Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$?

A related question is: Can we proved that $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ has a norm without the axiom of choice?

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    $\begingroup$ Yes, $\mathcal{C}^0(\mathbb{R},\mathbb{R}) = \{f : \mathbb{R} \to \mathbb{R} \ \text{continuous} \}$. $\endgroup$
    – Seirios
    Sep 18, 2012 at 17:00
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    $\begingroup$ You can find a bijection between $\mathcal C^0(\Bbb R,\Bbb R)$ and a subspace of the sequence of real numbers (giving the values of the map at rational points). So a sufficient condition for the problem to be solved would be an explicit formula for the sequences of real numbers. $\endgroup$ Sep 18, 2012 at 19:35
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    $\begingroup$ @DavideGiraudo: You can also do the embedding the other way, so the two problems are in fact equivalent. $\endgroup$ Sep 18, 2012 at 20:55
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    $\begingroup$ It wouldn't surprise me one bit if there were a variant of ZF set theory without the axiom of choice in which these spaces have no norm. $\endgroup$ Sep 18, 2012 at 20:56
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    $\begingroup$ @noobProgrammer The point is that norms are not allowed to take the value $\infty$, which is not so easy to achieve on this space by the standard examples. $\endgroup$
    – Erick Wong
    Mar 15, 2013 at 6:00

3 Answers 3


The answer is no. There is no explicit norm on $\mathcal{C}^0(\mathbb{R}, \mathbb{R})$; constructing any norm on this space requires the axiom of choice to be used in an essential way.

In my answer to the (newer) question Inner product on $C(\mathbb R)$, I show that it is consistent with ZF+DC that there does not exist a norm on the vector space $\mathcal{C}^0(\mathbb{R}, \mathbb{R})$ (called $C(\mathbb{R})$ in that question).

  • $\begingroup$ Finally. An answer to the question which is an answer to the question! $\endgroup$
    – Asaf Karagila
    Nov 13, 2014 at 16:52
  • $\begingroup$ @AsafKaragila: Meh. That's so cliché. $\endgroup$
    – tomasz
    Nov 13, 2014 at 17:03
  • $\begingroup$ @tomasz: Classics are not clichés, but I can understand the confusion... :-) $\endgroup$
    – Asaf Karagila
    Nov 13, 2014 at 17:28

Refining @Mebat's answer: the seminorms on $C^o(\mathbb R)$ (here meant to be continuous, real-valued functions on $\mathbb R$, with no decay or boundedness restrictions) given by $\nu_K(f)=\sup_{x\in K} |f(x)|$ for compact subsets $K$ of $\mathbb R$, give a Frechet-space (complete, locally convex, metric) structure on $C^o(\mathbb R)$. As @Mebat notes, there is a countable subset, e.g., $[-n,n]$ of compacts which give that topology. Then the usual trick of writing $$ d(f,g)=\sum_n {1\over 2^n}\cdot {\nu_{[-n,n]}(f-g)\over 1+\nu_{[-n,n]}(f-g)} $$ gives a (non-canonical) metric.

Significantly, this makes $C^o(\mathbb R)$ complete. We almost always want to "give" TVS's topologies with the best completeness properties possible.

But this is not a norm, only a metric.

There is a reasonable criterion for normability of TVS's (once a topology is given), namely, that every neighborhood of $0$ is "absorbing", meaning that sufficiently large dilates contain a given bounded set. In the present example, the fact that continuous functions can blow up arbitrarily fast enables construction of counter-examples to a claim of normability, with the natural topology.

