Let $W=\{W_t\}_{t\in[0;T]}$ be a Brownian motion, $\{F_t\}_{t\in [0;T]}$ the filtration generated by $W$, augmented with the nullsets. Let $X \in L^\infty(\mathcal F_T)$. I want to prove that the difference $$ \vert \mathbb E(X|\mathcal F_s) - \mathbb E(X|\mathcal F_t) \vert $$ is small (in whatever sense) if $\vert s-t \vert$ is small.
Intuitively, only very little information is added to the one $\sigma$-field in order to obtain the other one. Moreover, the Brownian motion has no jumps. This is why I think that the difference should be small, but I don't know how to approach that formally.
Some statement like $$ \lim_{h\to 0} \frac 1 h \mathbb E\Big( \big( \mathbb E(X|\mathcal F_{t+h}) - \mathbb E(X|\mathcal F_t) \big)^2 \Big) \in \mathbb R $$ would be great, but maybe less would also be enough, it depends.