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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2022; 62(4): 689-697

Published online December 31, 2022

### Dual Integrable Representations of Hypergroups

Department of Mathematics, University of Qom, Qom, Iran
e-mail : sm.tabatabaie@qom.ac.ir

Department of Basic Sciences, Qeshm Branch, Islamic Azad University, Qeshm, Iran
e-mail : soheilajokar2007@yahoo.com

Received: April 18, 2021; Revised: January 18, 2022; Accepted: January 24, 2022

### Abstract

In this paper, we study dual integrable representations of locally compact commutative hypergroups and give a sufficient and necessary condition for a dual integrable representation to have an admissible vector.

Keywords: locally compact hypergroup, locally compact group, dual integrable representation, left regular representation, frame

### 1. Introduction and Notations

Dual integrable representations of abelian locally compact groups were introduced in [6] (see also [1, 7]). Left regular representations are well known examples of such dual integrable representations. In this paper, we initiate the stucy of this concept for locally compact hypergroups and prove its basic properties. We find some some results which are novel even for the group case. In fact, we give an equivalence condition for a dual integrable representation to have an admissible vector, which is a version of our recent results in [9].

Throughout this paper K is a locally compact hypergroup. For the definitions and basic properties of hypergroups we refer to the monographs [8] (in which hypergroups are called "convo") and [2] (see also [5]). The space of all complex Radon measures on K is denoted by M(K), and the space of positive elements of M(K) is denoted by M+(K). B(H) denotes the space of all bounded operators on a Hilbert space H.

### 2. Main Results

In this section, first we recall the following definition from [8, 11.3].

Definition 2.1. Let H be a Hilbert space. A norm-decreasing *-homomorphism π:M(K)B(H) is called a representation of K on H if π(δe)=IdH and for each φ,ψH, the mapping μπ(μ)φ,ψ from M+(K) to is continuous, where M+(K) is equipped with the cone topology. For each x ∈ K, we denote πx:=π(δx). Let K be a commutative hypergroup with a left Haar measure m.

The left regular representation τ:M(K)B(L2(K)) is defined by τ(μ)f:=μf, where fL2(K) and μM(K). In particular, let H be a compact subgroup of a locally compact group G and G//H be the double coset hypergroup introduced in [8, 8.2] with quotient mapping q:GG//H defined by q(x):=HxH. Then, G//H has a left Haar measure m [8, 8.2B], and for each fL2(G//H,m) the left regular representation of G//H can be obtained by

τ(μ)f(q(x))=G//Hf(q(y1)q(x))dμ(q(y))=G//H G//H f (q(z))d(δq(y1)δq(x))(q(z))dμ(q(y))=G//H H f (q(y1tx))dσ(t)dμ(q(y)),

where σ is the normalized Haar measure on H and μM(G//H).

Definition 2.2. Let K be a commutative hypergroup with a Haar measure ω on K^ [8, 7.3I]. A representation π:M(K)B(Hπ) is called dual integrable if there exists a corresponding mapping ,π:Hπ×HπL1(K^,ω), called bracket product, such that for all φ,ψHπ and μM(K),

φ,π(μ)ψ=K^ φ,ψπ (ξ)μ^(ξ)¯dω(ξ),

where μ^ is the Fourier-Stieltjes transform of µ.

Remark 2.3. The bracket product of a dual integrable representation is conjugate symmetry and linear in the first argument. Also, for each x∈ K, putting μ=δx, we have

φ,πxψ=K^ φ,ψ π (ξ)ξ(x)dω(ξ)=φ,ψ^ π(x).

In particular,

φ,ψ=K^ φ,ψπ (ξ)dω(ξ).

For each φHπ, φ,φ=0 a.e. if and only if φ=0. In fact, if φ=0, then by (2.1) for each x∈ K we have

0=φ,πxφ=φ,φ^π(x),

and so, by [2, 2.1.6(vi)], φ,φ=0 a.e. The converse follows from (2.2).

Example 2.4. The left regular representation τ of a commutative hypergroup K is dual integrable. In fact, for each φ,ψL2(K) and μM(K) by [2, 2.2.26 and 2.2.34],

φ,τ(μ)ψL2(K)=Kφ (x)(μψ)(x)¯dm(x)= K^ φ^ (ξ)(μψ)^¯(ξ)dω(ξ)= K^ φ^ (ξ)μ^(ξ)¯ψ^(ξ)¯dω(ξ)= K^ φ,ψ τ μ^(ξ)¯dω(ξ),

with φ,ψτ(ξ):=φ^(ξ)ψ^(ξ)¯.

Remark 2.5. Same as the group case, for each representations (π,Hπ) and (π,H π ) of a hypergroup K, if π and π' are equivalence via a unitary operator U:HπH π and π is dual integrable, then π' is also dual integrable. In fact,

φ,π(μ)ψ=φ,Uπ(μ)U*ψ=U*φ,π(μ)U*ψ=K^ U * φ , U * ψ π (ξ)μ^(ξ)¯dω(ξ),

and so, φ,ψπ=U*φ,U*ψπ for all φ,ψH π , where U* is adjoint of the operator U.

