 Solution

If $X$ is reflexive, then $X^{\ast }$ is reflexive. The weak topology $\sigma (X^{\ast },X^{\ast \ast })$ and weak-star topology $\sigma (X^{\ast },X)$ therefore coincide under the canonical identification $X^{\ast \ast }=X$. Thus every weak-star convergent sequence in $X^{\ast }$ is weakly convergent, so $X$ is a Grothendieck space.

Conversely, suppose that $X$ is separable and Grothendieck. The Banach-Alaoglu theorem and weak-star metrizability of the dual ball make $B_{X^{\ast }}$ weak-star compact and metrizable, hence weak-star sequentially compact. Every convergent subsequence is weakly convergent by the Grothendieck property. Thus $B_{X^{\ast }}$ is weakly sequentially compact. The stated converse to part (a), equivalently the other direction of the Eberlein-Smulian theorem, makes $B_{X^{\ast }}$ weakly compact. Hence $X^{\ast }$ is reflexive, and therefore so is $X$.

Finally let $T:X\to Y$ be bounded and onto, with $X$ Grothendieck, and suppose $y_n^{\ast }\to y^{\ast }$ weak-star in $Y^{\ast }$. Thenweak-star in $X^{\ast }$, hence weakly. The open mapping theorem implies that $T^{\ast }:Y^{\ast }\to X^{\ast }$ is an isomorphism onto its closed range. Given $y^{\ast \ast }\in Y^{\ast \ast }$, the functionalis bounded on $T^{\ast }(Y^{\ast })$ and extends by the Hahn-Banach theorem to some $x^{\ast \ast }\in X^{\ast \ast }$. ThereforeThis is weak convergence in $Y^{\ast }$, so $Y$ is Grothendieck.