Let $M$ be a continuous martingale such that almost surely, the sample paths of $M$ are not constant.


**Question:** Is it true that $M$ is almost surely not differentiable?



Can an a.s. non constant continuous martingale be differentiable with nonzero probability?
Almost surely, we can write for every $t \ge 0$, $M\_t=M\_0+\beta\_{\langle M,M \rangle\_t}$, where $\beta$ is some Brownian motion.


By Kahane's theorem, almost surely, for every $s \ge 0$, $\limsup\_{\delta \to 0+} \delta^{-1/2}|\beta\_{t+\delta}-\beta\_t| \ge 1$.
<https://www.ams.org/journals/tran/1986-296-02/S0002-9947-1986-0846605-2/S0002-9947-1986-0846605-2.pdf>


Let $b > a \ge 0$. Almost surely, on the event $M$ is differentiable on the time interval $[a,b[$, $\langle M,M \rangle\_t$ and $M\_t$ do not depend on $t$ on the time interval $[a,b]$.


Indeed, for every $t \in [a,b[$, since $M$ is differentiable at $t$ whereas
$$\limsup\_{\delta \to 0+} \delta^{-1}|\beta\_{\langle M,M \rangle\_t+\delta}-\beta\_{\langle M,M \rangle\_t}| = +\infty,$$
we must have  

$$\liminf\_{h \to 0+} h^{-1}\big(\langle M,M \rangle\_{t+h}-\langle M,M \rangle\_t\big) = 0.$$
[Otherwise, the limsup as $h \to 0+$ of
$$\frac{M\_{t+h}-M\_t}{h} = \frac{M\_{t+h}-M\_t}{\langle M,M \rangle\_{t+h}-\langle M,M \rangle\_t} \times \frac{\langle M,M \rangle\_{t+h}-\langle M,M \rangle\_t}{h}$$
would be infinite, which would contradict the assumption.]


Given $\epsilon>0$, the set
$$S\_\epsilon := \sup\{t \in [a,b] : \langle M,M \rangle\_t - \langle M,M \rangle\_a \le \epsilon(t-a)\}$$
contains $a$ and bounded above by $b$. Moreover, if $S\_\epsilon$ contains $t \in [a,b[$, it contains $t+h$ for many arbitrarily small $h>0$. Hence $\sup S\_\epsilon = b$ and $b \in S\_\epsilon$ by left continuity.


As a result, $\langle M,M \rangle\_b - \langle M,M \rangle\_a \le \epsilon(b-a)$ for every $\epsilon>0$, so by monotonicity $\langle M,M \rangle\_b = \langle M,M \rangle\_a$ and $\langle M,M \rangle$ is constant on $[a,b]$.


