Linear maps preserving equivalence or asymptotic equivalence on Banach space

Zijie Qin; Lin Chen

doi:10.1515/math-2023-0125

Article Open Access

Linear maps preserving equivalence or asymptotic equivalence on Banach space

Zijie Qin and Lin Chen

Published/Copyright: October 10, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

Let X be a complex Banach space with dimension at least two and B ( X ) the algebra of all bounded linear operators on X . We show that a bijective linear map Φ preserves asymptotic equivalence if and only if it preserves equivalence, and in turn, if and only if there exist invertible bounded linear operators T and S such that either Φ ( A ) = T A S or Φ ( A ) = T A * S for all A ∈ B ( X ) .

Keywords: linear preservers; asymptotic equivalence; equivalence relations; Banach spaces

MSC 2010: 47B49

1 Introduction and main result

The problem of characterizing linear maps on operator algebras preserving certain properties, subsets, or relations has attracted the attention of many authors in the last few decades (see [1,2] and references therein). In this article, we will deal with linear maps preserving certain equivalence relations.

Throughout this article, all algebras and vector spaces will be over the complex field C . Let X be a complex Banach space with topological dual X * . We denote by B ( X ) the algebra of all bounded linear operators on X . Recall that two operators A and B are equivalent, denoted by A ≃ B , if A = T B S for some invertible operators T , S ∈ B ( X ) . A map Φ from B ( X ) into itself is said to be equivalence preserving if Φ ( A ) ≃ Φ ( B ) whenever A ≃ B . In the finite-dimensional case, Horn et al. [3] characterized linear maps preserving equivalence on the algebra of all n × n matrices. For the infinite-dimensional case, Petek and Radić [4] proved that if X is an infinite-dimensional reflexive complex Banach space, then linear bijections Φ : B ( X ) → B ( X ) preserve equivalence if and only if there exist bounded invertible linear operators T , S such that either Φ ( A ) = T A S , for all A ∈ B ( X ) , or Φ ( A ) = T A * S , for all A ∈ B ( X ) , where A * denotes the adjoint of A . Later, in [5], they showed that if X is an infinite-dimensional complex Banach space, then the relation A − B ≃ C if and only if Φ ( A ) − Φ ( B ) ≃ Φ ( C ) is enough to determine the structure of surjective maps Φ defined on B ( X ) . Recently, Radić [6] defined another equivalence relation on B ( X ) and refined the result in [4].

Maps preserving various types of equivalence relations have been considered in the last 20 years. For example, studies in [7–9] studied similarity; the article [10] studied asymptotic similarity; the article [11] studied unitary similarity; the first author of this article in [12] studied involution similarity. We now define another equivalence relation on B ( X ) . Let A and B be in B ( X ) . By O ( A ) , we denote the equivalence orbit of A , i.e.,

O ( A ) = { B ∈ B ( X ) : B ≃ A } .

It is easy to see that A ≃ B if and only if O ( A ) = O ( B ) . Let O ( A ) ¯ denote the norm closure of O ( A ) . We say that the two operators A and B are asymptotically equivalent, denoted by A ≃ a B , if O ( A ) ¯ = O ( B ) ¯ . It is obvious that asymptotic equivalence is an equivalence relation on B ( X ) . For A , B ∈ B ( X ) , if A ≃ B , then A ≃ a B . A map Φ : B ( X ) → B ( X ) is said to be asymptotic equivalence preserving if Φ ( A ) ≃ a Φ ( B ) whenever A ≃ a B . The purpose of this article is to obtain a complete classification of linear bijections that preserve equivalence or asymptotic equivalence and then generalize the result in [4] to the general Banach space case.

Our main result reads as follows.

Theorem 1.1

Let X be a complex Banach space with dim X ≥ 2 . Suppose Φ : B ( X ) → B ( X ) is a surjective linear map satisfying

A ≃ B ⇒ Φ ( A ) ≃ a Φ ( B ) ,

for all A , B ∈ B ( X ) . Then, one of the following statements holds.

