# Why does $\lim\limits_{ s \to 0} sB_{\frac{1}{s}}=0$ a.s.?

Let $$B$$ be the standard brownian motion: Why does $$\lim\limits_{ s \to 0} sB_{\frac{1}{s}}=0$$ a.s.?

I am thinking of applying the martingale convergence theorem here by some enumerated sequence $$(t_{n})_{n \in \mathbb N}$$ on $$Y_{n}:=B_{t_{n}}$$ such that

$$t_{n} \xrightarrow{n \to \infty} \infty$$. I am not sure whether the property $$\sup\limits_{n \in \mathbb N}E(Y_{n})_{+}<\infty$$ is satisfied, the best I could do was

$$E(Y_{n})_{+}\leq E[\lvert Y_{n} \rvert ]\leq \sqrt{E(Y_{n})^{2}}=\sqrt{t_{n}}$$ which clearly does not help me.

Assuming that it satisfies the necessary conditions for the martingale convergence theorem, then $$\lim Y_{n} = Y_{\infty}$$ almost surely where $$Y_{\infty}$$ is real-valued such that for a null set $$N$$

$$B_{t_{n}}(\omega)\xrightarrow{n \to \infty}Y_{\infty}(\omega)$$ for any $$\omega \in N^{c}$$

and hence $$\frac{1}{t_{n}}B_{t_{n}}\xrightarrow{ n \to \infty} 0$$ a.s.

Am I on the right track? If so, I need help to prove that $$\sup\limits_{n \in \mathbb N}E(Y_{n})_{+}<\infty$$

• $Y_n$ has same distribution as $\sqrt {t_n} X$ where $X$ has standard normal distribution. So $sup EY_n^{+}=\infty$ and your approach doesn't work. But you can find a proof of this result in many books. Aug 5, 2020 at 10:45

Note that this is equivalent to proving that $$\lim_{s\to \infty} \frac{1}{s} B_s=0$$ almost surely and by symmetry, it suffices to prove that $$\limsup_{s\to \infty} \frac{1}{s}B_s\leq 0$$ almost surely.
It's probably easiest to prove this via the Browian Reflection Principle. So that for any $$C>0$$ and any $$n$$, we have
$$\mathbb{P}(\sup_{0\leq t\leq n} B_t\geq Cn)=2\mathbb{P}(B_n\geq Cn)=2\mathbb{P}(\frac{B_n}{\sqrt{n}}\geq C\sqrt{n})$$ Since $$B_n/\sqrt{n}$$ is standard Gaussian, we can apply Markov's inequality to get that
$$\mathbb{P}(\frac{B_n}{\sqrt{n}}\geq C\sqrt{n})= \mathbb{P}(\frac{B_n^4}{n^2}\geq C^4n^2)\leq \frac{3}{C^4n^2}$$
Hence, we see that $$\sum_{n=1}^{\infty} \mathbb{P}(\sup_{0\leq t\leq n} B_t\geq C n)<\infty$$ and hence, by Borel-Cantelli's lemma, $$\limsup_{t\to\infty} \frac{B_t}{t}\leq C$$ almost surely. Letting $$C_k=\frac{1}{k},$$ we see that
$$\{\limsup_{t\to\infty} \frac{B_t}{t}\leq 0\}=\bigcap_{k=1}^{\infty}\{\limsup_{t\to\infty} \frac{B_t}{t}\leq C_k\}$$ is also almost sure and thus, we get the result.