[proofplan] We prove a stronger local obstruction: on every compact time interval, Brownian paths cannot be locally Lipschitz at any point. The core estimate is that three consecutive dyadic Brownian increments are all of order $O(2^{-n})$ with summable probability, because each increment has standard deviation $2^{-n/2}$. The first Borel-Cantelli lemma then rules out such triples for all sufficiently fine dyadic scales. A finite derivative would force exactly such small triples at every sufficiently fine scale, giving the contradiction. [/proofplan]

[step:Estimate the probability of three consecutive anomalously small dyadic increments] Fix $R \in \mathbb N$ and $m \in \mathbb N$. For $n \in \mathbb N$, set $\Delta_n := 2^{-n}$ and \begin{align*} N_{R,n} := \left\lceil (R+1)2^n \right\rceil. \end{align*} For each integer $j$ with $0 \le j \le N_{R,n}-1$, define the dyadic increment \begin{align*} I_{j,n}: \Omega &\to \mathbb R \\ \omega &\mapsto W_{(j+1)\Delta_n}(\omega)-W_{j\Delta_n}(\omega). \end{align*} Each $I_{j,n}$ is a real-valued Gaussian random variable with distribution $\mathcal N(0,\Delta_n)$, and the random variables $I_{j,n},I_{j+1,n},I_{j+2,n}$ are independent by the independent-increments property of [Brownian motion](/page/Brownian%20Motion). Define the event \begin{align*} A_{R,m,n}:=\left\{\omega \in \Omega: \begin{array}{l} \text{there exists }j \in \{0,\dots,N_{R,n}-3\}\text{ such that}\\ |I_{j,n}(\omega)|\le 6m\Delta_n,\ |I_{j+1,n}(\omega)|\le 6m\Delta_n,\ |I_{j+2,n}(\omega)|\le 6m\Delta_n \end{array} \right\}. \end{align*} We first bound the probability of $A_{R,m,n}$. If $X \sim \mathcal N(0,\Delta_n)$ and $a>0$, then the Gaussian density bound gives \begin{align*} \mathbb P(|X|\le a) &= \int_{-a}^{a}\frac{1}{\sqrt{2\pi\Delta_n}}\exp\left(-\frac{y^2}{2\Delta_n}\right)\,d\mathcal L^1(y) \\ &\le \int_{-a}^{a}\frac{1}{\sqrt{2\pi\Delta_n}}\,d\mathcal L^1(y) = \frac{2a}{\sqrt{2\pi\Delta_n}}. \end{align*} With $a=6m\Delta_n$, this yields \begin{align*} \mathbb P(|I_{j,n}|\le 6m\Delta_n)\le \frac{12m}{\sqrt{2\pi}}\Delta_n^{1/2}. \end{align*} Set $C_m:=12m/\sqrt{2\pi}$. By independence of the three consecutive increments, \begin{align*} \mathbb P\left(|I_{j,n}|\le 6m\Delta_n,\ |I_{j+1,n}|\le 6m\Delta_n,\ |I_{j+2,n}|\le 6m\Delta_n\right) \le C_m^3\Delta_n^{3/2}. \end{align*} Taking the union over at most $N_{R,n}\le (R+2)2^n$ possible starting indices gives \begin{align*} \mathbb P(A_{R,m,n}) \le (R+2)2^n C_m^3\Delta_n^{3/2} = (R+2)C_m^3 2^{-n/2}. \end{align*} Consequently, \begin{align*} \sum_{n=1}^{\infty}\mathbb P(A_{R,m,n}) \le (R+2)C_m^3\sum_{n=1}^{\infty}2^{-n/2} <\infty. \end{align*} [/step]

[step:Use Borel-Cantelli to exclude small triples at all sufficiently fine scales] The events $(A_{R,m,n})_{n\ge 1}$ are measurable because they are finite unions and intersections of inverse images of closed intervals under the random variables $I_{j,n}$. Since the probability series is finite, the [First Borel-Cantelli Lemma](/theorems/507) gives \begin{align*} \mathbb P\left(\limsup_{n\to\infty}A_{R,m,n}\right)=0. \end{align*} Equivalently, for each fixed pair $(R,m)$, with probability one only finitely many of the events $A_{R,m,n}$ occur. Define \begin{align*} \Omega_0:=\bigcap_{R=1}^{\infty}\bigcap_{m=1}^{\infty} \left(\Omega\setminus \limsup_{n\to\infty}A_{R,m,n}\right). \end{align*} This is a countable intersection of probability-one events, hence $\mathbb P(\Omega_0)=1$. For every $\omega\in\Omega_0$ and every fixed $R,m\in\mathbb N$, there exists $n_0=n_0(\omega,R,m)$ such that $\omega\notin A_{R,m,n}$ for all $n\ge n_0$. [/step]

[step:Show that a finite derivative would force forbidden small triples] Fix $\omega\in\Omega_0$. Suppose, for contradiction, that $w_\omega$ has a finite derivative at some $t>0$, or a finite right derivative at $t=0$. Choose $R\in\mathbb N$ with $t\in[0,R]$. Let $\ell\in\mathbb R$ denote the corresponding finite derivative. Choose $m\in\mathbb N$ with $m>|\ell|+1$. By the definition of derivative, there exists $\rho>0$ such that whenever $h$ satisfies $|h|<\rho$ and $t+h\ge 0$, we have \begin{align*} |w_\omega(t+h)-w_\omega(t)|\le m|h|. \end{align*} For $n$ large enough that $3\Delta_n<\rho$ and $(R+2)\Delta_n<1$, define \begin{align*} k_n:=\left\lfloor \frac{t}{\Delta_n}\right\rfloor,\qquad j_n:=\max\{k_n-1,0\}. \end{align*} For each $r\in\{0,1,2\}$, let \begin{align*} a_{r,n}:=(j_n+r)\Delta_n,\qquad b_{r,n}:=(j_n+r+1)\Delta_n. \end{align*} The points $a_{r,n}$ and $b_{r,n}$ are non-negative, lie in $[0,R+1]$ for all sufficiently large $n$, and satisfy \begin{align*} |a_{r,n}-t|+|b_{r,n}-t|\le 6\Delta_n. \end{align*} The local Lipschitz estimate around $t$ therefore gives \begin{align*} |W_{b_{r,n}}(\omega)-W_{a_{r,n}}(\omega)| &\le |W_{b_{r,n}}(\omega)-W_t(\omega)|+|W_{a_{r,n}}(\omega)-W_t(\omega)|\\ &\le m|b_{r,n}-t|+m|a_{r,n}-t| \le 6m\Delta_n. \end{align*} Thus $\omega\in A_{R,m,n}$ for every sufficiently large $n$, contradicting the defining property of $\Omega_0$. Hence no $\omega\in\Omega_0$ can have a finite derivative at any $t>0$ or a finite right derivative at $t=0$. Since $\mathbb P(\Omega_0)=1$, the theorem follows. [/step]