Amenability and harmonic $L^p$-functions on hypergroups

TL;DR

This paper explores the relationship between amenability and harmonic functions on hypergroups, establishing conditions under which harmonic functions are trivial or constant, and linking these properties to the hypergroup's structure.

Contribution

It proves the equivalence of Liouville property and amenability for second countable hypergroups and characterizes harmonic $L^p$-functions under various conditions.

Findings

Liouville property and amenability are equivalent for second countable hypergroups.

Harmonic $L^p$-functions are trivial for all $1 \\leq p < \\infty$.

Harmonic functions are constant if the hypergroup is compact and $\sigma$ is adapted.

Abstract

Let $K$ be a locally compact hypergroup with a left invariant Haar measure. We show that the Liouville property and amenability are equivalent for $K$ when it is second countable. Suppose that $\sigma$ is a non-degenerate probability measure on $K$, we show that there is no non-trivial $\sigma$-harmonic function which is continuous and vanishing at infinity. Using this, we prove that the space $H_\sigma^p(K)$ of all $\sigma$-harmonic $L^p$-functions, is trivial for all $1\leq p<\infty$. Further, it is shown that $H_\sigma^\infty(K)$ contains only constant functions if and only if it is a subalgebra of $L^\infty(K)$. In the case where $\sigma$ is adapted and $K$ is compact, we show that $H_\sigma^p(K)={\mathbb C}1$ for all $1\leq p\leq\infty$.

Equations

$${}_{x}f(y)=f_y(x)={\int }_{K}f(t)d({\delta }_x\ast {\delta }_y)(t)=f(x\ast y),$$

$$⟨\mu \ast \nu ,f⟩={\int }_{K}fd(\mu \ast \nu )={\int }_K{\int }_{K}f(x\ast y)d\mu (x)d\nu (y)(f\in C_0(K))$$

$$\mu \ast f(x)={\int }_{K}f(\check{y}\ast x)d\mu (y),f\ast \mu (x)={\int }_{K}f(x\ast \check{y})d\mu (y)(x\in K),$$

$${\delta }_x\ast \omega =\omega (x\in K).$$

$$\sigma \ast f=f$$

$$(f\ast g)(x)={\int }_{K}f(y)g(\check{y}\ast x)d\omega (y).$$

$$P_{\sigma }(f)=\underset{\mathcal{U}}{\lim }\frac{1}{n}\sum _{k=1}^n{\sigma }^k\ast f,$$

$$⟨\sigma \ast P_{\check{\sigma }}^{\ast }(f),g⟩$$

$$⟨P_{\sigma }{P}_{\check{\sigma }}^{\ast }(f),g⟩=⟨P_{\check{\sigma }}{P}_{\sigma }^{\ast }(g),f⟩$$

$$\Vert f\Vert \ge \inf \{\Vert f+g\Vert :g\in H_{\check{\sigma }}^q(K{)}^{\perp }\}$$

$$L_0^1(K)=\left\{f\in L^1(K):{\int }_{K}fd\omega =0\right\}$$

$$d(g,J_{\sigma })=\inf \{\Vert f-g\Vert :f\in J_{\sigma }\}<\epsilon .$$

$$\Vert f_{\alpha }\ast f-f_{\alpha }{\Vert }_1\to 0$$

$$K=\overline{\bigcup _{n=1}^{\infty }(\mathrm{supp}|\mu |{)}^n}=\overline{\bigcup _{n=1}^{\infty }\mathrm{supp}|\mu {|}^n},$$

$$K=\overline{\bigcup_{n=1}^{\infty}{\Big{(}}{\rm supp}|\mu|\cup({{\rm supp}|\mu|})\check{}{\Big{)}}^{n}}$$

$$⟨|\mu {|}^n,h⟩>0.$$

$$0<⟨|\mu {|}^n,fh⟩\le ⟨|\mu {|}^n,f⟩.$$

$${\int }_K(f(\check{y}\ast x)-f(x){)}^{2}d\sigma (y)$$

$$\underset{\alpha }{\lim }⟨f\ast \check{{\varphi }_{\alpha }},g⟩=\underset{\alpha }{\lim }⟨f,g\ast {\varphi }_{\alpha }⟩=⟨f,g⟩.$$

$$⟨g\cdot f,h⟩=⟨f,h\ast g⟩,⟨f\cdot g,h⟩=⟨f,g\ast h⟩(f\in L^{\infty }(K),g,h\in L^1(K)).$$

$$(g\cdot f)(x)={\int }_{K}g(y){(}_{x}f)(y)d\omega (y),(f\cdot g)(x)={\int }_{K}g(y)(f_x)(y)d\omega (y)$$

$$⟨f,g\ast h⟩$$

$${\int }_{K}f(\check{y}\ast x)d\sigma (y)$$

$${(}_{x}f)(y){=}_x(\sigma \ast f)(y)$$

$$(\sigma \ast (f\cdot g))(x)$$

$$⟨f,g\ast h⟩=⟨f\cdot g,h⟩=0,$$

$${\int }_K\mu \ast (1-\check{f})d\omega (x)=\mu (K){\int }_K(1-\check{f})d\omega (x)>0,$$

$$⟨{\sigma }^n,\mu \ast (1-\check{f})⟩>0.$$

$$⟨{\sigma }^n,\mu \ast (1-\check{f})⟩$$

$$\sigma \ast ({\sigma }_0\ast f)=(\sigma \ast {\sigma }_0)\ast f={\sigma }_0\ast f.$$

Full text

1. Introduction

Let $\sigma$ be a complex Borel measure on a locally compact group $G$. A Borel function $f$ on $G$ is called ${\sigma}$-harmonic if it satisfies the convolution equation $\sigma*f=f$. It is a well-known result of [4] that if $G$ is abelian, then the only bounded continuous $\sigma$-harmonic function are constant functions when the support of $\sigma$ generates a dense subgroup of $G$. Bounded harmonic functions have been investigated by several authors for various kinds of groups, e.g., nilpotent groups and compact groups [7, 8, 10, 11, 12]. Moreover, it was shown in [5] that for $1\leq p<\infty$, any $\sigma$-harmonic $L^{p}$-function associated to an adapted probability measure $\sigma$ on a locally compact group $G$ is trivial. Harmonic functions on groups play important roles in analysis, geometry and probability theory [6].

