Why $\frac{1}{nh}O(h)=o[(nh)^{-1}]$ and $O_p(h^2+(nh)^{-1/2})=o_p(1)$

Question

Suppose that we know: as $n\to\infty$, $h\to 0$ and $nh\to\infty$. Why does it follow that

1. $\frac{O(h)}{nh}=o[(nh)^{-1}]$,
2. $O_p(h^2+(nh)^{-1/2})=o_p(1)$?

I'm learning kernel density estimation from Li and Racine (2007). The above appeared on pages 9--12 of the text without any explanation. Could someone please help me figure them out?

Answer 1

Recall that $$ f=\mathcal{o}(g) \quad \text{as} \quad x \rightarrow +\infty \quad \text{if} \quad \frac{f}{g} \rightarrow 0. $$ Here, from your hypotheses, you have

1. $$ \frac{\frac{O(h)}{nh}}{(nh)^{-1}}=O(h)\rightarrow 0 \qquad \text{giving} \qquad \frac{O(h)}{nh}=o[(nh)^{-1}] $$

2. $$ \frac{O_p(h^2+(nh)^{-1/2})}{1}=O_p\left(\max (h^2,(nh)^{-1/2})\right)\rightarrow 0 $$

giving $$ O_p(h^2+(nh)^{-1/2})=o_p(1). $$