Throughout this paper all algebras and vector spaces are over the complex field $ \mathbb{C} $, and all algebras are associative with unity, unless indicated otherwise. Suppose that $ \mathcal{A} $ is a complex Banach $ * $-algebra, and $ \mathcal{X} $ is a Banach $ \mathcal{A} $-bimodule. We recall that a linear mapping $ D:\mathcal{A}\rightarrow \mathcal{X} $ is a derivation whenever $ D(ab)=D(a)b+aD(b), $ for every $ a, b\in \mathcal{A} $. J. Ringrose [1] extends S. Sakai's theorem [2] on automatic continuity of derivations on $ C^{*} $-algebras by proving that every derivation from a $ C^{*} $-algebra $ \mathcal{A} $ to a Banach $ \mathcal{A} $-bimodule is continuous.

In Section 2, we first consider a mapping $ T $ from an algebra $ \mathcal{A} $ into an $ \mathcal{A} $-bimodule $ \mathcal{X} $ that satisfies the following conditions:

$ ab=bc=0\Rightarrow aT(b)c+cT(b)a=0, $

and we give several applications of the conclusion.

Our results can be considered as extensions of some of the results in [3] and [4, 5] to more general classes of Banach algebras, as well as new applications of property $ \mathbb{B} $ in the sense of [6] for new types of preservers, and complementary results for [7, 8].

Spectrum-preserving linear mappings are studied for the first time by G. Frobenius [9]. In [10], B. Aupetit studies spectrum-preserving mappings on von Neumann algebras. In Section 3, we shall consider the spectrum-preserving mappings on algebras of $ \tau $-measurable operators, which is a version of a theorem known for von Neumann algebras, but dealing with algebras of unbounded opeators. Theorem 3.6 deals with a factor of type $ II_{1} $ and Theorem 3.7 with a finite von Neumann algebra. We prove such a mapping is a Jordan $ * $-isomorphism.

We recall that a linear mapping $ G $ from a Banach algebra $ \mathcal{A} $ into a Banach $ \mathcal{A} $-bimodule $ \mathcal{X} $ is said to be a generalized derivation if there exists $ \xi \in \mathcal{X}^{**} $ satisfying

$ G(ab)=G(a)b+aG(b)-a\xi b, (a, b\in \mathcal{A}). $

We shall say that a linear mapping $ G $ from a $ C^{*} $-algebra $ \mathcal{A} $ into a Banach $ \mathcal{A} $-bimodule $ \mathcal{X} $ is a Jordan derivation if $ G(a\circ b)=G(a)\circ b + a\circ G(b) $ for every $ a, b\in \mathcal{A} $, where the Jordan product is given by $ a \circ b :=\frac{1}{2}(ab+ba) $. $ G $ is called a generalized Jordan derivation if there exists $ \xi\in \mathcal{X}^{**} $ such that the identity $ G(a\circ b)=G(a)\circ b + a\circ G(b)- U_{a, b}(\xi), $ holds for every $ a, b $ in $ \mathcal{A} $, where $ U_{a, b}(z) := (a\circ z)\circ b+(b\circ z)\circ a-(a\circ b)\circ z. $ If $ \mathcal{A} $ is unital, every generalized (Jordan) derivation $ D : \mathcal{A}\rightarrow \mathcal{X} $ with $ D(1)=0 $ is a (Jordan) derivation, where $ U_{a, b}(x):= \frac{1}{2} (axb+bxa). $

We recall that every $ C^{*} $-algebra is a $ JB^{*} $-triple with respect to $ \{a, b, c\}=\frac{1}{2}(ab^{*}c+cb^{*}a). $ Whenever we use a triple product on a $ C^{*} $-algebra, it is always this triple product. More details about $ JB^{*} $-triple can be found in [11].

We recall that elements $ a, b $ in a $ JB^{*} $-triple $ \mathcal{E} $ are said to be $ orthogonal $ ($ a\perp b $ for short) if $ L(a, b)=0 $, where $ L(a, b) $ is the operator on $ \mathcal{E} $ given by $ L(a, b)x = \{a, b, x\} $. By [12, Lemma 1], we know that

$ a\perp b\Leftrightarrow \{a, a, b\}=0\Leftrightarrow \{b, b, a\}=0. $

When a $ C^{*} $-algebra $ \mathcal{A} $ is regarded as a $ JB^{*} $-triple, it is known that elements $ a, b $ in $ \mathcal{A} $ are orthogonal if and only if $ ab^{*}=0=b^{*}a $ ([13]). When $ \mathcal{A} $ is a commutative $ C^{*} $-algebra, $ a\perp b $ if and only if $ ab = 0. $

