Jordan Ideals of Vector-Valued Beurling Algebras and θ-Lau Product of Banach Algebras

Abstract

In this paper, we give a characterization of the closed Jordan ideals of vector-valued Beurling algebras $L^1(G,\mathcal{A},\omega )$ and the direct sum of Banach algebras. We also investigate Jordan ideals of the $\theta$-Lau product of Banach algebras and give some results about Banach algebras ${\mathcal{A}}^{♯}$ and $\mathcal{A}\oplus \mathcal{B}$. We also prove that every Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$ is an ideal if and only if every Jordan ideal of $\mathcal{A}$ and $\mathcal{B}$ is an ideal of $\mathcal{A}$ and $\mathcal{B}$, respectively, where $\mathcal{A}$ is a unital Banach algebra and $\mathcal{B}$ is any Banach algebra. Finally, we apply our results to some Banach algebras related to a locally compact group.

1. Introduction and Preliminaries

Let $(\mathcal{A},\parallel {\parallel }_{\mathcal{A}})$ be a Banach algebra, G be a locally compact group with a left Haar measure $\mu$ and the modular function $\Delta$. For any Borel measurable function f: $G\to \mathcal{A}$, let

$${\parallel f\parallel }_{\mathcal{A}}={\int }_G{\parallel f\left(x\right)\parallel }_{\mathcal{A}}d\mu \left(x\right).$$

We denote by $L^1(G,\mathcal{A})$ the Banach space of all Borel measurable functions f: $G\to \mathcal{A}$ such that ${\parallel f\parallel }_{\mathcal{A}}<\infty$. For each $f\in L^1(G,\mathcal{A})$, define $\overline{f}\left(x\right)={\parallel f\left(x\right)\parallel }_{\mathcal{A}}$. Then $f\in L^1(G,\mathcal{A})$ if and only if $\overline{f}\in L^1\left(G\right)$. Let us recall that the convolution on $L^1(G,\mathcal{A})$ can be defined by

$$(f\ast g)\left(x\right)={\int }_{G}f\left(xy\right)g\left(y^{-1}\right)d\mu \left(y\right)={\int }_{G}f\left(y\right)g\left(y^{-1}x\right)d\mu \left(y\right)$$

for all $f,g\in L^1(G,\mathcal{A})$. With convolution as multiplication, $L^1(G,\mathcal{A})$ becomes a Banach algebra under pointwise addition and scalar multiplication; this algebra is called the vector-valued group algebra [1]. Let us recall that a weight function on G is a Borel measurable function $\omega$: $G\to [1,\infty )$ such that $\omega \left(e\right)=1$ and

$$\omega \left(xy\right)\le \omega \left(x\right)\omega \left(y\right)(x,y\in G),$$

where e is the identity element of G. Thus we can define the vector-valued Beurling algebra

$$L^1(G,\mathcal{A},\omega )=\{f\in L^1(G,\mathcal{A}):{\int }_G{\parallel f\left(x\right)\parallel }_{\mathcal{A}}\omega \left(x\right)d\mu \left(x\right)<\infty \}.$$

In the case where $\mathcal{A}=\mathbb{C}$, we set $L^1(G,\omega )$ for $L^1(G,\mathcal{A},\omega )$. If also, $\omega =1$, then we write $L^1\left(G\right)$ for $L^1(G,\mathcal{A},\omega )$.

Let $\mathcal{B}$ be also a Banach algebra. Then the Cartesian product space $\mathcal{A}\times \mathcal{B}$ is a space with the usual operations. If we choose a linear functional $\theta$: $\mathcal{B}\to \mathbb{C}$ and equip $\mathcal{A}\times \mathcal{B}$ with the product

$$(a,b){·}_{\theta }(x,y)=(ax+\theta \left(b\right)x+\theta \left(y\right)a,by),\left((a,b),(x,y)\in \mathcal{A}\times \mathcal{B}\right).$$

and the norm

$${\parallel (a,b)\parallel }_{\mathcal{A}\times \mathcal{B}}={\parallel a\parallel }_{\mathcal{A}}+{\parallel b\parallel }_{\mathcal{B}},$$

then $\mathcal{A}\times \mathcal{B}$ is a Banach algebra. This Banach algebra is called the θ-Lau product of $\mathcal{A}$ and $\mathcal{B}$, denoted it by $\mathcal{A}{\times }_{\theta }\mathcal{B}$. The $\theta$-Lau products $\mathcal{A}{\times }_{\theta }\mathcal{B}$ were first introduced by Lau [2], for certain Banach algebras. Sangani Monfared [3] extended this product to arbitrary Banach algebras $\mathcal{A}$ and $\mathcal{B}$; see also [4,5].

A linear subspace $\mathcal{J}$ of $\mathcal{A}$ is called a Jordan ideal if for every $a\in \mathcal{A}$ and $b\in \mathcal{J}$, we have

$$a\circ b:=ab+ba\in \mathcal{J}.$$

Recall that a standard two-sided ideal requires both $ab\in \mathcal{J}$ and $ba\in \mathcal{J}$ for all $a\in \mathcal{A}$ and $b\in \mathcal{J}$. It is obvious that every such ideal is also a Jordan ideal. Note that the converse does not remain valid, in general.

