lipschitz mapping on (X, d), p(ϕ) = 1 and diam(X ) = ∞. Note that diam(ϕ(X )) = ∞ too. Moreover, ϕ2 (x) =

p(ϕ2 ) =

1 2

123

⎧ ⎨ −1 ⎩

x 2 1 2

x < −2, − 2 ≤ x < 1, x ≥ 1,

< 1, diam(ϕ2 (X )) < ∞ and Fix(ϕ2 ) = {0}.

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Example 2.7 Let a > 0, R > 1, X = (−Ra, −a) ∪ (−a, a) ∪ (a, Ra) and d be the Euclidean metric on X . Then (X, d) is an unconnected noncompact metric space. Define the map ϕ : X → X by ϕ(x) =

1 x (x ∈ X ). R

Then ϕ is a Lipschitz mapping on (X, d), p(ϕ) = and Fix(ϕ) = {0}.

1 R

< 1, diam(ϕ(X )) = 2a < ∞

is its completion and (Y, ρ) is a We recall that if (X, d) is a metric space and ( X , d) complete metric space, then every Lipschitz mapping ϕ from (X, d) into (Y, ρ) has a

into (Y, ρ) such that p( Lipschitz extension ϕ from ( X , d) ϕ ) = p(ϕ) [14, Proposition x ∈ X and {xn }∞ 1.7.1]. In fact ϕ ( x ) = limn→∞ ϕ(xn ), where n=1 is a sequence in X

x in ( X , d). such that limn→∞ xn = Note that if (X, d) is a metric space, then ( X , d α ) is the completion of (X, d α ) for each α ∈ (0, 1].

= lip(

= Lip( X , d α ), respecIn the rest of this paper we assume that A X , d α ) ( A α α tively), if A = Lip(X, d ) (A = lip(X, d ), respectively). The following lemma is useful and its proof is straightforward.

be the completion of (X, d) and A Lemma 2.8 Let (X, d) be a metric space, ( X , d) be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Then the following statements hold.

such that f  (i) If f ∈ A, then there exists a unique f ∈ A f |X = f ,  X =  f X

α and p( ( f ) = p ( f ). α (X,d ) X ,d )

and f = F| X , then f ∈ A and (ii) If F ∈ A f = F.

= { (iii) A f : f ∈ A}.

is an isometrical isomorphism from (iv) The map f  f :A→A

α (A,  · Lip(X,d ) ) onto ( A,  · Lip( X ,d α ) ). The following result is an immediate consequence of Lemma 2.8.

be the completion of (X, d) and Proposition 2.9 Let (X, d) be a metric space, ( X , d) α α A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d ) for α ∈ (0, 1).

: A

→ A

defined by T

(i) If T : A → A is a linear mapping and T f = Tf

( f ∈ A), then T is well-defined and a linear mapping.

→ A

is a linear mapping and Sˇ : A → A defined by Sˇ = S (ii) If S : A f | X ( f ∈ A), then Sˇ is a well-defined linear mapping. Moreover, Sˇ = S.

be the completion of (X, d) and Definition 2.10 Let (X, d) be a metric space, ( X , d) α α A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d ) for α ∈ (0, 1). For a linear mapping

: A

→ A

defined by T

T : A → A, we call the map T f = Tf ( f ∈ A), the extension

of T to A.

be the completion of (X, d), Proposition 2.11 Let (X, d) be a metric space, ( X , d) α α A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d ) for α ∈ (0, 1) and T : A → A be a linear mapping. Then the following statements hold.

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(i) (ii) (iii) (iv) (v) (vi)

n = (T

)n for each n ∈ N. T

is bounded and T

 = T . If T is bounded, then T

. If T is bounded, then T is bounded and T  = T

T is compact if and only if T is compact.

is an endomorphism of A

if and only if T is an endomorphism of A. T

T is unital if and only if T is unital.

by Proof Define the map : A → A

( f ) = f ( f ∈ A). By part (iv) of Lemma 2.8, is a unital isometrical isomorphism from (A, ·Lip(X,d α ) )

 · Lip( onto ( A, X ,d α ) ). It is clear that

= ◦ T ◦ −1 . T This implies that (i), (ii), (iii), (iv), (v) and (vi) hold.



Applying Propositions 2.9 and 2.11, we obtain the following result.

be the completion of (X, d) Proposition 2.12 Let (X, d) be a metric space, ( X , d) α and A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Then the following statements hold.

= {T

: T ∈ B(A)}. (i) B( A)

= {T

: T ∈ K(A)}. (ii) K( A)

be the completion of (X, d) Proposition 2.13 Let (X, d) be a metric space, ( X , d) and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let T : A → A

: A

→ A

be the extension of T to A.

Then the following be a linear mapping and T statements hold. (i) (ii) (iii) (iv)

)n e = T n e for each n ∈ N. (T

) = re (T ). re (T

T is quasicompact if and only if T is quasicompact.

is Riesz if and only if T is Riesz. T

− Proof Let n ∈ N. Clearly, (R − S) = R S for all linear mappings R, S : A → A. Considering this fact and parts (i), (ii) of Proposition 2.11, we obtain n −

n − S = T S, T n − S = (T n − S)  = T for each S ∈ K(A). This implies that (i) holds by part (ii) of Proposition 2.12. According to (i) and applying the formula of essential spectral radius, (ii) follows. Clearly, (iii) and (iv) hold by (ii). 

According to [5, Theorem 2.6] and part (ii) of Proposition 2.13, we generalize [5, Theorem 2.6] as the following.

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the completion of Theorem 2.14 Let (X, d) be a metric space such that ( X , d), α (X, d), is compact. Let α ∈ (0, 1), A be either Lip(X, d ) or lip(X, d α ), the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. If there is a decomposition of X into a finite number of mutually disjoint clopen subsets, say Y1 , Y2 ,…, Ym such that, for each ϕn i (Yi ) ⊆ Yi and p( ϕn i |Yi ) < 1, then i ∈ {1, . . . , m}, there exists n i ∈ N with ϕnn i |Yi )α/(nn i ) : i ∈ {1, . . . , m}}. re (T ) = max{limn→∞ p(

: A

−→ A

be the extension of T to A.

Clearly, T

is the composition Proof Let T

induced by endomorphism of A ϕ . Suppose that there exists a decomposition of X into a finite number of mutually disjoint clopen subsets, say Y1 , . . . , Ym , such that, for ϕn i (Yi ) ⊆ Yi and p( ϕn i )|Yi ) < 1. By each i ∈ {1, . . . , m}, there exists n i ∈ N with [5, Theorem 2.6],

) = max{ lim p( re (T ϕnn i |Yi )α/(nn i ) : i ∈ {1, . . . , m}}. n→∞

) by part (ii) of Proposition 2.13. Hence, the proof is On the other hand, re (T ) = re (T complete. 