A complex Banach algebra $ \mathcal{A} $ is said to have property $ \mathbb{B} $ if for every continuous bilinear mapping $ f: \mathcal{A}\times\mathcal{A}\rightarrow \mathcal{X} $ where $ \mathcal{X} $ is an arbitrary Banach space, the condition that for all $ x, y\in \mathcal{A}, $

$ xy=0\Rightarrow f(x, y)=0, $

implies that

$ f (xy, z)=f (x, yz) \ \ {\text{for\ all}}\ \ x, y, z \in \mathcal{A}. $

It is shown in [6] that many important examples of Banach algebras, including $ C^{*} $-algebras and group algebras $ L^{1}(G) $ where $ G $ is a locally compact group, have property $ \mathbb{B} $.

Recently, A. Essaleh and A. Peralta consider in [5] a linear preserver problem on maps which are triple derivable at orthogonal pairs. In this paper, we consider a weaker condition than that in [5].

In what follows, we denote by $ \mathcal{A}_{sa} $ the hermitian elements of a Banach $ * $-algebra $ \mathcal{A} $.

Before giving the next lemma, we first give the following definition, which appears in [4].

Definition 2.1 Let $ T: \mathcal{A}\rightarrow \mathcal{A} $ be a linear mapping on a $ C^{*} $-algebra $ \mathcal{A} $, and let $ z $ be an element in $ \mathcal{A} $. We shall say that $ T $ is a triple derivation at z if $ z=\{a, b, c\} $ in $ \mathcal{A} $ implies that

$ T(z)=\{T(a), b, c\}+\{a, T (b), c\}+\{a, b, T(c)\}. $

Lemma 2.1 Suppose that $ \mathcal{A} $ is a unital $ C^{*} $-algebra. Let $ T: \mathcal{A}\rightarrow \mathcal{A} $ be a linear mapping satisfying

$ a \perp b\perp c \Rightarrow \{a, T(b), c\}=0. $

Then the identity

$ \begin{equation} T(p)=pT(p)+T(p)p-pT(1)p \end{equation} $

(2.1)

holds for every idempotent $ p $ in $ \mathcal{A} $.

Proof Let $ a=p, b=1-p^{*}, c=p $, where $ p^{2}=p $ in $ \mathcal{A} $. According to the hypothesis, we have $ pT(1-p^{*})^{*}p=\{p, T(1-p^{*}), p\}=0 $, which gives $ pT (1)^{*}p =pT (p^{*})^{*}p $. By applying $ * $ to both sides, we get $ p^{*}T(1)p^{*} = p^{*}T(p^{*})p^{*} $ for every idempotent $ p $ in $ \mathcal{A} $. But $ p^{*} $ is an idempotent, so

$ pT (1)p = pT (p)p. $

By a similar method, let $ a=1-p^{*}, b=p, c=1-p^{*}. $ Then $ (1-p^{*})T(p)^{*}(1-p^{*})=0 $, so

$ T(p)=pT(p)+T(p)p-pT(p)p. $

Since $ pT(1)p=pT(p)p, $

$ T(p)=pT(p)+T(p)p-pT(1)p. $

Definition 2.2 A Banach $ \mathcal{A} $-bimodule $ \mathcal{M} $ is said to be essential if it is equal to the closed linear span of the set of elements of the form $ x\cdot m \cdot y $ with $ x, y\in \mathcal{A}, m\in \mathcal{M}. $

Definition 2.3 Let $ \mathcal{A} $ be a Banach algebra. A left approximate identity for $ \mathcal{A} $ is a net $ \{\rho_{i}\}_{i\in I} $ in $ \mathcal{A} $ such that

$ \lim\limits_{i}\rho_{i}x=x $

for every $ x\in\mathcal{A}. $ A right approximate identity for $ \mathcal{A} $ is defined similarly. An approximate identity for $ \mathcal{A} $ is a net $ \{\rho_{i}\}_{i\in I} $, which is both a left and a right approximate identity for $ \mathcal{A} $. A (left/right) approximate identity $ \{\rho_{i}\}_{i\in I} $ is bounded if for some positive $ K $ we have $ \|\rho_{i}\|\leq K $ for every $ i\in I. $

Next we give the following proposition, which plays a crucial role in our paper.