Example 1.

Let $\mathbb{F}$ be a field with $\mathrm{char}\left(\mathbb{F}\right)\ne 2$. Let $M_4\left(\mathbb{F}\right)$ be the associative algebra of $4\times 4$ matrices over $\mathbb{F}$. Denote by $E_{ij}$ the standard matrix units, so that $E_{ij}{E}_{kl}={\delta }_{jk}{E}_{il}$. Define three matrices in $M_4\left(\mathbb{F}\right)$ by

$$e:=E_{21}+E_{43},f:=E_{31}-E_{42},g:=E_{41}.$$

Let

$$\mathcal{A}:={\mathrm{span}}_{\mathbb{F}}\{e,f,g\}\subseteq M_4\left(\mathbb{F}\right),\mathcal{J}:={\mathrm{span}}_{\mathbb{F}}\{e,f\}\subseteq \mathcal{A}.$$

Take arbitrary

$$a=\alpha e+\beta f+\gamma g\in \mathcal{A},b=ue+vf\in \mathcal{J}.$$

Since $ef=g$ and $fe=-g$, we have $a\circ b=0\in \mathcal{J}$. Thus $\mathcal{A}\circ J\subseteq J$, and $\mathcal{J}$ is a Jordan ideal of $\mathcal{A}$. But $ef=g\notin \mathcal{J}$, this shows $\mathcal{J}$ is not an two-sided ideal of $\mathcal{A}$.

Many authors have been interested in finding conditions under which a non-zero Jordan ideal is an associative ideal [6,7,8,9,10,11,12]. It was proved that the Jordan ideals of all simple algebras [9], all full matrix algebras $M_n(\mathcal{A}$), where $\mathcal{A}$ is a unital algebra [10], all $W^{\ast }$-algebras [8] are associative ideals. Li and Lu [11] showed that a weakly closed Jordan ideal in a nest algebra on a Banach space, which satisfies certain conditions, is an associative ideal. In ref. [13], the authors recently characterized the closed Lie ideals of vector-valued group algebras.

In this paper, we investigate Jordan ideals of Banach algebras. In Section 2, we give a characterization of the closed Jordan ideal of vector-valued Beurling algebras $L^1(G,\mathcal{A},\omega )$ and show that a closed subspace $\mathcal{J}$ of $L^1(G,\mathcal{A},\omega )$ is a Jordan ideal if and only if

$$\Delta \left(x^{-1}\right)(f_{x^{-1}}·a)+a·{}_x\mathit{f}\in \mathcal{J}$$

for all $f\in \mathcal{J},x\in G$ and $a\in \mathcal{A}$. In Section 3, we study Jordan ideals of $\theta$-Lau products $\mathcal{A}{\times }_{\theta }\mathcal{B}$. In particular, we present a necessary and sufficient condition under which a subspace of the direct sum of Banach algebras is a Jordan ideal. For a unital Banach algebra $\mathcal{A}$, we establish that every Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$ is an ideal if and only if every Jordan ideal of $\mathcal{A}$ and $\mathcal{B}$ is an ideal of $\mathcal{A}$ and $\mathcal{B}$, respectively. Finally, we apply our results to finite group algebra $\mathbb{C}\left[G\right]$, as well as to the Banach algebras $\mathrm{LUC}{\left(G\right)}^{\ast }$ and $M_{\ast }{\left(G\right)}^{\ast }$.

2. Jordan Ideals of Vector-Valued Beurling Algebras

We denote by $C_c(G,\mathcal{A})$ the space of all continuous functions from G into $\mathcal{A}$ with compact support. If $\mathcal{A}=\mathbb{C}$, then we put $C_c\left(G\right):=C_c(G,\mathbb{C})$. It is well known that $C_c\left(G\right)$ is norm-dense in $L^1\left(G\right)$. To establish the density of this space in the vector-valued and weighted setting, we employ a decoupling approximation strategy. Specifically, we first treat $f\omega$ as an unweighted vector-valued function and approximate it by a continuous, compactly supported vector-valued function h. Subsequently, we locally approximate the weight $\omega$ by a continuous positive scalar function $\eta$. By constructing the quotient function $g=h/\eta$, we can effectively control the Bochner norm. This reasoning naturally leads to the following result.

Lemma 1.

Let G be a locally compact group and ω be a weight function on G. Then $C_c(G,\mathcal{A})$ is a norm dense subspace of $L^1(G,\mathcal{A},\omega )$.

By [1] (Lemma 1.3.6), the maps $x\mapsto {}_x\mathit{f}$ and $x\mapsto f_x$ from G into $L^1(G,\mathcal{A})$ are continuous, where ${}_x\mathit{f}\left(y\right)=f\left(x^{-1}y\right)$ and $f_x\left(y\right)=f\left(yx\right)$ for all $x,y\in G$. Similarly, since $\omega$ is locally bounded [1] (Lemma 1.3.3), with the necessary modifications in the proof, the result can also be established for $L^1(G,\mathcal{A},\omega )$ in place of $L^1(G,\mathcal{A})$.