By part (iii) of Proposition 2.13 we generalize [5, Theorem 2.10], as the following.

the completion of (X, d), Theorem 2.15 Let (X, d) be a metric space such that ( X , d), is compact. Let A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1), the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Then T is quasicompact if and only if there is a decomposition of X into a finite number of mutually disjoint clopen subsets, say Y1 , Y2 ,…, Ym such that, for each i ∈ {1, . . . , m}, there exists n i ∈ N with ϕn i |Yi ) < 1.

ϕn i (Yi ) ⊆ Yi and p(

: A

−→ A

be the extension of T to A.

Then the quasicompactness of T Proof Let T

is equivalent to the quasicompactness of T by part (iii) of Proposition 2.13. Since T

is the composition endomorphism of A induced by ϕ , the quasicompactness of T is equivalent to the property that there exists a decomposition of X into a finite number of mutually disjoint clopen subsets, say Y1 , . . . , Ym such that, for each i ∈ {1, . . . , m}, ϕn i (Yi ) ⊆ Yi and p( ϕn i )|Yi ) < 1 by [5, Theorem 2.10]. there exists n i ∈ N with Hence, the proof is complete. 

Now we give other sufficient conditions for the quasicompactness of a composition endomorphism of Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1), where (X, d) is a metric space not necessarily compact.

be the completion of (X, d) and Theorem 2.16 Let (X, d) be a metric space, ( X , d) α A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Suppose that the self-map ϕ : X → X is a Lipschitz mapping on (X, d) and T : A → A is the composition endomorphism of A induced by ϕ. If ϕn 0 (X ) is totally bounded in

and p(ϕn 0 ) < 1 for some n 0 ∈ N, then re (T ) ≤ limn→∞ ( p(ϕn )) αn and T is ( X , d) quasicompact.

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and p(ϕn 0 ) < 1 for some n 0 ∈ N. Let Proof Let ϕn 0 (X ) be totally bounded in ( X , d)

X , d). n ∈ N and define Yn = ϕnn 0 (X ), where ϕnn 0 (X ) is the closure of ϕnn 0 (X ) in (

Then Yn is a nonempty compact set in ( X , d) and Yn+1 ⊆ Yn . We claim that lim diam(Yn ) = 0.

n→∞

(2.4)

Let n ∈ N. Since

(n+1)n 0 (x), ϕ(n+1)n 0 (y)) = d(ϕ(n+1)n 0 (x), ϕ(n+1)n 0 (y)) d(ϕ ≤ p(ϕnn 0 )diam(ϕn 0 (X )) = p(ϕnn 0 )diam(Y1 ) ≤ ( p(ϕn 0 ))n diam(Y1 ), for all x, y ∈ X , we deduce that diam(ϕ(n+1)n 0 (X )) ≤ ( p(ϕn 0 ))n diam(Y1 ). Hence, diam(Yn+1 ) ≤ ( p(ϕn 0 ))n diam(Y1 ).

(2.5)

Since p(ϕn 0 ) < 1 and (2.5) holds for each n ∈ N, we conclude that (2.4) holds. and it is, in fact, a singleton by the Cantor intersection Therefore, ∞ n=1 Yn is nonempty  y}. We claim that y ∈ Fix( ϕn 0 ). Since ϕn 0 ( y) ∈ theorem. Assume that ∞ n=1 Yn := { ϕn 0 (Yn ) and ϕn 0 (ϕnn 0 (X )) ⊆ ϕn 0 (ϕnn 0 (X )) = ϕ(n+1)n 0 (X ) = Yn+1 , ϕn 0 (Yn ) =  y) ∈ ∞ y} and so our claim is justified. for each n ∈ N, we deduce that ϕn 0 ( n=2 Yn = {

: A

→ A

be the extension of T to A.

By part (v) of Proposition 2.11, T

Let T

It is easy to see that T

is an endomorphism of A. f = f ◦ ϕ for each f ∈ A where

Since

ϕ: X → X is the unique Lipschitz extension of ϕ on ( X , d). ϕ: X → X is a

diam( X )) = diam(Y1 ) < ∞, p( ϕn 0 ) = p(ϕn 0 ) < 1 Lipschitz mapping of ( X , d), ϕn 0 ( α

) ≤ limn→∞ ( p(

X , we conclude that re (T ϕn )) n and T and ϕn 0 has a fixed point in is quasicompact by Theorem 2.3. Sinceα p( ϕn ) = p(ϕn ) by [14, Proposition 1.7.1], it follows that re (T ) ≤ limn→∞ ( p(ϕn )) n and T is quasicompact by parts (ii) and (iii) of Proposition 2.13. 

be the completion of (X, d) and Corollary 2.17 Let (X, d) be a metric space, ( X , d) A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Suppose that the self-map ϕ : X → X is a Lipschitz mapping on (X, d) and T : A → A is the

X , d) composition endomorphism of A induced by ϕ. If ϕn 0 (X ) is totally bounded in ( 1

for some n 0 ∈ N and limn→∞ ( p(ϕn )) n = 0, then T is Riesz. The following example shows that in Theorem 2.16 it is not necessary to assume that ϕn 0 has a fixed point in X for some n 0 ∈ N.

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Example 2.18 Suppose that X = {x ∈ R : 0 < x < 1} and d is the Euclidean metric on X . Define the map ϕ : X → X by ϕ(x) =

x (x ∈ X ). 2

Then ϕ is a Lipschitz mapping on (X, d), p(ϕ) = 21 and ϕ(X ) = (0, 21 ) is totally

where bounded in ( X , d), X = [0, 1] and d is the Euclidean metric on X . Hence, the conditions of Theorem 2.16 hold, although ϕn has no fixed point for each n ∈ N. Lemma 2.19 Let (X, d) be a metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Suppose that the self-map ϕ : X → X is a Lipschitz mapping on (X, d) and T : A → A is the composition endomorphism of A induced by ϕ. Let j ∈ N∪{0}, A j = Lip(ϕ j (X ), d α ) if A = Lip(X, d α ) and A j = lip(ϕ j (X ), d α ) if A = lip(X, d α ). Let T j : A j → A j be the composition endomorphism of A j induced by ϕ|ϕ j (X ) . Then re (T j ) = re (T ). Proof Take M = 2 max{1, ( p(ϕ j ))α }. Suppose that S : A → A is a compact linear operator. Let f ∈ A j and F ∈ Lip(X, d α ) be the Sherbert extension of F to all of X by [13, Proposition 1.4]. Then FLip(X,d α ) ≤ 2 f Lip(ϕ j (X ),d α ) . and so F ◦ ϕ j Lip(X,d α ) ≤ M f Lip(ϕ j (X ),d α ) .

