# Brownian motion is differentiable nowhere

The elementary continuous functions we work with in calculus tend to be relatively well-behaved, in the sense that they are differentiable everywhere except some isolated set of points. The prototypical example is the map ${x\rightarrow|x|}$, which fails to be differentiable at zero since ${\frac{x-|x|}x = \frac x{|x|}}$ for ${x\ne 0}$, so ${\lim_{x\rightarrow 0}\frac{x-|x|}x}$ does not exist. More generally, let ${x_0 and ${y_0,\ldots,y_n\in\mathbb R}$ with

$\displaystyle \frac{y_{i+2}-y_{i+1}}{x_{i+2}-x_{i+1}} \ne \frac{y_{i+1}-y_n}{x_{i+1}-x_n}$

– that is, the slope of the line segment joining ${(x_i,y_i)}$ and ${(x_{i+1}, y_{i+1})}$ differs from the slope of the line segment joining ${(x_{i+1}, y_{i+1})}$ and ${(x_{i+2},y_{i+2})}$. Define ${f:[x_0,x_n]\rightarrow\mathbb R}$ such that ${f(x_i)=y_i}$ and the graph of ${f}$ is

$\displaystyle \bigcup_{i=0}^{n-1} \left\{\lambda(x_i,f(x_i))+(1-\lambda)(x_{i+1},f(x_{i+1}):\lambda\in[0,1] \right\}.$

That is, ${f}$ is a piecewise linear function connecting the points ${(x_i,y_i)}$ – like the functions ${B^{(n)}}$ used in the previous post. Then ${f}$ is differentiable on ${(x_0,x_n)\setminus\{x_1,\ldots,x_{n-1}}$. Moreover, if ${\{x_n\}}$ is a strictly increasing sequence with ${L:=\sum_{n=0}^{\infty}(x_{n+1}-x_n)<\infty}$ satisfying the aforementioned slope assumption, then defining ${f_n}$ as above, we see that ${\lim_{n\rightarrow\infty}f_n}$ is continuous on ${(x_0,x_0+L)}$ and differentiable at all but countably many points.

The paths of Brownian motion are even more pathological – they are with probability ${1}$ differentiable nowhere! Let ${W=\{W_t:t\in\mathbb R_+\}}$ be a standard Brownian motion and

$\displaystyle D = \{\omega:\exists t\in(0,1) \textrm{ s.t. } W \textrm{ differentiable at } t\}.$

We will show that ${\mathbb P(D)=0}$. For each integer ${n\geqslant2}$ and ${1\leqslant k\leqslant n-2}$, set

$\displaystyle M(k,n) = \max\left\{|W_{kn^{-1}}-W_{(k-1)n^{-1}}|,|W_{(k+1)n^{-1}}-W_{kn^{-1}}|,|W_{(k+2)n^{-1}}-W_{(k+1)n^{-1}}| \right\}$

and

$\displaystyle M_n = \min\{M(k,n): 1\leqslant k\leqslant n-2\}.$

Suppose ${\omega_0\in D}$, then since

$\displaystyle \lim_{s\rightarrow t}\frac{W(s,\omega_0)-W(t,\omega_0)}{s-t}$

exists, we may choose ${\delta>0}$, ${C>0}$ such that

$\displaystyle |W(t,\omega_0)-W(s,\omega_0)|< C|t-s|.$

Let ${N}$ be a positive integer with ${N > \frac3\delta}$, and put

$\displaystyle k = \max\left\{j\in\mathbb N: jN^{-1}\leqslant t\right\}.$

Then

$\displaystyle \max\left\{\left|jN^{-1}-t\right| : j\in\{k-1,k,k+1,k+2\} \right\}<\frac3N<\delta,$

so that

\displaystyle \begin{aligned} \left|W_{kN^{-1}}-W_{(k-1)N^{-1}} \right| &\leqslant \left| W_{kN^{-1}}-W_t\right|+ \left| W_{(k-1)N^{-1}}-W_t\right|\\ &\leqslant C\left|kN^{-1}-t| + C|(k-1)N^{-1}\right|\\ &\leqslant C(N^{-1} + 2N^{-1})\\ &= 3CN^{-1}, \end{aligned}

and similarly

$\displaystyle \left|W_{(k+1)N^{-1}}t-W_{kN^{-1}}\right|<3CN^{-1},\quad \left|W_{(k+2)N^{-1}}t-W_{(k+1)N^{-1}}\right|<3CN^{-1}.$

It follows that

$\displaystyle M_N(\omega_0)\leqslant M(k,N)(\omega_0)<3CN^{-1}.$

Now, the increments ${W_{(j+1)N^{-1}}-W_{jN^{-1}}}$ are i.i.d. with ${\mathcal N\left(0,N^{-1}\right)}$ distribution, so

$\displaystyle W_{(j+1)N^{-1}}-W_{jN^{-1}}\stackrel d= n^{-\frac12}W_1,$

from which we have

$\displaystyle \mathbb P(M(k,N) \leqslant 3CN^{-1}) = \mathbb P\left(|W_1|\leqslant 3CN^{-\frac12}\right)^3.$

It follows that

\displaystyle \begin{aligned} \mathbb P(M_N\leqslant 3CN^{-1}) &= \mathbb P\left(\bigcup_{j=1}^N \left\{M(j,N)\leqslant 3CN^{-1}\right\}\right)\\ &\leqslant \sum_{j=1}^N \mathbb P\left(M(j,N)\leqslant 3CN^{-1}\right)\\ &\leqslant N\mathbb P\left(M(k,N)\leqslant 3CN^{-1} \right)\\ &= N\left(\mathbb P\left(|W_1|\leqslant 3CN^{-\frac12} \right)\right)^3\\ &\leqslant N\left(6CN^{-\frac12}\right)^3\\ &= (6C)^3N^{-\frac12}\stackrel{n\rightarrow\infty}\longrightarrow0. \end{aligned}

Let ${E_n = \{\omega: M_m(\omega) \leqslant 3Cn^{-1}\}}$, then ${\omega_0\in\liminf_{n\rightarrow\infty} E_n}$ so ${D\subset\liminf_{n\rightarrow\infty} E_n}$. Fatou’s lemma then yields

$\displaystyle \mathbb P(D)\leqslant \mathbb P\left(\liminf_{n\rightarrow\infty} E_n\right) \leqslant\liminf_{n\rightarrow\infty} \mathbb P(E_n)=\lim_{n\rightarrow\infty}\mathbb P(E_n)=0,$

from which we conclude.