Our next goal is to characterize the closed Jordan ideals of vector-valued Beurling algebras. For every $a\in \mathcal{A}$ and $f\in L^1(G,\mathcal{A},\omega )$, consider the mappings $a·f,f·a$: $G\to \mathcal{A}$ given by

$$(a·f)\left(x\right)=af\left(x\right)\mathrm{and}(f·a)\left(x\right)=f\left(x\right)a$$

for all $x\in G$. It is easy seen that $a·f$ and $f·a\in L^1(G,\mathcal{A},\omega )$ for all $a\in \mathcal{A}$ and $f\in L^1(G,\mathcal{A},\omega )$. Laursen [14] established that, for a Banach algebra $\mathcal{A}$ with an approximate identity, a closed subspace $\mathcal{J}\subseteq L^1(G,\mathcal{A})$ is an ideal precisely when it is $\mathcal{A}$-translation invariant, i.e., ${}_x\mathit{f},f_x,a·f,f·a\in \mathcal{J}$ for all $f\in \mathcal{J},a\in \mathcal{A},x\in G$. This assertion also holds for vector-valued Beurling algebra $L^1(G,\mathcal{A},\omega )$; see [15]. We now state the main result of this section.

Theorem 1.

Let G be a locally compact group, ω a weight function on G, and $\mathcal{A}$ a Banach algebra. A closed subspace $\mathcal{J}$ of $L^1(G,\mathcal{A},\omega )$ is a Jordan ideal if and only if

$$\Delta \left(x^{-1}\right)(f_{x^{-1}}·a)+a·{}_x\mathit{f}\in \mathcal{J}$$

for all $f\in \mathcal{J},x\in G$, and $a\in \mathcal{A}$.

Proof. 

First, we denote by ${\chi }_K$ the characteristic function of K on G, where K is a subset of G. Let $f\in \mathcal{J},x\in G$ and $a\in \mathcal{A}$. Then for every $y\in G$, we have

$$\begin{array}{ccc}\hfill (\mu \left(K^{-1}\right)f_{x^{-1}}-f\ast {\chi }_K)\left(y\right)& =& \mu \left(K^{-1}\right)f_{x^{-1}}\left(y\right)-\int f\left(yz\right){\chi }_K\left(z^{-1}\right)dz\hfill \\ & =& {\int }_{K^{-1}}{f}_{x^{-1}}\left(y\right)dz-\int f\left(yz\right){\chi }_{K^{-1}}\left(z\right)dz\hfill \\ & =& {\int }_{K^{-1}}(f_{x^{-1}}\left(y\right)-f\left(yz\right))dz\hfill \\ & =& {\int }_{K^{-1}}(f_{x^{-1}}-f_z)\left(y\right)dz.\hfill \end{array}$$

This implies that

$$\begin{array}{c}\hfill \parallel \mu \left(K^{-1}\right)f_{x^{-1}}-f\ast {\chi }_K{\parallel }_{\omega }\le {\int }_{K^{-1}}{\parallel f_{x^{-1}}-f_z\parallel }_{\omega }dz.\end{array}$$

(1)

For every $f\in L^1(G,\mathcal{A},\omega )$, the maps $z\mapsto {}_z\mathit{f},f_z$ are continuous. Hence for every $ϵ>0$, we can choose a compact symmetric neighborhood V of e such that

$$\underset{z\in V}{sup}\parallel f_{x^{-1}}-f_{z^{-1}{x}^{-1}}{\parallel }_{\omega }<{\displaystyle \frac{ϵ}{2\parallel a\parallel \Delta \left(x^{-1}\right)}}\mathrm{and}\underset{z\in V}{sup}{\parallel {}_x\mathit{f}-{}_{xz}\mathit{f}\parallel }_{\omega }<{\displaystyle \frac{ϵ}{2\parallel a\parallel }}.$$

Taking $K=xV$ in (1), we have

$$\begin{array}{ccc}\hfill \parallel \mu \left(V{x}^{-1}\right)f_{x^{-1}}·a-f\ast (a·{\chi }_{xV}){\parallel }_{\omega }& \le & \mu \left(V{x}^{-1}\right)\parallel a\parallel \underset{z\in V{x}^{-1}}{sup}{\parallel f_{x^{-1}}-f_z\parallel }_{\omega }\hfill \\ & =& \Delta \left(x^{-1}\right)\mu \left(V\right)\parallel a\parallel \underset{z\in V}{sup}{\parallel f_{x^{-1}}-f_{z^{-1}{x}^{-1}}\parallel }_{\omega }\hfill \\ & <& ϵ/2\mu \left(V\right).\hfill \end{array}$$

This together with the fact $\mu \left(V{x}^{-1}\right)=\Delta \left(x^{-1}\right)\mu \left(V\right)$ shows that

$$\begin{array}{c}\hfill \parallel \Delta \left(x^{-1}\right)f_{x^{-1}}·a-{\displaystyle \frac{1}{\mu \left(V\right)}}f\ast (a·{\chi }_{xV}){\parallel }_{\omega }<ϵ/2.\end{array}$$

