# Approximate Preservers on Banach Algebras andC*-Algebras

Authors: M. Burgos; A. C. Márquez-García; A. Morales-Campoy
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The aim of the present paper is to give approximate versions of Hua’s theorem and other related results for Banach algebras andC*-algebras. We also study linear maps approximately preserving the conorm between unital C*-algebras.


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## Body

## 1. Preliminaries

A well known formulation of the celebrated Hua’s theorem [1] asserts that every bijective additive map T:K→K on a division ring K such that T(1)=1 and T(x-1)=T(x)-1 for every invertible element x is either an automorphism or an antiautomorphism. This result was later moved to matrix algebras in [2] and finally extended to Banach algebras in [3] (see also [4]). In [3], the author called the previous relation strongly preserving invertibility.LetA and B be unital Banach algebras. Recall that an additive map T:A→B is a Jordan homomorphism ifT(ab+ba)=T(a)T(b)+T(b)T(a)for every a,b∈A, or equivalently, T(a2)=T(a)2 for alla∈A. Obvious examples of Jordan homomorphisms are homomorphisms and anti-homomorphisms. It is well known that every unital (i.e.,T(1)=1) Jordan homomorphism strongly preserves invertibility. Reciprocally, one of the results in [4] proves that every additive map between Banach algebras strongly preserving invertibility is a multiple of a Jordan homomorphism. In particular, if the map is unital, the map itself is a Jordan homomorphism.There exist also other versions of Hua’s theorem involving some important kinds of generalized invertibility. Given a ringAand an element a∈A, a is said to be Drazin invertible if there exist b∈A and a nonnegative integer k such that
(1)bab=b,akba=ak,ab=ba.
Such b is unique whenever it exists. In this case, it is called the Drazin inverse of a and it is denoted by b=aD. This notion was introduced by Drazin in [5] and it has proved useful in many fields of pure and applied mathematics (see for instance [6, 7]). If the previous identities are satisfied for some b∈A and k=1, b is called the group inverse of a and we will denote it by aG. Let A-1, AD, and AG denote the sets of all invertible, Drazin invertible, and group invertible elements in A, respectively. Clearly
(2)A-1⊂AG⊂AD.
Linear or additive maps between unital Banach algebras strongly preserving group and Drazin invertibility were introduced in [3, 4] (see also [8], where the author described additive maps between operator algebras of infinite-dimensional Hilbert spaces strongly preserving Drazin invertibility). If T:A→B is a Jordan homomorphism between Banach algebras, it was shown in [3, Theorem 2.1] that (i)
T strongly preserves group invertibility, that is, T(aG)=T(a)G for every a∈AG,(ii)
T strongly preserves Drazin invertibility, that is, T(aD)=T(a)D for every a∈AD.Conversely, ifT:A→B is an additive map strongly preserving invertibility, group invertibility, or Drazin invertibility, and T(1)=1 (resp., T(1) is invertible or 1∈T(A)), then T (resp., T(1)T) is a unital Jordan homomorphism and T(1) commutes with the image of T, [4, Theorem 4.2]. In [9], the authors showed that the same holds without any hypothesis on T(1) and even whenBis not necessarily unital.Recall that an elementa∈A is called regular if there exists b∈A (not necessarily unique) such that a=aba and b=bab. Notice that the first equality a=aba is a necessary and sufficient condition forato be regular, and that, p=ab and q=ba are idempotents in A fulfilling aA=pA and Aa=Aq. For an element a∈A, let us consider the left and right multiplication operators La:x↦ax and Ra:x↦xa, respectively. If a is regular, then so are La and Ra, and thus its ranges aA=La(A) and Aa=Ra(A) are both closed.Regular elements in a unitalC*-algebra A were studied by Harte and Mbekhta in [10, 11].An elementa∈A has Moore-Penrose inverseb when a=aba, b=baband the associated idempotents ab and ba are self-adjoint. In the aforementioned papers by Harte and Mbekhta, it is shown that every regular element in a C*-algebra has a Moore-Penrose inverse, and that it is unique. For a regular element a in a C*-algebra A, a† will denote its Moore-Penrose inverse. The set of all Moore-Penrose invertible elements in a C*-algebra A will be denoted by A†.LetA and B be unital C*-algebras. A linear map T:A→Bstrongly preserves Moore-Penrose invertibility if T(a†)=T(a)† for every a∈A†. Every Jordan*-homomorphism strongly preserves Moore-Penrose invertibility, and the question is whether or not the converse holds. Some partial positive answers are given by Mbekhta in [3], and more recently by the authors of the present paper in [12], whenAhas a rich structure of projections. The problem for general C*-algebras remains open. However, we can consider an alternative approach.Recall that the class ofC*-algebras is contained in a wider class of Banach spaces, the so called JB*-triples, in which the concept of regularity extends the one given for C*-algebras. A JB*-triple is a complex Banach spaceEtogether with a continuous triple product{·,·,·}:E×E×E→E, which is conjugate linear in the middle variable, and symmetric and bilinear in the outer variables satisfying that(a)
L(a,b)L(x,y)=L(x,y)L(a,b)+L(L(a,b)x,y)-L(x,L(b,a)y), whereL(a,b)is the operator onEgiven byL(a,b)x={a,b,x};(b)
L(a,a) is an hermitian operator with nonnegative spectrum;(c)
∥L(a,a)∥=∥a∥2.For eachx in a JB*-triple E, Q(x) will stand for the conjugate linear operator on E defined by Q(x)(y)={x,y,x}.An elementa in a JB*-triple E is called von Neumann regular if there exists (a unique) b∈E such that Q(a)(b)=a, Q(b)(a)=b, and Q(a)Q(b)=Q(b)Q(a). The elementbis called the generalized inverse of a and it will be denoted by a∧. We will also note by E∧ the set of all elements in E with generalized inverse. We refer to [13–17] for basics results on von Neumann regularity in JB*-triples.EveryC*-algebra is a JB*-triple via the triple product given by
(3){x,y,z}=12(xy*z+zy*x).
For a C*-algebraA, it is well known thatA†=A∧and, for every regular elementainA, we havea∧=(a†)*.A linear mapT:E→F between JB*-triples is a triple homomorphism if
(4)T({x,y,z})={T(x),T(y),T(z)},
for every x,y,z∈E. Every triple homomorphism T:E→F between JB*-triples strongly preserves generalized invertibility; that is, T(x∧)=T(x)∧ for every x∈E∧. In [9], the authors characterized the triple homomorphism between C*-algebras as the linear maps strongly preserving generalized invertibility. As a consequence, it is proved that a self-adjoint linear map from a unital C*-algebraAinto a C*-algebra B is a triple homomorphism if and only if it strongly preserves Moore-Penrose invertibility [9, Theorem 3.5].The Hua-type descriptions belong to the framework of linear preserver problems. This has become an active research area in many topics of matrix theory, operator theory, and Banach algebras theory. Some of the most popular preserver problems are those dealing with determining the linear maps preserving properties related to invertibility. Every Jordan isomorphismT:A→B between unital Banach algebras is unital and preserves invertibility in both directions [18, Proposition 1.3], or equivalently, preserves the spectrum; that is, σ(T(a))=σ(a), for every a∈A. The celebrated Kaplansky’s conjecture, [19], reformulated by Aupetit in [20], states that every unital surjective linear map T between unital semisimple Banach algebras preserving invertibility in both directions is a Jordan isomorphism. Many partial positive results are known so far [18, 20–24], but the general problem is still open even in the class of C*-algebras. In the commutative setting, the classical Gleason-Kahane-Zelazko theorem (see [23]) states that a linear functional φ on a unital complex Banach algebra A is multiplicative if and only if φ(a)∈σ(a), for all a∈A. On the other hand, Jafarian and Sourour proved in [22] that every spectrum preserving surjective linear map T:ℬ(X)→ℬ(Y) is either an isomorphism or an anti-isomorphism, where ℬ(X) denotes the Banach algebra of all bounded linear operators on a Banach space X.The authors of [25, 26] consider the problem of characterizing the approximately multiplicative linear functionals among all linear functionals on a commutative Banach algebra in terms of spectra. More recently, in [27], (see also [28]) Alaminos, Extremera, and Villena investigate approximate versions of Kaplansky’s problem, by providing approximate formulations of [18, 22]. They considered linear maps that approximately preserve spectrum or spectral radius on operator algebras and established the relationship between approximately preserving spectrum (resp., spectral radius) and approximately being a Jordan homomorphism (resp., weighted Jordan homomorphism).LetA and B be Banach algebras and T:A→B be a bounded linear map. Following [27, 29], the multiplicativity, antimultiplicativity, and Jordan-multiplicativity of T can be measured by considering the following values:
(5)mult(T)=sup{∥T(ab)-T(a)T(b)∥:a,b∈A,∥a∥=∥b∥=1},amult(T)=sup{∥T(ab)-T(b)T(a)∥:a,b∈A,∥a∥=∥b∥=1},jmult(T)=sup{∥T(a2)-T(a)2∥:a∈A,∥a∥=1},
respectively. Obviously, T is a homomorphism (anti-homomorphism, Jordan homomorphism) if and only if mult(T)=0 (resp., amult(T)=0, jmult(T)=0).For a bounded linear mapT:A→B between C*-algebras, we define the triple multiplicativity and the self-adjointness of T, respectively, as the following quantities:
(6)tmult(T)=sup{∥T({a,b,c})-{T(a),T(b),T(c)}∥:∥a∥=∥b∥=∥c∥=1},sa(T)=sup{∥T(a*)*-T(a)∥:∥a∥=1}.Clearly,T:A→B is a triple homomorphism if and only if tmult(T)=0, and T is self-adjoint if and only if sa(T)=0.The aim of the present paper is to bring Hua type theorems into this framework. In order to make this possible, we have adapted some techniques from [27] involving ultraproducts of Banach algebras. Section 2 contains all the technical results about invertibility and coset representatives in ultraproducts of Banach algebras that we will need throughout the paper. Section 3 provides approximate versions of Hua’s theorem for invertibility and group invertibility in Banach algebras. We translate the strongly invertibility preserving condition T(a-1)=T(a)-1 into
(7)sup∥a∥=1,a∈A-1∥T(a-1)-T(a)-1∥<ɛ,
and the condition T(aG)=T(a)G into
(8)sup∥a∥=1,a∈AG∥T(aG)-T(a)G∥<ɛ,
for some ɛ>0. We prove that for every unital Banach algebras A and B, if ɛ→0 in (7) or (8) then jmult(T(1)T)→0, uniformly on any set of linear maps T:A→B with norms bounded above.Section4 includes an approximate formulation of [9, Theorem 3.5]. The condition T(a†)=T(a)† for every a∈A† is replaced by
(9)sup∥a∥=1,a∈A†∥T(a†)-T(a)†∥<ɛ.We show that for every unitalC*-algebras A and B if ɛ→0 in (9), then tmult(T)→0, uniformly on any set of linear maps whose norms are bounded above.In this section, we also study linear maps that approximately preserve the conorm. Recall that theconormc(a)of an element a in a unital Banach algebra is defined as the reduced minimum modulus of the left multiplication operator by a, c(a)=γ(La)[11]. For a bounded linear operator T on a complex Banach space X, its reduced minimum modulus is given by
(10)γ(T)=inf{∥Tx∥:dist(x,ker(T))≥1}.
It is well known that γ(T)>0 if and only if T has closed range. In [10], it is shown that an element a in a unital C*-algebra A is regular if and only if c(a)>0. In this case,
(11)c(a)=∥a†∥-1(see [11, Theorem 2]). In [30], the authors characterized the linear maps between unitalC*-algebras preserving the conorm. By [30, Theorem 3.1], if T:A→B is a unital linear map such that c(T(a))=c(a) for every a∈A, then T is an isometric Jordan*-homomorphism. Also, from [30, Theorem 3.2], if T:A→B is a surjective linear map such that c(T(a))=c(a) for every a∈A, then T is an isometric Jordan*-homomorphism multiplied by a unitary element. Hence, we replace the condition c(T(a))=c(a) by
(12)sup∥a∥=1|c(T(a))-c(a)|<ɛ,
in order to get approximate versions of [30, Theorems 3.1 and 3.2].


