Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 7 / 1

A separated dual pair consists of real vector spaces $E,F$ and a bilinear form $⟨x,y⟩$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.

The weak topology $\sigma (E,F)$ is the coarsest topology making all maps $x\mapsto ⟨x,y⟩$, $y\in F$, continuous. It makes $E$ a Hausdorff locally convex space, with topology generated by the seminorms $p_y(x)=|⟨x,y⟩|$. A neighbourhood base at zero consists of finite intersectionsSeparation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.

We determine the continuous dual of a weak topology. Suppose a linear functional $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell (x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell (x)=0$. Hence $\ell$ factors through the image ofExtend the resulting linear functional on $T(E)$ to ${\mathbb{R}}^m$. It has the form $z\mapsto {\sum }_j{a}_j{z}_j$, giving $\ell (x)=⟨x,{\sum }_j{a}_j{y}_j⟩$. Conversely every such evaluation is continuous by the definition of the weak topology. Since the pairing separates $F$, this identification is injective, andFor the remaining parts every topological assertion about $E^{\prime }$ or its subsets uses the stipulated weak-star topology $\sigma (E^{\prime },E)$.

True. Choose a countable dense sequence $(x_j)$ in the separable Banach space $E$. For $\varphi ,\psi$ in its closed dual unit ball, putThis is a metric: positivity follows because continuous functionals agreeing on a dense subset agree everywhere; symmetry is immediate; and the triangle inequality follows from subadditivity of $t/(1+t)$ for $t\ge 0$. The tail of the series is uniformly small, so convergence on these coordinates implies convergence for this metric, and conversely metric convergence controls each individual coordinate.

The uniform bound $\Vert \varphi -\psi \Vert \le 2$ upgrades coordinate convergence to pointwise convergence on all of $E$. For any $x\in E$, choose $x_j$ close to $x$ and useFor neighbourhoods involving finitely many $x$, choose finitely many such approximations. This proves equality of the induced topologies, including for arbitrary nets. Thus the weak-star metrizability of the dual ball gives

True. The Banach-Alaoglu theorem makes $B^{\prime }$ compact for the weak-star topology. Its compactness can be seen by embedding it in the product ${\prod }_{x\in E}[-\Vert x\Vert ,\Vert x\Vert ]$: the closed conditions expressing linearity identify a closed subset with $B^{\prime }$, and the product is compact by the Tychonoff theorem. The product topology is exactly pointwise convergence on $E$.

Part (i) supplies a compatible metric. Every compact metric space has a finite $1/n$-net for each positive integer $n$. The countable union of these finite nets is dense. ThereforeThe argument concerns topological separability, with the relative weak-star topology specified in the question.

True in the specified topology. Let $D\subseteq B^{\prime }$ be the countable dense set obtained in part (ii). Scaling by a positive integer is a homeomorphism for the weak-star topology, so $nD$ is dense in $n{B}^{\prime }$. Sincethe countable set ${\bigcup }_{n}nD$ is dense in $E^{\prime }$: a nonempty open set contains a point of some $n{B}^{\prime }$, so its relative open intersection with that ball meets $nD$. This proves weak-star separability of the entire dual.

ThusThere is no assertion of norm separability here. For example, the dual of the separable Banach space ${\ell }^1$ is ${\ell }^{\infty }$, whose binary sequences form an uncountable set with pairwise norm distance one. That example is nevertheless separable in its weak-star topology.

True. We prove weak-star topology on an entire infinite-dimensional Banach dual is not metrizable. Suppose instead that $E^{\prime }$ has a countable local base $(W_n)$ at zero. Choose a basic weak-star topology neighbourhood $U_n\subseteq W_n$, with its conditions involving a finite set $S_n\subseteq E$. The $U_n$ still form a local base.

For any $x\in E$, the set $\{\varphi :|\varphi (x)|<1\}$ is a neighbourhood of zero, so some $U_n$ is contained in it. Every functional annihilating $S_n$, and every scalar multiple of that functional, belongs to $U_n$. Consequently every such functional also annihilates $x$. This forcesIndeed a finite-dimensional span is norm closed, and the Hahn-Banach theorem supplies a bounded linear functional vanishing on it and nonzero at any point outside it.

It follows that $E={\bigcup }_n\mathrm{span}{S}_n$. Each span is a proper finite-dimensional closed vector subspace and has empty interior in the infinite-dimensional Banach space $E$. This contradicts the Baire category theorem. HenceThe uniform norm bound that made the metric work on $B^{\prime }$ is absent on the whole dual. The answers to (i), (ii), (iii) and (iv) are therefore all true.