There exist invertible bounded linear operators T , S : X → X such that:
Φ ( A ) = T A S ,
for all A ∈ B ( X ) .
There exist invertible bounded linear operators T : X * → X and S : X → X * such that:
Φ ( A ) = T A * S ,
for all A ∈ B ( X ) .
Φ ( F ) = 0 , for every finite-rank operator F in B ( X ) .

2 Preliminary results

In this section, we show some results that will be used to prove our main result. Let X be a complex Banach space. By ℱ ( X ) , we denote the ideal of all finite-rank operators in B ( X ) . For A ∈ B ( X ) , the notations ker ( A ) , ran ( A ) , and rank ( A ) will stand for the kernel, the range, and the rank of A , respectively. For nonzero vector x ∈ X and nonzero functional f ∈ X * , the rank-one operator x ⊗ f is defined as the map y ↦ f ( y ) x , y ∈ X .

The following lemma comes from [4, Lemma 2.2].

Lemma 2.1

Let X be a complex Banach space and A , B ∈ B ( X ) . Assume that A is of rank one and B is nonzero. If there exist two nonzero scalars λ 1 and λ 2 with λ 1 ≠ λ 2 such that rank ( A + λ i B ) = 1 , for i = 1 , 2 , then rank ( B ) = 1 .

The following result gives a characterization of rank-one non-increasing additive maps.

Proposition 2.2

[13] Let X be a complex Banach space with dim X ≥ 2 . Suppose Φ : ℱ ( X ) → ℱ ( X ) is an additive map. If Φ maps rank-one operators to operators of rank at most one, then one of the following statements holds.

There exist a functional g ∈ X * and an additive map τ : ℱ ( X ) → X such that Φ ( A ) = τ ( A ) ⊗ g , for every A ∈ ℱ ( X ) .
There exist a vector y ∈ X and an additive map φ : ℱ ( X ) → X * such that Φ ( A ) = y ⊗ φ ( A ) , for every A ∈ ℱ ( X ) .
There exist additive maps T : X → X and S : X * → X * such that Φ ( x ⊗ f ) = T x ⊗ S f , for every rank-one operator x ⊗ f ∈ ℱ ( X ) .
There exist additive maps T : X * → X and S : X → X * such that Φ ( x ⊗ f ) = T f ⊗ S x , for every rank-one operator x ⊗ f ∈ ℱ ( X ) .

Remark 2.3

If Φ is linear, it is clear that τ , φ , T , and S are linear. Moreover, if Φ is injective, then T and S are injective too.

Recall that a linear map Φ : B ( X ) → B ( X ) is said to be invertibility preserving if Φ ( A ) is invertible for every invertible operator A ∈ B ( X ) . The following proposition characterizes linear maps preserving invertibility on B ( X ) , which is crucial for proving Theorem 1.1.

Proposition 2.4

[14] Let X be a complex Banach space. Suppose that Φ : B ( X ) → B ( X ) is a linear bijective map. If Φ preserves invertibility, then one of the following statements holds.

There exist invertible bounded linear operators T , S : X → X such that Φ ( A ) = T A S , for all A ∈ B ( X ) .
There exist invertible bounded linear operators T : X * → X and S : X → X * such that Φ ( A ) = T A * S , for all A ∈ B ( X ) .

We will close this section with the following lemmas.

Lemma 2.5

[14] Let X be a complex Banach space and x ∈ X and f ∈ X * . Then, I + x ⊗ f is invertible in B ( X ) if and only if f ( x ) ≠ − 1 .

Lemma 2.6

Let X be a complex Banach space and A , B ∈ B ( X ) with A ≃ a B . If A is invertible, then B is invertible as well.

