Does there exist an almost surely differentiable martingale? Does there exist a continuous time martingale $X_t$ not a.s. constant in $t$ that is almost surely everywhere differentiable?
 A: Others have answered that there does not exist an a.s. differentiable martingale process that's not constant. One related fact (not asked for but interesting) is that an a.s. differentiable Markov process satisfies a deterministic ODE with random initial conditions. See https://www.ams.org/journals/notices/196808/196808FullIssue.pdf page 748
Somehow martingale/Markov don't work well with differentiability.
A: The answer is no.
Indeed, if a martingale is a.s. everywhere differentiable, then its quadratic variation is a.s $0$. So, by the Burkholder--Davis--Gundy inequality, the martingale is a.s. constant.

Details: Suppose that $X:=(X_t)_{t\in[0,1]}$ is an almost surely (a.s.) everywhere differentiable martingale. Replacing $X_t$ by $X_t-X_0$, without loss of generality let us assume that $X_0=0$. Take any real $a>0$ and consider the bounded martingale $X^a:=(X_t^a)_{t\ge0}$, where $X_t^a:=X_{\min(t,T_a)}$ and $T_a:=\inf\{t\in[0,1]\colon|X_t|=a\}$, with $\inf\emptyset:=\infty$.
For any real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula
$$[f]:=\limsup\sum_{j=1}^n(f(t_j)-f(t_{j-1}))^2,$$
where the $\limsup$ is taken over all "partitions" $0=t_0<\cdots<t_n=1$ of $[0,1]$  as $\max_{1\le j\le n}(t_j-t_{j-1})\to0$.
By the Burkholder--Davis--Gundy inequality,
\begin{equation}
    c\,E([X^a]^{1/2})\le EM^a\le C\,E([X^a]^{1/2}), \tag{BDG}
\end{equation}
where $c$ and $C$ are universal positive real constants and $M^a:=\max_{t\in[0,1]}|X_t^a|$.
By the first one of inequalities (BDG), $[X^a]<\infty$ a.s. So, using (i) Corollary 23 in this paper, (ii) the remark on line 3 of page 4228 of the same paper that $\mu_f=0$ iff $f\in V_2^0$, (iii) the obvious identity $\mu_f([0,1])=[f]$, and (iv) the definition of $V_2$ as the set of all functions $f$ with $[f]<\infty$, we conclude (as in this answer) that $[X^a]=0$ a.s. Therefore, by the second one of inequalities (BDG), $M^a=0$ a.s. for each $a$ and hence $X=0$ a.s., as desired.
A: This is also a direct consequence of Ito's formula. Let $T_n:=\inf\{t:|X_t|>n\}$ and define $Y^{(n)}_t:=X_{t\wedge T_n}$. As noted, $X$ has $0$ quadratic variation, hence so does $Y^{(n)}$. By Ito, the process $(Y^{(n)}_t-Y^{(n)}_0)^2$ is a bounded non-negative martingale with initial value $0$. Therefore $\Bbb E[(X_{t\wedge T_n}-X_0)^2]=\Bbb E[(Y^{(n)}_t-Y^{(n)}_0)^2]=0$ for each $t\ge 0$. By Fatou, $\Bbb E[(X_t-X_0)^2]=0$ for each $t\ge 0$.
