Remarkable fact about Brownian Motion #3: It is nowhere differentiable

A good problem to consider at the start of an introduction to analysis is whether continuous functions need to be differentiable on a large subset of the domain. Clearly from the definition, differentiability at a point is a stronger condition than continuity – consider the modulus function. Our intuition about what a continuous function looks like in general might suggest that a continuous can only be non-differentiable at ‘a few’ points, perhaps a set of measure 0?

But in fact there exist functions which are continuous but nowhere differentiable. The canonical example is due to Weierstrass: you create something like a fractal with a saw-tooth function as a base. Essentially, having a non-zero derivative at a point means the function is monotonic on an interval around that point, and this construction prevents this happening for any point. Eliminating the possibility that the derivative is 0 is reasonably straightforward. Informally, if you look at the function, it looks like the derivative ought to be +K or -K at every point (if it exists) where K is a constant determined by the precise details of the construction.

Having found one example, it then seems likely that the majority of continuous functions ought to be nowhere-differentiable, since you could take a limit of spiky functions in lots of ways. Proving that Brownian Motion almost surely has this property quantifies this statement. The Wiener measure associated with BM is the most natural measure on C[0,1] and so this will indeed show that almost all continuous functions have this property. The result was first shown by Paley, Wiener and Zigmund in the 30s, and the proof here is based on that of Dvoretsky, Erdos and Kakutani (1961) as paraphrased by Peres in some excellent notes on Brownian sample paths. Note that it is reasonably straightforward to show by the Strong Markov Property and Blumenthal’s 0-1 Law that BM is not differentiable at a given point (see previous post), but it is not possible to lift this to the whole line simultaneously because of uncountability.

Theorem: Brownian Motion is nowhere-differentiable almost surely.

Proof: We restrict our attention to right-differentiability. Heuristically, being differentiable at a point gives strong restrictions on behaviour on a neighbourhood of the point, but no information about the size of that neighbourhood. The aim is to use the triangle inequality to lift the condition about the point to any form of regularity condition on the whole domain, and hope that BM doesn’t have that condition.

Suppose B(t) is (right-)differentiable at T. This means that $\sup_h h^{-1}|B(T+h)-B(T)|$ is bounded by some integer M, where the sup is taken all h where what follows is defined. Later we will take a countable union over all such M. We look at how this influences the function on the dyadic intervals of width $2^{-n}$ . Concretely, choosing k so that $T\in[\frac{k-1}{2^n},\frac{k}{2^n})$ , then for all j on which it is meaningful, by application of the triangle inequality $|B(\frac{k+j}{2^n})-B(\frac{k+j-1}{2^n})|\leq 2^{-n}M(2j+1)$ . This looks promising, because the increments which we are looking at are independent. Hence, with n and k as chosen, and using Brownian rescaling, the probability that BM has this property is $\mathbb{P}(|B(1)|\leq 3.2^{-n/2}M)\mathbb{P}(|B(1)|\leq 5.2^{-n/2}M)\mathbb{P}(|B(1)|\leq 7.2^{-n/2}M)\ldots$ where this product is over the number of terms that stay within the domain. In particular, for all k except those within 2 of $2^n$ , it is $\leq\mathbb{P}(|B(1)|\leq 7.2^{-n/2}M)^3$ . B(1)~N(0,1) which has density function less than 1/2 at all points, so by comparing integrals this is $\leq 7.2^{-n/2}M$ .

This crude bound has given a factor of $n^{-3/2}$ which means if we call the event $A_{n,k}$ we can take a union over k to get $\mathbb{P}(\cup_k A_{n,k})\leq (7M)^32^{-n/2}$ . Ie, as n increases, the probability that this can happen in any way tends to 0 reasonably rapidly. In particular, the sum over n is finite, so by BC1, the probability that $A_n$ happens for infinitely many n is 0. But of course, if we have differentiability at T, this needs to hold for all n large enough that T is more than two $2^{-n}$ quantiles away from 1 (so the above three-interval argument makes sense). So almost surely there is no T which satisfies the supremum bound for the given M. Now take the countable union over M, so there is no T satisfying the bound for any M. But this was a necessary condition for right-differentiability, and so a.s. there is no T at which B is differentiable.

References

Dvoretsky, Erdos, Kakutani: Non-increase everywhere of the Brownian Motion process

Paley, Wiener, Zygmund: Notes on Random Functions

Peres: An Invitation to Sample Paths of Brownian Motion (link)

http://en.wikipedia.org/wiki/Weierstrass_function

3 thoughts on “Remarkable fact about Brownian Motion #3: It is nowhere differentiable”

Pingback: Advanced Probability Revision Summary | Eventually Almost Everywhere
Pingback: Ornstein-Uhlenbeck Process | Eventually Almost Everywhere
Bill on April 5, 2016 at 1:32 am said:

This is incredibly pedantic, but I’m pretty certain that in the proof it is supposed to be: [k/2^n,(k+1)/2^n), with k ranging from 0 to (2^n-1). Not that this changes the concept at all.

Reply ↓

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Remarkable fact about Brownian Motion #3: It is nowhere differentiable

3 thoughts on “Remarkable fact about Brownian Motion #3: It is nowhere differentiable”

Leave a Reply Cancel reply

Share this:

Like this:

Related

3 thoughts on “Remarkable fact about Brownian Motion #3: It is nowhere differentiable”

Leave a Reply Cancel reply