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 (Eberlein-Smulian theorem) $\endgroup$ – Yemon Choi Jan 27 '13 at 3:40
  • 4
    $\begingroup$ The examples you mention are very easy to recognize as non-reflexive. $X$ 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 non-reflexive, 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.