Motivated by these observations, bounded continuous harmonic functions on nilpotent, [IN] and central hypergroups have been studied in [1, 2].

In what follows, $K$ denotes a locally compact hypergroup with a left-invariant Haar measure. The purpose of this paper is to obtain some insight into the harmonic functions problem for the $L_{p}$-spaces, $1\leq p\leq\infty$, of $K$.

In Section 3, for given a complex Borel measure $\sigma$ on $K$ with $\|\sigma\|=1$, we first show that there is a contractive projection from $L^{p}(K)$, $1<p\leq\infty$, onto $H^{p}_{\sigma}(K)=\{f\in L^{p}(K):\sigma*f=f\}$. We also show that $K$ is necessarily amenable if it has the Liouville property; that is, there exists a probability measure $\sigma$ on $K$ such that all $\sigma$-harmonic $L^{\infty}$-functions on $K$ are constant. Further, we prove that a second countable hypergroup possesses the Liouville property if and only if it is amenable.

In Section 4, for the case that $\sigma$ is a non-degenerate probability measure on $K$, we show that the space of all $\sigma$-harmonic functions which are continuous and vanishing at infinity are trivial. Using this we prove that for $1\leq p<\infty$, any $\sigma$-harmonic $L^{p}$-function is trivial. For such a measure $\sigma$, we also prove that $H_{\sigma}^{\infty}(K)$ is a subalgebra of $L^{\infty}(K)$ if and only if $H_{\sigma}^{\infty}(K)=\mathbb{C}1$. In the case where $\sigma$ is adapted and $K$ is compact, we show that $H_{\sigma}^{p}(K)={\mathbb{C}}1$ for all $1\leq p\leq\infty$. These extend the results for the group case in [5].

2. Preliminaries

Let $K$ be a locally compact Hausdorff space. The space $K$ is a hypergroup if there exists a bilinear, associative, weakly continuous convolution $*$ on the Banach space ${M}(K)$ of all bounded regular complex valued Borel measures on $K$, such that $({M}(K),*)$ is an algebra and satisfies, for $x,y\in K$,

(i) $\delta_{x}*\delta_{y}$ is a probability measure on $K$ with compact support,

(ii) the mapping $K\times K\rightarrow{\mathcal{C}}(K)$ , $(x,y)\mapsto{\rm supp}(\delta_{x}*\delta_{y})$ is continuous with respect to the Michael topology on the space ${\mathcal{C}}(K)$ of nonvoid compact sets in $K$,

(ii) the mapping $K\times K\rightarrow{M}(K)$, $(x,y)\mapsto\delta_{x}*\delta_{y}$ is continuous,

(ii) there is an identity $e\in K$ with $\delta_{e}*\delta_{x}=\delta_{x}*\delta_{e}=\delta_{x}$,

(iv)there is a continuous involution $\check{}$ on $K$ such that $(\delta_{x}*\delta_{y})\check{}=\delta_{\check{y}}*\delta_{\check{x}}$ and $e\in{\rm supp}(\delta_{x}*\delta_{y})$ if and only if $x=\check{y}$. The image measure of $\mu\in{M}(K)$ under such involution is denoted by $\check{\mu}$.

Given a (complex) Borel function $f$ on $K$ and $x,y\in K$ the left translation ${}_{x}f$ and the right translation $f_{y}$ are defined by

[TABLE]

if the integral exists, where $f(x*y)=\int_{K}fd(\delta_{x}*\delta_{y})$. For a Borel function $f$ on $K$ the Borel function $\check{f}$ is defined by $\check{f}(x)=f(\check{x})$ for all $x\in K$. Given $\mu,\nu\in{M}(K)$, their convolution is given by

[TABLE]

and $\|\mu*\nu\|\leq\|\mu\|\nu\|$ which shows that $\mu*\nu\in{M}(K).$ Also for a measure $\mu\in M(K)$ and a Borel function $f$ on $K$, we define the convolutions $\mu*f$ and $f*\mu$ by

[TABLE]

if the integrals exist. Note that in this case $(\mu*f)\check{}=\check{f}*\check{\mu}$. Moreover, if $f$ is in $C_{b}(K)$, the Banach space of bounded complex continuous functions on $K$, then $\mu*f$ and $f*\mu$ are in $C_{b}(K)$ with $\|\mu*f\|_{\infty}\leq\|\mu\|\|f\|_{\infty}$ and $\langle\mu*\nu,f\rangle=\langle\nu,\check{\mu}*f\rangle$. We refer the reader to [3] for details of hypergroups.

