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I'm studying Conway's functional Analysis by myself. In page 132 of his book, for showing every Closed subspace M of a reflexive Banach space X is reflexive, he says $\sigma(X,X^*)_{|_{M}}=\sigma(M,M^*)$. But I can not understand how it is. Please regard me. Thanks in advance

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    $\begingroup$ We can apply the Hahn-Banach theorem to extend functionals on $M$ to $X$. Also by restricting functionals on X we get functionals on M...applying this on linear functionals we conclude that $$\sigma(X,X^*)_{|_{M}}, \sigma(M,M^*)$$ have the same continuous functionals. Therefore they are the same. $\endgroup$ – Fermat Jul 12 '14 at 6:44
  • $\begingroup$ Thanks.I did something like you. But I made a mistake. $\endgroup$ – niki Jul 12 '14 at 7:00
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You can use the criterion that reflexivity is equivalent to the unit ball being compact in the weak topology. Then the intersection of the unit ball with the closed subspace (which is still closed in the weak topology) is also compact, hence $E\subseteq X$ is reflexive.

Edit: If you want to do it directly, you can use the Hahn Banach theorem to do it directly, since a closed subspace is characterized by the linear functionals which vanish on it. Then if $E\subseteq X$ is a closed subspace, we see

$$E^*\cong Hom_{\Bbb F}(X,\Bbb F)/\{f\in X^* : f(E)=0\}$$

where here $\Bbb F$ is your field, either $\Bbb R$ or $\Bbb C$. Then dualizing this gives rise to the isomorphism since the map

$$\begin{cases}X\to X^{**}=Hom_{\Bbb F}(Hom_{\Bbb F}(X,\Bbb F), \Bbb F) \\ x\mapsto E_x\end{cases}$$

is surjective you conclude the same holds passing through the quotient.

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