(2.6)

Moreover, for each x, y ∈ X , | f (ϕ j (x)) − f (ϕ j (y))| |(F ◦ ϕ j )(x) − (F ◦ ϕ j )(y)| ≤ d α (x, y) d α (x, y) | f (ϕ j (x)) − f (ϕ j (y))| ( p(ϕ j ))α . ≤ d α (ϕ j (x), ϕ j (y)) Hence if f ∈ Lip(ϕ j (X ), d α ) ( f ∈ lip(ϕ j (X ), d α ), respectively), then F ◦ ϕ j ∈ Lip(X, d α ) (F ◦ ϕ j ∈ lip(X, d α ), respectively). Therefore, if f ∈ A j then F ◦ ϕ j ∈ A and so S(F ◦ ϕ j )|ϕ j (X ) ∈ A j . It is easy to see that if f ∈ A j and F, G ∈ Lip(X, d α ) with F|ϕ j (X ) = G|ϕ j (X ) = f , then F ◦ ϕ j = G ◦ ϕ j . Define the map S M, j : A j → A j by S M, j f =

1 S(F ◦ ϕ j )|ϕ j (X ) ( f ∈ A j ), M

where F ∈ Lip(X, d α ), F|ϕ j (X ) = f and FLip(X,d α ) ≤ 2 f Lip(ϕ j (X ),d α ) . Thus, S M, j is well-defined by Sherbert’s extension theorem and it is a linear operator. Moreover, S M, j is compact by (2.6) and the compactness of S. Let n ∈ N with n > j. We show that (T j )n − S M, j  ≤ M T n− j − S.

(2.7)

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Let f ∈ A j and F be the Sherbert extension of f to all of X by [13, Proposition 1.4]. Since for each x ∈ X we have       1 |((T j )n ( f ) − S M, j ( f ))(x)| = (T j )n ( f (x)) − S F ◦ ϕ j (x) M       1 = (T j )n− j ( f (ϕ j (x))) − S F ◦ ϕ j (x) M         1 1 =  M T n− j F ◦ ϕ j (x) − S F ◦ ϕ j (x) M M      1 n− j ≤ F ◦ ϕj  − S) (M T  , M X

it follows that      1 n− j  ((T j ) − S M, j )( f ) X ≤ (M T − S) F ◦ ϕj   . M n

X

On the other hand |((T j )n − S M, j ) f (x) − ((T j )n − S M, j ) f (y)|       1 1 F ◦ ϕ j (x) − S F ◦ ϕ j (x) =  M T n− j M M       1 1 F ◦ ϕ j (y) − S F ◦ ϕ j (y) − M T n− j M M    1 ≤ p(X,d α ) (M T n− j − S) F ◦ ϕj d α (x, y), M for all x, y ∈ X with x = y. Hence,    1 p(ϕ j (X ),d α ) ((T j )n − S M, j ) f ) ≤ p(X,d α ) ((M T n− j − S) . (2.8) F ◦ ϕj M Thus ((T j ) − S M, j )( f )Lip(ϕ j (X ),d α ) n

     1 n− j  F ◦ ϕj  ≤ (M T − S)  M

Lip(X,d α )

1 M T n− j − SF ◦ ϕ j )Lip(X,d α ) ≤ M ≤ M T n− j − S f Lip(ϕ j (X ),d α ) , by (2.6) and so (2.7) holds. By (2.7), we have (T j )n e ≤ M T n− j − S. Hence, (T j )n e ≤ M T n− j e = MT n− j e .

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(2.9)

Bull. Iran. Math. Soc.

Since (2.9) holds for each n ∈ N with n > j, we deduce that 1

1

1

re (T j ) = lim ((T j )n e ) n ≤ lim M n (T n− j e ) n . n→∞

n→∞

Therefore, re (T j ) ≤ re (T ).

(2.10)

Let S : A j → A j be a compact linear operator. Suppose that f ∈ A. It is clear that f |ϕ j (X ) ∈ Lip(ϕ j (X ), d α ) if f ∈ Lip(X, d α ) and f |ϕ j (X ) ∈ lip(ϕ j (X ), d α ) if f ∈ lip(X, d α ). Let F ∈ Lip(X, d α ) be the Sherbert extension of S ( f |ϕ j (X ) ) to all of X by [13, Proposition 1.4], then F ◦ ϕ j  X ≤ F X and S ( f |ϕ j (X ) )(x) − S ( f |ϕ j (X ) )(y) |F ◦ ϕ j (x) − F ◦ ϕ j (y)| = , d α (x, y) d α (x, y) for each x, y ∈ X with x = y. Hence, F ◦ ϕ j ∈ Lip(X, d α ) if f ∈ Lip(X, d α ) and F ◦ ϕ j ∈ lip(X, d α ) if f ∈ lip(X, d α ). Moreover, we can show that if f ∈ A and F, G ∈ Lip(X, d α ) with F|ϕ j (X ) = G|ϕ j (X ) = S ( f |ϕ j (X ) ), then F ◦ ϕ j = G ◦ ϕ j . Define the map S : A → A by S f = F ◦ ϕ j ( f ∈ A), where F ∈ Lip(X, d α ) with F|ϕ j (X ) = S ( f |ϕ j (X ) ) and FLip(X,d α ) ≤ 2S ( f |ϕ j (X ) )Lip(ϕ j (X ),d α ) . Then S is well-defined and it is a linear operator. We now show that S is compact. ∞ Let { f n }∞ n=1 be a bounded sequence in A. Then { f n |ϕ j (X ) }n=1 is a bounded sequence in A j . The compactness of S : A j −→ A j implies that there exist a subsequence ∞ { f n k |ϕ j (X ) }∞ k=1 of { f n |ϕ j (X ) }n=1 and a function f 0 in A j such that lim S ( f n k |ϕ j (X ) ) − f 0 Lip(ϕ j (X ),d α ) = 0.

k→∞

(2.11)

For each k ∈ N, let Fn k ∈ Lip(X, d α ) be the Sherbert extension of S ( f n k |ϕ j (X ) ) to all of X and F0 ∈ Lip(X, d α ) be the Sherbert extension of f 0 to all of X by [13, Proposition 1.4]. Let x ∈ X . Then |S( f n k )(x) − F0 ◦ ϕ j (x)| = |Fn k ◦ ϕ j (x) − F0 ◦ ϕ j (x)| = |S ( f n k |ϕ j (X ) )(ϕ j (x)) − f 0 (ϕ j (x))| ≤ S ( f n k |ϕ j (X ) ) − f 0 ϕ j (X ) ≤ MS ( f n k |ϕ j (X ) ) − f 0 ϕ j (X ) .