## 2. Ultraproducts of Banach Algebras: Basic Tools

Given a free ultrafilter𝒰 on ℕ and a sequence of Banach spaces {Xn}n∈ℕ, the so called ultraproduct of the sequence is defined as follows:
(13)(Xn)𝒰∶=ℓ∞(ℕ,Xn)𝒩𝒰,
where ℓ∞(ℕ,Xn) is the Banach space of all bounded sequences {xn}n∈ℕ, with xn∈Xn for all n∈ℕ, equipped with the ℓ∞ norm and
(14)𝒩𝒰∶={{xn}n∈ℕ∈ℓ∞(ℕ,Xn):lim𝒰∥xn∥=0}.
If the sequence {Xn}n∈ℕ={X} is constant, X𝒰∶=ℓ∞(ℕ,X)/𝒩𝒰, is called the ultrapower of X with respect to the ultrafilter 𝒰. We will denote by x=[xn] the equivalence class of the sequence {xn}n∈ℕ. The ultrapower of a Banach space is also a Banach space provided with the following norm:
(15)∥x∥∶=lim𝒰∥xn∥.
Of course, the ultrapower A𝒰 of a Banach algebra (resp., C*-algebra) is also a Banach algebra (resp., C*-algebra), with respect to the pointwise operations.Finally, for every Banach spacesX and Y, the canonical linear isometry ℬ(X,Y)𝒰→ℬ(X𝒰,Y𝒰) given by
(16)T(x)=[Tn(xn)],
for every T=[Tn]∈ℬ(X,Y)𝒰 and x=[xn]∈X𝒰, allows us to consider ℬ(X,Y)𝒰 as a closed subspace of ℬ(X𝒰,Y𝒰). For X=Y, the canonical map gives an isometric unital homomorphism from ℬ(X)𝒰 to ℬ(X𝒰). The reader can see [31] in order to find basic results on ultraproducts.LetA be a unital Banach algebra and 𝒰 a free ultrafilter on ℕ. The following proposition is devoted to the description of invertible elements in A𝒰 through certain coset representatives. The result is probably well known, but the lack of an adequate reference moves us to include it here.Proposition 1.
Leta∈A𝒰. The following assertions are equivalent.(1)
ais invertible.(2)
ahas a coset representative[un]such thatun∈A-1for alln∈ℕand{un-1}n∈ℕis bounded.Proof.
For(2)⇒(1), just note that [un-1]∈A𝒰 is an inverse for [un].
Reciprocally, assume thata=[an] is invertible. Then, there exists b=[bn]∈A𝒰 such that ab=ba=1. That is,
(17)lim𝒰∥anbn-1∥=0,lim𝒰∥bnan-1∥=0.
Fix0<δ<1. The above identities imply that
(18)R∶={n∈ℕ:∥anbn-1∥<δ}∈𝒰,L∶={n∈ℕ:∥bnan-1∥<δ}∈𝒰.
In particular, an is right invertible for every n∈R and an is left invertible for every n∈L. Thus, an is invertible for every n∈I∶=R∩L∈𝒰. Moreover,
(19)∥an-1∥=∥bn(anbn)-1∥≤∥bn∥∥(anbn)-1∥≤∥bn∥1-∥1-anbn∥≤∥bn∥1-δ,
which shows that {an-1:n∈I} is bounded. Therefore, we can assume without loss of generality that {an}n∈ℕ consists of invertible elements and {an-1}n∈ℕ is bounded. (Otherwise, we choose
(20)an′={an,ifn∈I,1,ifn∉I.
Clearly, [an]=[an′].)Remark 2.
It is clear that fora=[an]∈A𝒰 with ∥a∥=1, we can choose a coset representative a=[bn] such that ∥bn∥=1 for all n∈ℕ as follows:
(21)bn={an∥an∥,ifan≠0,1,ifan=0.
Hence, for every invertible element a in A𝒰, we can find a coset representative a=[an] fulfilling the conditions in Proposition 1 and satisfying ∥an∥=∥a∥ for all n∈ℕ. We will name this one a normalized representative for a.


