I have $$X\_i \sim N(0,1), \quad S\_n=X\_1+\cdots+X\_n,$$
$$ \mathscr{S}\_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\_{i=1}^{n} \left[ S\_{i-1} (\omega ) + n \left( t - \frac{i-1}{n} \right) X\_{i}(\omega) \right] \textbf{1}\_{ \left( \frac{i-1}{n}, \frac{i}{n} \right] } (t) $$ and let $$d(f,g) := \sup\_{x \in [0,1]}| f(x) - g(x) | .$$


Then $$ Y\_i^{(v)} := X\_i \textbf{1}\_{\{ |X\_i| \leq v \} }, 
 X\_i^{(v)} := \left( Y\_i^{(v)} - E [ Y\_1^{(v)} ] \right) $$ analogically we define
$$ S\_n^{(v)} = \sum\_{i=1}^n X\_i \textbf{1}\_{ \left\{ \left| X\_i \right| \leq v \right\} } - E \left[ \sum\_{i=1}^n X\_i \textbf{1}\_{ \left\{ \left| X\_i \right| \leq v \right\} } \right] $$ and $\mathscr{S}\_n^{(v)}$.
In the lemma 3.2 (lecture notes on Donsker's theorem Davar Khoshnevisan,p.6, [https://www.math.utah.edu/~davar/ps-pdf-files/donsker.pdf](https://www.math.utah.edu/%7Edavar/ps-pdf-files/donsker.pdf)) it's shown that


$$ \sup\_{n \geq 1} \| d \left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) \|\_2 \leq 2 \sqrt{E \left[ X\_1^2; |X\_1| > v \right] } \text{ for all } v>0.$$


But to prove the Donsker's theorem we also need that


$$ \lim\_{v \to \infty} \sup\_n P \left( d \left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) > \lambda \right)=0$$ for all $\lambda>0$ which should follow from the inequality above and Chebyshev’s inequality. I don't know how to show this.



Distance between trunctated random walk and its normal form
You have
$$ 
\begin{aligned}
&\sup\_{n\ge1}\sqrt{E[d\left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right)^2]} \\ 
&=\sup\_{n\ge1}\|d\left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) \|\_2 \leq 2 \sqrt{E \left[ X\_1^2; |X\_1| > v \right] } \text{ for all } v>0. 
\end{aligned}
$$
So, by the Chebyshev/Markov inequality,
$$ 
\begin{aligned}
&\lim\_{v \to \infty} \sup\_n P \left( d \left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right) > \lambda \right) \\ 
&\le\lim\_{v \to \infty} \sup\_n 
\frac{E[d\left( \mathscr{S}\_n^{(v) }, \mathscr{S}\_n \right)^2]}{\lambda^2} \\ 
&\le\lim\_{v \to \infty} 
\frac{4E \left[ X\_1^2; |X\_1| > v \right] } 
{\lambda^2} 
=0,
\end{aligned}
$$
because, by the dominated convergence theorem, $E \left[ X\_1^2; |X\_1| > v \right]\to0$ as $v\to\infty$.


