On superstability of derivations in Banach algebras

Lemma 2.1

Let ℒ be a linear mapping on a Banach algebra A and let τ and δ be mappings such that

If A is semiprime, then τ and δ satisfy

In particular, if A has a unit, the mappings τ and δ are linear.

Proof

It follows from (2.1) that for t ∈ C and all x , y ∈ A ,

Hence, we have

which gives that

By substituting w x in place of x in equation (2.2) and using (2.3), we obtain

that is, [ x , t δ ( y ) − δ ( t y ) ] lies in the right annihilator ran ( A ) of A . If A is semiprime, we have from (2.4) that [ x , t δ ( y ) − δ ( t y ) ] = 0 . So we arrive at δ ( t y ) − t δ ( y ) ∈ Z ( A ) . If A has a unit, set x = e in (2.2). Then δ ( t y ) = t δ ( y ) is true.

In view of (2.1), we have

Again, by (2.1) and additivity of ℒ , we obtain

By combining (2.5) and (2.6), we see that

By using (2.7), we obtain similar results to (2.3) and (2.4). As a result, if A is smiprime, then we have δ ( y + z ) − δ ( y ) − δ ( z ) ∈ Z ( A ) . Moreover, if A contains the unit, then we have δ ( y + z ) = δ ( y ) + δ ( z ) . It can be proved in a similar way for τ . The lemma is completely established.□

Theorem 2.2

Let A be a semiprime Banach algebra with no nonzero central ideals. Assume that mappings Φ : A × A → [ 0 , ∞ ) and φ : A × A → [ 0 , ∞ ) satisfy

1. σ ( x , y ) ≔ ∑ j = 0 ∞ 1 2 j Φ ( 2 j x , 2 j y ) < ∞ ( x , y ∈ A ) ,

2. lim n → ∞ 1 2 n φ ( 2 n x , y ) = lim n → ∞ 1 2 n φ ( x , 2 n y ) = 0 ( x , y ∈ A ) .

Then δ is a Jordan derivation. Moreover, δ is a derivation.

Proof

It follows from (2.8) and the Găvruta theorem [8] that there exists a unique additive mapping ℒ : A → A satisfying ‖ δ ( x ) − ℒ ( x ) ‖ ≤ 1 2 σ ( x , x ) for all x ∈ A . In this case, the mapping ℒ is defined as follows:

We then have by (2.9) that

which implies that

By (2.9), we see that

Hence, we obtain

So, by virtue of Lemma 2.1, we find that

By combining (2.11) and (2.12), we obtain

But then, we found that

This gives that

Considering (2.11) and (2.15), we are lead to

Therefore, with the help of (2.14) and (2.15), we figure out that

The expression (2.16) can be represented as follows:

By replacing y by y w in (2.18) and using (2.18), we obtain

That is, [ ℒ ( x ) − δ ( x ) , y ] lies in the left annihilator lan ( A ) of A . The semiprimeness of A implies that [ ℒ ( x ) − δ ( x ) , y ] = 0 . So we have

By applying (2.17) and (2.19), we yield that [ δ ( x ∘ y ) − δ ( x ) ∘ y − x ∘ δ ( y ) , z ] = 0 , that is,

Then, due to (2.13) and (2.20), we see that the mapping δ is a CE-Jordan derivation. So, based on [10, Lemma 3.4], δ is additive, hence the relation (2.10) forces to δ = ℒ . Therefore, by referring (2.17), we conclude that

Consequently, δ is a Jordan derivation. Moreover, since A is semiprime, δ is a derivation [11]. This proves the theorem.□

Corollary 2.3

Let A be a semisimple Banach algebra with no nonzero central ideals. Assume that mappings Φ : A × A → [ 0 , ∞ ) and φ : A × A → [ 0 , ∞ ) satisfy the assumptions of Theorem 2.2. Suppose that δ : A → A is a mapping such that

together with (2.9). Then δ is a continuous derivation.

Proof

We first consider t = 1 in (2.21). According to Theorem 2.2, we see that δ is a derivation.

Inequality (2.21) yields that for all x ∈ A and all t ∈ T ε ,

where the mapping ℒ is defined as (2.10). Hence, we have ℒ ( t x ) = t ℒ ( x ) . Then the mapping ℒ is linear (refer to [14]). Therefore, since δ = ℒ , the derivation δ is also linear. The semisimplicity of A ensures that δ is continuous. This completes the proof of the corollary.□

Lemma 3.1

Let A be a Banach algebra and let ℒ : A → A be a linear mapping. Suppose that τ : A → A and δ : A → A are mappings such that

If Z ( A ) = { 0 } holds, then the mappings τ and δ are linear.

