Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 106 / 2

Choose a norm-dense sequence $(f_n)$ in the unit sphere of the separable dual $X^{\ast }$. For every $n$, choose $x_n\in B_X$ with $|f_n(x_n)|>1/2$. If a functional $f\in X^{\ast }$ vanished on the closed linear span of the $x_n$, normalize it and choose $f_n$ with $\Vert f-f_n\Vert <1/4$; then $|f(x_n)|>1/4$, a contradiction. The Hahn-Banach theorem therefore shows that the span of $(x_n)$ is dense, so $X$ is separable.

Choose a norm-dense sequence $(u_n)$ in $B_X$. On $B_{X^{\ast }}$ defineWeak-star convergence implies convergence in this metric. Conversely, metric convergence gives convergence on the dense set $(u_n)$, and the uniform norm bound on the dual ball extends it to every $u\in X$. Thus $d_{\ast }$ metrizes the weak-star topology, proving weak-star metrizability of the dual ball.

Similarly, for a norm-dense sequence $(f_n)$ in $B_{X^{\ast }}$,metrizes the weak topology on $B_X$. Indeed, convergence against the dense functionals extends to every $f\in X^{\ast }$ because $x-y$ remains norm bounded.

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By definition of the weak topology, $x_n\rightharpoonup x$ exactly when $f(x_n)\to f(x)$ for every $f\in X^{\ast }$.

Let $(x_n)$ be bounded and choose a norm-dense sequence $(f_j)$ in $X^{\ast }$. Successive subsequences make $f_1(x_n),f_2(x_n),\dots$ converge, and the diagonal argument produces a single subsequence $(y_n)$ on which every $f_j$ converges. Uniform boundedness of $(y_n)$ and norm approximation of an arbitrary $f$ by the $f_j$ show that $f(y_n)$ is Cauchy for every $f\in X^{\ast }$. Thus $(y_n)$ is weakly Cauchy, andfor every $f$, so its difference sequence is weakly null.

For the countable family $(y_{m,n}{)}_n$, use the weak metric from part a. Delete a finite initial segment from the $m$th sequence so that every remaining term has weak distance less than $1/m$ from zero, and relabel that tail. Enumerate all these tails while preserving the order within each one. For every weak neighbourhood of zero, all terms from sufficiently large $m$ lie inside it, and only finitely many terms from each of the finitely many remaining sequences lie outside it. The resulting enumeration $(z_n)$ is weakly null and contains the relabelled $m$th sequence as a subsequence for every $m$. Equivalently, without relabelling, it contains a tail-subsequence of every original sequence, which is the form used below.

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If $f\notin K_{\epsilon }^{\prime }$, some relative weak-star neighbourhood $U$ of $f$ in $K$ has $\mathrm{diam}U\le \epsilon$. Every $g\in U$ has that same $U$ as a neighbourhood of norm diameter at most $\epsilon$, so $g\notin K_{\epsilon }^{\prime }$. Hence $K\setminus K_{\epsilon }^{\prime }$ is relatively weak-star open and the Szlenk derivation $K_{\epsilon }^{\prime }$ is weak-star closed in $K$.

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The set $K$ is weak-star compact by the Banach-Alaoglu theorem and weak-star metrizable by part a. Choose a decreasing countable neighbourhood base $(U_n)$ at $f$. Since $f\in K_{\epsilon }^{\prime }$, the set $U_n\cap K$ cannot lie inside the closed norm ball of radius $\epsilon /2$ about $f$, for then its diameter would be at most $\epsilon$. ChooseThe neighbourhood-base property gives $g_n\stackrel{w^{\ast }}{\to }f$.

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Start with the sequence from part ii. For each $n$, the definition of the operator norm gives $x_n\in B_X$ withBecause $g_n\stackrel{w^{\ast }}{\to }f$, pass recursively to a subsequence so far out that its $n$th member also satisfies

Part b supplies a weakly Cauchy subsequence $(x_{k_j})$. PutThen $(y_j)$ is weakly null, $h_j\stackrel{w^{\ast }}{\to }f$, and the two preceding estimates giveSince $f(y_j)\to 0$, discard finitely many terms to obtain $|h_j(y_j)|>\epsilon /16$ for every remaining $j$. Relabelling proves the claim.

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Part i makes $K_{\epsilon }^{\prime }$ a closed subset of the compact metrizable space $K$, so it is compact metrizable and therefore separable. Choose the stated dense sequence $(f_m)$ and, using part iii, the sequences $(g_{m,n})$ and $(y_{m,n})$. In constructing the universal weakly null sequence $(z_n)$ from part b, retain a tail of each $y_m$ and write $z_{j(m,n)}=y_{m,n}$ along the retained subsequence, where $j(m,n)\to \infty$.

Eachis weak-star closed. Since $z_n\rightharpoonup 0$, every fixed $f\in K$ belongs to some $A_N$, so $K={\bigcup }_N{A}_N$. The compact Hausdorff space $K$ is a Baire space, and the Baire category theorem implies that some $A_N$ has nonempty relative weak-star interior.

Suppose $K_{\epsilon }^{\prime }=K$. Choose a nonempty relatively open $O\subseteq A_N$ and then $f_m\in O$ by density. Since $g_{m,n}\stackrel{w^{\ast }}{\to }{f}_m$, eventually $g_{m,n}\in O$. For a sufficiently late retained index, also $j(m,n)\ge N$, and hencecontradicting $|g_{m,n}(y_{m,n})|>\epsilon /16$. Therefore the One-step Szlenk derivation for a separable dual gives $K_{\epsilon }^{\prime }⊊K$ whenever $K$ is nonempty.

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