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    $\begingroup$ There is still a huge distance from complete metric space to a normed space. After all, every non-empty set is a complete metric space using the discrete metric. (Granted, this metric seems a better candidate for being a norm, or somehow generate a norm than the discrete metric; but as I have learned time and time again for the past couple of years, the smallest $\varepsilon$ can be proved impassable when the context is completely counterintuitive. And in modern mathematics we use an intuition which comes from the axiom of choice.) $\endgroup$
    – Asaf Karagila
    Jul 22, 2013 at 15:53
  • $\begingroup$ @AsafKaragila. Indeed, complete metric is not normable. But/and part of my claim would be that we first want to ascertain a natural (complete, or anyway quasi-complete, locally convex) topology before asking about normability. That is, the AxCh issue is presumably not what is really intended in the question. If it is, then my answer is irrelevant, of course. $\endgroup$ Jul 22, 2013 at 15:57
  • $\begingroup$ It seems to me that the question is about the normability of the space in $\sf ZF$. I can't get my intuition to even begin and make an educated guess, but if I had to make one I'd say it's impossible in $\sf ZF$. $\endgroup$
    – Asaf Karagila
    Jul 22, 2013 at 16:02
  • $\begingroup$ @AsafKaragila, You may be right. But/and then it seems a bit un-natural, given the (standard!) functional analysis story here, thus my suspicion or presumption that it didn't really mean what it appeared to mean. :) $\endgroup$ Jul 22, 2013 at 16:10
  • $\begingroup$ paul, in my experience it's quite a standard $\sf AC$ question. Given something that we can do, but can't explicitly do. Can we still do it without $\sf AC$? $\endgroup$
    – Asaf Karagila
    Jul 22, 2013 at 16:29

Here some ideas to find an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$. If we define $\Vert\cdot\Vert$ like $$\Vert f\Vert=\sup_{x\in\mathbb{R}} \,\,\frac{\vert f(x)\vert}{1+\vert f(x)\vert}$$ for every $f\in\mathcal{C}_0(\mathbb{R},\mathbb{R})$, then:
(i) $\Vert f\Vert=0$ iff $f=0$
(ii)$\Vert f+g\Vert\leq\Vert f\Vert+\Vert g\Vert$
but (iii) $\Vert \alpha f\Vert=|\alpha|\Vert f\Vert$, is not satisfied.

To fix (iii) we can do the following. Consider on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ the relation $\sim$ defined as follows: for $f,g \in \mathcal{C}^0(\mathbb{R},\mathbb{R})$, $f\sim g $ iff $\exists c\in\mathbb{R}, c\neq0$ such that $f=cg$. This is an equivalence relation. Then we have a partition of $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ in classes. Denote by $[f]$ the class of $f\in\mathcal{C}^0(\mathbb{R},\mathbb{R})$. The class of the constant zero function is the singleton $\{0\}$. Notice that if $a\in \mathbb{R}$ is such that $f(a)\neq0$, then $g(a)\neq0$ for every $g\in[f]$. Hence, for every class $F=[f]$ different from $[0]$, we can choose $\alpha_F\in\mathbb{R}$ such that $f(\alpha_F)\neq0$ for each $f\in F$. Since we have chosen the numbers $\alpha_F$, now we can choose a representative function for each class in a unique way. For every class $F=[f]\neq[0]$ there is a unique function $\hat{f}\in F$ such that $\hat{f}(\alpha_F)=1$. Put $\hat{0}=0$. Now we can define $\Vert\cdot\Vert$ on $\mathcal{C}_0(\mathbb{R},\mathbb{R})$ by setting

$$\Vert f\Vert=\sup_{x\in\mathbb{R}} \,\,\frac{\vert f(x)\vert}{1+\vert \hat{f}(x)\vert}$$ for every $f\in\mathcal{C}_0(\mathbb{R},\mathbb{R})$.

This satisfies (i) and (iii) but we have lost (ii). If we could choose each $\hat{f}$ in such a way that $\vert\widehat{(f+g)}(x)\vert\leq\vert\hat{f}(x)\vert+\vert\hat{g}(x)\vert$ for all $x\in\mathbb{R}$ and for all $f,g\in\mathcal{C}_0(\mathbb{R},\mathbb{R})$, then the function $\Vert\cdot\Vert$ resultant will be a norm.

  • $\begingroup$ How did you choose the $\alpha$'s without the axiom of choice? $\endgroup$
    – Asaf Karagila
    Dec 17, 2013 at 10:55
  • $\begingroup$ In fact I used the axiom of choice to choose the $\alpha$'s. I do not see a way to do it without the AC. $\endgroup$
    – Chilote
    Dec 17, 2013 at 19:25
  • $\begingroup$ Yes, so this answer is not that useful after all. $\endgroup$
    – Asaf Karagila
    Dec 17, 2013 at 19:26

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