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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?

Seirios
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  • This notation just names the continuous functions from $\mathbb{R}$ to itself, right? – Harald Hanche-Olsen Sep 18 '12 at 15:42
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    Yes, $\mathcal{C}^0(\mathbb{R},\mathbb{R}) = {f : \mathbb{R} \to \mathbb{R} \ \text{continuous} }$. – Seirios Sep 18 '12 at 17:00
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    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. – Davide Giraudo Sep 18 '12 at 19:35
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    @DavideGiraudo: You can also do the embedding the other way, so the two problems are in fact equivalent. – Harald Hanche-Olsen Sep 18 '12 at 20:55
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    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. – Harald Hanche-Olsen Sep 18 '12 at 20:56
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    Is there any restriction on the norm? Should it be compatible with some sort of convergence? Should it be compatible with some operator? – Asaf Karagila Mar 08 '13 at 23:54
  • @AsafKaragila: A priori, no restriction is required. – Seirios Mar 09 '13 at 00:02
  • This is just off the top of my head, but would the inner product $<f,g> = \int_{-\infty}^{+\infty}fg$ work? The norm would then just be $\sqrt{<f,f>}$. Since $f$ is continuous it is Riemann integrable, and so is it's square. – noobProgrammer Mar 15 '13 at 03:17
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    @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. – Erick Wong Mar 15 '13 at 06:00
  • Well, I can't think of a counterexample yet. But I am certain it exists. – Asaf Karagila Mar 15 '13 at 08:56
  • It may be good to note that under any such norm all but finitely many of the evaluation mappings $\pi_a: f \longmapsto f(a)$ must be discontinuous. – nullUser Mar 15 '13 at 15:51
  • @nullUser: If we could have proved that the norm is complete that would have been great, because there are models where linear functionals from Banach spaces are automatically continuous. Perhaps we can show that the completion of such norm would have to have discontinuous evaluation functionals. That would imply contradiction. – Asaf Karagila Mar 16 '13 at 10:23
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    Apologies if the question is dumb, but is there some particular topology we want this norm to induce, or do we just want any norm? If the former, then is uniform convergence on compact sets what we're after? – Jakub Konieczny Apr 23 '13 at 08:07
  • @Feanor: I believe anything is allowed – Thomas Rot May 02 '13 at 13:59
  • Is thisrelated? – Torsten Hĕrculĕ Cärlemän Jul 14 '13 at 20:30
  • @AnuragPallaprolu: Your link adds an additional condition on the norm. But thank you for the reference, it is interesting. – Seirios Jul 15 '13 at 07:56

3 Answers3

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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).

Community
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Nate Eldredge
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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.

paul garrett
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    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.) – Asaf Karagila Jul 22 '13 at 15:53
  • @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. – paul garrett Jul 22 '13 at 15:57
  • 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 $\sf ZF$. – Asaf Karagila Jul 22 '13 at 16:02
  • @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. :) – paul garrett Jul 22 '13 at 16:10
  • 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 $\sf AC$? – Asaf Karagila Jul 22 '13 at 16:29
  • @AsafKaragila, you're probably right, AC-wise. But functional-analysis-wise, such a norm, with or without AC, is not what we want for other purposes, I think. Should I delete my answer? – paul garrett Jul 22 '13 at 16:37
  • Perhaps you are right about that. But I feel that if the question was about locally convex topology induced by a complete metric, then the question would have said that. I think that the answer should join to the other answers, yes. I also think that any actual answer to this question would probably merit a publication (although not as much as Theo's question about the open mapping theorem; if I'd solve that one, I'd submit it to a very reputable journal). – Asaf Karagila Jul 22 '13 at 16:55
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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.

Chilote
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