3

As in title, I was wondering whether $C(\mathbb R)$ was reflexive (here $C(\mathbb R)$ is meant as the space of continuous functions on $\mathbb R$, without any other condition). This question is generated by the following well-known result:

Proposition. $(C(K), \| \, \|_\infty)$, $K$ infinite compact metric space, is nonreflexive.

This a simple consequence of the James' theorem. But what if we dropped compactness assumption?

My considerations. In order to exploit usual "positive" (Kakutani's theorem) and "negative" (James' theorem) characterizations of reflexivity, we have to endow $C(\mathbb R)$ with a norm that makes it a Banach space. Now, if I'm not wrong, any linear space has at least one norm. On the other hand, not all linear spaces can be endowed with a norm that makes them Banach: the space of polynomials on $[0,1]$ is a counterexample. Given a norm, even without showing completeness, another possibility is to determine the dual of $C(\mathbb R)$ and then consider that a Banach space $X$ is reflexive if and only if its dual $X'$ is. Iterating, if $X'$ is reflexive, then $X''$ is such (and viceversa), and $X''$ is automatically complete, being a dual space. So $X$ is Banach reflexive. Whether this program is succesfull or not, it depends on who the dual space is. (it could be equally or more difficult show it is reflexive or not.)

Remark. It could be the context has to move from that of Banach spaces into that of locally convex topological vector spaces.

The point is that I'm not able to establish any of preceeding conditions. Does anyone know whether such program can be worked out? And, first of all, whether the question in the title has an answer (in the affermative or in the negative)?

user91126
  • 2,326
  • 1
    is the question whether there is a norm on $C(\mathbb R)$ which makes is to a reflexive Banach space? No further conditions on the norm? As a vector space $C(\mathbb R)$ is isomorphic to a Hilbert space ... – user8268 May 26 '14 at 20:55
  • Yeah, no further conditions on the norm. Really is it isomorphic to a Hilbert space? I didn't realize... – user91126 May 26 '14 at 20:58

1 Answers1

5

In the usual topology - that of locally uniform convergence - $C(\mathbb{R})$ is a non-normable Fréchet space. That Fréchet space is not reflexive, since it contains the Banach space $\mathscr{D}^0([-2,2])$ of continuous functions vanishing outside the interval $[-2,2]$ as a closed subspace, and a closed subspace of a reflexive Fréchet space is itself reflexive, but $C([-1,1])$ is a non-reflexive quotient of the Banach space $\mathscr{D}^0([-2,2])$, and every quotient of a reflexive Banach space is reflexive. Thus $\mathscr{D}^0([-2,2])$ is not reflexive, and hence $C(\mathbb{R})$ is (with the topology of locally uniform convergence) not reflexive.

If we can endow the space with any topology we wish, we note that its (algebraic) dimension is $2^{\aleph_0}$, like the dimension of $L^p(\mathbb{R})$ for any $1 < p < \infty$, so any isomorphism to such a space endows the space with a reflexive Banach space structure.

Daniel Fischer
  • 206,697
  • Thanks for your answer. I need a little clarification: who is the space respect to which we are taking the quotient? Doesn't it need to be closed, in order the quotient be reflexive? (i.e., $X$ reflexive, $M < X$ closed $\implies X / M$ reflexive) – user91126 May 26 '14 at 21:18
  • 1
    The space with respect to which we take the quotient is the kernel of the restriction $\rho \colon \mathscr{D}^0([-2,2]) \to C([-1,1])$, that is $\ker\rho = { f\in \mathscr{D}^0([-2,2]) : f\lvert_{[-1,1]} \equiv 0}$. The restriction is surjective by Tietze's extension theorem. – Daniel Fischer May 26 '14 at 21:23