Proposition 2.1 Suppose that $ \mathcal{A} $ is a Banach algebra satisfying property $ \mathbb{B} $ and having a bounded approximate identity $ \{\rho_{i}\}_{i\in I} $. Let $ \mathcal{M} $ be an essential Banach $ \mathcal{A} $-bimodule, and let $ T: \mathcal{A}\rightarrow \mathcal{M} $ be a continuous linear mapping satisfying

$ \begin{equation} ab=bc=0\Rightarrow aT(b)c+cT(b)a=0. \end{equation} $

(2.2)

Then $ T $ is a generalized Jordan derivation.

Proof Fix $ a, b \in \mathcal{A} $ with $ ab=0 $. Define a continuous bilinear map $ \varphi: \mathcal{A} \times \mathcal{A} \rightarrow \mathcal{M} $ given by

$ \varphi(x, y)=aT(bx)y+yT(bx)a. $

When $ xy=0 $ in $ \mathcal{A} $, we have $ abx=0=bxy $. Hence $ \varphi(x, y)=0 $ whenever $ ab=0, a, b \in \mathcal{A}. $ According to the hypothesis

$ \varphi(xy, z)=\varphi(x, yz) $

for any $ x, y, z\in \mathcal{A} $, that is,

$ aT(bxy)z+zT(bxy)a=aT(bx)yz+yzT(bx)a, $

for all $ x, y, z, a, b\in \mathcal{A} $ with $ ab=0. $ Fix $ x, y, z\in \mathcal{A} $ and define a continuous bilinear mapping on $ \mathcal{A} $ by

$ \Phi(a, b)=aT(bxy)z+zT(bxy)a-aT(bx)yz+yzT(bx)a. $

Hence, $ ab=0 $ in $ \mathcal{A} $ implies $ \Phi(a, b)=0 $. It follows from the hypothesis on $ \mathcal{A} $ that

$ \Phi(ab, c)=\Phi(ab, c) $

for any $ a, b, c\in \mathcal{A} $, that is,

$ abT(cxy)z+zT(cxy)ab-abT(cx)yz-yzT(cx)ab $

$ =aT(bcxy)z+zT(bcxy)a-aT(bcx)yz-yzT(bcx)a, $

for all $ x, y, z, a, b, c\in \mathcal{A} $.

First let $ a=\rho_{i} $, we get

$ \rho_{i}bT(cxy)z+zT(cxy)\rho_{i}b-\rho_{i}bT(cx)yz-yzT(cx)\rho_{i}b $

$ =\rho_{i}T(bcxy)z+zT(bcxy) \rho_{i}-\rho_{i}T(bcx)yz-yzT(bcx)\rho_{i} $

which converges to

$ bT(cxy)z+zT(cxy)b-bT(cx)yz-yzT(cx)b $

$ =T(bcxy)z+zT(bcxy)-T(bcx)yz-yzT(bcx) $

with respect to the norm topology. On the other hand, let $ x=\rho_{i} $, we get

$ bT(c\rho_{i}y)z+zT(c\rho_{i}y)b-bT(c\rho_{i})yz-yzT(c\rho_{i})b $

$ =T(bc\rho_{i}y)z+zT(bc\rho_{i}y)-T(bc\rho_{i})yz-yzT(bc\rho_{i}), $

which converges to

$ bT(cy)z+zT(cy)b-bT(c)yz-yzT(c)b $

$ =T(bcy)z+zT(bcy)-T(bc)yz-yzT(bc) $

with respect to the norm topology.

In the next steps, we consider $ z=\rho_{i}, c=\rho_{i} $.

Since $ T $ is a bounded linear mapping, $ \{\rho_{i}\}_{i\in I} $ is bounded, $ \{T\rho_{i}\}_{i\in I} $ is bounded too, and we can assume that $ \{T\rho_{i}\}_{i\in I} $ converges to an element $ \xi $ in $ \mathcal{M}^{**} $ with respect to the $ w^{*} $-topology, then $ \{y\cdot T(\rho_{i})\cdot b\}_{i\in I} $ converges to $ y\cdot \xi \cdot b $ with respect to the $ w^{*} $-topology. Hence

$ 2T(by)=T(b)y+bT(y)+T(y)b+yT(b)-y\xi b-b\xi y. $

Since the right-hand-side of the above identity is symmetric on $ b $ and $ y $, we deduce that $ T(by)=T(yb) $.

$ T(by+yb)=T(b)y+yT(b)+T(y)b+bT(y)-y\xi b-b\xi y. $

Hence $ T $ is a generalized Jordan derivation.

