In the study of weak convergence in $C[0,1]$, a common example is always being considered: $$X\_{n}(t)=nt1\_{[0,1/n]}(t)+(2-nt)1\_{(1/n,2/n]}(t).$$ This example serves a counter-example to show that the weak convergence of the finite dimensional distribution of a process $X\_{n}(t)$ does not imply the weak convergence in $C[0,1]$.


I want to start from here to construct a counterexample showing that weak convergence the f.d.d of a Gaussian process does not imply the weak convergence in $C[0,1]$.


My idea is to construct some similar $X\_{n}(t)$ so that it is Gaussian. Then if I showed that $\mathbb{E}X\_{n}(t)\longrightarrow\mathbb{E}X\_{\infty}(t)$ and $cov(X\_{n}(t), X\_{n}(s))\longrightarrow cov(X\_{\infty}(t),X\_{\infty}(t)),$ we see that the f.d.d. converges weakly. Then, I try to argue that they do not converge weakly in $C[0,1]$.


My first thought was to multiply the standard normal random variable: $$X\_{n}(t):=nt\xi\_{1}(t)1\_{[0,1/n]}(t)+(2-nt)\xi\_{2}(t)1\_{(1/n,2/n]}(t),$$ where $\xi\_{1}(t)$ and $\xi\_{2}(t)$ are independent standard normal. 


However, I found that this process is not even continuous at $t=\frac{1}{n}$, then I got stuck..


Also, what extra condition I can add (in addition to the convergence of mean and covariance function) so that the Gaussian process do converge weakly in $C[0,1]$? 


Any idea? Thank you!



The weak convergence of finite dimensional distribution of Gaussian process does not imply the weak convergence in $C[0,1]$
The process given by 
$$X\_n(t):=[nt1\_{[0,1/n]}(t)+(2-nt)1\_{(1/n,2/n]}(t)]Z,$$
where $Z\sim N(0,1)$, is a Gaussian process in $C[0,1]$ whose finite-dimensional distributions weakly converge to those of the zero process. 


However, the process $X\_n(\cdot)$ does not converge in distribution to any process. Indeed, since the finite-dimensional distributions of a process in $C[0,1]$ determine the distribution of the process, the limit process here can only be the zero process, given by $X(t)=0$. By one of the equivalent definitions, the convergence $X\_n(\cdot)$ to $X(\cdot)$ in distribution means 
$$Ef(X\_n(\cdot))\to Ef(X(\cdot))$$
for any bounded continuous function $f\colon C[0,1]\to\mathbb R$. Take now the bounded continuous function $f\colon C[0,1]\to\mathbb R$ defined by 
$$f(x):=\min(1,\max\_{t\in[0,1]}|x(t)|)$$
for $x\in C[0,1]$. Then 
$$Ef(X\_n(\cdot))=E\min(1,|Z|)\not\to0=Ef(X(\cdot)).$$
So, $X\_n(\cdot)$ does not converge to $X(\cdot)$ in distribution. 