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Hence, S( f n k ) − F0 ◦ ϕ j  X ≤ MS ( f n k |ϕ j (X ) ) − f 0 ϕ j (X ) . On the other hand, |(S( f n k ) − F0 ◦ ϕ j )(x) − (S( f n k ) − F0 ◦ ϕ j )(y)| = |Fn k (ϕ j (x)) − F0 (ϕ j (x)) − Fn k (ϕ j (y)) − F0 (ϕ j (y))|

= |(S ( f n k |ϕ j (X ) ) − f 0 )(ϕ j (x)) − (S ( f n k |ϕ j (y) ) − f 0 )(ϕ j (y))|

≤ p(ϕ j (X ),d α ) (S ( f n k |ϕ j (X ) ) − f 0 )d α (ϕ j (x), ϕ j (y)) ≤ p(ϕ j (X ),d α ) (S ( f n k |ϕ j (X ) ) − f 0 )( p(ϕ j ))α d α (x, y) ≤ M p(ϕ j (X ),d α ) (S ( f n k |ϕ j (X ) ) − f 0 )d α (x, y), for each x, y ∈ X and so p(X,d α ) (S( f n k ) − F0 ◦ ϕ j ) ≤ M p(ϕ j (X ),d α ) (S ( f n k |ϕ j (X ) ) − f 0 ). Thus, S( f n k ) − F0 ◦ ϕ j Lip(X,d α ) ≤ MS ( f n k |ϕ j (X ) ) − f 0 Lip(ϕ j (X ),d α ) . Hence, by (2.11) we have lim S( f n k ) − F0 ◦ ϕ j Lip(X,d α ) = 0

k→∞

and so our claim is justified. Let n ∈ N with n > j and f ∈ A. We show that (T n − S) f Lip(X,d α ) ≤ M((T j )n− j − S )( f |ϕ j (X ) )Lip(ϕ j (X ),d α ) . Suppose that F is the Sherbert extension of S ( f |ϕ j (X ) ) to all of X . Since |(T n − S)( f )(x)| = |T n f (x) − S f (x)| = | f ◦ ϕn− j (ϕ j (x)) − F ◦ ϕ j (x)| = | f |ϕ j (X ) (ϕn− j (ϕ j (x))) − F(ϕ j (x))| = |(T j )n− j ( f |ϕ j (X ) )(ϕ j (x)) − S ( f |ϕ j (X ) )(ϕ j (x))| ≤ (T j )n− j ( f |ϕ j (X ) ) − S ( f |ϕ j (X ) )ϕ j (X ) ≤ M((T j )n− j − S )( f |ϕ j (X ) )ϕ j (X ) for each x ∈ X , we deduce that (T n − S) f  X ≤ ((T j )n− j − S )( f |ϕ j (X ) )ϕ j (X ) ,

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(2.12)

Bull. Iran. Math. Soc.

Since, f (ϕn (x)) = ( f |ϕ j (X ) ◦ ϕn− j )(ϕ j (x)) for each x ∈ X , we have |(T n − S) f (x) − (T n − S) f (y)| = | f (ϕn (x)) − F(ϕ j (x)) − f (ϕn (y)) − F(ϕ j (y))| = |((T j )n− j − S )( f |ϕ j (X ) )(ϕ j (x)) − ((T j )n− j − S )( f |ϕ j (X ) )(ϕ j (y))| ≤ p(ϕ j (X ),d α ) (((T j )n− j − S )( f |ϕ j (X ) ))d α (ϕ j (x), ϕ j (y)) ≤ p(ϕ j (X ),d α ) (((T j )n− j − S )( f |ϕ j (X ) ))( p(ϕ j ))α d α (x, y) ≤ M p(ϕ j (X ),d α ) (((T j )n− j − S )( f |ϕ j (X ) ))d α (x, y), for each x, y ∈ X and so p(X,d α ) ((T n − S) f ) ≤ ( p(ϕ j ))α p(ϕ j (X ),d α ) (((T j )n− j − S )( f |ϕ j (X ) )). Hence, (2.12) holds. By (2.12), we have T n e ≤ M(T j )n− j − S  since f |ϕ j (X ) ∈ A j and  f |ϕ j (X ) Lip(ϕ j (X ),d α ) ≤  f Lip(X,d α ) , for each f ∈ A. Thus, T n e ≤ M(T j )n− j e .

(2.13)

Since (2.13) holds for each n ∈ N with n > j, we conclude that 1

1

1

re (T ) = lim (T n e ) n ≤ lim M n ((T j )n− j e ) n . n→∞

n→∞

Therefore, re (T ) ≤ re (T j ).

(2.14)

From (2.10) and (2.14), we have re (T j ) = re (T ).



be the completion of (X, d) and Theorem 2.20 Let (X, d) be a metric space, ( X , d) α A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Suppose that j ∈ N ∪ {0} such that ϕ j (X ) is an

and p((ϕ|ϕ j (X ) )n 0 ) < 1 for some n 0 ∈ N. infinite set and totally bounded in ( X , d), α

(i) If α ∈ (0, 1), then re (T ) = limn→∞ ( p(ϕn |ϕ j (X ) ) n and T is quasicompact. 1

(ii) If α = 1, then re (T ) ≤ limn→∞ ( p(ϕn |ϕ j (X ) ) n and T is quasicompact. Proof Let A j = Lip(ϕ j (X ), d α ) if A = Lip(X, d α ) and A j = lip(ϕ j (X ), d α ) if A = lip(X, d α ). Assume that T j : A j → A j is the composition endomorphism of A j

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induced by the Lipschitz mapping ψ = ϕ|ϕ j (X ) : ϕ j (X ) → ϕ j (X ). By Lemma 2.19, we have re (T ) = re (T j ).

(2.15)

Assume that Y = ϕ j (X ) and ρ = d|Y ×Y . Then (Y , d| Y ×Y ) is the completion of (Y, ρ),

and so Y

= Y and ρ

X , d)

= d| which Y is the closure of Y in ( Y ×Y . By part (ii) of Proposition 2.13, we have re (T j ) = re (T j ),

(2.16)

where T j :  Aj →  A j is the extension of T j to  A j . Since Y is totally bounded in ( X , d),

is compact in (

Hence T j is the composition endomorphism of we deduce that Y X , d).