Proposition 2.6. Let G1 and G2 be abelian locally compact groups and π and σ be dual representations of G1 and G2 respectively. Then, πσ is a dual integrable representation of G1×G2.

Proof. For each φ1,φ2Hπ and ψ1,ψ2Hσ we define

φ1ψ1,φ2ψ2πσ(ξ,η):=φ1,φ2π(ξ)ψ1,ψ2σ(η),  (ξG1^,ηG2^).

Then, by [4, Proposition 4.6] and [4, Section 7.3], for each x ∈ G1 and y ∈ G2 we have

φ1ψ1,(πσ)(x,y)(φ2ψ2)=φ1ψ1,πxφ2σyψ2=φ1,πxφ2ψ1,σyψ2=G1^ φ 1, φ 2 π (ξ)ξ(x)¯dμ(ξ)G2^ ψ 1, ψ 2 σ (η)η(y)¯dν(η)=G1^ G 2 ^ φ1 ,φ2 π (ξ)ψ1,ψ2ση(y)¯dμ(ξ)dν(η)=G1×G2^ φ 1 ψ 1, φ 2 ψ 2 πσ (ξ,η)(ξ,η)(x,y)¯d(μ×ν)(ξ,η),

and the proof is complete.

Corollary 2.7. Let π be a dual integrable representation of a locally compact abelian group G1 and τ be the left regular representation of a locally compact abelian group G2. Then, the induced representation indG1G1×G2 is dual integrable.

Proof. The proof is obtained from Proposition , Example 2.3 and [4, Proposition 7.26].

In spite of the group case, for a commutative hypergroup K we only have

Δ(L1(K))=Xb(K):={ξCb(K):ξ(xy)=ξ(x)ξ(y) for all x,yK},

while in general K^Xb(K).

There are hypergroups with K^Xb(K) (see [8, Example 9.5]). Although, for a large class of hypergroups, like orbit hypergroups, we have K^=Xb(K) (for more details see [3]). The following theorem holds for hypergroups with K^=Xb(K).

Theorem 2.8. Let K be a commutative hypergroup with Xb(K)=K^ and π be a representation of K on Hπ. Then, there is a projection-valued Borel measure T on K^ such that

π(μ):=K^ μ^(ξ)¯dT(ξ),(μM(K)).

Also, for each φ,ψHπ, there is a complex Borel measure μφ,ψ on K^ such that

μφ,ψ(E)=T(E)φ,ψ=φ,T(E)ψ=T(E)φ,T(E)ψ

for all Borel set EK^, and

φ,π(μ)ψ=K^ μ^(ξ)¯dμφ,ψ(ξ)(μM(K)).

Proof. By [8, 6.1G], M(K) is a commutative Banach *-algebra. Therefore, by [4, 1.54], there exists a unique regular projection-valued measure T on K^ such that

π(μ):=K^ μ^(ξ)¯dT(ξ)

for each μM(K). Then, for each φ,ψHπ,

φ,π(μ)ψ=K^ μ^(ξ)¯dφ,T(ξ)ψ.

The mapping μφ,ψ defined by

μφ,ψ(E):=φ,T(E)ψ

for all Borel set EK^, is a meaure on K^ and

φ,π(μ)ψ=K^ μ^(ξ)¯dμφ,ψ(ξ)(μM(K)).

As the group case, by [2, 2.1.6(vi)], we can conclude:

Corollary 2.9. Let (π,Hπ) be a representation of a commutative hypergroup K and ω be the Haar measure on K^. Then, the following statements are equivalent:

• (i) π is dual integrable.

• (ii) For each φ,ψHπ, the measure μφ,ψ is absolutely continuous with respect to ω.

Corollary 2.10. Let (π,Hπ) be a dual integrable representation of a commutative hypergroup K. Then, for all φ,ψHπ:

• (i) φ,φπ0 a.e.

• (ii) φ,ψπφ,φπ1/2.ψ,ψπ1/2 a.e.

• (iii) φ,ψπ1φψ.

Corollary 2.11. Let (π,Hπ) be a dual integrable representation of K and φHπ. Put Aφ:={ξK^;φ,φπ(ξ)>0}. Then, for all ψHπ, φ,ψπ=φ,ψπχAφ in L1(K^).

Proof. By the Cauchy-Schwartz inequality,

0K^Aφ φ,ψπ (ξ)dω(ξ)K^Aφ ( φ,φπ (ξ))1/2(ψ,ψ π(ξ))1/2dω(ξ)=0,

and the proof is complete.

Corollary 2.12. Let (π,Hπ) be a dual integrable representation of a commutative hypergroup K. Then, for each μM(K) and φ,ψHπ, we have

π(μ)φ,ψπ=φ,π(μ)ψπ=μ^[φ,ψπ).

Proof. By dual integrability of π, [2, 2.1.6(iv)] and [8, 12.1E], for all μ,νM(K) and φ,ψHπ we have

π(μ)φ,π(ν)ψ=φ,π(μ)*π(ν)ψ=φ,π(μ*ν)ψ=K^ φ,ψ π (ξ)(μ*ν)^ ¯(ξ)dω(ξ)=K^ φ,ψ π (ξ)μ^(ξ)ν ^ ¯(ξ)dω(ξ).

