Hermitian operators on Banach algebras of Lipschitz functions

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 142, Number 10, October 2014, Pages 3469–3481 S 0002-9939(2014)12048-X Article electronically published on May 30, 2014 HERMITIAN OPERATORS ON BANACH ALGEBRAS OF LIPSCHITZ FUNCTIONS FERNANDA BOTELHO, JAMES JAMISON, A. JIM´ ENEZ-VARGAS, AND MOIS´ ES VILLEGAS-VALLECILLOS (Communicated by Thomas Schlumprecht) Abstract. For compact metric spaces (X, d), we show that the Lipschitz spaces Lip(X, d) and the little Lipschitz spaces lip(X, dα)with0<α<1, equipped with the sum norm, support only trivial hermitian operators, that is, real multiples of the identity operator. 1. Introduction Let Abe a complex Banach algebra with unity Iand let A∗be its dual space. Given a∈A, recall that the algebraic numerical range V(a)isgivenby V(a)={F(a): F∈A∗,F=F(I)=1}. An element a∈Ais said to be hermitian if V(a)⊂R.Itisknownthata∈Ais hermitian if and only if exp(ita)=1forallt∈R;see[3]. Let Ebe a complex Banach space and B(E) the Banach algebra of all bounded linear operators on Eequipped with the operator norm. It is well-known that an operator T∈B(E) is hermitian if and only if exp(itT)isanisometryforeach t∈R; see [6, Theorem 5.2.6]. The set of hermitian operators on Eis a real subspace of B(E) which contains all operators of the form λI,whereλis a real number. A hermitian operator is said to be trivial if it is a real multiple of the identity operator. Some important Banach spaces only support trivial hermitian operators, as for example, the Bergman spaces Lp a(Δ) (1 ≤p<∞,p= 2) [9, Corollary 5.4] and the Hardy spaces Hp(Δ) (1 ≤p<∞,p= 2) [1]. Also, the hermitian operators on several spaces of scalar-valued continuous functions defined on the interval [0,1] are known to be just real scalar multiples of the identity. Such spaces include the space of continuously differentiable functions C1[0,1]; the space of absolutely continuous functions AC[0,1]; and the spaces of Lipschitz functions: Lip[0,1] and lip α,0<α<1. We recall that lip αconsists of all period 1 functions on R satisfying |f(x)−f(y)|=o(|x−y|α) uniformly as |x−y|→0; cf. [2, Theorem 3.1]. Received by the editors February 6, 2012 and, in revised form, September 5, 2012; September 13, 2012; and October 9, 2012. 2010 Mathematics Subject Classification. Primary 46E15, 47B15, 47B38. Key words and phrases. Spaces of Lipschitz functions, hermitian operator, derivation, bicircular projection. The third and fourth authors were partially supported by Junta de Andaluc´ıa grants FQM-3737 and FQM-194 and by MICINN project MTM 2010-17687. c 2014 American Mathematical Society Reverts to public domain 28 years from publication 3469 Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 3470 FERNANDA BOTELHO ET AL. In this paper we investigate the hermitian operators on spaces of Lipschitz functions defined on a compact metric space. More precisely, for a compact metric space (X,d) and a positive real parameter α∈(0,1], we consider the space of all α-Lipschitz functions f:X→Csuch that pα(f):=sup|f(x)−f(y)| d(x, y)α:x, y ∈X, x =y<∞, and also the subspace of all α-Lipschitz functions f:X→Csatisfying the additional local flatness condition: lim d(x,y)→0 |f(x)−f(y)| d(x, y)α=0. These two spaces with the standard