→ Y

on the compact metric space (Y

, ρ 

:Y

). A j induced by the Lipschitz mapping ψ

)n and p(ψ n ) = p(ψn ) for each n ∈ N, we deduce that p((ψ

)n ) = n = (ψ Since ψ

n 0 ) = p(ψn 0 ). Hence, p(ψ

n 0 ) < 1 since p(ψn ) for each n ∈ N. In particular, p(ψ p(ψn 0 ) < 1. Suppose that α ∈ (0, 1). Then α

n ) n < 1, re (T j ) = lim p(ψ n→∞

(2.17)

by Theorem 2.2 and [5, Corollary 2.4(i)]. From (2.15), (2.16) and (2.17), we get α

n ) n < 1. re (T ) = lim p(ψ n→∞

(2.18)

n ) = p(ψn ) and ψn = (ϕ|ϕ j (X ) )n = ϕn |ϕ j (X ) Hence, T is quasicompact. Since p(ψ for each n ∈ N, by (2.18), we deduce that α

re (T ) = lim p(ϕn |ϕ j (X ) ) n . n→∞

Suppose that α = 1. Then

n ) n < 1, re (T j ) ≤ lim p(ψ 1

n→∞

(2.19)

by Proposition 2.9 and [5, Corollary 2.4(i)]. From (2.15), (2.16) and (2.19), we deduce that

n ) n < 1. re (T ) ≤ lim p(ψ 1

n→∞

This implies that T is quasicompact and 1

re (T ) ≤ lim p(ϕn |ϕ j (X ) ) n . n→∞

Hence, the proof is complete.

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Note that Theorem 2.20 extends Theorem 2.2, Proposition 2.9 and [5, Corollary 2.4(i)].

be the completion of (X, d) and Corollary 2.21 Let (X, d) be a metric space, ( X , d) α α A be either Lip(X, d ) for α ∈ (0, 1] or lip(X, d ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Suppose that j ∈ N ∪ {0} such that ϕ j (X ) is an

infinite set and totally bounded in ( X , d). 1

(i) If limn→∞ ( p(ϕn |ϕ j (X ) )) n = 0, then T is Riesz. 1

(ii) If 0 < α < 1 and T is Riesz, then limn→∞ ( p(ϕn |ϕ j (X ) )) n = 0. Note that Corollary 2.21 is a generalization of [5, Corollary 2.5]. We now generalize Theorem 2.2 for possibly unconnected X . For this purpose, we shall need Lemma 2.1 and a result due to Feinstein and Kamowitz [4, Theorem 3.2]. Theorem 2.22 Let (X, d) be a metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. If T is quasicompact, then there is a decomposition of X into a finite number of mutually disjoint clopen subsets, say X 1 , X 2 ,…, X m such that, for each i ∈ {1, . . . , m}, ϕ N (X i ) ⊆ X i for some N ∈ N and limn→∞ diam(ϕn (X i )) = limk→∞ p(ϕk N | X i ) = 0. Proof Suppose that T is quasicompact. Since (A, ·Lip(X,d α ) ) is a unital commutative semiprime Banach algebra, by [4, Theorem 3.2], there exists N ∈ N such that the quasicompact unital endomorphism T N of A has the following properties. (i) The eigenspace of T N corresponding to eigenvalue 1 is a finite-dimensional, unital subalgebra of A isomorphic to Cm for some m ∈ N and hence spanned by m orthogonal idempotents, say e1 , . . . , em . (ii) Set Ai = ei A (i ∈ {1, . . . , m}). Then (under an equivalent norm) each Ai is a unital commutative semiprime complex Banach algebra, with identity ei and A=

m 

Ai .

i=1

(iii) For each i ∈ {1, . . . , m}, T N | Ai is a quasicompact unital endomorphism of Ai . The sequence {T k N | Ai }∞ k=1 converges in operator norm to a rank-1 unital endomorphism of Ai , say Si . (iv) The sequence {T k N }∞ k=1 converges in operator norm to a rank-m unital endomorphism S of A given by Sg =

m 

Si (ei g) (g ∈ A).

i=1

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We define X i = {y ∈ X : ei (y) = 1},

i ∈ {1, . . . , m}.

and ei = χ X i 1 X , where χ X i is Then, for each i ∈ {1, . . . , m}, X i is a clopen set in X m X i , X i ∩ X j = ∅ if i, j ∈ N the characteristic function of X i . Moreover, X = i=1 with 1 ≤ i < j ≤ m and ϕ N (X i ) ⊆ X i for each i ∈ {1, . . . , m}. Let i ∈ {1, . . . , m}. We claim that Ai is semisimple. Let ei f ∈ rad(Ai ), where f ∈ A and rad(Ai ) is the radical of Ai . If x ∈ X \ X i , then (ei f )(x) = ei (x) f (x) = 0 f (x) = 0. If x ∈ X i , then δx | Ai ∈ M Ai and so (ei f )(x) = δx (ei f ) = δx | Ai (ei f ) = 0. Hence, (ei f )(x) = 0 for each x ∈ X and so ei f = 0. Therefore, rad(Ai ) ⊆ {0} and so our claim is justified. Suppose that  · i is the norm on Ai which is equivalent to the  · Lip(X,d α ) that obtained from (ii). Then there exist positive constants M1 and M2 such that M1 ei f i ≤ ei f Lip(X,d α ) ≤ M2 ei f i , for all f ∈ A. Since Ai is a unital commutative semisimple complex Banach algebra with identity ei and Si : Ai → Ai is a rank-1 unital endomorphism, there exists θi ∈ M Ai such that Si (ei f ) = θi (ei f )ei ( f ∈ A). Let x ∈ X i . Then δx | Ai ∈ M Ai , (δx | Ai ) ◦ Si ∈ M Ai and (δx | Ai ) ◦ Si = θi . Sine for each k ∈ N, we have (δϕk N (x) )| Ai (ei f ) = (ei f )(ϕk N (x)) = (T k N (ei f ))(x) = δx (T k N (ei f )) = (δx ◦ T k N )| Ai (ei f ) for each f ∈ A, we conclude that δx ◦ T k N | Ai = δϕk N (x) | Ai for each k ∈ N. Therefore, for each k ∈ N we have |(δϕk N (x) | Ai − θi )(ei f )| = |(δx ◦ T k N | Ai )(ei f ) − (δx | Ai ◦ Si )(ei f )| = |(T k N | Ai (ei f ) − Si (ei f ))(x)| ≤ T k N | Ai (ei f ) − Si (ei f ) X ≤ (T k N | Ai − Si )(ei f )Lip(X,d α ) ≤ M2 (T k N | Ai − Si )(ei f )i

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≤ M2 T k N | Ai − Si (ei f )i , for all f ∈ A. This implies that for each k ∈ N we have δϕk N (x) | Ai − θi  ≤ M2 T k N | Ai − Si .

