Since there seems to be no progress in this interesting question, I took the liberty to reformulate it in a way, that is easier to understand. Moreover, my answer shows that the question is related to amenable groups. Therefore I changed the title to make it more attractive to group theorists.

$C_b(\mathbb{R})$ resp. $C_0(\mathbb{R})$ are the continuous functions $\mathbb{R} \to \mathbb{R}$ that are bounded resp. vanish at infinity.

Question: Given two homoeomorphisms $g_1,g_2: \mathbb{R} \to \mathbb{R}$ and a continuous linear functional $L: C_b(\mathbb{R}) \to \mathbb{R}$ with the following properties:

  1. The restriction of $L$ to $C_0(\mathbb{R})$ is zero
  2. $L$ is invariant under $g_1,g_2$, i.e. $L(f \circ g_i) = L(f)$ for all $f \in C_b(X)$

What are conditions on $g_1,g_2$ that enforce $L=0$ ?

Original formulation:

I have next situation:

Let $C_b(\mathbb R)=$ { $ f:\mathbb R\rightarrow \mathbb R $ } is the subset of $C(\mathbb R)$ consisting of all bounded continuous functions with a norm $\|\cdot\|$ defined as $\|f\|_\infty=\sup\|f(x)\|$.

Let $C_0(\mathbb R)=$ { $ f:\mathbb R\rightarrow \mathbb R $ } is the subset of $C(\mathbb R)$ consisting of functions such that for every $ε > 0$, there is a compact set $K⊂\mathbb R$ such that $|f(x)| < ε$ for all $x \in \mathbb R\setminus K $. This is usually called the space of functions vanishing at infinity.

There are two homeomorphisms of the line $g_1, g_2$ and a continuous linear positive functional $l$ on $C_b(\mathbb R)$ which is invariant with respect to $g_1, g_2$. Also this functional is permanent: $l([C_0(\mathbb R)])=0$, so $l$ is "concentrate at infinity".

Then we make a Stone-Čech compactification of the $\mathbb R $ designated as $\beta \mathbb R $.

After Stone-Čech compactification of the line, the homeomorphism still will be a homeomorphism and I can show that it will transfer $\mathbb R$ to $\mathbb R$ and the remainder $\mathbb R^* $ to $\mathbb R^* $ ($\mathbb R^* = \beta\mathbb R\setminus\mathbb R $). By the Riesz representation theorem, for our linear functional (already on $\beta\mathbb R$ and still invariant) there is a unique regular countably additive Borel measure $\mu$ on $\beta\mathbb R$. I can show that this measure will be trivial zero at $\mathbb R$. I need to understand under which conditions on the homeomorphisms this measure will be trivial zero at $\mathbb R^* $. I will be very grateful for links on this problem.

UPDATE [12.06.2012] There is a potential result: let $g_1, g_2\in Homeo_+(\mathbb R)$. $g_1$ can be represented as a line $y=x+k$, where $k>0$, and $g_2$ is such that there are two points $t_1,t_2$, for which following conditions are fulfilled $ t_1 < t_2; g_2(t_1)=t_1; g_2(t_2)=t_2; g_2(t)>t$ (or $ g_2(t) < t $ ) for $t \in (t_1, t_2); g_1(t_1) \in (t_1, t_2); g_2(t)$ is an arbitrary monotone increasing curve for $ t\in (-\infty, t_1) \cup (t_2,+\infty) $. Also we have a group $G =< g_1,g_2 >$ with two generators.

If $L$ is a continuous linear functional, which is invariant under $g_1, g_2$, and $L$ was "lowered" from the group $G$ to the $\mathbb R$, and after "lowering" it appears to be permanent (the restriction of $L$ to $C_0(\mathbb R)$ is zero), then $L=0$ and the group $G$ is not amenable.

The definition of "lowering" functional from the group to $\mathbb R$ can be found in section §3.1 of the paper http://arxiv.org/abs/1112.1942 and the proof of the statement is the whole paper. This paper has not yet verified.

