An Extension of the Spectral Mapping Theorem

Abstract

We give an extension of the spectral mapping theorem on hypergroups and prove that if 𝐾 is a commutative strong hypergroup with 𝐾=𝑋_𝑏(𝐾) and 𝜅 is a weakly continuous representation of 𝑀(𝐾) on a 𝑊^∗-algebra such that for every 𝑡∈𝐾, 𝜅_𝑡 is an ∗-automorphism, 𝑠𝑝𝜅 is a synthesis set for 𝐿¹(𝐾) and 𝜅(𝐿¹(𝐾)) is without order, then for any 𝜇 in a closed regular subalgebra of 𝑀(𝐾) containing 𝐿¹(𝐾), 𝜎(𝜅(𝜇))=𝜇(𝑠𝑝𝜅), where 𝑠𝑝𝜅 is the Arveson spectrum of 𝜅.

1. Introduction and Notation

Hypergroups were introduced in a series of papers by Jewett [1], Dunkle [2], and Spector [3] in 70's. They are in fact extensions of topological groups with the difference that hypergroups do not have necessarily an algebraic structure. Roughly speaking, the product of two elements of a hypergroup is a probability measure. A hypergroup is a locally compact space which has enough structure so that a convolution on the space of finite regular Borel measures can be defined. Therefore, the extension of Fourier analysis on hypergroups is made with more difficulties and usually with different proofs in the group case. Classical examples of hypergroups are locally compact groups, the space of conjugacy classes of a compact group, spaces of orbits in the group of automorphisms, and double-cosets of certain nonnormal closed subgroups of a compact group. We will state definition and some basic properties of hypergroups in Section 2. Throughout this paper, 𝐾 is a commutative strong hypergroup with 𝐾=𝑋_𝑏(𝐾). In Section 4, we give some examples of this type of hypergroups. We denote by 𝑀 (𝐾) the space of all regular complex Borel measures on 𝐾, by 𝑀⁺(𝐾) the subset of positive measures in 𝑀 (𝐾), and by 𝛿_𝑥 the Dirac measure at the point 𝑥.

A 𝐶^∗-algebra 𝑀 is called 𝑊^∗-algebra if for a Banach algebra 𝑀_∗, (𝑀_∗)^∗=𝑀. Any 𝑊^∗-algebra is unitary (with unit 1_𝑀). The famous examples of 𝑊^∗-algebras are von Neumann algebras. We can consider 𝜎(𝑀,𝑀_∗) (i.e., 𝑤𝑘^∗-) topology on 𝑀 [4]. In this paper, 𝑀 is always a 𝑊^∗-algebra. We denote by 𝐵_𝜎(𝑀) the set of all (𝜎(𝑀,𝑀_∗),𝜎(𝑀,𝑀_∗))-continuous operators on 𝑀.

Let 𝜅∶𝑀(𝐾)→𝐵_𝜎(𝑀) be a norm-decreasing algebra-homomorphism. For any 𝑡∈𝐾, we denote 𝜅_𝑡=𝜅(𝛿_𝑡). Suppose that 𝜅 has the following properties.

(1)For any 𝑡∈𝐾, 𝜅_𝑡∶𝑀→𝑀 is an ∗-automorphism. (Then by [5, Theorem 4.8, page 253] any 𝜅_𝑡 is an isometry.)(2)For any 𝑥∈𝑀 and 𝜌∈𝑀_∗, the function 𝑡↦⟨𝜅_𝑡(𝑥),𝜌⟩ is continuous.(3)𝜅_𝑒=𝐼_𝑀, where 𝑒 is the identity of 𝐾 and 𝐼_𝑀 is the identity mapping on 𝑀.

