Brownian Motion Is Nowhere Differentiable

Brownian motion is continuous, but its paths oscillate too violently to have a tangent line. The right derivative is controlled by the secant quotients

$$\begin{equation*}\frac{B(t+h)-B(t)}{h}, \qquad h>0.\end{equation*}$$

For smooth functions these quotients settle as $h\downarrow0$. For Brownian motion they almost surely fail to settle at every time.

PALEY, WIENER AND ZYGMUND

Almost surely, Brownian motion is nowhere differentiable.

The main theorem

The upper and lower differential

DEFINITION

For a function $f$, we define the upper and lower right derivatives

$$\begin{equation*}D^* f(t) = \limsup_{h\downarrow0}\frac{f(t+h)-f(t)}{h},\end{equation*}$$

and

$$\begin{equation*}D_* f(t) = \liminf_{h\downarrow0}\frac{f(t+h)-f(t)}{h}.\end{equation*}$$

PALEY, WIENER AND ZYGMUND

Almost surely, Brownian motion is nowhere differentiable. Furthermore, almost surely, for every $t$,

$$\begin{equation*}D^*B(t)=+\infty \quad \text{or} \quad D_*B(t)=-\infty \quad \text{or both},\end{equation*}$$

where

$$\begin{equation*}D^*B(t)=\limsup_{h\downarrow0}\frac{B(t+h)-B(t)}{h}, \qquad D_*B(t)=\liminf_{h\downarrow0}\frac{B(t+h)-B(t)}{h}.\end{equation*}$$

The proof

It is enough to prove the statement on $[0,1]$. Suppose that for some $t_0\in[0,1]$ both one-sided derivative bounds are finite. Since Brownian motion is bounded, this implies that for some finite $M$,

$$\begin{equation*}\sup_{0<h\leq1}\frac{|B(t_0+h)-B(t_0)|}{h}\leq M.\end{equation*}$$

We show that this event has probability zero for every fixed $M$.

Fix $n$ and let $t_0$ lie in the dyadic interval

$$\begin{equation*}t_0 \in \left( \frac{k-1}{2^{n}},\,\frac{k}{2^{n}}\right].\end{equation*}$$

For $j=1,2,3$, the triangle inequality gives

$$\begin{align*}\left|B(\frac{k+j}{2^{n}})-B(\frac{k+j-1}{2^{n}})\right| & =\left|B(\frac{k+j}{2^{n}})-B(t_0)+ B(t_0) - B(\frac{k+j-1}{2^{n}}) \right| \\&\leq \left|B(\frac{k+j}{2^{n}})-B(t_0)\right| + \left|B(t_0)-B(\frac{k+j-1}{2^{n}})\right| \\&\leq M \frac{2j+1}{2^{n}}.\end{align*}$$

Define the event $\Omega_{n,k}$ that three neighboring increments are all that small,

$$\begin{equation*}\Omega_{n,k}:=\left\{\left|B((k+j)2^{-n})-B((k+j-1)2^{-n})\right|\leq M(2j+1)2^{-n}\text{ for } j=1,2,3\right\}.\end{equation*}$$

By independence of Brownian increments and Brownian scaling,

$$\begin{align*}\mathbb{P}(\Omega_{n,k}) \leq\prod_{j=1}^{3} \mathbb{P}\left\{ |B(1)|\leq M \frac{2j+1}{2^{n/2}}\right\}\leq \mathbb{P}\left\{|B(1)|\leq \frac{7M}{2^{n/2}}\right\}^{3}.\end{align*}$$

Since the standard normal density is bounded by $1/2$,

$$\begin{equation*}\mathbb{P}(\Omega_{n,k})\leq (7M2^{-n/2})^3.\end{equation*}$$

Therefore

$$\begin{equation*}\mathbb{P}\left(\bigcup_{k=1}^{2^n-3}\Omega_{n,k}\right)\leq2^n(7M)^3 2^{-3n/2} = (7M)^3 2^{-n/2},\end{equation*}$$

which is summable over all $n$. Hence, by the Borel-Cantelli lemma,

$$\begin{equation*}\mathbb{P}\left\{\text{there is } t_0\in[0,1]\text{ with }\sup_{0<h\leq1}\frac{|B(t_0+h)-B(t_0)|}{h}\leq M\right\} \leq\mathbb{P}\left(\bigcup_{k=1}^{2^n-3}\Omega_{n,k}\text{ for infinitely many } n\right) =0.\end{equation*}$$

Taking the countable union over integer $M$ proves the theorem. $\Box$

References

• [PaleyEtAl1933]Paley, R. E. A. C., N. Wiener, and A. Zygmund. ``Notes on random functions.'' Mathematische Zeitschrift 37 (1933): 647–668. PDF.
• [DvoretzkyEtAl1961]Dvoretzky, A., P. Erdős, and S. Kakutani. ``Nonincrease everywhere of the Brownian motion process.'' In Proceedings of the Fourth Berkeley Symposium on Mathematical Statistics and Probability, volume 2, 103–116. University of California Press, 1961. PDF.
• [MortersPeres2010]Mörters, Peter, and Yuval Peres. Brownian Motion. First edition. Cambridge University Press, 2010. PDF.