Corollary 2.2 Suppose that $ \mathcal{A} $ is a commutative Banach algebra with the property $ \mathbb{B} $ and having a bounded approximate identity $ \{\rho_{i}\}_{i\in I} $. Let $ \mathcal{M} $ be an essential Banach $ \mathcal{A} $-bimodule. Then the following conditions are equivalent:

(1) $ T: \mathcal{A}\rightarrow \mathcal{M} $ is a generalized Jordan derivation;

(2) $ aT(b)c+cT(b)a=0 $ when $ ab=bc=0, a, b, c \in \mathcal{A} $.

Proof $ (2)\Rightarrow (1) $ is clear from Proposition 2.1.

$ (1)\Rightarrow (2) $ If $ T $ is a generalized Jordan derivation, then $ T(b)=d(b)+\xi b $, for any $ b \in \mathcal{A} $, where $ d $ is a Jordan derivation from $ \mathcal{A} $ to $ \mathcal{M}^{**} $, $ \xi \in \mathcal{M}^{**} $. Hence

$ aT(b)c+cT(b)a=a(d(b)+\xi b)c+c(d(b)+\xi b)a. $

For the Jordan derivations $ d $, we have

$ d(abc+cba)=d(a)bc+ad(b)c+abd(c)+d(c)ba+cd(b)a+cbd(a) $

for every $ a, b, c $ in $ \mathcal{A} $. Hence, by the commutativity of $ \mathcal{A} $, we get

$ ad(b)c+cd(b)a=0 $

for all $ a, b, c $ with $ ab=bc=0. $ By using the commutativity of $ \mathcal{A} $, and $ a\xi bc+c\xi ba=0 $. Hence

$ aT(b)c+cT(b)a=0 $

when $ ab=bc=0, a, b, c \in \mathcal{A} $. The proof is completed.

Suppose that $ \mathcal{A} $ is a $ * $-algebra, an $ \mathcal{A} $-bimodule $ \mathcal{M} $ is called an $ \mathcal{A} $-$ * $-bimodule if $ \mathcal{M} $ is equipped with a $ * $-mapping from $ \mathcal{M} $ into itself, such that

$ (\alpha m+\beta n)^{*}=\overline{\alpha} m^{*}+ \overline{\beta} n^{*}, (am)^{*}=m^{*}a^{*}, (ma)^{*}=a^{*}m^{*} \ \text{and}\ (m^{*})^{*}=m $

whenever $ a $ in $ \mathcal{A} $, $ m, n $ in $ \mathcal{M} $ and $ \alpha, \beta $ in $ \mathbb{C}. $

Corollary 2.3 Suppose that $ \mathcal{A} $ is a commutative Banach $ * $-algebra with property $ \mathbb{B} $ and having a bounded approximate identity $ \{\rho_{i}\}_{i\in I} $, $ \mathcal{M} $ is an essential Banach $ \mathcal{A} $-$ * $-bimodule. Let $ T $ be a continuous linear mapping from $ \mathcal{A} $ to $ \mathcal{M} $ which satisfies

$ aT(b)^{*}c+cT(b)^{*}a=0 $

whenever $ a, b, c\in \mathcal{A} $ with $ a\perp b\perp c. $ Then $ T $ is a generalized Jordan derivation.

Proof If $ \mathcal{A} $ is commutative, $ a\perp b\perp c $ is equivalent to $ ab^{*}=b^{*}c=0 $. Let $ d=b^{*} $, then the conditions in the corollary can be replaced by

$ ad=dc=0\Rightarrow aT(d^{*})^{*}c+cT(d^{*})^{*}a=0. $

By defining $ \tau(d)=T(d^{*})^{*} $, we get that

$ a\tau(d)c+c\tau(d)a=0, $

when $ ad=dc=0. $ So, we can apply Corollary 2.2 to deduce that $ \tau $ is a generalized Jordan derivation. Hence, according to the definition of $ \tau $, $ T $ is a generalized Jordan derivation.

Remark 1 We can not deduce that $ T $ is a symmetric mapping or a $ * $-mapping. If $ T $ is a inner derivation, it satisfies the above equation. However there are inner derivations which are not $ * $-derivations. In particular, $ T $ need not to be a local triple derivation, since each local triple derivation preserves the adjoint ([14, Lemma 9]).

We give the following results which contain some new generalizations of [15, Proposition 3.4], [3, Theorem 2.11], [4, Lemma 2.8] with slightly weaker hypotheses.

