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This question comes from Space of bounded functions is reflexive if the domain is finite.

Let $(X, d)$ be a metric space. For $x \in X$ we define $\delta_x : C_b(X) \rightarrow \mathbb{K}, \ f \mapsto f(x)$. Prove that the space $C_b(X)$ is reflexive iff $X$ is finite.

If $X$ is infinite, then there is an injective mapping $\mathbb{N} \rightarrow X, \ n \mapsto x_n$, such that

$T: l^1 \rightarrow C_b(X)', \ \ (\alpha_n)_n \mapsto \sum_{n=1}^{\infty} \alpha_n \delta_{x_n}$.

I want to prove that $T$ is a well-defined isometric homomorphism. I always have a problem with this kind of questions. First of all how do I prove that $T$ is well-defined?

MathStudent
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2 Answers2

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Let $\|\mu\|$ denote the total variation norm of $\mu$.

$T$ is well defined because $\|\sum \alpha_n \delta_{x_n}\| \leq \sum |\alpha_n|=\|(\alpha_n)\|$ since $\|\delta_{x_n}\|=1$. [In any Banach space $\sum \|y_n|| <\infty$ implies that the series $\sum y_n$ converges].

Also $\|\sum \alpha_n \delta_{x_n}\|\geq \sum_k | (\sum_n\alpha_n \delta_{x_n}) (E_k)|$ where $E_k=\{x_k\}$ so $\|T(\alpha_n)\|=\sum |\alpha _n|=\|(\alpha_n)\|$ and $T$ is an isometry.

The fact that $\ell^{1}$ is isometrically isomorphic to a subspace of $C_b(X)$ implies that $C_b(X)$ is not reflexive. Hence $C_b(X)$ is not reflexive whenever $X$ is an infinite set. If $X$ is finite it is clear that $C_b(X)$ is reflexive.

Kavi Rama Murthy
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  • please show me how to prove that $T$ is a homomorphism – MathStudent Jan 27 '20 at 10:07
  • To answer the question in the title you only need the fact that $T$ is a linear isometry. In what sense do you want to show that $T$ is a homomorphism? Are you consideing $\ell^{1}$ as an algebra under convolution? @MathStudent – Kavi Rama Murthy Jan 27 '20 at 10:13
  • My bad. In a metric space a homomorphism is just a linear map. I take my question back. Sorry. And thank you – MathStudent Jan 27 '20 at 10:20
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$T$ is well-defined:

From the description $(\alpha_n)_n \mapsto \sum_{n=1}^{\infty} \alpha_n \delta_{x_n}$ it is not immediately clear that $T$ is actually a function from $\ell^1$ to $C_b(X)'$. Thus, for $T$ being well-defined you have to show that $ \sum_{n=1}^{\infty} \alpha_n \delta_{x_n}\in C_b(X)'$ for all $(\alpha_n)_n\in\ell^1$.

supinf
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