http://scma.maragheh.ac.ir
ON STRONGLY JORDAN ZERO-PRODUCT PRESERVING MAPS
ALI REZA KHODDAMI1
Abstract. In this paper, we give a characterization of strongly Jordan zero-product preserving maps on normed algebras as a gen-eralization of Jordan zero-product preserving maps. In this di-rection, we give some illustrative examples to show that the no-tions of strongly zero-product preserving maps and strongly Jordan zero-product preserving maps are completely different. Also, we prove that the direct product and the composition of two strongly Jordan zero-product preserving maps are again strongly Jordan zero-product preserving maps. But this fact is not the case for tensor product of them in general. Finally, we prove that every
∗−preserving linear map from a normed∗−algebra into aC∗−algebra
that strongly preserves Jordan zero-products is necessarily contin-uous.
1. Introduction and preliminaries
The author of the current study recently has introduced and inves-tigated the notions of strongly product and strongly Jordan zero-product preserving maps on a class of normed algebras [4]. These no-tions are generalization of the concepts “ product and Jordan zero-product preserving maps ” respectively.
In this direction, the basic properties of strongly zero-product pre-serving maps were investigated in [5] on general normed algebras. A linear mapϕ:A−→B between two algebrasAand B, over a fieldF is said to be a zero-product preserving map if, ϕ(a)ϕ(c) = 0 whenever
2010Mathematics Subject Classification. 15A04, 46H99, 47B48.
Key words and phrases. Strongly zero-product preserving map, Strongly Jordan zero-product preserving map, Zero-product preserving map, Jordan zero-product pre-serving map, Tensor product.
Received: 31 July 2015, Accepted: 2 January 2016.
ac= 0, (a, c∈A). Also,ϕis said to be a Jordan zero-product preserv-ing map if, ϕ(a)◦ϕ(c) = 0 whenever a◦c = 0, where “ ◦ ” denotes the Jordan product a◦c = ac+ca. The notions of zero-product and Jordan zero-product preserving maps are different notions and one does not imply the other in general (see Example 3.1 in this paper). Some useful references in the field of zero-product and Jordan zero-product preserving maps are [1, 2].
Let A and B be two normed algebras over a field F. We say that a linear map ϕ : A −→ B is a strongly zero-product preserving map if, for any two sequences {an}n,{cn}n in A, ϕ(an)ϕ(cn) −→ 0 whenever ancn−→0.
Also, we say thatϕis a strongly Jordan zero-product preserving map if, for any two sequences {an}n,{cn}n in A,ϕ(an)◦ϕ(cn)−→0 whenever an◦cn−→0.
The following Remark similar to the [5, Remark 3.4] holds.
Remark 1.1. (i) Let A and B be normed algebras and let B be unital. Then every surjective strongly Jordan zero-product pre-serving map is continuous.
(ii) Let A and B be two unital normed algebras with the units 1A and 1B, respectively. Also letϕ:A−→B be a strongly Jordan zero-product preserving map such thatϕ(1A) = 1B. Then ϕis continuous.
For the normed algebras A and B over a field F, we will denote by A⊗B the algebraic tensor product of A and B. It is well known that A⊗Bis a normed algebra with the following projective cross norm given by
∥u∥= inf
{k=n ∑
k=1
∥ak∥∥bk∥, u= k=n
∑
k=1
ak⊗bk, ak∈A, bk∈B, n∈N
}
,
for all u∈A⊗B.
2. Examples
Example 2.1. (i) Every continuous homomorphism between normed algebras is strongly zero-product and also strongly Jor-dan zero-product preserving map.
normed algebra (for the basic properties of this algebra see [3]). We denote it byVf. Define ϕ:Vf −→ Vf such that ϕ(e1) = 0 and ϕ(en) = 2ne2 for all n ≥2. Since f◦ϕ≡0; it is obvious thatϕis a strongly product and also strongly Jordan zero-product preserving map. We show thatϕis neither a continuous map nor a homomorphism onVf. To this end let an= enn. So,
∥an∥= 1
n∥en∥= 1
n −→0.
But
lim
n→∞∥ϕ(an)∥= limn→∞ 2n
n∥e2∥
= lim n→∞
2n n =∞.
This shows thatϕis not a continuous map. Also, 4e2 =ϕ(e2) = ϕ(e1e2)̸=ϕ(e1)ϕ(e2) = 0. So, ϕis not a homomorphism.
