Given $S \in B(Y^{*}, X^{*})$, does there exist $T\in B(X,Y)$ such that $S=T^{*}$? Let $X, Y$ be Banach spaces, $S \in B(Y^{*}, X^{*})$. Does such operator $T \in B(X, Y)$ exist so that $T^{*}=S$?
I suppose that the answer should be - no. Are there any hints that might help in constructing a counterexample?
Any help would be much appreciated.
 A: For a normed space $E$, let $\Phi_E \colon E \to E^{\ast\ast}$ denote the canonical embedding. The key to the following is the identity
$$\Phi_Y \circ T = T^{\ast\ast} \circ \Phi_X\tag{1}$$
for all $T \in B(X,Y)$. So we have $T^{\ast\ast}(\Phi_X(X)) \subseteq \Phi_Y(Y)$ and find the necessary condition
$$S^{\ast}(\Phi_X(X)) \subseteq \Phi_Y(Y)\tag{2}$$
for $S\in B(Y^{\ast}, X^{\ast})$ to be the transpose of some $T\in B(X,Y)$.
We note that $(2)$ is also sufficient for the existence of a $T$ with $S = T^{\ast}$, using $(1)$ we can then define $T = \Phi_Y^{-1} \circ S^{\ast} \circ \Phi_X$. It is a routine verification that $T$ is then a well-defined continuous linear operator with $T^{\ast} = S$.
Hence there always is such a $T$ if $Y$ is a reflexive Banach space or $X = \{0\}$.
If $Y$ is not reflexive and $X \neq \{0\}$, we can construct an $S\in B(Y^{\ast}, X^{\ast})$ that isn't a transpose by choosing an $\eta \in Y^{\ast\ast}$ and a $\xi \in X^{\ast}\setminus \{0\}$, and setting
$$S(\lambda) = \eta(\lambda)\cdot \xi.\tag{3}$$
For $S^{\ast}$, we find that
$$S^{\ast}(\psi) = \psi(\xi)\cdot \eta.\tag{4}$$
Since $\xi \neq 0$, there is an $x\in X$ with $\xi(x) \neq 0$, and then we have
$$S^{\ast}(\Phi_X(x)) = \Phi_X(x)(\xi)\cdot \eta = \xi(x)\cdot \eta \notin \Phi_Y(Y),$$
so $S$ cannot be the transpose of any $T\in B(X,Y)$.
A: Such an operator exists if and only if $S$ is weakly*-to-weakly* continuous. This is always possible if $X$ and $Y$ are reflexive.
As for concrete examples of operators that do not have a "pre-joint", let $X$ be a separable, reflexive Banach space. Recall that $c_0^*\cong \ell_1$ and take any bounded linear surjection $S\colon (c_0)^*\to X^*$. It does not have a pre-joint because if it had, it would be an isomorphic embedding, but a reflexive Banach space cannot embed into $c_0$.
Here is another, perhaps more exciting example. Note that $C[0,1]^*$ contains a complemented copy of $L_1$. Let $P\colon C[0,1]^*\to C[0,1]^*$ be a projection onto any such subspace. If there were an operator $T$ such that $T^*=P$ then $T$ would be a projection too (that is $T^2=T$), but $L_1$ is not isomorphic to a dual space because separable dual spaces have the Radon–Nikodym property which $L_1$ clearly does not have.
If you do not like this example, note that $C[0,1]^*$ contains a copy of the non-separable space $\ell_1([0,1])$ (the closed linear span of Dirac delta measures) which is complemented. Again, the projection onto $\ell_1([0,1])$ does not have a pre-joint because if it had, it would be a projection with non-separable range.
There are indecomposable spaces with dual isomorphic to $\ell_1$. A space is indecomposable if the only complemented subspaces are either finite- or cofinite-dimensional. Now, take any projection on $\ell_1$ whose range is infinite-dimensional and has infinite codimension. In this case there is no chance for a pre-joint.
