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?
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asked Apr 25, 2012 at 5:27
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4$\begingroup$ This is what Mick Jagger was singing about. $\endgroup$– copper.hatApr 25, 2012 at 6:04
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4$\begingroup$ Cue the AC guys... Oh wait, that's me. Let me write something up. :-) $\endgroup$– Asaf Karagila ♦Apr 25, 2012 at 7:43
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$\begingroup$ 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. $\endgroup$– Asaf Karagila ♦Apr 25, 2012 at 9:10
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$\begingroup$ @Asaf Karagila: Some papers by Lorenz Halbeisen at iam.unibe.ch/~halbeis/publications/publications.html may be relevant (perhaps not directly, but the references might be helpful), such as #5, 9, 21. $\endgroup$– Dave L. RenfroApr 25, 2012 at 15:35
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1$\begingroup$ @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? $\endgroup$– Nate EldredgeMay 29, 2012 at 12:47
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3 Answers
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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.
answered May 29, 2012 at 13:56
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1$\begingroup$ That is probably the best explanation I've seen as to why not to usel Hamel bases in real/complex function spaces. +1! $\endgroup$ May 29, 2012 at 22:41
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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:
- Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.
answered May 29, 2012 at 15:27
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$\begingroup$ Very nice! So in fact one certainly needs the full axiom of choice, and DC is not enough. $\endgroup$ May 29, 2012 at 15:56
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$\begingroup$ @Nate: Indeed. DC itself is often not enough. Shelah even showed that DC($\aleph_1$) is not enough to have sets without the Baire property! $\endgroup$– Asaf Karagila ♦May 29, 2012 at 16:00
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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,
answered Apr 25, 2012 at 5:32
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3$\begingroup$ just a remark on your last claim - that's not literally true since you need to take convergent infinite "linear combinations" whereas in the definition for a basis for a vector space, finite linear combinations give you everything. $\endgroup$– user29743Apr 25, 2012 at 5:34
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$\begingroup$ Sorry for the lack of clarity in my comment ; indeed, all what you said was correct, and was what I meant, I just wrote it in a hurry with my mobile phone. Now that I am at home in front of my computer I can see that I wrote too sketchy =) Thanks for the comments. $\endgroup$ Apr 25, 2012 at 6:10
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5$\begingroup$ The "topological" basis is called Schauder basis. $\endgroup$– Asaf Karagila ♦Apr 25, 2012 at 9:07