Does sample-path continuity imply mean square continuity?

Question

Let $(X_t)_{t>0}$ be a square-integrable stochastic process on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

I'm well aware that in general almost sure convergence does not imply mean square convergence, unless some additional conditions (such as the ones of the dominated convergence theorem) are met.

I've also come across this related question where the OP is interested in the case of (the weaker) almost sure continuity at a given time.

However, in the stronger case of sample path continuity (i.e. $\forall \omega \in \Omega, t \to X(t, \omega)$ is a continuous function) I am curious as to whether the process would also be mean square continuous. If so, I'd appreciate any pointer to the proof. If not, can you please provide a counter-example? The counter-example provided in the question above does not have continuous paths.

Please note that these lecture notes, more specifically the remark following Definition 72, seem to suggest that not only is it true in general, but the proof is also obvious.

Thanks,

Answer 1

No, this is, in general, not true as the following counterexample shows:

Consider $((1,\infty),\mathcal{B}((1,\infty))$ endowed with the probability measure $$\mathbb{P}(dx) = c \frac{1}{x^2} \, dx$$

(where $c>0$ is chosen such that $\mathbb{P}((1,\infty))=1$). Choose a continuous function $\chi : [0,\infty) \to [0,\infty)$ such that $\chi(x)=0$ for all $x \notin [2,5]$ and $\chi(x) \geq 1$ for all $x \in [3,4]$. Note that $\chi$ is bounded, i.e. $\|\chi\|_{\infty}<\infty$. Define

$$X_t(x) := x \chi \left( tx \right).$$

Obviously, $(X_t)_{t \geq 0}$ has continuous sample paths, $X_0 = 0$ and

$$\begin{align*} \mathbb{E}(X_t^2) = \int_{(1,\infty)} x^2 \chi(tx)^2 \frac{1}{x^2} \, dx &= \int_{(1,\infty)} \chi^2(tx) \, dx \\ &\leq \|\chi\|_{\infty}^2 \int_{2/t}^{5/t} \, dx < \infty \end{align*}$$

i.e. $X_t \in L^2(\mathbb{P})$ for each $t>0$. On the other hand, we find by a similar calculation

$$\begin{align*} \mathbb{E}(X_t^2) = \int_{(1,\infty)} \chi(tx)^2 \, dx \geq \int_{(3/t,4/t)} 1 \, dx = \frac{4}{t}- \frac{3}{t} = \frac{1}{t} \end{align*}$$

where we have used that $\chi$ is non-negative and $\chi(tx) \geq 1$ for all $x \in [3/t,4/t]$. This shows that $\mathbb{E}(X_t^2)$ does not converge to $\mathbb{E}(X_0^2)=0$ as $t \to 0$. Consequently, $(X_t)_{t \geq 0}$ is not mean-square continuous.