(2)

Similarly, one can prove that

$$(\mu {\left(V\right)}_{x}f-{\chi }_{xV}\ast f)\left(y\right)={\int }_V{(}_{x}f{-}_{xz}f)\left(y\right)dz$$

for all $y\in G$. Hence we have

$$\parallel \mu \left(V\right)\left(a{·}_{x}f\right)-(a·{\chi }_{xV})\ast {f\parallel }_{\omega }\le \parallel a\parallel {\int }_V{\parallel }_{x}f{-}_{xz}f\parallel dz<\mu \left(V\right)ϵ/2.$$

But,

$$\Delta \left(x^{-1}\right)\parallel \mu \left(V\right)\left(a{·}_{x}f\right)-(a·{\chi }_{xV})\ast {f\parallel }_{\omega }=\mu \left(V{x}^{-1}\right)\parallel a·{}_x\mathit{f}-{\displaystyle \frac{1}{\mu \left(V\right)}}{a}_{xV}\ast {f\parallel }_{\omega }.$$

Thus

$$\begin{array}{c}\hfill \parallel a·{}_x\mathit{f}-{\displaystyle \frac{1}{\mu \left(V\right)}}{a}_{xV}\ast {f\parallel }_{\omega }<ϵ/2.\end{array}$$

(3)

From (2) and (3) we see that

$$\parallel \Delta \left(x^{-1}\right)(f_{x^{-1}}·a)+a·{}_x\mathit{f}-{\displaystyle \frac{1}{\mu \left(V\right)}}\left(f\ast (a·{\chi }_{xV})+(a·{\chi }_{xV})\ast f\right){\parallel }_{\omega }<ϵ.$$

This implies that if $\mathcal{J}$ is a closed Jordan ideal of $L^1(G,\mathcal{A},\omega )$, then

$$\Delta \left(x^{-1}\right)(f_{x^{-1}}·a)+a·{}_x\mathit{f}\in \mathcal{J}.$$

Conversely, assume that $\varphi \in C_c(G,\mathcal{A})$ and $K=\mathrm{supp}\left(\varphi \right)\cup \mathrm{supp}{\left(\varphi \right)}^{-1}$, where $\mathrm{supp}{\left(\varphi \right)}^{-1}=\{y^{-1}$: $y\in \mathrm{supp}\left(\varphi \right)\}$. Since the inversion map in the topological group G is continuous and $\mathrm{supp}\left(\varphi \right)$ is compact, the set K is also a compact subset of G. To approximate the integral, we partition the compact set K into a finite number of mutually disjoint Borel measurable sets. Let $K={\cup }_{i=1}^n{K}_i$, where $K_i\cap K_j=\varnothing$ for $i\ne j$. Then for every $f\in \mathcal{J},y_i\in K_i$ and $x\in G$, we have

$$\begin{array}{ccc}\hfill (\varphi \ast f-\sum _{i=1}^n\mu \left(K_i\right)\varphi {\left(y_i\right)}_{y_i}f)\left(x\right)& =& {\int }_K\varphi \left(y\right)f\left(y^{-1}x\right)dy-\sum _{i=1}^n{\int }_{K_i}\varphi {\left(y_i\right)}_{y_i}f\left(x\right)dy\hfill \\ & =& \sum _{i=1}^n{\int }_{K_i}(\varphi {\left(y\right)}_{y}f\left(x\right)-\varphi {\left(y_i\right)}_{y_i}f\left(x\right))dy.\hfill \end{array}$$

A similar argument shows that

$$\begin{array}{cc}\hfill (f\ast \varphi -\sum _{i=1}^n\mu \left(K_i\right)\Delta \left(y_i^{-1}\right)f_{y_i^{-1}}\varphi \left(y_i\right))\left(x\right)& ={\sum }_{i=1}^n{\int }_{K_i}\Delta \left(y^{-1}\right)f_{y^{-1}}\varphi \left(y\right)\\ & -\Delta \left(y_i^{-1}\right)f_{y_i}\left(x\right)\varphi \left(y_i\right).\end{array}$$

Hence

$$\begin{array}{ccc}\hfill \parallel \varphi \ast f+f\ast \varphi & -& \sum _{i=1}^n\mu \left(K_i\right)\left(\varphi \left(y_i\right){}_{yi}\mathit{f}+\Delta \left(y_i^{-1}\right)f_{y_i^{-1}}\varphi \left(y_i\right)\right){\parallel }_{\omega }\hfill \\ & \le & \sum _{i=1}^n{\int }_{K_i}\parallel \varphi \left(y\right){}_y\mathit{f}-\varphi \left(y_i\right){}_{yi}\mathit{f}+\Delta \left(y^{-1}\right)f_{y^{-1}}\varphi \left(y\right)\hfill \\ & -& \Delta \left(y_i^{-1}\right)f_{y_i^{-1}}\left(x\right)\varphi \left(y_i\right){dy)\parallel }_{\omega }dy.\hfill \end{array}$$

