# The Fixed Point Property of a Banach Algebra Generated by an Element with Infinite Spectrum

Authors: P. Thongin; W. Fupinwong
Journal: Journal of Function Spaces
 (2018)

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## Abstract

A Banach spaceX is said to have the fixed point property if for each nonexpansive mapping T:E→E on a bounded closed convex subset E of X has a fixed point. Let X be an infinite dimensional unital Abelian complex Banach algebra satisfying the following: (i) condition (A) in Fupinwong and Dhompongsa, 2010, (ii) if x,y∈X is such that τx≤τy, for each τ∈Ω(X), then x≤y, and (iii) inf⁡{r(x):x∈X,x=1}>0. We prove that there exists an element x0 in X such that 〈x0〉R=∑i=1kαix0i:k∈N,αi∈R¯ does not have the fixed point property. Moreover, as a consequence of the proof, we have that, for each element x0 in X with infinite spectrum and σ(x0)⊂R, the Banach algebra 〈x0〉=∑i=1kαix0i:k∈N,αi∈C¯ generated by x0 does not have the fixed point property.


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## Body

## 1. Introduction

A Banach spaceX is said to have the fixed point property if for each nonexpansive mapping T:E→E on a bounded closed convex subset E of X has a fixed point, to have the weak fixed point property if for each nonexpansive mapping T:E→E on a weakly compact convex subset E of X has a fixed point.In 1981, D. E. Alspach [1] showed that there is an isometry T:E→E on a weakly compact convex subset E of the Lebesgue space L1[0,1] without a fixed point. Consequently, L1[0,1] does not have the weak fixed point property.In 1983, J. Elton, P. K. Lin, E. Odell, and S. Szarek [2] proved that C(α,R)) has the weak fixed point property, if α is a compact ordinal with α<ωω.In 1997, A. T. Lau, P. F. Mah, and Ali Ülger [3] proved the following theorem.Theorem 1.
LetX be a locally compact Hausdorff space. If C0(X) has the weak fixed point property, then X is dispersed.Moreover, by using Theorem1, they proved the following results.Corollary 2.
LetG be a locally compact group. Then the C∗-algebra C0(G) has the weak fixed point property if and only if G is discrete.Corollary 3.
A von Neumann algebraM has the weak fixed point property if and only if M is finite dimensional.In 2005, Benavides and Pineda [4] studied the concept of ω-almost weak orthogonality in the Banach lattice C(K) and proved the following results.Theorem 4.
LetX be a ω-almost weakly orthogonal closed subspace of C(K) where K is a metrizable compact space. Then X has the weak fixed point property.Theorem 5.
LetK be a metrizable compact space. Then, the following conditions are all equivalent:(1)
C(K) is ω-almost weakly orthogonal,(2)
C(K) is ω-weakly orthogonal,(3)
K(ω)=∅.Corollary 6.
LetK be a compact set with K(ω)=∅. Then C(K) has the weak fixed point property.IfX is a complex Banach algebra, condition (A) is defined by the following.(A)For eachx∈X,there exists an elementy∈Xsuch thatτ(y)=τ(x)¯,for eachτ∈Ω(X).It can be seen that eachC∗-algebra satisfies condition (A).In 2010, W. Fupinwong and S. Dhompongsa [5] proved that each infinite dimensional unital Abelian real Banach algebra X with Ω(X)≠∅ satisfying (i) if x,y∈X is such that |τ(x)|≤|τ(y)| for each τ∈Ω(X) then x≤y and (ii) inf⁡{r(x):x∈X,x=1}>0 does not have the fixed point property. Moreover, they proved the following theorem.Theorem 7.
LetX be an infinite dimensional unital Abelian complex Banach algebra satisfying condition (A) and each of the following:(i)
Ifx,y∈X is such that |τ(x)|≤|τ(y)| for each τ∈Ω(X), then x≤y.(ii)
inf{r(x):x∈X,x=1}>0.
ThenX does not have the fixed point property.In 2010, D. Alimohammadi and S. Moradi [6] used the above result to obtain sufficient conditions to show that some unital uniformly closed subalgebras of C(Ω), where Ω is a compact space, do not have the fixed point property.In 2011, S. Dhompongsa, W. Fupinwong, and W. Lawton et al. [7] showed that a C∗-algebra has the fixed point property if and only if it is finite dimensional.In 2012, W. Fupinwong [8] show that the unitality in Theorem 7 proved in [5] can be omitted.In 2016, by using Urysohn’s lemma and Schauder-Tychonoff fixed point theorem, D. Alimohammadi [9] proved the following result.Theorem 8.
LetΩ be a locally compact Hausdorff space. Then the following statements are equivalent:(i)
Ω is infinite set.(ii)
C0(Ω) is infinite dimensional.(iii)
C0(Ω) does not have the fixed point property.In 2017, J. Daengsaen and W. Fupinwong [10] showed that for each infinite dimensional real Abelian Banach algebra X with Ω(X)≠∅ satisfying (i) if x,y∈X is such that |τ(x)|≤|τ(y)| for each τ∈Ω(X) then x≤y and (ii) inf{r(x):x∈X,x=1}>0 does not have the fixed point property.In this paper, letX be an infinite dimensional unital Abelian complex Banach algebra satisfying (i) condition (A), (ii) if x,y∈X is such that |τ(x)|≤|τ(y)|, for each τ∈Ω(X), then x≤y, and (iii) inf⁡{r(x):x∈X,x=1}>0. We prove that there exists an element x0 in X such that (1)x0R=∑i=1kαix0i:k∈N,αi∈R¯does not have the fixed point property. Our result is a generalization of Theorem 7. And, as a consequence of the proof, we have that, for each element x0 in X with infinite spectrum and σ(x0)⊂R, the Banach algebra 〈x0〉=∑i=1kαix0i:k∈N,αi∈C¯ generated by x0 does not have the fixed point property.


