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Additive mappings on ${C}^{\ast}$algebras subpreserving absolute values of products
Journal of Inequalities and Applications volume 2012, Article number: 161 (2012)
Abstract
Let $\mathcal{A}$ be a ${C}^{\ast}$algebra of real rank zero and $\mathcal{B}$ be a ${C}^{\ast}$algebra with unit I. It is shown that if $\varphi :\mathcal{A}\u27f6\mathcal{B}$ is an additive mapping which satisfies $\varphi (A)\varphi (B)\le \varphi (AB)$ for every $A,B\in {\mathcal{A}}_{+}$ and $\varphi (A)=I$ for some $A\in {\mathcal{A}}_{s}$ with $\parallel A\parallel \le 1$, then the restriction of mapping ϕ to ${\mathcal{A}}_{s}$ is a Jordan homomorphism, where ${\mathcal{A}}_{s}$ denotes the set of all selfadjoint elements. We will also show that if ϕ is surjective preserving the product and an absolute value, then ϕ is a $\mathbb{C}$linear or $\mathbb{C}$antilinear ∗homomorphism on $\mathcal{A}$.
MSC:47B49, 46L05, 47L30.
1 Introduction and preliminaries
In recent years, the subject of linear preserver problems is the focus of attention of many mathematicians, and much research has been going on in this area. Here we refer to the articles [1–6, 9–16].
In what follows, let $\mathcal{A}$ and $\mathcal{B}$ be two ${C}^{\ast}$algebras with unit I. We say that a mapping $\varphi :\mathcal{A}\u27f6\mathcal{B}$ is preserving (resp. subpreserving) absolute values of a product if $\varphi (A)\varphi (B)=\varphi (AB)$ (resp. $\varphi (A)\varphi (B)\le \varphi (AB)$) for every $A,B\in \mathcal{A}$, where ${A}^{2}={A}^{\ast}A$. By a ∗homomorphism we just mean a map $\varphi :\mathcal{A}\u27f6\mathcal{B}$ which preserves the ring structure and for which $\varphi ({A}^{\ast})=\varphi {(A)}^{\ast}$ for every $A\in \mathcal{A}$. A map $\varphi :\mathcal{A}\to \mathcal{B}$ is said to be a Jordan ∗homomorphism if it is linear, $\varphi ({A}^{\ast})=\varphi {(A)}^{\ast}$ and $\varphi {(A)}^{2}=\varphi ({A}^{2})$ for all $A\in \mathcal{A}$. We also say a map $\varphi :\mathcal{A}\to \mathcal{B}$ is unital if $\varphi (I)=I$. The class of all selfadjoint elements in a ${C}^{\ast}$algebra $\mathcal{A}$ is denoted by ${\mathcal{A}}_{s}$. We define $A\circ B=\frac{1}{2}[AB+BA]$ for all $A,B\in {\mathcal{A}}_{s}$. It is known that $({\mathcal{A}}_{s},+,\circ )$ is a ${C}^{\ast}$algebra which is called a Jordan algebra.
In [11], Molnar considered bijective mappings ϕ from a von Neumann algebra which is a factor onto a von Neumann algebra which preserves a product and an absolute value. He showed ϕ is of the form
where ψ is either a linear or conjugate linear *algebra isomorphism and is a scalar function of Modulus 1.
It is the aim of this paper to continue this work by studying additive mappings ϕ from a ${C}^{\ast}$algebra of real rank zero into a ${C}^{\ast}$algebra that subpreserve product and absolute value. In fact, we show that if the mapping ϕ which is an additive subpreserving product absolute value map from a ${C}^{\ast}$algebra $\mathcal{A}$ into a ${C}^{\ast}$algebra then ϕ is a contraction. Moreover, if $\mathcal{A}$ is a ${C}^{\ast}$algebra of real rank zero and $\varphi (A)=I$ for some A in the closed unit ball ${({\mathcal{A}}_{s})}_{1}$, then the restriction of mapping ϕ to ${\mathcal{A}}_{s}$ is a Jordan homomorphism. We will also show that if ϕ is a surjective preserving product and absolute value and $\mathcal{A}$ is a ${C}^{\ast}$algebra of real rank zero, then ϕ is a linear or antilinear ∗homomorphism on $\mathcal{A}$. All we need about ${C}^{\ast}$algebras and von Neumann algebras can be found in [7, 8].