## 3. Approximate Preservers in Banach Algebras

LetA and B be unital Banach algebras. Recall that, Boudi and Mbekhta proved in [4, Theorem 2.2] that an additive map T:A→B strongly preserves invertibility if and only if T(1)T is a unital Jordan homomorphism and T(1) commutes with the range of T. Hence, for a bounded linear map T:A→B between unital Banach algebras, we consider the unit-commutativity of T, defined as
(22)ucomm(T)=sup∥a∥=1∥T(a)T(1)-T(1)T(a)∥,
in order to measure how close is our “approximately preserving invertibility” map to fulfilling that property.Obviously, every bounded linear map satisfiesucomm(T)≤2∥T∥2. The next lemma shows the good behaviour of this concept with the ultraproduct of operators.Some arguments in this section are inspired in [27].Lemma 3.
Let{Tn}n∈ℕ be a bounded sequence of linear maps between Banach algebras A and B, where A is supposed to be unital. Consider T=[Tn]:A𝒰→B𝒰. Then,
(23)lim𝒰ucomm(Tn)=ucomm(T).Proof.
Givena∈A𝒰 with ∥a∥=1, we can choose a=[an] with ∥an∥=1 for every n∈ℕ. Therefore,
(24)∥T(a)T(1)-T(1)T(a)∥=lim𝒰∥Tn(an)Tn(1)-Tn(1)Tn(an)∥≤lim𝒰ucomm(Tn),
and hence,
(25)ucomm(T)≤lim𝒰ucomm(Tn).
Reciprocally, for each n∈ℕ, there exists an∈A with ∥an∥=1 such that
(26)ucomm(Tn)-1n<∥Tn(an)Tn(1)-Tn(1)Tn(an)∥.
Taking limit along 𝒰 we obtain
(27)lim𝒰ucomm(Tn)≤lim𝒰∥Tn(an)Tn(1)-Tn(1)Tn(an)∥=∥T(a)T(1)-T(1)T(a)∥≤ucomm(T).Our first main result provides an approximate version of Hua’s theorem for Banach algebras [4, Theorem 2.2] in the above mentioned.Theorem 4.
LetA and B be unital Banach algebras and K,ɛ>0. Then, there exists δ>0 such that for every linear map T:A→B with ∥T∥<K, the condition
(28)sup∥a∥=1,a∈A-1∥T(a-1)-T(a)-1∥<δ
implies that
(29)jmult(T(1)T)<ɛ,ucomm(T)<ɛ.Proof.
Suppose that the assertion of the theorem is false. Then, we can findK0,ɛ0>0 and a sequence {Tn}n∈ℕ of linear maps from A to B such that, for every n∈ℕ,(i)
∥Tn∥<K0,(ii)
sup∥a∥=1∥Tn(a-1)-Tn(a)-1∥<1/n,(iii)
jmult(Tn(1)Tn)≥ɛ0 orucomm(Tn)≥ɛ0.
Consider thatT=[Tn]:A𝒰→B𝒰. We claim that T strongly preserves invertibility. Indeed, let a∈A𝒰 be an invertible element. We can suppose, without loss of generality, that ∥a∥=1. Let [an] be its normalized representative, with ∥an-1∥<α, for some α>0 (see Proposition 1 and Remark 2). As
(30)∥Tn(an-1)∥≤∥Tn∥∥an-1∥<K0α,∥Tn(an-1)-Tn(an)-1∥<1,
we get
(31)∥Tn(an)-1∥<K0α+1,
for all n∈ℕ. Hence, T(a) is invertible and [Tn(an)-1] is its inverse. This yields
(32)∥T(a-1)-T(a)-1∥=lim𝒰∥Tn(an-1)-Tn(an)-1∥≤lim𝒰1n=0.
Thus, T(a-1)=T(a)-1 for every invertible element a∈A𝒰. By [4, Theorem 2.2], T(1)T(a2)=(T(1)T(a))2 and T(1)T(a)=T(a)T(1), for every a∈A𝒰. We apply [27, Lemmas 3.4] and Lemma 3 to obtain, respectively, the following:
(33)0=jmult(T(1)T)=lim𝒰jmult(Tn(1)Tn),0=ucomm(T)=lim𝒰ucomm(Tn).
Consequently,
(34)I={n∈ℕ:jmult(Tn(1)Tn)<ɛ0}∈𝒰,J={n∈ℕ:ucomm(Tn)<ɛ0}∈𝒰.
Finally, I∩J∈𝒰gives us the desired contradiction.Our goal now is to achieve a group invertibility version for the previous theorem. Recall that given an additive mapT:A→B from a unital Banach algebra A into a Banach algebra B, by [9, Theorem 2.4], if T strongly preserves group invertibility, then T(1)T is a Jordan homomorphism and T(1) commutes with the range of T. In order to take advantage of Proposition 1, our first step is to improve [9, Theorem 2.4] by showing that all the information required is located in A-1.Recall that the so-called Hua’s identity asserts that, ifa,b, and a-b-1 are invertible elements in a ring, then
(35)(a-1-(a-b-1)-1)-1=a-aba.Theorem 5.
LetA and B be Banach algebras, A being unital, and T:A→B be an additive map such thatT(a-1)=T(a)Gfor alla∈A-1. Then, T(1)Tis a Jordan homomorphism and T(1) commutes withT(A).Proof.
A look to the arguments employed in [9, Lemma 2.1], allows us to show that T preserves the cubes of the invertible elements. Indeed, given u∈A-1 and λ∈ℚ with 0<|λ|<∥u-1∥-2, as λ-1u and u-λu-1 are invertible elements, we can apply Hua’ s identity to obtain
(36)(u-1-(u-λu-1)-1)-1=u-u(λ-1u)u=u-λ-1u3.
Let us assume thatT(u)≠0. SinceT(u)∈BG, it follows thatT(u)is invertible in the unital Banach algebrapBpforp=T(u)T(u)G, with inverseT(u)G. Identity (36) applied forT(u)and0<|λ|<∥T(u)G∥-2gives
(37)T(u)-λ-1T(u)3=(T(u)G-(T(u)-λT(u)G)G)G.
Hence, for every λ∈ℚ such that 0<|λ|<min{∥u-1∥-2,∥T(u)G∥-2}, we get
(38)T(u)-λ-1T(u)3=(T(u-1)-T(u-λu-1)G)G=T(u-1-(u-λu-1)-1)G=T((u-1-(u-λu-1)-1)-1)=T(u-λ-1u3)=T(u)-λ-1T(u3).
Hence, T(u3)=T(u)3, as desired. From this last identity, reasoning as in [9, Proposition 2.3], we deduce that the following equalities hold for every x∈A:
(39)3T(x)=T(1)2T(x)+T(x)T(1)2+T(1)T(x)T(1),3T(x2)=T(x)2T(1)+T(1)T(x)2+T(x)T(1)T(x).
Finally, it only remains to repeat the arguments in(2)⇒(3)in [9, Theorem 2.4] to conclude the proof.Now, we can state the following result.Theorem 6.
LetA and B be Banach algebras where A is unital and K,ɛ>0. Then, there exists δ>0 such that for every linear map T:A→B with ∥T∥<K, the condition
(40)sup∥a∥=1,a∈AG∥T(aG)-T(a)G∥<δ
implies that
(41)jmult(T(1)T)<ɛ,ucomm(T)<ɛ.Proof.
First, notice that ifb∈A𝒰 has a coset representative b=[bn] where bn is group invertible for every n∈ℕ and {bnG}n∈ℕ is bounded, then bis group invertible and bG=[bnG]. Hence, the same arguments used in Theorem 4 produce an operatorT=[Tn]:A𝒰→B𝒰satisfyingT(a-1)=T(a)Gfor every invertible element a∈A𝒰. Now, Theorem 5 proves thatT(1)Tis a Jordan homomorphism andT(1)commutes withT(A𝒰). Again, the final argument in Theorem 4 completes the proof.In [32, Proposition 2.5], the authors proved, in particular, that if an additive mapT:A→Bbetween unital Banach algebras satisfies
(42)T(a)T(a-1)=T(1)2,for everya∈A-1,
and T(1)is invertible, thenT(1)-1Tis a Jordan homomorphism andT(1)commutes withT(A). It is clear now that for a sequence of linear operatorsTn:A→Bsatisfying that∥Tn∥,∥Tn(1)-1∥<Kfor all n∈ℕ, and
(43)∥Tn(a)Tn(a-1)-Tn(1)2∥<1n∀n∈ℕ,
its ultrapoductT:A𝒰→B𝒰fulfillsT(a)T(a-1)=T(1)2for every invertiblea∈A𝒰. Therefore,T(1)-1Tis a Jordan homomorphism andT(1)commutes withT(A𝒰). This leads us to the following approximate formulation of [32, Proposition 2.5].Theorem 7.
LetA and Bbe unital Banach algebras and K,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥,∥T(1)-1∥<K, the following condition:
(44)supa∈A-1∥T(a)T(a-1)-T(1)2∥<δ
implies that
(45)jmult(T(1)-1T)<ɛ,ucomm(T)<ɛ.


