If an operator is weak$^*$-to-weak$^*$ continuous, will it induce an operator? Suppose that $X$ and $Y$ are Banach spaces and have predual, say $X_*$ and $Y_*$. Define an operator $T:X \rightarrow Y$. 
If $T$ is weak$^*$-to-weak$^*$ continuous, will $T$ induce an operator $T_* : X_* \rightarrow Y_*$? 
I think the answer is yes as the authors in this paper used the 'result' above. (at the end of the proof of Proposition $1$)
 A: No, but an operator $T_*\colon Y_* \to X_*$. Let $y_* \in Y_*$. Consider the functional $$ x^{*} \colon X \to \mathbf R, x \mapsto (Tx)(y_*) $$
As $T$ is weak$^*$ to weak$^*$ continuous, for a weak$^*$ convergence net $x_i \to x$, we have $Tx_i \to Tx$ (weak$^*$) and hence $x^*(x_i) \to x^*(x)$. So $x^*$ is weak$^*$-continuous, and can hence be represented by a unique element $x_* \in X_*$, that is 
$$ x^*(x) = x(x_*), \qquad x \in X$$
We define $T_*y_* := x_*$.

Addition: Regarding the existence/uniqueness of $x_*$: For uniqueness, note, that if for $x_{1,*}, x_{2,*} \in X_*$ we have $x(x_{1,*}) = x(x_{2,*})$ for all $x \in X$, then $x_{1,*} = x_{2,*}$ by Hahn-Banach (the dual space $X = (X_*)^*$ of $X_*$ seperates points). For existence: Let $Y$ be any Banach space and $\phi \colon Y^* \to \mathbf R$ be weak$^*$-continuous (in our case $Y = X_*$). Hence, there are $C \ge 0$ and $y_1, \ldots, y_n \in Y$ such that 
$$ |\phi(y^*)| \le C\max_i |y^*(y_i)| $$
Hence, $\ker \phi \subseteq \bigcap_i \ker j(y_i)$, where $j \colon Y \to Y^{**}$ denotes the natural inclusion. Therefore, there are $a_i \in \mathbf R$ such that $\phi = \sum_i a_i \cdot j(y_i)$, that is $\phi = j(\sum_i a_i y_i) \in j[Y]$. Hence, with $y := \sum_i a_i y_i$ 
$$ \phi(y^*) = y^*(y), \qquad y^* \in Y^* $$
and $\phi$ is represented by an $y \in Y$, as wanted.
