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I have problems understanding the proof of why a Brownian motion $(B_t)_{t\geq 0}$ is not differentiable wrt $t$. The proof is as follows:


Let $(\Omega,\mathcal F, P)$ be a probability space, and $(B_t)_{t\geq 0}$ be a Brownian motion. The set $D\subseteq\Omega$ denotes the set of all $\omega$ for which the map $t\mapsto B_t(\omega)$ is differentiable. Furthermore, let $L\subseteq $ be the set of all $\omega$ for which the map $t\mapsto B_t(\omega)$ is Lipschitz-continuous. Since every differentiable function is Lipschitz, we have that $D\subseteq L$. Define $$M_{n,k} := \bigvee_{j=1}^3\vert B_{(k+j)2^{-n}} - B_{(k+j-1)2^{-n}}\vert,$$ and $$ E_{n,k} := \big\{\omega\in\Omega : M_{n,k}(\omega)\leq n2^{-n}\big\}$$ for each $n\in\mathbb N$ and $k\in[0,n2^n]\cap\mathbb Z$. Here $\bigvee$ denotes the maximum operator. (whats the point in defining these random variables? how does these random variable help me proving what I want to show?)

Since a Brownian motion has independent increments, it holds that $$P(E_{n,k}) = P\left(\big\{\omega\in\Omega : M_{n,k}(\omega)\leq n2^{-n}\big\}\right) = P\left(\bigcap_{j=1}^3\big\{\omega\in\Omega : \vert B_{(k+j)2^{-n}}(\omega) - B_{(k+j-1)2^{-n}}(\omega)\vert \leq n2^{-n}\big\}\right) \\ =\prod_{j=1}^3P\left(\big\{\omega\in\Omega : \vert B_{(k+j)2^{-n}}(\omega) - B_{(k+j-1)2^{-n}}(\omega)\vert \leq n2^{-n}\big\}\right) \\ = \prod_{j=1}^3P\left(\big\{\omega\in\Omega : \vert B_{2^{-n}}(\omega)\vert \leq n2^{-n}\big\}\right) \\ = P\left(\big\{\omega\in\Omega : \vert B_{2^{-n}}(\omega)\vert \leq n2^{-n}\big\}\right)^3.$$

(these bounds make no sense to me; why not chose $\exp(-n)$ instead of $2^{-n}$? And why $n$ and not $n^{-1}$ or something else?) Moreover, since the increments are normally distributed with mean zero and variance $2^{-n}$, it holds that $$P(E_{n,k}) = P\left(\big\{\omega\in\Omega : \vert B_{2^{-n}}(\omega)\vert \leq c_n\big\}\right)^3 \leq \left(2\sqrt{2^n}n2^{-n}\right)^3 = 8n^32^{-3/2n}.$$ Now we define $\tilde M_n = \bigwedge_{k\in[0,2^n-3]\cap\mathbb Z}M_{n,k}$, and $F_n := \big\{\omega\in\Omega : \tilde M_n(\omega)\leq n2^{-n}\big\}$ for each $n\in\mathbb N$ (again what is the purpose in defining these variables? I don't understand why we take the minimum here. And I also don't understand why $n2^n-3$ is chosen). Then $$F_n\subseteq\bigcup_{k=0}^{n2^n-3} E_{n,k}$$ (why?) and hence $$P\left(\big\{\omega\in\Omega : \tilde M_n(\omega)\leq n2^{-n}\big\}\right)\leq (n2^n)(8n^32^{-3/2n}) = 8n^4\sqrt{2^{-n}}.$$ This proves that $P(\liminf_{n\rightarrow\infty} F_n) = 0$ (why?). To show that $L\subseteq\liminf_{n\rightarrow\infty} F_n$ pick $\omega\in L$, i.e. there exists a $t_0\geq 0$ and $C>0$ and $\delta>0$ such that $$\vert B_{t_0+h}(\omega) - B_{t_0}(\omega)\vert\leq Ch$$ for all $s\in[0,\delta]$ (why? I assume because we are in the set of Lipschitz continuity. But this definition appears to be very different from what is usually taken as definition of Lipschitz continuity). For $N\in\mathbb N$ large enough, this implies that $$\vert B_{(k+j)2^{-n}}(\omega) - B_{(k+j-1)2^{-n}}(\omega)\vert\leq 8C2^{-N},$$ (why?) and therefore both $M_{N,k}\leq 8C2^{-N}$ and $\tilde M_{N}(\omega)\leq 8C2^{-N}$. Consequently, $\omega\in F_n$ for all $n\geq N$ (why?). Thus, $\omega\in\liminf_{n\rightarrow\infty} F_n$ (why?).