(2.20)

Let 0 < ε ≤ 1 be given. Since {T k N | Ai }∞ k=1 converges in operator norm to Si , there exists ki ∈ N such that T ki N | Ai − Si  <

εM1 . 4M2 ei Lip(X,d α )

(2.21)

εM1 . 4ei Lip(X,d α )

(2.22)

From (2.20) and (2.21) we have δϕki N (x) | Ai − θi  < Let x, y ∈ X i . Since |(δϕki N (x) − δϕki N (y) )(F)| = |F(ϕki N (x)) − F(ϕki N (y))| = |ei F(ϕki N (x)) − ei F(ϕki N (y))| = |(δϕki N (x) | Ai − δϕki N (y) | Ai )(ei F)| ≤ δϕki N (x) | Ai − δϕki N (y) | Ai ei Fi 1 δϕki N (x) | Ai − δϕki N (y) | Ai ei FLip(X,d α ) M1 ei Lip(X,d α ) ≤ δϕki N (x) | Ai − δϕki N (y) | Ai FLip(X,d α ) , M1 ≤

for all F ∈ A, it follows that δϕki N (x) − δϕki N (y)  ≤

ei Lip(X,d α ) δϕki N (x) | Ai − δϕki N (y) | Ai . M1

(2.23)

By (2.22), (2.23) and Lemma 2.1, we deduce that d α (ϕki N (x), ϕki N (y)) ≤ 2δϕki N (x) − δϕki N (y)  2ei Lip(X,d α ) (δϕki N (x) | Ai − θi  + δϕki N (y) | Ai − θi ) M1 2ei Lip(X,d α ) εM1 < = ε. M1 2ei Lip(X,d α )

≤

Therefore (diam(ϕki N (X i )))α < ε. Since ϕn (X i ) ⊆ ϕki N (X i ) for all n ≥ ki N , it follows that limn→∞ diam(ϕn (X i )) = 0.

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To prove the next assertion of the theorem we take ki large enough so that diam(ϕki N (X i )) < 1. Now let x ∈ X i and k > ki . (i) Let A = Lip(X, d α ), where α ∈ (0, 1]. We define f : ϕki N (X i ) → R by f (t) = d α (t, ϕk N (x)). It is clear that f ∈ Lip(ϕki N (X i ), d α ) and  f Lip(ϕki N (X i ),d α ) ≤ 2. Let F be the Sherbert extension of f to all of X by [13, Proposition 1.4]. Clearly, F ∈ A and FLip(X,d α ) ≤ 2. (ii) Let A = lip(X, d α ), where α ∈ (0, 1). We take β > α and define f : ϕki N (X i ) → R by f (t) = d β (t, ϕk N (x)). Clearly  f ϕki N (X i ) ≤ 1, f ∈ Lip(ϕki N (X i ), d β ) and p(ϕk N (X i ),d β ) ≤ 1. By Sherbert extension theorem, there exists F ∈ Lip(X, d β ) i such that F|ϕki N (X i ) = f , F X ≤ 1 and p(X,d β ) (F) ≤ 1. Therefore, FLip(X,d β ) ≤ 1 + 1 = 2. Moreover, F ∈ Lip(X, d α ), since Lip(X, d β ) ⊆ lip(X, d α ). On the other hand, one can see that FLip(X,d α ) ≤ 3. Hence, for both cases (i) and (ii), 1 ei FLip(X,d α ) M1 ei Lip (X, d α ) ≤ FLip(X,d α ) M1 3ei Lip (X, d α ) ≤ . M1

ei Fi ≤

Consequently the following inequalities hold for both cases (i) and (ii), whenever x, y ∈ X i and x = y: 3M2 ei Lip(X,d α ) k N T | Ai − Si  ≥ M2 ei Fi T k N | Ai − Si  M1 ≥ M2 T k N | Ai (ei F) − Si (ei F)i ≥ T k N | Ai (ei F) − Si (ei F)Lip(X,d α ) ≥ p(X,d α ) (ei F ◦ ϕk N ) |F(ϕk N (x)) − F(ϕk N (y))| ≥ d α (x, y) | f (ϕk N (x)) − f (ϕk N (y))| . = d α (x, y) Therefore, in the case (i) we have 3M2 ei Lip(X,d α ) k N d α (ϕk N (x), ϕk N (y)) ≤ T | Ai − Si , d α (x, y) M1 and in the case (ii) we have 3M2 ei Lip(X,d α ) k N d β (ϕk N (x), ϕk N (y)) ≤ T | Ai − Si . d α (x, y) M1

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Since β (β > α) is arbitrary, we may assume that β → α. Hence, for both cases we conclude that ( p(ϕk N | X i ))α ≤

3M2 ei Lip(X,d α ) k N T | Ai − Si , M1

for all k > ki . Since limk→∞ T k N | Ai − Si  = 0, it follows that limk→∞ p(ϕk N | X i ) = 0. 

3 Spectra of Quasicompact and Riesz Endomorphisms Let X be a nonempty set, m ∈ N and ϕ : X → X be a self-map of X . Then a point x0 ∈ X is called a fixed point of ϕ of order n ∈ N if ϕn (x0 ) = x0 and ϕk (x0 ) = x0 for all k = 1, . . . , n − 1. We denote by σ p (T ) and σ (T ) the set of eigenvalues and the spectrum of a linear operator T on a Banach space E, respectively. The following result is a modification of [7, Proposition 2.7] which is useful in the sequel. Theorem 3.1 Let (X, d) be a metric space and A be either Lip(X, d α ) for α ∈ (0, 1] or lip(X, d α ) for α ∈ (0, 1). Let the self-map ϕ : X → X be a Lipschitz mapping and T : A → A be the composition endomorphism of A induced by ϕ. If ϕ has a fixed point of order n ∈ N and λ ∈ C satisfies λn = 1, then λ ∈ σ (T ). Golbaharan and Mahyar obtained a relation for the spectrum and the set of eigenvalues of a quasicompact endomorphism of Lipschitz algebras on compact metric spaces [5, Theorem 2.12]. Applying Theorem 3.1, we obtain a modification of their result as the following. Theorem 3.2 Let (X, d) be a compact metric space, α ∈ (0, 1) and A be either Lip(X, d α ) or lip(X, d α ). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Suppose that T is quasicompact and L is the set of all n ∈ N such that ϕ has a fixed point of order n. Then L is finite and    n (i) σ p (T ) ⊆ {λ ∈ C : |λ| ≤ re (T )} ∪ {λ ∈ C : λ = 1} , n∈L  n {λ ∈ C : λ = 1} ⊆ σ (T ), (ii) n∈L  (iii) σ (T ) ⊆ {λ ∈ C : |λ| ≤ re (T )} ∪ ( {λ ∈ C : λn = 1}). n∈L

Proof Clearly, (ii) holds by Theorem 3.1. According to the proof of [5, Theorem 2.10], there exists a finite number of mutually disjoint clopen subsets of X , say X 1 , X 2 ,…, X m , with union X such that for each i ∈ {1, 2, . . . , m}, there exists n i ∈ N with ϕn i (X i ) ⊆ X i and limk→∞ p(ϕkn i | X i ) = 0. Let i ∈ {1, 2, . . . , m}. It is easy to see that diam(ϕ(k+1)n i (X i )) ≤ p(ϕkn i | X i )diam(ϕn i (X i )).

