Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2008 / A / 3

The weak-star topology on $X^{\ast }$ is the coarsest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous. Finite intersections of evaluation inequalities form a neighborhood basis. Distinct functionals differ on some $x$, so disjoint disks around their evaluation values give disjoint neighborhoods; thus the topology is Hausdorff. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this topology.

If $f\in X^{\ast \ast }$ is weak-star continuous, some $x_1,\dots ,x_n$ and $\epsilon >0$ make $|f(\varphi )|<1$ whenever $|\varphi (x_i)|<\epsilon$ for every $i$. If all evaluations vanish, every scalar multiple of $\varphi$ lies in this neighborhood, forcing $f(\varphi )=0$. Thus $f$ vanishes on the kernel of the map $E(\varphi )=(\varphi (x_1),\dots ,\varphi (x_n))$. It factors linearly through the finite-dimensional range of $E$. Extending that factor algebraically to ${\mathbb{K}}^n$ gives scalars $a_i$ withConsequently $f=J_{X}x$ for $x={\sum }_i{a}_i{x}_i$.

For the two induced topologies, let $B$ be the common norm-closed unit ball and write ${\tau }_j=\sigma (Y,X_j)$. Under either topology $B$ is compact by Banach-Alaoglu, and the ambient space is Hausdorff. If the two topologies are comparable, the identity map from $B$ with the finer topology to $B$ with the coarser one is a continuous bijection from a compact space to a Hausdorff space. Such a map is a homeomorphism, since it carries closed compact sets to closed sets. Thus in the comparable case the topologies agree on $B$, giving alternative (a).

If the two ambient topologies are not comparable, neither is contained in the other. By the definition of inclusion of topologies, there is a ${\tau }_1$-open set absent from ${\tau }_2$, and a ${\tau }_2$-open set absent from ${\tau }_1$. These are the requested $U_1,U_2$, giving alternative (b). Together with the comparable case, this proves that at least one of the two alternatives always holds; it does not assume that the alternatives are mutually exclusive.

TakeIt is evaluation against the first coordinate vector of $c_0$, hence continuous for $\sigma ({\ell }^1,c_0)$. The preceding continuous-dual result says that a functional continuous for $\sigma ({\ell }^1,c)$ must have coefficient sequence $({\gamma }_{\infty },{\gamma }_1,{\gamma }_2,\dots )$ for a convergent sequence $\gamma$. To represent ${\alpha }_1$, it would require ${\gamma }_{\infty }=1$ but every ${\gamma }_j=0$, impossible.

Directly, $e_1-e_n\to 0$ in $\sigma ({\ell }^1,c)$ because its pairing is ${\gamma }_{\infty }-{\gamma }_{n-1}\to 0$, whereas ${\psi }_1(e_1-e_n)=1$ for $n\ge 2$.

TakeThe series defines a bounded linear functional of norm one. It corresponds under ${\Phi }_2$ to the constant sequence ${\gamma }_j=1$, so is $\sigma ({\ell }^1,c)$-continuous. Its coefficient sequence is constantly one, which is not in $c_0$, so the continuous-dual result makes it noncontinuous for $\sigma ({\ell }^1,c_0)$. Directly $e_n\to 0$ in that topology but ${\psi }_2(e_n)=1$.