operations of addition, multiplication and scalar multiplication are complex algebras, and when equipped with the norm fα=pα(f)+f∞ become Banach algebras. These two algebras are denoted by Lip(X,dα)and lip(X,dα), respectively. It is important to observe that Lip(X,dα) and lip(X, dα) are unital semi-simple commutative complex Banach algebras, and lip(X,dα) is a closed subalgebra of Lip(X,dα). Notice that lip(X,d) may contain only constant functions, for example lip[0,1] with the usual metric. When X=[0,1] or X=Twith the usual metrics, Lip(X,dα) and lip(X, dα) are among the classical algebras considered by de Leeuw in [4]. These algebras were first studied by Sherbert in [14,15]. In [2], it was shown that the hermitian operators on the Lipschitz spaces Lip[0,1] and lip α,0<α<1, are real multiples of the identity operator. In this paper we prove that the same property holds for the spaces Lip(X,d) and the spaces lip(X,dα)with0<α<1, for (X,d) a compact metric space. This generalizes the aforementioned result. We also mention the natural connection between hermitian operators and the class of bi-circular projections. A projection Pon a complex Banach space is bicircular if eisP+eit(I−P) is an isometry for all s, t ∈R. This type of projection was introduced by Stach´o and Zalar in [17]. Jamison [10] showed that these projections are exactly the hermitian projections. Our result implies that the only bi-circular projections on Lip(X, d) and lip(X,dα)with0<α<1 are the trivial projections, 0andI. 2. Preliminaries In this section we give a representation for all surjective linear isometries on Lip(X,d) or lip(X, dα)(0<α<1) that fix the constant function everywhere equal to 1. Then we characterize the hermitian elements of Lip(X,d) and lip(X,dα) (0 <α<1). The last result provides a useful description of the continuous linear functionals on both spaces. Throughout this paper (X,d) is a compact metric space, 1Xdenotes the constant function equal to 1 on X,IXrepresents the identity function on Xand Iis the identity operator on Lip(X, d) or lip(X,dα), 0 <α<1. For each x∈X,δxstands for the evaluation functional at the point xdefined on Lip(X, d) or lip(X,dα), 0<α<1. Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use HERMITIAN OPERATORS 3471 Our approach requires that the surjective linear isometries on the spaces Lip(X, d) and lip(X,dα)(0<α<1) have a suitable representation. Rao and Roy [13] proved that any surjective linear isometry of Lip[0,1] can be expressed as a weighted composition operator f→ τf ◦ϕ(f∈Lip[0,1]) where τis a scalar of modulus 1 and ϕ is a surjective isometry of [0,1]. They asked whether every isometry on the Banach spaces Lip(X,d) and lip(X,dα)(0<α<1) are induced by the isometries of the metric space X. Next we derive a characterization of surjective linear isometries on these spaces that fix 1X. This characterization follows from a theorem due to Jarosz in [11], a theorem in [8] (page 144) and a result by Sherbert in [14]. Theorem 2.1. Let Xbe a compact metric space. Then T: Lip(X,d)→Lip(X,d) is a surjective linear isometry such that T(1X)=1 Xif and only if there exists a surjective isometry ϕ:X→Xsuch that Tis of the form T(f)=f◦ϕfor all f∈Lip(X,d). The same characterization holds for a surjective linear isometry T on