# What does the unique mean on weakly almost periodic functions look like?

There is a unique invariant mean $m$ on WAP functions on any discrete group (see definitions below, theorem of ?). However, the proofs I found are fairly non-explicit on how to obtain this invariant mean.

By "explicit", I would like to know (preferably, for the free group on two generators, but other non-amenable groups are welcome), if there is a sequence of functions $f_n \in \ell^1\Gamma$ so that $f_n$ tends weak$^*$ to the mean $m$?

There is of course bound to be a big amount of non-explicitness due to the fact that, ultimately, elements of $(\ell^\infty \Gamma)^* \setminus \ell^1\Gamma$ require [a weak form of] the axiom of choice (the "Hahn-Banach axiom").

Here is an example of what I mean by explicit. When the group is amenable, uniqueness implies that this means coincides with any other invariant mean on $\ell^\infty\Gamma$. So it would suffice to pick a sequence $F_n$ of Folner sets, and let $f_n = \chi_{F_n} /|F_n|$ be the normalised characteristic functions of those sets.

As for the definitions...

1. $\Gamma$ acts on $\ell^\infty \Gamma$ by translation by the left: $(\gamma \cdot f)(x) = f(\gamma^{-1}x)$. (Likewise, it acts on the right.) A function $f$ is WAP if the weak closure of its orbit under this action is weakly compact. (WAP is a subspace of $\ell^\infty$ which contains $c_0$.)

2. An invariant mean on a subspace $X$ of $\ell^\infty \Gamma$ is a linear map $m\colon X \to \mathbb{R}$ so that $m(\chi_\Gamma) = 1$ (where $\chi_\Gamma$ is the constant function taking value $1$) $m(f) \geq 0$ if $f \geq 0$ and $m( \gamma \cdot f) = m(f)$ (likewise on the right). Linearity, positivity and $m(\chi_\Gamma) =1$ automatically implies that $m$ is bounded, i.e. $m \in (\ell^\infty\Gamma)^*$.

3. A [non-explicit] characterisation of the invariant mean on WAP functions goes as follows. Let $f$ be WAP. Then there is a unique constant function in the weak closure of its orbit under translations. The value that this constant function takes is the value of the mean.

• As an update (pointed out by the answer of @user89334 below): I am actually also happy if the answer with an explicit $\textit{net}$ rather than a sequence. Also, I forgot to mention that the finitely supported measures are weak$^*$ dense in the space of means. The space of means is the space of linear maps $\ell^\infty \Gamma \to \mathbb{R}$ satisfying $m(\chi_\Gamma) =1$ and $m(f) \geq 0$ if $f \geq 0$. It is a subspace of $(\ell^\infty \Gamma)^*$. – ARG Apr 18 '16 at 7:48

I don't think you will ever get what you want. I will explain here the way I like to think of this invariant mean. I hope you will find it nice, but I doubt you will find it explicit.

There are two natural classes of compactifications which we associate with a group $\Gamma$ that I would like to recall. A group compactification is just a group homomorphism from $\Gamma$ into a compact topological group. A semi-group compactification is a semi-group homomorphism from $\Gamma$ into a compact semi-topological semi-group (a semi-group endowed with a topology for which both left and right multiplication are (separately) continuous). Both types of compactifications could be seen as objects of corresponding categories in which morphisms are continuous homomorphisms between the target (semi-)groups making the obvious diagram commute. It is not hard to see that both categories have initial objects. The universal group compactification is called the Bohr compactification and the universal semi-group compactification is called the WAP compactification. Not surprisingly, when you pull back to $\Gamma$ the continuous functions on these compact objects you get correspondingly the algebras of Almost Periodic and Weakly Almost Periodic functions.

Let me remark here that AP / WAP functions are the matrix coefficients of finite dimensional / reflexive Banach representations. I will not elaborate on that here, though it is crucial for a better understanding.

Since every group is a semi-group there is a canonical homomorphism $\text{WAP}(\Gamma)\to \text{Bohr}(\Gamma)$. It turns out that there is a section to this map, a homomorphism $\text{Bohr}(\Gamma)\to \text{WAP}(\Gamma)$. Its image is called the Sushkevich kernel of $\text{WAP}(\Gamma)$. It is the unique minimal two sided ideal in $\text{WAP}(\Gamma)$ and it is usually denoted by $K(\Gamma)$. So $K(\Gamma)$ is an isomorphic copy of $\text{Bohr}(\Gamma)$ which happens to live inside $\text{WAP}(\Gamma)$.

As I mentioned before, every WAP functions on $\Gamma$ extends canonically to a continuous function on the compact semi-group $\text{WAP}(\Gamma)$. You can now take this extended function and restrict it to the subgroup $K(\Gamma)$. The resulting function on $K(\Gamma)$ you can average wrt the Haar measure of $K(\Gamma)$. This process takes a WAP function on $\Gamma$ and gives back a scalar. This is the invariant mean you wanted to understand.

Of course, since $\Gamma$ is dense in $\text{WAP}(\Gamma)$, the finitely supported probability measures on $\Gamma$ are w*-dense in $\text{Prob}(\text{WAP}(\Gamma))$, so in particular you could find a net of finitely supported functions in $\ell^1(\Gamma)$ that will converge to the Haar measure on $K(\Gamma)$. But I seriously doubt that you can find a sequence with this property. In general these objects are not metrizable.

• Wow, great answer. – Nik Weaver Apr 14 '16 at 22:11
• Thanks for the great answer! It's always nice to have more perspective on this mean (even non-explicit ones). I would also be happy with an explicit net (for any non-amenable $\Gamma$) if sequences never do the trick. Actually, in the picture you give, it is not at all clear (to me! and I do not [yet] claim that I took the time to understand fully the description you gave) why the uniform measure on any Folner sequence should converge to the Haar measure on $K(\Gamma)$. – ARG Apr 15 '16 at 15:22
• @Nik Weaver, thanks. Antoine, I keep thinking of your question. I don't yet have a better answer. Regrding the question in the comment: (asuming $\Gamma$ is amenable) the sequence of uniform measures on Folner sets will converge to the unique invariant mean (in the space of means on WAP functions) by the standard subsubnet argument. Viewd as measures on $\text{WAP}(\Gamma)$ you get the convrence to the Haar measure on $K( \Gamma)$. – Uri Bader Apr 16 '16 at 19:06
• @user89334 thanks, keep me/us posted! As for the comment-question: it just felt strange that a measure on the Folner sets (which may be taken to spread nicely all over $\Gamma$) converges to the Haar measure on the [closed?] ideal $K(\Gamma)$, given that $\Gamma$ is dense in $\mathrm{WAP}(\Gamma)$. But as I wanted to say, this has to do with my lack of understanding of the phenomenon at hand, not with any mathematical problem. – ARG Apr 18 '16 at 7:43