Let 𝑥∈𝑀 and 𝜌∈𝑀_∗. Obviously for any measure 𝜐∈𝑀(𝐾) with finite support, we have ⟨𝜅(𝜐)(𝑥),𝜌⟩=∫_𝐾⟨𝜅_𝑡(𝑥),𝜌⟩𝑑𝜐(𝑡). Let 𝜇∈𝑀(𝐾). Since the set 𝐸 containing all measures in 𝑀(𝐾) that have finite support is dense in 𝑀(𝐾), there exists a net (𝜐_𝛽)⊆𝐸 such that 𝜐_𝛽→𝜇 in 𝑀(𝐾). Then by [1, Lemma 2.2C], we have ∫_𝐾𝑓𝑑𝜐_𝛽→∫_𝐾𝑓𝑑𝜇, where 𝑓(𝑡)=⟨𝜅_𝑡(𝑥),𝜌⟩(𝑡∈𝐾). On the other hand, by continuity of 𝜅, ∫_𝐾𝑓𝑑𝜐_𝛽=⟨𝜅(𝜐_𝛽)(𝑥),𝜌⟩→⟨𝜅(𝜇)(𝑥),𝜌⟩. Then, for any 𝜇∈𝑀(𝐾) we have𝜅(𝜇)(𝑥),𝜌=_𝐾𝜅_𝑡(𝑥),𝜌𝑑𝜇(𝑡), (1.1)where 𝑥∈𝑀 and 𝜌∈𝑀_∗.

2. Basic Properties of Hypergroups

First, we recall the definition and basic properties of a hypergroup. The main references are [1, 6].

Definition 2.1. Let 𝐾 be a locally compact Hausdorff space. The space 𝐾 is a hypergroup if there exists a binary mapping (𝑥,𝑦)↦𝛿_𝑥∗𝛿_𝑦 from 𝐾×𝐾 into 𝑀⁺(𝐾) satisfying the following conditions.
(1) The mapping (𝛿_𝑥,𝛿_𝑦)↦𝛿_𝑥∗𝛿_𝑦 extends to a bilinear associative operator ∗ from 𝑀(𝐾)×𝑀(𝐾) into 𝑀 (𝐾) such that_𝐾𝑓𝑑(𝜇∗𝜈)=_𝐾𝐾𝐾𝑓𝑑𝛿_𝑥∗𝛿_𝑦𝑑𝜇(𝑥)𝑑𝜈(𝑦) (2.1)for all continuous functions 𝑓 on 𝐾 vanishing at infinity.(2)For each 𝑥,𝑦∈𝐾 the measure 𝛿_𝑥∗𝛿_𝑦 is a probability measure with compact support.(3)The mapping (𝜇,𝜈)↦𝜇∗𝜈 is continuous from 𝑀⁺(𝐾)×𝑀⁺(𝐾) into 𝑀⁺(𝐾); the topology on 𝑀⁺(𝐾) being the cone topology.(4) There exists an 𝑒∈𝐾 such that 𝛿_𝑒∗𝛿_𝑥=𝛿_𝑥=𝛿_𝑥∗𝛿_𝑒, for all 𝑥∈𝐾.(5)There exists a homeomorphism involution 𝑥↦𝑥⁻ from 𝐾 onto 𝐾 such that, for all 𝑥,𝑦∈𝐾, we have (𝛿_𝑥∗𝛿_𝑦)⁻=𝛿_𝑦−∗𝛿_𝑥−, where for 𝜇∈𝑀(𝐾), 𝜇⁻ is defined by_𝐾𝑓(𝑡)𝑑𝜇⁻(𝑡)=_𝐾𝑓𝑡⁻𝑑𝜇(𝑡), (2.2)and also,𝑒∈supp𝛿_𝑥∗𝛿_𝑦iff𝑦=𝑥⁻, (2.3)where supp(𝛿_𝑥∗𝛿_𝑦) is the support of the measure 𝛿_𝑥∗𝛿_𝑦.(6)The mapping (𝑥,𝑦)↦supp(𝛿_𝑥∗𝛿_𝑦) from 𝐾×𝐾 into the space 𝐂(𝐾) of compact subsets of 𝐾 is continuous, where 𝐂(𝐾) is given the topology whose subbasis is given by all𝐂_𝑈,𝑉=𝐴∈𝐂(𝐾)∶𝐴∩𝑈≠∅,𝐴⊆𝑉, (2.4)where 𝑈, 𝑉 are open subsets of 𝐾.