Theorem 2.4 Suppose that $ \mathcal{A} $ is a $ C^{*} $-algebra, and let $ T: \mathcal{A}\rightarrow \mathcal{A} $ be a bounded linear mapping. Then the following statements are equivalent:

(1) $ \{a, T(b), c\}=0, $ when $ a\perp b\perp c, \ a, b, c\in \mathcal{A} $;

(2) $ T $ is a generalized derivation.

Proof $ (1)\Rightarrow (2) $. If $ a, b, c \in \mathcal{A}_{sa} $ and $ ab=bc=0 $. Hence $ ab^{*}=b^{*}a=0, bc^{*}=c^{*}b=0 $. So we obtain

$ aT(b)^{*}c+cT(b)^{*}a=0. $

Then applying $ * $ to both sides. We get

$ aT(b)c+cT(b)a=0. $

By [4], so $ T $ is a generalized derivation.

$ (2)\Rightarrow (1) $. By the definition of generalized derivation, $ T(b)=D(b)+\xi b $ for all $ b\in \mathcal{A} $, where $ D $ is a derivation from $ \mathcal{A} $ to $ \mathcal{A}^{**} $, $ \xi \in \mathcal{A}^{**} $. If $ ab^{*}=b^{*}a=0, b^{*}c=cb^{*}=0 $, $ D(a^{*}b)=D(a^{*})b+a^{*}D(b)=0 $, then

$ b^{*}D(a^{*})^{*}+D(b)^{*}a=0, \ cb^{*}D(a^{*})^{*}+cD(b)^{*}a=0, $

so

$ cD(b)^{*}a=0. $

By a similar calculation,

$ aD(b)^{*}c=0. $

So

$ aT(b)^{*}c+cT(b)^{*}a=a(D(b)^{*}+b^{*}\xi^{*})c+c(D(b)^{*}+b^{*}\xi^{*})a= $

$ aD(b)^{*}c+ab^{*}\xi^{*}c+ cD(b)^{*}a+cb^{*}\xi^{*}a=0. $

Hence $ T $ satisfies (1).

Before proving the next main result, we give the following lemma whose proof is contained in the proof of [3, Theorem 2.1].

Lemma 2.5 Suppose that $ \mathcal{A} $ is a $ C^{*} $-algebra, $ \mathcal{M} $ is a Banach $ \mathcal{A} $-bimodule, and let $ T $ be a linear mapping from $ \mathcal{A} $ to $ \mathcal{M} $. If there exists a $ \xi \in \mathcal{M}^{**} $ such that $ T $ satisfies

$ T(a^{2})=T(a)a+aT(a)-a\xi a, $

for all $ a\in \mathcal{A}_{sa}, $ then $ T $ is a generalized Jordan derivation.

Proposition 2.2 Suppose that $ \mathcal{A} $ is a $ C^{*} $-algebra, $ \mathcal{M} $ is an essential Banach $ \mathcal{A} $-$ * $-bimodule, and let $ T: \mathcal{A}\rightarrow \mathcal{M} $ be a bounded linear mapping satisfying the following conditions:

$ \begin{equation} aT(b)^{*}c+cT(b)^{*}a=0, \ a\perp b\perp c, \ a, b, c\in \mathcal{A}. \end{equation} $

(2.3)

Then $ T $ is a generalized Jordan derivation.

Proof Let $ \mathcal{B} $ denote the abelian $ C^{*} $-subalgebra of $ \mathcal{A} $ generated by a self-adjoint element $ a $ of $ \mathcal{A} $. According to Corollary 2.3, we see that $ T|_{\mathcal{B}}: \mathcal{B}\rightarrow \mathcal{M} $ is a generalized Jordan derivation. Hence

$ T(a^{2})= T(a)a+aT(a)-a\xi a $

where $ \xi \in \mathcal{M}^{**} $. For any $ a\in \mathcal{A} $, $ a=a_{1}+ia_{2} $, where $ a_{1}, a_{2}\in \mathcal{A}_{sa} $, according to Lemma 2.5, we obtain that

$ T(ab+ba)= T(a)b+aT(b)+T(b)a+bT(a)-a\xi b-b\xi a $

for any $ a, b\in \mathcal{A}. $ So $ T $ is a generalized Jordan derivation.

We conclude this section with some results about homomorphisms.

