Weak-* closed subspaces and Preduals, a la von Neumann Algebras Let $X$  be a Banach Space. Suppose that $M \leq X^*$ is a weak-* closed subspace. Is it true that $M$ has a predual? According to my understanding, taking the pre-annihilator $M^\perp = \{x \in X \vert  \forall a \in M, \, a(x)=0\}$ we should get that $M_* = X/M^\perp$ satisfies $(M_*)^* \simeq M$. I couldn't prove it. Further, what is really bothering me is if one looks at the restriction of the functionals $X \subset X^{**}$ under the canonical injection, to be functionals on $M$, is it true that this space can be identified as Banach spaces with $M_*$?
I am asking this since this somehow should be the situation for von Neumann Algebras, where the above translates to Why can we identify the predual $M_*$ of a a von Neuman Algebra both as the space of ultraweakly continuous functionals on $M$, and as the quotient Banch space $L^1(\mathcal{B}(\mathcal{H}))/M^\perp$ of the trace class operators?
 A: Note that $M^\perp$ is closed, as such $X/M^\perp$ is a normed space and also Banach. Further every element $a\in M$ induces a map $X/M^\perp\to\Bbb C$, $[x]\mapsto a(x)$. This map has the same norm as $a$ as can be checked, hence you may identify $M$ with a sub-space of $(X/M^\perp)^*$. What is left to check is that every element of $(X/M^\perp)^*$ comes from an element of $M$; here is where the weak* closedness of $M$ will enter.
Specifically if $M$ is weak* closed and $V\subseteq X/M^\perp$ is finite dimensional and $q:V\to\Bbb C$ linear then there is an $a\in M$ with $a\lvert_V=q$. We'll do this proof for completeness.
If $\dim(V)=1$ this is clear, as there must be an $a\in M$ with $a\lvert_V\neq0$, else $\pi^{-1}(V)\subseteq M^\perp$ which is a contradiction ($\pi:X\to X/M^\perp$ the projection). For $\dim(V)>1$ do an induction, assume that for each strict subspace of $V$ we can find an $a$ agreeing with $q$ on that subspace.
So let $e_1,...,e_n$ be a basis of $V$, there must be some $b\in M$ with $b(e_1)=...=b(e_{n-1})=0$ and $b(e_n)\neq0$, as otherwise whenever two elements of $M$ agree on $\mathrm{span}(e_1,...,e_{n-1})$ they agree on $\mathrm{span}(e_1,...,e_n)$ and there must be a linear formula so that $a(e_n)=\sum_i x_i\,a(e_i)$, hence $a(e_n-\sum_ix_i e_i)=0$ for all $a\in M$ and $e_n-\sum_i x_ie_i$ is $0$ in $X/M^\perp$, contradicting that $e_1,...,e_n$ is a basis.
So if $a\in M$ with $a\lvert_{\mathrm{span}(e_1,...,e_{n-1}})=q\lvert_{\mathrm{span}(e_1,...,e_{n-1}})$, then $[a + (q(e_n)-a(e_n))b]\lvert_V =q$ completing the induction.
Now if $q\in (X/M^\perp)^*$ let $\mathcal V$ denote the directed set of finite dimensional sub-spaces of $X/M^\perp$ and for each $V\in\mathcal V$ let $a_V\in M$ be such that $a_V\lvert_V=q\lvert_V$. Then $a_V$ converges pointwise to $q$ on $X/M^\perp$, by weak* closure you get that $q\in M$.
(Small remark: In the end I am a bit sloppy with identifications. For $q\in X/M^\perp$ the above procedure will give a net $a_V\in M$ so that $a_V \to q\circ \pi$ as elements of $X^*$, giving a pre-image of $q$ in $M$ under the identification of $M$ with a sub-space of $X/M^\perp$.)
