Let $\{\omega_i\}_{i\in I}$ be a non-empty set of increasing (not necessarily strictly) continuous functions preserving $0$. Then, for each $i \in I$ define the space $$ C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d):= \left\{ f \in (\mathbb{R}^n,\mathbb{R}^d):\, \|f\|_{\omega_i,\infty}<\infty \right\} \mbox{ where } \|f\|_{\omega_i,\infty}:= \sup_{x \in \mathbb{R}^n} (\omega_i(\|x\|)+1)^{-1}\|f(x)\|. $$ Then this is a Banach space since the map $f \mapsto f (\omega_i(\|x\|)+1)$ is clearly an isometry with $C_0(\mathbb{R}^n,\mathbb{R}^d)$ for the sup-norm, and the maps between $C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d)$ and $C_{\omega_j}(\mathbb{R}^n,\mathbb{R}^d)$ can be defined similarly by rescaling analogously. This makes $I$ into a poset with $$ i\leq j \mbox{ if and only if } \sup_{x \in \mathbb{R}^n}\omega_i(\|x\|) \leq \sup_{x \in \mathbb{R}^n}\omega_j(\|x\|). $$ Thus, we can define the LCS colimit of this co-cone $\left\{C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d)\right\}$.

Now for the question, if $\{\omega_i\}_i$ is taken to be the collection of all monotonically increasing and continuous functions identifying $0$ (i.e.: $\omega(0)=0$) then does $\operatorname{co-lim}_i C_{\omega_i}(\mathbb{R}^n,\mathbb{R}^d)$ contain all uniformly continuous functions?

Note: Here the colimit is in the LCS sense.


1 Answer 1


Maybe, I miss something, but the answer seems to be easy: If $f:\mathbb R^n\to\mathbb R^d$ is continuous with $f(0)=0$ you can define the weight function $$\omega(r)=\sup\{\|f(x)\|: \|x\|\le r\}$$ which is obviously increasing with $\omega(0)=0$. Moreover, it is continuous at $r\ge 0$ because of the uniform continuity of $f$ on the compact set $\{x\in\mathbb R^n:\|x\|\le r+1\}$. Then $f\in C_\omega(\mathbb R^n, \mathbb R^d)$ holds. The extra assumption $f(0)=0$ can be removed by applying this to $\tilde f(x)=f(x)-f(0)$.

This shows that $C(\mathbb R^n, \mathbb R^d)$ is the co-limit of $\{C_\omega(\mathbb R^n, \mathbb R^d)\}$ in the category of vector spaces. Whether this holds in LCS is a different question.

  • $\begingroup$ You mean in the category of topological vector spaces? In that case $C(\mathbb{R}^n,\mathbb{R}^d)$ has compact-convergence topology or some (non-metrizable) final topology? Also, are these examples where, $I$ is say countable and fully ordered (besides the space of functions of moderate growth?) $\endgroup$
    – ABIM
    Jul 31, 2020 at 8:57
  • $\begingroup$ You wrote: Does the co-limit contain all uniformly continuous functions? -- This sounds very much like an algebraic question (where the co-limit is the union). $\endgroup$ Jul 31, 2020 at 10:31
  • $\begingroup$ Ah, I see the issue; my error. But I'm mostly looking in the LCS category (I made an edit to help clarify), I'm sorry abou the confusion $\endgroup$
    – ABIM
    Jul 31, 2020 at 10:46
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    $\begingroup$ Note that a uniformly continuous function on a normed space has a linear growth, $\|f(x)\|\le a\|x\|+b$. So you don't need such a wide class of weights $\{\omega_i\}$. 0n the other hand, adding $1$ in the denominator in the definition of the norms means you do not care to much about being uniformly continuous or just continuous. $\endgroup$ Jul 31, 2020 at 14:38

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