## 4. Approximate Preservers inC*-Algebras

The aim of this section is twofold. On the one hand, we prove that linear maps approximately preserving generalized invertibility inC*-algebras are close to be triple homomorphisms. On the other hand, we study linear maps approximately preserving the conorm.


### 4.1. Approximate Preservers of the Moore-Penrose Inverse and the Generalized Inverse

Given aJB*-tripleE, the triple cube of an elementx∈Eis defined asx[3]∶={x,x,x}. An element satisfyinge[3]=eis called a tripotent. The following polarization identity allows us to write the triple product as linear combination of triple cubes:
(46){x,y,z}=∑α2=1‍∑β4=1‍αβ(x+αy+βz)[3],forx,y,z∈E.
Hence, a linear map between JB*-triples is a triple homomorphism if and only if it preserves triple cubes.Each tripotente in E gives rise to the so-called Peirce decomposition of E associated to e, that is,
(47)E=E2(e)⊕E1(e)⊕E0(e),
where for i=0,1,2,Ei(e)is thei/2eigenspace of L(e,e). The peirce spaceE2(e)is a JB*-algebra with product x·y∶={x,e,y} and involutionx♯∶={e,x,e}. Moreover, the triple product induced onE2(e)by this Jordan *-algebra structure coincides with its original triple product.It is proved in [16, Lemma 3.2] (compare with [13, Theorem 3.4]) that for every regular elementain a JB*-tripleE, there exists a tripotente∈Esuch thatais a self-adjoint invertible element in the JB*-algebraE2(e). Ifais invertible, its inverse is denoted as usual bya-1. Moreover, ifaandbare invertible elements in the Jordan algebraJ=(J,∘)such thata-b-1is also invertible, thena-1+(b-1-a)-1 is invertible and the Hua identity
(48)(a-1+(b-1-a)-1)-1=a-Ua(b)
holds, whereUa(x)=2a∘(a∘x)-a2∘x(see [33], (11)).LetAbe a unitalC*-algebra, andu∈A∧∖{0}. Then, there exists a unique partial isometrye, such thatuis self-adjoint and invertible in the Jordan algebra(EA)2(e)=ee*Ae*e, with inverseu∧.Hence, for everyλ∈ℂwith0<|λ|<∥u∧∥-2, the elementu-λu∧is invertible inee*Ae*e. Reciprocally, the inverses ofu-λu∧andu∧-(u-λu∧)∧inee*Ae*eare their generalized inverses inA. By the Hua identity (48), we obtain
(49)u-λ-1u[3]=(u∧-(u-λu∧)∧)∧.Theorem 8.
LetAandBbe C*-algebras,Abeing unital, andT:A→Ba bounded linear map satisfyingT(x-1)=T(x)∧for every self-adjoint invertible elementx∈A. Then, Tis a triple homomorphism.Proof.
Arguing as in [9, Lemma 3.1], pick a self-adjoint invertible elementu∈A. We may assume thatT(u)≠0. Then, givenλ∈ℂwith0<|λ|<min{∥u-1∥-2,∥T(u)∧∥-2}, from identities (36) and (49), we get
(50)T(u)-λ-1T(u)[3]=(T(u)∧-(T(u)-λT(u)∧)∧)∧=T(u)-λ-1T(u3),
which shows thatT(u3)=T(u)[3]. Once we have proved thatTpreserves cubes of self-adjoint invertible elements, given a self-adjoint elementa∈A, andα∈ℝwith|α|>∥a∥, as the elementa+αis self-adjoint and invertible, we getT((a+α)3)=T(a+α)[3]. Expanding this las equation we obtain
(51)T(a3)+3αT(a2)+3α2T(a)+α3T(1)=T(a)[3]+α3T(1)+2α{T(a),T(a),T(1)}+α{T(a),T(1),T(a)}+2α2{T(1),T(1),T(a)}+α2{T(1),T(a),T(1)},
for everya∈A, and|α|>∥a∥. From this, we deduce thatT(a3)=T(a)[3]for everya∈A. That is,Tpreserves triple cubes of self-adjoint elements. By [34, Theorem 20],Tis a triple homomorphism.Recall that we measure how close is a linear mapT:A→Bbetween C*-algebras to being a triple homomorphism or self-adjoint by the triple multiplicativity and the self-adjointness ofT, respectively as follows:
(52)tmult(T)∶=sup{∥T({a,b,c})-{T(a),T(b),T(c)}∥:∥a∥=∥b∥=∥c∥=1},sa(T)∶=sup∥a∥=1∥T(a*)*-T(a)∥.Remark 9.
LetAandBbe C*-algebras. It is clear that every Jordan*-homomorphismT:A→Bis a triple homomorphism. We ask whetherjmult(T)and sa(T)being small implytmult(T)being small.
LetT:A→Bbe a bounded linear map. Define
(53)TJ(a,b)∶=T(a∘b)-T(a)∘T(b),TT(a,b,c)∶=T({a,b,c})-{T(a),T(b),T(c)},T*(a)∶=T(a*)*-T(a),
for everya,b,c∈A. Then,
(54)T({a,b,c})=T((a∘b*)∘c)+T(a∘(b*∘c))-T((a∘c)∘b*)=T(a∘b*)∘T(c)+TJ(a∘b*,c)+T(a)∘T(b*∘c)+TJ(a,b*∘c)-T(a∘c)∘T(b*)-TJ(a∘c,b*)=(T(a)∘T(b*))∘T(c)+TJ(a,b*)∘T(c)+TJ(a∘b*,c)+T(a)∘(T(b*)∘T(c))+T(a)∘TJ(b*,c)+TJ(a,b*∘c)-(T(a)∘T(c))∘T(b*)-TJ(a,c)∘T(b*)-TJ(a∘c,b*)={T(a),T(b*)*,T(c)}+TJ(a,b*)∘T(c)+TJ(a∘b*,c)+T(a)∘TJ(b*,c)+TJ(a,b*∘c)-TJ(a,c)∘T(b*)-TJ(a∘c,b*).
Therefore,
(55)TT(a,b,c)={T(a),T*(b),T(c)}+TJ(a,b*)∘T(c)+TJ(a∘b*,c)+T(a)∘TJ(b*,c)+TJ(a,b*∘c)-TJ(a,c)∘T(b*)-TJ(a∘c,b*).
This implies that
(56)tmult(T)≤∥T∥2sa(T)+3(∥T∥+1)jmult(T).As in Lemma3, it can be shown that, for an operatorT=[Tn], the following holds:
(57)lim𝒰tmult(Tn)=tmult(T),lim𝒰sa(Tn)=sa(T).We will omit the proof of the next result in order to avoid repetition. The argument is analogous to the one used in the proofs of Theorems4 and 6: assuming the contrary, we can construct a mapT=[Tn]:A𝒰→B𝒰between the ultrapowers fulfillingT(x-1)=T(x)∧for every self-adjoint invertiblex∈A𝒰. By Theorem 8,Tis a triple homomorphism.Theorem 10.
LetAandBbe C*-algebras whereAis unital andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<K, the condition
(58)sup∥a∥=1,a=a*∥T(a∧)-T(a)∧∥<δ
implies that
(59)tmult(T)<ɛ.Remark 11.
Notice that in Theorem8 it can be also obtained thatT(1)*Tis a Jordan*-homomorphism. Hence, the hypothesis of Theorem 10 also yields tojmult(T(1)*T)<ɛandsa(T(1)*T)<ɛ.Corollary 12.
LetAandBbe C*-algebras whereAis unital andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<K, the conditions
(60)sup∥a∥=1,a∈A†∥T(a†)-T(a)†∥<δ,sa(T)<δ
imply that
(61)tmult(T)<ɛ.Proof.
Let us briefly sketch the proof: assuming the contrary, there existK0,ɛ0>0and a sequence{Tn}n∈ℕof linear maps from A toBsuch that, for everyn∈ℕ,
(62)∥Tn∥<K0,sup∥a∥=1∥Tn(a†)-Tn(a)†∥<1n,sa(Tn)<1n,tmult(Tn)≥ɛ0.
Then, T=[Tn]:A𝒰→B𝒰is a self-adjoint map such thatT(a-1)=T(a)†, for every invertible elementa∈A𝒰. In particular,T(a-1)=T(a)∧, for every invertible self-adjoint elementa∈A𝒰. From Theorem 8,Tis a triple homomorphism (Contradiction).