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    $\begingroup$ Hi Mariarty. I for one don't understand the question. What "initial problem" are you referring to? I guess it's another question you asked here, in which case it would be helpful to provide a link. Also, it would help to define some of your notation (such as C_b), and to split your question into paragraphs. $\endgroup$ – Tom Leinster Mar 5 '12 at 13:22
  • $\begingroup$ PS - in case you hadn't noticed, there's an "edit" button with which you can do all this. $\endgroup$ – Tom Leinster Mar 5 '12 at 13:22
  • $\begingroup$ @Tom Leinster I think that subject of the "initial problem" doesn't matter. I could not find any references related to measures on remainder and I described the whole problem without expecting help in solving but in hope that someone met something like this in literature. $\endgroup$ – Mariarty Mar 5 '12 at 15:27
  • $\begingroup$ Is this question related to things mentioned here? mathoverflow.net/questions/26821/is-thompsons-group-f-amenable/… $\endgroup$ – Yemon Choi Mar 5 '12 at 15:50
  • 1
    $\begingroup$ To my understanding, if $\mu$ is zero on $\mathbb{R}$ and on $\beta \mathbb{R} \setminus \mathbb{R}$ then $\mu=0$ and hence $l=0$. Conversely, if $l \in C_b(\mathbb{R})^\ast = C(\beta\mathbb{R})^\ast$ is zero, then the corresponding Radon measure is also zero. So, if I'm not missing something, you are looking for conditions on $g_i$ that imply $l=0$ ? $\endgroup$ – Ralph Mar 5 '12 at 15:58

A necessary condition is that the subgroup of homoeomorphisms of $\mathbb{R}$ generated by $g_1,g_2$ is not amenable.

As an example concerning $g_1,g_2$ in the OP's comment above, take $$g_1(t) = t+2\hspace{140pt}$$ $$g_2(t)= g(t-2n) + 2n\;\;\; \text{ if }\;\;\; 2n \le t \le 2n+2$$ where $g:[0,2] \to [0,2]$ is any homeo. with $g(0)=0,g(2) =2$. Then $g_1,g_2$ commute. Hence $\langle g_1,g_2\rangle$ is abelian and therefore amenable. Now apply:

If $G$ is an amenable group of homeomorphisms of $\mathbb{R}$, there is a continuous functional $L$ with $L(1) = 1$ such that $L|C_0(\mathbb{R})=0$ and $L(f \circ g)=L(f)$ for all $g \in G$.

Proof: Suppose $L_0: C_b(\mathbb{R}) \to \mathbb{R}$ is a contiunous functional with $L_0(1) = 1$ and $L_0|C_0(\mathbb{R}) = 0$. Since $G$ is amenable, there is a finitely additive $G$-invariant measure $\mu$ with $\mu(G) = 1$. For each $f \in C_b(\mathbb{R})$, the function $G \to \mathbb{R},\; g \mapsto L_0(f \circ g)$ is bounded. Hence we can define $$L: C_b(\mathbb{R}) \to \mathbb{R}; \; f \mapsto \int_G L_0(f \circ g)\; d\mu(g).$$ It's easy to see that $L$ has the desired properties.

In order to obtain $L_0$ let $C_l(\mathbb{R})$ be the space of the continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that $\lim_{t \to +\infty}f(t)$ exists. Then $$L_0: C_l(\mathbb{R}) \to \mathbb{R},\; f \mapsto \lim_{t \to +\infty}f(t)$$ is continous, linear with $L_0|C_0(\mathbb{R}) = 0$ and $L_0(1)=1$. Finally, since $C_l(\mathbb{R})$ is a closed subspace of $C_b(\mathbb{R})$, $L_0$ can be continuously extended to $C_b(\mathbb{R})$ by the Hahn-Banach theorem.

  • $\begingroup$ probably in the statement you mean $L|C_0(\mathbb R)=0$. $\endgroup$ – Valerio Capraro Mar 22 '12 at 7:42
  • $\begingroup$ @Ralph Great thanks for realized calculations. But your result is exactly what I called as "initial problem". I've already done the way back from your result to my question. But this is good news: I'm on the right path. Thank you again, question can be closed. $\endgroup$ – Mariarty Mar 22 '12 at 18:47
  • $\begingroup$ Good to hear that you are on the right path. Would you mind sharing your insights with us, maybe in form of an arxiv-link or so, when you finished your work on the problem ? $\endgroup$ – Ralph Mar 22 '12 at 19:18
  • $\begingroup$ Сertainly I will $\endgroup$ – Mariarty Mar 23 '12 at 14:33
  • $\begingroup$ I've updated the main question with potential result. $\endgroup$ – Mariarty Jun 12 '12 at 6:21

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