A standard result in linear functional analysis is that any closed subspace of a reflexive space is reflexive. However, we find that some textbooks provide quite complicated proofs. After discussing with KKK and Ken, we arrive at the following simple proof of this result.
Theorem (KKK-Ken’s Theorem, 2011) Let be a reflexive space and be a closed subspace of . Then is reflexive.
We recall that a Banach space is reflexive if the map given by is surjective.
Proof. We want to show for any , there exists such that for all . To prove this, given , define by . By the reflexivity of , there exists such that
for any . It remains to show . Assume to the contrary that . Then apply the Hahn-Banach Theorem (note is a closed subspace of ) to get an such that and . Putting this into (1), we see that
which is a contradiction. Therefore and the proof is complete.
Remark. Since is a subspace of , there is an inclusion map . Then we can construct the adjoint and the double adjoint and . Well, one can easily check that the map in the above proof is indeed .