Function of mean square continuous process

Question

I have been asked to prove that, if $\{X_t\}$ is a ($n$-dimensional) mean square continuous process and $f:\mathbb{R}^n \rightarrow \mathbb{R}^d$ is a Lipschitz function, the process $\{f(X_t)\}$ is mean square continuous.

I was trying to show that $\{f(X_t)\}$ satisfies the definition of mean square continuous, i.e.

• $ \sup_{t\le T} \mathbb{E}\big(~\big|f(X_t)\big|^2 \big) < +\infty $
• $ \lim_{s \rightarrow t } \mathbb{E}\big( \big|f(X_t) - f(X_s)\big|^2 \big) =0 ~~~ \forall s,t \in [0,T]$

but I immediately got stuck: I thought I could apply the mean value theorem in the integral form as

$\mathbb{E}\big(\big|f(X_t)\big|^2\big) = \int_\Omega \big|f(X(t,\omega) \big|^2 d\mathbb{P} = \big|f(X(t,\omega_t) \big|^2 |\Omega| = \big|f(X(t,\omega_t) \big|^2$

for some $\omega_t \in \Omega$. Then exploit the lipschitzianity of $f$ to conclude that the $\sup_{t \le T}$ is bounded.

Actually I have many doubts on the soundness of this proof. Can I reason this way or is there a better way to proceed?

Answer 1

No, your proof is not correct There does in general not exist $\omega_t$ such that the equality

$$\mathbb{E}(f(X_t)^2) = |f(X(t,\omega_t))| |\Omega|$$

holds.

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Hints:

1. Use the estimate $$\begin{align*} |f(X_t)|^2 &= |(f(X_t)-f(0))+f(0)|^2 \\ &\leq 2 |f(X_t)-f(0)|^2 + 2 |f(0)|^2 \\ &\leq 2 C^2 |X_t-0|^2 + 2 |f(0)|^2 \end{align*}$$ to obtain $$\sup_{t \leq T} \mathbb{E}(f(X_t)^2)< \infty.$$
2. In order to show $\lim_{s \to t} \mathbb{E}(|f(X_t)-f(X_s)|^2)=0$ use that $$|f(X_t)-f(X_s)|^2 \leq C^2 |X_t-X_s|^2$$ and the mean-square continuity of $(X_t)_{t \geq 0}$.