Let $G$ be a group acting on a set $X$ such that the Reiter property holds, i.e. $$ \forall S\subset G \text{ finite, }\forall \varepsilon>0,\ \exists\ \varphi\in\mathcal{l}^1(X),\ \forall s\in S: \|s\varphi-\varphi\|_1<\varepsilon\|\varphi\|_1, $$ where $\mathcal{l}^1(X):=\{\psi:X\to\mathbb{R}\ |\ \sum_{x\in X}|\psi(x)|<+\infty\}$ and where if $g\in G$ and $\psi\in\mathcal{l}^1(X)$ we define $g\psi:X\to\mathbb{R},\ x\mapsto \psi(g^{-1}x)$.
I want to show that this implies the Folter property, i.e. $$ \forall S\subset G \text{ finite, }\forall \varepsilon>0,\ \exists\ A\subset X \text{ finite, }\forall s\in S: |sA\Delta A|<\varepsilon|A|. $$
My thoughts: without loss of generality we may assume that $\varphi\geq 0$ and $\|\varphi\|_1=1$ (by replacing it with $|\varphi|/\|\varphi\|_1$). Then for $\delta>0$, we can take $A_\delta\subset X$ finite such that $\varphi(a)>\varphi(x)$ for all $a\in A$, $x\notin A$, and such that $\sum_{a\in A_\delta}\varphi(a)>1-\delta$. Then if $m=\min \varphi(A_\delta)$ and $s=\max \varphi(X\backslash A_\delta)$, we have: $$ (m-s)|sA_\delta\Delta A_\delta|\leq \sum_{x\in sA_\delta\Delta A_\delta}|s\varphi(x)-\varphi(x)|\leq \|s\varphi-\varphi\|_1<\varepsilon $$ So if we could somehow controle the quantity $(m-s)|A_\delta|$, we could finish. Is it possible? How to proceed?
Have a look at Theorem 1 of these notes: https://terrytao.wordpress.com/2009/04/14/some-notes-on-amenability/. (Terry normalizes $\varphi$ so that $\Vert \varphi \Vert_1 = 1$, but otherwise condition (ii) in Theorem 1 is the same as your version of the Reiter property.)