 Solution

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.