How is this property equivalent to the Reiter Property?

Question

We have the Reiter Property $(R_2)$ for an action of a group G on a set X: For any $\epsilon>0$, any finite subset $S$ of G, there exists $\phi\in{\ell^2(X)}$ such that $\|s\phi-\phi\|_{\ell^2}<\epsilon{\|\phi\|_{\ell^2}}$ for all $s\in{S}$. I am trying to show this is equivalent to the alternative property $(R_2)'$: for any $\epsilon>0$, any finite subset $S$ of G, there exists $\phi\in{\ell^2(X)}$ such that $$\left\|\frac{1}{|S|}\sum_{s\in{S}}{s\phi}\right\|_{\ell^2}>(1-\epsilon)\|\phi\|_{\ell^2}$$ but I am completely stuck. I have tried using some uniform convexity since $\ell^2$ has an inner product, but can only get anything out of it when $|S|=2$, I have heard from someone else that this is related to adjoint operators, so I have tried defining $T:\ell^2(X)\rightarrow\ell^2(X)$ by $T(\phi)=\frac{1}{|S|}\sum_{s\in{S}}{s\phi}$ and can deduce that it has norm 1 and, if we extend S to also contain the inverses of all its elements, is self adjoint, but I can't see how this could be helpful to solve the problem.

Many thanks.

Answer 1

It is enough to prove the equivalence for finite sets $S$ such that $S = S^{-1}$ and $1 \in S$. Given a such a set, let $T := \frac{1}{|S|} \sum_{s \in S} s \phi$. As you point out, this is a self-adjoint operator of norm at most $1$, and $(R_2)'$ is equivalent to: $\| T \| = 1$ for all $S$.

Suppose $(R_2)$. Then for all $S$ there exists a sequence $\phi_i$ such that $\| \phi_i \| = 1$ and $\| s \phi_i - \phi_i \| \to 0$ for all $s \in S$. It follows that $\| T \phi_i - \phi_i \| \to 0$. Therefore $\| T \phi_i \| \geq \| \phi_i \| - \| T \phi_i - \phi_i \| \to 1$, and $(R_2)'$ follows.

Suppose $(R_2)'$. Since $T$ is self-adjoint and has norm $1$, we have $1 = \| T \| = \sup_{\| \phi \| = 1} | \langle T \phi, \phi \rangle |$. So there exists a sequence $\phi_i$ with $\| \phi_i \| = 1$ and $|\langle T \phi_i, \phi_i \rangle | \to 1$. Developing and using the Cauchy-Schwarz inequality:

$$|\langle T \phi_i, \phi_i \rangle| \leq \frac{1}{|S|}\sum_{s \in S} |\langle s \phi_i, \phi_i \rangle| \leq \frac{1}{|S|} \sum_{s \in S} \| s \phi_i \| \cdot \|\phi_i \| = 1.$$

Since the left-hand-side converges to $1$, this implies that $|\langle s \phi_i, \phi_i \rangle| \to 1$ for all $s \in S$.

The idea now is to use an approximate version of the equality case from the Cauchy-Schwarz inequality. More precisely, (see e.g. Proof 2 of Cauchy-Schwarz from wikipedia), if $u$ and $v$ have norm $1$, then we have: $|\langle u, v \rangle^2 - 1| = \| u - \langle u, v \rangle v \|^2$. So in our case taking $u = s \phi_i$ and $v = \phi_i$, and letting $\lambda_s = \lim_{i \to \infty} \langle s \phi_i, \phi_i \rangle = \pm 1$ we obtain $$\lim_{i \to \infty} \| s \phi_i - \lambda_s \phi_i \| = \lim_{i \to \infty} \| s \phi_i - \langle s \phi_i, \phi_i \rangle \phi_i \| = 0.$$ We are left to take care of the sign. First, note that $\lim_{i \to \infty} \langle T \phi_i, \phi_i \rangle = \frac{1}{|S|} \lambda_s$, and since the left-hand side is $\pm 1$ this shows that all of the $\lambda_s$ are equal. But clearly $\lambda_1 = 1$ (recall that we are assuming $1 \in S$) and so $\lambda_s = 1$ for all $s \in S$.