Proof

Since A ∈ O ( B ) ¯ , there exist sequences { T n } n = 1 ∞ and { S n } n = 1 ∞ of invertible operators in B ( X ) such that A = lim n → ∞ T n B S n . As A is invertible and bounded, it is clear that ‖ A − 1 ‖ > 0 . Now, we can observe that there exists a positive integer N such that ‖ A − T n B S n ‖ < ‖ A − 1 ‖ − 1 , for every n > N . Note that

‖ A − 1 ( A − T N + 1 B S N + 1 ) ‖ ≤ ‖ A − 1 ‖ ⋅ ‖ A − T N + 1 B S N + 1 ‖ < 1 ,

T N + 1 B S N + 1 = A − ( A − T N + 1 B S N + 1 ) = A ( I − A − 1 ( A − T N + 1 B S N + 1 ) )

is invertible. Hence, B is invertible. The proof is complete.□

Lemma 2.7

Let X be a complex Banach space. Assume that { T n } n = 1 ∞ is a sequence of rank-one operators in B ( X ) . If { T n } n = 1 ∞ converges to a nonzero operator T ∈ B ( X ) , then T is of rank one.

Proof

We first claim that there exists a positive integer N such that m < ‖ T n ‖ < M , for all n > N , where 0 < m < M . Actually, since lim n → ∞ T n = T , we have lim n → ∞ ‖ T n ‖ = ‖ T ‖ . Note that T ≠ 0 . So, for ‖ T ‖ > 0 , there exists a positive integer N such that

∣ ‖ T n ‖ − ‖ T ‖ ∣ < 1 2 ‖ T ‖ ,

for all n > N , which further implies that 1 2 ‖ T ‖ < ‖ T n ‖ < 3 2 ‖ T ‖ , for all n > N . Set m = 1 2 ‖ T ‖ and M = 3 2 ‖ T ‖ , establishing the claim.

Now suppose that T n = x n ⊗ f n , where x n ∈ X and f n ∈ X * with ‖ f n ‖ = 1 . By the claim, there exist positive numbers m and M such that m < ‖ x n ‖ < M , for all n > N . Assume, on the contrary, that rank ( T ) ≥ 2 . Then, there exist unit vectors y 1 , y 2 ∈ X such that T y 1 and T y 2 are linearly independent. Since lim n → ∞ T n = T , we see that lim n → ∞ T n y i = T y i , for i = 1 , 2 . It follows that

(2.1) lim n → ∞ f n ( y i ) x n = T y i , i = 1 , 2 .

Note that ‖ x n ⊗ f n ‖ < M . So, for i = 1 , 2 , { f n ( y i ) } n = 1 ∞ is a bounded sequence. Hence, it has a convergent subsequence. Without loss of generality, we may assume that there exists a nonzero scalar λ i such that lim n → ∞ f n ( y i ) = λ i , for i = 1 , 2 . This together with (2.1) gives { T y 1 , T y 2 } is a linearly dependent set, a contradiction. The proof is complete.□

3 Proof of the main result

In this section, we will complete the proof of Theorem 1.1. Throughout this section, X is a complex Banach space with dim X ≥ 2 and Φ : B ( X ) → B ( X ) is assumed to satisfy the hypotheses in Theorem 1.1. For clarity, we will organize the proof into a series of lemmas.

Lemma 3.1

Φ maps rank-one operators to operators of rank at most one.

Proof

Let B ∈ B ( X ) be of rank one. By the surjectivity of Φ , there exists an A ∈ B ( X ) such that Φ ( A ) = B . Since A ≠ 0 , there exists an x ∈ X such that A x ≠ 0 . Take a nonzero functional f ∈ X * such that f ( x ) = 0 . Let λ ∈ C be arbitrary. Then, I − λ x ⊗ f is invertible by Lemma 2.5. It follows that

A ≃ A ( I − λ x ⊗ f ) = A − λ A x ⊗ f .

This gives us

(3.1) O ( B ) ¯ = O ( B − λ Φ ( A x ⊗ f ) ) ¯ .

