Suppose I have a sequence of stochastic processes $(X_n(t))_{t\in [0,1]}$ such that $X_n(t) \in H$, where $H$ is some separable Hilbert space and $X_n \in C([0,1,], H)$.
Goal. I want to show that $X_n$ converges weakly to a stochastic process $X$, i.e. $\mathrm{Law}(X_n) \rightharpoonup \mathrm{Law}(X)$ as probability measures on $C([0,1],H)$.
Question. Suppose I know that for any dual elements $\phi_1,\ldots,\phi_k \in H'$, where $k$ is finite but arbitrary, it holds that the sequence of processes $(\langle{\phi_1,X_n(t)}\rangle,\ldots,\langle{\phi_k,X_n(t)}\rangle)_{t\in [0,1]} \in C([0,1], \mathbb{R}^k)$ converges weakly to $(\langle{\phi_1,X(t)}\rangle,\ldots,\langle{\phi_k,X(t)}\rangle)_{t\in [0,1]}$. Does this imply that $X_n$ weakly converges to $X$?
If it makes a difference, the desired limit $X$ is a Gaussian process. I suspect the answer is yes by some sort of approximation argument, but I couldn't find a reference in the usual suspects as far as texts on infinite-dimensional stochastic processes. Any help would (in particular, a specific reference if it's true) is greatly appreciated!
Here is a (very degenerate) counterexample.
Let $(e_n)_{n\in\mathbb N}$ be a Hilbert basis of $H$. As is well-known, the sequence $(e_n)_{n\in\mathbb N}$ converges to the zero vector of $H$ for the weak topology $\sigma(H,H')$ but not for the strong topology $\beta(H,H')$ defined by the norm $\lVert\cdot\rVert_H$ of $H$ because $\lVert e_n\rVert_H=1$ for every $n\in\mathbb N$. Such a sequence makes it is possible to obtain a counterexample using only constant processes.
For every $n\in\mathbb N$ and every $t\in[0,1]$, let $X_n(t)$ be everywhere equal to $e_n$ and let $X(t)$ be everywhere equal to the zero vector of $H$. Every path of every $X_n$ and every path of $X$ is continuous (because constant), and $X$ is Gaussian if you acccept that Dirac measures are Gaussian measures with variance zero.
From the weak convergence of $(e_n)_{n\in\mathbb N}$ to the zero vector of $H$, it is easy to see that for every $\phi_1,\dots,\phi_k\in H$, the law of $(\langle\phi_1,X_n\rangle,\dots,\langle\phi_k,X_n\rangle)$ converges to that of $(\langle\phi_1,X\rangle,\dots,\langle\phi_k,X\rangle)$ as $n\to\infty$.
Now consider the continuous bounded function $$f:\begin{array}{ccc}C([0,1],H)&\longrightarrow&\mathbb R\\ \mathbf g&\longmapsto& \inf\Bigl\lbrace 1,\displaystyle\sup_{t\in[0,1]}\lVert\mathbf g(t)\rVert_H\Bigr\rbrace\end{array}.$$ Since $\mathbb E(f(X_n))=1$ for every $n\in\mathbb N$ and $\mathbb E(f(X))=0$, the law of $X_n$ does not converge to that of $X$ as $n\to\infty$.