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Let $C_g:C(\mathbb{R}^d;\mathbb{R}^d)\rightarrow C(\mathbb{R}^d;\mathbb{R}^d)$ be defined by $C_g(f)\triangleq f\circ g$ for some fixed $g \in C(\mathbb{R}^d;\mathbb{R}^d)$.

What are examples of Borel probability measures on $C(\mathbb{R}^d;\mathbb{R}^d)$ which are invariant with respect to $C_g$ and are non-atomic (preferably also locally positive: ie: positive value on non-empty open subsets).

Note: Here, $C(\mathbb{R}^d;\mathbb{R}^d)$ is equipped with the compact-open topology; ie: uniform convergence on compacts.

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  • $\begingroup$ I'm not sure this can have a meaningful answer in this generality. Of course it's trivial to give examples in special cases such as $g(x)=0$ or $g(x)=x+1$ (or anything that has a periodic orbit for that matter), but I suppose that's not what you wanted. $\endgroup$
    – Christian Remling
    Nov 6, 2019 at 17:46
  • $\begingroup$ (Though I find the general case most interesting) and example which I'm also particularly curious about is the case where $g(x)=Ax+b$ and $A\in Mat_{d\times d},\, b \in \mathbb{R}^d$. $\endgroup$
    – ABIM
    Nov 6, 2019 at 18:16
  • $\begingroup$ By the way, the extra assumptions on the measures that you added don't really rule out the trivial examples from periodic orbits (for something like $g(x)=x+1$, say) since you can take averages of measures supported by periodic orbits. $\endgroup$
    – Christian Remling
    Nov 6, 2019 at 18:19
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    $\begingroup$ if g is in the orthogonal group I suppose you can take any proba on R^d and average it by the compact group. $\endgroup$
    – sanette
    Nov 8, 2019 at 9:42
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    $\begingroup$ I'd phrase it in a different way: the distribution of the Ornstein-Uhlenbeck process is shift-invariant. It also satisfies your hypothesis of locality (full support in continuous functions). $\endgroup$
    – D. Thomine
    Nov 9, 2019 at 10:37

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Let us assume that $g$ is invertible. Then everything depends on the "size" of the closed subgroup $K_g$ generated by $g$. If $K_g$ is compact, then its Haar measure and all its right convolutions are invariant with respect to the left $g$-translations. If $G$ is non-compact, then there are no probability $g$-invariant measures. I presume that a similar dichotomy should hold in the non-invertible case as well.

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    $\begingroup$ Isn't $\delta_c$ invariant for any constant function $c$? I think your argument only gives the non-existence of invariant probability measures on $Homeo (\mathbb{R}^d)$ (or similar groups)... $\endgroup$
    – D. Thomine
    Nov 8, 2019 at 21:21
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    $\begingroup$ I said "Let us assume that $g$ is invertible" at the very beginning $\endgroup$
    – R W
    Nov 8, 2019 at 22:15
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    $\begingroup$ Yes, but $f$ needs not be. $\endgroup$
    – D. Thomine
    Nov 9, 2019 at 10:29
  • $\begingroup$ You are right - I was too much fixated on the group case. Concerning your Ornstein-Uhlenbeck example - any stationary process will actually do (either with continuous or with discrete time; in the latter case one can just interpolate between the integer time moments). $\endgroup$
    – R W
    Nov 9, 2019 at 13:30

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