= Solution
If $X$ is reflexive, then $X^*$ is reflexive. The weak topology $\sigma(X^*,X^{**})$ and <weak-star topology> $\sigma(X^*,X)$ therefore coincide under the canonical identification $X^{**}=X$. Thus every weak-star convergent sequence in $X^*$ 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 $B_{X^*}$ weak-star compact and metrizable, hence weak-star sequentially compact. Every convergent subsequence is weakly convergent by the Grothendieck property. Thus $B_{X^*}$ is weakly sequentially compact. The stated converse to part (a), equivalently the other direction of the <Eberlein-Smulian theorem>, makes $B_{X^*}$ weakly compact. Hence $X^*$ is reflexive, and therefore so is $X$.
Finally let $T:X\to Y$ be bounded and onto, with $X$ Grothendieck, and suppose $y_n^*\to y^*$ weak-star in $Y^*$. Then
$$
T^*y_n^*\longrightarrow T^*y^*
$$
weak-star in $X^*$, hence weakly. The <open mapping theorem> implies that $T^*:Y^*\to X^*$ is an isomorphism onto its closed range. Given $y^{**}\in Y^{**}$, the functional
$$
T^*y^*\longmapsto y^{**}(y^*)
$$
is bounded on $T^*(Y^*)$ and extends by the <Hahn-Banach theorem> to some $x^{**}\in X^{**}$. Therefore
$$
y^{**}(y_n^*)=x^{**}(T^*y_n^*)longrightarrow x^{**}(T^*y^*)=y^{**}(y^*).
$$
This is weak convergence in $Y^*$, so $Y$ is Grothendieck.
Back to article page
