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Weak compactness of the adjoint forces weak compactness of the original operator. Assume $T^{\ast }:Y^{\ast }\to X^{\ast }$ is weakly compact. Apply the already proved $(i)⟹(ii)$ to this operator, rather than assuming the adjoint equivalence in advance. It givesTake $\psi \in Y^{\ast \ast \ast }$ annihilating $J_{Y}Y$, so $\psi (J_{Y}y)=0$ for every $y\in Y$. Write $T^{\ast \ast \ast }\psi =J_{X^{\ast }}f$ for some $f\in X^{\ast }$. For every $x\in X$,Hence $f=0$ and $T^{\ast \ast \ast }\psi =0$. It follows thatfor every $\psi$ annihilating $J_{Y}Y$.

The vector subspace $J_{Y}Y$ is closed in the norm topology in $Y^{\ast \ast }$ because $Y$ is complete and $J_Y$ is an isometry. The Hahn-Banach theorem says that any point outside a closed linear subspace can be separated from it by a bounded linear functional vanishing on that subspace. Applying this in $Y^{\ast \ast }$ shows that an element annihilated by all such $\psi \in Y^{\ast \ast \ast }$ must belong to $J_{Y}Y$. Therefore $T^{\ast \ast }F\in J_{Y}Y$ for every $F$, proving $(iv)⟹(ii)$. The reverse implications above now establish all four equivalences, including weak compactness of an operator and its adjoint:

Finally, if $X$ is a reflexive Banach space, every $F\in X^{\ast \ast }$ is $J_{X}x$ for some $x$, and $T^{\ast \ast }F=J_{Y}Tx\in J_{Y}Y$. If $Y$ is a reflexive Banach space, $Y^{\ast \ast }=J_{Y}Y$, so the same range condition holds automatically. In either case, $(ii)⟹(i)$ proves