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∞ Hence, limk→∞ diam(ϕkn i (X i )) = 0. Therefore, ∞ k=1 ϕkn i (X i ) is a singleton by the Cantor intersection theorem. Assume that k=1 ϕkn i (X i ) = {xi }. Since ϕn i (xi ) ∈  ∞ k=1 ϕkn i (X i ), it follows  that ϕn i (xi ) = xi . One can see that if yi is a fixed point of ϕn i | X i , then yi ∈ ∞ k=1 ϕkn i (X i ) = {x i }. Hence x i is the unique fixed point of ϕn i | X i . This implies that L is finite. Let λ ∈ σ p (T ) with λn = 1 for all n ∈ L and li be the order of xi for ϕ whenever i ∈ {1, 2, . . . , m}. There is 0 = f ∈ A such that f ◦ ϕ = λ f , and so, f (xi ) = ( f ◦ ϕli )(xi ) = λli f (xi ) for each i ∈ {1, . . . , m}. Thus, f (xi ) = 0 for each i ∈ {1, 2, . . . , m}. Since f = 0, there exists a point x ∈ X i for some i ∈ {1, 2, . . . , m} such that f (x) = 0. The rest of the proof is the same as in the 

proof of [5, Theorem 2.12] by putting n i li instead of n. Remark 3.3 If X is connected in Theorem 3.2, then L = {1}. Theorem 3.4 Let (X, d) be a metric space, α ∈ (0, 1) and A be either Lip(X, d α ) or lip(X, d α ). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Suppose that ϕ j (X )

for some j ∈ N ∪ {0} and L is the set of all n ∈ N such is totally bounded in ( X , d) that ϕ has a fixed point of order n. If T is quasicompact, then L is finite and    (i) σ p (T ) ⊆ {λ ∈ C : |λ| ≤ re (T )} ∪ {λ ∈ C : λn = 1} , n∈L  (ii) {λ ∈ C : λn = 1} ⊆ σ (T ), n∈L    {λ ∈ C : λn = 1} . (iii) σ (T ) ⊆ {λ ∈ C : |λ| ≤ re (T )} ∪ n∈L

Proof Clearly, (ii) holds by Theorem 3.1. Suppose that 0 = λ ∈ σ p (T ). Then f ◦ ϕ = λ f for some 0 = f ∈ A. We set Y = ϕ j (X ), ρ = d|Y ×Y and A j = Lip(Y, ρ α ) if x0 ∈ X is a fixed point of A = Lip(X, d α ) or A j = lip(Y, ρ α ) if A = lip(X, d α ). If

ϕ of order k, for some k ∈ N, then ϕk j ( x0 ) = lim ϕk j (xn ),

x0 = n→∞

where, {xn }∞ x0 ) = 0. Hence, x0 ∈ Y , n=1 is a sequence in X such that lim n→∞ d(x n ,

which Y is the closure of Y in ( X , d). This implies that L = {n ∈ N : ϕ |Y has a fixed point of order n}.

(3.1)

Let S : A j → A j be the composition endomorphism of A j induced by ψ = ϕ|Y . Then S is quasicompact and re (S) = re (T ),

(3.2)

by Lemma 2.19. Let g = f |Y . Then g ∈ A j and λg = g ◦ ψ. Moreover, since f (y0 ) = 0 for some y0 ∈ X and g(ϕ j (y0 )) = f ◦ ϕ j (y0 ) = λ j f (y0 ),

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we deduce that g = 0, and so λ ∈ σ p (S). We first assume that Y is finite. Then (Y, ρ) is a compact metric space, A j = C b (Y ) = Lip(Y, ρ) and S is a compact operator. Hence, by [7, Theorem 1.4], λ∈



{λ ∈ C : λn = 1},

(3.3)

n∈L 0

has a fixed point of order n} is a finite set. Since Y = Y , we where, L 0 = {n ∈ N : ψ

. Hence, L = L 0 by (3.1). Therefore, L is finite and deduce that ϕ |Y = ϕ|Y = ψ = ψ  σ p (T ) ⊆ {0} ∪



 {λ ∈ C : λ = 1} , n

n∈L

by (3.3) and so (i) holds. As in the proof of [5, Theorem 2.12], one can show that if λ = 0 and ker(λI − T ) = {0}, then λ ∈ / σ (T ). Hence, (iii) holds.

We now assume that Y is infinite. It is easy to see that (Y , d| Y ×Y ) is the completion



, ρ of (Y, ρ), and so Y = Y and ρ

= d|Y ×Y . Let A j = Lip(Y , ρ

α ) or  A j = lip(Y

α ) and S: Aj →  A j be the extension of S to  A j . Then S is quasicompact and S) = re (S), re (

(3.4)

= by Proposition 2.13. One can show that g = 0 and λ g = g◦ψ S( g ). Thus, S) and so λ ∈ σ p ( σ p (T ) \ {0} ⊆ σ p ( S).

(3.5)

we deduce that (Y

, ρ Since Y is totally bounded in ( X , d),

) is a compact metric space. Hence, by Theorem 3.2, we get ⎛ σ p ( S) ⊆ {λ ∈ C : |λ| ≤ re ( S)} ∪ ⎝



⎞ {λ ∈ C : λn = 1}⎠ ,

(3.6)

n∈L 1

has a fixed point of order n} is a finite set. It is easy to see where L 1 = {n ∈ N : ψ

that ψ = ϕ |Y . Hence, L1 = L ,

(3.7)

by (3.1). Therefore, L is finite and  σ p (T ) ⊆ {λ ∈ C : |λ| ≤ re (T )} ∪



 {λ ∈ C : λ = 1} , n

n∈L

by (3.2), (3.4), (3.5), (3.6) and (3.7), and so (i) holds.

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As in the proof of [5, Theorem 2.12], one can show that if |λ| > re (T ) and ker(λI − T ) = {0}, then λ ∈ / σ (T ). Hence,  σ (T ) ⊆ {λ : |λ| ≤ re (T )} ∪



 {λ ∈ C : λ = 1} . n

n∈L



Thus, (iii) holds.