lip(X, dα)(0<α<1) such that T(1X)=1 X. Proof. It is straightforward to check that an operator Tof the form described in the theorem is a surjective isometry. Then we just prove the reversed implication. Let A(X) represent either Lip(X,d) or lip(X, dα)with0<α<1 and let C(X) be the algebra of continuous complex-valued functions on X.Wefirstobservethat A(X) is a regular subspace of C(X) and the sum norm is a p-norm for the norm on R2given by p(s, t)=|s|+|t|. Let us recall (see [11]) that given a compact Hausdorff space X, a complex linear subspace Aof C(X) that contains the function 1X,is said to be regular if for any ε>0, any x0∈ChAand any open neighborhood Uof x0,thereisanf∈Awith f∞≤1+ε,f(x0) = 1, and |f(x)|<εfor x∈X\U.ChAdenotes the set of extreme points Fof the unit ball of (A, ·∞)∗ such that F(1X) = 1, and we identify ChAwith a subset of X. Suppose that Tis a surjective linear isometry on A(X) such that T(1X)=1 X. An application of the main theorem in [11] to A(X) yields that Tis a surjective isometry on (A(X),·∞). Next we quote a theorem from Hoffman’s book [8, p. 44]: Let Xbe a compact Hausdorff space and let Bbe a complex linear subalgebra of C(X) that contains the function 1X. Suppose that Sis a linear map of Bonto Bsuch that S(f)∞= f∞for all f∈B.IfS(1X)=1 X,thenSis multiplicative. Therefore Tis an automorphism of A(X). By Sherbert’s theorem [14, Corollary 5.2], every automorphism Tof Lip(X,d)thatcarries1 Xinto 1Xis of the form T(f)=f◦ϕ,whereϕ:X→Xis a homeomorphism. Similarly, we can prove that this is also true for those automorphisms of lip(X, dα)(0<α<1) that fix 1X. We now show that ϕis an isometry of X. Observe that given any α∈(0,1], we have pα(T(f)) = pα(f) for all f∈A(X)sinceTis an isometry for both norms ·αand ·∞. For the case A(X)=Lip(X,d), fix y∈Xand define fy:X→Rby fy(z)= d(z,ϕ(y)) for all z∈X. Clearly, fy∈Lip(X,d)andp1(fy)≤1. For any x, y ∈X, we have d(ϕ(x),ϕ(y)) = |fy(ϕ(x)) −fy(ϕ(y))| =|T(fy)(x)−T(fy)(y)| ≤p1(T(fy))d(x, y) ≤d(x, y). Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 3472 FERNANDA BOTELHO ET AL. For the case A(X) = lip(X,dα)(0<α<1), fix x, y ∈X,x=y,choose β∈(α, 1) and define fxy(z)=d(z,ϕ(y))β−d(z,ϕ(x))β 2d(ϕ(x),ϕ(y))β−α,∀z∈X. It is not hard to check that fxy ∈lip(X,dα)andpα(fxy) = 1 (see, for example, [12, p. 62]). An easy calculation gives d(ϕ(x),ϕ(y))α=|fxy(ϕ(x)) −fxy(ϕ(y))| =|T(fxy)(x)−T(fxy)(y)| ≤pα(T(fxy))d(x, y)α =d(x, y)α. In either case we have d(ϕ(x),ϕ(y)) ≤d(x, y) for all x, y ∈X. Since T−1is also a surjective linear isometry on A(X) such that T−1(1X)=1 X, the same argument used above implies the existence of a homeomorphism φ:X→ Xsuch that T−1(f)=f◦φfor all f∈A(X). Therefore d(φ(x),φ(y)) ≤d(x, y)for all x, y ∈X.Givenx∈X,wehave f(ϕ−1(x)) = T(T−1(f))(ϕ−1(x)) = T−1(f)(x)=f(φ(x)) for all f∈A(X). Since A(X) separates the points of X,thisimpliesthatϕ−1=φ. Consequently, ϕis a surjective isometry. This completes the proof of the theorem.  