Since left and right translations, inverse function and module function are continuous, for any $ϵ>0$ we can choose finitely many measurable disjoint subsets $K_i(i=1,\dots n)$ of K such that

$$\parallel \varphi \left(y\right){}_y\mathit{f}-\Delta \left(y^{-1}\right)f_{y^{-1}}\varphi \left(y\right)+\varphi \left(z\right){}_z\mathit{f}-\Delta \left(z^{-1}\right)f_{z^{-1}}\varphi \left(z\right){\parallel }_{\omega }\le {\displaystyle \frac{ϵ}{\mu \left(K\right)}}$$

for all $y,z\in K_i$, $1\le i\le n$. The proof will be complete if we only recall that $C_c(G,\mathcal{A})$ is dense in $L^1(G,\mathcal{A},\omega )$. □

As an immediate consequence of the preceding theorem, we obtain the following corollary.

Corollary 1.

Let G be a locally compact group and ω a weight function on G. A closed subspace $\mathcal{J}$ of Beurling algebra $L^1(G,\omega )$ is a Jordan ideal if and only if

$$\Delta \left(x^{-1}\right)f_{x^{-1}}+{}_x\mathit{f}\in \mathcal{J}$$

for all $f\in \mathcal{J},x\in G$.

The following example shows that one-sided invariance does not imply the Jordan ideal property; the symmetric condition in Corollary 1 is essential.

Example 2.

Let $G=\left\{e\right\}$ be the trivial group and let $\omega \equiv 1$. Then, in the isometric isomorphism sense,

$$L^1(G,A,\omega )\cong A.$$

Take

$$A=M_2\left(\mathbb{C}\right),\mathcal{J}=\left\{\left(\begin{array}{cc}a& 0\\ b& 0\end{array}\right):a,b\in \mathbb{C}\right\}.$$

For any

$$x=\left(\begin{array}{cc}{x}_{11}& x_{12}\\ x_{21}& x_{22}\end{array}\right)\in A,y=\left(\begin{array}{cc}a& 0\\ b& 0\end{array}\right)\in \mathcal{J},$$

we have

$$xy=\left(\begin{array}{cc}{x}_{11}a+x_{12}b& 0\\ x_{21}a+x_{22}b& 0\end{array}\right)\in \mathcal{J}.$$

Hence $\mathcal{J}$ is a closed left ideal of A. However, J is not a Jordan ideal. Let $E_{ij}$ be the matrix units in $M_2\left(\mathbb{C}\right)$; we have

$$E_{21}{E}_{12}+E_{12}{E}_{21}=E_{22}+E_{11}=I,$$

but $I\notin \mathcal{J}$.

3. Jordan Ideals of $\mathit{\theta }$-Lau Product Banach Algebras

Let $\mathcal{A}$ and $\mathcal{B}$ be Banach algebras. For linear functionals $\theta$ and $\gamma$ on $\mathcal{B}$, a subspace $\mathcal{J}$ of $\mathcal{A}\times \mathcal{B}$ is called a $(\theta ,\gamma )$-Jordan ideal if for every $(a,b)\in \mathcal{A}\times \mathcal{B}$ and $(x,y)\in \mathcal{J}$, we have

$${\displaystyle \frac{1}{2}}[(a,b){·}_{\theta }(x,y)+(x,y){·}_{\gamma }(a,b)]\in \mathcal{J}.$$

In the following of this section, we assume that $\mathcal{J}$ is a $(\theta ,\gamma )$-Jordan ideal and put

$${\mathcal{J}}_1=\{a\in \mathcal{A}:(a,y)\in \mathcal{J}\mathrm{for}\mathrm{some}y\in \mathcal{B}\}$$

and

$${\mathcal{J}}_2=\{b\in \mathcal{B}:(x,b)\in \mathcal{J}\mathrm{for}\mathrm{some}x\in \mathcal{A}\}.$$

Note that if $a\in {\mathcal{J}}_1$, then there is $y\in {\mathcal{J}}_2$ such that $(a,y)\in \mathcal{J}$. This is true for elements of ${\mathcal{J}}_2$. Hence $\mathcal{J}\subseteq {\mathcal{J}}_1\times {\mathcal{J}}_2$. The following example shows that there are $(\theta ,\gamma )$-Jordan ideals $\mathcal{J}\ne {\mathcal{J}}_1\times {\mathcal{J}}_2$.

Example 3.

Let $\mathcal{A}=\mathcal{B}=C_0\left(\mathbb{R}\right)$, the space of all continuous functions on real numbers filed $\mathbb{R}$ vanishes at infinity. Define $\theta :\mathcal{B}\to \mathbb{C}$ by $\theta \left(f\right)=f\left(0\right)$. Then

$$\mathcal{J}=\left\{\right(f,g)\in \mathcal{A}\times \mathcal{B}:f(0)=-g(0\left)\right\}$$

is a $(\theta ,\theta )$-Jordan ideal and ${\mathcal{J}}_1={\mathcal{J}}_2=C_0\left(\mathbb{R}\right)$. Clearly, $\mathcal{J}\ne {\mathcal{J}}_1\times {\mathcal{J}}_2$.