The part (ii) of Example 2.1 shows that strongly zero-product pre-serving maps and strongly Jordan zero-product prepre-serving maps are not continuous maps or homomorphisms in general.
Remark 2.2. It is obvious that every strongly zero-product (strongly Jordan product) preserving map is a product (Jordan zero-product) preserving map. But the converse is not the case in general. The following example shows this fact.
Example 2.3. Let V be an infinite dimensional normed vector space with the basis β = {e1, e2, e3, . . .} such that ∥en∥ = 1 for all n ≥ 1 . Also let f ∈ V∗ be a continuous linear functional satisfying f(e
1) = 1 and f(en) = 0 for all n ≥ 2. So, kerf = span{e2, e3, e4, . . .}. Define ϕ :Vf −→ Vf such that ϕ(a) = f(a)e1+θ(a), where θ : Vf −→ kerf is a linear map such that θ(e1) = 0 and θ(en) = 2ne2 for all n ≥2. It is obvious that ϕ(kerf) ⊆kerf. So, by [4, Theorems 2.1 and 2.2 ], ϕ is a zero-product (Jordan zero-product) preserving map. But we show that ϕ is not a strongly Jordan zero-product (strongly zero-product) preserving map. To this end, let an = en1 and cn = en+1. Clearly,
∥an ◦ cn∥ −→ 0 but limn→∞∥ϕ(an) ◦ ϕ(cn)∥ = limn→∞ 2
n+1
n ∥e2∥ = limn→∞2
n+1
3. the notions of strongly zero-product and strongly Jordan zero-product preserving maps are different
In this section, we give some illustrative examples to show that the notions of strongly zero-product and strongly Jordan zero-product pre-serving maps are different and one does not imply the other.
Example 3.1. (i) LetA=
{[ α β 0 0 ]
α, β ∈C
}
andB =M2×2(C).
It is obvious thatA and B with the usual addition and multi-plication and with the norm
[ α β γ δ ]
= 2 max{
|α|, |β|, |γ|, |δ|}
,
are Banach algebras. Defineϕ:A−→B such that
ϕ ([ α β 0 0 ]) = [ α α β 0 ] .
The linearity of ϕ is obvious. We show that ϕ is a strongly
Jordan zero-product preserving map. For eacha=
[ α β 0 0 ] and c= [ λ µ 0 0 ] ∈A,
∥ϕ(a)◦ϕ(c)∥=
[
2αλ+αµ+λβ 2αλ
βλ+αµ βλ+αµ
] ≤2∥a◦c∥.
It follows that ϕ is a strongly Jordan zero-product preserving map. But ϕ is not a strongly zero-product preserving map.
Indeed, let an =
[
0 n 0 0
]
and cn =
[
n 0
0 0
]
. Clearly, ancn −→0
but
∥ϕ(an)ϕ(cn)∥=
[ 0 0
n2 n2
]
= 2n2 −→ ∞.
(Also note thatϕis not a zero-product preserving map).
(ii) Let A =
{[
α β
−β¯ α¯
]
α, β∈C
}
and B =M2×2(C). It is
obvi-ous thatAandBare normed algebras over the real fieldF =R,
with the mentioned norm in part (i). Also every non-zero ele-ment ofAis invertible. Define ϕ:A−→B such that
ϕ
([
α β
−β¯ α¯
Clearly, ϕ is an R−linear map. We show that ϕ is a strongly
zero-product preserving map.
Leta=
[
α β
−β¯ α¯
]
andc=
[
λ µ
−µ¯ λ¯
]
be two elements ofA. It is
obvious that
ac=
[
αλ−βµ¯ αµ+βλ¯
−βλ¯ −α¯µ¯ −βµ¯ + ¯αλ¯
]
=
[
s t
−¯t s¯
]
where,
{
αλ−βµ¯=s αµ+βλ¯=t, (3.1)
Letc̸= 0. So,|λ|2+|µ|2̸= 0. From (3.1) we have
β = −µ
|λ|2+|µ|2s+ λ
|λ|2+|µ|2t.
So,
βλ= −λµ
|λ|2+|µ|2s+ λ2
|λ|2+|µ|2t and
βµ= −µ
2
|λ|2+|µ|2s+ λµ
|λ|2+|µ|2t. It follows that
|βλ| ≤ | −λµ| |λ|2+|µ|2|s|+
|λ2| |λ|2+|µ|2|t|
≤ 1
2|s|+|t|
≤ ∥ac∥.