On the other hand,

π(μ)φ,π(ν)ψ=K^ π(μ)φ,ψπ (ξ)ν^¯(ξ)dω(ξ).

So,

K^ φ,ψπ (ξ)μ^(ξ)ν ^ ¯(ξ)dω(ξ)=K^ π(μ)φ,ψπ (ξ)ν ^ ¯(ξ)dω(ξ).

Finally, by uniqueness in Rieze Representation Theorem and [8, 7.3H],

π(μ)φ,ψπ=μ^[φ,ψπ).

Similarly,

φ,π(μ)ψπ=μ^[φ,ψπ).

Definition 2.13. Let π be a dual integrable representation of a commutative hypergroup K on Hπ. For each φ,ψHπ, we say φ and ψ are π-orthogonal, and write φπψ, if φ,ψπ=0 a.e. on K^. A subset EHπ is called π-orthogonal if for each distinct φ,ψE, φπψ.

Corollary 2.14. Let (π,Hπ) be a dual integrable representation of a Pontrjagin hypergroup K. Then, φ,ψHπ are π-orthogonal if and only if φπ(M(K))ψ¯ in Hπ.

Proof. Since π is dual integrable, φπ(M(K)ψ¯ if and only if

K^ φ,ψπ (ξ)μ^(ξ)¯dω(ξ)=0,

for all μM(K). Thus, for each x∈ K, setting μ=δx, we have

φ,ψ^π(x)= K^ φ,ψ π (ξ)ξ(x)¯dω(ξ)=0.

This implies that φ,ψπ=0 a.e. in K^. Clearly, by (2.3), if φ,ψπ=0 a.e. in K^, then φπ(M(K)ψ¯.

Proposition 2.15. Let π be a dual integrable representation of a commutative hypergroup K on Hπ. The subset EHπ is π-orthogonal if and only if the set {π(μ)φ:φE,μM(K)} is orthogonal.

Proof. For each φ,ψE and μ,νM(K), by Corollary we have

π(μ)φ,π(ν)ψ=K^ π(μ)φ,ψπ (ξ)ν^(ξ)¯dω(ξ)
=K^ φ,ψπ (ξ)μ^(ξ)ν^(ξ)¯dω(ξ).

Hence, if E is π-orthogonal, then π(μ)φ,π(ν)ψ=0 for all distinct φ,ψHπ and μ,νM(K).

Conversely, if {π(μ)φ:φE,μM(K)} is orthogonal, then putting ν=δe and μ=δx(xK) in (5), we have

K^ φ,ψπ (ξ)ξ(x)¯dω(ξ)=0

for all distinct φ and ψHπ, and so φ,ψ^π=0 a.e. Finally, by [2, 2.1.6(vi)] we have φ,ψπ=0 in L1(K^), and the proof is complete.

Corollary 2.16. Let (π,Hπ) be a dual integrable representation of a commutative hypergroup K and φ,ψHπ. The following statements are equivalent:

• (i) φπψ

• (ii) φπ(M(K))ψ¯

• (iii) π(M(K))φ¯π(M(K))ψ¯.

Proposition 2.17. Let (π,Hπ) be a dual integrable representation of a strong commutative compact hypergroup K and φHπ. If φ,φπ=1 in L1(K^), then the set {πxφ:xK} is an orthogonal system in Hπ.

Proof. Since π is dual integrable, for each x,y∈ K we have

πxφ,πyφ=K^ φ,φ π (ξ)ξ(xy)dω(ξ).

Therefore, if φ,φπ=1 in L1(K^), then by [2, proof of Theorem 2.2.9(ii)] and [2, Theorem 2.4.3], πxφ,πyφ=0.

We recall the next definition from [9].

Definition 2.18. Let K be a hypergroup with a (left) Haar measure m, H be a subhypergroup of K, and π:M(K)B(Hπ) be a representation of K on a Hilbert space Hπ, and VHπ. A vector h0Hπ is called a (π,V)-admissible vector with respect to H if there are constant numbers A,B>0 such that for every h ∈ V,

Ah2H|πx(h0),h|2dmH(x)Bh2,

where mH is a left Haar measure on H.

If A=B=1, h0 is called Parseval admissible.

Proposition 2.19. Let K be a Pontrjagin hypergroup with a left Haar measure m, π:M(K)B(Hπ) be a dual integrable representation of K corresponding to a Plancherel measure ω, and VHπ. A vector h0Hπ is a Parseval (π,V)-admissible vector if and only if for every h∈ V, h=h0,h2.

Proof. Let h0Hπ. Since π is dual integrable, for every h∈ V we have

πx(h0),h=K^ h 0 ,h π(ξ)δ^x (ξ)¯dω(ξ)=K^ h 0 ,h πξ(x)¯dω(ξ)=h0,h^ π(x).

So by Plancherel Theorem, h0 is a Parseval (π,V)-admissible vector if and only if for each h ∈ V,

h2=K|h0 ,h^ π(x)|2dmK(x)= K^ |h0,hπ(ξ)|2dω(ξ)=h0,h2.

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