Existence left invariant mean implies existence right invariant mean
Denote by $l^{infty}(G)$ the Banach space of bounded real functions on the group $G$. The left-action of $G$ on $l^{infty}(G)$ is defined by $g cdot f(t)=f(gt)$, for all $gin G$ and $f in l^{infty}(G)$. A mean on the group $G$ is a linear functional $L$ on $l^{infty}(G)$ such that $Lf geq 0$ for all $f ≥ 0$ and $Lmathbb{I}=1$, with $mathbb{I}(x)=1$, for all $x in G$.
$L$ is called $G$-invariant if $L(g cdot f)=Lf$ for all $f in l^{infty}(G)$ and $g in G$. Note that, one can also consider the right-action of $G$ on $l^{infty}(G)$. This action is defined by $f(t) cdot g = f(tg)$.
It's well known that the existence of a left-action from a group on a set implies the existence of right-action and vice versa. I was wondering: does the existence of a left $G$-invariant mean imply the existence of a right $G$-invariant mean? Perhaps this question has a trivial explanation that I'm missing...
functional-analysis group-actions
asked Mar 10 at 19:30
J.Bosser
329210
add a comment |
Denote by $l^{infty}(G)$ the Banach space of bounded real functions on the group $G$. The left-action of $G$ on $l^{infty}(G)$ is defined by $g cdot f(t)=f(gt)$, for all $gin G$ and $f in l^{infty}(G)$. A mean on the group $G$ is a linear functional $L$ on $l^{infty}(G)$ such that $Lf geq 0$ for all $f ≥ 0$ and $Lmathbb{I}=1$, with $mathbb{I}(x)=1$, for all $x in G$.
$L$ is called $G$-invariant if $L(g cdot f)=Lf$ for all $f in l^{infty}(G)$ and $g in G$. Note that, one can also consider the right-action of $G$ on $l^{infty}(G)$. This action is defined by $f(t) cdot g = f(tg)$.
It's well known that the existence of a left-action from a group on a set implies the existence of right-action and vice versa. I was wondering: does the existence of a left $G$-invariant mean imply the existence of a right $G$-invariant mean? Perhaps this question has a trivial explanation that I'm missing...
functional-analysis group-actions
asked Mar 10 at 19:30
J.Bosser
329210
add a comment |
Denote by $l^{infty}(G)$ the Banach space of bounded real functions on the group $G$. The left-action of $G$ on $l^{infty}(G)$ is defined by $g cdot f(t)=f(gt)$, for all $gin G$ and $f in l^{infty}(G)$. A mean on the group $G$ is a linear functional $L$ on $l^{infty}(G)$ such that $Lf geq 0$ for all $f ≥ 0$ and $Lmathbb{I}=1$, with $mathbb{I}(x)=1$, for all $x in G$.
$L$ is called $G$-invariant if $L(g cdot f)=Lf$ for all $f in l^{infty}(G)$ and $g in G$. Note that, one can also consider the right-action of $G$ on $l^{infty}(G)$. This action is defined by $f(t) cdot g = f(tg)$.
It's well known that the existence of a left-action from a group on a set implies the existence of right-action and vice versa. I was wondering: does the existence of a left $G$-invariant mean imply the existence of a right $G$-invariant mean? Perhaps this question has a trivial explanation that I'm missing...
functional-analysis group-actions
asked Mar 10 at 19:30
J.Bosser
329210
Denote by $l^{infty}(G)$ the Banach space of bounded real functions on the group $G$. The left-action of $G$ on $l^{infty}(G)$ is defined by $g cdot f(t)=f(gt)$, for all $gin G$ and $f in l^{infty}(G)$. A mean on the group $G$ is a linear functional $L$ on $l^{infty}(G)$ such that $Lf geq 0$ for all $f ≥ 0$ and $Lmathbb{I}=1$, with $mathbb{I}(x)=1$, for all $x in G$.
$L$ is called $G$-invariant if $L(g cdot f)=Lf$ for all $f in l^{infty}(G)$ and $g in G$. Note that, one can also consider the right-action of $G$ on $l^{infty}(G)$. This action is defined by $f(t) cdot g = f(tg)$.
It's well known that the existence of a left-action from a group on a set implies the existence of right-action and vice versa. I was wondering: does the existence of a left $G$-invariant mean imply the existence of a right $G$-invariant mean? Perhaps this question has a trivial explanation that I'm missing...
functional-analysis group-actions
functional-analysis group-actions
asked Mar 10 at 19:30
J.Bosser
329210
asked Mar 10 at 19:30
J.Bosser
329210
asked Mar 10 at 19:30
J.Bosser
329210
asked Mar 10 at 19:30
J.Bosser
329210
asked Mar 10 at 19:30
J.Bosser
329210
329210
add a comment |
add a comment |
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Let $L:l^infty(G) rightarrow mathbb{C}$ be a left invariant mean. Then define $R: l^infty (G) rightarrow mathbb{C}$ by $R(f):=L(tilde{f})$, where $tilde{f}in l^infty (G)$ is defined by $tilde{f}(g):=f(g^{-1})$. Now if we write down the left invariance of $L$ in terms of $R$, we will get the right invariance of $R$.
answered Nov 29 at 14:46
Surajit
5689
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1 Answer
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1 Answer
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Let $L:l^infty(G) rightarrow mathbb{C}$ be a left invariant mean. Then define $R: l^infty (G) rightarrow mathbb{C}$ by $R(f):=L(tilde{f})$, where $tilde{f}in l^infty (G)$ is defined by $tilde{f}(g):=f(g^{-1})$. Now if we write down the left invariance of $L$ in terms of $R$, we will get the right invariance of $R$.
answered Nov 29 at 14:46
Surajit
5689
add a comment |
Let $L:l^infty(G) rightarrow mathbb{C}$ be a left invariant mean. Then define $R: l^infty (G) rightarrow mathbb{C}$ by $R(f):=L(tilde{f})$, where $tilde{f}in l^infty (G)$ is defined by $tilde{f}(g):=f(g^{-1})$. Now if we write down the left invariance of $L$ in terms of $R$, we will get the right invariance of $R$.
answered Nov 29 at 14:46
Surajit
5689
add a comment |
Let $L:l^infty(G) rightarrow mathbb{C}$ be a left invariant mean. Then define $R: l^infty (G) rightarrow mathbb{C}$ by $R(f):=L(tilde{f})$, where $tilde{f}in l^infty (G)$ is defined by $tilde{f}(g):=f(g^{-1})$. Now if we write down the left invariance of $L$ in terms of $R$, we will get the right invariance of $R$.
answered Nov 29 at 14:46
Surajit
5689
Let $L:l^infty(G) rightarrow mathbb{C}$ be a left invariant mean. Then define $R: l^infty (G) rightarrow mathbb{C}$ by $R(f):=L(tilde{f})$, where $tilde{f}in l^infty (G)$ is defined by $tilde{f}(g):=f(g^{-1})$. Now if we write down the left invariance of $L$ in terms of $R$, we will get the right invariance of $R$.
answered Nov 29 at 14:46
Surajit
5689
answered Nov 29 at 14:46
Surajit
5689
answered Nov 29 at 14:46
Surajit
5689
answered Nov 29 at 14:46
Surajit
5689
5689
add a comment |
add a comment |
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