## 2. Preliminaries

LetF be the field R or C. Let X be a Banach space over F. We say that a mapping T:E⊂X→X is nonexpansive if (2)Tx-Ty≤x-yfor each x,y∈E, where E is a nonempty subset of X. A Banach space X over F is said to have the fixed point property if for each nonexpansive mapping T:E→E on a nonempty bounded closed convex subset E of X has a fixed point.We define the spectrum of an elementx of a unital Banach algebra X over F to be the set (3)σx=σXx=λ∈F:λ1-x∉InvX,where Inv(X) is the set of all invertible elements in X.The spectral radius ofx is defined to be (4)rx=supλ∈σx⁡λ.We say that a mappingτ:X→F is a character on an algebra X over F if τ is a nonzero homomorphism. We denote by Ω(X) the set of all characters on X. If X is a unital Abelian Banach algebra over F, it is known that Ω(X) is compact.IfX is a complex Banach algebra, condition (A) is defined by the following.(A)For eachx∈X,there exists an elementy∈Xsuch thatτ(y)=τ(x)¯,for eachτ∈Ω(X).We denote byCF(S) the unital Banach algebra of continuous functions from a topological space S to F where the operations are defined pointwise and the norm is the sup-norm.The following Theorem is known as the Stone-Weierstrass approximation theorem forCR(S).Theorem 9.
LetA be a subalgebra of CR(S) satisfying the following conditions:(i)
A separates the points of S.(ii)
A annihilates no point ofS.
ThenA is dense in CR(S).LetX be an Abelian Banach algebra over F. The Gelfand representation φ:X→CF(Ω(X)) is defined by x↦x^, where x^ is defined by (5)x^τ=τx,for each τ∈Ω(X). If X is unital and Abelian, then σ(x)={τ(x):τ∈Ω(X)}, for each x∈X. It is known that r(x)=x^∞,X if X is Abelian, where (6)x^∞,X=supτ∈ΩXx^τ.The Jacobson radicalJ(X) of a Banach algebra X over F is the intersection of all regular maximal left ideals of X. It is known that if X is a unital complex Banach algebra and x∈J(X) then the spectral radius r(x) of x is equal to zero. A Banach algebra X over F is said to be semisimple if J(X)={0}.


