# Completion of $C_c$ with respect to $\|\cdot\|_\Psi$

I'm doing the second half of the following exercise in my lecture notes:

"Let $C_c(R)$ be the vector space of continuous functions $f : R \to R$ with $\mathrm{supp}(f)=\overline{ \{x \in R \mid f(x)\neq 0 \} }$ compact, with the norm $\|\cdot\|_\infty$. Show that this space is not complete, and find a Banach space containing $C_c(R)$ so that the induced norm obtained by restriction is $\|\cdot\|_\infty$. Can you do the same for the norm $\|f\|_\Psi =\|f \Psi \|_\infty$ where $\Psi : R \to R_{>0}$ is a fixed continuous function (for example, $\Psi (x) = e^{x^2}$ )?"

Assuming "do the same" means that I want to show that

1) $C_c$ is incomplete with respect to $\|\cdot\|_\Psi$ and then

2) find a Banach space containing it:

1) For a Cauchy sequence $f_n$ in $(C_c, \|\cdot\|_\infty)$ with limit not in $C_c$ we get a Cauchy sequence $g_n$ in $(C_c, \|\cdot\|_\Psi)$ with limit not in $C_c$ by taking $g_n = f/\Psi$. I guess one should argue that $\|f\|_\Psi$ defines a norm on $C_c$: but that's clear since the product of continuous functions is continuous and the support of $f \Psi$ is $\mathrm{supp}(f) \cap \mathrm{supp}(\Psi) \subset \mathrm{supp}(f)$.

Unfortunately, $\|\cdot\|_\Psi$ does not define a norm on $B(X)$, the space of bounded functions because $\Psi$ is not bounded. So we are looking for a smaller space: $C_0(X)$ is probably the space we're looking for. It contains $C_c$, $\|\cdot\|_\Psi$ defines a norm on it and it's complete w.r.t. $\|\cdot\|_\Psi$: if $g_n$ is Cauchy w.r.t. $\|\cdot\|_\Psi$ we can use that $\|g_n\Psi\|_\infty \geq \|g_n\|_\infty$ so $g_n$ is Cauchy w.r.t. the sup norm hence its uniform limit $g$ is in $C_0$ and hence $g \Psi$ is, too.

Apparently the construction $g_n = f_n \Psi$ only works if $\Psi$ doesn't have compact support. If it does, $f \Psi$ will have compact support and hence $\lim_{n \to \infty} g_n$ will be in $C_c$ and hence $C_c$ will already be complete w.r.t. $\|\cdot\|_\Psi$ in this case. So a necessary condition on $\Psi$ is that it has non-compact support.

What's a sufficient condition on $\Psi$ so that $C_c$ is incomplete w.r.t. ?

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1. $C_0(\mathbb R)$ will (in general) not be closed under multiplication with $\Psi$ (if, for example $\Psi(x) = \exp(x^2)$, then $f(x) = \exp(-x^2)$ has $f \in C_0$, $f\Psi \not\in C_0$. ... 2. $\Psi$ is positive, so its support is $\mathbb R$. – martini Jul 12 '12 at 8:20
As you know, that $C_0(\mathbb R)$ is the completion of $C_c(\mathbb R)$ w. r. t. $\|\|_\infty$, you can consider $X = \{f \in C(\mathbb R) \mid f\Psi \in C_0(\mathbb R)\}$ and show that $(X, \|\|_\Psi)$ is complete and contains $C_c(\mathbb R)$ as a dense subspace ... – martini Jul 12 '12 at 8:52

(1) Consider the function $f_n$ given by the following graph: $\{(x,e^{-x^2}):x\in[-n,n]\}\cup$ the straight line segment from $(-n,e^{n^2})$ to $(-n-e^{n^2},0)$ $\cup$ the straight line segment from $(n,e^{-n^2})$ to $(n+e^{-n^2},0)$ Then extend along the real line on each side where you left off.
Check $\{\frac{f_n}\Psi\}$ is a Cauchy sequence. Let $f\in C_c$. I would like to argue $\frac{f_n}\Psi$ does not converge to $\frac f\Psi$. The support of $f$ and hence $\frac f\Psi$ lies in $[-m,m]$ for some large enough $m$. But then for all $n>m$, $\|\frac{f_n}\Psi-\frac f\Psi\|_\Psi\ge|f_n(m)-f(m)|=|f_n(m)|=|e^{-m^2}|$ showing that indeed $f_n$ is distant from $\frac f\Psi$. Since any $g$ with compact support can be written as $\frac h\Psi$, $h\in C_r(\mathbb R)$ this proves the incompleteness!
As a hint for (2), $C^0(\mathbb(R))\text{(let)}=\{f: f(x)\rightarrow0\text{ as }x\rightarrow\pm\infty\}$ lies in the completion for $C_c(\mathbb R)$ under $\|.\|_\infty$ as you can concoct a sequence of compactly supported functions analogous to above construction. Is $C^0(\mathbb R)$ complete? – Host-website-on-iPage Jul 12 '12 at 8:56
Sorry the set should be $\{f:f(x)\Psi(x)\rightarrow0\text{ as }x\rightarrow\pm\infty\}$ – Host-website-on-iPage Jul 12 '12 at 9:09