Isometric isomorphism and annihilators of annihilators

Question

The following exercise is taken from lecture notes on Funtional Analysis by Prof. Mueger that I'm reading. For the sake of completion, I have added at the end the relevant definitions.

Exercise: If $V$ is Banach and $Z \subset V^*$ a closed subspace, construct an isometric isomorphism $$V^*/Z \cong (Z^{\top})^*$$

where $Z^{\top}$ is the (pre-)annhilator of $Z$ (see below). The symbol $\cong$ is used to mean that the two spaces are isometrically isomorphic.

And here is my question: Given $\varphi, \varphi' \in V^*$, $\varphi \sim \varphi'$ (mod $Z$) $\Leftrightarrow \varphi-\varphi' \in Z$ and thus $ \varphi_{|_{Z^{\top}}} = \varphi'_{|_{Z^{\top}}}.$

Hence, the map
\begin{align*} \beta \colon V^*/Z &\to (Z^{\top})^* \\ q \in V^*/Z &\mapsto \varphi_{|_{Z^{\top}}} \in (Z^{\top})^* \end{align*} (where $\varphi \in V^*$ is an arbitrary representative of $q$, $q = [\varphi]$) is well-defined.

$\beta$ is linear, and Hahn-Banach guarantees that $\beta$ is surjective. But I cannot prove that such map is injective. To do so would require proving that $(Z^{\top})^{\perp} = Z$, and such a claim is unwarranted. (Obviously, $Z \subseteq (Z^{\top})^{\perp}$, but not necessarily the other way around)

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Definition of annihilators: Let $V$ be a normed space.

• For any $W \subset V$ , the annihilator of $W$ is $$W^\bot = \{φ \in V^* \mid φ(x) = 0, \forall x \in W\} \subset V^*.$$

• For any $\Phi \subset V^*$, the (pre-)annihilator of $\Phi$ is $$\Phi^{\top} = \{x \in V \mid φ(x) = 0, \forall φ \in \Phi \} \subset V.$$

Answer 1

The result is true iff $Z\subset V^*$ is weak-star closed. The map $β$ is indeed injective if and only if $Z=(Z^\top)^\bot$. Since $(Z^\top)^\bot = \overline{Z}^{w^*}$ you see that $β$ is injective iff $Z$ is weak-star closed. Notice that this result is usually applied to $Z= W^\bot$ for some $W \subset V$, which is weak-star closed.

Here is a counter example: Consider $X$ a non-reflexive Banach space, $V=X^*$ and $Z= j(X) \subset V^*=X^{**}$ where $j \colon X \to X^{**}$ is the canonical embedding. Then $Z$ is weak-star dense in $V^*$ and since $X$ is not reflexive, $Z \neq V^*$ and thus $Z$ is not weak-star closed. Now, $$Z^\top = \{x^* \in X^* \colon \langle x^*,x \rangle =0 , \quad \forall x \in X \}= \{0\} $$ so that $$(Z^\top)^* = \{0\}$$ can not be isomorphic to $V^*/Z = X^{**}/j(X)$ since the latter quotient space contains atleast one non zero element i.e. there exists $x^{**} \in X^{**}$ such that $x^{**} \notin j(X)$ (since $X$ is not reflexive).

In fact, one can create many counter-examples by taking $Z \subset V^*$ to be any weak-star dense proper subspace, then $Z^\top =\{0\}$ so that $(Z^\top)^* = \{0\}$.

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Here is a proof of $\overline{Z}^{w^*} = (Z^\top)^\bot$. Evidently, $Z \subset (Z^\top)^\bot$ and since the latter is weak-star closed, $\overline{Z}^{w^*} \subset (Z ^\top)^\bot$. Arguing by contradiction suppose that there exists $ x^*_0 \in X^* $ such that $ x^*_0 \in ( Z ^\top )^\bot \setminus \overline{Z}^{w^*} $. By the strong separation theorem for $(X^*,w^*)$ there exists (a weak-star continuous functional) $x\in X$ such that
$$ \sup_{x^* \in \overline{Z }^{w^*} } \langle x^*,x\rangle < \langle x_0^*, x \rangle . $$

Now consider the two cases: If $ x\in Z^\top $ then since $\overline{Z}^{w^*} \subset ( Z^\top )^\bot $, you get that $0<0$ which is absurd. On the other hand, if $ x \notin Z^\top $ then there exists $x^* \in Z$ with $ \langle x^*,x \rangle \neq0$. Hence, $$ \frac{ \langle x^*_0,x \rangle } { \langle x^*,x \rangle } \langle x^*,x \rangle = \Big \langle \frac{ \langle x^*_0,x \rangle } { \langle x^*,x \rangle } x^*,x \Big \rangle < \langle x_0^*,x \rangle $$ which is, again, impossible. This leads to a contradiction.