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 $g\in G$ 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 $L\mathbb{I}=1$, 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...
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 $R$.