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Is there a fixed-point theorem that implies the following result?

Let $F$ be a nonempty convex set of functions on a discrete group with values in $[0,1]$. Suppose $F$ is invariant with respect to left shifts and closed with respect to the pointwise convergence. Then $F$ contains a constant function.

This statement looks like Ryll-Nardzewski fixed point theorem, but it does not seem to follow from the theorem.

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    $\begingroup$ For amenable groups, the result should follow from a theorem by Mahlon Day; if $X$ is a locally convex topological vector space and $G$ is an amenable group that acts affinely and continuously on a compact convex subset $C$ of $X$, then this action has at least one fixed point. $\endgroup$
    – Joseph Van Name
    Sep 12, 2021 at 22:23
  • $\begingroup$ My group is free; in particular, it is not amenable. $\endgroup$
    – Anton Petrunin
    Sep 12, 2021 at 22:26
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    $\begingroup$ This doesn't seem true. Let $F$ be the free group on $\{a,b\}$ and $X\subset F$ be the words whose last letter is $a$ or $b$. Let $\xi=\delta_a+\delta_b-\delta_{a^{-1}}-\delta_{b^{-1}}\in\ell_1(F)$. Then for any $s\in F$, one has $\langle 1_X,s\xi\rangle\geq1$. Hence the pointwise-topology closed convex hull of $F1_X$ cannot contain constant functions. $\endgroup$
    – Narutaka OZAWA
    Sep 12, 2021 at 23:05
  • $\begingroup$ @NarutakaOZAWA yes, looks like a counterexample -- I need to think about it. BTW, is it a standard example? $\endgroup$
    – Anton Petrunin
    Sep 12, 2021 at 23:58
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    $\begingroup$ @Anton Petrunin: It's my standard way of doing it via the Hahn-Banach theorem. It suggests to look at some $G$-orbit in $\ell_1(G)\cap\{1\}^\perp$. I tried the simplest among the non-amenable. $\endgroup$
    – Narutaka OZAWA
    Sep 13, 2021 at 0:19

2 Answers 2

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The claim does not hold. Let $F$ be the free group on $\{a,b\}$ and $X⊂F$ be the words whose last letter is $a$ or $b$. Let $\xi=\delta_a+\delta_b−\delta_{a^{-1}}−\delta_{b^{-1}}\in\ell_1(F)$. Then for any $s\in F$, one has $\langle 1_X,s\xi\rangle\geq1$. (To see this, view $F$ as the $4$-regular tree and $s\xi$ as a signed characteristic function of the $1$-neighborhood of $s\in F$.) Hence the pointwise-topology closed convex hull of $F1_X$ in $\ell_\infty(F)$ cannot contain constant functions.

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  • $\begingroup$ Narutaka, excuse me, "whose last letter is $a$ or $b$" - doesn't this mean that $X=F$? $\endgroup$
    – Sergei Akbarov
    Sep 13, 2021 at 5:02
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    $\begingroup$ @Sergei Akbarov: I mean not $a^{-1}$ nor $b^{-1}$ (nor $\emptyset$). $\endgroup$
    – Narutaka OZAWA
    Sep 13, 2021 at 6:40
  • $\begingroup$ Ah, OK, thank you! $\endgroup$
    – Sergei Akbarov
    Sep 13, 2021 at 6:53
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    $\begingroup$ @Anton Petrunin: Very interesting. Perhaps, you could post another problem. I do not know which pairs $H \le G$ has the property that every nonempty $G$-invariant pointwise-closed convex set of functions on $\ell_\infty(G/H)$ contains a constant function. By following a standard recipe, one can find several equivalent conditions, but none seems easy to check. $\endgroup$
    – Narutaka OZAWA
    Sep 20, 2021 at 6:48
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    $\begingroup$ I posted it. mathoverflow.net/questions/404616/… $\endgroup$
    – Narutaka OZAWA
    Sep 23, 2021 at 10:15
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The claim does not hold, as the nice example by Naratuka Ozawa shows. The purpose of this answer (or rather, extended comment) is to share a related fixed point theorem.

Theorem: Let $X$ be a non-empty set endowed with two structures:

  1. The structure of a compact convex set, that is a convex structure and a compatible compact topology $T$ such that continuous affine functionals separate the points.
  2. The structure of a separable metric space $(X,d)$.

Assume the following compatibility of the two structures: $d$-balls in $X$ are measurable with respect to the Borel $\sigma$-algebra associated with $T$.

If $G$ is a group that acts on $X$ preserving both structures - the convex structure, the topology $T$ and the metric $d$ - then $G$ has a fixed point in $X$.

Let me illustrate how the theorem above implies the fixed point theorem of Ryll-Nardzewski: for a non-empty weakly compact subset $C$ in a Banach space, there is a point in $C$ which is fixed by all affine isometries of $C$.

It is enough to show that $C^G$ is non-empty for every countable group of affine isometries $G$. Given such a group $G$, fix a point $c\in C$ and set $X$ to be the norm closure of the convex hull of the orbit $Gc$. Take $T$ to be the weak topology and $d$ to be the norm metric on $X$. We are now in a position to apply the above theorem.

Sketch op the proof: As above, we assume as we may that $G$ is countable. We fix a fully supported probability measure $\mu$ on $G$ and denote by $B$ the corresponding Furstenberg-Poisson boundary. Then the $G$ action on $B$ is amenable and metrically ergodic. By amenability and by 1 there exists a measurable, defined a.e, $G$-equivariant map $B\to X$. By the compatibility of 1 and 2, this map is $d$-measurable, thus by metric ergodicity and by 2 this map must be essentially constant. Its essential image is a fixed point.

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  • $\begingroup$ When you say "preserving both structures" it seems literally that $G$ preserves the topology $T$ and the metric $d$ (while not required to preserve the affine structure induced by an embedding as in 1). Is this what you mean? $\endgroup$
    – YCor
    Sep 13, 2021 at 9:57
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    $\begingroup$ @YCor thanks for your comment! In fact, when I wrote up the theorem I had in mind preservation of the convex structure as well, but now when I think of it, it might not be needed by using Tychonoff fixed-point theorem. I will check up the details and update the answer accordingly. Also, I will add a proof, to be safe... $\endgroup$
    – Uri Bader
    Sep 13, 2021 at 11:12
  • $\begingroup$ I clarified and added a proof. The convex structure sure needs to be preserved, e.g there is a fixed point free action of $\mathbb{Z}/2 \ltimes \mathbb{Z}$ on $[0,1]$. $\endgroup$
    – Uri Bader
    Sep 14, 2021 at 7:30
  • $\begingroup$ Are you assuming that $d$ is preserved? (i.e. the action on $(X,d)$ is isometric)? $\endgroup$
    – YCor
    Sep 14, 2021 at 8:08
  • $\begingroup$ @YCor yes. Isn't it clear from the context? $\endgroup$
    – Uri Bader
    Sep 14, 2021 at 8:42

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