I am reading a proof where it is assumed that $$ \lim_{n \to \infty} \sup_{0<s\leq s_0}\left| \frac{t_n(s)}{s}-1 \right|=0 , \tag{1}$$ 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 $$\lim_{n \to \infty} \sup_{0<s\leq s_0} \left| \frac{W(t_n(s))-W(s)}{t_n(s)^{1/2-\epsilon}} \right|=0 \quad \text{a.s.} \tag{3}$$

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 $\{s: 0<s\leq s_0\}$ only.


1 Answer 1


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

$$\sup_{0<s \leq s_0} \left| \frac{W(t_n(s))-W(s)}{t_n(s)^{1/2-\epsilon}} \right| \leq \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| + \delta.$$

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 $$\sup_{0<s \leq r} \left| \frac{W(s)}{s^{1/2-\epsilon}} \right| \leq \gamma.$$

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).


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