Definition 2.4 Let $ \mathcal{A} $ and $ \mathcal{B} $ be Banach algebras. A Jordan homomorphism from $ \mathcal{A} $ into $ \mathcal{B} $ is a linear mapping $ T: \mathcal{A}\rightarrow \mathcal{B} $ such that

$ T(a\circ b)=T(a)\circ T(b)\ \ (a, b\in \mathcal{A}), $

where the symbol $ \circ $ denotes the Jordan product on $ \mathcal{A} $, i.e.

$ a\circ b=\frac{1}{2}(ab + ba)\ \ (a, b \in \mathcal{A}). $

We make full use of the powerful property $ \mathbb{B} $ to characterize homomorphisms on unital Banach algebras satisfying this property.

Proposition 2.3 Suppose that $ \mathcal{A} $ is a unital Banach algebra satisfying the property $ \mathbb{B} $, let $ T: \mathcal{A}\rightarrow \mathcal{A} $ be a continuous linear mapping satisfying

$ ab=bc=0\Rightarrow T(a)T(b)T(c)+T(c)T(b)T(a)=0. $

If $ T(1)=1 $, then $ T $ is a Jordan homomorphism.

Proof Fix $ a, b \in \mathcal{A} $ with $ ab=0 $. Define a continuous bilinear mapping

$ \varphi: \mathcal{A} \times \mathcal{A} \rightarrow \mathcal{A}, $

such that

$ \varphi(x, y)=T(a)T(bx)T(y)+T(y)T(bx)T(a). $

When $ xy=0 $, we have $ abx=0=bxy $. Hence $ \varphi(x, y)=0 $, when $ xy=0, a, b \in \mathcal{A}. $ By property $ \mathbb{B} $,

$ \varphi(x, 1)=\varphi(1, x), $

for any $ x\in \mathcal{A} $, that is, $ T(a)T(bx)+T(bx)T(a)=T(a)T(b)T(x)+T(x)T(b)T(a) $.

Define a continuous bilinear mapping given by

$ \Phi(a, b)=aT(bx)+T(bx)a-aT(b)x-xT(b)a. $

By the previous paragraph, $ ab=0\Rightarrow \Phi(a, b)=0 $. So, by property $ \mathbb{B} $, we get

$ \Phi(a, 1)=\Phi(1, a), $

for any $ a\in \mathcal{A} $.

So

$ T(a)T(x)+T(x)T(a)-T(a)T(1)T(x)-T(x)T(1)T(a) $

$ =T(ax)+T(ax)-T(a)T(x)-T(x)T(a), $

i.e.

$ 2T(ax)=T(a)T(x)+T(a)T(x)+T(x)T(a)+T(x)T(a)- $

$ T(a)T(1)T(x)-T(x)T(1)T(a). $

Since $ T(ax)=T(xa) $ and $ T(1)=1 $,

$ T(ax+xa)=T(a)T(x)+T(a)T(x). $

Hence $ T $ is a Jordan homomorphism.

Let $ \mathcal{M} $ be a semifinite von Neumann algebra with a faithful semifinite normal trace $ \tau $ acting on a Hilbert space $ \mathcal{H} $. We denote by $ \mathcal{P}(\mathcal{M}) $ the collection of all projections in $ \mathcal{M} $, by $ S(\mathcal{M}, \tau) $ the collection of all $ \tau $-measurable operators with respect to $ \mathcal{M} $. More details about $ \tau $-measurable operators can be found in [16].

We recall the definition of the measure topology $ t_{\tau} $ on the algebra $ S(\mathcal{M}, \tau ) $. For every $ \epsilon, \delta> 0 $, we define the set

$ \begin{equation*} U (\epsilon, \delta) = \{X\in S(\mathcal{M}, \tau ): \ \text{there exists} \ P \in \mathcal{P} (\mathcal{M})\ \text{such that} \ \|X(I-P)\|\leq \epsilon, \tau(P)\leq \delta\}. \end{equation*} $

The topology generated by the sets $ U (\epsilon, \delta), \epsilon, \delta>0, $ is called the measure topology $ t_{\tau} $ on $ S(\mathcal{M}, \tau) $. It is well known that the algebra $ S(\mathcal{M}, \tau) $ equipped with the measure topology is a complete metrizable topological algebra.

Definition 3.5 Suppose that $ T $ is a closed densely defined linear operator on a Hilbert space $ \mathcal{H} $ with domain $ D(T) $. The spectrum $ \sigma(T) $ of $ T $ is the set of those complex numbers $ \lambda $ such that $ T-\lambda I $ is not a one-to-one mapping of $ D(T) $ onto $ \mathcal{H} $.

