sufficient conditions for an operator on a Banach space to be a dual operator?

Question

If $X$ is a Banach space, we denote by $\mathcal{L}(X)$ the algebra of continuous linear operators acting on $X$. Suppose $T\in\mathcal{L}(X^*)$, where $X^*$ is the dual space to $X$. I am interested in conditions on $T$ to guarantee that it is a dual operator, i.e. so that there exists $S\in\mathcal{L}(X)$ with $S^*=T$.

Clearly, if $X$ is reflexive then every $T\in\mathcal{L}(X^*)$ is a dual operator. So we can assume $X$ is nonreflexive.

It is well-known that $T$ is a dual operator if and only if it is weak*-to-weak* continuous. However, this is not always easy to check, so I am interested in "nicer" sufficient conditions for $T$ to be a dual operator. One question that immediately comes to mind is this:

Question 1. Suppose $T\in\mathcal{L}(X^*)$ is compact. Is it necessarily a dual operator?

(UPDATE: The answer to Q1 is obviously not; see comments below.)

I am especially interested in the case where $X=c_0$ and hence $X^*=\ell_1$. So:

Question 2. What are some "nice" sufficient conditions to guarantee that $T\in\mathcal{L}(\ell_1)$ is a dual operator?

Actually, I only need $T$ to be a dual operator up to finite-rank perturbation. So:

Question 3. If $T\in\mathcal{L}(X^*)$, what are some "nice" sufficient conditions to guarantee that $T=S^*+F$ for some $S\in\mathcal{L}(X)$ and some finite-rank $F\in\mathcal{F}(X^*)$?

Note: In the context I am working, I can assume that $X^{**}/X$ is infinite-dimensional and that $X^*$ is separable.

Thanks guys!

Answer 1

I don't know any sufficient conditions. But I know that not even finite-rank is sufficient.

Denote the canonical basis of $\ell_1$ (and of $c_0$) by $\{e_n\}$. Let $F:\ell_1\to\ell_1$ be given by $$ Fx=(\sum x_n)\,e_1. $$ We have that $e_n\to 0$ weak$^*$, because if $y\in c_0$ we have $\langle e_n,y\rangle = y_n\to0$. But $Te_n=e_1$ for all $n$, so $T$ is not weak$^*$-weak$^*$ continuous.