Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 106 / 1

First suppose that $X$ is separable. Choose a norm-dense sequence $(x_n)$ in $B_X$. On every norm-bounded subset of $X$, a countable norm-dense subset $(f_n)$ of $B_{X^{\ast }}$ in the weak-star topology separates points, andmetrizes the weak topology. Indeed, convergence for all $f_n$, boundedness, and weak-star density imply convergence for every $f\in X^{\ast }$. Thus a weakly compact set $K$ is a compact metric space and hence is sequentially compact.

For general $X$, take a sequence $(x_n)$ in $K$ and letThe space $Y$ is separable and norm closed. The Hahn-Banach theorem shows both that $Y$ is weakly closed in $X$ and that its own weak topology is the subspace topology inherited from $X$. Hence $K\cap Y$ is weakly compact and, by the separable case, contains a weakly convergent subsequence of $(x_n)$. Its limit lies in $K\cap Y$. Therefore every weakly compact subset of a Banach space is weakly sequentially compact, which is one direction of the Eberlein-Smulian theorem.

A bounded sequence $(x_n)$ lies in a scalar multiple of the closed unit ball of the reflexive Banach space $X$, which is weakly compact. Part (a) gives a subsequence $x_{k_n}\rightharpoonup z$ for some $z\in X$.

The compact operator $T$ is weak-to-weak continuous, so $T{x}_{k_n}\rightharpoonup Tz$. We claim that the convergence is in norm. Otherwise some further subsequence would satisfy $\Vert T{x}_{k_n}-Tz\Vert \ge \epsilon$. Compactness supplies a norm-convergent subsubsequence; its norm limit must also be its weak limit $Tz$, a contradiction. Hence

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.