Non-Differentiability of Paths of Brownian Motion

A function $f$ is left-differentiable at $t$ if the left limit of the quotient of $f$ has a real limit. That is, $\lim_{h \downarrow 0} R(t)(h) = \frac{f(t-h) - f(t)}{h} = f(t-)$ for some real number $f(t-)$ . This implies that $\lim_{n \rightarrow \infty} R(t)(n^{-1})=f(t-)$ . if for any $k > 0$ , $\lim_{n \rightarrow \infty} R(t)(n^{-1}) \in (-\infty, -k) \cup (k,\infty)$ then no such value exists.

lemma: Brownian motion is almost-surely not left-differentiable for any $t > 0$ .

Let $B$ be a brownian motion. For a given $t > 0$ , $\frac{B(t-h) - B(t)}{h^{1/2}}$ is normal(0,1) distributed. And so $q(h) = \frac{B(t-h)-B(t)}{h}$ is normal (0, $h^{-1/2}$ ) distributed. Fix a given $k > 0$ . The probability that for all $N$ there is an $n \ge N$ such that $q(n^{-1}) \in (-\infty,-k) \cup (k,\infty)$ is the probability that for all $N$ there is an $n \ge N$ such that $n^{-1/2} q(n^{-1}) \in A_{n,k} = (-\infty,-kn^{-1/2}) \cup (kn^{-1/2},\infty)$ .

Since $\left< A_{n,k} \right>$ is non-increasing in $k$ .

$P(\cap_k \cap_N \cup_{n \ge N} A_{n,k}) = \lim_k P(\cap_N \cup_{n \ge N} A_{n,k}) = \lim_k \lim_N P(\cup_{n \ge N} A_{n,k})$

Since $\left< A_{n,k} \right>$ is non-decreasing in $n$ ,

$P(\cup_{n \ge N} A_{n,k}) \ge P(\cap_{n \ge N} A_{n,k}) \ge P(A_{N,k})$ , it follows that $\lim_N P(\cup_{n \ge N} A_{n,k}) \ge \lim_N P(A_{N,k}) = 1$ since $\lim_N [(1 - N(kN^{-1/2})) + N(-kN^{-1/2}) ] = 1$ .

Hence the left derivative of $B$ almost surely does not exist.

Dually, it is possible to show the right derivative of $B$ almost surely does not exist. And so Brownian motion is almost surely not differentiable.

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