What is a basis for the vector space of continuous functions?

Question

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?

Answer 1

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.

Answer 2

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.

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Bibliography:

1. Kechris, A. Classical Descriptive Set Theory. Springer-Verlag, 1994.

Answer 3

I'll use faith to believe we are in one of those situations describe 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 $n \in \mathbb Z$ are a basis of the space $L^2([a,b])$ of $\mathbb C$-valued functions. For real valued functions, take the functions $\sin(nx)$ and $\cos(nx)$ as your basis.

Someone help me here ; I know the notion of basis I speak of here is well-defined, i.e. that we say that an infinite series represents the element of the vector space if the infinite series converges. This is clearly not an Hamel basis, where we require the linear combination to be finite. Is there a name for this kind of space? At first I thought the name Hamel was given to this infinite-convergence-notion of basis, but now I realize that the Hamel name was given to the algebraic basis.

Hope that helps,