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?
Best Answer
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.