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A natural vector space is the set of continuous functions on $\mathbb{R}$. Is there a nice basis for this vector space? Or is this one of those situations where we're guaranteed a basis by invoking the Axiom of Choice, but are left rather unsatisfied?

Martin Sleziak
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I. J. Kennedy
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    This is what Mick Jagger was singing about. – copper.hat Apr 25 '12 at 06:04
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    Cue the AC guys... Oh wait, that's me. Let me write something up. :-) – Asaf Karagila Apr 25 '12 at 07:43
  • Hmmm. Since I don't see how the continuous functions make a Polish group, I can't really use the "usual" tools for this sort of proof. I think it is very unlikely that a Hamel basis exists without the aid of the axiom of choice here. If I can come up with a clean argument I'll post it later today. – Asaf Karagila Apr 25 '12 at 09:10
  • @Asaf Karagila: Some papers by Lorenz Halbeisen at http://www.iam.unibe.ch/~halbeis/publications/publications.html may be relevant (perhaps not directly, but the references might be helpful), such as #5, 9, 21. – Dave L. Renfro Apr 25 '12 at 15:35
  • @Dave: Thanks for the reference! I actually ran into some of those before. Is the space of continuous functions from $\mathbb R$ to $\mathbb R$ is a Banach space at all? What is the metric? I know it is true if the domain was compact, but alas $\mathbb R$ is not compact and $\mathbb Q$ is even less compact than that. – Asaf Karagila Apr 25 '12 at 15:45
  • @Asaf Karagila: Off-hand, I don't know about the case for $\mathbb R$ to $\mathbb R,$ but I suppose one would need to restrict attention to bounded functions (continuous, or whatever). I don't know much functional analysis, but it seems to me that for topological issues you could just use the sup norm for the arctangents of the functions. But maybe I'm talking nonsense . . . – Dave L. Renfro Apr 25 '12 at 15:54
  • @Dave: The question here, however, asks about all the continuous functions so we have to have all of them. Proving that a subspace does not have a basis is not enough because it is possible to have a space which has a basis and a subspace without a basis. – Asaf Karagila Apr 25 '12 at 16:05
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    @AsafKaragila: How about if we equip $C(\mathbb{R})$ with the topology of uniform convergence on compact sets. This gives us a separable Frechet space and in particular a Polish vector space. Now can you tell us about your "usual" tools? – Nate Eldredge May 29 '12 at 12:47
  • @Nate: Well, we know that Polish groups make a nice object in these contexts; and correct me if my understanding is lacking this is even a Banach space (which makes things even easier). I will consider it in depth in a while, I have to run and give a lecture now. Thanks for the idea! – Asaf Karagila May 29 '12 at 12:56
  • @AsafKaragila: No, it isn't Banach. Every nonempty open set in this topology contains a line. – Nate Eldredge May 29 '12 at 13:09
  • @copper.hat, "you can't always get what you want" or "I can get no satisfaction"? Both apply! – lhf May 29 '12 at 14:38
  • @Nate: Well, a Polish groups gives me "enough" to work with. I'll post some answer soon. – Asaf Karagila May 29 '12 at 14:40

3 Answers3

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There is, in a fairly strong sense, no reasonable basis of this space. Zoom in on a neighborhood at any point and note that a finite linear combination of functions which have various kinds of nice behavior in that neighborhood also has that nice behavior in that neighborhood (differentiable, $C^k$, smooth, etc.). So any basis necessarily contains, for every such neighborhood, a function which does not behave nicely in that neighborhood. More generally, but roughly speaking, a basis needs to have functions which are at least as pathological as the most pathological continuous functions.

(Hamel / algebraic) bases of most infinite-dimensional vector spaces simply are not useful. In applications, the various topologies you could put on such a thing matter a lot and the notion of a Schauder basis becomes more useful.

Qiaochu Yuan
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Using Nate Eldredge's comment we have that $C(\mathbb R)$ is a Polish vector space.

Consider a Solovay model, that is ZF+DC+"All sets have the Baire property". In such model all linear maps into separable vector spaces are continuous, this is a consequence of [1, Th. 9.10].

It is important to remark that a continuous function (from $\mathbb R$ to $\mathbb R$) from a compact set is uniformly continuous is a result which do not require any form of choice, and I believe that Dependent Choice (DC) ensures that uniform converges on compact sets is well behaved.

Suppose that there was a Hamel basis, $B$, it has to be of cardinality $\frak c$. So it has $2^\frak c$ many permutations, which induce $2^\frak c$ different linear automorphisms.

However every linear automorphism is automatically continuous, so it is determined completely by the countable dense set, and therefore there can only be $\frak c$ many linear automorphisms which is a contradiction to Cantor's theorem since $\mathfrak c\neq 2^\frak c$.

This is essentially the same argument as I used in this answer.

Bibliography:

  1. Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.
Community
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Asaf Karagila
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I'll use faith to believe we are in one of those situations described by the axiom of choice ; had one discovered a useful basis for this vector space, it'd be known all over the place. The best we have as a basis right now, (and the word "best" means 'to my belief, the one that looks the prettiest') is the fact that the functions $e^{inx}$ with $n \in \mathbb Z$ form a Schauder basis of the Hilbert space $L^2([a,b])$ of $\mathbb C$-valued functions (modulo functions which are zero almost everywhere on $[a,b]$ ; note that this is not a Hamel basis). For real valued functions, take the functions $\sin(nx)$ and $\cos(nx)$ as your basis.

Hope that helps,

Patrick Da Silva
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