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

(1)

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 .

I am very good with conway’s proof in his functional analysis.

But this one is identical to the one in Lax.

I see. I haven’t read the one in Lax’s book.

Thanks, intuitive solution

I think you should still prove that every functional $f \in F’$ is actually the restriction of some functional $g \in X’$. And that would give the result. Or is this unnecessary?