So B − λ Φ ( A x ⊗ f ) ∈ O ( B ) ¯ . Then, there exist sequences { T n } n = 1 ∞ and { S n } n = 1 ∞ of invertible operators in B ( X ) such that B − λ Φ ( A x ⊗ f ) = lim n → ∞ T n B S n . Note that B is of rank one. By Lemma 2.7, rank ( B − λ Φ ( A x ⊗ f ) ) = 1 for every λ ∈ C . Actually, if there exists λ ∈ C such that B − λ Φ ( A x ⊗ f ) = 0 , by equation (3.1), B = 0 , a contradiction. Now, we consider two cases.

Case 1: Φ ( A x ⊗ f ) ≠ 0 . By Lemma 2.1, Φ ( A x ⊗ f ) is of rank one. For any rank-one operator F ∈ B ( X ) , we have F ≃ A x ⊗ f . It follows that O ( Φ ( F ) ) ¯ = O ( Φ ( A x ⊗ f ) ) ¯ . By a similar argument, we obtain that rank ( Φ ( F ) ) = 1 .

Case 2: Φ ( A x ⊗ f ) = 0 . For every rank-one operator F ∈ B ( X ) , we have F ≃ A x ⊗ f . This leads to Φ ( F ) ≃ a Φ ( A x ⊗ f ) = 0 , which implies that Φ ( F ) = 0 , for every rank-one operator F ∈ B ( X ) .

So, Φ maps rank-one operators to operators of rank at most one. The proof is complete.□

Note that every finite-rank operator in B ( X ) can be written as a sum of rank-one operators in B ( X ) . By the proof of Lemma 3.1, we have either Φ is rank-one preserving or Φ ( ℱ ( X ) ) = 0 . In what follows, we may assume that Φ is a rank-one preserving map.

Lemma 3.2

Φ is injective.

Proof

Suppose that Φ ( A ) = 0 , for some operator A ∈ B ( X ) . If A ≠ 0 , then there exists a nonzero vector x ∈ X such that A x ≠ 0 . Since dim X ≥ 2 , we can take a nonzero functional f ∈ X * such that f ( x ) = 0 . By Lemma 2.5, I + x ⊗ f is invertible in B ( X ) . It follows from A ≃ A ( I + x ⊗ f ) that

0 = Φ ( A ) ≃ a Φ ( A + A x ⊗ f ) = Φ ( A x ⊗ f ) ,

which implies that Φ ( A x ⊗ f ) = 0 , which is a contradiction. So, A = 0 .□

In the sequel, we assume that Φ is injective.

Lemma 3.3

One and only one of the following statements holds.

There exist injective linear maps T : X → X and S : X * → X * such that Φ ( x ⊗ f ) = T x ⊗ S f , for every rank-one operator x ⊗ f ∈ B ( X ) .
There exist injective linear maps T : X * → X and S : X → X * such that Φ ( x ⊗ f ) = T f ⊗ S x , for every rank-one operator x ⊗ f ∈ B ( X ) .

Proof

By Lemma 3.1, one of the four cases in Proposition 2.2 holds. It is sufficient to show that Case 1 and Case 2 do not hold. For this, we first assume that Case 1 holds. Then, there exist a nonzero functional g ∈ X * and a linear map τ : ℱ ( X ) → X such that Φ ( x ⊗ f ) = τ ( x ⊗ f ) ⊗ g , for every rank-one operator x ⊗ f ∈ B ( X ) . Take and fix g 0 ∈ X * such that g 0 and g are linearly independent and then choose a nonzero vector x 0 ∈ X . By the surjectivity of Φ , there exists an A ∈ B ( X ) such that Φ ( A ) = x 0 ⊗ g 0 . Note that rank ( A ) ≥ 2 . Hence, there exist x 1 , x 2 ∈ X such that A x 1 and A x 2 are linearly independent. Take nonzero functionals f 1 , f 2 ∈ X * such that f 1 ( x 1 ) = f 2 ( x 2 ) = 0 . Then,

A ≃ A ( I + x i ⊗ f i ) = A + A x i ⊗ f i , i = 1 , 2 .

It follows that

O ( x 0 ⊗ g 0 ) ¯ = O ( x 0 ⊗ g 0 + τ ( A x i ⊗ f i ) ⊗ g ) ¯ , i = 1 , 2 .