Corollary 3.5 Let (X, d) be a metric space, α ∈ (0, 1) and A be either Lip(X, d α ) or lip(X, d α ). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Suppose that ϕ j (X )

for some j ∈ N ∪ {0} and L is the set of all n ∈ N such is totally bounded in ( X , d) that ϕ has a fixed point of order n.  (i) If T is Riesz, then L is finite and σ (T ) = {0} ∪ ( n∈L {λ ∈ C : λn = 1}). (ii) If T is quasicompact and σ (T ) ⊆ {0} ∪ {λ ∈ C : |λ| = 1}, then T is Riesz. According to Remark 3.3 and the proof of Theorem 3.4, we obtain the following result. Corollary 3.6 Let (X, d) be a metric space, α ∈ (0, 1) and A be either Lip(X, d α ) or lip(X, d α ). Let the self-map ϕ : X → X be a Lipschitz mapping on (X, d) and T : A → A be the composition endomorphism of A induced by ϕ. Suppose that ϕ j (X )

for some j ∈ N ∪ {0}. is an infinite connected totally bounded set in ( X , d) (i) If T is quasicompact, then σ (T ) ⊆ {λ ∈ C : |λ| ≤ re (T )} ∪ {1}. (ii) If T is Riesz then σ (T ) = {0, 1}. Example 3.7 Let X = (0, 1) ∪ (1, 2) and d be the Euclidean metric on X . Define the map ϕ : X → X by ϕ(x) =

1 2 x 2

x ∈ (0, 1), x ∈ (1, 2).

Then ϕ is a Lipschitz mapping, p(ϕ) = 21 , ϕ(X ) is an infinite set and ϕ2 (X ) = { 21 }. Example 3.8 Let X = (0, 21 ) ∪ (1, 2) and d be the Euclidean metric on X . Define the map ϕ : X → X by  ϕ(x) =

1+x 1 4x

  x ∈ 0, 21 , x ∈ (1, 2).

X = [0, 21 ] ∪ [1, 2], d is the Then ϕ is a Lipschitz mapping and p(ϕ) ≤ 25 . Moreover,

is a compact metric space and Euclidean metric on X , ( X , d) ϕ defines by 

ϕ (x) =

123

1+x 1 4x

  x ∈ 0, 21 , x ∈ [1, 2].

Bull. Iran. Math. Soc.

Suppose that Y1 = [0, 21 ] and Y2 = [1, 2]. Then Y1 , Y2 are mutually disjoint clopen subsets of X, ϕ2 (Y1 ) ⊆ Y1 and ϕ2 (Y2 ) ⊆ Y2 . Furthermore, 

ϕ2n+1 (x) =

1 4

1 + 41 + · · · + 41n + 4xn x + 412 + · · · + 41n + 4n+1

x ∈ Y1 , x ∈ Y2 ,

and 

ϕ2n (x) =

+ 412 + · · · + 41n + 4xn 1 1 + 41 + · · · + 4n−1 + 4xn 1 4

x ∈ Y1 , x ∈ Y2 ,

for all n ∈ N. Hence, for each n ∈ N we have p( ϕ2n |Yi ) = 41n where i ∈ {1, 2}. Let α α α ∈ (0, 1), A = Lip(X, d ) or A = lip(X, d ) and T : A → A be the composition endomorphism of A induced by ϕ. Then α

re (T ) = maxi∈{1,2} lim p( ϕ2n |Yi ) 2n = lim ( n→∞

n→∞

1 α ) 2n = 2−α , 4n

by Theorem 2.14 and so T is quasicompact. On the other hand, we can show that Fix( ϕ2n−1 ) = ∅ and Fix( ϕ2n ) = { 13 , 43 } for all n ∈ N. Hence {−1, 1} ⊆ σ (T ) ⊆ {λ ∈ C : |λ| ≤ 2−α } ∪ {−1, 1}, by Theorem 3.4. Acknowledgements The authors would like to thank the referee for her/his very insightful suggestions and comments, in particular, for calling their attention to the Lemma 2.8 of the reference [6], resulted in improving and generalizing the Theorems 2.2 and 2.22.

References 1. Behrouzi, F.: Riesz and quasi-compact endomorphisms of Lipschitz algebras. Houst. J. Math. 36(3), 793–802 (2010) 2. Dales, H.G.: Banach algebras and automatic continuity. In: London Mathematical Society Monograph, vol. 24. The Clarendon Press, Oxford (2000) 3. Feinstein, J.F., Kamowitz, H.: Quasicompact and Riesz endomorphisms of Banach algebras. J. Funct. Anal. 225, 427–438 (2005) 4. Feinstein, J.F., Kamowitz, H.: Quasicompact endomorphisms of commutative semiprime Banach algebras. Banach Center Publ. 91, 159–167 (2010) 5. Golbaharan, A., Mahyar, H.: Essential spectral radius of qusicompact endomorphisms of Lipschitz algebras. Rocky Mt. J. Math. 45(4), 1149–1164 (2015) 6. Golbaharan, A., Mahyar, H.: Linear operators of Banach spaces with range in Lipschitz algebras. Bull. Iran. Math. Soc. 42(1), 69–78 (2016) 7. Jiménez-Vargas, A., Villegas-Vallecillos, M.: Compact composition operators on noncompact Lipschitz spaces. J. Math. Anal. Appl. 398, 221–229 (2013) 8. Jiménez-Vargas, A., Lacruz, M., Villegas-Vallecillos, M. : Essential norm of composition operators on Banach spaces of Hölder functions. In: Abstract and Applied Analysis. Hindawi Publishing Corporation (2011) (article ID 590853) 9. Kamowitz, H., Scheinberg, S.: Some properties of endomorphisms of Lipschitz algebras. Stud. Math. 24(3), 255–261 (1990)

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Bull. Iran. Math. Soc. 10. Mahyar, H., Sanatpour, A.H.: Quasicompact endomorphisms of Lipschitz algebras of analytic functions. Publ. Math. Debr. 76/1–2, 135–145 (2010) 11. Sanatpour, A.H.: Quasicompact composition operators and power-contractive selfmaps. Annal. Funct. Anal. 7(2), 281–289 (2016) 12. Sherbert, D.R.: Banach algebras of Lipschitz functions. Pac. J. Math. 13, 1387–1399 (1963) 13. Sherbert, D.R.: The structure of ideals and point derivations in Banach algebras of Lipschitz functions. Trans. Am. Math. Soc. 111, 240–272 (1964) 14. Weaver, N.: Lipschitz Algebras. World Scientific, River Edge (1999)

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