Similarly,|βµ| ≤ ∥ac∥. So,
∥ϕ(a)ϕ(c)∥=
[
0 0
βλ βµ
] ≤2∥ac∥.
This shows that ϕ is a strongly zero-product preserving map. We shall show that ϕ is not a strongly Jordan zero-product
preserving map. Indeed, letan=
[
0 1
−1 0
]
andcn=
[
ni 0
0 −ni
]
Clearly,an◦cn= 0 so limn−→∞an◦cn= 0. But
lim
n−→∞∥ϕ(an)◦ϕ(cn)∥= limn−→∞
[
0 0 ni 0
]
= lim n−→∞2|ni| =∞.
This shows that ϕ is not a strongly Jordan zero-product pre-serving map. (Also note that ϕ is not a Jordan zero-product preserving map).
Remark 3.2. Example 3.1 shows that the notions of zero-product and Jordan zero-product preserving maps are completely different and one does not imply the other.
4. Main Results
In this section, we give a characterization of strongly Jordan zero-product preserving maps on normed algebras. Also, we prove that every
∗−preserving linear map from a normed ∗−algebra into a C∗−algebra that strongly preserves Jordan zero-products is necessarily continuous.
Theorem 4.1. Let A and B be normed algebras. Then a linear map
ϕ : A −→ B is a strongly Jordan zero-product preserving map if and only if there exists M >0 such that
∥ϕ(a)◦ϕ(c)∥ ≤M∥a◦c∥,(a, c∈A).
Proof. Similar to the proof of [5, Theorem 3.1], by contradiction, suppose thatϕis a strongly Jordan zero-product preserving map and the desired inequality is not true for allM >0. So, forM = 1 there exista1, c1 ∈A such that,
∥ϕ(a1)◦ϕ(c1)∥>∥a1◦c1∥.
For M = ∥ϕ(a 2
1)◦ϕ(c1)∥ there exist a2, c2 ∈A such that ∥ϕ(a2)◦ϕ(c2)∥>
2
∥ϕ(a1)◦ϕ(c1)∥
∥a2◦c2∥.
It follows that
a2
∥ϕ(a2)◦ϕ(c2)∥
◦c2
< ∥ϕ(a1)◦ϕ(c1)∥
2 .
A similar argument can be applied to show that, for
M = n
there exist an, cn∈A such that
an
∥ϕ(an)◦ϕ(cn)∥
◦cn
< ∥ϕ(a1)◦ϕ(c1)∥
n .
Let a′n= an
∥ϕ(an)◦ϕ(cn)∥ and
c′n=cn. Asa′n◦c′n−→0, it follows that
ϕ(a′n)◦ϕ(c′n)−→0.
That is a contradiction. The converse is obvious. □
Definition 4.2. LetAand B be two∗−algebras. We say that a linear map ϕ:A−→B is ∗−preserving if,ϕ(a∗) =ϕ(a)∗, (a∈A).
Proposition 4.3. LetAbe a normed∗−algebra andB be aC∗−algebra. Also letϕ:A−→Bbe a∗−preserving linear map that strongly preserves Jordan zero-products. Then, ϕis continuous.
Proof. Let ϕ : A −→ B be a ∗−preserving linear map that strongly preserves Jordan zero-products. Also, Let{an}nbe a sequence inAsuch that an −→ 0. So, an = bn+icn, where bn = a
n+a∗n
2 and cn = a
n−a∗n
2i . Obviously,bn and cn are self adjoint elements such that
lim
n−→∞bn= limn−→∞cn= 0.
It follows that bn◦bn −→0 and cn◦cn −→ 0. So, ϕ(bn)◦ϕ(bn) −→0 and ϕ(cn)◦ϕ(cn)−→0. So
∥ϕ(bn)∥2=∥ϕ(bn)ϕ(bn)∗∥ =∥ϕ(bn)ϕ(b∗n)∥ =∥ϕ(bn)ϕ(bn)∥
= 1
2∥ϕ(bn)◦ϕ(bn)∥ −→0.
