Can an a.s. non constant continuous martingale be differentiable with nonzero probability?

Question

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?

Answer 1

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 $[a,b]$.