3. Amenability and Liouville property

Throughout of this paper, let $K$ be a locally compact hypergroup with a left-invariant Haar measure $\omega$; that is, a non-zero positive Radon measure on $K$ such that

[TABLE]

Let $C_{0}(K)$ be the Banach space of complex continuous functions on $K$ vanishing at infinity. Then its dual identifies, via the Riesz representation theorem, with the space $M(K)$. Let $L^{p}(K)$ be the complex Lebesgue spaces with respect to $\omega$, for $1\leq p\leq\infty$. Given a Borel measure $\sigma$ on a hypergroup $K$, a Borel function $f$ on $K$ satisfying the convolution equation

[TABLE]

is called ${\sigma}$-harmonic. For $1\leq p\leq\infty$ define $H_{\sigma}^{p}(K)$ to be the set of all ${\sigma}$-harmonic $L^{p}$-functions; that is, $H_{\sigma}^{p}(K)=\{f\in L^{p}(K):\sigma*f=f\}$. For Borel functions $f$ and $g$ at least one of which is $\sigma$-finite, define the convolution $f*g$ on $K$ by

[TABLE]

We commence with the following lemma whose proof is similar to those given in [5]. For completeness, we present the argument here.

Lemma 3.1**.**

Let $\sigma\in M(K)$ with $\|\sigma\|=1$ and let $1<p\leq\infty$. Then there is a contractive projection $P_{\sigma}:L^{p}(K)\rightarrow L^{p}(K)$ with $P_{\sigma}(L^{p}(K))=H_{\sigma}^{p}(K)$. Moreover, if $1<p,q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$, then $P_{\sigma}$ is the dual map of the projection $P_{\check{\sigma}}:L^{q}(K)\rightarrow L^{q}(K)$.

Proof.

Let ${\mathcal{U}}$ be a free ultra-filter on $\mathbb{N}$, and define $P_{\sigma}:L^{p}(K)\rightarrow L^{p}(K)$ by the weak∗ limit

[TABLE]

where $\sigma^{k}$ is the $k$-times convolution of $\sigma$ with itself. It is easy to see that $P_{\sigma}(f)=f$ for all $f\in H_{\sigma}^{p}(K)$. Moreover, if $f\in L^{p}(K)$, then it is easily verified that $\sigma*P_{\sigma}(f)=P_{\sigma}(f)$ and so $P_{\sigma}(f)\in H_{\sigma}^{p}(K)$. These show that $P_{\sigma}^{2}=P_{\sigma}$ and $P_{\sigma}(L^{p}(K))=H_{\sigma}^{p}(K)$.

Suppose now that $1<p<\infty$. Then it is not hard to check that $\sigma*P_{\sigma}(f)=P_{\sigma}(\sigma*f)$ for all $f\in L^{p}(K)$. Therefore, for each $g\in L^{q}(K)$, we have

[TABLE]

This shows that $\sigma*P_{\check{\sigma}}^{*}(f)=P_{\check{\sigma}}^{*}(f)$ for all $f\in L^{p}(K)$. Similarly, we can show that $\check{\sigma}*P_{{\sigma}}^{*}(g)=P_{{\sigma}}^{*}(g)$ for all $g\in L^{q}(K)$. Consequently, for each $f\in L^{p}(K)$ and $g\in L^{q}(K)$, we have

[TABLE]

This shows that $P^{*}_{\check{\sigma}}(f)=P_{\sigma}P_{\check{\sigma}}^{*}(f)=P_{\sigma}(f)$ for all $f\in L^{p}(K)$, as required. ∎

Remark 3.2*.*

Let $\sigma\in M(K)$ with $\|\sigma\|=1$ and let $1<p,q<\infty$ be such that $\frac{1}{p}+\frac{1}{q}=1$. Then we have linear isometric isomorphisms $H_{\sigma}^{p}(K)\cong L^{p}(K)/H_{\check{\sigma}}^{q}(K)^{\perp}\cong H_{\check{\sigma}}^{q}(K)^{*}$, where the first isometry is given by $H_{\sigma}^{p}(K)\ni f\mapsto f+H_{\check{\sigma}}^{q}(K)^{\perp}\in L^{p}(K)/H_{\check{\sigma}}^{q}(K)^{\perp}$. Indeed, for each $f\in H_{\sigma}^{p}(K)$, we have

[TABLE]

Recall that the hypergroup $K$ is called amenable if there exists a topological left invariant mean on $L^{\infty}(K)$; that is, there exists $m\in L^{\infty}(K)^{*}$ such that $\|m\|=m(1)=1$ and $m(g*f)=(\int_{K}g~{}d\omega)m(f)$ for all $f\in L^{\infty}(K)$ and $g\in L^{1}(K)$. A topological right invariant mean on $L^{\infty}(K)$ is a functional $m\in L^{\infty}(K)^{*}$ such that $\|m\|=m(1)=1$ and $m(f*\check{g})=(\int_{K}g~{}d\omega)m(f)$ for all $f\in L^{\infty}(K)$ and $g\in L^{1}(K)$. It is known that the involution on $L^{1}(K)$ can be canonically extended to a linear involution $\star$ on $L^{1}(K)^{**}$; see [9, Chapter 2]. Clearly, $m\in L^{\infty}(K)^{*}$ is a topological left invariant mean if and only if $m^{\star}$ is a topological right invariant mean. Therefore, the existence of a topological right invariant mean on $L^{\infty}(K)$ is equivalent to $K$ being amenable.

Theorem 3.3**.**

Let $K$ be a hypergroup with the Liouville property; that is, there exists a probability measure $\sigma$ on $K$ such that $H_{\sigma}^{\infty}(K)=\mathbb{C}1$. Then $K$ is amenable.

Proof.

