Help understanding a proof of non-differentiablity of Brownian motion

Question

The following statement and proof are taken from the book Brownian Motion by Peter Morters and Yuval Peres. Since I initially didn't fully understand the proof I added some clarifications and I was hoping someone could confirm their correctness. Additionally there's one crucial step of the proof I still don't quite understand. Apologies for length, I wanted to be very detailed.

Theorem 1.3: Almost surely, Brownian motion is nowhere differentiable. Futhermore, almost surely for all $t$: $$\text{either}\qquad D^{*}B\left(t\right)=+\infty\quad\text{or}\quad D_{*}B\left(t\right)=-\infty\qquad\text{or both}$$

Proof: Suppose that there is a $t_{0}\in\left[0,1\right]$ such that $$\tag{1} -\infty<D_{*}B\left(t_{0}\right)\leq D^{*}B\left(t_{0}\right)<\infty$$ Then $$\tag{2} \limsup_{h\downarrow0}\frac{\left|B\left(t_{0}+h\right)-B\left(t_{0}\right)\right|}{h}<\infty$$ and, using the boundedness of Brownian motion on $\left[0,2\right]$ , this implies that for some finite constant $M$ there exists $t_{0}$ such that $$\tag{3} \sup_{h\in\left[0,1\right]}\frac{\left|B\left(t_{0}+h\right)-B\left(t_{0}\right)\right|}{h}\leq M$$ Clarifictation 1: The $t_{0}$ for which (2) and (3) hold is the same one we assumed (1) holds for?

It suffices to show that this event has probability zero for any $M$ . From now on, fix $M$ . If $t_{0}$ is contained in the binary interval $\left[\frac{k-1}{2^{n}},\frac{k}{2^{n}}\right]$ for $n>2$ then for all $1\leq j\leq2^{n}-k$ the triangle inequality gives $$\left|B\left(\frac{k+j}{2^{n}}\right)-B\left(\frac{k+j-1}{2^{n}}\right)\right|\leq\left|B\left(\frac{k+j}{2^{n}}\right)-B\left(t_{0}\right)\right|+\left|B\left(t_{0}\right)-B\left(\frac{k+j-1}{2^{n}}\right)\right|\overbrace{\leq}^{\dagger}M\left(\frac{2j+1}{2^{n}}\right)$$ Claification 2: the phrasing says "if $t_{0}$ is contained in the binary interval $\left[\frac{k-1}{2^{n}},\frac{k}{2^{n}}\right]$ for $n>2$ ", does it imply that there necessarily are $k$ and $n>2$ such that $t_{0}$ is contained in such an interval ?

Clarification 3: initially I didn't quite understand the last inequality, I'm wondering whether the following reasoning is the correct explanation. From equation (3) we got: $$\frac{\left|B\left(t_{0}+\left(\frac{k+j}{2^{n}}-t_{0}\right)\right)-B\left(t_{0}\right)\right|}{\left(\frac{k+j}{2^{n}}-t_{0}\right)}\leq M\Longrightarrow\left|B\left(\frac{k+j}{2^{n}}\right)-B\left(t_{0}\right)\right|\leq M\left(\frac{k+j}{2^{n}}-t_{0}\right)$$ $$\frac{\left|B\left(t_{0}\right)-B\left(t_{0}+\left(\frac{k+j-1}{2^{n}}-t_{0}\right)\right)\right|}{\left(\frac{k+j-1}{2^{n}}-t_{0}\right)}\leq M\Longrightarrow\left|B\left(\frac{k+j}{2^{n}}\right)-B\left(t_{0}\right)\right|\leq M\left(\frac{k+j-1}{2^{n}}-t_{0}\right)$$ Combining these two results and the fact $t_{0}\geq\frac{k-1}{2^{n}}$ we get:$$\left|B\left(\frac{k+j}{2^{n}}\right)-B\left(t_{0}\right)\right|+\left|B\left(t_{0}\right)-B\left(\frac{k+j-1}{2^{n}}\right)\right|\leq M\left[\left(\frac{k+j}{2^{n}}-t_{0}\right)+\left(\frac{k+j-1}{2^{n}}-t_{0}\right)\right] =M\left[\frac{2k+2j-1}{2^{n}}-2t_{0}\right]\leq M\left[\frac{2k+2j-1}{2^{n}}-2\frac{\left(k-1\right)}{2^{n}}\right]=M\left(\frac{2j+1}{2^{n}}\right)$$

Is this correct?

