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Let $K$ be a compact set of $\mathbb{R}^n$ $(n\geq 1)$ and $v\in\mathcal{C}^1(K,\mathbb{R}^n)$ be a speed field on $K$ such that for any initial condition $x_0\in K$, the following dynamical system

\begin{equation} \begin{cases} x(0)=x_0 \\ \dot{x}(t)=v\big(x(t)\big),\>t\geq 0 \end{cases} \end{equation}

has a solution $x$ defined on $\mathbb{R}_+$. Hence, there exists a probability measure on $K$, denoted $\mu_{|x_0}$, which is invariant for this dynamical system, and the pointwise ergodic theorem states that for any $f\in\mathcal{C}^0(K,\mathbb{R})$

\begin{equation} \lim\limits_{T\rightarrow+\infty}\frac{1}{T}\int_0^T f\big(x(t)\big)\,\mathrm{d}t=\int_K f\,\mathrm{d}\mu_{|x_0}. \end{equation}

Given $f$, if $v$ is nowhere-vanishing, is this convergence uniform toward the choice of $x_0\in K$?

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  • $\begingroup$ This is not what the usual ergodic theorem says. Rather, there is a fixed measure $\mu$ (independent of $x_0$), and you have convergence for almost all $x_0$. en.wikipedia.org/wiki/Ergodic_theory#Ergodic_theorems $\endgroup$
    – Christian Remling
    Commented Dec 3, 2020 at 16:57
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    $\begingroup$ In any event, the convergence is not uniform in $x_0$. Consider for example a 1D system on $K=[0,1]$, with $v(0)=v(1)=0$ and $v>0$ otherwise. So you move from $x=0$ to $x=1$, and $\mu=\delta_1$, but this will take very long if you start close to $0$. (If you want an example with $v\not=0$, you can make this two-dimensional and approach periodic orbits instead of points.) $\endgroup$
    – Christian Remling
    Commented Dec 3, 2020 at 17:00
  • $\begingroup$ If $\mu$ is invariant with support $\text{supp}(\mu)$ and $v$ is still nowhere vanishing, does this convergence happen uniformly toward the choice of $x_0\in\text{supp}(\mu)$? $\endgroup$
    – G. Panel
    Commented Dec 3, 2020 at 22:52
  • $\begingroup$ No, it doesn't (e.g. you can have mutually singular invariant measures with full support). You basically get uniform convergence only in the case where the system has a unique invariant probability measure (unique ergodicity), but that's a pretty specific property. $\endgroup$
    – D. Thomine
    Commented Dec 4, 2020 at 6:32
  • $\begingroup$ Can you give a hint for exhibiting such a counter example please? (still considering that the $\mathcal{C}^1$ speed field is non-vanishing on a compact) $\endgroup$
    – G. Panel
    Commented Dec 4, 2020 at 16:25

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