I have marked my questions with boldface at the corresponding spots. I am grateful for any help

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1 Answer 1

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I'll try to go through your questions in order:

  1. "What's the point of the $M_{n,k}$ and $E_{n,k}$?" We are trying to show that, for almost every $\omega$, $t \mapsto B_t(\omega)$ is nowhere differentiable. To do this, we will show that $P(t \mapsto B_t(\omega) \text{ is differentiable at some }t > 0) = 0$. Since a function is differentiable at some point $t$ implies it is Lipschitz in some neighborhood of that point (maybe not true, see lemma below), we will show $t \mapsto B_t(\omega)$ is not Lipschitz in any neighborhood. If $t \mapsto B_t(\omega)$ is Lipschitz with constant $L$ on $[(k+j)2^{-n},(k+j+1)2^{-n}]$, then $\vert B_{(k+j)2^{-n}} - B_{(k+j-1)2^{-n}}\vert \le L2^{-n}$, so $E_{n,k}$ contains the event "$t \mapsto B_t(\omega)$ is Lipschitz with constant $n$ on the three consecutive intervals $[k2^{-n},(k+1)2^{-n}]$, $[(k+1)2^{-n},(k+2)2^{-n}]$, $[(k+2)2^{-n},(k+3)2^{-n}]$." The reason we need three consecutive intervals instead of two or four comes up later in the proof. I think the proof is easier to understand if we use a different Lipschitz constant than $n$ here, so I'm going to let $$E_{n,k,L} := \{\omega : M_{n,k}(\omega) \le L 2^{-n}\}$$ be the event "$t \mapsto B_t(\omega)$ is Lipschitz with constant $L$ on the three consecutive intervals $[k2^{-n},(k+1)2^{-n}]$, $[(k+1)2^{-n},(k+2)2^{-n}]$, $[(k+2)2^{-n},(k+3)2^{-n}]$."
  2. "Why these bounds? Why not $e^{-n}$ instead of $2^{-n}$ or $n^{-1}$ instead of $n$?" We chose $2^{-n}$ because we are looking at partitioning the interval with the Dyadic rationals. We could partition the interval with something besides Dyadic rationals, but the Dyadic rationals are convenient because they are a refining partition and easy to write out. In terms of our event $E_{n,k,L}$, this step would show $$P(E_{n,k,L}) = P\left(\big\{\omega\in\Omega : M_{n,k}(\omega)\leq L2^{-n}\big\}\right) \\ = P\left(\bigcap_{j=1}^3\big\{\omega\in\Omega : \vert B_{(k+j)2^{-n}}(\omega) - B_{(k+j-1)2^{-n}}(\omega)\vert \leq L2^{-n}\big\}\right) \\ =\prod_{j=1}^3P\left(\big\{\omega\in\Omega : \vert B_{(k+j)2^{-n}}(\omega) - B_{(k+j-1)2^{-n}}(\omega)\vert \leq L2^{-n}\big\}\right) \\ = \prod_{j=1}^3P\left(\big\{\omega\in\Omega : \vert B_{2^{-n}}(\omega)\vert \leq L2^{-n}\big\}\right) \\ = P\left(\big\{\omega\in\Omega : \vert B_{2^{-n}}(\omega)\vert \leq L2^{-n}\big\}\right)^3 \\ \le 8L^3 2^{-3n/2}$$ The use of $n$ in the original proof is because it shows that $t \mapsto B_t(\omega)$ is not Lipschitz with constant $n$ for any $n$; here we're going to show it's not Lipschitz with constant $L$ for any $L$. I agree that reusing $n$ is somewhat confusing, which is why I introduced $L$ even though it ends up playing the same role.
  3. "What is the point of $\bar M_n := \wedge_{k \in [0,n2^n-3] \cap \mathbb{N}} M_{n,k}$ and $F_n$?" Here $\bar M_n$ is the smallest of the $M_{n,k}$, so $F_n$ contains the event "$t \mapsto B_t(\omega)$ is Lipschitz with constant $n$ on the three consecutive intervals $[k2^{-n},(k+1)2^{-n}]$, $[(k+1)2^{-n},(k+2)2^{-n}]$, $[(k+2)2^{-n},(k+3)2^{-n}]$ for ANY $k \in [0,n2^n-3]$." I'm completely lost on why they chose $n2^n-3$ for the upper bound here, but the point is that it's just a finite bound that goes to $\infty$ when $n$ goes to $\infty$ so that we eventually are looking at all of $[0,\infty)$. In terms of the $L$ we were using above, we would set $$F_{n,L} := \{ \omega : \bar M_n(\omega) \le L 2^{-n}\}$$
  4. "Why is $F_n\subseteq\bigcup_{k=0}^{n2^n-3} E_{n,k}$?" I think this is addressed in the point above: $F_n$ is the event that the bound from ANY one of the $E_{n_k}$s hold. In terms of the $L$ I've been using, we would have $$F_{n,L}\subseteq\bigcup_{k=0}^{n2^n-3} E_{n,k,L}.$$
  5. "Why is $P(\liminf F_n) = 0$?" They showed $P(F_n) \le 8n^4 2^{-n/2}$*(see below), so $$\sum_{n=1}^\infty P(F_n) < \infty.$$ By the Borel-Cantelli lemma, this implies $P(\limsup F_n) = 0$ and therefore $P(\liminf F_n) \le P(\limsup F_n) = 0$. In terms of $L$, we have $P(F_{n,L}) \le 8 L^4 2^{-n/2}$ so $P(\liminf_{n \rightarrow \infty} F_{n,L}) = 0$.

