Weak*-closure of finite rank operators on dual space

Question

Given a Banach space $X$, we consider the space $B(X^*)$ of bounded, linear operators on $X^*$ with the weak*-topology from its canonical predual $B(X^*)_*=X^*\hat{\otimes}X$. What is $\overline{F(X^*)}^{wk*}$, the weak*-closure of the finite rank operators on $X^*$? Since this is rather vague, here are some concrete questions:

Q1: Do we always have $K(X^*)\subseteq\overline{F(X^*)}^{wk*}$, i.e., is every compact operator in the weak*-closure of finite-rank operators?

Q2: Is there a characterization when $B(X^*)=\overline{F(X^*)}^{wk*}$, i.e., when the finite-rank operators are weak*-dense?

Q3: Is there a characterization when $B(X^*)=\overline{K(X^*)}^{wk*}$, i.e., when the compact operators are weak*-dense?

Considering $B(X)$ as a subalgebra of $B(X^*)$ in the usual way, we may ask related questions in connection with $F(X)$ and $K(X)$. The principle of local reflexivity implies $\overline{F(X)}^{wk*}=\overline{F(X^*)}^{wk*}$ in $B(X^*)$. However, it is not clear to me if we always have $\overline{K(X)}^{wk*}=\overline{K(X^*)}^{wk*}$ (I guess not). Therefore, we may also ask:

Q4: Do we always have $K(X)\subseteq\overline{F(X^*)}^{wk*}$?

Answer 1

Regarding Q2: If and only if $X$ has the approximation property. I'll use Ryan's book "Introduction to tensor products of Banach spaces" as a reference, see Prop 4.6 (but this is all standard stuff).

Theorem: $X$ has the approximation property if and only if, whenever $u=\sum_{n\geq 1} \mu_n\otimes x_n \in X^*\hat\otimes X$ is such that $ \sum_n \mu_n(x) x_n = 0$ for all $x\in X$, then $u=0$.

As we have $\sum_n \|\mu_n\| \|x_n\|<\infty$, the condition is equivalent to $\sum_n \mu_n(x) \mu(x_n)=0$ for all $x\in X,\mu\in X^*$ (Hahn-Banach) and hence also equivalent to $\sum_n \mu(x_n)\mu_n = 0$ for all $\mu\in X^*$, and so finally also equivalent to $\sum_n \mu(x_n) f(\mu_n) = 0$ for all $\mu\in X^*, f\in X^{**}$.

This in turn is equivalent to $\langle u, F \rangle=0$ for all finite rank operators $F$ on $X^*$, under your dual pairing between $X^*\hat\otimes X$ and $F(X^*)$.

Finally, observe that $F(X^*)$ is weak$^*$ dense in $B(X^*)$ if and only if the only element of $X^*\hat\otimes X$ which annihilates all of $F(X^*)$ is $0$.

There is a related definition of the "compact approximation property". If I recall it correctly, then you can adapt the proof, and get

Thm: $X$ has the compact approximation property if and only if, whenever $u=\sum_{n\geq 1} \mu_n\otimes x_n \in X^*\hat\otimes X$ is such that $ \sum_n \langle\mu_n,T(x_n)\rangle = 0$ for all $T\in K(X)$, then $u=0$.

Thus, if $X$ has the compact approximation property, but not the approximation property (I think there is an example due to Willis) then we can find $u\in X^*\hat\otimes X$ which annihilates all of $F(X^*)$ but is not zero. There is then $T\in K(X)\subseteq K(X^*)$ with $\langle T,u\rangle \not=0$, and as $$ \overline{F(X^*)}^{wk^*} = \{ T\in B(X^*) : \langle T,u\rangle=0 \text{ for all }u\in X^*\hat\otimes X\text{ with } \langle S,u\rangle=0 \text{ for all } S\in F(X^*) \} $$ we conclude that $K(X)$ is not contained in the weak$^*$-closure of $F(X^*)$. So Q4 is a negative.