Why is $\lim_{t\to 0}W(t)=0$ equivalent to $\forall n\ge 1 \; \exists m\ge 1\; \forall r\in \mathbb{Q}\cap (0,\frac{1}{m}]:|W(r)|\le \frac{1}{n}$

Question

Let $t \mapsto W_t, t>0$ be continuous. We want to show that $\lim_{t\downarrow 0}W(t)=W(0)=0.$ $W$ here is actually the reflected Brownian motion $tB(1/t)$. It says in the text that the limit $\lim_{t\to 0}W(t,w)=0$ if, and only if, $$\forall n\ge 1 \; \exists m\ge 1\; \forall r\in \mathbb{Q}\cap (0,\frac{1}{m}]:|W(r)|=|rB(\frac{1}{r})|\le \frac{1}{n}.$$

I'm not sure why we only need to consider rational points here. What we want to show is that $\forall \epsilon>0$ there is some $\delta>0$ such that $|t|<\delta$ then $|W(t)|<\epsilon$. But how is this a consequence, or equivalent to the above statement? I tried showing this as follows:

We can choose $n$ such that $1/n<\epsilon$, and $\delta<1/m$. Then we want to show that $|W(t)|<\epsilon$ for $|t|<\delta$, so we can use triangle inequality $|W(t)|\le |W(r)|+|W(t)-W(r)|$. However, since we don't have uniform continuity, I can't bound $|W(t)-W(r)|$ uniformly in this interval $(0,\delta)$ so that $\delta$ does not depend on the specific $t$. How can I prove this? I would greatly appreciate any help.

Answer 1

Lemma Let $(B_t)_{t \geq 0}$ be a 1-dimensional Brownian motion. Then $$\mathbb{P} \left( \limsup_{t \to \infty} \left| \frac{B_t}{t} \right |\leq R \right) = 1$$ for any $R>0$.

Proof: By Doob's maximal inequality, we have

$$\begin{align*} \mathbb{P} \left( \sup_{n \leq t \leq n+1} \left| \frac{B_t}{t} \right| \geq R \right) &\leq \mathbb{P} \left( \sup_{n \leq t \leq n+1} |B_t|^4 \geq n^4 R^4 \right) \\ &\leq \frac{4}{n^4 R^4} \mathbb{E} \left( |B_{n+1}|^4 \right) \\ &\leq \frac{24}{n^2 R^4} \end{align*}$$

for any fixed $R>0$. Applying the Borel-Cantelli lemma we find

$$\mathbb{P} \left( \sup_{n \leq t \leq n+1} \left| \frac{B_t}{t} \right| \geq R \, \, \text{for infinitely many $n$} \right) = 0$$

implying

$$\mathbb{P} \left( \limsup_{t \to \infty} \left| \frac{B_t}{t} \right| \leq R \right) = 1.$$

---

Since $R>0$ is arbitrary, the above Lemma implies that

$$\mathbb{P} \left( \lim_{t \to \infty} \left| \frac{B_t}{t} \right| = 0 \right) = 1.$$

For the reflected Brownian motion $W_t := t B_{1/t}$ this means that

$$\mathbb{P} \left( \lim_{t \to 0} |W_t| = 0 \right) = 1.$$

Remark: The Lemma shows, in particular, that $\sup_{t \geq 1} |B_t/t| < \infty$ almost surely. This, in turn, implies that

$$\sup_{s \in (0,t)} |W_s| < \infty$$

almost surely for any $t>0$. Using that the mapping $[1,\infty) \ni t \mapsto t B(1/t)$ is continuous, it is not difficult to see that

$$\sup_{s \in (0,t)} |W_s| = \sup_{s \in (0,t) \cap \mathbb{Q}} |W_s|.$$

Therefore

$$\lim_{t \to 0} |W_t| = 0$$

is equivalent to

$$\lim_{t \to 0} \sup_{s \in (0,t) \cap \mathbb{Q}} |W_s| = 0,$$

i.e.

$$\forall n \geq 1 \, \, \exists m \geq 1 \, \, \forall s \in (0,1/m) \cap \mathbb{Q}: \, \, |W_s| \leq \frac{1}{n}.$$