Let $P_{\sigma}:L^{\infty}(K)\rightarrow H_{\sigma}^{\infty}(K)$ be the contractive projection as defined in Lemma 3.1. Then there is a unique functional $m\in L^{\infty}(K)^{*}$ such that $P_{\sigma}(f)=m(f)1$ for all $f\in L^{\infty}(K)$. Since $\sigma*(f*\check{g})=(\sigma*f)*\check{g}$ for all $f\in L^{\infty}(K)$ and $g\in L^{1}(K)$, it follows that $P_{\sigma}(f*\check{g})=P_{\sigma}(f)*\check{g}$. Moreover, since the projection $P_{\sigma}$ is positive and $P_{\sigma}(1)=1$, we conclude that $\|m\|=m(1)=1$. This shows that $m$ is a topological right invariant mean on $L^{\infty}(K)$, which implies that $K$ is amenable.

∎

For a a locally compact hypergroup $K$ consider the closed two sided ideal

[TABLE]

in $L^{1}(K)$ and for each $\sigma\in M(K)$ let $J_{\sigma}$ be the norm closure of $\{f-\check{\sigma}*f:~{}f\in L^{1}(K)\}$ in $L^{1}(K)$. It is well known that $L^{1}_{0}(K)$ has codimension one in $L^{1}(K)$ and if $\sigma$ is a probability measure, then $J_{\sigma}\subseteq L^{1}_{0}(K)$. Moreover, it is easy to see that $J_{\sigma}^{\perp}=\{f\in L^{\infty}(K):\sigma*f=f\}=H_{\sigma}^{\infty}(K)$ and hence $H^{\infty}_{\sigma}(K)=({L^{1}(K)}/{J_{\sigma}})^{*}$

We have the following lemma whose proof is similar to those given in [14, Lemma 1.1 and Remark 3, p.210] for locally compact groups. Thus, we omit the proof.

Lemma 3.4**.**

Let $K$ be a locally compact hypergroup and ${\mathcal{S}}$ be a norm closed, convex subsemigroup of probability measuers on $K$. Let $I$ be a separable, closed subspace of $L^{1}(K)$ such that

(i)* ${J}_{\sigma}\subseteq I$ for every $\sigma\in{\mathcal{S}}$; and*

(ii)* for each $\varepsilon>0$ and $g\in I$ there is $\sigma\in{\mathcal{S}}$ such that*

[TABLE]

Then there is $\sigma\in{\mathcal{S}}$ such that $I={J}_{\sigma}$.

Corollary 3.5**.**

Let $K$ be a second countable locally compact hypergroup. Then the following conditions are equivalent.

(i)* $K$ is amenable.*

(ii)* There exists a probability measure $\sigma$ on $K$ such that $L^{1}_{0}(K)=J_{\sigma}$.*

(iii)* $K$ has the Liouville property.*

Proof.

(i)$\Rightarrow$(ii). Suppose that $K$ is amenable. Then by [13, Corollary 4.2], there is a net $(f_{\alpha})$ in $P_{1}(K):=\{f\in L^{1}(K):\|f\|_{1}=\int_{K}f~{}d\omega=1\}$ such that

[TABLE]

for all $f\in P_{1}(K)$. In particular, for each $g\in L^{1}_{0}(K)$ we have $\|f_{\alpha}*g\|_{1}\rightarrow 0$. Moreover, $f_{\alpha}*g-g\in J_{\check{\sigma}_{\alpha}}$ for all $\alpha$, where $\sigma_{\alpha}=f_{\alpha}\omega$. This shows that the condition (ii) of Lemma 3.4 is satisfied. Since $L^{1}(K)$ is separable, we give that $L^{1}_{0}(K)=J_{\sigma}$ for some probability measure $\sigma$ on $K$.

(iii)$\Rightarrow$(i). This follows from Theorem 3.3.

(ii)$\Leftrightarrow$(iii). This follows from the inclusion $J_{\sigma}\subseteq L^{1}_{0}(K)$ with the fact that $L^{1}_{0}(K)^{\perp}={\mathbb{C}}1$. ∎

Proposition 3.6**.**

Let $\sigma$ be a probability measure on $K$ and let $1<p\leq\infty$. Then $H_{\sigma}^{p}(K)$ is generated by its non-negative elements.

Proof.

Suppose that $f\in H_{\sigma}^{p}(K)$. Then $\sigma*\overline{f}=\overline{(\sigma*f)}=\overline{f}$. This shows that $H_{\sigma}^{p}(K)$ is self-adjoint and consequently is generated by its real function parts. Now let $f\in H_{\sigma}^{p}(K)$ be a real function and let $f=f_{+}-f_{-}$, where $f_{+},f_{-}$ are non-negative functions in $L^{p}(K)$. Since $\sigma$ is positive, the projection $P_{\sigma}$, as defined in Theorem 3.1, is positive. It follows that $P_{\sigma}(f_{+}),P_{\sigma}(f_{-})\in H_{\sigma}^{p}(K)$ are non-negative. Moreover, $f=P_{\sigma}(f)=P_{\sigma}(f_{+})-P_{\sigma}(f_{-})$, and this completes the proof. ∎

4. Harmonic $L^{p}$-functions

Let $\mu$ be a complex Borel measure on a locally compact hypergroup $K$. We say that $\mu$ is non-degenerate if

[TABLE]

where $|\mu|^{n}$ is the $n$-fold convolution of $|\mu|$ and ${\rm supp}|\mu|^{n}$ equals the closure of $({\rm supp}|\mu|)^{n}$. If $\mu$ satisfies the weaker condition that

[TABLE]

then we say that $\mu$ is adapted.

