Consider the von Neumann algebra $M:=L^\infty(\mathbb{R})$, which consists of (classes) of essentially bounded measurable functions $\mathbb{R}\to \mathbb{C}$. Here $\mathbb{R}$ has the classical Lebesgue measure.
In a paper I'm reading, one talks about an invariant mean on $M$. This is supposed to be some state $m:M \to \mathbb{C}$, but what exactly does it mean that $m$ is invariant?
I'm aware of a notion of invariance when we have a comultiplication $L^\infty(\mathbb{R})$, so I was thinking to consider a comultiplication $$\Delta: L^\infty(\mathbb{R})\to L^\infty(\mathbb{R}) \otimes L^\infty(\mathbb{R}) \cong L^\infty(\mathbb{R}\times \mathbb{R}):f \mapsto ((s,t) \mapsto f(s+t))$$ and then require that $m (\omega \otimes \iota)\Delta = m\omega(1) = m(\iota \otimes \omega)\Delta$. Is this the correct notion of invariant mean in this context? Does such a mean always exist?
Despite my earlier comment that the existence of an invariant mean on $L^\infty (\mathbb R)$ is a non-trivial result, let me give a reasonably simple proof of this fact.
A Fölner sequence is, by definition, a sequence $\{F_n\}_{n\in {\mathbb N}}$, where each $F_n$ is a measurable subset of $\mathbb R$, with finite positive measure, and such that $$ \lim_{n\to \infty } \frac{|(F_n + s)\mathop{\Delta } F_n|}{|F_n|} = 0, $$ for every $s$ in ${\mathbb R}$.
Here I am using "$|\cdot|$" to denote Lebesgue measure, and by "$\Delta $" I mean the symmetric difference of sets, namely $$ A\mathop{\Delta }B = (A\setminus B)\cup (B\setminus A). $$
The sets in a Folner sequence should be thought of as being asymptotically invariant under translation.
As an example, notice that the sets $F_n=(-n,n)$ form a Folner sequence.
Given any Folner sequence $\{F_n\}_{n\in {\mathbb N}}$, consider for each $n$, the linear functional $\varphi _n$ on $L^\infty (\mathbb R)$ defined by $$ \varphi _n(f) = \frac 1 {|F_n|}\int_{F_n}f(t)\, dt. $$
It is easy to see that each $\varphi _n$ is a positive linear functional with norm 1 (aka a state).
Given any real number $s$, let $$ \tau _s:L^\infty (\mathbb R)\to L^\infty (\mathbb R) $$ be the translation operator defined by $$ \tau _s(f)|_t = f(t-s), \quad \forall f \in L^\infty (\mathbb R). $$
For $f$ in $L^\infty (\mathbb R)$ notice that $$ \varphi _n(\tau _s(f)) - \varphi _n(f) = \frac 1 {|F_n|}\Big(\int_{F_n}f(t-s)\, dt - \int_{F_n}f(t)\, dt\Big) = $$$$ = \frac 1 {|F_n|}\Big(\int_{F_n+s}f(t)\, dt - \int_{F_n}f(t)\, dt\Big) = $$$$ = \frac 1 {|F_n|}\Big(\int_{(F_n+s)\setminus F_n}f(t)\, dt - \int_{F_n\setminus (F_n+s)}f(t)\, dt\Big), $$ from where it easily follows that $$ |\varphi _n(\tau _s(f)) - \varphi _n(f)| \leq \frac{|(F_n + s)\mathop{\Delta } F_n|}{|F_n|}\Vert f\Vert . \tag 1 $$
Thanks to Alaoglu's Theorem, the set of all states is compact for the weak$^*$ topology, so there is some state $m$ which is a cluster point of the set $\{\varphi _n:n\in {\mathbb N}\}$. By (1) one then concludes that $m$ satisfies $$ m(\tau _s(f)) = m(f), $$ which is to say that $m$ is an invariant state.