Question on invariant mean.

Question

Given a group G, consider the algebra $l^{\infty}(G)$. A $\textit{mean}$ over $l^{\infty}(G)$ is a a non-negative linear functional $\lambda:l^{\infty}(G) \rightarrow \mathbb{C}$ such that $\lambda(1) = 1$. We say that $\lambda$ is invariant if for all $f \in l^{\infty}(G)$ and $g \in G$ we have $\lambda(f)(g^{-1}h) = \lambda(f)(h)$ for all $h \in G$.

I want to understand some exemples. Is there any caracterization of all invariant means over $l^{\infty}(\mathbb{Z})$?

Answer 1

Let $G$ be a locally compact Abelian group. A function $f \in L^{\infty} (G)$ is called almost periodic whenever the collection of translates $\{f (x y^{-1}) \}_{y \in G}$ is relatively compact in $L^{\infty} (G)$. The collection of all almost periodic functions on $G$ is denoted by $AP(G)$.

There is the following fundamental result on the existence and uniqueness of a mean on $AP(G) \subset L^{\infty} (G)$:

Let $G$ be a locally compact Abelian group. Then there exists a translation-invariant mean $M$ on $AP(G)$. Moreover, if $N : AP(G) \to \mathbb{C}$ is a linear functional satisfying

1. $N(1_G) = 1$;
2. $N(f) \leq 0$ for $f \leq 0$
3. $N(f(\cdot y^{-1})) = N(f(\cdot))$ for $y \in G$

Then $M = N$.

A more concrete expression of the mean can be given for the case that $G$ is assumed to be $\sigma$-compact, which your specific case $G = \mathbb{Z}$ clearly is. Under this additional assumption, there is the following result, which can be found as Theorem 18.10 in Abstract Harmonic Analysis by Hewitt & Ross.

Let G be a locally compact, $\sigma$-compact, Abelian group. Then there exists an increasing sequence $\{H_n \}_{n \in \mathbb{N}}$ in $G$ of relatively compact, open sets such that $$ M(f) = \lim_{n \to \infty} \mu_G (H_n)^{-1} \int_{H_n} f(x) \; d\mu_G (x) $$ for all $f \in AP(G)$.

In the special case $G = \mathbb{R}$, a mean on $AP(G)$ is given by $$ M(f) = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^T f(x) dx, $$ and for $G = \mathbb{Z}$, it is given by $$ M(f) = \lim_{n \to \infty} \frac{1}{2n+1} \sum_{k = - n}^n f(k). $$ Both examples are discussed in Example 18.15 of the aforementioned book by Hewitt & Ross.