 Solution

Weak compactness forces the second adjoint to take values in $Y$. Suppose $C={\overline{T(B_X)}}^{\Vert \cdot \Vert }$ is weakly compact in $Y$. By the first observation, $J_{Y}C$ is closed for the weak-star topology in $Y^{\ast \ast }$. The second adjoint $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star-to-weak-star continuous, by the adjoint-continuity argument applied to $T^{\ast }$.

Thus $(T^{\ast \ast }{)}^{-1}(J_{Y}C)$ is closed for the weak-star topology in $X^{\ast \ast }$. The identity $T^{\ast \ast }{J}_X=J_{Y}T$ makes it contain $J_X(B_X)$. The Goldstine theorem therefore givesScaling any $F\in X^{\ast \ast }$ into the closed unit ball provesThis is the forward direction of the bidual characterization of weakly compact operators.