We say that a random process $X_t$ is mean-square differentiable at time $t_0$ if there is a random process $X_{t_0}'$ such that $$\lim_{t\to t_0} \mathbb{E}\Big[\big(X_{t_0}' - \frac{X_t - X_{t_0}}{t-t_0}\big)^2\Big]=0$$ We say that a random process $X_t$ is mean-square continuous at time $t_0$ if $$\lim_{t\to t_0} \mathbb{E}\big[(X_{t} - X_{t_0})^2\big]=0$$
From calculus, we know that differentiability implies continuity. Does the same statement hold in stochastic calculus?
More formally, suppose a process is mean-square differentiable for all $t\in \mathbb{R}$. Is this process mean-square continuous for all $t\in\mathbb{R}$?
Yes, that's true.
Using the elementary inequality $$(x+y)^2 \leq 2x^2+2y^2, \qquad x,y \in \mathbb{R}$$ we find
$$\begin{align*} (X_t-X_{t_0})^2 &= (t-t_0)^2 \left( \frac{X_t-X_{t_0}}{t-t_0} - X_{t_0}' + X_{t_0}' \right)^2 \\ &\leq 2 (t-t_0)^2 \left( \frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' \right)^2 + 2(t-t_0)^2 (X_{t_0}')^2. \end{align*}$$
Taking expectations on both sides we get
$$\begin{align*} \mathbb{E}((X_t-X_{t_0})^2) &\leq 2(t-t_0)^2 \underbrace{\mathbb{E} \left[ \left( \frac{X_t-X_{t_0}}{t-t_0}-X_{t_0}' \right)^2 \right]}_{\xrightarrow[]{t \to t_0} 0} + 2 \underbrace{(t-t_0)^2}_{\xrightarrow[]{t \to t_0} 0} \mathbb{E}((X_{t_0}')^2) \\ &\xrightarrow[]{t \to t_0} 0. \end{align*}$$