A norm to define uniform convergence on every bounded interval.

Question

Consider sequence of functions $f_n$ in $\mathcal{C}(\mathbb{R})$. If we define convergence as: $f_n \to f$ if the sequence converges to $f$ uniformly on every bounded interval $[a,b]$

Is there a norm on $\mathcal{C}(\mathbb{R})$ that can define such convergence?

The sup-norm would guarantee the uniform convergence on every bounded interval, but it is too restrictive. If we consider $f_n(x) = \frac{x}{n}$, then $f_n$ converges to 0 for every bounded interval, however, it would not work under the sup-norm.

Answer 1

Compact convergence is uniform convergence on compact subsets.

Answer 2

As mentioned in another answer, the type of convergence you want is compact convergence: $f_n\to f$ compactly if $f_n\to f$ uniformly on all compact subsets of the domain. There is no single norm able to describe this kind of convergence. But there is a family of seminorms which can. Define $$\vert f\vert_K:=\sup_{x\in K}\Vert f(x)\Vert,$$ where $K$ is a compact subset of $\mathbb R$. Define the open $K$-ball with radius $r$ around $f_0$ as $B_r^K(f_0):=\{f\in\mathcal C(\mathbb R)~\vert~\vert f-f_0\vert_K<r\}$. Then the family of seminorms $(\vert\cdot\vert_K)_{K\subset\mathbb R\textrm{ compact}}$ induces the topology of compact convergence, that is, $A\subseteq\mathcal C(\mathbb R)$ is open if for every $f$ there is a finite number of open balls $B_{\varepsilon_1}^{K_1}(f),\dots,B_{\varepsilon_n}^{K_n}(f)$ whose intersection is included in $A$, that is, $$\bigcap_{i=1}^nB_{\varepsilon_i}^{K_i}(f)\subseteq A.$$ This is very similar to the topology of a normed vector space, with the difference that there is just one seminorm (a norm, even) instead of many, so we don't need to consider intersections of balls, just simple balls.

Convergence with respect to this topology is compact convergence.

Answer 3

The topological space you describe is $C(\mathbb{R})$ and is not normable. But there is a countable family of seminorm which makes $C(\mathbb{R})$ a complete Frechet space. And this is metrizable by a translation-invariant metric.

If you look here,

https://en.wikipedia.org/wiki/Fr%C3%A9chet_space#Constructing_Fr%C3%A9chet_spaces

you can easily guess which is the family of seminorms: If $K_n$ is an exhaustion of $\mathbb{R}$ by compact subsets that the family $\mathcal{S}$ of seminorm is given by $$\vert f\vert_{K_n}:=\sup_{x\in K_n}\Vert f(x)\Vert.$$ After that, one can show that it is not normable through the following argument. Suppose that $C(\mathbb{R})$ is normable, then there exists a norm $\|\cdot\|$ on $C(\mathbb{R})$ which induces the same topology. This means that the identity map $i:(C(\mathbb{R}),\mathcal{S})\to (C(\mathbb{R}),\|\cdot\|)$ is continuous. This implies the existence of a constant such that $$ \|f\|\leq c \cdot \vert f \vert_{K_n} $$ for every $f\in C(\mathbb{R})$. This cannot be the case because if you consider a function supported on the complement of $K_n$ then, by the previous inequality $\|f\|=0$ and $f\neq 0$.