Remark 4.1*.*

Let $\mu\in M(K)$. Then it is not hard to check that non-degeneracy of $\mu$ is equivalent to that $\sum_{n=1}^{\infty}\langle|\mu|^{n},h\rangle>0$ for every non-zero $h\in C_{c}(K)^{+}$, or equivalently there exists $n\in\mathbb{N}$ such that

[TABLE]

Therefore, if $f\in C_{b}(K)^{+}$ is non-zero, then we may find $h\in C_{c}(K)^{+}$ such that $\|h\|_{\infty}=1$ and $fh\neq 0$. It follows that $fh\in C_{c}(K)^{+}$ and $fh\leq f$. Therefore, there exists $n\in\mathbb{N}$ such that

[TABLE]

Theorem 4.2**.**

Let $\sigma$ be a probability measure on $K$. Then the following statements are equivalent.

(i)* $H^{\infty}_{\sigma}(K)$ is a subalgebra of $L^{\infty}(K)$.*

(ii)* $H^{\infty}_{\sigma}(K)$ is a von Neumann subalgebra of $L^{\infty}(K)$..*

(iii)* H^{\infty}_{\sigma}(K)=\{f\in L^{\infty}(K):\forall x\in K,f(\check{y}*x)=f(x),~{}\hbox{for \sigma-a.e.}~{}y\in K\}.*

Proof.

Since $H^{\infty}_{\sigma}(K)$ is a weak∗ closed operator system, (i) implies that $H^{\infty}_{\sigma}(K)$ is a von Neumann subalgebra of $L^{\infty}(K)$. The implication (iii)$\Rightarrow$(i) is trivial. We need to prove (ii)$\Rightarrow$(iii). Let $x\in K$ and $f\in H^{\infty}_{\sigma}(K)$. Without loss of generality assume that $f$ is real valued. Since $(f-f(x))^{2}\in H^{\infty}_{\sigma}(K)$, it follows that

[TABLE]

This implies that $f(\check{y}\ast x)=f(x)$ for $\sigma-$almost every $y\in K$. ∎

For a hypergroup $K$, we denote by $LUC(K)$ to be the Banach space of all bounded left uniformly continuous complex functions on $K$, consisting of bounded continuous function $f$ on $K$ such that the map $K\ni x\mapsto f_{x}\in C_{b}(K)$ is continuous.

Lemma 4.3**.**

Let $\sigma\in M(K)$. Then $H_{\sigma}^{\infty}(K)\cap LUC(K)$ is weak∗ dense in $H_{\sigma}^{\infty}(K)$.

Let $(\phi_{\alpha})$ be a bounded approximate identity for $L^{1}(K)$ and $f\in H^{\infty}_{\sigma}(K)$. Then $f*\check{\phi_{\alpha}}\in H^{\infty}_{\sigma}(K)\cap LUC(K)$ for all $\alpha$, by [13, Lemma 2.2]. Moreover, for each $g\in L^{1}(K)$, we have

[TABLE]

Thus, $H_{\sigma}^{\infty}(K)\cap LUC(K)$ is weak∗ dense in $H_{\sigma}^{\infty}(K)$.

Corollary 4.4**.**

Let $\sigma$ be a non-degenerate probability measure on $K$. Then the following conditions are equivalent.

(i)* $H_{\sigma}^{\infty}(K)$ is a subalgebra of $L^{\infty}(K)$.*

(ii)* $H_{\sigma}^{\infty}(K)=\mathbb{C}{1}$.*

Proof.

Suppose that (i) holds. Given $f\in H_{\sigma}^{\infty}(K)\cap LUC(K)$, by Theorem 4.2 for each $n\in\mathbb{N}$, we have $\check{f}(y)=\check{f}(e)$ for all $y\in(\rm{supp}\sigma)^{n}$. It follows from non-degeneracy of $\sigma$ and continuity of $f$ that $f$ is constant. Since $H_{\sigma}^{\infty}(K)\cap LUC(K)$ is weak∗ dense in $H_{\sigma}^{\infty}(K)$, we give that $H_{\sigma}^{\infty}(K)=\mathbb{C}{1}$. ∎

A subspace $X$ of $L^{p}(K)$, $1\leq p\leq\infty$, is called left (resp. right) translation invariant if ${}_{x}f\in X$ (resp. $f_{x}\in X$) for all $f\in X$ and $x\in K$. The subspace $X$ is called translation invariant if it is left and right translation invariant. It is easy to check that for each $\sigma\in M(K)$ the space $H^{\infty}_{\sigma}(K)$ is a right translation invariant subspace of $L^{\infty}(K)$. Since $\int_{K}(g_{x})(t)f(t)d\omega(t)=\frac{1}{\Delta(x)}\int_{K}g(t)(f_{\check{x}})(t)d\omega(t)$ for all $f\in L^{\infty}(K)$, $g\in L^{1}(K)$ and $x\in K$, it follows that $J_{\sigma}$ is also right translation invariant in $L^{1}(K)$, where $\Delta$ is the modular function on $K$. We recall that $L^{\infty}(K)$ is naturally a Banach $L^{1}(K)$-bimodule by the following module actions

[TABLE]

It is easily verified that

[TABLE]

for all $x\in K$.

Proposition 4.5**.**

Let $\sigma\in M(K)$. Then the following conditions are equivalent.