Definition 3.6 Suppose that $ \mathcal{A}, \mathcal{B} $ are algebras over the complex filed $ \mathbb{C} $, and $ \phi $ is a linear mapping from $ \mathcal{A} $ to $ \mathcal{B} $. If $ \phi $ satisfies $ \sigma(\phi (a))=\sigma(a) $, for every $ a\in \mathcal{A} $, we shall say $ \phi $ is a spectrum-preserving linear mapping.

Proposition 3.4 If $ h=h^{*}\in S(\mathcal{M}, \tau) $, then $ h $ is the limit of a sequence of linear combinations of mutually orthogonal projections in measure topology ($ S(\mathcal{M}, \tau)=\overline{\mathcal{P}(\mathcal{M})}^{t_{\tau}} $).

Proof By [17, Theorem 5.6.18], $ h $ is affiliated with an abelian von Neumann subalgebra $ \mathcal{R} $ of $ \mathcal{M} $. Hence $ h $ belongs to the $ S(\mathcal{R}, \tau|_{\mathcal{R}}) $. For an abelian von Neumann algebra, it is well known that $ \mathcal{R} $ can be uniformly approximated by finite linear combinations of mutually orthogonal projections ([2, Proposition1.3.1 and Lemma 1.7.5]) i.e. $ \mathcal{R}=\overline{\mathcal{P}(\mathcal{R})}^{\|\cdot\|} $. With the consideration that $ S(\mathcal{R}, \tau|_{\mathcal{R}})=\overline{\mathcal{R}}^{t_{\tau|_{\mathcal{R}}}} $, it follows that $ S(\mathcal{R}, \tau|_{\mathcal{R}})=\overline{\overline{\mathcal{P}(\mathcal{R})}^{\|\cdot\|}}^{t_{\tau| _{\mathcal{R}}}}=\overline{\mathcal{P}(\mathcal{R})}^{t_{\tau|_{\mathcal{R}}}}. $ Hence for any $ h=h^{*}\in S(\mathcal{M}, \tau) $, there exists a von Neumann subalgebra $ \mathcal{R} $ of $ \mathcal{M} $ such that $ h\in \overline{\mathcal{P}(\mathcal{R})}^{t_{\tau |_{\mathcal{R}}}}. $

Proposition 3.5 Let $ \mathcal{M}_{1} $ and $ \mathcal{M}_{2} $ be finite von Neumann algebras, and $ \Phi: \mathcal{M}_{1}\rightarrow \mathcal{M}_{2} $ be a unital $ * $-anti-homomorphism. If $ \Phi $ is normal, then $ \Phi $ is Cauchy-continuous for the measure topologies on $ \mathcal{M}_{1} $ and $ \mathcal{M}_{2}. $

Proof Let $ \tau_{2} $ be a normal tracial state on $ \mathcal{M}_{2} $. Since $ \Phi $ is normal, we note that $ \tau_{1}:= \tau_{2}\circ\Phi $ is a normal tracial state on $ \mathcal{M}_{1} $. For $ \epsilon, \delta>0 $, $ A\in U(\tau_{1}, \epsilon, \delta) $, there is a projection $ E $ in $ \mathcal{M}_{1} $ such that $ \|AE\| \leq\epsilon $ and $ \tau_{1}(I-E)\leq\delta $. First we note that if $ \Phi $ is a $ * $-anti-homomorphism, then $ \Phi(F)\Phi(A)\Phi(E)=\Phi(EAF) $ for any $ A, E, F\in \mathcal{M}_{1} $, and $ \Phi(E) $ is a projection if $ E $ is a projection.

Let $ E\in \mathcal{P}(\mathcal{M}_{1}), A\in \mathcal{M}_{1}, $

$ \begin{equation*} \begin{aligned} \|\Phi(A)\Phi(E)\| &=\|(1-\Phi(E))\Phi(A)\Phi(E)+\Phi(E)\Phi(A)\Phi(E)\|\\ &\leq\|(1-\Phi(E))\Phi(A)\Phi(E)\|+\|\Phi(E)\Phi(A)\Phi(E)\|\\ &=\|\Phi(EA(1-E))\|+\|\Phi(EAE)\|\\ &\leq\|AE(1-E)\|+\|EAE\|\\ &\leq 2\|AE\|\\ &\leq 2\epsilon, \end{aligned} \end{equation*} $

and $ \tau_{2} (1-\Phi(E))=\tau_{1}(I-E)\leq\delta. $ Consequently,

$ \Phi(U(\tau_{1}, \epsilon, \delta))\subseteq U(\tau_{2}, 2\epsilon, \delta). $

Thus if a net $ \{A_{i}\} $ in $ \mathcal{M}_{1} $ is Cauchy in measure topology, then the net $ \{\Phi(A_{i})\} $ in $ \mathcal{M}_{2} $ is also Cauchy in measure topology. We conclude that $ \Phi $ is Cauchy-continuous for the measure topologies on $ \mathcal{M}_{1} $ and $ \mathcal{M}_{2}. $

