Weak*-closed subspaces as dual spaces

Let $X$ be a Banach space and $M$ be a subspace of $X^*$ closed under the weak* topology. We show that M can be identified as the dual space of some Banach space.

Let $M_\perp = \{x \in X: f(x) = 0 \,\, \forall f \in M\} = \bigcap_{f \in M} \ker(f)$. This is a (weakly) closed subspace of $X$. Then we can form the quotient Banach space $X/M_\perp$ with the norm $\|x + M_\perp\| = \inf_{m \in M_\perp} \|x - m\|$.

Theorem. $M$ is isometrically isomorphic to $(X/M_\perp)^*$.
Proof: Define a map $\Phi: M \rightarrow (X/M_\perp)^*$ by $[\Phi(f)](x + M_\perp) = f(x)$ for all $f \in M$. Firstly, $\Phi(f)$ is a well-defined linear functional on $X/M_\perp$ because $f$ vanishes on $M_\perp$ whenever $f \in M$. Secondly, $\Phi(f)$ is bounded as \[ |\Phi(f)(x + M_\perp)| = |f(x)| = |f(x - m)| \leq \|f\|\|x - m\| \,\, \forall m \in M_\perp, \] so $|\Phi(f)(x + M_\perp)| \leq \|f\|\|x + M_\perp\|$. Hence $\Phi$ is well-defined.
Clearly $\Phi$ is linear. From the above we have $\|\Phi(f)\| \leq \|f\|$. Let $\epsilon > 0$. Choose some $x \in X$ with $\|x\| \leq 1$ and $|f(x)| > \|f\| - \epsilon$. Then $\|x + M_\perp\| \leq 1$ and $|[\Phi(f)](x + M_\perp)| = |f(x)|  > \|f\| - \epsilon$. Thus $\|\Phi(f)\| = \|f\|$. It follows that $\Phi$ is an isometry.
It remains to show that $\Phi$ is surjective. Let $\ell \in (X/M_\perp)^*$ be given. Define $f \in X^*$ by $f(x) = \ell(x + M_\perp)$, i.e. $f = \ell \circ \pi$ where $\pi$ is the quotient map. Then $f$ vanishes on $M_\perp$ by definition, hence $f \in M$ by the following lemma and it is clear that $\Phi(f) = \ell$.

Lemma. If $f \in X^*$ vanishes on $M_\perp$, then $f \in M$.
Proof: Suppose $f \in X^* \backslash M$. By Hahn-Banach Theorem and weak*-closedness we can find some $x \in X$ such that $f(x) = 1$ and $g(x) = 0$ whenever $g \in M$. But then $x \in M_\perp$ and so $f(x) = 0$, contradiction.

This result implies the following important observation in the theory of operator algebras.

Theorem. Every von Neumann algebra is the dual space of some Banach space.
Proof: Let $M$ be a von Neumann algebra on a Hilbert space $\mathcal{H}$. Then $M$ is a $\sigma$-weakly closed subalgebra of $B(\mathcal{H})$, i.e. it is a weak*-closed subalgebra of $B(\mathcal{H})$ where the weak* topology of $B(\mathcal{H})$ comes from the fact that the dual space of the trace-class operators $L^1(\mathcal{H})$ on $\mathcal{H}$ can be identified with $B(\mathcal{H})$. By our main theorem, $M \cong (L^1(\mathcal{H})/M_\perp)^*$.

A predual of a von Neumann algebra $M$ is a Banach space $X$ such that $M = X^*$. The above result says that every von Neumann algebra has a predual. In fact, the predual is unique by a result of Sakai. Furthermore, the property that having a predual characterizes von Neumann algebras among the class of C*-algebras.

Theorem (Sakai).  Suppose $A$ is a W*-algebra, i.e. $A$ is a C*-algebra and the dual space of some Banach space. Then $A$ is *-isomorphic to a von Neumann algebra over some Hilbert space.