## 3. Lemmas

First of all, we study the relationship between the sup-norm and the spectral radius, and we prove some properties of the spectral radius on a complex unital Banach algebra satisfyinginf⁡{x^∞,X:x∈X,x=1}>0.Lemma 10.
LetX be a complex unital Banach algebra satisfying (7)inf⁡x^∞,X:x∈X,x=1>0.Then (i)
inf{r(x):x∈X,x=1}>0.(ii)
X is semisimple.Proof.
LetX be a complex unital Banach algebra satisfying (8)inf⁡x^∞,X:x∈X,x=1>0.
(i) From{τ(x):τ∈Ω(X)}⊂σ(x), for each x∈X, it follows that x^∞,X≤r(x), for each x∈X. Therefore, (9)0<inf⁡x^∞,X:x∈X,x=1≤inf⁡rx:x∈X,x=1.
(ii) From (i), we have(10)inf⁡rx:x∈X,x=1>0,and hence, for each x∈X, r(x)=0 implies x=0. Since r(x)=0, for each x∈J(X), J(X)={0}. Therefore, X is semisimple.The following lemma was proved in [11].Lemma 11.
LetX be an infinite dimensional semisimple complex Banach algebra. Then there exists an element with an infinite spectrum.As a consequence of condition (A), Lemma10 and Lemma 11, we obtain the following results, Lemma 12 and Lemma 13, immediately.Lemma 12.
LetX be an infinite dimensional complex unital Abelian Banach algebra with condition (A) and satisfy (11)inf⁡rx:x∈X,x=1>0.Then there exists x0∈X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X).Proof.
SinceX is Abelian, (12)0<inf⁡rx:x∈X,x=1=inf⁡x^∞,X:x∈X,x=1.It follows from Lemma 10 that X is semisimple. From Lemma 11, there exists an element x in X with infinite spectrum. From condition (A), there exists y∈X such that (13)τy=τx¯,for each τ∈Ω(X). Hence (14)τxy=τxτy=τxτx¯∈R,for each τ∈Ω(X).Lemma 13.
LetX be an infinite dimensional complex unital Banach algebra, and let x0 be an element in X with infinite spectrum. Then {x0n:n∈N} is linearly independent.Proof.
Assume that(15)∑i=1kαix0i=0,where αi∈C. Let (16)λ1,λ2,λ3,…⊆σx0with λi≠λj for each i≠j.
From∑i=1kαi(σx0)i=σ∑i=1kαix0i={0}, it follows that (17)∑i=1kαiλji=0,for each j∈N. Hence αi=0, for each i∈{1,2,3,…,k}.Lemma 14.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A), and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (18)Z=∑i=0kαix0i:k∈N,αi∈R¯.Then Z is an infinite dimensional real unital Abelian Banach algebra with Ω(Z)≠∅.Proof.
From Lemma13, {x0n:n∈N} is linearly independent in X, so Z is infinite dimensional. To show that Ω(Z)≠∅. Let τ∈Ω(X), and define ω:Z→R by (19)x⟼τx.Indeed, ω is real-valued since τ(x0)∈R, for each τ∈Ω(X). Obviously, ω is a nonzero homomorphism on Z. So Ω(Z)≠∅.Similarly, one can prove the following lemma.Lemma 15.
LetX be an infinite dimensional complex unital Banach algebra satisfying condition (A), and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (20)x0R=∑i=1kαix0i:k∈N,αi∈R¯.Then 〈x0〉R is an infinite dimensional real nonunital Abelian Banach algebra.Some useful properties of the real unital Abelian Banach algebraZ is shown in the following lemma.Lemma 16.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A), and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (21)Z=∑i=0kαix0i:k∈N,αi∈R¯.If X satisfies (22)inf⁡x^∞,X:x∈X,x=1>0,then Z is a real unital Abelian Banach algebra satisfying the following conditions:
