Show that a Banach space $E$ is reflexive $\iff$ for all closed subspace $F$ of $E$ we have $F^{\perp \perp} = J(F)$

functional-analysishilbert-spacesproof-verification

Problem: Let $E$ be a normed space over $\mathbb{C}$.

  1. Show that we are able to embed $E$ into the second dual space (bidual) $E''$ of $E$ by a linear isometry i.e we can consider $E$ as a subspace of $E''$.
  2. For each subset $A$ of $E$, let $A^{\perp} = \{f \in E' : f|_A = 0\}$. Show that a Banach space $E$ is reflexive $\iff$ for all closed subspace $F$ of $E$ we have $F^{\perp \perp} = J(F)$ in which $J$ is canonical embedding $E$ into $E'$.

My attempt:

  1. Let $E$ be a normed space. For each $x \in E$, let $(Jx)(u) = u(x)$ for all $u \in E'$.

We show that $J: E \rightarrow E''$ is a linear isometry. Clearly, $Jx$ is a linear form on $E'$. Since $(Jx)(u) = |u(x)| \le ||u||$ $||x|| $, $Jx$ is continuous on $E'$and $||Jx|| \le ||x||$. Therefore, $Jx \in E''$. It is routine to show that $J$ is a linear map from $E$ into $E''$. Take any $x \in E$. There is $v \in E'$ such that $||v|| = 1$ and $v(x) = ||x||$. Hence
$$||Jx|| = \sup\{|(Jx)(u)|:||u|| \le 1\}$$
$$= \sup\{|u(x)|: u \in E', ||u|| \le 1\} \ge v(x) = ||x||.$$
Therefore $J$ is a linear isometry.
Anybody can help me to solve question number $2$? Thank all!

Best Answer

The implication $\Leftarrow$ direction is very simple. Since $E^{\perp\perp}=E^{\prime\prime}$, it follows from $E^{\perp\perp}=J(E)$ that $J$ is surjective.

For the converse implication let $F$ be a closed subspace of $E$. If $\phi\in E^\perp$, then $0=\phi(x)=Jx(\phi)$ for all $x\in F$. Thus $J(F)\subset F^{\perp\perp}$ (this inclusion does not require reflexivity or closedness). On the other hand let $\psi\in F^{\perp\perp}$. By assumption there exists $y\in E$ such that $Jy=\psi$. Then $0=\psi(u)=u(y)$ for all $u\in F^\perp$. It follows from the Hahn-Banach theorem that $y\in F$ (here one needs the closedness of $F$).

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