# The topological vector space that is not metrizable.

Let $C_0(\mathbb{R})$ denote the vector space of continuous functions on the real line with compact support. For any positive function $\rho$ let $$||f||_{\rho}:=\sup_x\rho(x)|f(x)| \ \ .$$ 1) I could show that $C_0(\mathbb{R})$ is a topological vector space.

2) I could show that for given any countable sequence of $\{\rho_j\}$, there exist $\rho$ so that $\displaystyle\lim_{x\to\pm\infty}\dfrac{\rho}{\rho_j}=\infty \ \text{for all j} .$

But by using (2), I couldn't prove that $C_0(\mathbb{R})$ is not metrizable.

How can I show that $C_0(\mathbb{R})$ is not metrizable? Hint: Use (2)

## 1 Answer

A topological vector space is metrizable if and only if there exists a countable basis of the system $\mathcal U_0$ of $0$-neighbourhoods, that is, a sequence $U_n \in \mathcal U_0$ such that every $U\in \mathcal U_0$ contains some $U_n$.

In your case, every $U_n$ contains some ball $B_n=\lbrace \|f\|_{\rho_n}\le c_n\rbrace$ for some $\rho_n$ and $C_n>0$. If $\rho$ is as in 2. the ball $B_\rho = \lbrace \|f\|_{\rho}\le 1\rbrace$ belongs to $\mathcal U_0$ but does not contain any $U_n$.

EDIT. One wants to prove $B_n \subseteq B_\rho$ $\Rightarrow$ $\rho \le c_n^{-1} \rho_n$ but for this one needs some regularity for the weights which I, implicitly, assumed to be continuous. You already need this (or at least something) to prove that $\|f\|_\rho$ is real-valued. In the continuous case, if $\rho(x_0) > c_n^{-1} \rho_n(x_0)$ then this is true in an open interval containing $x_0$ and you can find a continuous $f$ with support in that interval which belongs to $B_n$ but not to $B_\rho$.

• I do not find out why $U=\{f\in C_0(\mathbb{R}), \ ||f||_{\rho}\leq 1\}\in \mathcal{U}_0$ does not contain any $U_n$ . Sep 1 '13 at 5:08
• 1) Why "If $B_n\subseteq B_{\rho}$ then $\rho\le c_n^{-1}\rho_n$" ?? 2) How do you find $f$ that not to $B_{\rho}$ ? 3) If $\rho$ do not be continuous function, What are you doing ? Sep 4 '13 at 5:57