(i)* $H^{\infty}_{\sigma}(K)$ is translation invariant.*

(ii)* $J_{\sigma}$ is translation invariant.*

(iii)* $J_{\sigma}$ is an ideal in $L^{1}(K)$.*

(iv)* $\int_{K}f(\check{y}*x)d\sigma(y)=\int_{K}f(x*\check{y})d\sigma(y)$ for all $f\in H^{\infty}_{\sigma}(K)\cap LUC(K)$.*

(v)* $H^{\infty}_{\sigma}(K)$ is a sub-$L^{1}(K)$-bimodule of $L^{\infty}(K)$.*

Proof.

(i)$\Leftrightarrow$(ii). Since $\int_{K}({{}_{x}}g)(t)f(t)d\omega(t)=\frac{1}{\Delta(x)}\int_{K}g(t)(_{\check{x}}f)(t)d\omega(t)$ for all $f\in L^{\infty}(K)$, $g\in L^{1}(K)$ and $x\in K$, it follows that $H^{\infty}_{\sigma}(K)$ is left translation invariant if and only if $J_{\sigma}$ is.

(ii)$\Rightarrow$(iii). It suffices to show that $J_{\sigma}$ is a left ideal in $L^{1}(K)$. To prove this, given $g\in L^{1}(K)$, $h\in J_{\sigma}$ and $f\in H_{\sigma}^{\infty}(K)$, we have

[TABLE]

which implies that $g*h\in J_{\sigma}$.

(iii)$\Rightarrow$(ii). Let $(\phi_{\alpha})$ be a bounded approximate identity for $L^{1}(K)$ and let $g\in J_{\sigma}$. Since $((_{x}\phi_{\alpha})*g)=_{x}(\phi_{\alpha}*g)$ and ${}_{x}\phi_{\alpha}\in L^{1}(K)$ for all $\alpha$ and $x\in K$, the proof follows from the fact that $\phi_{\alpha}*g\rightarrow g$.

(i)$\Rightarrow$(iv). Suppose that $f\in H^{\infty}_{\sigma}(K)\cap LUC(K)$. Since $H^{\infty}_{\sigma}(K)$ is left translation invariant, we obtain that

[TABLE]

(iv)$\Rightarrow$(i). Suppose that $f\in H^{\infty}_{\sigma}(K)\cap LUC(K)$ and $x\in K$. Then $f_{x}\in H^{\infty}_{\sigma}(K)\cap LUC(K)$, by right translation invariance of $H^{\infty}_{\sigma}(K)\cap LUC(K)$. Moreover, for each $y\in K$, we have

[TABLE]

This shows that $H^{\infty}_{\sigma}(K)\cap LUC(K)$ and hence $H^{\infty}_{\sigma}(K)$ is left translation invariant by weak∗ density of $H^{\infty}_{\sigma}(K)\cap LUC(K)$ in $H^{\infty}_{\sigma}(K)$.

(i)$\Rightarrow$(v). Let $f\in H_{\sigma}^{\infty}(K)$ and $g\in L^{1}(K)$. As $g\cdot f=f*\check{g}$ , we give that $g\cdot f\in H^{\infty}_{\sigma}(K)$. Moreover, by assumption ${}_{t}f\in H_{\sigma}^{\infty}(K)$ for all $t\in K$. Thus,

[TABLE]

This implies that $f\cdot g\in H^{\infty}_{\sigma}(K)$.

(v)$\Rightarrow$(iii). It suffices to show that $J_{\sigma}$ is a left ideal in $L^{1}(K)$. Indeed, given $f\in H_{\sigma}^{\infty}(K)$, $g\in L^{1}(K)$ and $h\in J_{\sigma}$, we have

[TABLE]

which yields that $g*h\in J_{\sigma}$, as required. ∎

Remark 4.6*.*

It is obvious that under each of above equivalent conditions in Proposition 4.5, the quotient space $L^{1}(K)/J_{\sigma}$ is a Banach algebra. This implies that $H^{\infty}_{\sigma}(K)^{*}=({L^{1}(K)}/{J_{\sigma}})^{**}$ is a Banach algebra with respect to the two Arens products.

Theorem 4.7**.**

Let $\sigma$ be a non-degenerate probability measure on $K$. Then every bounded continuous $\sigma$-harmonic function on $K$ vanishing at infinity is constant.

Proof.

Let $f\in{H}_{\sigma}^{\infty}(K)\cap C_{0}(K)$ be real-valued. Without loss of generality assume that $\|f\|_{\infty}=1$. Therefor, we can find a probability measure $\mu$ on $K$ such that $\|f\|_{\infty}=\langle\mu,f\rangle$. If $f\neq 1$, then the function $1-f$ is also non-negative and non-zero $\sigma$-harmonic function in $C_{b}(K)$. It is well known from [3, Proposition 1.2.16] that $\mu*(1-\check{f})$ is a non-negative and bounded continuous function. Moreover,

[TABLE]

which implies that $\mu*(1-\check{f})$ is non-zero. Since $\sigma$ is non-degenerate, by Remark 4.1, there exists $n\in\mathbb{N}$ such that

[TABLE]

On the other hand, since $f$ is $\sigma$-harmonic, $\sigma^{n}*f=f$. Therefore,

[TABLE]

which is a contradiction. Since ${H}_{\sigma}^{\infty}(K)\cap C_{0}(K)$ is generated by its non-negative elements, the proof is complete. ∎

Corollary 4.8**.**

Let $K$ be non-compact and let $\sigma$ be a non-degenerate probability measure on $K$. Then the sequence $\left(\frac{1}{n}\sum_{k=1}^{n}\sigma^{k}\right)$ is weak∗-convergent to [math].

Proof.

