$\lim \sup_{t\rightarrow \infty} \frac{W_t}{\sqrt{t}}$ question

Question

Why does this hold ? :

$$\lim \sup_{t\rightarrow \infty} \frac{W_t}{\sqrt{t}} \geq \lim \sup_{t\rightarrow \infty} \frac{W_{\lfloor{t}\rfloor}}{\sqrt{\lfloor{t}\rfloor}} $$

With $W_t$ a Brownian Motion

I can see that $\frac{1}{\sqrt{t}}\geq \frac{1}{\sqrt{\lfloor{t}\rfloor}}$ but what do we know in order to compare $W_{\lfloor{t}\rfloor}$ and $W_t$ ? Is that when $t\rightarrow \infty$ we consider $W_{\lfloor{t}\rfloor}$ and $W_t$ equal? if it's the case why?

Answer 1

This isn't really to do with Brownian motion but rather to do with the definition of $\lim \sup$. One definition of $\lim \sup_{t \to \infty} f(t)$ is $\lim \sup_{t \to \infty} f(t) = \sup E$ where $E$ $$E = \{x : \mbox{ there is an increasing sequence } t_n \to \infty \mbox{ such that } f(t_n) \to x \}$$ That is $E$ is the set of all subsequential limits of the sequence $f(t)$. One can define the $\lim \sup$ of a sequence in terms of subsequential limits similarly.

The desired inequality is then clear since the set of subsequential limits of the right hand side is contained in the set of subsequential limits of the left hand side.

Answer 2

This has nothing to do with the properties of $W_t$.

It follows from the fact that for $a_t$ a real valued sequence $\limsup_{t\rightarrow \infty}a_t\geq \limsup_{n \rightarrow \infty} a_n$.

Since, if $a$ is an accumulation point of $a_n$, then it also an accumulation point for $a_t$.