Proof

For all x , y ∈ A and all t ∈ C ,

Then δ satisfies the equation [ x , δ ( t y ) − t δ ( y ) ] = 0 , that is,

We now have from (3.1) that

On the other hand, if we calculate in a different way, we obtain

By combining (3.2) and (3.3) gives [ x , δ ( y + z ) − δ ( y ) − δ ( z ) ] = 0 , which forces to

It can be similarly proved that τ also satisfies the following relation:

Therefore, if Z ( A ) = { 0 } , then τ and δ are linear, which is the assertion of Lemma.□

Theorem 3.2

Let A be a Banach algebra. Assume that mappings Φ : A × A → [ 0 , ∞ ) , φ 0 : A × A → [ 0 , ∞ ) and φ 1 : A × A → [ 0 , ∞ ) satisfy

1. σ ( x , y ) ≔ ∑ j = 0 ∞ 1 2 j Φ ( 2 j x , 2 j y ) < ∞ ( x , y ∈ A ) ,

2. lim n → ∞ 1 2 n φ k ( 2 n x , y ) = lim n → ∞ 1 2 n φ k ( x , 2 n y ) = 0 ( k = 0 , 1 ; x , y ∈ A ) .

and the following inequalities

If Z ( A ) = { 0 } , then δ 0 is a generalized Lie 2-derivation associated with the Lie derivation δ 1 .

Proof

Employing the same method as in the proof of Corollary 2.3, for each k = 0 , 1, there exists a unique linear mapping ℒ k : A → A satisfying ‖ δ k ( x ) − ℒ k ( x ) ‖ ≤ 1 2 σ ( x , x ) for all x ∈ A , where the mapping ℒ k is defined as (2.10).

It follows from (3.5) that

This means that

Moreover, we have by (3.5) that

which implies that

It follows from (3.5) that

Hence, we have

Expressions (3.8), (3.9), and Lemma 3.1 ensure that δ 0 and δ 1 are linear. Then by (2.10), we are lead to δ 0 = ℒ 0 and δ 1 = ℒ 1 . Thus, we obtain by (3.7) that

The following equation δ 1 ( [ x , y ] ) = [ δ 1 ( x ) , y ] + [ x , δ 1 ( y ) ] can be proved in a similar way to the one proved above. Hence, δ 1 is a Lie derivation. Therefore, δ 0 is a generalized Lie 2-derivation associated with the Lie derivation δ 1 . This completes the proof.□

Theorem 3.3

Let A be a semisimple Banach algebra. Assume that mappings Φ : A × A → [ 0 , ∞ ) and φ : A × A → [ 0 , ∞ ) satisfy the assumptions of Theorem 3.2. Suppose that δ : A → A is a mapping subjected to the inequality (2.21) and

for all x , y ∈ A . If Z ( A ) = { 0 } , then δ is a continuous Lie derivation.

Proof

By similar approach to the proof of Corollary 2.3 and Theorem 3.2, we can see that δ is a Lie derivation. Also, it is known that every Lie derivation on a semisimple Banach algebra A is continuous if Z ( A ) = 0 (refer to [17]). Therefore, we complete the proof.□

Theorem 3.4

Let A be either a semiprime Banach algebra or a unital Banach algebra. Assume that mappings Φ : A × A → [ 0 , ∞ ) , φ 0 : A × A → [ 0 , ∞ ) and φ 1 : A × A → [ 0 , ∞ ) satisfy the assumptions of Theorem 3.2. If δ 0 : A → A and δ 1 : A → A are mappings such that the inequality (3.4) and

are fulfilled, then δ 0 is a generalized Lie derivation associated with the derivation δ 1 .

Proof

As we did in the proof of Corollary 2.3, for each k = 0,1 , there exists a unique linear mapping ℒ k : A → A satisfying ‖ δ k ( x ) − ℒ k ( x ) ‖ ≤ 1 2 σ ( x , x ) for all x ∈ A . Here, the mapping ℒ k is defined as (2.10).

It follows from (3.11) that

Hence, we figure out that

On the other hand, in view of (3.12), we see that

which implies that

Then we have by [18, Lemma 2.1] that δ 1 is linear. Considering (2.10) gives δ 1 = ℒ 1 and then, δ 1 is a derivation.

Equation (3.13) can be expressed as follows:

Hence, by linearity of δ 1 , we obtain

for all x , y ∈ A and all t ∈ C . Furthermore, due to (3.14), we have

for all x , y ∈ A and all t ∈ C . By combining (3.15) and (3.16), we are lead to

This means that δ 0 ( t x ) − t δ 0 ( x ) lies in the left annihilator lan ( A ) of A . If A is semiprime, expression (3.17) ensures that δ 0 ( t x ) = t δ 0 ( x ) . If A has a unit, set y = e in (3.17). Then we arrive at δ 0 ( t x ) = t δ 0 ( x ) .

But then, we have by (3.14) and by additivity of δ 1 , that

Another way to compute this is