Theorem 3.6 Suppose that $ \mathcal{M} $ is a factor of type $ II_{1} $, and let $ \phi $ be a spectrum-preserving linear mapping from $ S(\mathcal{M}, \tau) $ onto itself. Then $ \phi $ is a $ * $-isomorphism or a $ * $-anti-isomorphism.

Proof It can be easily seen that if $ \phi $ satisfies $ \sigma(\phi (a))=\sigma(a) $, then $ \phi $ is a positive mapping from $ S(\mathcal{M}, \tau) $ onto itself. Hence $ \phi $ is self-adjoint, i.e. $ \phi(a^{*})=\phi(a)^{*} $, for every $ a\in S(\mathcal{M}, \tau) $, and if $ a\in \mathcal{M} $, we can deduce that $ \phi(a)\in \mathcal{M} $. It follows that the restriction of $ \phi $ on $ \mathcal{M} $, denoted by $ \phi\rvert_{\mathcal{M}} $, is a spectrum-preserving mapping. According to [10, Theorem 1.3], $ \phi\rvert_{\mathcal{M}} $ is a Jordan isomorphism. By [18, Corollary 11], $ \phi\rvert_{\mathcal{M}} $ is a $ * $-isomorphism or a $ * $-anti-isomorphism. Hence $ \phi\rvert_{\mathcal{M}} $ is normal. By [19, Theorem 4.9] and Proposition 3.5, $ \phi\rvert_{\mathcal{M}} $ is continuous in measure topology. Let $ h=h^{*}\in S(\mathcal{M}, \tau) $, by Proposition 3.4, $ h $ is the limit of a sequence of linear combinations of orthogonal idempotents Consequently, by [10, Theorem 1.2], $ \phi(h) $ is the limit of a sequence of linear combinations of orthogonal idempotents. By continuity of $ \phi $, taking the limits of these sequences we conclude that $ \phi(h^{2})=\phi(h)^{2}. $ Taking $ h, k $ self-adjoint in $ S(\mathcal{M}, \tau) $ we get

$ \begin{equation*} \begin{aligned} (\phi(h+k))^{2} &=(\phi(h)+\phi(k))^{2}=\phi(h)^{2}+\phi(k)^{2}+\phi(h)\phi(k)+\phi(k)\phi(h) \\ &=\phi((h+k)^{2})=\phi(h^{2})+\phi(k^{2})+\phi(hk+kh). \end{aligned} \end{equation*} $

Thus $ \phi(kh+kh)=\phi(h)\phi(k)+\phi(k)\phi(h) $, for every $ h, k $ self-adjoint elements. Let $ x\in S(\mathcal{M}, \tau) $, then $ x=h+ik $ where $ h=(x+x^{*})/2 $ and $ k=(x-x^{*})/2i $ are self-adjoint elements. Hence

$ \begin{equation*} \begin{aligned} \phi(x^{2}) &= \phi(h^{2}-k^{2}+i(hk+kh))=\phi(h^{2})-\phi(k)^{2}+i(\phi(h)\phi(k)+\phi(k)\phi(h)) \\ &= (\phi(h)+i\phi(k))^{2}=\phi(x)^{2}. \end{aligned} \end{equation*} $

Hence, $ \phi $ is a Jordan $ * $-isomorphism. It follows that $ \phi $ is a $ * $-isomorphism or a $ * $-anti-isomorphism.

Theorem 3.7 Suppose that $ \mathcal{M} $ is a finite von Neumann algebra, and let $ \phi $ be a spectrum-preserving linear mapping from $ S(\mathcal{M}, \tau) $ onto itself. Then $ \phi $ is a Jordan $ * $-isomorphism.

Proof In the proof, we need the [18, Theorem 10] instead of [18, Corollary 11]. The remainder of the proof is similar to that of Theorem 3.6.