Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 106 / 2 / b

For each $t\in [0,1]$, evaluation ${\delta }_t:f\mapsto f(t)$ is a bounded linear functional of norm one. Thus weak convergence to zero gives $f_n(t)\to 0$ at every $t$. The set $\{f_n:n\ge 1\}$ is a weakly bounded set, since every scalar sequence $\ell (f_n)$ converges, and part (a) supplies a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$.

The constant $M$ is integrable for Lebesgue measure on $[0,1]$. Applying the dominated convergence theorem to $|f_n|$ gives the weakly null continuous functions converge in L1 conclusionPointwise convergence alone would not provide the needed uniform dominating function; it is the weak boundedness argument that supplies it.

Put $d=\inf \{\Vert x\Vert :x\in C\}$. The nonemptiness of $C$ gives $0\le d<\infty$. By Mazur theorem, a norm-closed convex set is weakly closed. This applies to both $C$ and every closed norm ball.

For $n\ge 1$, the sets $C_n=C\cap (d+1/n)B_X$ are nonempty by the definition of the infimum, weakly closed and nested. They all lie in the weakly compact set $(d+1)B_X$, by the weak compactness characterization of reflexivity. Hence they have the finite intersection property, and compactness gives $x_0\in {\bigcap }_n{C}_n$. Then $x_0\in C$ and $\Vert x_0\Vert \le d+1/n$ for every $n$, soThis proves the norm minimizer in a closed convex subset of a reflexive Banach space assertion without assuming sequential weak compactness. It includes $d=0$; uniqueness is not asserted without an additional condition such as strict convexity of the norm.

The coordinate evaluations $e_n:x\mapsto x_n$ belong to the dual unit ball of the l-infinity sequence space and separate its points. Therefore the countable separating family metrizes a weakly compact set result makes any nonempty weakly compact $K\subset {\ell }_{\infty }$ a compact metric space in its weak topology. Choose finite $1/n$-nets and take their countable union $D$; it is weakly dense in $K$.

It follows that $K\subset {\overline{D}}^w\subset {\overline{\mathrm{conv}D}}^w$. By Mazur theorem, the last set is ${\overline{\mathrm{conv}D}}^{\Vert \cdot \Vert }$. Finite convex combinations of members of $D$ with nonnegative rational coefficients summing to one form a countable norm-dense subset of $\mathrm{conv}D$: for each fixed finite list, approximate its coefficients in the simplex by rational coefficients, and use the norm triangle inequality. Thus its norm closure is a separable metric space containing $K$.

Every subset of a separable metric space is separable: balls centred on a countable dense set with positive rational radii form a countable base; intersect this base with the subset and choose one point from each nonempty intersection. Consequently the weakly compact subsets of l-infinity are norm separable conclusion isThe empty case is immediate. The convexification matters: the chosen weakly dense set $D$ itself need not be norm dense in a nonconvex $K$.