Probability measures on [0,1] are characterized by their moments, so we do get an order relation.

Anyways, this looks like a concentration property: if a sequence of positive random variables $(Z_n)$ satisfies $Z_n/a_n \to 1$ in probabiliity for some sequence $(a_n)$ of positive numbers, if furthermore the sequences $(Z_n/a_n)$ and $(a_n/Z_n)$ are uniformly integrable, then $E[Z_n]/a_n \to 1$ and $a_nE[1/Z_n] \to 1$, so $E[Z_n]E[1/Z_n] \to 1$.

Do you assume that $x$ is centered?

@colin Can we have a good statement of the problem? If $(X_1,\ldots,X_n)$ is assumed to be a gaussian centered vector with covariance matrix $\Sigma$ , write it explicitely. The assumption positive definite should be relaxed into positive semidefinite.

@Iosif Pinelis. In the question, $\Sigma$ is assumed to be positive definite (positive semi-definite would be better for $f$ to have a maximum and not only a least upper bound), and not to have non-negative entries.

I am very surprised by this answer, although I am not surprised by Slepian's lemma. My guess, is that when $n$ is even, the maximum will be achieved when $X_1,...,X_{n/2}$ are i.i.d. and $(X_{n/2+1},...,X_n)=-(X_1,...,X_{n/2})$. For example, $E[\max(X_1,-X_1)] = \sqrt{2/\pi}$ whereas $E[\max(X_1,X_2)]=\sqrt{1/\pi}$ for $X_1,X_2$ i.i.d. $\mathcal{N}(0,1)$.

I corrected the typo, by replacing $f$ with $\sqrt{}$. Anyway, you do not know at the beginning what the range of the $y_n$ can be. It has no reason to be contained in $[0,1]$, that is why I wrote that $f_n$ should be defined on $\mathbb{R}_+$ (and we still can have explosion in finite time for small values of $n$). As you note, it is sufficient to assume that $f_n$ converges uniformly to $\sqrt{}$ on some interval $[0,1/4+\delta]$, so the interval $[0,1]$ works. It is because we are lucky. In general, it is not always possible to take the `spatial' interval equal to the time interval.

The question would be less trivial if we ask whether $X \ge_{st} Y$ (stochastic order). But my intuition is that the answer will still be negative, even the distribution of $X$ and $Y$ are determined by their moments.

@OXbadfood Other possible assumption that may work: $c$ Hölder with exponent $\alpha$ and $E[d(x,X_t)^\alpha] \to 0$ as $t \to 0$ under $P_x$. I do not see how to avoid global assumptions on $c$.

@Iosif Pinellis Yes, you are right. I will remove the comment.

Thank you. I believe more on your second argument. $X-X_0$ is not independent of $X_0$ in general since $d\langle X \rangle_t/dt$ equals $\sigma(0,X_0)^2$ at time $0$. A more complete argument could be that the quadratic variation computed under $P$ is still the quadratic variation computed or under $P[\cdot|A]$ when $A \in \sigma(X_0)$ has positive probability. Calling $\tau_\cdot$ the inverse of the quadratic variation, we derive that $B := X_{\tau_\cdot}$ is a Brownian motion under $P$ and also under $P[\cdot|A]$ when $A \in \sigma(X_0)$ has positive probability.