I have a question in some middle steps of a proof of some theorem.
Here is the condition we need:
Let $B(t)$ be an $\mathbb{R}^m$-valued Brownian motion. Consider the $\mathbb{R}^n$-valued SIE,
$$X_t=x+\int_a^t f(s,X_s)ds+\int_a^t \sigma(s,X_s)dB(s),\quad a\leq t\leq b$$
where the matrix valued $\sigma(t,x)$ and vector valued $f(t,x)$ are measurable and satisfy linear growth and Lipschitz condition. Assume that $\sigma(t,x)$ and $f(t,x)$ are continuous on $[a,b] \times \mathbb{R}^n$.
I am trying to show that $$\mathbb{E}\left( \int_{t}^{t+\epsilon}f_i(s,X_s)ds \int_{t}^{t+\epsilon}f_j(s,X_s)ds \right)=O(\epsilon^2)$$
My attempt:
\begin{align} & \mathbb{E}\left( \int_{t}^{t+\epsilon}f_i(s,X_s)ds \int_{t}^{t+\epsilon}f_j(s,X_s)ds \right) \\ & =\mathbb{E}\left( \sqrt{\left[\int_{t}^{t+\epsilon} f_i(s,X_s)ds\right]^2} \sqrt{\left[\int_{t}^{t+\epsilon} f_j(s,X_s)ds\right]^2} \right) \\ &\leq \mathbb{E}\left( \left[\epsilon\int_{t}^{t+\epsilon} f_i^2(s,X_s)ds\right]^{1/2} \left[\epsilon\int_{t}^{t+\epsilon} f_j^2(s,X_s)ds\right]^{1/2} \right) \quad \text{Cauchy-Schwartz} \\ & =\mathbb{E}\left( \epsilon\left[\int_{t}^{t+\epsilon} f_i^2(s,X_s)ds\right]^{1/2} \left[\int_{t}^{t+\epsilon} f_j^2(s,X_s)ds\right]^{1/2} \right) \\ & =\epsilon\mathbb{E}\left( \left[\int_{t}^{t+\epsilon} f_i^2(s,X_s)ds\right]^{1/2} \left[\int_{t}^{t+\epsilon} f_j^2(s,X_s)ds\right]^{1/2} \right) \\ & =\epsilon \left( \left[\int_{t}^{t+\epsilon} \mathbb{E}\left(f_i^2(s,X_s)\right)ds\right]^{1/2} \left[\int_{t}^{t+\epsilon} \mathbb{E}\left(f_j^2(s,X_s)\right)ds\right]^{1/2} \right) \\ \end{align}
First question: I am unsure about the validity of the last equality. Why $f_i$ and $f_j$ are independent to each other? Is it because $f$ and $\sigma$ are deterministic functions?
If it's correct, then I may continue as follows:
\begin{align} & =\epsilon \left( \left[\int_{t}^{t+\epsilon} \mathbb{E}\left(f_i^2(s,X_s)\right)ds\right]^{1/2} \left[\int_{t}^{t+\epsilon} \mathbb{E}\left(f_j^2(s,X_s)\right)ds\right]^{1/2} \right) \\ & \leq \epsilon \left( \left[\int_{t}^{t+\epsilon} C_1(1+\mathbb{E}(|X_s|^2)) ds\right]^{1/2} \left[\int_{t}^{t+\epsilon} C_2(1+\mathbb{E}(|X_s|^2)) ds\right]^{1/2} \right) \\ & \leq \epsilon \left( \left[\int_{t}^{t+\epsilon} C_1' ds\right]^{1/2} \left[\int_{t}^{t+\epsilon} C_2' ds\right]^{1/2} \right) \\ & =O(\epsilon^2) \end{align}
Second question: For the first inequality above, I used the condition that the $f$ satisfies linear growth condition: $|f(t,x)|^2\leq C(1+|x|^2)$. But then how do I show $\mathbb{E}(|X_s|^2)$ is finite? I know it may be related to quadratic variation property of Brownian motion, but how do I write out the proof rigorously?
\begin{align} & \mathbb{E}\left( \int_{t}^{t+\epsilon}f_i(s,X_s)ds \int_{t}^{t+\epsilon}f_j(s,X_s)ds \right) \\ \leq & \sqrt{ \mathbb{E}\left(\int_{t}^{t+\epsilon}f_i(s,X_s)ds\right)^2 \mathbb{E}\left(\int_{t}^{t+\epsilon}f_j(s,X_s)ds \right)^2 } \\ \leq & \sqrt{ \mathbb{E}\left(\epsilon\int_{t}^{t+\epsilon}f_i^2(s,X_s)ds\right) \mathbb{E}\left(\epsilon\int_{t}^{t+\epsilon}f_j^2(s,X_s)ds \right)} \\ \leq & \epsilon\int_{t}^{t+\epsilon}\mathbb{E}\left(1+|X_s|^2\right)ds \\ = & O(\epsilon^2) \end{align}