### 4.2.  Maps Approximately Preserving the Conorm

LetAandBbe unitalC*-algebras. Kadison proved in [35] that a surjective linear mapT:A→Bis an isometry if and only ifTis a Jordan*-isomorphism multiplied by a unitary element inB. In [30], the authors address the question of characterizing surjective linear maps preserving some spectral quantities. Given an elementaof a Banach algebraA, the minimum modulus and the surjectivity modulus ofaare defined, respectively, by
(63)m(a)∶=inf{∥ax∥:x∈A,∥x∥=1},q(a)∶=inf{∥xa∥:x∈A,∥x∥=1}.
Obviously,m(a)=0(resp.,q(a)=0) if and only ifais a left (resp., right) topological divisor of zero. It is well known that for any invertible elementa∈A,
(64)m(a)=q(a)=∥a-1∥-1.
Moreover,
(65)m(a)m(b)≤m(ab)≤∥a∥m(b),|m(a)-m(b)|≤∥a-b∥,
for everya,b∈A. IfAis aC*-algebra, thenm(a)=q(a*). In particular,m(a)>0(resp.,q(a)>0) if and only ifais left (resp., right) invertible.Recall also from the introduction that theconorm of an elementain a Banach algebraA, is defined as
(66)c(a)∶={inf{∥ax∥:dist(x,ker(La))≥1},ifa≠0,∞,ifa=0.
For a regular elementain aC*-algebraA,
(67)c(a)=∥a†∥-1.LetAandBbe unital C*-algebras. By Theorems 3.1 and  3.2 in [30], ifT:A→Bis a linear map preserving any of these spectral quantities, thenTis an isometric Jordan *-homomorphism wheneverTis unital, andTis an isometric Jordan *-homomorphism multiplied by a unitary element, wheneverTis surjective. In the next results, we show that the same holds if we just impose the preserving condition for invertible elements. Notice that we focus our attention on the conorm but identical results can be established for the minimum and surjective modulus.Theorem 13.
LetAandBbe unital C*-algebras andT:A→Ba unital linear map satisfying c(T(x))=c(x)for allx∈A-1. Then, Tis a Jordan *-homomorphism.Proof.
First, let us prove thatTis injective. Takea0∈Asuch thatT(a0)=0and letα∈ℂbe sufficiently small so that1+αa0is invertible. Then,
(68)1=c(T(1))=c(T(1+αa0))=c(1+αa0).
In particular, we get
(69)1≤c(1+ita0),1≤c(1-ta0),
ast→0. Hence, by [30, Lemma 4.1], bothia0anda0are self-adjoint and, consequently,a0=0.
We claim thatTis positive. Indeed, given a self-adjoint elementa∈A, we know that
(70)1+o(t)≤c(1+ita),(ast→0).
Since1+itais invertible fort∈ℝsmall enough, it follows that
(71)1+o(t)≤c(1+ita)=c(1+itT(a)),(ast→0).
This implies thatT(a)is self-adjoint.
Moreover, givenx∈Aandλ∉σ(x), there exists a neighborhoodUλofλsuch thatUλ∩σ(x)=∅. Ifλ∉σ(x), thenλ∉σK(x), whereσK(x)denotes the Kato spectrum ofxas follows:
(72)σK(a)∶={λ∈ℂ:limμ→λc(a-μ)=0}.
As forμ∈Uλ, the elementx-μis invertible; then, we havec(T(x)-μ)=c(x-μ)for everyμ∈Uλ. Consequently,
(73)limμ→λc(T(x)-μ)>0,
andλ∉σK(T(x)). Since∂σ(a)⊆σK(a)⊆σ(a)for everya∈A(see [36, Sections 12,13]), we have just proved that
(74)∂σ(T(x))⊆σK(T(x))⊆σ(x),
for everyx∈A. BeingTself-adjoint, this implies thatTis positive and hence,∥T∥=1.
Arguing as in [30, Theorem 5.1], given a self-adjoint elementa∈Aandtsufficiently small so thatu=eitais a unitary element with spectrum strictly contained in the unit circle𝕋, since
(75)∂σ(T(u))⫋𝕋,∥T(u)∥≤1,∥T(u)-1∥=c(T(u))-1=c(u)-1=1,
the elementT(u)is unitary. From
(76)1=T(u)T(u)*=T(u)T(u*)=(1+itT(a)-12t2T(a2)+⋯)×(1-itT(a)-12t2T(a2)+⋯),
we deduce thatT(a2)=T(a)2as desired.Theorem 14.
LetAandBbe unital C*-algebras and T:A→B a surjective linear map satisfyingc(T(x))=c(x)for allx∈A-1. Then, Tis a Jordan *-homomorphism multiplied by a unitary element inB.Proof.
First, let us prove thatb=T(1)is invertible. Sincec(T(1))=c(1)=1,bis regular. Lety=1-bb†, andx∈Asuch thaty=T(x). Notice thatb*y=y*b=0.
Forα∈ℂsufficiently small such that1+αx∈A-1,
(77)c(1+αx)2=c(b+αy)2=c(bb*+|α|2yy*).
Hence,
(78)lim|α|→0c(bb*+|α|2yy*)=lim|α|→0c(1+αx)2=1=c(bb*).
Reasoning in a similar way to [30, Theorem 6.2], we get
(79)c(1+αx)2=c(bb*+|α|yy*)≥1-|α|2∥y∥2,
and therefore
(80)c(1+itx)≥(1-t2∥y∥2)1/2,c(1-tx)≥(1-t2∥y∥2)1/2,
for small enough t∈ℝ. From these inequalities, we get, respectively, thatxandixare self-adjoint. This shows thatx=0and thusy=0. Consequently,1=bb†, that is,bis right invertible. Similarly it can be proved thatbis left invertible.
Note that, as in the previous theorem,Tis injective. ThereforeS∶=b-1Tis a unital and bijective linear map satisfying
(81)m(S(x))=m(b-1T(x))≤∥b-1∥m(T(x))=m(T(x))≤c(T(x))=c(x),∀x∈A-1.
Letybe a self-adjoint element inBandt∈ℝsmall such that1+itS-1(y)is invertible. Takingx=1+itS-1(y)in the previous identity, we have
(82)m(1+ity)≤c(1+itS-1(y)).
It follows thatS-1is self-adjoint and so isS.
We claim thatSis positive. Note that for everyx∈A†andu∈A-1, it is clear thatux∈A†, with
(83)(ux)(x†u-1)(ux)=ux,(x†u-1)(ux)(x†u-1)=x†u-1.
This implies, by [11, Theorem 2], the following:
(84)1∥x†u-1∥≤c(ux)≤∥x†u-1ux∥∥uxx†u-1∥∥x†u-1∥≤∥u∥∥u-1∥∥x†u-1∥.
Hence, for everyx∈A-1, we have
(85)c(S(x))=c(b-1T(x))≥1∥T(x)†b∥≥1∥b∥∥T(x)†∥=∥b∥-1c(T(x)),c(S(x))=c(b-1T(x))≤∥b-1∥∥T(1)∥∥T(x)†b∥≤∥b∥∣∥b-1∥2c(T(x)).
So, we have shown so far the following:
(86)∥b∥-1c(x)≤c(S(x))≤∥b∥∥b-1∥2c(x),∀x∈A-1.
The first inequality can be used to show the following:
(87)∂σ(S(x))⊂σK(S(x))⊆σ(x),
in a similar way as in the previous theorem. As a consequence,Sis positive. In order to conclude thatSis an isometric Jordan*-isomorphism, it suffices to prove thatS-1is also positive (see for instance [35, Corollary 5]). So, leth=S(a)be a positive element. AsS-1is self-adjoint,ais self-adjoint. We can therefore writea=x-y, wherex and yare positive elements andxy=yx=0. For everyμ∈ℂ, we have
(88)∂σ(S(x)+μS(y))⊂σK(S(x)+μS(y))⊂σ(x+μy)⊂ℝ∪μℝ.
(Recall that ifwz=zw=0, thenσ(w+z)∖{0}=(σ(w)∖{0})∪(σ(z)∖{0}).) The previous spectral inclusion gives
(89)σ(S(x)+μS(y))⊂ℝ∪μℝ.
By LemmasB and C in [37], we getS(y)=0and soy=0. Consequently,ais positive as desired. We conclude the proof by showing thatbis unitary. Indeed, sinceSis a Jordan*-isomorphism andT(x)=bS(x), it is clear that, for everyx∈A-1,T(x)is invertible with inverseT(x)-1=S(x-1)b-1. Moreover,c(T(x))=c(x); that is,
(90)∥T(x)-1∥=∥x-1∥,
for everyx∈A-1. SinceSis an isometry,
(91)∥S(x-1)∥=∥x-1∥=∥S(x-1)b-1∥,
or equivalently,
(92)∥y∥=∥yb-1∥,
for everyy∈B-1. This yields thatbis unitary.Note that, for every invertible elementa∈A𝒰, wherea=[an]is a normalized representative,
(93)c(a)=∥a-1∥-1=lim𝒰∥an-1∥-1=lim𝒰c(an).
We are now in position to prove the main results in this section.Theorem 15.
LetAandBbe unital C*-algebras andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<K, the following conditions:
(94)sup∥a∥=1,a∈A-1|c(T(a))-c(a)|<δ,∥T(1)-1∥<δ
imply that
(95)jmult(T)<ɛ,sa(T)<ɛ.Proof.
Note that ifb∈B𝒰has a coset representativeb=[bn]wherebnhas Moore-Penrose inverse for everyn∈ℕand{bn†}n∈ℕis bounded, thenbis Moore-Penrose invertible andb†=[bn†].
As we did above, suppose that the assertion is false, that is, we can findK0,ɛ0>0and a sequence{Tn}n∈ℕof linear maps fromAtoBsatisfying(i)
∥Tn∥≤K0,(ii)
sup∥a∥=1,a∈A-1|c(Tn(a))-c(a)|<1/n, ∥Tn(1)-1∥<1/n,(iii)
jmult(Tn)≥ɛ0orsa(Tn)≥ɛ0,
for everyn∈ℕ. ConsiderT=[Tn]:A𝒰→B𝒰. We claim thatTis unital and preserves the conorm of invertible elements. On the one hand,
(96)∥T(1)-1∥=lim𝒰∥Tn(1)-1∥≤lim𝒰1n=0,
soT(1)=1. On the other hand, for an invertible elementa∈A𝒰with norm 1, let[an]be its normalized representative:∥an∥=1and∥an-1∥<αfor some positiveα. We know that
(97)c(an)-1n<c(Tn(an)),for everyn∈ℕ.
HenceTn(an)∈A†, with∥Tn(an)†∥<2α, forn>2α. That is,{Tn(an)}is Moore-Penrose invertible almost everywhere, with their respective Moore-Penrose inverses uniformly bounded in norm. This ensures thatT(a)is Moore-Penrose invertible andc(T(a))=lim𝒰c(Tn(an)). Finally, Consider the following:
(98)|c(T(a))-c(a)|=lim𝒰|c(Tn(an))-c(an)|≤lim𝒰1n=0.
By Theorem 13,Tis a Jordan *-homomorphism, which gives the contradiction.LetX and Y be complex Banach spaces, and letℬ(X,Y)be the algebra of all bounded linear operators fromXtoY. Recall that the surjectivity modulus of T is given byq(T)∶=sup{ɛ≥0:ɛBY⊆T(BX)}, whereas usualBXdenotes the closed unit ball ofX. Note thatq(T)>0 if and only if T is surjective, andq(T)=inf{∥ST∥:S∈ℬ(X),∥S∥=1} (see [36, Theorem II.9.11]).Theorem 16.
LetAandBbe unital C*-algebras andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<Kand q(T)>K-1, the following condition:
(99)sup∥a∥=1,a∈A-1|c(T(a))-c(a)|<δ
implies that
(100)jmult(T(1)*T)<ɛ,sa(T(1)*T)<ɛ.Proof.
If we assume the contrary, hence, as above we findK0,ɛ0>0and a sequence{Tn}n∈ℕof linear maps fromAtoBsatisfying(i)
∥Tn∥≤K0,(ii)
sup∥a∥=1,a∈A-1|c(Tn(a))-c(a)|<1/n, q(Tn)≥K0-1,(iii)
jmult(Tn(1)*Tn)≥ɛ0orsa(Tn(1)*Tn)≥ɛ0,
for everyn∈ℕ.
As in the previous theorem, the mapT=[Tn]preserves the conorm of all invertible elements. Moreover,
(101)q(T)=lim𝒰q(Tn)≥K0-1>0,
and thusTis surjective. By Theorem 14,T(1)is unitary andT(1)*Tis a unital Jordan *-isomorphism.


