Order of A Random Sequence Given that $\sup_{x\in\mathcal{X}}|\widehat{f}_n(x)-f(x)|=O_P(a_n)$ for $a_n\to0$, to characterize the order of $\sup_{x\in\mathcal{X}}\left|\frac{1}{\widehat{f}_n(x)}-\frac{1}{f(x)}\right|$, we can do Taylor expansion to get
\begin{align*}
\sup_{x\in\mathcal{X}}\left|\frac{1}{\widehat f_n(x)}-\frac{1}{f(x)}\right|
&=\sup_{x\in\mathcal{X}}\left|\frac{-1}{\overline f_n^2(x)}\left(\widehat f_n(x)-f(x)\right)\right|\\
&\le\sup_{x\in\mathcal{X}}\left|\frac{1}{\overline f_n^2(x)}\right|\sup_{x\in\mathcal{X}}\left|\widehat f_n(x)-f(x)\right|,
\end{align*}
where $\overline f_n(x)$ is between $\widehat f_n(x)$ and $f(x)$. My question is that what assumptions do we need to control the term $\sup_{x\in\mathcal{X}}\left|\frac{1}{\overline f_n^2(x)}\right|$ so that we have the right hand side to be $O_P(a_n)$? And how do we show this under those assumptions?
 A: 
If $\inf_{x\in\mathcal X} f(x) \geq C > 0$ is true, then 
  $$ \sup_{x\in\mathcal X} \left\vert \frac{1}{\hat{f}_n(x)} - \frac{1}{f(x)} \right\vert = O_P(a_n). $$

Proof: By assumptions, for each $\varepsilon>0$ there exists $M>0$ such that for all $n$ large enough
$$ P\left( \sup_{x\in\mathcal X} \left\vert f(x) - \hat{f}_n(x) \right\vert > M a_n \right) < \varepsilon.$$
If $f$ is bounded from below as above,
$$ \sup_{x\in\mathcal X} \left\vert f(x) - \hat{f}_n(x) \right\vert \geq \left\vert \inf_{x\in\mathcal X} f(x) - \inf_{x\in\mathcal X}\hat{f}_n(x) \right\vert \geq C - \inf_{x\in\mathcal X}\hat{f}_n(x),$$
and for $n$ large enough for $C - a_n M \geq C/2$
$$ P\left( \inf_{x\in\mathcal X} \hat{f}_n(x) \geq C/2 \right) \geq P\left( \sup_{x\in\mathcal X} \left\vert f(x) - \hat{f}_n(x) \right\vert \leq a_n M\right) \geq 1 - \varepsilon $$
for $n$ large enough. Thus, for $n$ large enough,
$$ P\left( \sup_{x\in\mathcal X} \frac{1}{\bar{f}^2_n(x)} \leq \frac{4}{C^2} \right) \geq 1 - \varepsilon. $$
Now we can bound 
$$ P\left( \sup_{x\in\mathcal X} \frac{1}{\bar{f}^2_n(x)} \sup_{x\in\mathcal X} \left\vert \frac{1}{\hat{f}_n(x)} - \frac{1}{f(x)} \right\vert > M a_n \right) \leq P\left( \frac{4}{C^2} \sup_{x\in\mathcal X} \left\vert \frac{1}{\hat{f}_n(x)} - \frac{1}{f(x)} \right\vert > M a_n \right) + \varepsilon, $$
and since $O_P(a_n)$ implies $O_P(C^2 a_n/4)$, the last expression can be for $n$ large enough bounded by $2\varepsilon$, yielding the result.
