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

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How do you define $O^n$? –  Nate Eldredge Jul 19 '12 at 19:23
Continuous functions with property $lim_{x\to\infty}{|x|^nf(x)}=0$ –  user33869 Jul 19 '12 at 19:26
$O^n\supset O^{n+1}$ so $O=O^1$. Did you mean to take the intersection? –  Colin McQuillan Jul 19 '12 at 19:38
Thank you. Mistake corrected. I actually wonder if there any spaces approximating $C_0$. I think that would help me understand that space better. –  user33869 Jul 19 '12 at 19:46
I don't get why add the absolute value in a limit which is practically always coming from the positive side. We approach $\infty$ so without loss of generality $x>42$ and therefore $x=|x|$. –  Asaf Karagila Jul 19 '12 at 21:01

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.

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For the third question: $1/\log(10+|x|)$ is in $C_0$ but not in $C^{1/n}$ for any $n$. –  Colin McQuillan Jul 19 '12 at 19:55
That's an example of $C_0 \backslash O^n$, but it doesn't say anything about the closedness of $O_n$/$\cap_{n=1}^{\infty}{O^n}$/$\cup_{n=1}^{\infty}{O^{1/n}}$. –  user33869 Jul 19 '12 at 19:57
user33869: the second paragraph of my answer shows that the closure of $O^n$ is $C_0$. –  Colin McQuillan Jul 19 '12 at 20:10
I didn't think it through. Thank you very much. –  user33869 Jul 20 '12 at 1:49

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 $n$.

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Thank you...... –  user33869 Jul 20 '12 at 1:49