We will next characterize the hermitian elements of the spaces Lip(X,d)and lip(X,dα), 0 <α<1. Lemma 2.2. Let (X, d)be a compact metric space and h∈Lip(X,d)(or lip(X,dα), 0<α<1). Then his a hermitian element in Lip(X,d)(or lip(X,dα))ifandonly if his a real constant function. Proof. Assume that his hermitian in Lip(X, d). Then F(h)∈V(h)⊂Rfor all F∈Lip(X,d)∗such that F=F(1X) = 1. In particular, h(x)=δx(h)∈Rfor all x∈X,andsohis real-valued. Using that ea−eb≤|a−b|exp (max {|a|,|b|}) for all a, b ∈C, we deduce that exp(ih) is a function in Lip(X,d). We also have that, for each t∈R,exp(ith)1=1. Sinceexp(ith)∞= 1, it follows that p1(exp(ith)) = 0. Hence exp(ith) is a constant function on Xfor all t∈Rwhich implies that his constant. Conversely, if his a real constant function, then his a real multiple of 1X. Therefore his hermitian in Lip(X,d). The same proof works for lip(X,dα), 0 < α<1.  Following an idea of de Leeuw [4], we embed the Banach spaces Lip(X,d)and lip(X,dα)(0<α<1) isometrically into some suitable spaces of complex-valued continuous functions. Let Xbe a compact metric space and let  Xbe the set (x, y)∈X2:x=y. It is easy to check that  Xis completely regular; we denote by β Xthe Stone-ˇ Cech compactification of  X.LetC(X∪β X) denote the Banach space of all complexvalued continuous functions on X∪β X, under the norm f=f|X∞+  f|β  X  ∞ (f∈C(X∪β X)), Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use HERMITIAN OPERATORS 3473 and let C0(X∪ X) denote the Banach space of all complex-valued continuous functions on X∪ Xvanishing at infinity, endowed with the norm f=f|X∞+ f| X ∞(f∈C0(X∪ X)). We now recall that the Riesz representation theorem states that the map μ→ Fμ, given by Fμ(f)=X∪β  X fdμ (f∈C(X∪β X)), defines an isometric isomorphism from the Banach space M(X∪β X) of all complexvalued regular Borel measures on X∪β Xequipped with the norm of total variation: μ=|μ|(X∪β X)(μ∈M(X∪β X)) onto the dual space of (C(X∪β X),·∞). Similarly, the map ν→ Gνdefined by Gν(f)=X∪  X fdν (f∈C0(X∪ X)) is an isometric isomorphism from the Banach space M(X∪ X) with the norm ν=|ν|(X∪ X)(ν∈M(X∪ X)) onto the dual space of (C0(X∪ X),·∞). For each f∈Lip(X, d)orf∈lip(X, dα), 0 <α<1, we set  f: X→Cto be the map given by  f(x, y)=f(x)−f(y) d(x, y)α,∀(x, y)∈ X, where α=1whenf∈Lip(X,d). It is easy to show that  fis continuous on  Xand    f  ∞ =pα(f)(0<α≤1). Hence there exists a unique continuous function β f on β Xsuch that (β f) X= fand   β f  ∞ =   f  ∞ .Furthermore,iff∈lip(X,dα), then  fvanishes at infinity on  X. The maps Φ: Lip(X,d)→C(X∪β X)and Ψ: lip(X,dα)→C0(X∪ X), defined by (1) Φ(f)(w)=⎧ ⎨ ⎩ f(w)ifw∈X, β f(w)ifw∈β X, and (2) Ψ(f)(w)=⎧ ⎨ ⎩ f(w)ifw∈X,  f(w)ifw∈ X, are isometric linear embeddings from Lip(X,d) with the norm ·1into C(X∪β X), and from lip(X, dα) with the norm ·αinto C0(X∪ X), respectively. The Hahn–Banach theorem and the Riesz representation theorem yield the following lemma. Lemma 2.3. Let (X,d)be a compact metric space. Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 3474 FERNANDA BOTELHO ET AL. (1) For each F∈Lip(X, d)∗,thereexistsμ∈M(X∪β X)with F≤μ satisfying F(f)=X∪β  X Φ(f)(w)dμ(w),∀f∈Lip(X, d). (2) Let α∈(0,1).ForeachG∈lip(X, dα)∗,thereexistsν∈M(X∪ X)with G≤νsuch that G(f)=X∪  X Ψ(f)(w)dν(w),∀f∈lip(X,dα). Proof. Let F∈Lip(X,d)∗. The functional T: Φ(Lip(X, d)) →C, defined by T(Φ(f)) = F(f) for all f∈Lip(X, d), is linear, continuous and T=F.Bythe Hahn–Banach theorem, there exists a linear continuous functional  T:C(X∪β X)→ Csuch that  T(Φ(f)) = T(Φ(f)) for all f∈Lip(X,d)and   T  =T. Since g≤2g∞for all g∈C(X∪β X), it follows that the linear functional  Tis continuous on the space C(X∪β X) equipped with the norm ·∞.Wedenote by ·∗ ∞the norm on the dual Banach space of C(X∪β X),·∞.BytheRiesz representation theorem, there exists μ∈M(X∪β X) with    T   ∗ ∞ =μsatisfying  T(g)=X∪β  X g(w)dμ(w),∀g∈C(X∪β X). Since g∞≤gfor all g∈C(X∪β X), we have    T  ≤   T   ∗ ∞ ,andsoF≤μ. Moreover, F(f)=T(Φ(f)) =  T(Φ(f)) = X∪β  X Φ(f)(w)dμ(w) for all f∈Lip(X,d), as we wanted. Similarly, we prove statement (2).  