Proposition 1.

Let θ and γ be linear functionals on $\mathcal{B}$ such that ${\mathcal{J}}_2\subseteq \ker (\theta +\gamma )$. Then the following statements hold.

(i) If $\mathcal{A}$ has an approximate identity and $\mathcal{J}$ is closed, then $\mathcal{J}={\mathcal{J}}_1\times {\mathcal{J}}_2$.

(ii) If $\mathcal{A}$ has an identity, then $\mathcal{J}={\mathcal{J}}_1\times {\mathcal{J}}_2$.

Proof. 

(i) Let ${\left\{e_{\alpha }\right\}}_{\alpha }$ be an approximate identity for $\mathcal{A}$. Let $x\in {\mathcal{J}}_1$ and $y\in {\mathcal{J}}_2$. Then $(x,z),(u,y)\in \mathcal{J}$ for some $u\in {\mathcal{J}}_1$ and $z\in {\mathcal{J}}_2$. Since ${\mathcal{J}}_2$ is the subspace of $\ker (\theta +\gamma )$, it follows that

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2}}[(e_{\alpha },0){·}_{\theta }(x,z)+(x,z){·}_{\gamma }(e_{\alpha },0)]& ={\displaystyle \frac{1}{2}}\left[(e_{\alpha }x+x{e}_{\alpha }+e_{\alpha }(\theta +\gamma )\left(z\right),0)\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}(e_{\alpha }x+x{e}_{\alpha },0)\to (x,0).\hfill \end{array}$$

Therefore, $(x,0)\in \mathcal{J}$. Similarly, $(u,0)\in \mathcal{J}$. Consequently, $(u,y)-(u,0)\in \mathcal{J}$. This shows that $(x,y)=(x,0)+(0,y)\in \mathcal{J}$.

(ii) The proof is similar to (i). □

It is easy to see that ${\mathcal{J}}_2$ is a Jordan ideal of $\mathcal{B}$. It is natural to ask: Is ${\mathcal{J}}_1$ a Jordan ideal of $\mathcal{A}$?

Proposition 2.

Let θ and γ be linear functionals on $\mathcal{B}$. Then ${\mathcal{J}}_1$ is a Jordan ideal if and only if ${\mathcal{J}}_1=\mathcal{A}$ or ${\mathcal{J}}_2\subseteq \mathit{ker}(\theta +\gamma )$.

Proof. 

Let ${\mathcal{J}}_2\subseteq \ker (\theta +\gamma )$. If $x\in {\mathcal{J}}_1$, then $(x,y)\in \mathcal{J}$ for some $y\in {\mathcal{J}}_2$. So, for every $a\in \mathcal{A}$, we have

$$\begin{array}{cc}\hfill (a,0){·}_{\theta }(x,y)+(x,y){·}_{\gamma }(a,0)& =(ax+xa+a(\theta +\gamma \left)\right(y),0)\hfill \\ \hfill & =(ax+xa,0)\in \mathcal{J}.\hfill \end{array}$$

Thus $ax+xa\in {\mathcal{J}}_1$. That is, ${\mathcal{J}}_1$ is a Jordan ideal.

Conversely, let ${\mathcal{J}}_1$ be a Jordan ideal and ${\mathcal{J}}_2\cancel{\subseteq }\ker (\theta +\gamma )$. Choose $y\in {\mathcal{J}}_2$ with $(\theta +\gamma )\left(y\right)\ne 0$. Then there exists $x\in {\mathcal{J}}_1$ such that $(x,y)\in \mathcal{J}$. If $a\in \mathcal{A}$, then

$$(x,y){·}_{\theta }(a,0)+(a,0){·}_{\gamma }(x,y)=(xa+ax+a(\theta +\gamma )\left(y\right),0)\in \mathcal{J}.$$

Hence $xa+ax+a(\theta +\gamma )\left(y\right)\in {\mathcal{J}}_1$. This shows that $a(\theta +\gamma )\left(y\right)\in {\mathcal{J}}_1$. Therefore, $a\in {\mathcal{J}}_1$. □

In the following, let ${\mathcal{A}}^{♯}$ be the unitization of $\mathcal{A}$.

Corollary 2.

Let $\mathcal{J}$ be a Jordan ideal of ${\mathcal{A}}^{♯}$ and ${\mathcal{J}}_2\ne 0$. If ${\mathcal{J}}_1$ is a Jordan ideal of $\mathcal{A}$, then ${\mathcal{J}}_1=\mathcal{A}$ and ${\mathcal{J}}_2=\mathbb{C}$.

Proof. 

