Is variance continuous for Gaussian martingale with continuous paths?

Question

Let $X: [0, \infty) \times \Omega \to \mathbb{R}$ be a Gaussian process, I.e. all $X_t$ lie in the same Gaussian space, a closed sub space of $L^2(\Omega)$ containing only Gaussian random variables. Assume $X$ also has continuous sample paths and $X_0=0$. Is this enough to conclude that the function $f: t\mapsto var(X_t)$ is continuous? What if X is also assumed to be a martingale? The reason I ask is I’m asked to prove that that under these conditions, $f: t\mapsto var(X_t)$ is the quadratic variation of $X$. I can prove that $f$ is nondecreasing and that $X^2_t-f(t)$ is a martingale, but I need to show $f$ is continuous. If pointwise convergence in a Gaussian space implied L^2 convergence, that would do it, but of course that is stronger than what I need.

Answer 1

Assuming $X$ is a martingale, we have $\mathbb{E}[X_t] = \mathbb{E}[X_0] = 0$. Since each $X_t$ is Gaussian, $\mathbb{E}[X_t^4] = 3 \mathbb{E}[X_t^2]^2 = 3 f(t)^2 \le 3 f(T)^2$ where $T > t$. This implies $(X_t^2)_{t \in [0,T]}$ is a uniformly integrable set of random variables, and therefore $$\lim_{s \rightarrow t} f(s) = \lim_{s \rightarrow t} \mathbb{E}[X_s^2] = \mathbb{E}\left[ \lim_{s \rightarrow t} X_s^2 \right] = \mathbb{E}[X_t^2] = f(t)$$ so we conclude $f$ is continuous.