I am working on exercise 2.3(b) from "Essential Results of Functional Analysis" by Zimmer and couldn't find a similar question asked here.
A mean $m$ on a measure space $(X,\mu)$ is a continuous linear functional $L^\infty(X) \rightarrow \mathbb{C}$ satisfying
(i) $m(1_X) = 1$
(ii) $f \geq 0$ (that is, $f$ is real-valued and nonnegative) implies $m(f) \geq 0$
(iii) $m(\overline{f})=\overline{m(f)}$
For the case ($X = \mathbb{R}^n$,Lebesgue), we say $m$ is invariant if for all $t \in \mathbb{R}^n$ and all $f$, $$ m(f_t) = m(f) $$ where $f_t(x):=f(x-t)$.
My question is: why does an invariant mean exist? I have tried using the Hahn-Banach theorem after defining $m$ on the one-dimensional subspace of constant functions, but it is not clear to me that the extension should satisfy the invariance property. I have also tried defining $m$ using the average value of $f$ over larger and larger balls about the origin, but this limit may not be defined.
TIA!
Fix $m$ to be a mean. Let's write $m_j$ to be the functional defined by $m_j(f) = m(f_{jt})$. Consider the sums $\frac{1}{2N+1}\sum_{|j| \leq N} m_j =: m'_N$, for $N \in \mathbb{N}$. Each of these is a mean, simply by checking the definitions. Now, $\{m'_N\}_N$ is a sequence of uniformly bounded linear functionals, hence belonging to some dilation of the unit ball in $L^{\infty}(X)^{\ast}$, so by the Banach-Alaoglu theorem, there exists some convergent subsequence of these linear functionals. Let $M$ be a limit point of such a sequence. Then $M(f_t) = M(f)$ by definition: indeed, if our convergent subsequence is $\{n_j\}_j$, since we have \begin{equation*} \bigg|m'_{n_j}(f)-m'_{n_j}(f_t)\bigg| = \frac{1}{2n_j+1} \left(\bigg|\sum_{k = -n_j}^{n_j} m(f_{kt}) - \sum_{k = -n_j+1}^{n_j+1} m(f_{kt})\bigg |\right) \leq \frac{2\|f\|}{2n_j+1}, \end{equation*} $\{m'_{n_j}(f)\}_j$ and $\{m'_{n_j}(f_t)\}_j$ have the same limit function, $g = M(f) = M(f_t)$. Thus, $M$ is invariant.