Note that 𝛿_𝑥∗𝛿_𝑦 is not necessarily a Dirac measure. A hypergroup 𝐾 is commutative if 𝛿_𝑥∗𝛿_𝑦=𝛿_𝑦∗𝛿_𝑥, for all 𝑥,𝑦 in 𝐾. Each commutative hypergroup 𝐾 carries a Haar measure 𝑚 such that 𝛿_𝑥∗𝑚=𝑚, for all 𝑥∈𝐾, as shown by Spector [7]. In any commutative hypergroup 𝐾, we have 𝑚=𝑚⁻ (see [1, Section 5.3]). Let 𝑓,𝑔 be Borel functions on 𝐾 and 𝜇∈𝑀(𝐾). For any 𝑥,𝑦∈𝐾 we denote 𝑓_𝑥(𝑦)=𝑓(𝑥∗𝑦)∶=∫_𝐾𝑓𝑑(𝛿_𝑥∗𝛿_𝑦). Also we define(𝜇∗𝑓)(𝑥)∶=_𝐾𝑓𝑦⁻∗𝑥𝑑𝜇(𝑦),(𝑓∗𝑔)(𝑥)∶=_𝑘𝑓(𝑥∗𝑦)𝑔𝑦⁻𝑑𝑚(𝑦), (2.5)where 𝑥∈𝐾. If 𝑥,𝑦∈𝐾 and 𝐴,𝐵⊆𝐾, we denote {𝑥}∗{𝑦}=supp(𝛿_𝑥∗𝛿_𝑦), and 𝐴∗𝐵=⋃_{𝑥∈𝐴,𝑦∈𝐵}{𝑥}∗{𝑦}.

A complex-valued continuous function 𝜉 on 𝐾 is said to be multiplicative if 𝜉(𝑥∗𝑦)=𝜉(𝑥)𝜉(𝑦) holds for all 𝑥,𝑦∈𝐾. The space of all multiplicative functions on 𝐾 is denoted by 𝑋_𝑏(𝐾). A nonzero multiplicative function 𝜉 on 𝐾 is called a character if 𝜉(𝑥⁻)=𝜉(𝑥), for all 𝑥 in 𝐾. The dual 𝐾 of 𝐾 is the locally compact Hausdorff space of all characters with the topology of uniform convergence on compacta. In general, 𝐾 is not necessarily a hypergroup. A hypergroup 𝐾 is called strong if its dual 𝐾 is also a hypergroup with complex conjugation as involution, pointwise product as convolution, that is𝐾𝜂(𝑥)𝜒(𝑥)=𝜉(𝑥)𝑑𝛿_𝜂∗𝛿_𝜒(𝜉), (2.6)for all 𝜂,𝜒∈𝐾 and 𝑥∈𝐾, and has the constant function 1 as the identity element.

For any 𝑝>0, we denote 𝐿^𝑝(𝐾)=𝐿^𝑝(𝐾,𝑚) and 𝐿^𝑝(𝐾)=𝐿^𝑝(𝐾,𝜋), where 𝜋 is the Plancherel measure on 𝐾 associated with 𝑚. The structure space Δ(𝐿¹(𝐾)) of Banach algebra 𝐿¹(𝐾) doses not necessarily equal with 𝐾 and we only have Δ(𝐿¹(𝐾))=𝑋_𝑏(𝐾), while 𝐾⊆𝑋_𝑏(𝐾).

For any 𝑓∈𝐿¹(𝐾) and 𝜇∈𝑀(𝐾), the Fourier-Stieltjes transform 𝜇 of 𝜇 and the Fourier transform 𝑓 of 𝑓 are defined by𝜇(𝜉)=_𝐾𝜉(𝑡)𝑑𝜇(𝑡),𝑓(𝜉)=_𝐾𝜉(𝑡)𝑓(𝑡)𝑑𝑚(𝑡), (2.7)where 𝜉∈𝐾. For any 𝑓,𝑔∈𝐿²(𝐾), we have 𝑓,𝑔∈𝐿²(𝐾) and 𝑓∗𝑔=𝑓𝑔 [6].

3. The Main Result

Recall that for any complex commutative algebra 𝐴, Δ(𝐴), the structure space of 𝐴, is a locally compact Hausdorff space with Gelfand topology and for each 𝑎∈𝐴, the Gelfand transform 𝑎 is in 𝐶₀(Δ(𝐴)). For an ideal 𝐼 of 𝐴, the ℎ𝑢𝑙𝑙 of 𝐼 is defined by ℎ(𝐼)=𝜉∈Δ(𝐴)∶𝑎(𝜉)=0 for every 𝑎∈𝐼}. For each closed 𝐹⊆Δ(𝐴), we denote 𝑘(𝐹)={𝑎∈𝐴∶𝑎(𝜉)=0 for every 𝜉 in 𝐹} and 𝐽₀(𝐹)={𝑎∈𝐴∶𝑎=0 on some neighborhood of 𝐹, and supp(𝑎) is compact}, where supp(𝑎) is the support of the function 𝑎. We denote 𝐽(𝐹)=𝐽₀(𝐹), the closure of 𝐽(𝐹). For any Banach algebra 𝐴, the spectrum of any 𝑎∈𝐴 is denoted by 𝜎_𝐴(𝑎).