This shows that ϕ(bn) −→ 0. Similarly ϕ(cn) −→ 0. So, ϕ(an) −→ 0
and equivalently ϕis continuous. □
5. Hereditary Properties
Proposition 5.1. Let A, B, C, D be normed algebras and let ϕ:A−→
B and ψ : C −→ D be two strongly Jordan zero-product preserving maps. Then, ϕ⊕ψ:A⊕C−→B⊕Dis a strongly Jordan zero-product preserving map.
Proof. As ϕ and ψ are strongly Jordan zero-product preserving maps, there exist M, N >0 such that∥ϕ(a)◦ϕ(a′)∥ ≤M∥a◦a′∥and
∥ψ(c)◦ψ(c′)∥ ≤N∥c◦c′∥, (a, a′ ∈A, c, c′ ∈C). So,
∥(ϕ⊕ψ)(a, c)◦(ϕ⊕ψ)(a′, c′)∥=∥(ϕ(a), ψ(c))◦(ϕ(a′), ψ(c′))∥
=∥(ϕ(a)◦ϕ(a′), ψ(c)◦ψ(c′))∥
=∥ϕ(a)◦ϕ(a′)∥+∥ψ(c)◦ψ(c′)∥
≤M∥a◦a′∥+N∥c◦c′∥ ≤M(∥a◦a′∥+∥c◦c′∥)
+N(∥a◦a′∥+∥c◦c′∥)
= (M+N)(∥a◦a′∥+∥c◦c′∥) = (M+N)∥(a◦a′, c◦c′)∥
= (M+N)∥(a, c)◦(a′, c′)∥.
Applying Theorem 4.1 shows thatϕ⊕ψis a strongly Jordan zero-product
preserving map. □
Proposition 5.2. LetA, B andC be normed algebras and letϕ:A−→
B and ψ : B −→ C be two strongly Jordan zero-product preserving maps. Then,ψ◦ϕ:A−→Cis a strongly Jordan zero-product preserving map.
Proof. As ϕ and ψ are strongly Jordan zero-product preserving maps, there exist M, N >0 such that∥ϕ(a)◦ϕ(a′)∥ ≤M∥a◦a′∥ and
∥ψ(b)◦ψ(b′)∥ ≤N∥b◦b′∥, (a, a′∈A, b, b′∈B). So,
∥(ψ◦ϕ)(a)◦(ψ◦ϕ)(a′)∥=∥ψ(ϕ(a))◦ψ(ϕ(a′))∥
≤N∥ϕ(a)◦ϕ(a′)∥
≤M N∥a◦a′∥, (a, a′∈A).
This shows thatψ◦ϕis a strongly Jordan zero-product preserving map.
□
Example 5.3. Let
A=
{[
α β 0 0
]
α, β ∈C
}
and B =M2×2(C). Defineϕ:A−→B such that
ϕ
([
α β 0 0
])
=
[
α α
β 0
]
.
By Example 3.1 ϕ is a strongly Jordan zero-product preserving map. We shall show that ϕ⊗ϕ:A⊗A−→ B⊗B is not a strongly Jordan
zero-product preserving map. To this end, letx=
[
0 1 0 0
] ⊗
[
1 0 0 0
]
and
y=
[
1 0 0 0
] ⊗
[
0 1 0 0
]
. One can simply verify that x◦y= 0. But
(ϕ⊗ϕ)(x)◦(ϕ⊗ϕ)(y) =
[
0 0 1 1
] ⊗
[
1 0 0 0
]
+
[
1 0 0 0
] ⊗
[
0 0 1 1
]
̸
= 0.
This shows thatϕ⊗ϕis not a Jordan zero-product preserving map which implies thatϕ⊗ϕis not a strongly Jordan zero-product preserving map.
Acknowledgment. The author would like to thank the referees for careful reading of the paper.
References
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2. H. Ghahramani, Zero product determined triangular algebras, Linear Multilin-ear Algebra, 61 (2013), 741–757.
3. A.R. Khoddami and H.R.E. Vishki, The higher duals of a Banach algebra in-duced by a bounded linear functional, Bull. Math. Anal. Appl. 3 (2011), 118–122. 4. A.R. Khoddami,Strongly zero-product preserving maps on normed algebras in-duced by a bounded linear functional, Khayyam J. Math., 1 (2015), no. 1, 107– 114.
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1
Department of Pure Mathematics, University of Shahrood, P. O. Box 3619995161-316, Shahrood, Iran.
