Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
Yes, $\mathcal{C}^0(\mathbb{R},\mathbb{R}) = \{f : \mathbb{R} \to \mathbb{R} \ \text{continuous} \}$. –  Seirios Sep 18 '12 at 17:00
5  
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
3  
@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
4  
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
3  
@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 6:00

2 Answers 2

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.

share|improve this answer
3  
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

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.

share|improve this answer
    
How did you choose the $\alpha$'s without the axiom of choice? –  Asaf Karagila Dec 17 '13 at 10:55
    
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. –  Chilote Dec 17 '13 at 19:25
    
Yes, so this answer is not that useful after all. –  Asaf Karagila Dec 17 '13 at 19:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.