2 The main results
Firstly, we need some auxiliary lemmas to prove our main result.
Lemma 2.1 Let$\mathcal{A}$and$\mathcal{B}$be two unital${C}^{\ast}$algebras with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is a map satisfying
then ϕ preserves positive elements. Moreover, if$\varphi (0)=0$, then for all$A,B\in {\mathcal{A}}_{+}$we have
Proof If A is a positive element in $\mathcal{A}$, then $\varphi (A)\varphi (I)\le \varphi (A)$. This means $\varphi (A)$ is positive and preserves positive elements.
Let $A,B\in {\mathcal{A}}_{+}$ and $AB=BA=0$. By the assumption $\varphi (A)\varphi (B)\le \varphi (AB)=0$. Thus $\varphi (A)\varphi (B)=0$ and hence $\varphi (A)\varphi (B)=\varphi (B)\varphi (A)=0$. □
Lemma 2.2 Let$\mathcal{A}$and$\mathcal{B}$be two unital${C}^{\ast}$algebras with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is an additive mapping satisfying (2.1), then ϕ is order preserving and contraction (i.e. $\parallel \varphi (A)\parallel \le \parallel A\parallel $) on$({\mathcal{A}}_{s},+,\circ )$.
Proof By Lemma 2.1 ϕ preserves positive elements. Hence additivity of ϕ implies that ϕ is order preserving. And also, since every selfadjoint element is the difference of two positive elements, ϕ preserves selfadjoint elements. Indeed, we show that ϕ maps the part of positive (resp. negative) of A to the part of positive (resp. negative) of $\varphi (A)$. In fact, $\varphi ({A}_{+})=\varphi {(A)}_{+}$ and $\varphi ({A}_{})=\varphi {(A)}_{}$, where, $A={A}_{+}{A}_{}$ and ${A}_{+}{A}_{}=0={A}_{}{A}_{+}$. We just need to show $\varphi ({A}_{+})\varphi ({A}_{})=0=\varphi ({A}_{})\varphi ({A}_{+})$ because the decomposition of $\varphi (A)$ is unique and ϕ preserves positives. Applying Lemma 2.1 and the equation ${A}_{+}{A}_{}=0={A}_{}{A}_{+}$, we get the assertion.
The proof of linearity of ϕ is similar to the first step of the proof of [10], Theorem 1]. The details are omitted.
Now, we show ϕ is contraction on $({\mathcal{A}}_{s},+,\circ )$. If A is a selfadjoint element. We can write $A={A}_{+}{A}_{}$, where ${A}_{+}$ and ${A}_{}$ are positive elements. We have
Since ϕ is order preserving, observe
because ${A}^{2}\le \parallel {A}^{2}\parallel I$. It implies that
since $\parallel \varphi (I)\parallel \le 1$. Taking square root, we obtain $\parallel \varphi (A)\parallel \le \parallel A\parallel $, which yields ϕ is a contraction on $({\mathcal{A}}_{s},+,\circ )$. □
Lemma 2.3 Let$\mathcal{A}$and$\mathcal{B}$be two unital${C}^{\ast}$algebras with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is an additive map satisfying (2.1) and
then ϕ is unital.
Proof By the hypothesis, there exists an operator $U\in {({\mathcal{A}}_{s})}_{1}$ such that $\varphi (U)=I$. Since ϕ is order preserving by Lemma 2.2 and $A\le \parallel A\parallel I$ for every $A\in \mathcal{A}$, we have
On the other hand, we have
because ϕ is contraction by Lemma 2.2. Therefore, $\varphi (I)=I$. □
The following example shows that the condition (2.3) in Lemma 2.3 is necessary.
