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<x_1<\ldots<x_n} 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.

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