Onward, define the following events $$\Omega_{n,k}=\left\{ \left|B\left(\frac{k+j}{2^{n}}\right)-B\left(\frac{k+j-1}{2^{n}}\right)\right|\leq M\left(\frac{2j+1}{2^{n}}\right)\;\mbox{for }j=1,2,3\right\} $$ Then by independence of the increments and the scaling property of Brownian motion for $1\leq k\leq2^{n}-3$ we got $$\mathbb{P}\left(\Omega_{n,k}\right)\leq\prod_{j=1}^{3}\mathbb{P}\left\{ \left|B\left(\frac{k+j}{2^{n}}\right)-B\left(\frac{k+j-1}{2^{n}}\right)\right|\leq M\left(\frac{2j+1}{2^{n}}\right)\right\} \overbrace{\leq}^{\dagger}\mathbb{P}\left\{ \left|B\left(1\right)\right|\leq\frac{7M}{\sqrt{2^{n}}}\right\} ^{3}$$ Clarification 4: Doesn't the independence of the increments mean that the first inequality is actually an equality? additionally is it correct that choosing $j=1,2,3$ is what allowed the variation in $k$ such that we can take $1\leq k\leq2^{n}-3$ ?

Clarification 5: to understand the marked inequality I wanted to add some intermediate steps. The scaling propery states that if $\left\{ B\left(t\right),\, t\geq0\right\}$ is a standard Brownian motion then for all $a>0$ the process $\left\{ X\left(t\right),\, t\geq0\right\} $ defined by $X\left(t\right)=\frac{1}{a}B\left(a^{2}t\right)$ is also a standard Brownian motion. Specifically here the application is taken with $a=\frac{1}{\sqrt{2^{n}}}$ thus for $j=1,2,3$ and for every aforementioned $k$ we get that $$\mathbb{P}\left\{ \left|B\left(\frac{k+j}{2^{n}}\right)-B\left(\frac{k+j-1}{2^{n}}\right)\right|\leq M\left(\frac{2j+1}{2^{n}}\right)\right\} =\mathbb{P}\left\{ \left|\sqrt{2^{n}}B\left(\frac{1}{2^{n}}\left(k+j\right)\right)-\sqrt{2^{n}}B\left(\frac{1}{2^{n}}\left(k+j-1\right)\right)\right|\leq M\left(\frac{2j+1}{\sqrt{2^{n}}}\right)\right\} \leq\mathbb{P}\left\{ \left|\sqrt{2^{n}}B\left(\frac{1}{2^{n}}\left(k+j\right)\right)-\sqrt{2^{n}}B\left(\frac{1}{2^{n}}\left(k+j-1\right)\right)\right|\leq M\frac{7M}{\sqrt{2^{n}}}\right\} $$ Now since $X\left(t\right)=\sqrt{2^{n}}B\left(\frac{1}{2^{n}}t\right)$ is a standard Brownian motion we know that $\sqrt{2^{n}}B\left(\frac{1}{2^{n}}\left(k+j\right)\right)-\sqrt{2^{n}}B\left(\frac{1}{2^{n}}\left(k+j-1\right)\right)$ is actually a standard normal variable and thus we can replace it with $B\left(1\right)$ which is also a standard normal variable. Thus yielding the final inequality by replacing the product with the exponent. Correct?