Definition 3.1. The Arveson spectrum of 𝜅 is defined by𝑠𝑝𝜅∶=ℎ𝑓∈𝐿¹=(𝐾)∶𝜅(𝑓)=0𝜉∈𝐾∶forany𝑓∈𝐿¹(𝐾),𝜅(𝑓)=0implies𝑓(𝜉)=0. (3.1)The Arveson spectrum and spectral subspaces on hypergroups have been studied in [8]. In this paper, we concentrate on an extension of the spectral mapping theorem to hypergroups. The spectral mapping theorem gives the relation 𝜎(𝜅(𝜇))=𝜇(𝑠𝑝𝜅) for some measures 𝜇∈𝑀(𝐾). In the case that 𝐾 is a locally compact Abelian group, Connes proved the spectral mapping theorem for every Dirac measure 𝜇 [9]. Then, D'Antoni et al. proved that the spectral mapping theorem holds for those measures whose continuous part belongs to 𝐿¹(𝐺) [10]. Furthermore, Eschmeier proved the spectral mapping theorem for 𝜅 as the translation group representation, in the case that 𝑀 is Banach algebra 𝐿¹(𝐺) or 𝑀(𝐺) and the convolution operator induced by 𝜇 has the weak 2-SDP [11]. Takahasi and Inoue proved the spectral mapping theorem for any regular subalgebra of 𝑀(𝐺) in the case that G is compact [12]. Also, Seferoğlu proved that 𝜎(𝜅(𝜇))=𝜇(spec(𝜅)) holds when 𝜅(𝐿¹(𝐺)) is without order and 𝑠𝑝𝜅 is a synthesis set for 𝐿¹(𝐺) [13]. (A subset 𝐶 of Banach algebra 𝐴 is called without order if for all 𝑎∈𝐴, 𝑎.𝐶={0} implies 𝑎=0 [14].)

A Banach algebra 𝐴 is called regular if for any closed subset 𝐸 of Δ(𝐴) and any 𝜑∈Δ(𝐴)⧵𝐸, there is an 𝑎∈𝐴 such that 𝑎(𝜑)=1 and 𝑎≡0 on 𝐸. For any commutative and strong hypergroup 𝐾 with 𝐾=𝑋_𝑏(𝐾), 𝐿¹(𝐾) is a regular Banach algebra [15]. Also, since the Fourier transform ̂∶𝐿¹(𝐾)→𝐶₀(𝐾) is a norm-decreasing ∗-homomorphism, 𝐿¹(𝐾) is semisimple. If 𝐴 is a regular commutative Banach algebra, then for any closed 𝐹⊆Δ(𝐴), ℎ(𝐽(𝐹))=𝐹. If 𝐴 is also semisimple, then 𝑘(𝐹) and 𝐽(𝐹) are the largest closed ideal and the smallest ideal with hull equal to 𝐹, respectively [16]. 𝐹 is called a set of spectral synthesis if and only if 𝐽(𝐹)=𝑘(𝐹). Then, for such sets, 𝑘(𝐹) is the only closed ideal in 𝐴 such that its hull is 𝐹.

It is well known that the correspondence 𝐹↦ℎ(𝑘(𝐹)) is a proper closure operation and that the topology on Δ(𝐴) determined by this operation, called hull-kernel topology, is weaker than the Gelfand topology [17]. A commutative Banach algebra 𝐴 is regular if and only if the Gelfand and the hull-kernel topologies coincide on Δ(𝐴) (see [18, Theorem 3.2.10]). Another property of commutative regular semisimple Banach algebras appears in the following lemma. It is given here for the convenience of the reader.

Lemma 3.2. Let 𝐴 be a commutative regular semisimple Banach algebra, let 𝐵 be a unitary Banach algebra (with unit 1_𝐵), and let 𝜅∶𝐴→𝐵 be a continuous one-to-one homomorphism.
(i) If 𝐴 is unitary with unit 1_𝐴 and 𝜅(1_𝐴)=1_𝐵, then for any 𝑎∈𝐴,𝜎_𝐵𝜅(𝑎)=𝑎Δ(𝐴). (3.2)(ii) If 𝐴 is without unit, then for any 𝑎∈𝐴,𝜎_𝐵𝜅(𝑎)=𝑎Δ(𝐴). (3.3)

Proof. See [13, Lemmas 1 and 2] for the proof.