## 4.1. Approximate Preservers of the Moore-Penrose Inverse and the Generalized Inverse

Given aJB*-tripleE, the triple cube of an elementx∈Eis defined asx[3]∶={x,x,x}. An element satisfyinge[3]=eis called a tripotent. The following polarization identity allows us to write the triple product as linear combination of triple cubes:
(46){x,y,z}=∑α2=1‍∑β4=1‍αβ(x+αy+βz)[3],forx,y,z∈E.
Hence, a linear map between JB*-triples is a triple homomorphism if and only if it preserves triple cubes.Each tripotente in E gives rise to the so-called Peirce decomposition of E associated to e, that is,
(47)E=E2(e)⊕E1(e)⊕E0(e),
where for i=0,1,2,Ei(e)is thei/2eigenspace of L(e,e). The peirce spaceE2(e)is a JB*-algebra with product x·y∶={x,e,y} and involutionx♯∶={e,x,e}. Moreover, the triple product induced onE2(e)by this Jordan *-algebra structure coincides with its original triple product.It is proved in [16, Lemma 3.2] (compare with [13, Theorem 3.4]) that for every regular elementain a JB*-tripleE, there exists a tripotente∈Esuch thatais a self-adjoint invertible element in the JB*-algebraE2(e). Ifais invertible, its inverse is denoted as usual bya-1. Moreover, ifaandbare invertible elements in the Jordan algebraJ=(J,∘)such thata-b-1is also invertible, thena-1+(b-1-a)-1 is invertible and the Hua identity
(48)(a-1+(b-1-a)-1)-1=a-Ua(b)
holds, whereUa(x)=2a∘(a∘x)-a2∘x(see [33], (11)).LetAbe a unitalC*-algebra, andu∈A∧∖{0}. Then, there exists a unique partial isometrye, such thatuis self-adjoint and invertible in the Jordan algebra(EA)2(e)=ee*Ae*e, with inverseu∧.Hence, for everyλ∈ℂwith0<|λ|<∥u∧∥-2, the elementu-λu∧is invertible inee*Ae*e. Reciprocally, the inverses ofu-λu∧andu∧-(u-λu∧)∧inee*Ae*eare their generalized inverses inA. By the Hua identity (48), we obtain
(49)u-λ-1u[3]=(u∧-(u-λu∧)∧)∧.Theorem 8.
LetAandBbe C*-algebras,Abeing unital, andT:A→Ba bounded linear map satisfyingT(x-1)=T(x)∧for every self-adjoint invertible elementx∈A. Then, Tis a triple homomorphism.Proof.
Arguing as in [9, Lemma 3.1], pick a self-adjoint invertible elementu∈A. We may assume thatT(u)≠0. Then, givenλ∈ℂwith0<|λ|<min{∥u-1∥-2,∥T(u)∧∥-2}, from identities (36) and (49), we get
(50)T(u)-λ-1T(u)[3]=(T(u)∧-(T(u)-λT(u)∧)∧)∧=T(u)-λ-1T(u3),
which shows thatT(u3)=T(u)[3]. Once we have proved thatTpreserves cubes of self-adjoint invertible elements, given a self-adjoint elementa∈A, andα∈ℝwith|α|>∥a∥, as the elementa+αis self-adjoint and invertible, we getT((a+α)3)=T(a+α)[3]. Expanding this las equation we obtain
(51)T(a3)+3αT(a2)+3α2T(a)+α3T(1)=T(a)[3]+α3T(1)+2α{T(a),T(a),T(1)}+α{T(a),T(1),T(a)}+2α2{T(1),T(1),T(a)}+α2{T(1),T(a),T(1)},
for everya∈A, and|α|>∥a∥. From this, we deduce thatT(a3)=T(a)[3]for everya∈A. That is,Tpreserves triple cubes of self-adjoint elements. By [34, Theorem 20],Tis a triple homomorphism.Recall that we measure how close is a linear mapT:A→Bbetween C*-algebras to being a triple homomorphism or self-adjoint by the triple multiplicativity and the self-adjointness ofT, respectively as follows:
(52)tmult(T)∶=sup{∥T({a,b,c})-{T(a),T(b),T(c)}∥:∥a∥=∥b∥=∥c∥=1},sa(T)∶=sup∥a∥=1∥T(a*)*-T(a)∥.Remark 9.
LetAandBbe C*-algebras. It is clear that every Jordan*-homomorphismT:A→Bis a triple homomorphism. We ask whetherjmult(T)and sa(T)being small implytmult(T)being small.
LetT:A→Bbe a bounded linear map. Define
(53)TJ(a,b)∶=T(a∘b)-T(a)∘T(b),TT(a,b,c)∶=T({a,b,c})-{T(a),T(b),T(c)},T*(a)∶=T(a*)*-T(a),
for everya,b,c∈A. Then,
(54)T({a,b,c})=T((a∘b*)∘c)+T(a∘(b*∘c))-T((a∘c)∘b*)=T(a∘b*)∘T(c)+TJ(a∘b*,c)+T(a)∘T(b*∘c)+TJ(a,b*∘c)-T(a∘c)∘T(b*)-TJ(a∘c,b*)=(T(a)∘T(b*))∘T(c)+TJ(a,b*)∘T(c)+TJ(a∘b*,c)+T(a)∘(T(b*)∘T(c))+T(a)∘TJ(b*,c)+TJ(a,b*∘c)-(T(a)∘T(c))∘T(b*)-TJ(a,c)∘T(b*)-TJ(a∘c,b*)={T(a),T(b*)*,T(c)}+TJ(a,b*)∘T(c)+TJ(a∘b*,c)+T(a)∘TJ(b*,c)+TJ(a,b*∘c)-TJ(a,c)∘T(b*)-TJ(a∘c,b*).
Therefore,
(55)TT(a,b,c)={T(a),T*(b),T(c)}+TJ(a,b*)∘T(c)+TJ(a∘b*,c)+T(a)∘TJ(b*,c)+TJ(a,b*∘c)-TJ(a,c)∘T(b*)-TJ(a∘c,b*).
This implies that
(56)tmult(T)≤∥T∥2sa(T)+3(∥T∥+1)jmult(T).As in Lemma3, it can be shown that, for an operatorT=[Tn], the following holds:
(57)lim𝒰tmult(Tn)=tmult(T),lim𝒰sa(Tn)=sa(T).We will omit the proof of the next result in order to avoid repetition. The argument is analogous to the one used in the proofs of Theorems4 and 6: assuming the contrary, we can construct a mapT=[Tn]:A𝒰→B𝒰between the ultrapowers fulfillingT(x-1)=T(x)∧for every self-adjoint invertiblex∈A𝒰. By Theorem 8,Tis a triple homomorphism.Theorem 10.
LetAandBbe C*-algebras whereAis unital andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<K, the condition
(58)sup∥a∥=1,a=a*∥T(a∧)-T(a)∧∥<δ
implies that
(59)tmult(T)<ɛ.Remark 11.
Notice that in Theorem8 it can be also obtained thatT(1)*Tis a Jordan*-homomorphism. Hence, the hypothesis of Theorem 10 also yields tojmult(T(1)*T)<ɛandsa(T(1)*T)<ɛ.Corollary 12.
LetAandBbe C*-algebras whereAis unital andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<K, the conditions
(60)sup∥a∥=1,a∈A†∥T(a†)-T(a)†∥<δ,sa(T)<δ
imply that
(61)tmult(T)<ɛ.Proof.
Let us briefly sketch the proof: assuming the contrary, there existK0,ɛ0>0and a sequence{Tn}n∈ℕof linear maps from A toBsuch that, for everyn∈ℕ,
(62)∥Tn∥<K0,sup∥a∥=1∥Tn(a†)-Tn(a)†∥<1n,sa(Tn)<1n,tmult(Tn)≥ɛ0.
Then, T=[Tn]:A𝒰→B𝒰is a self-adjoint map such thatT(a-1)=T(a)†, for every invertible elementa∈A𝒰. In particular,T(a-1)=T(a)∧, for every invertible self-adjoint elementa∈A𝒰. From Theorem 8,Tis a triple homomorphism (Contradiction).


