This is part of the proof of Lemma 3.1 from Rene Schilling's Brownian Motion.(The full proof is attached at the bottom of my question.)
Consider the Hilbert space $L^2(dt)=L^2([0,1],dt)$ with scalar product $\langle f,g \rangle_{L^2} = \int_0^1 f(t)g(t)dt$, and assume that $(\phi_n)_{n \ge 0}$ is any complete ONS and let $(G_n)_{n \ge 0}$ be a sequence of real-valued iid Gaussian $N(0,1)$ random variables on the probability space $(\Omega, \mathscr{A},P)$. Set $$W_N(t) := \sum_{n=0}^{N-1} G_n \langle 1_{[0,t)}, \phi_n \rangle_{L^2} = \sum_{n=0}^{N-1} G_n \int_0^t \phi_n(s) ds.$$
Then the limit $W(t):= \lim_{N \to \infty} W_N(t) $ exists for every $t \in [0,1]$ in $L^2(P)$ and the process $W(t)$ satisfies the properties of Brownian Motion.
Proof.
The proof first shows that using the independence of $G_n$ and Parseval's identity wee get for every $t \in [0,1]$ $E[W_N(t)]^2 = t$ and $W(t) = L^2-\lim_N W_N(t)$ exists.
An analogous calculation yields for $s<t $ and $u<v$
$$E(W(t)-W(s))(W(v)-W(u)) = \sum_{n=0}^\infty \langle 1_{[0,t)} - 1_{[0,s)}, \phi_n \rangle_{L^2} \langle 1_{[0,v)} - 1_{[0,u)}, \phi_n \rangle_{L^2} = \langle 1_{[s,t)} , 1_{[u,v)}\rangle_{L^2},$$ and we see that $E(W(t)-W(s))(W(v)-W(u)) = (v \wedge t - u \vee s)^+$, so $0$ if $[s,t) \cap [u,v) = \emptyset$.
Question.
I have a question about the next line of the proof. It says in the text that:
With this calculation we find for all $0 \le s < t \le u < v$ and $\xi , \eta \in \mathbb{R}$
$$E[\exp(i \xi ( W(t)-W(s)) + i \eta (W(v)-W(u)))] = \lim_N E[\exp(i \sum_{n=0}^{N-1} (\xi \langle 1_{[s,t)}, \phi_n \rangle + \eta 1_{[u,v)}, \phi_n \rangle ) G_n)].$$
I can't figure out how the above calculation is used to get this identity. What exactly allows us to take the limit outside of the exponent and the expectation when we have a $L^2$ limit?
An argument I came up with is we can consider $g$ to be the bounded continuous function $g(x) = \exp(i ( \xi f(x) + \eta h(x)))$, where $f_n \to f$ in $L^2$ and $h_n \to h$ in $L^2$ ( Take $f_n = W_n(t) - W_n(s)$ and $h_n = W_n(v)-W_n(u)$. ) Then by Vitali's generalized dominated convergence theorem, we would get $\lim_n \exp(i(\xi f_n(x)+\eta h_n(x)))=g(x)$, which gives the identity above.
However, this argument does not use the calculation $E(W(t)-W(s))(W(v)-W(u)) = (v \wedge t - u \vee s)^+$. So I don't think this is what the author intended.
I would greatly appreciate a justification of this limiting argument.
I attach below the full proof.
Here is how I understand it, though I am not exactly sure what the author actually intended. First, for every $t \in [0,1]$, the sequence $(W_N(t))$ converges to $W(t)$ in $L^2(\mathbb{P})$. Therefore, for $0 \leq s < t \leq u < v$, the sequence $(W_N(t)-W_N(s),W_N(v)-W_N(u))$ converges to $(W_N(t)-W_N(s),W_N(v)-W_N(u))$ in $L^2(\mathbb{P})$, which means the characteristic functions converge: for any $\xi, \eta \in \mathbb{R}$, $$ \mathbb{E} \left [ e^{i \xi (W_N(t) - W_N(s)) + i \eta (W_N(v) - W_N(u))} \right ] \to \mathbb{E} \left [ e^{i \xi (W(t) - W(s)) + i \eta (W(v) - W(u))}\right ]. $$ The rest of the computation shows that the left-hand side is $$ \exp \left [ - \frac12 \sum_{n=0}^{N-1} \left ( \xi^2 \langle 1_{[s,t)}, \phi_n \rangle^2 + \eta^2 \langle 1_{[u,v)}, \phi_n \rangle^2 \right ) \right ]. $$ But $$ \sum_{n=0}^{N-1} \langle 1_{[s,t)}, \phi_n \rangle^2 \to \langle 1_{[s,t)}, 1_{[s,t)} \rangle = t - s $$ in $L^2([0,1])$ since $(\phi_n)$ is a complete ONS. Similarly for the other part, and as $L^2$ convergence is preserved by bounded continuous functions, this yields that $$ \exp \left [ - \frac12 \sum_{n=0}^{N-1} \left ( \xi^2 \langle 1_{[s,t)}, \phi_n \rangle^2 + \eta^2 \langle 1_{[u,v)}, \phi_n \rangle^2 \right ) \right ] \to \exp \left [ - \frac12 \xi^2 (t-s) + \eta^2 (v-u) \right ], $$ and thus finally $$ \mathbb{E} \left [ e^{i \xi (W(t) - W(s)) + i \eta (W(v) - W(u))}\right ] = e^{- \frac12 \xi^2 (t-s) + \eta^2 (v-u)}. $$ But indeed, this does not use the computation of the second moment, which is definitely not enough to compute characteristic functions.