Onward, the next statement is that $\mathbb{P}\left\{ \left|B\left(1\right)\right|\leq\frac{7M}{\sqrt{2^{n}}}\right\} ^{3}$ is at most $\left(7M2^{-\frac{n}{2}}\right)^{3}$ since the normal density is bounded by $\frac{1}{2}$ . If I'm not mistaken this is a result of $B\left(1\right)\sim N\left(0,1\right)$ and the following simple calculation $$\mathbb{P}\left\{ \left|B\left(1\right)\right|\leq\frac{7M}{\sqrt{2^{n}}}\right\} =\int_{-7M2^{-\frac{n}{2}}}^{7M2^{-\frac{n}{2}}}f\left(t\right)dt\leq\int_{-7M2^{-\frac{n}{2}}}^{7M2^{-\frac{n}{2}}}\frac{1}{2}dt=7M2^{-\frac{n}{2}} $$ where $f\left(t\right)$ is the standard normal density. From this follows that $$\mathbb{P}\left(\bigcup_{k=1}^{2^{n}-3}\Omega_{n,k}\right)\leq\left(2^{n}-3\right)\left(7M2^{-\frac{n}{2}}\right)^{3}\leq2^{n}\left(7M2^{-\frac{n}{2}}\right)^{3}=\left(7M\right)^{3}2^{-\frac{n}{2}}$$ Since $\sum_{n=1}^{\infty}2^{-\frac{n}{2}}<\infty$ we know from Borel-Cantelli's lemma that: $$\mathbb{P}\left(\bigcup_{k=1}^{2^{n}-3}\Omega_{n,k}\,\mbox{for infinitely many }n\right)=0$$ Clarification 6: now for the part I just don't understand, the proof concludes with $$\mathbb{P}\left\{ \mbox{there is }t_{0}\in\left[0,1\right]\,\mbox{such that}\:\sup_{h\in\left[0,1\right]}\frac{\left|B\left(t_{0}+h\right)-B\left(t_{0}\right)\right|}{h}\leq M\right\} \overbrace{\leq}^{\dagger}\mathbb{P}\left(\bigcup_{k=1}^{2^{n}-3}\Omega_{n,k}\,\mbox{for infinitely many }n\right)=0$$ I just don't understand why the marked inequality holds.

Clarification 7: Finally, assuming everything stated thus far is correct, this shows that Brownian motion is nowhere differentiable in $\left[0,1\right]$ and to expand this conclusion to the entire real line one would need to use the scaling and time-inversion properties of Brownian motion?

Thanks in advance to anyone who helps!

Answer 1

• C.1: Yes, exactly.
• C.2: Note that the sets $$\left[ \frac{k-1}{2^n}, \frac{k}{2^n} \right], \qquad k = 1,\ldots,2^n,$$ are a partition of the interval $[0,1]$. In particular, for any $t_0 \in [0,1]$ and $n \in \mathbb{N}$, we can pick $k \in \{1,\ldots,2^n\}$ such that $$t_0 \in \left[ \frac{k-1}{2^n},\frac{k}{2^n} \right].$$
• C.3: I guess you mean $t_0 \geq \frac{k-1}{2^n}$ instead of $t_0 \geq \frac{k-2}{2^n}$ but except for that your argumentation is correct.
• C.4: Yes, the first inequality is actually an equality. The condition $1 \leq k \leq 2^n-3$ is used in order to ensure that $\frac{k+j}{2^n} \leq 1$.
• C.5: Lots of typos here, I would say. The $\sqrt{2^n}$ should read $\sqrt{2^n}$, right? Again: Except for that, your reasoning is fine.
• C.6: It was shown that $$\sup_{h \in [0,1]} \frac{|B(t_0+h)-B(t_0)|}{h} \leq M$$ for $t_0 \in [0,1]$ implies $$ \left| B \left( \frac{k+j}{2^n} \right)-B \left( \frac{k+j-1}{2^n} \right) \right| \leq M \frac{2j+1}{2^n}$$ for all $j \leq 2^n-k$, i.e. $$\left\{ \sup_{h \in [0,1]} \frac{|B(t_0+h)-B(t_0)|}{h} \leq M \right\} \subseteq \bigcap_{j=1}^3 \left\{ \left| B \left( \frac{k+j}{2^n} \right)-B \left( \frac{k+j-1}{2^n} \right) \right| \leq M \frac{2j+1}{2^n} \right\} = \Omega_{n,k}$$ (where $k=k(t_0)$ is as in C.2). Now the claim follows from the fact that $\mathbb{P}(A) \leq \mathbb{P}(B)$ for $A \subseteq B$.
• C.7: Not necessarily. It suffices to note that $W_t^n := B_{t+n}-B_n$ defines a Brownian motion and that therefore $(W_t^n)_{t \in [0,1]}$ is a.s. nowhere differentiable. This shows that $$\begin{align*}\mathbb{P}( B_t \, \text{differentiable for some} \, t>0) &= \mathbb{P}(\exists n \in \mathbb{N}, t \in (0,1]: W_t^n \, \text{is differentiable}) \\ &\leq \sum_{n \geq 0} \mathbb{P} ( W_t^n \, \text{differentiable for some} \, t \in (0,1]) = 0. \end{align*}$$