Let $\theta =\gamma ={\mathrm{id}}_{\mathcal{B}}$, the identity map on $\mathcal{B}$. Then ${\mathcal{A}}^{♯}=\mathcal{A}{\times }_{\theta }\mathbb{C}$ and $\mathcal{J}$ is a $(\theta ,\gamma )$-Jordan ideal. Since ${\mathcal{J}}_2$ is a Jordan ideal of $\mathbb{C}$, it follows that it is an ideal of $\mathbb{C}$ and so ${\mathcal{J}}_2=0$ or $\mathbb{C}$. Thus ${\mathcal{J}}_2=\mathbb{C}$. Hence ${\mathcal{J}}_2\cancel{\subseteq }\ker (\theta +\gamma )$. Now, if ${\mathcal{J}}_1$ is a Jordan ideal of $\mathcal{A}$, then by Proposition 2, ${\mathcal{J}}_1=\mathcal{A}$. □

Lemma 2.

Let ${\mathcal{I}}_1$ and ${\mathcal{I}}_2$ be Jordan ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively, and let θ and γ be linear functionals on $\mathcal{B}$. Then ${\mathcal{I}}_1\times {\mathcal{I}}_2$ is a $(\theta ,\gamma )$-Jordan ideal if and only if ${\mathcal{I}}_1=\mathcal{A}$ or ${\mathcal{I}}_2\subseteq \mathit{ker}(\theta +\gamma )$.

Proof. 

Let $\mathcal{J}={\mathcal{I}}_1\times {\mathcal{I}}_2$ be a $(\theta ,\gamma )$-Jordan ideal. Then ${\mathcal{J}}_1={\mathcal{I}}_1$ and ${\mathcal{J}}_2={\mathcal{I}}_2$. Since ${\mathcal{J}}_1$ is a Jordan ideal, by Proposition 2, ${\mathcal{J}}_1=\mathcal{A}$ or ${\mathcal{J}}_2\subseteq \ker (\theta +\gamma )$. The converse is trivial. □

Theorem 2.

The following statements are valid.

(i) If $\mathcal{J}$ is a Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$, then ${\mathcal{J}}_1$ and ${\mathcal{J}}_2$ are Jordan ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively.

(ii) Let $\mathcal{A}$ be unital. Then $\mathcal{J}$ is a Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$ if and only if ${\mathcal{J}}_1$ and ${\mathcal{J}}_2$ are Jordan ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively, and $\mathcal{J}={\mathcal{J}}_1\times {\mathcal{J}}_2$.

(iii) Let $\mathcal{A}$ be unital. If $\mathcal{A}$ and $\mathcal{B}$ are both simple, then the Jordan ideals of $\mathcal{A}\oplus \mathcal{B}$ are 0, $\mathcal{A}$, $\mathcal{B}$ and $\mathcal{A}\oplus \mathcal{B}$.

Proof. 

Let $\theta =\gamma =0$, and let $\mathcal{J}$ be a Jordan ideal of $\mathcal{A}\oplus \mathcal{B}=\mathcal{A}{\times }_{\theta }\mathcal{B}$. Then $\mathcal{J}$ is a $(\theta ,\gamma )$-Jordan ideal. So, ${\mathcal{J}}_2$ is a Jordan ideal of $\mathcal{B}$ and $\mathcal{J}\subseteq \ker (\theta +\gamma )$. Hence, by Proposition 2, ${\mathcal{J}}_1$ is a Jordan ideal. That is, (i) holds. Assume now that $\mathcal{A}$ has identity. By Proposition 1 and (i), ${\mathcal{J}}_1$ and ${\mathcal{J}}_2$ are Jordan ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively, and $\mathcal{J}={\mathcal{J}}_1\times {\mathcal{J}}_2$. The converse follows from Lemma 2. Therefore, (ii) holds. The statement (iii) follows from (ii). □

Theorem 3.

Let $\mathcal{A}$ be unital. Then every Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$ is an ideal if and only if every Jordan ideal of $\mathcal{A}$ and $\mathcal{B}$ is an ideal of $\mathcal{A}$ and $\mathcal{B}$, respectively.

Proof. 

Let $\mathcal{J}$ be a Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$. Then by Theorem 2 (ii), ${\mathcal{J}}_1$ and ${\mathcal{J}}_2$ are Jordan ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively, and $\mathcal{J}={\mathcal{J}}_1\times {\mathcal{J}}_2$. If every Jordan ideal of $\mathcal{A}$ and $\mathcal{B}$ is an ideal of $\mathcal{A}$ and $\mathcal{B}$, respectively, then ${\mathcal{J}}_1$ and ${\mathcal{J}}_2$ are ideals. Hence, $\mathcal{J}$ is an ideal of $\mathcal{A}\oplus \mathcal{B}$. Conversely, let ${\mathcal{I}}_1$ and ${\mathcal{I}}_2$ be Jordan ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively. Then $\mathcal{J}={\mathcal{I}}_1\times {\mathcal{I}}_2$ is a Jordan ideal of $\mathcal{A}\oplus \mathcal{B}$. Hence $J$ is an ideal of $\mathcal{A}\oplus \mathcal{B}$. Therefore ${\mathcal{I}}_1$ and ${\mathcal{I}}_2$ are ideals of $\mathcal{A}$ and $\mathcal{B}$, respectively. □

