Consider the interpolated and scaled random walk generated by the independent random variables $(\xi_i)_{i \ge 1}$ with mean zero and finite variance:
$$ X^{(n)}_t := \sum_{i=1}^{\lfloor tn \rfloor } \xi_i + (tn - \lfloor tn \rfloor )\xi_{\lfloor tn \rfloor +1}$$
There is the known invariance principle that states that $( X^{(n)}_t )_{t \ge 0}$ converges weakly to a Brownian motion.
I have the following question: if one considers one path $ X^{(n)}_t(\omega)$ of $ X^{(n)}_t $, as $n $ increases, it seams that this path is continuously changing as an element of $C[0,1]$ (as a function in $t $) and thus never converges - except for the case when for some $N $
$$\frac{1}{\sqrt n}\sum_{i \ge N}^n \xi_i(\omega) \ \overset{n} \to 0 ,$$
in which case it converges to the constant zero function.
Is it possible to give a short argument that almost surely no path of $ X^{(n)}_t $ converges to any function in $C[0,1]$ in the metric $\sup_{0 \le t 1}\| \cdot \|$?
Most grateful for any help provided!
You can use the Kolmogorov or the Hewitt-Savage 0-1 law. Namely, if the limit of $X_t^n$ existed, it would be, thanks to any of these 0-1 laws, a nonrandom function. This would imply that $X_t^n$ converges to this nonrandom function also in law, contradicting the invariance principle.
Actually, it is possible to show that the closure of $\{X^n,n\ge 1\}$ is almost surely the whole $C_0[0,1]$.
I didn't find a reference to the fact, but here is a simple argument through the 0-1 law. Take a dense countable set $D\subset C_0[0,1]$ and let $\mathcal U$ be the collection of open balls centered at elements of $D$ with rational radii. It is not hard to see that for any $U\in \mathcal U$,
$$ A_U:=\{X^n \in U \text{ infinitely often}\} = \limsup_{n\to\infty}\{X^n\in U\} $$ is stable under finite permutations of $(\xi_n,n\ge1)$. Therefore, by the Hewitt-Savage 0-1 law, $\mathrm P(A_U)\in \{0,1\}$. If $\mathrm P(A_U)=0$, then, using the portmanteau theorem and the Fatou lemma for probabilities, we get that $$ \mathrm P(W \in U) \le \limsup_{n\to\infty} \mathrm P(X^n\in U) \le \mathrm P\Bigl(\limsup_{n\to\infty}\{X^n\in U\}\Bigr) = \mathrm P(A_U)= 0. $$ This, however, contradicts the well-known fact that the support of the distribution of $W$ is $C_0[0,1]$. As a result, for any $U\in\mathcal U$, $\mathrm P(A_U)=1$. Since $\mathcal U$ is countable, this implies that $\mathrm P\left(\bigcap_{U\in \mathcal U} A_U\right)=1$, whence it easily follows that $$ \mathrm P\bigl(\text{the closure of }\{X^n\}\text{ is } C_0[0,1] \bigr) = 1. $$
Here is an explanation why $A_U$ is stable under finite permutations of $\xi_n$ for any $U\in \mathcal U$.
Let $U = B(f,q)$ with $f\in D$ and $q\in \mathbb Q\cap (0,\infty)$. There exists some $a>0$ such that $|f(x)|<q/2$ for $x\in [0,a]$. Set $\delta = \min(a,q/2)$.
Fix some $m\ge 1$. For $n> \delta^{-2}\left(\sum_{i=1}^m|\xi_i|\right)^2$, $\max_{t\in[0,m/n]} |X_t^n|<\delta$. And if additionally $n\ge m/\delta$, then $$\max_{t\in[0,m/n]} |X_t^n - f(t)|\le \max_{t\in[0,m/n]} |X_t^n| + \max_{t\in[0,\delta]}|f(t)| <q.$$ Therefore, for such $n$, the fact $X^n\in U$ depends only on its values on $[m/n,1]$. However, permuting $\xi_1,\dots, \xi_m$ will not change the values of $X^n$ on $[m/n,1]$, implying the invariance of $A_U$ under any permutations of the first $m$ elements.