Lemma 3.3. Suppose that 𝜅∶𝑀(𝐾)→𝐵(𝑀) is the function introduced in Section 1. Let 𝜅(𝐿¹(𝐾)) be without order and let 𝑠𝑝𝜅 be a spectral synthesis set for 𝐿¹(𝐾). Then, 𝜅(𝜇)=0 if and only if 𝜇|_𝑠𝑝𝜅=0.

Proof. Let 𝜇∈𝑀(𝐾) and 𝜅(𝜇)=0. For any 𝑓∈𝐿¹(𝐾), 𝜇∗𝑓∈𝐿¹(𝐾) and 𝜅(𝜇∗𝑓)=𝜅(𝜇)𝜅(𝑓)=0. Then𝜇∗𝑓∈𝑘ℎ{𝜇∗𝑓}⊆𝑘ℎ𝑓∈𝐿¹(𝐾)∶𝜅(𝑓)=0=𝑘(𝑠𝑝𝜅). (3.4)So 𝜇(𝜉)𝑓(𝜉)=𝜇∗𝑓(𝜉)=0, where 𝜉∈𝑠𝑝𝜅. For each 𝜉∈𝐾, there exists a 𝑔∈𝐶⁺_𝑐(𝐾) such that 𝜉∈supp(𝑔). We can consider ℎ∈𝐶_𝑐(𝐾) such that ℎ≥0 and ℎ>𝑔 on supp(𝑔) (see [6, Proposition 2.2.5]). In other words, the set {𝑓∶𝑓∈𝐶_𝑐(𝐾)} separates the points of 𝐾. Then, for any 𝜉∈𝑠𝑝𝜅 we have 𝜇(𝜉)=0.
Conversely, let 𝜇∈𝑀(𝐾) and 𝜇|_𝑠𝑝𝜅≡0. Then, for each 𝜉∈𝐾 and 𝑓∈𝐿¹(𝐾) we have 𝜇∗𝑓(𝜉)=𝜇(𝜉)𝑓(𝜉)=0. Now, since 𝑘(𝑠𝑝𝜅) is the only closed ideal in 𝐿¹(𝐾) whose hull is 𝑠𝑝𝜅,𝜇∗𝑓∈𝑘(𝑠𝑝𝜅)=𝑘ℎ𝑓∈𝐿¹(𝐾)∶𝜅(𝑓)=0=𝑓∈𝐿¹(𝐾)∶𝜅(𝑓)=0. (3.5)Therefore, 𝜅(𝜇)⋅𝜅(𝐿¹(𝐾))=0. Since 𝜅(𝐿¹(𝐾)) is without order, 𝜅(𝜇)=0.
Let 𝑀₀(𝐾) be a closed subalgebra of 𝑀(𝐾) which contains 𝐿¹(𝐾). Then, 𝐾 can be considered as a subset of Δ(𝑀₀(𝐾)) and for every 𝜇 in 𝐾, the restriction of the Gelfand transform 𝑇𝜇 of 𝜇 to 𝐾 coincides with the Fourier-Stieltjes transform 𝜇 of 𝜇. This implies that 𝑀₀(𝐾) is a semisimple algebra.

Theorem 3.4. Suppose that 𝜅∶𝑀(𝐾)→𝐵_𝜎(𝑀) is the function introduced in Section 1. Let 𝜅(𝐿¹(𝐾)) be without order and let 𝑠𝑝𝜅 be a spectral synthesis set. Then, for any 𝜇∈𝑀₀(𝐾),𝜎_𝐵(𝑀)𝜅(𝜇)=𝜇(𝑠𝑝𝜅), (3.6)where 𝑀₀(𝐾) is a closed regular subalgebra of 𝑀(𝐾) containing 𝐿¹(𝐾).

