Fix a filtered probability space $(\Omega,\mathscr{A}, (\mathscr{A}_t)_{t\geqslant 0},\mathbb{P})$ and let $(X_t)_{t\geqslant 0}:\Omega\times\mathbb{R}_+\to\mathbb{R}^d$ be an adapted stationary continuous time stochastic process (initialised at its stationary measure). Consider a (countable) class $\mathcal{F}\subseteq L^2$ and let $Z_T(f)$ be the empirical process \begin{equation} Z_T(f)=\frac{1}{\sqrt{T}}\int_0^Tf(X_t)\mathrm{d}t-\sqrt{T}\mathbb{E}f(X_0),\quad T>0,\quad f\in \mathcal{F}. \end{equation} Question: Are there known conditions (e.g., from books or articles) on the process $(X_t)_{t\geqslant 0}$ such that $Z_T$ converges in $\ell^\infty(\mathcal{F})$ to a tight Borel measurable map $Z$? I am especially interested in the case where $(X_t)_{t\geqslant 0}$ is mixing, or when $X$ is an Itô semimartingale.
Indeed, replacing $(X_t)_{t\geqslant 0}$ with an i.i.d. sequence (or a mixing sequence with appropriate mixing conditions), then $Z_n(f)=\frac{1}{\sqrt{n}}\sum_{i=1}^nf(X_i)-\mathbb{E}f(X_i)$ converges weakly to a Gaussian process in $\ell^\infty(\mathcal{F})$ granted $\mathcal{F}$ is sufficiently regular in terms of its metric entropy. Hence, if $(X_t)_{t\geqslant 0}$ is also mixing, then $Y_n=\int_{n-1}^nX_t\mathrm{d}t$ defines a stationary sequence which is also mixing, and so if $Z_n^*(f)=\frac{1}{\sqrt{n}}\sum_{i=1}^nf(Y_i)-\mathbb{E}f(Y_i)$ converges weakly, then I hypothesise that $Z_T$ perhaps does too, but I am unable to come up with a strategy to (dis)prove this claim.