By Lemma 2.7, we have rank ( x 0 ⊗ g 0 + τ ( A x i ⊗ f i ) ⊗ g ) = 1 , for i = 1 , 2 . Since g 0 and g are linearly independent, both { x 0 , τ ( A x 1 ⊗ f 1 ) } and { x 0 , τ ( A x 2 ⊗ f 2 ) } are linearly dependent, which implies that Φ ( A x 1 ⊗ f 1 ) and Φ ( A x 2 ⊗ f 2 ) are linearly dependent. Because Φ is injective and linear, A x 1 and A x 2 are linearly dependent, which is a contradiction. By a similar way, we show that Case 2 does not hold. The proof is complete.□

According to Lemma 3.3, we further assume that there exist injective linear maps T : X → X and S : X * → X * such that

(3.2) Φ ( x ⊗ f ) = T x ⊗ S f ,

for every rank-one operator x ⊗ f ∈ B ( X ) .

Lemma 3.4

T and S are surjective.

Proof

We only show that S is surjective. Suppose, on the contrary, that there exists a nonzero functional f 0 ∈ X * \ ran ( S ) . Fix a nonzero vector x 0 ∈ X . Since Φ is surjective, there exists an A ∈ B ( X ) such that Φ ( A ) = x 0 ⊗ f 0 . Note that rank ( A ) ≥ 2 . So, there exist x 1 , x 2 ∈ X such that A x 1 and A x 2 are linearly independent. Take f 1 , f 2 ∈ X * such that f 1 ( x 1 ) = f 2 ( x 2 ) = 1 . Then,

A ≃ A ( I + x i ⊗ f i ) = A + A x i ⊗ f i , i = 1 , 2 .

This together with equation (3.2) gives us

O ( x 0 ⊗ f 0 ) ¯ = O ( x 0 ⊗ f 0 + T A x i ⊗ S f i ) ¯ , i = 1 , 2 .

By Lemma 2.7, rank ( x 0 ⊗ f 0 + T A x i ⊗ S f i ) = 1 , for i = 1 , 2 . Since both { f 0 , S f 1 } and { f 0 , S f 2 } are linearly independent sets, both { x 0 , T A x 1 } and { x 0 , T A x 2 } are linearly dependent sets, which implies that T A x 1 and T A x 2 are linearly dependent. It follows from the linearity and the injectivity of T that A x 1 and A x 2 are linearly dependent, which is a contradiction. By a similar argument, we show that T is surjective too. The proof is complete.□

Since Φ is surjective, in the sequel, we assume that Φ ( U ) = I , for some nonzero operator U ∈ B ( X ) .

Lemma 3.5

Let x ∈ X and f ∈ X * . Then,

(3.3) ( S f ) ( T U x ) = f ( x )

and

(3.4) ( S U * f ) ( T x ) = f ( x ) .

Proof

Take any nonzero vector x ∈ X and nonzero functional f ∈ X * such that f ( x ) = 0 . Let λ be in C . Then, I + λ x ⊗ f is invertible by Lemma 2.5. It follows that

U ≃ U ( I + λ x ⊗ f ) = U + λ U x ⊗ f .

This together with equation (3.2) gives

I ≃ a I + λ T U x ⊗ S f .

By Lemma 2.6, I + λ T U x ⊗ S f is invertible. So, we have λ ( S f ) ( T U x ) ≠ − 1 by Lemma 2.5. Hence, by the arbitrariness of λ , we have

(3.5) ( S f ) ( T U x ) = 0 ,

for every x ∈ X and f ∈ X * with f ( x ) = 0 .

