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 $\lim_{x\to\infty}{|x|^nf(x)}=0$