Example 2.4 Define an additive mapping $\varphi :C[0,1]\u27f6C[0,1]$ by
for all $f\in C[0,1]$, where $r(t)=\frac{1+t}{2}$. Note that r is positive and ${r}^{2}\le r$. Obviously, ϕ is an additive mapping satisfying
for every $f,g\in C[0,1]$, but clearly there is not any such that $\varphi (I)=\lambda I$ because $\varphi (I)=r$.
Lemma 2.5 Let$\mathcal{A}$be a${C}^{\ast}$algebra of real rank zero and$\mathcal{B}$be a unital${C}^{\ast}$algebra with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is an additive mapping satisfying (2.1), then

(i)
for every selfadjoint operator $A\in \mathcal{A}$, we have
$$\varphi {(A)}^{2}=\varphi \left({A}^{2}\right)\varphi (I)=\varphi (I)\varphi \left({A}^{2}\right).$$(2.4) 
(ii)
$N=\{A\in {\mathcal{A}}_{s}:\varphi (A)=0\}$ is a closed ideal of $({\mathcal{A}}_{s},+,\circ )$.
Proof (i) Let E and F be mutually orthogonal projections. By Lemma 2.1 $\varphi (E)\varphi (F)=\varphi (F)\varphi (E)=0$, in particular, $\varphi (IE)\varphi (E)=\varphi (E)\varphi (IE)=0$. That is, $\varphi {(E)}^{2}=\varphi (E)\varphi (I)=\varphi (I)\varphi (E)$.
Assume that $A\in \mathcal{A}$ is of the form ${\sum}_{j=1}^{n}{\lambda}_{j}{E}_{j}$ for some scaler and finitely many mutually orthogonal projections ${E}_{j}$, then
Now, assume A is an arbitrary selfadjoint element. Since $\mathcal{A}$ is a ${C}^{\ast}$algebra with real rank zero, its every selfadjoint element can be approximated by the elements of the above form. Hence the continuity of ϕ entails
for every selfadjoint operator $A\in \mathcal{A}$.

(ii)
Let $A\in \mathcal{A}$ be a selfadjoint element such that $\varphi (I)\varphi (A)=0=\varphi (A)\varphi (I)$. We show that $\varphi (A)=0$.
Multiplying through equation (2.4) by $\varphi (I)$ (on the left) we get $0=\varphi (I)\varphi {(A)}^{2}=\varphi {(I)}^{2}\varphi ({A}^{2})$. Since $\varphi (I)$ and $\varphi ({A}^{2})$ commute and by the assumption, we have $0=\varphi (I)\varphi {(A)}^{2}=\varphi {(I)}^{2}\varphi ({A}^{2})\ge \varphi {(I)}^{2}\varphi {(A)}^{2}$. This implies that
Since ${A}^{2}\le \parallel A\parallel A$, by using the order preserving property of ϕ we yield $\varphi {(A)}^{2}=\varphi (I)\varphi ({A}^{2})=\varphi (I)\varphi ({A}^{2})\le \varphi (I)\varphi (\parallel A\parallel A)=\parallel A\parallel \varphi (I)\varphi (A)=0$. So $\varphi (A)=0$ because $\varphi (A)$ is a selfadjoint element.
It follows N is a closed ideal of $({\mathcal{A}}_{s},+,\circ )$ by the step 4 of [17], Theorem 2.1]. □
In the following theorem we would like to characterize the Jordan homomorphisms ϕ which are additive mappings subpreserving a product and an absolute value.
Theorem 2.6 Let$\mathcal{A}$be a${C}^{\ast}$algebra of real rank zero and$\mathcal{B}$be a unital${C}^{\ast}$algebra with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is an additive mapping satisfying (2.1) and (2.3), then the restriction of the map ϕ to${\mathcal{A}}_{s}$is a Jordan homomorphism.
Proof According to Lemma 2.3 and Lemma 2.5(i) we yield the statement. □
Now we also show that the condition (2.3) in Theorem 2.6 is necessary.