In what follows, we will show that there exists a constant μ ∈ { 0 , 1 } such that

(3.6) ( S f ) ( T U x ) = μ f ( x ) ,

for every x ∈ X and f ∈ X * . Choose x 1 ∈ X and f 1 ∈ X * such that f 1 ( x 1 ) = 1 . For every λ 1 ∈ C \ { 1 } , I − λ 1 x 1 ⊗ f 1 is invertible by Lemma 2.5. Set μ = ( S f 1 ) ( T U x 1 ) . By the same method as for the proof of equation (3.5), where, instead of I + λ x ⊗ f , we use invertible operator I − λ 1 x 1 ⊗ f 1 , we obtain that λ 1 μ ≠ 1 for all λ 1 ∈ C \ { 1 } . This implies that μ = 0 or μ = 1 .

Now, we shall consider two cases according to the dimension of X .

Case 1: X is finite-dimensional. Assume that dim X = n ( 2 ≤ n < ∞ ). Let { x 1 , x 2 , … , x n } and { f 1 , f 2 , … , f n } be the bases of X and X * satisfying f i ( x j ) = δ i j , where δ i j denotes the Kronecker delta. For i ≠ j , since ( f i + f j ) ( x i − x j ) = 0 , applying equation (3.5) gives ( S ( f i + f j ) ) ( T U ( x i − x j ) ) = 0 . This together with ( S f i ) ( T U x j ) = 0 gives us ( S f i ) ( T U x i ) = ( S f j ) ( T U x j ) = μ . For any x ∈ X , f ∈ X * , we can write x = ∑ i = 1 n α i x i and f = ∑ j = 1 n β j f j , where α 1 , … , α n , β 1 , … , β n ∈ C . If follows that

( S f ) ( T U x ) = S ∑ j = 1 n β j f j T U ∑ i = 1 n α i x i = μ f ( x ) .

Case 2: X is infinite-dimensional. Since T and S are the linear maps, it is sufficient to show that ( S f ) ( T U x ) = μ for every x ∈ X and f ∈ X * with f ( x ) = 1 . To do this, take a nonzero vector y ∈ X such that f ( y ) = f 1 ( y ) = 0 and y ∉ span { x , x 1 } . Then, we can find a functional g ∈ X * such that g ( y ) = 1 and g ( x ) = g ( x 1 ) = 0 . Since f 1 ( y ) = g ( x 1 ) = ( f 1 + g ) ( x 1 − y ) = 0 , by equation (3.5), we see that ( S g ) ( T U y ) = μ . By a similar argument, we obtain that ( S f ) ( T U x ) = ( S g ) ( T U y ) , which yields that ( S f ) ( T U x ) = μ .

So far, equation (3.6) has been established.

Finally, we will show that μ = 1 . Assume, on the contrary, that μ = 0 . Then, for all x ∈ X and f ∈ X * , ( S f ) ( T U x ) = 0 . Since S is surjective, we have f ( T U x ) = 0 , for all x ∈ X and f ∈ X * . This implies that T U x = 0 , for all x ∈ X . By the injectivity of T , U x = 0 , for all x ∈ X , which contradicts the fact that U ≠ 0 .□

The following lemma comes from Step 5 in [6]. For the sake of completeness, we give the details here.

Lemma 3.6

T and S are bounded.

Proof

Let Φ ( U ) = I . We first show that T U is bounded. Let { x n } n = 1 ∞ ⊆ X be such that lim n → ∞ x n = 0 and lim n → ∞ T U x n = y 0 , where y 0 ∈ X . It follows from equation (3.3) that for every f ∈ X * , we have S f ( y 0 ) = 0 . Since S is surjective by Lemma 3.4, y 0 = 0 . By the closed graph theorem, T U is bounded.

Now, we show that S is bounded. By the bijectivity of S and equation (3.3), we obtain S − 1 f ( x ) = f ( T U x ) , for every x ∈ X and f ∈ X * . Since T U is bounded, we have

∣ S − 1 f ( x ) ∣ = ∣ f ( T U x ) ∣ ≤ ‖ f ‖ ⋅ ‖ T U ‖ ⋅ ‖ x ‖ ,

for every x ∈ X and f ∈ X * . If follows that ‖ S − 1 f ‖ ≤ ‖ T U ‖ ⋅ ‖ f ‖ , for all f ∈ X * . So ‖ S − 1 ‖ ≤ ‖ T U ‖ , and hence, S is bounded. By the similar method, we show that T is bounded by equation (3.4).□

Lemma 3.7

U is invertible.

