I quote Mörters/Peres (2010).
First, consider the following result contained here in Lemma 1.1 (p.2) (Scaling invariance property of Brownian motion). Then
For a function $f$, we define the upper and lower right derivatives $$D^* f(t)=\limsup\limits_{h\downarrow0}\frac{f(t+h)-f(t)}{h}\tag{1}$$ $$D_* f(t)=\liminf\limits_{h\downarrow0}\frac{f(t+h)-f(t)}{h}\tag{2}$$ [...]
Given a standard Brownian motion $(B(n))_n$, by Lemma 1.1 (p.2) we construct a further Brownian motion $(X(n))_n$ (as stated there, $X(n)=\frac{1}{a}B(a^2 n)$). Then: $$D^* X(0)\color{red}{\ge}\limsup_{n\to\infty}\frac{X\left(\frac{1}{n}\right)-X(0)}{\frac{1}{n}}\color{red}{\ge}\limsup_{n\to\infty}\sqrt{n}X\left(\frac{1}{n}\right)\tag{3}$$ [...]
I cannot understand the two $\color{red}{\text{red}}$ inequalities above in $(3)$:
- As to the first one: I would expect, by definition $(1)$, that $D^* X(0)\color{red}{=}\limsup_{n\to\infty}\frac{X\left(\frac{1}{n}\right)-X(0)}{\frac{1}{n}}$ and not that $D^* X(0)\color{red}{\ge}\limsup_{n\to\infty}\frac{X\left(\frac{1}{n}\right)-X(0)}{\frac{1}{n}}$. Why is it that way?
- As to the second one: why does it hold true that $\limsup_{n\to\infty}\frac{X\left(\frac{1}{n}\right)-X(0)}{\frac{1}{n}}\color{red}{\ge}\limsup_{n\to\infty}\sqrt{n}X\left(\frac{1}{n}\right)$? How can one make a comparison between the $\limsup$ of $\frac{X\left(\frac{1}{n}\right)-X(0)}{\frac{1}{n}}$ and the $\limsup$ of $\sqrt{n}X\left(\frac{1}{n}\right)$?