I need to know if a certain Banach space I stumbled upon is reflexive or not. I need to know what are the state of the art techniques to determine if a Banach space is reflexive or not. For example, how does one prove that the space of bounded continuous functions is not reflexive. Same question for Linfinity. A description of the methods involved or a reference would both be appreciated.

6$\begingroup$ Can you give any further details about your space? This might help people to suggest likely techniques $\endgroup$ – Yemon Choi Jan 27 '13 at 3:39

$\begingroup$ For the examples you give, my first thought would be to locate a bounded sequence which has no weakly convergent subsequence (EberleinSmulian theorem) $\endgroup$ – Yemon Choi Jan 27 '13 at 3:40

4$\begingroup$ The examples you mention are very easy to recognize as nonreflexive. $X$ is reflexive iff $X^\ast$ is reflexive and closed subspaces of reflexive spaces are reflexive. Identify $(c_0)^\ast = \ell_1$ and $(c_0)^{\ast\ast} = \ell_\infty$. Thus, $c_0$ is not reflexive. It follows that $\ell_1$ and $\ell_\infty$ are not reflexive either. Now show that you can embed $c_0$ as a closed subspace into a space of (continuous) bounded functions or $\ell_1$ into the dual of such a space and this covers all of your examples. $\endgroup$ – Martin Jan 27 '13 at 5:06
A good reference for functional analysis in general is the book by John B. Conway. This book states that if $X$ is a compact space(or more generally a completely regular space), then $C(X)$ is reflexive if and only if $X$ is finite(p. 90). See This link for proofs or you can easily construct an element in $C(X)^{**}\setminus C(X)$ using the Riesz representation theorem. In particular, by Gelfand duality, every infinite dimensional commutative $C^{*}$algebra is nonreflexive, so the space $L^{\infty}(\mu)$ is reflexive if and only if $L^{\infty}(\mu)$ is finite dimensional.

$\begingroup$ Every reflexive space is weakly sequentially complete, $C(K)$ spaces contain copies of $c_0$ which is not WSC (and this property is hereditary). $\endgroup$ – Bojan Kwitek Jan 27 '13 at 12:37