# Understanding a proof, Brownian motion convergence

I am reading a proof where it is assumed that $$\lim_{n \to \infty} \sup_{0 where $$t_n(\cdot)$$ is some sequence of functions. Furthermore, we know that $$\lim_{s \to 0^{+}}s^{-1/2+\epsilon}W(s)=0 \quad \text{a.s.}\tag{2}$$ for $$W(s)$$ a Brownian motion. The author then wants to prove that $$\lim_{n \to \infty} \sup_{0

The author then writes something along the lines of:

Take a sequence $$s_n \to s_0 \geq 0, n \to \infty$$". Then by (1) also $$t_n(s_n) \to s_0$$. For $$s_0>0$$ by continuity of Brownian motion (3) is true. For $$s_0=0$$ we use (2)

I have trouble understanding this proof, why are we looking at a sequence $$s_n$$? I suppose the mentioned continuity of the Brownian motion here is in fact uniform continuity because we are looking at $$\{s: 0 only.

## 1 Answer

Fix $$\delta>0$$. By the very definition of the supremum, there exists for any $$n \in \mathbb{N}$$ some $$s_n \in (0,s_0]$$ such that

$$\sup_{0

If we can show that

$$\limsup_{n \to \infty} \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| =0 \tag{4}$$

then this proves the assertion.

Proof of $$(4)$$: There exists a subsequence $$(s_n')_{n \in \mathbb{N}}$$ of $$(s_n)_{n \in \mathbb{N}}$$ such that the $$\limsup$$ in $$(4)$$ is attained, i.e.

$$\lim_{n \to \infty} \left| \frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}} \right| = \limsup_{n \to \infty} \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| \tag{5}$$

Since $$(s_n')_{n \in \mathbb{N}}$$ is contained in the compact interval $$[0,s_0]$$, we may assume without loss of generality that $$u := \lim_{n \to \infty} s_n'$$ exists (otherwise we take another subsequence). In order to prove $$(4)$$, we will show that the left-hand side of $$(5)$$ is zero, and to this end we consider two cases separately.

Case 1: $$u>0$$. Since $$s_n' \to 0$$ implies, by $$(1)$$, $$t_n(s_n') \to u$$ it follows from the continuity of the sample paths of Brownian motion that

$$\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}} \xrightarrow[]{n \to \infty} \frac{W(u)-W(u)}{u^{1/2-\epsilon}} = 0.$$

Case 2: $$u=0$$. Fix some $$\gamma>0$$. Because of $$(2)$$, there exists $$r>0$$ such that $$\sup_{0

As in Case 1 we have $$t_n(s_n') \to u$$ and so $$t_n(s_n') \to 0$$. In particular, we can choose $$N \in \mathbb{N}$$ such that $$|t_n(s_n')| \leq r \quad \text{and} \quad |s_n'| \leq r$$ for all $$n \geq N$$. Hence,

\begin{align*} \left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| &\leq \left| \frac{W(t_n(s_n'))}{t_n(s_n')^{1/2-\epsilon}} \right| + \frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \left| \frac{W(s_n')}{|s_n'|^{1/2-\epsilon}} \right| \\ &\leq \gamma + \frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \gamma \end{align*}

for all $$n \geq N$$. It is immediate from $$(1)$$ that

$$\frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \leq 1+ \gamma$$

for $$n$$ sufficiently large, and therefore we conclude that

$$\left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| \leq \gamma + (1+\gamma) \gamma$$

for all $$n$$ sufficiently large. As $$\gamma>0$$ was arbitrary, this proves

$$\lim_{n \to \infty} \left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| =0.$$

Remark: Note that we have actually shown that $$(4)$$ holds for any sequence $$(s_n)_{n \in \mathbb{N}} \subseteq (0,s_0]$$ (... and not only for the sequence which we picked at the very beginning of this answer).