(i) The Gelfand representationφ from Z into CR(Ω(Z)) is a bounded isomorphism.
(ii) The inverseφ-1 is also a bounded isomorphism.Proof.
(i) It follows thatker(φ)={0}. So φ is injective. We have φ(Z) is a subalgebra of CR(Ω(Z)) separating the points of Ω(Z) and annihilating no point of Ω(Z). Moreover, φ(Z) is complete, so φ(Z) is closed. In fact, if {zn^} is a Cauchy sequence in φ(Z), assume to the contrary that {zn} is not Cauchy. So there exist ε0>0 and subsequences {zn′} and {zn′′} of {zn} such that (23)zn′-zn′′≥ε0,for each n∈N. Let yn=(zn′-zn′′)/ε0. So yn≥1, for each n∈N. Since {zn^} is Cauchy, limyn^=0^. Hence (24)0<infx^∞,X:x∈X,x=1≤infn∈Nynyn^∞,X≤infn∈Nynyn^∞,Z=0,which is a contradiction. So we conclude that {zn} is a Cauchy sequence. Then {zn} is a convergent sequence in Z, say limzn=z0∈Z. Therefore, (25)lim⁡zn^-z0^∞,Z=0,since for each n∈N, (26)zn^-z0^∞,Z=φzn-z0∞,Z≤zn-z0.So φ(Z) is complete. It follows from the Stone-Weierstrass theorem that φ is surjective.
(ii) is a consequence of the open mapping theorem.Lemma 17.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A), and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (27)Z=∑i=0kαix0i:k∈N,αi∈R¯.If there exists an element x in 〈x0〉R with infinite spectrum σ(x) and σ(x)⊂R, then there exists y∈〈x0〉R satisfying the following conditions:
(i)1∈σ(y)⊂[0,1].
(ii) There exists a strictly decreasing sequence inσ(y).Proof.
LetX be an infinite dimensional complex unital Banach algebra. Assume that there exists an element x in 〈x0〉R with infinite spectrum σ(x) and σ(x)⊂R.
SinceZ is unital and Abelian, the spectrum of x is x^(Ω(Z)). Hence (28)σx2rx2=τx2rx2:τ∈ΩZ⊂0,1.
Let{an} be an infinite sequence in σx2/r(x2). We may assume that {an} is strictly increasing and a1>0.
Define a continuous functiong:[0,1]→[0,1] by(29)
g
t
=
t
a
1
,
if
t
∈
0
,
a
1
,
1
-
t
1
-
a
1
,
if
t
∈
a
1
,
1
.So g is joining the points (0,0) and (a1,1), and g(1)=0. Let (30)z^=g∘y^.It follows from Lemma 16 that z∈Z. Since g(0)=0, z∈〈x0〉R. We have that {g(an)} is a strictly decreasing sequence in σ(z). Moreover, 1=g(a1)∈σ(z)⊂[0,1].We next give the following two lemmas which are important tools for proving the main result.Lemma 18.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A) and (31)inf⁡x^∞,X:x∈X,x=1>0,and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (32)Z=∑i=0kαix0i:k∈N,αi∈R¯.Then there exists a sequence {zn} in 〈x0〉R such that {τ(zn):τ∈Ω(Z)}⊂[0,1], for each n∈N, and {(zn^)-1{1}} is a sequence of nonempty pairwise disjoint subsets of Ω(Z).Proof.
Obviously,(33)inf⁡z^∞,Z:z∈Z,z=1>0,since inf⁡{x^∞,X:x∈X,x=1}>0. From Lemma 14 and Lemma 2.10 (iii) in [5], there exists ∑i=0kαix0i=z0∈Z such that {τ(z0):τ∈Ω(Z)} is infinite. We have ∑i=1kαix0i=z1∈〈x0〉R such that σ(z1) is infinite. From Lemma 17, we may assume without generality that z1 satisfies (34)1∈σz1⊂0,1and there exists a strictly decreasing sequence of real number in σ(z1), say {an}. Moreover, we may assume that a1<1.
Define a continuous functiong1:[0,1]→[0,1] by(35)g1t=ta1,ift∈0,a1,1+g1a2-1t-a121-a1,ift∈a1,1.So g1 is joining the points (0,0) and (a1,1), and g1(1)∈(g1(a2),1).