Proof. Since 𝐿¹(𝐾)⊆𝑀₀(𝐾)⊆𝑀(𝐾), 𝐾≅Δ(𝐿¹(𝐾)) is a subset of Δ(𝑀₀(𝐾)). For each 𝜇∈𝑀₀(𝐾), we denote by 𝑇𝜇 the Gelfand transform of 𝜇. We have 𝑇𝜇∈𝐶₀(Δ(𝑀₀(𝐾))) and 𝐾𝜇=𝑇𝜇|. The function 𝜅∶𝑀₀(𝐾)/𝐼_𝜅→𝐵(𝑀) naturally defined by 𝜅(𝜇+𝐼_𝜅)∶=𝜅(𝜇) is a continuous one to one homomorphism, where 𝐼_𝜅={𝜇∈𝑀₀(𝐾)∶𝜅(𝜇)=0}. (𝐼_𝜅 clearly is a closed ideal of 𝑀₀(𝐾).)
Since 0→𝐼_𝜅→𝑀₀(𝐾)→𝑀₀(𝐾)/𝐼_𝜅→0 is an exact sequence, by [18, page 324] we have Δ(𝑀₀(𝐾)/𝐼_𝜅)≅ℎ(𝐼_𝜅). Consider 𝑠𝑝𝜅 as a subset of Δ(𝑀₀(𝐾)) and denote by 𝑠𝑝𝜅^𝑔 the closure of 𝑠𝑝𝜅 in Gelfand topology. By regularity of 𝑀₀(𝐾), Gelfand and hull-kernel topologies coincide on Δ(𝑀₀(𝐾)). So 𝑠𝑝𝜅^𝑔 is closed in hull-kernel topology. But the only hull-kernel-closed subsets of Δ(𝑀₀(𝐾)) are hull-sets. Then, there exists a subset 𝐼⊆𝑀₀(𝐾) such that 𝑠𝑝𝜅^𝑔=ℎ(𝐼). Therefore,ℎ𝑘𝑠𝑝𝜅^𝑔=ℎ𝑘ℎ(𝐼)=ℎ(𝐼)=𝑠𝑝𝜅^𝑔. (3.7)On the other hand, for every 𝜇∈𝑀₀(𝐾),𝜇∈𝑘(𝑠𝑝𝜅^𝑔) if and only if 𝑇𝜇=0 on 𝑠𝑝𝜅^𝑔. This is also equivalent to vanishing 𝜇 on 𝑠𝑝𝜅. By Lemma 3.3, this holds if and only if 𝜅(𝜇)=0. Therefore, ℎ(𝐼_𝜅)=𝑠𝑝𝜅^𝑔.
By the regularity of 𝑀₀(𝐾), the Banach algebra 𝑀₀(𝐾)/𝐼_𝜅 is also regular. For any 𝜇∈𝑀₀(𝐾), 𝜇+𝐼_𝜅∈rad(𝑀₀(𝐾)/𝐼_𝜅) if and only if 𝜇+𝐼_𝜅=0 on Δ(𝑀₀(𝐾)/𝐼_𝜅). This is also equivalent to vanishing 𝑇𝜇 on ℎ(𝐼_𝜅). By the previous paragraph, this holds if and only if 𝜅(𝜇)=0, that is, 𝜇∈𝐼_𝜅. Therefore, 𝑀₀(𝐾)/𝐼_𝜅 is a semisimple algebra.
In the sequel, we consider two cases.
Case 1. Let 𝑠𝑝𝜅^𝑔 be compact. There exists 𝜐∈𝑀₀(𝐾) such that 𝑇𝜐=1 on a neighborhood 𝑠𝑝𝜅^𝑔. Then, for any 𝜇∈𝑀₀(𝐾) we have𝑇(𝜇∗𝜐−𝜇)=𝑇𝜇𝑇𝜐−𝑇𝜇=0on𝑠𝑝𝜅^𝑔. (3.8)So 𝜇∗𝜐−𝜇=0 on 𝑠𝑝𝜅 and by Lemma 3.3𝜇∗𝜐−𝜇∈𝐼_𝜅(⋆). Then, for any 𝜇∈𝑀₀(𝐾), (𝜇+𝐼_𝜅)(𝜐+𝐼_𝜅)=𝜇+𝐼_𝜅. By (⋆), for any 𝑓∈𝐿¹(𝐾),0=𝜅(𝜐∗𝑓−𝑓)=𝜅(𝜐)𝜅(𝑓)−𝜅(𝑓)=𝜅(𝜐)−𝐼_𝐵(𝑀)⋅𝜅(𝑓). (3.9)That is (𝜅(𝜐)−𝐼_𝐵(𝑀))⋅𝜅(𝐿¹(𝐾))=0. Now, since 𝜅(𝐿¹(𝐾)) is without order, 𝜅(𝜐)=𝐼_𝐵(𝑀). Therefore, 𝜅(𝜐+𝐼_𝜅)=𝐼_𝐵(𝑀). Now, we are in a position using Lemma 3.2(i) for 𝜅,𝑀₀(𝐾)/𝐼_𝜅 and 𝐵(𝑀). Then, for every 𝜇∈𝑀₀(𝐾),𝜎_𝐵(𝑀)𝜅𝜇+𝐼_𝜅=𝜇+𝐼_𝜅Δ𝑀₀(𝐾)𝐼_𝜅, (3.10)and so 𝜎_𝐵(𝑀)𝜅(𝜇)=𝑇𝜇𝑠𝑝𝜅^𝑔. (⋆⋆)Now compactness of 𝑠𝑝𝜅^𝑔 in Δ(𝑀₀(𝐾)) and continuity of 𝑇𝜇 on Δ(𝑀₀(𝐾)) imply that 𝑇𝜇(𝑠𝑝𝜅^𝑔) is compact and so is closed. This fact by using relation 𝜇(𝑠𝑝𝜅)=𝑇𝜇(𝑠𝑝𝜅)⊆𝑇𝜇(𝑠𝑝𝜅^𝑔) shows that𝜇(𝑠𝑝𝜅)⊆𝑇𝜇𝑠𝑝𝜅^𝑔. (3.11)On the other hand, since 𝑇𝜇 is continuous, we have𝑇𝜇𝑠𝑝𝜅^𝑔⊆𝑇𝜇(𝑠𝑝𝜅)=𝜇(𝑠𝑝𝜅). (3.12)Thus,𝜇(𝑠𝑝𝜅)=𝑇𝜇𝑠𝑝𝜅^𝑔. (3.13)Now by (⋆⋆), 𝜎_𝐵(𝑀)(𝜅(𝜇))=𝜇(𝑠𝑝𝜅).
Case 2. Let 𝑠𝑝𝜅^𝑔 be noncompact. Then, by Δ(𝑀₀(𝐾)/𝐼_𝜅)≅𝑠𝑝𝜅^𝑔,𝑀₀(𝐾)/𝐼_𝜅 is not unitary. So by Lemma 3.2(ii), for any 𝜇∈𝑀₀(𝐾),𝜎_𝐵(𝑀)𝜅𝜇+𝐼_𝜅=𝜇+𝐼_𝜅Δ𝑀₀(𝐾)𝐼_𝜅. (3.14)Now the inclusion 𝜇(𝑠𝑝𝜅)⊆𝑇𝜇(𝑠𝑝𝜅^𝑔) implies 𝜇(𝑠𝑝𝜅)⊆𝑇𝜇(𝑠𝑝𝜅^𝑔). Also since 𝑇𝜇 is continuous, we have 𝑇𝜇(𝑠𝑝𝜅^𝑔)⊆𝜇(𝑠𝑝𝜅), and so that 𝑇𝜇(𝑠𝑝𝜅^𝑔)⊆𝜇(𝑠𝑝𝜅). Therefore,𝜎_𝐵(𝑀)𝜅(𝜇)=𝑇𝜇𝑠𝑝𝜅^𝑔=𝜇(𝑠𝑝𝜅). (3.15)

4. Examples

In this section, we give some examples of hypergroups 𝐾 that are commutative strong hypergroups and 𝐾=𝑋_𝑏(𝐾) (we refer to these conditions by notation (℘)). Observe that any locally compact Abelian group has these properties. Also, if 𝐺 is a locally compact Abelian group and 𝐻 is a compact subgroup of Aut(𝐺), then the space 𝐺_𝐻 containing all 𝐻-orbits is a commutative hypergroup satisfying in (℘). In fact, 𝐺_𝐻≅(𝐺)_𝐻. We refer to [19] for more details. As another example, let 𝐺 be a group such that 𝐺/𝑍 is compact, where 𝑍={𝑥∈𝐺∶forany𝑦∈𝐺,𝑥𝑦=𝑦𝑥}. If 𝐾 is the hypergroup containing all conjugacy classes of 𝐺, then 𝐾 and its dual 𝐾 satisfy in (℘) [6]. On the other hand, an interesting example of Naimark given in [1, Section 9.5] does not satisfy in conditions (℘).