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For the purpuse of this question, a group is amenable iff there exists a Foelner sequence.

Dixmier unitarisability problem asks whether a (countable discrete) group G is amenable iff every bounded representation of G on a Hilbert space is unitarisable.

The problem is currently not solved, but at least argument "=>" is not very difficult: let G be amenable and let $\rho\colon G \curvearrowright H$ be a bounded representation. We need to make a new scalar product on $H$ for which $\rho$ is unitary, and such that vectors of norm one in the old product have norms bounded from above in the new product (so that the identity mapping on $H$ is continuous) and bounded from below in the new product (so that $H$ with the new product is complete).

Let $F$ be a mean on $G$. For a pair of vectors $v,w$ consider a function $f_{vw}$ on $G$: $g \mapsto \langle gv, gw \rangle$, define the new scalar product as $\langle v, w \rangle_{new} := F(f_{vw})$. It's easy to check that this new product has the desired properties.

But in the proof we used axiom of choice, because we use the mean on $G$. I tried to figure out a proof which uses less than a full or almost full axiom of choice, but so far I failed.

Question. Is the above fact implication true in a set of axioms in which also the statement "every subset of R is measurable" is true?

Maybe it's easier to answer this question specifically for the infinite cyclic group Z. Note that Z is amenable, according to our definition of amenability, in Zermelo-Fraenkel set theory.

Even if the anser is no, I'd be interested to learn about the proof of the above fact which uses less than Boolean prime ideal theorem (AFAIU using only BPIM is easy, because to prove existence of a mean one uses Banach-Aleoglu theorem, which uses Tychonoff theorem for Hasudorff space, which, according to wikipedia, uses only BPIM)

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For the purpuse of this question, a group is amenable iff there exists a Foelner sequence.

Dixmier unitarisability problem asks whether a (countable discrete) group G is amenable iff every bounded representation of G on a Hilbert space is unitarisable.

The problem is currently not solved, but at least argument "=>" is not very difficult: let G be amenable and let $\rho\colon G \curvearrowright H$ be a bounded representation. We need to make a new scalar product on $H$ for which $\rho$ is unitary, and such that vectors of norm one in the old product have norms bounded from above in the new product (so that the identity mapping on $H$ is continuous) and bounded from below in the new product (so that $H$ with the new product is complete).

Let $F$ be a mean on $G$. For a pair of vectors $v,w$ consider a function $f_{vw}$ on $G$: $g \mapsto \langle gv, gw \rangle$, define the new scalar product as $\langle v, w \rangle_{new} := F(f_{vw})$. It's easy to check that this new product has the desired properties.

But in the proof we used axiom of choice, because we use the mean on $G$. I tried to figure out a proof which uses less than a full or almost full axiom of choice, but so far I failed.

Question. Is the above fact true in a set of axioms in which also the statement "every subset of R is measurable" is true?

Maybe it's easier to answer this question specifically for the infinite cyclic group Z. Note that Z is amenable, according to our definition of amenability, in Zermelo-Fraenkel set theory.

Even if the anser is no, I'd be interested to learn about the proof of the above fact which uses less than Boolean prime ideal theorem (AFAIU using only BPIM is easy, because to prove existence of a mean one uses Banach-Aleoglu theorem, which uses Tychonoff theorem for Hasudorff space, which, according to wikipedia, uses only BPIM)

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Is every bounded representation of Z unitarisable when all sets are measurable?

Dixmier unitarisability problem asks whether a (countable discrete) group G is amenable iff every bounded representation of G on a Hilbert space is unitarisable.

The problem is currently not solved, but at least argument "=>" is not very difficult: let G be amenable and let $\rho\colon G \curvearrowright H$ be a bounded representation. We need to make a new scalar product on $H$ for which $\rho$ is unitary, and such that vectors of norm one in the old product have norms bounded from above in the new product (so that the identity mapping on $H$ is continuous) and bounded from below in the new product (so that $H$ with the new product is complete).

Let $F$ be a mean on $G$. For a pair of vectors $v,w$ consider a function $f_{vw}$ on $G$: $g \mapsto \langle gv, gw \rangle$, define the new scalar product as $\langle v, w \rangle_{new} := F(f_{vw})$. It's easy to check that this new product has the desired properties.

But in the proof we used axiom of choice, because we use the mean on $G$. I tried to figure out a proof which uses less than a full or almost full axiom of choice, but so far I failed.

Question. Is the above fact true in a set of axioms in which also the statement "every subset of R is measurable" is true?

Maybe it's easier to answer this question specifically for the infinite cyclic group Z.

Even if the anser is no, I'd be interested to learn about the proof of the above fact which uses less than Boolean prime ideal theorem (AFAIU using only BPIM is easy, because to prove existence of a mean one uses Banach-Aleoglu theorem, which uses Tychonoff theorem for Hasudorff space, which, according to wikipedia, uses only BPIM)