Does there exist a continuous time martingale $X\_t$ not a.s. constant in $t$ that is almost surely everywhere differentiable?



Does there exist an almost surely differentiable martingale?
The answer is no.


Indeed, if a martingale is a.s. everywhere differentiable, then its [quadratic variation](https://en.wikipedia.org/wiki/Quadratic_variation#Definition) is a.s $0$. So, by the [Burkholder--Davis--Gundy inequality](https://en.wikipedia.org/wiki/Quadratic_variation#Martingales), the martingale is a.s. constant.




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**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](https://en.wikipedia.org/wiki/Quadratic_variation#Martingales),
\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](https://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05209-8/S0002-9947-2011-05209-8.pdf), (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](https://mathoverflow.net/a/398042/36721)) 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.


