Closed subspaces of a reflexive space

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 X be a reflexive space and Y be a closed subspace of X. Then Y is reflexive.

We recall that a Banach space X is reflexive if the map X \rightarrow X^{**} given by x \mapsto (x^* \mapsto \langle x^*,x \rangle) is surjective.

Proof. We want to show for any y^{**}\in Y^{**}, there exists y_0\in Y such that \langle y^{**},y^* \rangle= \langle y^*,y_0 \rangle for all y^*\in Y^*. To prove this, given y^{**}, define x^{**}\in X^{**} by \langle x^{**},x^*\rangle=\langle y^{**},x^*|_Y \rangle. By the reflexivity of X, there exists x_0\in X such that

\langle y^{**},x^*|_Y \rangle = \langle x^{**},x^* \rangle =\langle x^*,x_0 \rangle            (1)

for any x^*\in X^*. It remains to show x_0\in Y. Assume to the contrary that x_0\notin Y. Then apply the Hahn-Banach Theorem (note Y is a closed subspace of X) to get an x^*_0\in X^* such that x^*_0 |_Y \equiv 0 and \langle x^*_0,x_0 \rangle >0. Putting this x^*_0 into (1), we see that

0=\langle y^{**},x^*_0|_Y \rangle = \langle x^{**},x^*_0 \rangle =\langle x^*_0,x_0 \rangle>0

which is a contradiction. Therefore x_0\in Y and the proof is complete.

Remark. Since Y is a subspace of X, there is an inclusion map i:Y\rightarrow X. Then we can construct the adjoint and the double adjoint i^*: X^* \rightarrow Y^* and i^{**}: Y^{**}\rightarrow X^{**}. Well, one can easily check that the map y^{**} \mapsto x^{**} in the above proof is indeed i^{**}.

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5 Responses to Closed subspaces of a reflexive space

  1. Anonymous says:

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

  2. Anonymous says:

    But this one is identical to the one in Lax.

  3. Pepek says:

    Thanks, intuitive solution

  4. Anonymous says:

    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?

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