Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 106 / 1

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.