Letz2^=g1∘z1^, and define a continuous function g2:[0,1]→[0,1] by(36)
g
2
t
=
t
g
1
a
2
,
if
t
∈
0
,
g
1
a
2
,
1
+
g
2
g
1
a
3
-
1
t
-
g
1
a
2
2
1
-
g
1
a
2
,
if
t
∈
g
1
a
2
,
1
.So g2 is joining the points (0,0) and (g1(a2),1), and g2(1)∈(g2(g1(a3)),1).
Letz3^=g2∘z2^. Continuing in this manner, we get a sequence of points {zn} in Z with 1∈{τ(zn):τ∈Ω(Z)}⊂[0,1], for each n∈N, and {(zn^)-1{1}} is a sequence of nonempty pairwise disjoint subsets of Ω(Z).
Moreover,zn∈〈x0〉R, for each n∈N, since gn(0)=0, for each n∈N.Lemma 19.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A) and (37)inf⁡x^∞,X:x∈X,x=1>0,and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (38)Z=∑i=0kαix0i:k∈N,αi∈R¯.Assume that there exists a bounded sequence {yn} in Z which contains no convergent subsequences such that {τ(yn):τ∈Ω(Z)} is finite for each n∈N. Then there exists an element z0∈〈x0〉R such that {τ(z0):τ∈Ω(Z)} is equal to {0,1,1/2,2/3,3/4,…} or {0,1,1/2,1/3,1/4,…}.Proof.
It follows form Lemma14 and Lemma 16 that Z is an infinite dimensional real unital Abelian Banach algebra with Ω(Z)≠∅ and homeomorphic to CR(Ω(Z)). Assume that there exists a bounded sequence {yn} in Z which contains no convergent subsequences and such that {τ(yn):τ∈Ω(Z)} is finite for each n∈N. From the proof of Lemma 2.10 (ii) in [5], we have (39)ΩZ=∪n∈NGn∪F,where F is a closed set in Ω(Z), Gn is closed and open for each n∈N, and {F,G1,G2,…} is a partition of Ω(Z). Define τZ:Z→R by (40)τZ∑i=0kαix0i=α0,for each ∑i=0kαix0i∈Z. So τZ is a character on Z. There are two cases to be considered. If τZ is in F, define ψ:Ω(Z)→R by(41)ψτ=1,ifτ∈G1,1n,ifτ∈Gn,n≥2,0,ifτ∈F.If τZ is in Gn0, for some n0∈N, we may assume without loss of generality that n0=1, and define ψ:Ω(Z)→R by(42)ψτ=0,ifτ∈G1,n-1n,ifτ∈Gn,n≥2,1,ifτ∈F.For each case, the inverse image of each closed set in ψ(Ω(Z)) is closed, so ψ∈CR(Ω(Z)). Let φ:Z→CR(Ω(Z)) be the Gelfand representation. Therefore, φ-1(ψ) is an element in Z, say z0, such that {τ(z0):τ∈Ω(Z)} is equal to {0,1,1/2,2/3,3/4,…} or {0,1,1/2,1/3,1/4,…}. Moreover, z0∈〈x0〉R since τZ(z0)=ψ(τZ)=0.Lemma 20.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A) and (43)inf⁡x^∞,X:x∈X,x=1>0,and let x0 be an element in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Define (44)Z=∑i=0kαix0i:k∈N,αi∈R¯.and let x∈〈x0〉R with (x^)-1{1}≠∅, and 0≤τ(x)≤1, for each τ∈Ω(Z). Define (45)E=z∈x0R:0≤τz≤1for  eachτ∈ΩZ,andτz=1ifτ∈A,where A=(x^)-1{1}, and define T:E→E by (46)z⟼xz.Assume that X satisfies the following condition.
Ifx,y∈X is such that τx≤τy, for each τ∈Ω(X), then x≤y.
ThenE is a nonempty bounded closed convex subset of 〈x0〉R and T:E→E is a nonexpansive mapping.Proof.
Obviously,E is closed and convex. E is nonempty since x∈E
Letz∈E. Hence (47)0<inf⁡x^∞,X:x∈X,x=1≤zz^∞,X≤zz^∞,Z≤1z.Therefore, E is bounded.
Letω∈Ω(X), and let z,z′∈E. Define τ:Z→R by (48)τ∑i=0kαix0i=ω∑i=1kαix0i+α0,for each ∑i=0kαix0i∈Z. So τ∈Ω(Z).
Then(49)ωTz-Tz′=τxz-xz′,=τxτz-z′,≤ωz-z′.From (i), we have (50)Tz-Tz′≤z-z′.So T is nonexpansive.