Example 2.7 The same as in Example 2.4, let $\mathcal{A}$ be a von Neumann algebra with a non trivial center. Define an additive mapping $\varphi :\mathcal{A}\u27f6\mathcal{A}$ by
where is invertible, ${U}^{2}<U$ and . Obviously, ϕ is an additive mapping that satisfies in (2.1), but no nonzero multiple of ϕ is a Jordan homomorphism, because if $\psi (A)=\lambda \varphi (A)$ with is a Jordan homomorphism, then we obtain ${U}^{2}={\lambda}^{2}$, that is a contradiction.
In the following theorem we show that if $\varphi (I)$ is an injective operator and ϕ is an additive map which satisfies in (2.1), then the restriction of ϕ is a Jordan homomorphism multiplied by $\varphi (I)$.
Theorem 2.8 Let$\mathcal{A}$be a${C}^{\ast}$algebra of real rank zero and$\mathcal{B}$be a unital${C}^{\ast}$algebra with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is an additive mapping satisfying$\varphi (A)\varphi (B)\le \varphi (AB)$for every$A,B\in {\mathcal{A}}_{+}$and$\varphi (I)$is an injective operator, then the restriction of mapping$\psi :\mathcal{A}\u27f6\mathcal{B}$is defined by$\varphi (A)=\varphi (I)\psi (A)$, to${\mathcal{A}}_{s}$is a Jordan homomorphism.
Proof Injectivity of $\varphi (I)$ implies ψ is well defined. Let $A\in {\mathcal{A}}_{s}$. By applying Lemma 2.5, we can show $\varphi (A)\varphi (I)=\varphi (I)\varphi (A)$. By the definition of ψ, we have
This means $\varphi (I)$ commutes with $\psi (A)$ for every selfadjoint operator $A\in \mathcal{A}$. Again, by using Lemma 2.5, we yield
This completes the proof. □
In [11], Molnar considered bijective mappings ϕ from a von Neumann algebra which is a factor onto a von Neumann algebra which preserves a product and an absolute value. He showed ϕ is of the form
where, ψ is either a linear or conjugate linear ∗algebra isomorphism and is a scalar function of Modulus 1. Below, we present the result where we do not assume injectivity but ϕ is an additive map from ${C}^{\ast}$algebra $\mathcal{A}$ onto a ${C}^{\ast}$algebra of real rank zero which preserves a product and an absolute value, and it is shown ϕ is a linear or antilinear ∗homomorphism.
Theorem 2.9 Let$\mathcal{A}$and$\mathcal{B}$be two unital${C}^{\ast}$algebra with unit I. If$\varphi :\mathcal{A}\u27f6\mathcal{B}$is an additive mapping satisfying$\varphi (A)\varphi (B)=\varphi (AB)$for every$A,B\in \mathcal{A}$and$\varphi (A)=I$for some$A\in \mathcal{A}$, then ϕ is unital and the restriction of mapping ϕ to${\mathcal{A}}_{S}$is a Jordan homomorphism. Moreover, if ϕ is surjective and$\mathcal{B}$be a${C}^{\ast}$algebra of real rank zero then, ϕ is alinear orantilinear ∗homomorphism.
Proof By Lemma 2.2 ϕ preserves selfadjoint element. So $\varphi {(A)}^{2}=\varphi ({A}^{2})$, for all of selfadjoint elements A, in particular, ϕ preserves projection. By the hypothesis, there is an element $U\in \mathcal{A}$ such that $\varphi (U)=I$. Then,
Thus without loss of generality we can assume U is a positive element. Now, we have
So ϕ is a unital map.
And also we have
Therefore, ϕ is a linear or antilinear ∗homomorphism on $\mathcal{A}$ by [17], Theorem 2.5]. □
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Acknowledgement
This research is partially supported by the Research Center in Algebraic Hyperstructures and Fuzzy Mathematics, University of Mazandaran, Babolsar, Iran.
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Taghavi, A. Additive mappings on ${C}^{\ast}$algebras subpreserving absolute values of products. J Inequal Appl 2012, 161 (2012). https://doi.org/10.1186/1029242X2012161
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Keywords
 ${C}^{\ast}$algebra
 $\mathbb{C}$linear
 $\mathbb{C}$antilinear
 homomorphism
 linear preserver problem
 real rank zero