Let $\sigma_{0}$ be a weak∗ cluster point of $(\frac{1}{n}\sum_{k=1}^{n}\sigma^{k})$ in $M(K)$. Then $\sigma*\sigma_{0}=\sigma_{0}$, which implies that $\sigma_{0}$ is an idempotent probability measure in $M(K)$. Moreover, for each $f\in C_{0}(G)$ we have

[TABLE]

This shows that $\sigma_{0}*f$ is $\sigma$-harmonic and so it is constant by Theorem 4.7. Since $K$ is non-compact, we must have $\sigma_{0}*f=0$ for all $f\in C_{0}(K)$. This implies that $\sigma_{0}=0$. Thus, [math] is the only weak∗ cluster point of $\left(\frac{1}{n}\sum_{k=1}^{n}\sigma^{k}\right)$ in $M(K)$. By weak∗ compactness of the unit ball of $M(K)$ we conclude that $\frac{1}{n}\sum_{k=1}^{n}\sigma^{k}\stackrel{{\scriptstyle w^{*}}}{{\longrightarrow}}0$. ∎

Corollary 4.9**.**

The hypergroup $K$ is compact if and only if there is a non-degenerate idempotent probability measure on $K$.

Theorem 4.10**.**

Let $K$ be non-compact and let $\sigma$ be a non-degenerate probability measure on $K$. Then we have $H_{\sigma}^{p}(K)=\{0\}$ for all $1<p<\infty$.

Proof.

Let $1<p<\infty$ and let $f$ be a non-negative function in $H_{\sigma}^{p}(K)$ with $\|f\|_{p}=1$. Consider the probability measure $\sigma_{0}$ on $K$ defined by

[TABLE]

It is easy to see that $\sigma_{0}*f=f$. Moreover, we may find $g\in L^{q}(K)$ with $\|g\|_{q}\leq 1$ such that $\langle f,g\rangle=\|f\|_{p}=1$, where $1<q<\infty$ and $\frac{1}{p}+\frac{1}{q}=1$. It follows from [3, (1.4.11), (1.4.12)] that $g*\check{f}\in C_{0}(K)$ and $\|g*\check{f}\|_{\infty}\leq\|g\|_{q}\|f\|_{p}\leq 1$. Therefore, $1-g*\check{f}\geq 0$ and

[TABLE]

It follows from non-degeneracy of $\sigma$ that $1=g*\check{f}\in C_{0}(K)$, contradicting $K$ being non-compact. Hence, $f=0$. Thus, Proposition 3.6 implies that $H_{\sigma}^{p}(K)=\{0\}$. ∎

We use the following result which is proved in [2, Theorem 3.8] to show that if $\sigma$ is an adapted probability measure on compact hypergroup $K$, then each $\sigma$-harmonic $L^{p}$-function is trivial for all $1\leq p\leq\infty$.

Theorem 4.11**.**

If $\sigma$ is an adapted probability measure on a compact hypergroup $K$, then each $\sigma$-harmonic continuous function on $K$ is constant.

Theorem 4.12**.**

Let $K$ be a compact hypergroup and let $\sigma$ be an adapted probability measure on $K$. Then for $1\leq p\leq\infty$, we have $H_{\sigma}^{p}(K)=\mathbb{C}1$.

Proof.

Let $K$ be compact. Then we have $L^{\infty}(K)\subseteq L^{p}(K)\subseteq L^{1}(K)$ for all $1\leq p\leq\infty$. Thus, it suffices to prove the assertion for the case $p=1$. Let $f\in{H}_{\sigma}^{1}(K)$ and let $(\phi_{\alpha})$ be a bounded approximate identity for $L^{1}(K)$ such that $\phi_{\alpha}$ is a bounded continuous function with compact support for all $\alpha$; see [3, Theorem 1.6.15]. Then [13, Lemma 2.2(i)] implies that $f*\phi_{\alpha}$ is continuous and bounded for all $\alpha$. Moreover, it is clear that $f*\phi_{\alpha}$ is $\sigma$-harmonic, and is therefore constant by Theorem 4.11. This shows that $f$ is also constant, as desired. ∎

Corollary 4.13**.**

Let $K$ be a compact hypergroup and let $\sigma$ be a non-degenerate probability measure on $K$. Then any weak∗ cluster point $\sigma_{0}$ of the sequence $\left(\frac{1}{n}\sum_{k=1}^{n}\sigma^{k}\right)$ in $M(K)$ is the normalized Haar measure on $K$.

Proof.

Let $\sigma_{0}$ be a weak∗ cluster point of $(\frac{1}{n}\sum_{k=1}^{n}\sigma^{k})$ in $M(K)$. Then we have $\sigma_{0}$ is an idempotent probability measure on $K$ satisfying $\sigma*\sigma_{0}=\sigma_{0}$, and therefore $\sigma_{0}*f\in{H}_{\sigma}^{\infty}(K)$ for all $f\in C_{b}(K)$. It follows from Theorem 4.11 and the non-degeneracy of $\sigma$ that for each $f\in C_{b}(K)$ there exists $\lambda_{f}\in{C}$ such that $\sigma_{0}*f=\lambda_{f}1$. Let $\omega$ be the normalized Haar measure on $K$. Then for each $f\in C_{b}(K)$,

[TABLE]

Moreover,

[TABLE]

This shows that $\check{\sigma_{0}}=\omega$. Since $\check{\omega}=\omega$, we conclude that $\sigma_{0}=\omega$. ∎

Let $\sigma\in M(K)$. We say that a measure $\mu\in M(K)$ is $\sigma$-harmonic if it satisfies the convolution equation $\sigma*\mu=\mu$. Define ${H}_{\sigma}(K)$ to be the set of all ${\sigma}$-harmonic measures.

