Could someone check my solution for the following problem please? Or maybe propose a smarter/shorter solution.
Consider a stochastic process $X=(X_t)_{t \in [0,1]}$ defined in a filtred probability space $(\Omega,\mathcal F , (\mathcal F ^W_t)_{t\in [0,1]}, \mathbb P)$ (where $W$ is a standard brownian motion and $(\mathcal F ^W_t)_{t\in [0,1]}$ is its natural filtration) satisfying
$$ X_t = X_0+\int _0^t b_s ds + \int _0^t \sigma_s dW_s \quad \mathbb P-\text{a.e.}\ , \forall t \in [0,1]$$
where $b=(b_s (\omega))_{t\in [0,1]}$ and $\sigma=(\sigma_s (\omega))_{t\in [0,1]}$ are defined on the same filtered probability space , progressively mesuables and verify:
(H1) (Uniformly bounded) There is $C>0$ such that $\sup_{s\in [0,1]}(|b_s(\omega)|+|\sigma_s(\omega)|)\leq C$,
(H2) $\mathbb E [(\sigma_s^2- \sigma_t^2)^2]\leq \varpi(t-s), \quad \varpi(h) \rightarrow 0 \text{ as } h \rightarrow 0$
In this context we observe $(X_0,X_{1/n}, \cdots, X_{i/n}, \cdots , X_1)$ ($n \in \mathbb N ^*)$ and we are interested in estimates the linear functional
$$ \Lambda(\phi)_t := \int _0 ^t \phi(s) \sigma^2_s ds, \quad t\in [0,1]$$
whrere $\phi \in \mathcal C^0 ([0,1]; \mathbb R)$ by $$\widehat\Lambda_n(\phi)_t := \sum_{i=1} ^{[nt]}\phi(\frac{i-1}{n})(X_{i/n}-X_{(i-1)/n})^2$$
Now I want to show that indeed $\widehat\Lambda_n(\phi)_t \overset{\mathbb P}\longrightarrow \Lambda(\phi)_t $.
I started by supposing $b\equiv 0$ in other to focus on the main problem and thy to show that it converges in $\mathbb L ^2(\mathbb P)$
$\widehat\Lambda_n(\phi)_t - \Lambda(\phi)_t =\sum_{i=1} ^{[nt]}\left[\phi(\frac{i-1}{n})(X_{i/n}-X_{(i-1)/n})^2 -\int _{(i-1)/n} ^{i/n} \phi(s) \sigma^2_s ds\right] + \int ^{t} _{[nt]/n} \phi(s) \sigma^2_s ds$
Then consider
$$ M^{i,n} := \phi(\frac{i-1}{n})(X_{i/n}-X_{(i-1)/n})^2 -\int _{(i-1)/n} ^{i/n} \phi(s) \sigma^2_s ds$$ Ito's lemma implies that $$ M^{i,n} = \int _{(i-1)/n} ^{i/n} \left( \int ^s _{(i-1)/n}2\phi(\frac{i-1}{n})\sigma_u dW_u \right)\sigma_s dW_s + \int _{(i-1)/n} ^{i/n} \phi(\frac{i-1}{n})\sigma^2_s ds -\int _{(i-1)/n} ^{i/n} \phi(s) \sigma^2_s ds$$
then let be
$$A^{i,n} := \int _{(i-1)/n} ^{i/n} \left( \int ^s _{(i-1)/n}2\phi(\frac{i-1}{n})\sigma_u dW_u \right)\sigma_s dW_s $$
$$B^{i,n} :=\int _{(i-1)/n} ^{i/n} \phi(\frac{i-1}{n})\sigma^2_s ds -\int _{(i-1)/n} ^{i/n} \phi(s) \sigma^2_s ds$$
Thus we have $ (M^{i,n})^2 \leq 2 \left( A^{i,n}\right)^2 + 2\left(B^{i,n}\right)^2$ and taking expected values both sides
\begin{align} \mathbb E[ (M^{i,n})^2 ]&\leq 2 \mathbb E\left[\left( A^{i,n}\right)^2\right] + 2\mathbb E\left[\left(B^{i,n}\right)^2\right] \leq \frac{K}{n^2} \end{align}
whith $K = 12C^4 (\sup_{s\in [0,1]}\phi(s))^2$, since \begin{align} \mathbb E\left[\left( A^{i,n}\right)^2\right] &= \mathbb E\left[\int _{(i-1)/n} ^{i/n} \left( \int ^s _{(i-1)/n}2\phi(\frac{i-1}{n})\sigma_u dW_u \right)^2 \sigma_s^2 ds\right] \\ &=\int _{(i-1)/n} ^{i/n} \mathbb E\left[\left( \int ^s _{(i-1)/n}2\phi(\frac{i-1}{n})\sigma_u \sigma_s dW_u \right)^2 \right]ds \\ &=\int _{(i-1)/n} ^{i/n} \mathbb E\left[ \int ^s _{(i-1)/n}4\phi^2(\frac{i-1}{n})\sigma_u^2 \sigma_s^2 du \right]ds \\ &\leq 4 C^4 (\sup_{s\in [0,1]}\phi(s))^2 \int _{(i-1)/n} ^{i/n} \int ^s _{(i-1)/n} du ds = \frac{2C^4 (\sup_{s\in [0,1]}\phi(s))^2} {n^2} [i^2 -(i-1)^2 -2(i-1)]\\ &\leq \frac{2C^4 (\sup_{s\in [0,1]}\phi(s))^2} {n^2} \end{align}
(by Ito's isometry in §1,3, Fubini's theorem in §2 and (H1) in §4) and
$$B^{i,n} \leq \frac{2 \sup_{s \in [0,1]}\phi(s) C^2}{n}$$ (by (H1)).
We conclude by remarking that
\begin{align} \left|\widehat\Lambda_n(\phi)_t - \Lambda(\phi)_t \right|^2 &\leq 2\sum_{i=1} ^{[nt]}\left( M^{i,n}\right)^2 +2 \left(\int ^{t} _{[nt]/n} \phi(s) \sigma^2_s ds\right)^2\\ &\leq 2\sum_{i=1} ^{[nt]}\left( M^{i,n}\right)^2 + 2 C^4 \left(\sup_{s\in [0,1]} \phi(s)\right)^2 \frac{(nt -[nt])}{n}\\ &\leq 2\sum_{i=1} ^{[nt]}\left( M^{i,n}\right)^2 + \frac{2 C^4 \left(\sup_{s\in [0,1]} \phi(s)\right)^2}{n} \end{align} and taking expected value both sides
\begin{align} \parallel\widehat\Lambda_n(\phi)_t - \Lambda(\phi)_t\parallel^2_{\mathbb L^2(\mathbb P)} =\mathbb E \left[\left|\widehat\Lambda_n(\phi)_t - \Lambda(\phi)_t\right|^2 \right] \leq \frac{K'}{n} \longrightarrow 0 \text{ as } n\rightarrow \infty \end{align}
where $ K' = 14 C^4 \left(\sup_{s\in [0,1]} \phi(s)\right)^2$. The convergence in probability is consequence of the convergence in $\mathbb L^2(\mathbb P)$