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I have $X_i \sim N(0,1)$, $S_n=X_1+\cdots+X_n$, $ \mathscr{S}_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\limits_{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\limits_{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\limits_{i=1}^n X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } - E \left[ \sum\limits_{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) 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\limits_{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.

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) 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.

I have $X_i \sim N(0,1)$, $S_n=X_1+...+X_n$, $ \mathscr{S}_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\limits_{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's $ d(f,g):=\sup\limits_{x \in [0,1]}| f(x) - g(x) | $.

Then $ Y_{i}^{\left( v \right)} := X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} }, X_{i}^{\left( v \right)} := \left( Y_{i}^{\left( v \right)} - E [ Y_{1}^{\left( v \right)} ] \right) $ analogically we define $ S_{n}^{(v)} = \sum\limits_{i=1}^{n} X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } - E \left[ \sum\limits_{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) it's shown that

$ \sup_{n \geq 1} || d \left( \mathscr{S}_{n}^{\left( v \right) }, \mathscr{S}_{n} \right) ||_2 \leq 2 \sqrt{E \left[ X_{1}^{2}; |X_1| > v \right] } $ for all $v>0$.

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

$ \lim\limits_{v \to \infty} \sup_{n} P \left( d \left( \mathscr{S}_{n}^{\left( v \right) }, \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.

I have $X_i \sim N(0,1)$, $S_n=X_1+\cdots+X_n$, $ \mathscr{S}_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\limits_{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\limits_{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\limits_{i=1}^n X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } - E \left[ \sum\limits_{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) 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\limits_{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.

I have $X_i \sim N(0,1)$, $S_n=X_1+...+X_n$, $ \mathscr{S}_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\limits_{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's $ d(f,g):=\sup\limits_{x \in [0,1]}| f(x) - g(x) | $.

Then $ Y_{i}^{\left( v \right)} := X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} }, X_{i}^{\left( v \right)} := \left( Y_{i}^{\left( v \right)} - E [ Y_{1}^{\left( v \right)} ] \right) $ analogically we define $ S_{n}^{(v)} = \sum\limits_{i=1}^{n} X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } - E \left[ \sum\limits_{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) it's shown that

$ \sup_{n \geq 1} || d \left( \mathscr{S}_{n}^{\left( v \right) }, \mathscr{S}_{n} \right) ||_2 \leq 2 \sqrt{E \left[ X_{1}^{2}; |X_1| > v \right] } $ for all $v>0$.

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

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

I have $X_i \sim N(0,1)$, $S_n=X_1+...+X_n$, $ \mathscr{S}_n (t, \omega) := \frac{1}{ \sqrt{n} } \sum\limits_{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's $ d(f,g):=\sup\limits_{x \in [0,1]}| f(x) - g(x) | $.

Then $ Y_{i}^{\left( v \right)} := X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} }, X_{i}^{\left( v \right)} := \left( Y_{i}^{\left( v \right)} - E [ Y_{1}^{\left( v \right)} ] \right) $ analogically we define $ S_{n}^{(v)} = \sum\limits_{i=1}^{n} X_i \textbf{1}_{ \left\{ \left| X_i \right| \leq v \right\} } - E \left[ \sum\limits_{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) it's shown that

$ \sup_{n \geq 1} || d \left( \mathscr{S}_{n}^{\left( v \right) }, \mathscr{S}_{n} \right) ||_2 \leq 2 \sqrt{E \left[ X_{1}^{2}; |X_1| > v \right] } $ for all $v>0$.

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

$ \lim\limits_{v \to \infty} \sup_{n} P \left( d \left( \mathscr{S}_{n}^{\left( v \right) }, \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.