## 4. Main Result

Theorem 21.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A) and the following conditions:(i)
Ifx,y∈X is such that |τ(x)|≤|τ(y)|, for each τ∈Ω(X), then x≤y,(ii)
inf⁡{r(x):x∈X,x=1}>0.
Then there exists an elementx0 in X such that (51)x0R=∑i=1kαix0i:k∈N,αi∈R¯does not have the fixed point property.Proof.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying (i), (ii), and condition (A). It follows from Lemma 12 that there exists an element x0 in X with infinite spectrum and τ(x0)∈R, for each τ∈Ω(X). Let (52)Z=∑i=0kαix0i:k∈N,αi∈R¯.From Lemma 14, Z is an infinite dimensional real unital Abelian Banach algebra with Ω(Z)≠∅. From Lemma 18, it follows that there is a sequence {zn} in 〈x0〉R such that {τ(zn):τ∈Ω(Z)}⊂[0,1] for each n∈N, and (z1^)-1{1}, (z2^)-1{1}, (z3^)-1{1},… are nonempty pairwise disjoint.
WriteAn=(zn^)-1{1}, define (53)En=z∈x0R⊂Z:0≤τz≤1for  eachτ∈ΩZ,andτz=1ifτ∈An,and define Tn:En→En by (54)z⟼znz.From Lemma 20, En is a nonempty bounded closed convex subset in 〈x0〉R and Tn is nonexpansive for each n∈N.
Assume to the contrary that〈x0〉R has the fixed point property. For each n∈N, since En is a nonempty bounded closed convex subset in 〈x0〉R, Tn has a fixed point in En, say yn. Since yn is a fixed point of Tn, yn=znyn. Then yn^=zn^yn^, and then(55)yn^τ=0,ifτis  not  inAn,1,ifτis  inAn,for each n∈N. Since A1,A2,A3,… are pairwise disjoint, ym^-yn^∞,Z=1, if m≠n. Thus {yn^} has no convergent subsequences. Since Z and CR(Ω(Z)) are homeomorphic, {yn} has no convergent subsequences. From Lemma 19, there exists an element z0 in 〈x0〉R such that {τ(z0):τ∈Ω(Z)} is equal to {0,1,1/2,2/3,3/4,…} or {0,1,1/2,1/3,1/4,…}.
WriteA0=(z0^)-1{1}, define (56)E0=z∈x0R⊂Z:0≤τz≤1for  eachτ∈ΩZ,andωz=1ifτ∈A0,and define T0:E0→E0 by (57)z⟼z0z.It follows from Lemma 20 that T0 is a nonexpansive mapping on a nonempty bounded closed convex subset E0 in 〈x0〉R. So T0 has a fixed point in E0, say y0. There are two cases to be considered.Case 1 ({τ(z0):τ∈Ω(Z)}={0,1,1/2,2/3,3/4,…}).
Hencey0^=z0^y0^. Then(58)y0^τ=0,ifτis  not  inA0,1,ifτis  inA0.So (59)y0^-11=z0^-11=A0and (60)ΩZ∖A0=y0^-10=⋃n=0∞z0^-1nn+1.
It follows from(61)τz0:τ∈ΩZ=0,1,12,23,34,…that (z0^)-1{n/(n+1)}:n∈N⋃(z0^)-1{0} is a nonempty pairwise disjoint open covering of the compact set Ω(Z)∖A0, which is a contradiction.Case 2 (τz0:τ∈ΩZ={0,1,1/2,1/3,1/4,…}).
(62)E=z+1=∑i=1kαix0i+1∈Z:0≤τz+1≤1for  eachτ∈ΩZ,andτz+1=1ifτ∈A,where A=(-z0+1^)-1{1}.
It can be seen thatE is a nonempty bounded closed convex subset of Z.
DefineT:E→E by (63)z+1⟼-z0+1z+1,for each z+1∈E. We have (64)τ-z0+1:τ∈ΩZ=0,1,12,23,34,….
DefineS:Z→Z by (65)∑i=0kαix0i⟼∑i=0k-αix0i+1.It follows from (i) that STS:SE→SE is a nonexpansive mapping on a nonempty bounded closed convex subset SE of 〈x0〉R.
S T S has a fixed point since 〈x0〉R has the fixed point property. It follows that T has a fixed point, say y0+1. Then(66)y0+1^τ=0,ifτis  not  inA,1,ifτis  inA.So (67)y0+1^-11=-z0+1^-11=Aand (68)ΩZ∖A=y0+1^-10=⋃n=0∞-z0+1^-1nn+1.
It follows from(69)τ-z0+1:τ∈ΩZ=0,1,12,23,34,…that (-z0+1^)-1{n/(n+1)}:n∈N⋃(-z0+1^)-1{0} is a nonempty pairwise disjoint open covering of the compact set Ω(Z)∖A, which is a contradiction.
So we conclude that〈x0〉R does not have the fixed point property.By following the proof of the above theorem, we obtain some corollaries.Corollary 22.
LetX be an infinite dimensional complex unital Abelian Banach algebra satisfying condition (A) and the following conditions:
(i) Ifx,y∈X is such that |τ(x)|≤|τ(y)|, for each τ∈Ω(X), then x≤y,
(ii)inf{r(x):x∈X,x=1}>0.
Ifx0 is an element in X with infinite spectrum and σ(x)⊂R, then the Banach algebra 〈x0〉=∑i=1kαix0i:k∈N,αi∈C¯ generated by x0 does not have the fixed point property.Corollary 23.
Letx0 be a self-adjoint element in a unital Abelian C∗-algebra X, then the algebra 〈x0〉=∑i=1kαix0i:k∈N,αi∈C¯ generated by x0 does not have the fixed point property.Finally, we pose an interesting problem.Problem 24.
Can condition (A) be removed from Theorem21?



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*Source: 9045790-2018-06-10.xml*