Let us recall that for a finite group G, the complex group algebra $\mathbb{C}\left[G\right]:=\{{\sum }_{s\in G}{a}_{s}s$: $a_s\in \mathbb{C}\}$ is a *-algebra equipped with the following operations:

$${\left(\sum _{s\in G}{a}_{s}s\right)}^{\ast }=\sum _{s\in G}{\overline{a}}_s{s}^{-1},\sum _{s\in G}{a}_{s}s+\sum _{s\in G}{b}_{s}s=\sum _{s\in G}(a_s+b_s)s$$

and

$$\left(\sum _{s\in G}{a}_{s}s\right)\left(\sum _{s\in G}{b}_{s}s\right)=\sum _{s\in G}\left(\sum _{t\in G}{a}_t{b}_{t^{-1}s}\right)s.$$

Consider the left regular representation $\lambda :G\to \mathcal{U}\left({\ell }^2\left(G\right)\right)$, defined by ${\lambda }_x\left(\xi \right)\left(y\right)=\xi \left(x^{-1}y\right)$, where $x,y\in G$ and $\xi \in {\ell }^2\left(G\right)$, extends to an injective ∗-representation $\lambda$: $\mathbb{C}\left[G\right]\to B\left({\ell }^2\left(G\right)\right)$. This endows $\mathbb{C}\left[G\right]$ with a $C^{\ast }$-norm defined by $\parallel u\parallel =\parallel {\lambda }_u\parallel$ for $u\in \mathbb{C}\left[G\right]$. As a finite-dimensional $C^{\ast }$-algebra, $\mathbb{C}\left[G\right]$ admits a ∗-isomorphism $\psi$: $\mathbb{C}\left[G\right]\to {⨁}_{j=1}^k{M}_{n_j}$ by the standard structure theorem, where $M_{n_j}$ is the algebra of $n_j\times n_j$ complex matrices and $n_j$ is the dimension of the corresponding irreducible ∗-representation. This fact, together with Theorem 3, yields the following result.

Corollary 3.

Let G be a finite group. Then every Jordan ideal of $\mathbb{C}\left[G\right]$ is an ideal.

Let G be a locally compact group. Denote by $C_b\left(G\right)$ the space of all bounded continuous functions on G, and by $\mathrm{LUC}\left(G\right)$ the subspace of all $f\in C_b\left(G\right)$ for which the map $x\mapsto {}_x\mathit{f}$ from G into $C_b\left(G\right)$ is continuous. It is well known that $\mathrm{LUC}{\left(G\right)}^{\ast }$ is a Banach algebra with respect to the following product.

$$\langle m·n\rangle =\langle m,nf\rangle ,$$

where $\langle nf,x\rangle =\langle n,{}_x\mathit{f}\rangle$ for all $m,n\in \mathrm{LUC}{\left(G\right)}^{\ast }$, $f\in \mathrm{LUC}\left(G\right)$ and $x\in G$. One can prove that $\mathrm{LUC}{\left(G\right)}^{\ast }$ is unital and

$$\mathrm{LUC}{\left(G\right)}^{\ast }=M\left(G\right)\oplus C_0{\left(G\right)}^{\perp },$$

where $M\left(G\right)$ is the Banach algebra of all regular Borel measure on G and

$$C_0{\left(G\right)}^{\perp }=\{m\in \mathrm{LUC}{\left(G\right)}^{\ast }:\langle m,f\rangle =0\mathrm{for}\mathrm{all}f\in C_0\left(G\right)\};$$

see for example [16].

Corollary 4.

Let G be an abelian locally compact group. Then every Jordan ideal of $\mathit{LUC}{\left(G\right)}^{\ast }$ is an ideal if and only if every Jordan ideal of $C_0{\left(G\right)}^{\perp }$ is an ideal.

Proof. 

Let G be abelian. Then $M\left(G\right)$ is commutative. So every Jordan ideal of $M\left(G\right)$ is an ideal of $M\left(G\right)$. Now apply Theorem 3. □

Remark 1.

Let $M_{\ast }\left(G\right)$ be the subspace of $M{\left(G\right)}^{\ast }$ consisting of all linear functionals that vanish at infinity. It is proved that $C_0\left(G\right)$ is a subspace of $M_{\ast }\left(G\right)$. Also, $M_{\ast }{\left(G\right)}^{\ast }$ is a unital Banach algebra; for more details see [17,18,19,20]. So

$$M_{\ast }{\left(G\right)}^{\ast }=M\left(G\right)\oplus M_0\left(G\right),$$

where

$$M_0\left(G\right)=\{m\in M_{\ast }{\left(G\right)}^{\ast }:\langle m,f\rangle =0\mathrm{for}\mathrm{all}f\in C_0\left(G\right)\}.$$

By an argument similar to that given in the proof of Corollary 4, we prove that for an abelian locally compact group G, every Jordan ideal of $M_{\ast }{\left(G\right)}^{\ast }$ is an ideal if and only if every Jordan ideal of $M_0\left(G\right)$ is an ideal.