## 4.2.  Maps Approximately Preserving the Conorm

LetAandBbe unitalC*-algebras. Kadison proved in [35] that a surjective linear mapT:A→Bis an isometry if and only ifTis a Jordan*-isomorphism multiplied by a unitary element inB. In [30], the authors address the question of characterizing surjective linear maps preserving some spectral quantities. Given an elementaof a Banach algebraA, the minimum modulus and the surjectivity modulus ofaare defined, respectively, by
(63)m(a)∶=inf{∥ax∥:x∈A,∥x∥=1},q(a)∶=inf{∥xa∥:x∈A,∥x∥=1}.
Obviously,m(a)=0(resp.,q(a)=0) if and only ifais a left (resp., right) topological divisor of zero. It is well known that for any invertible elementa∈A,
(64)m(a)=q(a)=∥a-1∥-1.
Moreover,
(65)m(a)m(b)≤m(ab)≤∥a∥m(b),|m(a)-m(b)|≤∥a-b∥,
for everya,b∈A. IfAis aC*-algebra, thenm(a)=q(a*). In particular,m(a)>0(resp.,q(a)>0) if and only ifais left (resp., right) invertible.Recall also from the introduction that theconorm of an elementain a Banach algebraA, is defined as
(66)c(a)∶={inf{∥ax∥:dist(x,ker(La))≥1},ifa≠0,∞,ifa=0.
For a regular elementain aC*-algebraA,
(67)c(a)=∥a†∥-1.LetAandBbe unital C*-algebras. By Theorems 3.1 and  3.2 in [30], ifT:A→Bis a linear map preserving any of these spectral quantities, thenTis an isometric Jordan *-homomorphism wheneverTis unital, andTis an isometric Jordan *-homomorphism multiplied by a unitary element, wheneverTis surjective. In the next results, we show that the same holds if we just impose the preserving condition for invertible elements. Notice that we focus our attention on the conorm but identical results can be established for the minimum and surjective modulus.Theorem 13.
LetAandBbe unital C*-algebras andT:A→Ba unital linear map satisfying c(T(x))=c(x)for allx∈A-1. Then, Tis a Jordan *-homomorphism.Proof.
First, let us prove thatTis injective. Takea0∈Asuch thatT(a0)=0and letα∈ℂbe sufficiently small so that1+αa0is invertible. Then,
(68)1=c(T(1))=c(T(1+αa0))=c(1+αa0).
In particular, we get
(69)1≤c(1+ita0),1≤c(1-ta0),
ast→0. Hence, by [30, Lemma 4.1], bothia0anda0are self-adjoint and, consequently,a0=0.
We claim thatTis positive. Indeed, given a self-adjoint elementa∈A, we know that
(70)1+o(t)≤c(1+ita),(ast→0).
Since1+itais invertible fort∈ℝsmall enough, it follows that
(71)1+o(t)≤c(1+ita)=c(1+itT(a)),(ast→0).
This implies thatT(a)is self-adjoint.
Moreover, givenx∈Aandλ∉σ(x), there exists a neighborhoodUλofλsuch thatUλ∩σ(x)=∅. Ifλ∉σ(x), thenλ∉σK(x), whereσK(x)denotes the Kato spectrum ofxas follows:
(72)σK(a)∶={λ∈ℂ:limμ→λc(a-μ)=0}.
As forμ∈Uλ, the elementx-μis invertible; then, we havec(T(x)-μ)=c(x-μ)for everyμ∈Uλ. Consequently,
(73)limμ→λc(T(x)-μ)>0,
andλ∉σK(T(x)). Since∂σ(a)⊆σK(a)⊆σ(a)for everya∈A(see [36, Sections 12,13]), we have just proved that
(74)∂σ(T(x))⊆σK(T(x))⊆σ(x),
for everyx∈A. BeingTself-adjoint, this implies thatTis positive and hence,∥T∥=1.
Arguing as in [30, Theorem 5.1], given a self-adjoint elementa∈Aandtsufficiently small so thatu=eitais a unitary element with spectrum strictly contained in the unit circle𝕋, since
(75)∂σ(T(u))⫋𝕋,∥T(u)∥≤1,∥T(u)-1∥=c(T(u))-1=c(u)-1=1,
the elementT(u)is unitary. From
(76)1=T(u)T(u)*=T(u)T(u*)=(1+itT(a)-12t2T(a2)+⋯)×(1-itT(a)-12t2T(a2)+⋯),
we deduce thatT(a2)=T(a)2as desired.Theorem 14.
LetAandBbe unital C*-algebras and T:A→B a surjective linear map satisfyingc(T(x))=c(x)for allx∈A-1. Then, Tis a Jordan *-homomorphism multiplied by a unitary element inB.Proof.
First, let us prove thatb=T(1)is invertible. Sincec(T(1))=c(1)=1,bis regular. Lety=1-bb†, andx∈Asuch thaty=T(x). Notice thatb*y=y*b=0.
Forα∈ℂsufficiently small such that1+αx∈A-1,
(77)c(1+αx)2=c(b+αy)2=c(bb*+|α|2yy*).
Hence,
(78)lim|α|→0c(bb*+|α|2yy*)=lim|α|→0c(1+αx)2=1=c(bb*).
Reasoning in a similar way to [30, Theorem 6.2], we get
(79)c(1+αx)2=c(bb*+|α|yy*)≥1-|α|2∥y∥2,
and therefore
(80)c(1+itx)≥(1-t2∥y∥2)1/2,c(1-tx)≥(1-t2∥y∥2)1/2,
for small enough t∈ℝ. From these inequalities, we get, respectively, thatxandixare self-adjoint. This shows thatx=0and thusy=0. Consequently,1=bb†, that is,bis right invertible. Similarly it can be proved thatbis left invertible.
Note that, as in the previous theorem,Tis injective. ThereforeS∶=b-1Tis a unital and bijective linear map satisfying
(81)m(S(x))=m(b-1T(x))≤∥b-1∥m(T(x))=m(T(x))≤c(T(x))=c(x),∀x∈A-1.
Letybe a self-adjoint element inBandt∈ℝsmall such that1+itS-1(y)is invertible. Takingx=1+itS-1(y)in the previous identity, we have