Proof

Let us first see that U has dense range. It suffices to show that U * is injective. For this, let U * f = 0 and f ∈ X * . By equation (3.4), f ( x ) = 0 , for all x ∈ X , i.e., f = 0 . Hence, U * is injective.

Now, we show that U is bounded below. Choose any nonzero x ∈ X . Since S is bijective, we can take f x ∈ X * such that ‖ S − 1 f x ‖ = 1 and ( S − 1 f x ) ( x ) = ‖ x ‖ . Then, ‖ f x ‖ = ‖ S S − 1 f x ‖ ≤ ‖ S ‖ . By Lemma 3.6, we obtain

‖ x ‖ = ∣ ( S − 1 f x ) ( x ) ∣ = ∣ f x ( T U x ) ∣ ≤ ‖ f x ‖ ⋅ ‖ T U x ‖ ≤ ‖ S ‖ ⋅ ‖ T ‖ ⋅ ‖ U x ‖ .

By the arbitrariness of x , U is bounded below, which completes the proof.□

The proof of Theorem 1.1

Let A ∈ B ( X ) be invertible. Then, A − 1 A U = U . Since U is invertible by Lemma 3.7, A ≃ U . It follows that Φ ( A ) ≃ a I . Then, by Lemma 2.6, Φ ( A ) is invertible. Applying Proposition 2.4, we complete the proof.

4 Applications

In this section, we will give some applications of Theorem 1.1. First, we generalize the result in [4] to the general Banach space case.

Corollary 4.1

Let X be a complex Banach space with dim X ≥ 2 . Then, a linear bijection Φ : B ( X ) → B ( X ) preserves equivalence if and only if it is of the form either (1) or (2) in Theorem 1.1.

Proof

The sufficiency is obvious. It remains to show the necessity. For this, suppose that A ≃ B , for A , B ∈ B ( X ) . It follows that Φ ( A ) ≃ Φ ( B ) . This leads to Φ ( A ) ≃ a Φ ( B ) . Since Φ is bijective, applying Theorem 1.1, we complete the proof.□

The following corollary characterizes bijective linear maps preserving asymptotic equivalence on Banach space.

Corollary 4.2

Let X be a complex Banach space with dim X ≥ 2 . Then, a linear bijection Φ : B ( X ) → B ( X ) preserves asymptotic equivalence if and only if it is of the form either (1) or (2) in Theorem 1.1.

Proof

The sufficiency is obvious. Now, we show the necessity. Let A ≃ B and A , B ∈ B ( X ) . Then, A ≃ a B . Since Φ preserves asymptotic equivalence, we see that Φ ( A ) ≃ a Φ ( B ) . Applying Theorem 1.1, we complete the proof.□

Applying Corollaries 4.1 and 4.2, we have

Corollary 4.3

Let X be a complex Banach space with dim X ≥ 2 . Suppose Φ : B ( X ) → B ( X ) is a bijective linear map. Then, the following statements are equivalent.

Φ preserves equivalence.
Φ preserves asymptotic equivalence.
One of the following statements holds.
1. There exist invertible bounded linear operators T , S : X → X such that Φ ( A ) = T A S , for all A ∈ B ( X ) .
2. There exist invertible bounded linear operators T : X * → X and S : X → X * such that Φ ( A ) = T A * S , for all A ∈ B ( X ) .

Acknowledgement

The authors would like to thank the referee for a very thorough reading of this article and many helpful comments that improved this article.

Funding information: This work was supported by the National Natural Science Foundation of China (No. 12061018).
Conflict of interest: The authors state no conflict of interest.

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Received: 2023-02-17

Revised: 2023-09-07

Accepted: 2023-09-07

Published Online: 2023-10-10

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2023-0125

Keywords for this article

linear preservers; asymptotic equivalence; equivalence relations; Banach spaces

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