The answer is yes for separable Hilbert spaces.

If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\langle e_n,e_m \rangle_{\rm new}$ for $n,m \in \mathbb N$. Assuming Folner's condition, you know that

$$\frac{1}{n} \sum_{i=1}^n \langle \pi(i)e_n,\pi(i)e_m \rangle$$

is a bounded sequence which defines an asymptotically $\mathbb Z$-invariant inner product. So, using the Axiom of Dependent Choice, which allows you to make a countable number of dependent choice (like -- for example - if you choose a convergent subsequences in a bounded sequence), you can now choose a sequence of integers $(n_k)_{k \in \mathbb N}$ such that

$$\frac{1}{n_k} \sum_{i=1}^{n_k} \langle \pi(i)e_n,\pi(i)e_m \rangle$$

will converge for all $n,m \in \mathbb N$ as $k \to \infty$. This of course needs the usual diagonalization procedure, but no futher set theoretic complications.

Solovay's model of ZF + Dependent Choice has the property that all subsets of $\mathbb R$ are Lebesgue measurable. Hence, in this world uniformly bounded representations of $\mathbb Z$ (and in fact any amenable group $G$) on separable Hilbert spaces are unitarisable and at the same time, all subsets of the real numbers are Lebesgue measurable.

Now, one may ask whether separability is an issue or not, but that is surprisingly unclear. It (even with AC) is not clear if you can decompose every uniformly bounded representation into cyclic representations or even in somewhat smaller pieces. In particular, it is unclear whether the above result implies something for arbitrary uniformly bounded representations. In some sense, this is not surprising, since one is limiting oneself to situations with additional countability assumptions.

The answer is yes for separable Hilbert spaces.

If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\langle e_n,e_m \rangle_{\rm new}$ for $n,m \in \mathbb N$. Assuming Folner's condition, you know that

$$\frac{1}{n} \sum_{i=1}^n \langle \pi(i)e_n,\pi(i)e_m \rangle$$

is a bounded sequence which defines an asymptotically $\mathbb Z$-invariant inner product. So, using the Axiom of Dependent Choice, which allows you to make a countable number of dependent choice (like -- for example - if you choose a convergent subsequences in a bounded sequence), you can now choose a sequence of integers $(n_k)_{k \in \mathbb N}$ such that

$$\frac{1}{n_k} \sum_{i=1}^{n_k} \langle \pi(i)e_n,\pi(i)e_m \rangle$$

will converge for all $n,m \in \mathbb N$ as $k \to \infty$. This of course needs the usual diagonalization procedure, but no futher set theoretic complications.

Solovay's model of ZF + Dependent Choice has the property that all subsets of $\mathbb R$ are Lebesgue measurable. Hence, in this world uniformly bounded representations of $\mathbb Z$ (and in fact any amenable group $G$) on separable Hilbert spaces are unitarisable and at the same time, all subsets of the real numbers are Lebesgue measurable.

Now, one may ask whether separability is an issue or not, but that is surprisingly unclear. It (even with AC) is not clear if you can decompose every uniformly bounded representation into cyclic representations or even in somewhat smaller pieces. In particular, it is unclear whether the above result implies something for arbitrary uniformly bounded representations. In some sense, this is not surprising, since one is limiting oneself to situations with additional countability assumptions.

The answer is yes for separable Hilbert spaces.

If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\langle e_n,e_m \rangle_{\rm new}$ for $n,m \in \mathbb N$. Assuming Folner's condition, you know that

$$\frac{1}{n} \sum_{i=1}^n \langle \pi(i)e_n,\pi(i)e_m \rangle$$

is a bounded sequence which defines an asymptotically $\mathbb Z$-invariant inner product. So, using the Axiom of Dependent Choice, which allows you to make a countable number of dependent choice (like -- for example - if you choose a convergent subsequences in a bounded sequence), you can now choose a sequence of integers $(n_k)_{k \in \mathbb N}$ such that

$$\frac{1}{n_k} \sum_{i=1}^{n_k} \langle \pi(i)e_n,\pi(i)e_m \rangle$$

will converge for all $n,m \in \mathbb N$ as $k \to \infty$. This of course needs the usual diagonalization procedure, but no futher set theoretic complications.

Solovay's model of ZF + Dependent Choice has the property that all subsets of $\mathbb R$ are Lebesgue measurable. Hence, in this world uniformly bounded representations of $\mathbb Z$ (and in fact any amenable group $G$) on separable Hilbert spaces are unitarisable and at the same time, all subsets of the real numbers are Lebesgue measurable.

Now, one may ask whether separability is an issue or not, but that is surprisingly unclear. It is not clear if you can decompose every uniformly bounded representation into cyclic representations or even in somewhat smaller pieces. In particular, it is unclear whether the above result implies something for arbitrary uniformly bounded representations. In some sense, this is not surprising, since one is limiting oneself to situations with additional countability assumptions.

The answer is yes for separable Hilbert spaces.

If the Hilbert space is separable with basis $\lbrace e_n \mid n \in \mathbb N\rbrace$, you only have to fix countably many inner products and define $\langle e_n,e_m \rangle_{\rm new}$ for $n,m \in \mathbb N$. Assuming Folner's condition, you know that

$$\frac{1}{n} \sum_{i=1}^n \langle \pi(i)e_n,\pi(i)e_m \rangle$$

is a bounded sequence which defines an asymptotically $\mathbb Z$-invariant inner product. So, using the Axiom of Dependent Choice, which allows you to make a countable number of dependent choice (like -- for example - if you choose a convergent subsequences in a bounded sequence), you can now choose a sequence of integers $(n_k)_{k \in \mathbb N}$ such that

$$\frac{1}{n_k} \sum_{i=1}^{n_k} \langle \pi(i)e_n,\pi(i)e_m \rangle$$

will converge for all $n,m \in \mathbb N$ as $k \to \infty$. This of course needs the usual diagonalization procedure, but no futher set theoretic complications.

Solovay's model of ZF + Dependent Choice has the property that all subsets of $\mathbb R$ are Lebesgue measurable. Hence, in this world uniformly bounded representations of $\mathbb Z$ (and in fact any amenable group $G$) on separable Hilbert spaces are unitarisable and at the same time, all subsets of the real numbers are Lebesgue measurable.

Now, one may ask whether separability is an issue or not, but that is surprisingly unclear. It (even with AC) is not clear if you can decompose every uniformly bounded representation into cyclic representations or even in somewhat smaller pieces. In particular, it is unclear whether the above result implies something for arbitrary uniformly bounded representations. In some sense, this is not surprising, since one is limiting oneself to situations with additional countability assumptions.