(82)m(1+ity)≤c(1+itS-1(y)).
It follows thatS-1is self-adjoint and so isS.
We claim thatSis positive. Note that for everyx∈A†andu∈A-1, it is clear thatux∈A†, with
(83)(ux)(x†u-1)(ux)=ux,(x†u-1)(ux)(x†u-1)=x†u-1.
This implies, by [11, Theorem 2], the following:
(84)1∥x†u-1∥≤c(ux)≤∥x†u-1ux∥∥uxx†u-1∥∥x†u-1∥≤∥u∥∥u-1∥∥x†u-1∥.
Hence, for everyx∈A-1, we have
(85)c(S(x))=c(b-1T(x))≥1∥T(x)†b∥≥1∥b∥∥T(x)†∥=∥b∥-1c(T(x)),c(S(x))=c(b-1T(x))≤∥b-1∥∥T(1)∥∥T(x)†b∥≤∥b∥∣∥b-1∥2c(T(x)).
So, we have shown so far the following:
(86)∥b∥-1c(x)≤c(S(x))≤∥b∥∥b-1∥2c(x),∀x∈A-1.
The first inequality can be used to show the following:
(87)∂σ(S(x))⊂σK(S(x))⊆σ(x),
in a similar way as in the previous theorem. As a consequence,Sis positive. In order to conclude thatSis an isometric Jordan*-isomorphism, it suffices to prove thatS-1is also positive (see for instance [35, Corollary 5]). So, leth=S(a)be a positive element. AsS-1is self-adjoint,ais self-adjoint. We can therefore writea=x-y, wherex and yare positive elements andxy=yx=0. For everyμ∈ℂ, we have
(88)∂σ(S(x)+μS(y))⊂σK(S(x)+μS(y))⊂σ(x+μy)⊂ℝ∪μℝ.
(Recall that ifwz=zw=0, thenσ(w+z)∖{0}=(σ(w)∖{0})∪(σ(z)∖{0}).) The previous spectral inclusion gives
(89)σ(S(x)+μS(y))⊂ℝ∪μℝ.
By LemmasB and C in [37], we getS(y)=0and soy=0. Consequently,ais positive as desired. We conclude the proof by showing thatbis unitary. Indeed, sinceSis a Jordan*-isomorphism andT(x)=bS(x), it is clear that, for everyx∈A-1,T(x)is invertible with inverseT(x)-1=S(x-1)b-1. Moreover,c(T(x))=c(x); that is,
(90)∥T(x)-1∥=∥x-1∥,
for everyx∈A-1. SinceSis an isometry,
(91)∥S(x-1)∥=∥x-1∥=∥S(x-1)b-1∥,
or equivalently,
(92)∥y∥=∥yb-1∥,
for everyy∈B-1. This yields thatbis unitary.Note that, for every invertible elementa∈A𝒰, wherea=[an]is a normalized representative,
(93)c(a)=∥a-1∥-1=lim𝒰∥an-1∥-1=lim𝒰c(an).
We are now in position to prove the main results in this section.Theorem 15.
LetAandBbe unital C*-algebras andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<K, the following conditions:
(94)sup∥a∥=1,a∈A-1|c(T(a))-c(a)|<δ,∥T(1)-1∥<δ
imply that
(95)jmult(T)<ɛ,sa(T)<ɛ.Proof.
Note that ifb∈B𝒰has a coset representativeb=[bn]wherebnhas Moore-Penrose inverse for everyn∈ℕand{bn†}n∈ℕis bounded, thenbis Moore-Penrose invertible andb†=[bn†].
As we did above, suppose that the assertion is false, that is, we can findK0,ɛ0>0and a sequence{Tn}n∈ℕof linear maps fromAtoBsatisfying(i)
∥Tn∥≤K0,(ii)
sup∥a∥=1,a∈A-1|c(Tn(a))-c(a)|<1/n, ∥Tn(1)-1∥<1/n,(iii)
jmult(Tn)≥ɛ0orsa(Tn)≥ɛ0,
for everyn∈ℕ. ConsiderT=[Tn]:A𝒰→B𝒰. We claim thatTis unital and preserves the conorm of invertible elements. On the one hand,
(96)∥T(1)-1∥=lim𝒰∥Tn(1)-1∥≤lim𝒰1n=0,
soT(1)=1. On the other hand, for an invertible elementa∈A𝒰with norm 1, let[an]be its normalized representative:∥an∥=1and∥an-1∥<αfor some positiveα. We know that
(97)c(an)-1n<c(Tn(an)),for everyn∈ℕ.
HenceTn(an)∈A†, with∥Tn(an)†∥<2α, forn>2α. That is,{Tn(an)}is Moore-Penrose invertible almost everywhere, with their respective Moore-Penrose inverses uniformly bounded in norm. This ensures thatT(a)is Moore-Penrose invertible andc(T(a))=lim𝒰c(Tn(an)). Finally, Consider the following:
(98)|c(T(a))-c(a)|=lim𝒰|c(Tn(an))-c(an)|≤lim𝒰1n=0.
By Theorem 13,Tis a Jordan *-homomorphism, which gives the contradiction.LetX and Y be complex Banach spaces, and letℬ(X,Y)be the algebra of all bounded linear operators fromXtoY. Recall that the surjectivity modulus of T is given byq(T)∶=sup{ɛ≥0:ɛBY⊆T(BX)}, whereas usualBXdenotes the closed unit ball ofX. Note thatq(T)>0 if and only if T is surjective, andq(T)=inf{∥ST∥:S∈ℬ(X),∥S∥=1} (see [36, Theorem II.9.11]).Theorem 16.
LetAandBbe unital C*-algebras andK,ɛ>0. Then, there existsδ>0such that for every linear mapT:A→Bwith∥T∥<Kand q(T)>K-1, the following condition:
(99)sup∥a∥=1,a∈A-1|c(T(a))-c(a)|<δ
implies that
(100)jmult(T(1)*T)<ɛ,sa(T(1)*T)<ɛ.Proof.
If we assume the contrary, hence, as above we findK0,ɛ0>0and a sequence{Tn}n∈ℕof linear maps fromAtoBsatisfying(i)
∥Tn∥≤K0,(ii)
sup∥a∥=1,a∈A-1|c(Tn(a))-c(a)|<1/n, q(Tn)≥K0-1,(iii)
jmult(Tn(1)*Tn)≥ɛ0orsa(Tn(1)*Tn)≥ɛ0,
for everyn∈ℕ.
As in the previous theorem, the mapT=[Tn]preserves the conorm of all invertible elements. Moreover,
(101)q(T)=lim𝒰q(Tn)≥K0-1>0,
and thusTis surjective. By Theorem 14,T(1)is unitary andT(1)*Tis a unital Jordan *-isomorphism.



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*Source: 757646-2013-12-08.xml*