Suppose $\{X(t),t\in T\}$ is a real-valued stochastic process, $\{X(t),t\in T\}$ is mean-square differentiable on $T$ with its mean-square derivative $X'(t)=0$, prove that $X(t)=X$ where $X$ is some random variable.
I know it seems obvious but I failed to prove it. In real analysis, the similar conclusion can be proved using Lagrange's mean value theorem, however there is no corresponding theorem in a mean-square derivative fashion. Could anyone help me out here?
Suppose without loss of generality that $X(0) = 0$.
Start by verifying that a mean-square differentiable process is mean-square continuous.
Set $f(t) = E[X(t)^2]$, so $f(0)=0$. Note for any $t$ that $$\begin{align*}\left|\frac{f(t+\epsilon) - f(t)}{\epsilon} \right|&= \left|E\left[\frac{X(t+\epsilon)^2 - X(t)^2}{\epsilon}\right]\right| \\ &= \left|E\left[\frac{X(t+\epsilon)-X(t)}{\epsilon} (X(t+\epsilon)+X(t))\right]\right| \\ &\le \sqrt{E\left[\left(\frac{X(t+\epsilon)-X(t)}{\epsilon} \right)^2\right]E\left[ (X(t+\epsilon)+X(t))^2\right]} \\ &\to \sqrt{E[X'(t)^2] \cdot E[(2 X(t))^2]} = 0 \end{align*}$$ as $\epsilon \to 0$, using mean-square differentiability and continuity. Hence $f$ is differentiable with $f'(t) = 0$ for all $t$. By the usual mean value theorem, $f(t) =0 $ for all $t$. This is to say that $X(t) = 0$ almost surely, for every $t$.