Theorem 4.14**.**

Let $\sigma\in M(K)$ with $\|\sigma\|=1$. Then there is a contractive projection $P_{\sigma}:M(K)\rightarrow M(K)$ with $P_{\sigma}(M(K))={H}_{\sigma}(K)$.

Proof.

Let $\mathcal{U}$ be a free ultra-filter on $\mathbb{N}$, and define $P_{\sigma}:M(K)\rightarrow M(K)$ by the weak∗ limit

[TABLE]

It is easy to see that $P_{\sigma}(\mu)=\mu$ for all $\mu\in{H}_{\sigma}(K)$. Moreover, if $\mu\in M(K)$, then it is easily verified that $\sigma*P_{\sigma}(\mu)=P_{\sigma}(\mu)$ and hence $P_{\sigma}(\mu)\in{H}_{\sigma}(K)$. These show that $P_{\sigma}^{2}=P_{\sigma}$ and $P_{\sigma}(M(K))={H}_{\sigma}(K)$. ∎

Recall that a measure $\mu\in M(K)$ is non-negative if $\langle\mu,f\rangle\geq 0$ for all $f\in C_{0}(K)^{+}$.

Proposition 4.15**.**

Let $\sigma$ be a probability measure on $K$. Then ${H}_{\sigma}(K)$ is generated by its non-negative elements.

Proof.

Suppose that $\mu\in{H}_{\sigma}(K)$. Then $\sigma*\overline{\mu}=\overline{(\sigma*\mu)}=\overline{\mu}$. This shows that ${H}_{\sigma}(K)$ is generated by its real measure parts. Now let $\mu\in{H}_{\sigma}(K)$ be a real measure and let $\mu=\mu_{+}-\mu_{-}$, where $\mu_{+},\mu_{-}$ are non-negative measures in $M(K)$. Since $\sigma$ is positive, the measures $P_{\sigma}(\mu_{+}),P_{\sigma}(\mu_{-})\in{H}_{\sigma}(K)$ are non-negative. Moreover, $\mu=P_{\sigma}(\mu)=P_{\sigma}(\mu_{+})-P_{\sigma}(\mu_{-})$, which completes the proof. ∎

Theorem 4.16**.**

Let $K$ be non-compact and let $\sigma$ be a non-degenerate probability measure on $K$. Then we have ${H}_{\sigma}(K)=\{0\}$.

Proof.

Suppose that $\mu\in H_{\sigma}(K)$ is non-zero. By Proposition 4.15, we can assume that $\mu$ is positive. Consider the probability measure $\sigma_{0}$ defined by $\sigma_{0}:=\sum_{n=1}^{\infty}\frac{1}{2^{n}}\sigma^{n}$. It is clear that $\sigma_{0}\ast\mu=\mu$. By [3, Theorem 1.6.9], we have that $\delta_{x}\ast\mu=\mu$ for all $x\in{\rm supp}\sigma_{0}=K$. This shows that $\mu$ is the left Haar measure that is finite on $K$, which is a contradiction with non-compactness of $K$.

∎

Corollary 4.17**.**

Let $K$ be non-compact and let $\sigma$ be a non-degenerate probability measure on $K$. Then we have ${H}_{\sigma}^{1}(K)=\{0\}$.

Proof.

Suppose that $f\in H_{\sigma}^{1}(K)$. Define $\mu:=f\omega$, where $\omega$ is the left Haar measure on $K$. Then we have

[TABLE]

Therefore, $\mu=0$ by Theorem 4.16 and hence $f=0$. ∎

Theorem 4.18**.**

let $K$ be compact and let $\sigma$ be an adabted probability measure on $K$. Then we have ${H}_{\sigma}(K)=\mathbb{C}\omega$, where $\omega$ is the normalized Haar measure on $K$.

Proof.

Because $K$ is compact, the Haar measure $\omega$ is in $M(K)$ and we have $\sigma\ast\omega=\omega$. Therefore, $\mathbb{C}\omega_{K}\subseteq{H}_{\sigma}(K)$. To prove the converse, suppose that $\mu\in{H}_{\sigma}(K)$. Then $\sigma\ast\mu=\mu$. Let $(\phi_{\alpha})$ be a bounded approximate identity for $L^{1}(K)$ such that $\phi_{\alpha}\in C_{c}^{+}(K)$ for every $\alpha$ and $\phi_{\alpha}\stackrel{{\scriptstyle w^{*}}}{{\longrightarrow}}\delta_{e}$; see [3, Theorem 1.6.15]. Then $\mu\ast\phi_{\alpha}\stackrel{{\scriptstyle w^{*}}}{{\longrightarrow}}\mu$ and $\mu\ast\phi_{\alpha}\in{H}_{\sigma}^{1}(K)$ for all $\alpha$. Thus, Theorem 4.12 implies that $\mu\ast\phi_{\alpha}$ is constant for all $\alpha$. Hence, for every $\alpha$ there is $\lambda_{\alpha}\in\mathbb{C}$ such that $\mu\ast\phi_{\alpha}=\lambda_{\alpha}{1}$. It follows that for each $f\in C_{b}(K)$, we have

[TABLE]

Therefore, $\langle\lambda_{\alpha}\omega_{K},f\rangle\longrightarrow\langle\mu,f\rangle$ for all $f\in C_{b}(K)$. This shows that there exists $\lambda\in\mathbb{C}$ such that $\mu=\lambda\omega$. It follows that ${H}_{\sigma}(K)=\mathbb{C}\omega$. ∎