Such a μis called a representing measure for F(analogously, νfor G). We should note that a representing measure for For Gis not always determined uniquely. 3. The main result In this section we describe all the hermitian operators on Lip(X, d) or lip(X,dα) with 0 <α<1. We proceed with the statement and proof of our main result. Theorem 3.1. Let (X,d)be a compact metric space. A bounded linear operator T: Lip(X,d)→Lip(X,d)is hermitian if and only if Tis a real multiple of the identity operator on Lip(X, d). An analogous assertion holds for T: lip(X,dα)→ lip(X,dα)with 0<α<1. Before proving this theorem we set notation and prove some preliminary lemmas. Let A(X) denote either Lip(X, d) or lip(X,dα), 0 <α<1. Recall that α=1in the case A(X)=Lip(X,d). Lemma 3.2. If T:A(X)→A(X)is a hermitian bounded linear operator, then the following statements hold: (i) There exists λ∈Rsuch that T(1X)=λ1X. (ii) For each t∈R,exp(it(T−λI)) is a surjective linear isometry on A(X) fixing 1X. Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use HERMITIAN OPERATORS 3475 (iii) For each t∈R, there exists a surjective isometry ϕton Xsuch that exp(it(T−λI))(f)(x)=f(ϕt(x)),∀f∈A(X),∀x∈X. (iv) {ϕt}t∈Ris a one-parameter group of surjective isometries on Xsuch that, for each x∈X,themapt→ ϕt(x)from Rto Xis continuous. (v) For every f∈A(X), (3) lim t→0(f◦ϕt−f)(x)=0,∀x∈X, and (4) lim t→0 (f◦ϕt−f)(x)−(f◦ϕt−f)(y) d(x, y)α=0,∀(x, y)∈ X. Proof. (i) For each F∈A(X)∗with F=F(1X) = 1, define ΦF:B(A(X)) →C by ΦF(S)=F(S(1X)),∀S∈B(A(X)). It is easy to check that ΦFis a linear functional on B(A(X)), and since |ΦF(S)|=|F(S(1X))|≤FS(1X)α≤S1Xα=S, for all S∈B(A(X)), then ΦFis continuous and ΦF≤1. Moreover, ΦF(I)= F(1X) = 1; hence ΦF≥|ΦF(I)|=1andthusΦF=Φ F(I)=1. Since T∈B(A(X)) is hermitian, it follows that F(T(1X)) = ΦF(T)∈V(T)⊂R for all F∈A(X)∗such that F=F(1X) = 1. This means that T(1X)isa hermitian element in A(X). Then, according to Lemma 2.2, there exists λ∈R such that T(1X)=λ1X. (ii) By (i), we have (T−λI)(1X) = 0 and so exp(it(T−λI))(1X)=1 Xfor all t∈R. Indeed, since exp(it(T−λI)) = I+ ∞  n=1 intn(T−λI)n, it follows that exp(it(T−λI))(1X)=1 X+ ∞  n=1 intn(T−λI)n(1X)=1 X. Since Tand λI are hermitian operators in B(A(X)), it is easily seen that so is T−λI. Indeed, using the fact that exp(it(T−λI)) = exp(itT)exp(−itλI) for all t∈R,wehave 1=1Xα=exp(it(T−λI))(1X)α≤exp(it(T−λI)) ≤exp(itT)exp(−itλI)=1. Therefore, for each t∈R,exp(it(T−λI)) is a linear isometry from A(X)onto itself, fixing 1X. (iii) In view of (ii), by applying Theorem 2.1, for each t∈Rthere exists a surjective isometry ϕton Xsuch that (5) exp(it(T−λI))(f)(x)=f(ϕt(x)),∀f∈A(X),∀x∈X. Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 3476 FERNANDA BOTELHO ET AL. (iv) Using the fact that A(X) separates the points of X, it is easily derived that ϕ(s+t)=ϕs◦ϕtfor all s, t ∈Rand ϕ0=IX. More precisely, given f∈A(X)and x∈X,wehave f(ϕs+t(x)) = exp(i(s+t)(T−λI))(f)(x) =exp(i(t+s)(T−λI))(f)(x) =exp(it(T−λI))exp(is(T−λI))(f)(x) =exp(it(T−λI))(exp(is(T−λI))(f))(x) =exp(is(T−λI))(f)(ϕt(x)) =f(ϕs(ϕt(x))) =f(ϕs◦ϕt(x)) and f(ϕ0(x)) = exp(i0(T−λI))(f)(x)=exp(0)(f)(x)=I(f)(x)=f(x). We next prove that for each x∈X,themapt→ ϕt(x)fromRto Xis continuous. Note first that δ:X→A(X)∗defined by δ(x)=δxis a Lipschitz bijection from (X,dα)ontoδ(X). Indeed, δis injective since A(X) separates points; and given x, y ∈X,wehave |(δ(x)−δ(y)) (f)|=|f(x)−f(y)|≤fαd(x, y)α for all f∈A(X). Hence δ(x)−δ(y)≤d(x, y)α.SinceXis compact, we deduce that δ−1:δ(X)→Xis continuous. Notice that δ−1(δx)=xfor all x∈X. Fix x∈X. The maps t→ exp(it(T−λI)) from Rto B(A(X)), U→ U∗ from B(A(X)) to B(A(X)∗)andS→ S(δ(x)) from B(A(X)∗)toA(X)∗are clearly continuous. From (5), we deduce that (6) (exp(it(T−λI)))∗(δ(x)) = δ(ϕt(x)) (t∈R,x∈X). Since ϕt(x)=δ−1(exp(it(T−λI)))∗(δ(x)) (t∈R,x∈X), we conclude that t→ ϕt(x)fromRto Xis continuous. (v) Let f∈A(X). Given x∈X, we have limt→0(f◦ϕt)(x)=(f◦ϕ0)(x)=f(x) by (iv), and thus limt→0(f◦ϕt−f)(x)=0. Usingthis,for(x, y)∈ X, we deduce that lim t→0 (f◦ϕt−f)(x)−(f◦ϕt−f)(y) d(x, y)α=0.  We recall that for f∈A(X)themap  f: X→Cis defined to be  f(x, y)=(f(x)−f(y))/d(x, y)α. We recall that β Xrepresents the Stone-ˇ Cech compactification of  X. This entails that every bounded, continuous and scalar-valued map defined on  Xhas a unique continuous extension to β X. Lemma 3.3. If f∈A(X),then (7) lim t→0β ft(w)=0,∀w∈β X, where, for each t∈R,ftdenotes the function f◦ϕt−f. Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use HERMITIAN OPERATORS 3477 Proof. We define g:[−1,1] × X→Cby g(t, (x, y)) = (f◦ϕt−f)(x)−(f◦ϕt−f)(y) d(x, y)α. The function gis continuous and bounded. In fact, we have |g(t, (x, y))|≤pα(f)+pα(f)=2pα(f) for all t∈[−1,1] and (x, y)∈ X. For the continuity of g, define σ:[−1,1] × X→ A(X)∗by σ(t, (x, y)) = δ(ϕt(x)) −δ(x)−(δ(ϕt(y)) −δ(y)) d(x, y)α and notice that g(t, (x, y)) = σ(t, (x, y))(f)(t∈[−1,1],(x, y)∈ X). Taking into account the equality (6), for any t, s ∈[−1,1] and x, y ∈X,wehave δ(ϕt(x)) −δ(ϕs(y))=(exp(it(T−λI)))∗(δ(x)) −(exp(is(T−λI)))∗(δ(y)) ≤(exp(it(T−λI)))∗δ(x)−δ(y) +(exp(it(T−λI)))∗−(exp(is(T−λI)))∗δ(y) ≤d(x, y)+exp(it(T−λI)) −exp(is(T−λI))δ(y). Let us recall now that if Aand Bare bounded commuting operators on a Banach algebra, then exp(iA)−exp(iB)≤A−Bexp (max {A,B}). Applying this formula to A=t(T−λI)andB=s(T−λI), we obtain (8) exp(it(T−λI)) −exp(is(T−λI))≤|t−s|k, where k=T−λIexp(T−λI)isaconstant,andso δ(ϕt(x)) −δ(ϕs(y))≤d(x, y)α+k|t−s|. Therefore, for every t, s ∈[−1,1] and x, y ∈X,wehave δ(ϕt(x)) −δ(x)−(δ(ϕs(y)) −δ(y))≤2d(x, y)α+k|t−s|. Hence the mapping (t, (x, y)) → δ(ϕt(x)) −δ(x), defined on [−1,1] × Xand with values in A(X)∗, is continuous. Since (x, y)→ d(x, y)αfrom  Xto Ris continuous, it follows that σis continuous. Hence, given ε>0and(t0,(x0,y 0)) ∈[−1,1] ×  X, there is a neighborhood Vof (t0,(x0,y 0)) such that if (t, (x, y)) ∈V,then σ(t, (x, y)) −σ(t0,(x0,y 0))<ε/(1 + fα). Therefore, for every (t, (x, y)) ∈V, we have |g(t, (x, y)) −g(t0,(x0,y 0))|<ε 1+fα fα<ε, and this proves that gis continuous. Licensed to University de Almeria. Prepared on Thu Jul 31 21:03:52 EDT 2014 for download from IP 150.214.156.17. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use