Is the space of continuous function with given order of decrease closed? For example, denote $O^1$ the space of continuous function with property $\lim_{x\to\infty}{|x|f(x)}=0$ or $f(x)=o\left(|x|^{-1}\right)$ as $x\to\infty$. It's obviously an intermediate vector space between $C_c$ and $C_0$. Is it closed?
Assume knowledge of completeness of $C_0$, then for a convergent sequence $f_n\to f$, we know $f\in C_0$. Now we prove $f\in O^1$. $\forall\varepsilon,\exists N_1,\text{s.t. }d_{\sup}{(f,f_{N_1})}<\varepsilon /2,\exists N_2,\text{s.t. } |xf_{N_1}(x)|<\varepsilon /2\text{ if }|x|>N_2,\text{ then }...$ Ah, I see the argument break down here.
$O^1$ can't be closed since if it's closed, it will contradict the fact that $C_0=\overline{C_c}$.
My question is: is there any example of a sequence of continuous functions with given order of decrease uniformly convergent to a limit function with less order of decrease?
I have an additional question: Is space $O=\cap_{n=1}^{\infty}{O^n}$ closed? Any proof or example?
Third question: Is space $\cup_{n=1}^{\infty}{O^{1/n}}=C_0$?
PS: $C_c$ means continuous function with compact support, $C_0$ means continuous function vanishing at infinity. $O^n$ is the space of continuous functions with property $\lim_{x\to\infty}{|x|^nf(x)}=0$
 A: We can show that $O$ is dense in $C_0$ for the supremum norm. Let $\chi_k$ a continuous function, $\chi_k(x)=1$ if $-k\leq x\leq k$, $\chi_k(x)=0$ if $|x|\geq k+1$ and $0\leq \chi\leq 1$. If $f\in C_0$, we have $f\chi_k\in O$. Now we check uniform convergence: 
\begin{align}
\sup_{x\in\Bbb R}|f(x)-f(x)\chi_k(x)|&\leq\sup_{|x|\geq k+1}|f(x)|+\sup_{k\leq |x|\leq k+1}f(x)|(1-\chi_k(x))\\
&\leq \sup_{|x|\geq k}|f(x)|.
\end{align}
For a fixed $\varepsilon>0$, fix $k$ such that $|f(x)|\leq \varepsilon$ if $|x|\geq k$ to get the result. 
In particular, this proves that $O^1$ is dense in $C^0$, and it's strict since $x\mapsto\frac 1{1+\sqrt{|x|}}$ is in $C_0$ but not in $O^1$. 
Let $V:=\bigcup_{n\geq 1}O^{1/n}$. It's not equal to $C_0$ since the map 
$$f(x):=\frac 1{1+(\log x)^2}$$
is in $C_0$ but not in $O^{1/n}$ for any $n$.
A: The closure of $O^n$ under uniform convergence is precisely $C_0$. There are functions in $C_0\setminus O^n$, such as $1/(1+|x|^{1/2})$.
Given a function $f$, let $f_n$ be the function obtained by multiplying pointwise by the indicator function of $[-n,n]$, then convolving with a standard mollifier. You can show that $f\in C_0$ implies $f_n\to f$ uniformly.
