We start with Kolmogorov continuity theorem which lies in the central of the continuous martingale theory.
Let \(\{X_{t}\}_{t\geq 0}\) be a real-valued stochastic process such that there exist \(\alpha, \beta , K\) satisfying \begin{equation*} \E \bigl[|X_{t}-X_{s}|^{\alpha}\bigr]\leq K |t-s|^{1+\beta}, \end{equation*} for any \(t,s\geq 0\). Then \(X\) has a \(\gamma\)–Hölder continuous modification for any \(0 <\gamma <\beta/\alpha \).
By noticing the ultracontractivity of Gaussion measures, we immediately derive that fractional Brownian motion $ B^{H} $ has a $\gamma$–Hölder modification where $0< \gamma < H$. More generally, we can show for any centered random Gaussian field $\{X(x)\}_{x\in\mathbb{R}^{d}}$ with covariance $C$ has a $\gamma$–Hölder modification where $0< \gamma < H$ if \begin{equation} \label{eq-C} C(x,x)+C(y,y)-2C(x,y)\leq K|x-y|^{2H}. \end{equation} Here, $C$ is given by $C(x,y)=\E[X(x)X(y)]$. This, of course, implies the continuity of $C$. But, can we circumvent using Kolmogorov theorem?
Let \(C:\mathbb{R}^{d}\times\mathbb{R}^{d}\to\mathbb{R}\) be symmetric, locally bounded, and positive definite in the sense that for any finite collection \(\{x_{i}\}_{i=1}^{m}\) in \(\mathbb{R}^{d}\), the martix \(M_{ij}=C(x_{i},x_{j})\) is positive definite. Then \eqref{eq-C} implies the jointly continuity of \(C\).
By assmuption we take positive definite matrix \begin{equation*} M= \begin{pmatrix} C(y,y) &C(y,x_{1}) & C(y,x_{2}) \\ C(x_{1},y)&C(x_{1},x_{1}) & C(x_{1},x_{2}) \\ C(x_{2},y)& C(x_{2},x_{1}) & C(x_{2},x_{2}) \end{pmatrix}\end{equation*} and vecter \begin{equation*} \alpha =\begin{pmatrix} C(x_{1},x_{1})+C(x_{2},x_{2})-2C(x_{1},x_{2}) \\ C(x_{2},y)-C(x_{1},y)\\ C(x_{1},y)-C(x_{2},y)\end{pmatrix}.\end{equation*} Then we notice $\alpha^{\intercal} M\alpha \geq 0$ is equivalent to \begin{equation*} [C(x_1,y)-C(x_2,y)]^2 \leq 4 C(y,y)[C(x_{1},x_{1})+C(x_{2},x_{2})-2C(x_{1},x_{2})]. \end{equation*} By triangle inequality we conclude the proof.