The remainder of the proof I think is addressed by my method of going through everything in terms of a different constant $L$. Since $L$ was arbitrary, and we showed that $P(t \mapsto B_t(\omega)\text{ is Lipschitz with constant }L\text{ in any neighborhood}) = 0$, we conclude that $t \mapsto B_t(\omega)$ is not Lipschitz on any neighborhood and therefore nowhere differentiable. I'm a little confused by your question on Lipschitz continuity near the end: It looks like the standard definition to me. What is the definition of Lipschitz continuity you are working with?

EDIT: Technically we showed that $$P(t \mapsto B_t(\omega)\text{ is Lipschitz with constant }L\text{ in any neighborhood of the form } [k2^{-n},(k+1)2^{-n}] ) = 0$$, but from the density of the Dyadic rationals any neighborhood $(a,b)$ must contain some neighborhood of the form $[k2^{-n},(k+1)2^{-n}]$ so this implies $$P(t \mapsto B_t(\omega)\text{ is Lipschitz with constant }L\text{ in any neighborhood}) = 0$$

EDIT 2: *This is why we needed three consecutive intervals: It gave us a factor of $2^{-3n/2}$ in $P(E_{n,k})$, so when we multiply by $2^n$ when computing $P(F_n)$ the series is still summable. We could have chosen more than three intervals, but could not have chosen less.

To address the question of why this is enough to show it's not differentiable at any point, we have the following lemma, adapted from Exercise 1.2.9 in Revuz and Yor's Continuous Martingales and Brownian Motion:

Lemma: If a function $f$ is differentiable at a point $x$ then there exists a constant $L$ and $N \in \mathbb{N}$ such that for $n \ge N$ we have $$|f((k+j)2^{-n}) - f((k+j - 1)2^{-n})| < L 2^{-n}$$ for $j = 1,2,3$ and $k \in \mathbb{N}$ satisfying $(k-1)2^{-n} < x < k2^{-n}$ (so $k = [2^n x] + 1$).

Proof: By the definition of differentiability, there exists $\delta > 0$ such that if $|x-y| < \delta$ then $|f(x)-f(y)-f'(x)(x-y)| < |x-y|$. Choose $n$ large enough so that $4 \cdot 2^{-n} < \delta$ and let $k$ and $j$ be as above. For $y = (k+j)2^{-n}$ and $z = (k+j-1)2^{-n}$ we have $|x-y| = y-x < 4 \cdot 2^{-n} < \delta$ and similarly $|x-z| = z-x < \delta$. We compute \begin{align*} |f(y)-f(z)| &= |f(y)-f(x)+f(x)-f(z)| \\ &= |f(y)-f(x)-f'(x)(y-x) + f(x)-f(z) - f'(x)(x-z) + f'(x)(y-z)| \\ &\le |f(y)-f(x)-f'(x)(y-x)| + |f(x)-f(z) - f'(x)(x-z)| + |f'(x)|2^{-n} \\ &\le |y-x| + |x-z| + |f'(x)| 2^{-n} \\ &\le 4 \cdot 2^{-n} + 4 \cdot 2^{-n} + |f'(x)| 2^{-n} \\ &= (8+|f'(x)|) \cdot 2^{-n}, \end{align*} so let $L = 8 + |f'(x)|$.

So our proof showed that we don't have $|B((k+j)2^{-n}) - B((k+j - 1)2^{-n})| < L 2^{-n}$ for any $k,L \in \mathbb{N}$ and hence $B$ cannot be differentiable at any point.

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    $\begingroup$ what a great answer! Very well explained! thank you! I think I now fully understand the proof and what is going on in the proof (I even think I know why we chose $n2^n-3$ as we a partitioning $[0,n]$ and if I plug in $n2^n-3$ für $k$ I end up with $n$ as expected). However, there is just a minor thing which I did not understand: I claimed that every differentiable function is Lipschitz (which is not true; math.stackexchange.com/questions/2368526/…). how do I justify non-Lipschitz $\implies$ non-differentiable? $\endgroup$
    – lmaosome
    May 5, 2021 at 16:26
  • $\begingroup$ thanks for the edit! $\endgroup$
    – lmaosome
    May 5, 2021 at 17:29
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    $\begingroup$ That's a very good point. I added a lemma expanding on that a bit. I don't know that we can conclude non-Lipschitz on any neighborhood implies nowhere differentiable, but here we have some additional control on how "spaced out" the $y$ and $z$ are in terms of how far they are from $x$ $\endgroup$ May 5, 2021 at 17:30

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