Existence of Folner Sequence Implies Left Invariant Mean

amenabilityfunctional-analysishahn-banach-theorem

Let $G$ be a countable group. Let $A_1, A_2, \dots$ be finite subsets of $G$ such that for each $x \in G, \frac{|(x \cdot A_n) \Delta A_N|}{|A_n|} \to 0$ (a Folner sequence). Our goal is to show that $G$ has a left invariant-mean $\lambda$, i.e., a linear functional on $\ell^{\infty}(G)$ that is

  1. Non-negative, if $f$ is a bounded, complex valued function on $G$ such that $f(s) \geq 0$ for every $s \in G$, then $\lambda(f) \geq 0$.
  2. $\lambda(1) = 1$.
  3. $\lambda(f) = \lambda(\tau_xf)$, for every $x \in G$, where $(\tau_xf)(s) := f(x^{-1}s)$ (left invariance).

I am trying to follow the presentation by Tao here in the proof of theorem $1$. "Use Hahn-Banach Theorem to select an "infinite mean" $\rho \in \ell^{\infty}(\mathbb{N})^*\setminus\ell^1(\mathbb{N})$, and define,

$$
\lambda(m) = \rho((\langle m,\frac{1}{|A_n|}1_{A_N}\rangle)_{n\in \mathbf{N}}),
$$
where the inner product of $f,g \in \ell^{\infty}(G)$ is $\sum_{s \in G}f(s)g(s)$, when it converges.

My progress and Question
I can see that $\lambda$ does define a linear functional on $\ell^{\infty}(G)$ by using the finite-ness of the $A_n$'s.

  • By Hahn-Banach, I may choose $\rho$ so that $\rho((1,1,1,…)) = 1$, which in turn ensures $\lambda(1) = 1$ (property 2 above).
  • We may be able to use the Geometric Hahn Banach somehow. If $F \subset \ell^{\infty}(\mathbb{N})$ is the set of positive functions, then $F$ is convex, is a cone, and closed. My question Can I take $A = \{\}$, which is open, convex and $A \cap F = \{\}$, and say by the Hahn Banach Separation theorem that there is some functional $\phi$ with $\phi(f) \geq 0$ for $f \in F$? Scaling this $\phi$ would also satisfy $\phi(1) = 1$. But I am not sure if we can get left-invariance from here.

Any solution/hint to construct the mean $\lambda$ would be appreciated!

Best Answer

I don't really understand what $A$ is supposed to be in your attempt. However, let me explain what I think is a natural approach here.

The following is a classical exercise in functional analysis (the proof uses Hahn-Banach)

Exercise: Show that there exist a Banach limit on $\ell^\infty$, i.e. a bounded linear functional of norm one, namely $\phi\in(\ell^\infty)^*$ with the following properties:

  1. if $\ell^\infty\ni x\ge0$, i.e. $x=(x_n)$ with $x_n\ge0$, then $\phi(x)\ge0$
  2. If $x=(x_n)\in\ell^\infty$ is convergent, then $\phi(x)=\lim_{n\to\infty}x_n$.
  3. If $x=(x_n)\in\ell^\infty$, then $\phi(x_1,x_2,x_3,\dots)=\phi(x_2,x_3,\dots)$ (the sequence where we just ignored the 1st term of $x$).

Now suppose that you have solved the exercise; I do not include a proof as there are tons of proof online, just search for existence of Banach limits if you cannot solve it yourself (I would advise you to try it, it's a fun exercise; here is a hint though: you might need to use this consequence of Hahn-Banach). You can get your invariant mean for free: define $\lambda:\ell^\infty(G)\to\mathbb{R}$ by $$\lambda(f)=\phi\bigg(\bigg\{\frac{1}{|A_n|}\sum_{g\in A_n}f(g)\bigg\}_{n=1}^\infty\bigg)$$

It is trivially verified that $\lambda$ is a bounded linear functional, $\lambda(1)=1$ and $f\ge0\implies\lambda(f)\ge0$. So all one needs to verify is that $\lambda$ is translation invariant. For that we will use property (3) of $\phi$.

Fix $s\in G$ and $f\in\ell^\infty(G)$. For any $n\ge1$, do the estimate: $$\bigg|\frac{1}{|A_n|}\sum_{g\in A_n}f(g)-\frac{1}{|A_n|}\sum_{g\in A_n}\tau_s(f)(g)\bigg|=\frac{1}{|A_n|}\bigg|\sum_{g\in A_n}f(g)-\sum_{g\in s^{-1}A_n}f(g)\bigg|\le\frac{|A_n\triangle s^{-1}A_n|}{|A_n|}\|f\|_\infty $$ (convince yourself that the above estimate is true). Now given $\varepsilon>0$, find $N$ large enough so that, for all $n\ge N$ we have $$\frac{|A_n\triangle s^{-1}A_n|}{|A_n|}\|f\|_\infty<\varepsilon$$ And, finally, by iterating property (3) of $\phi$ $N$ times, you get that $$\lambda(f)=\phi\bigg(\bigg\{\frac{1}{|A_n|}\sum_{g\in A_n}f(g)\bigg\}_{n=N}^\infty\bigg)$$ and that $$\lambda(\tau_s(f))=\phi\bigg(\bigg\{\frac{1}{|A_n|}\sum_{g\in A_n}\tau_s(f)(g)\bigg\}_{n=N}^\infty\bigg)$$ and thus $$|\lambda(f)-\lambda(\tau_s(f))|\leq\sup_{n\ge N}\frac{|A_n\triangle s^{-1}A_n|}{|A_n|}\|f\|_\infty<\varepsilon$$ As $\varepsilon$ is arbitrary, allowing it to approach $0$ yields $\lambda(f)=\lambda(\tau_s(f))$.

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