Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 3

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ for which every bounded linear functional in $X^{\ast }$ is continuous. Thus every $f\in X^{\ast }$ is weakly continuous. Conversely, if a linear functional $f$ is weakly continuous at zero, some basic weak neighbourhood gives $f_1,\dots ,f_n\in X^{\ast }$ and $\epsilon >0$ such thatIt follows that ${\bigcap }_j\ker f_j\subseteq \ker f$, and elementary linear algebra then gives $f\in \mathrm{span}\{f_1,\dots ,f_n\}\subseteq X^{\ast }$.

To prove Mazur theorem, let $K$ be norm-closed and convex and let $x\notin K$. The Hahn-Banach separation theorem strictly separates $x$ from $K$ by some member of $X^{\ast }$. The corresponding open half-space is weakly open, contains $x$, and misses $K$. Thus $K$ is weakly closed.

If $X$ is reflexive, the Banach-Alaoglu theorem makes $B_{X^{\ast \ast }}$ weak-star compact, and the canonical identification transports this to weak compactness of $B_X$. Conversely, if $B_X$ is weakly compact, then $J(B_X)$ is weak-star compact and hence weak-star closed in $B_{X^{\ast \ast }}$. Goldstine theorem says it is weak-star dense there, so it equals $B_{X^{\ast \ast }}$ and $X$ is reflexive.

When $X$ is reflexive, weak and weak-star topologies coincide on $X^{\ast }$, so Banach–Alaoglu makes $B_{X^{\ast }}$ weakly compact and $X^{\ast }$ is reflexive. If $Y\subseteq X$ is closed, then $B_Y$ is a weakly closed subset of $B_X$, hence weakly compact. The quotient map sends a suitable weakly compact ball of $X$ onto the unit ball of $X/Y$, which is therefore weakly compact. Thus $Y$ and $X/Y$ are reflexive as well.

For each $n$, define $J{x}_n:X^{\ast }\to \mathbb{R}$ by $J{x}_n(f)=f(x_n)$. Weak convergence makes $(J{x}_n(f){)}_n$ bounded for every $f\in X^{\ast }$. The Uniform boundedness principle gives

Let $K$ be the closed convex hull of the $x_n$. Given a sequence in $K$, approximate its terms in norm by finite convex combinations of the $x_n$. A diagonal subsequence makes every coefficient converge. Any loss of total coefficient mass is assigned to zero, which belongs to $K$ by Mazur theorem because $x_n\rightharpoonup 0$. Since $f(x_n)\to 0$ for every $f\in X^{\ast }$, splitting each sum into a finite head and a uniformly small tail proves weak convergence of this subsequence to the corresponding convex combination. Hence $K$ is weakly sequentially compact and, by the stated theorem, weakly compact.

DefineIt is bounded because $(x_n)$ is norm bounded, and its values lie in $c_0$ because $x_n\rightharpoonup 0$. Its adjoint-on-preduals map isIf $K$ had nonempty norm interior, then $K-K$ would contain a ball about zero. The quantitative open-mapping argument applied to the convex combinations above would make $T_{\ast }$ surjective, and hence make $T$ bounded below. Its range $R$ would be a closed infinite-dimensional subspace of $c_0$ whose unit ball is compact for coordinatewise convergence, since it lies in the coordinatewise compact image of a weak-star compact ball of $X^{\ast }$.

By the stated structural theorem, $R$ contains a closed subspace isomorphic to $c_0$. The bounded partial sums of the image of the standard $c_0$ basis would then have a coordinatewise convergent subnet. Uniform boundedness turns coordinatewise convergence in $c_0$ into weak convergence, and norm-closed subspaces are weakly closed. Pulling the limit back would make the partial sums of the standard basis converge weakly in $c_0$, impossible because their coordinate values force the putative limit to be the constant-one sequence. This contradiction proves that $K$ has empty norm interior.