Could you give an explicit construction of a sequence $\mathbb P_n$ of probability measures on $C[0,\infty]$ which converges in the sense of finite dimensional distributions BUT does not converge weakly?
Also why is the converse always true how to show it?
Thanks a lot!
Take $x_n\in C[0,\infty)$ piecewise linear, with $x_n(n^{-1})=1$, $x_n(0)=0$, $x_n(t)=0$ for $t\geqslant 2n^{-1}$. Take $\mu_n:=\delta_{x_n}$. The finite-dimensional distributions converge weakly to those of $\delta_{\mathbf 0}$.
Indeed, let $t_1,\dots,t_d\in (0,+\infty)$. Then $\mu_n\pi_{t_1,\dots,t_d}^{—1}=\delta_0$ when $n$ is large enough. Gluing $0$ to $t_1,\dots,t_d$ is not a problem.
But these is not weak convergence (because $(x_n,n\geqslant 1)$ doesn't converge uniformly to $0$). Indeed, one can show that if $(X,d)$ is a metric space and $(x_n,n\geqslant 1)$ is a sequence in $X$ and $x\in X$, then $\delta_{x_n}\to \delta_x$ in distribution if and only if $d(x_n,x)\to 0$.
However, for general $\mu_n$ and $\mu$, if $\mu_n\to \mu$ weakly, since the finite-dimensional projection are continuous, we have $\mu_n\pi^{-1}_{t_1,\dots,t_d}\to \mu\pi^{-1}_{t_1,\dots,t_d}$ for all integer $d$ and $t_1,\dots,t_d\in [0,\infty)$.
What was missing in the first example is tightness.