Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 6 / 4

Let $p$ be the Minkowski functional of $E$:The contained ball makes $E$ absorbing, so $p$ is finite, nonnegative and satisfies $p(v)\le \Vert v\Vert /ϵ$. It is positively homogeneous. To check subadditivity, if $v\in sE$ and $w\in tE$, the defining property of a convex set gives $(v+w)/(s+t)\in E$, whence $p(v+w)\le s+t$; let $s,t$ decrease to the respective infima. Thus $p$ is a sublinear functional. We have $p(e)\le 1$ for every $e\in E$. Also $p(x)\ge 1$: otherwise $x\in tE$ for some $0<t<1$, which would put $x$ in $E$ because $E$ is a convex set and $0\in E$.

We prove the dominated form of the Hahn-Banach theorem needed here. Suppose $g$ is linear on a subspace $M$ and $g\le p$ there. To extend it to $M+\mathbb{R}v$ with $v\notin M$, choose the value $c$ at $v$ betweenEvery lower candidate is at most every upper candidate, becauseTaking $m=n=0$ among the candidates shows the two endpoints are finite: the lower supremum is at least $-p(-v)$ and at most $p(v)$, and the upper infimum lies between them and $p(v)$. Define $\tilde{g}(m+tv)=g(m)+tc$. For $t>0$, the upper inequality applied to $m/t$ gives domination by $p(m+tv)$. For $t<0$, apply the lower inequality to $m/(-t)$ and multiply by $-t$. At $t=0$ domination already holds. Hence this is a dominated one-dimensional extension.

Order all dominated extensions by inclusion of their domains. Chain unions preserve linearity and domination, so Zorn's lemma gives a maximal extension. The one-dimensional argument shows its domain is the whole vector space. This proves the requisite dominated Hahn-Banach theorem, rather than invoking an unproved separation form.

On the line $\mathbb{R}x$, start with $g(tx)=tp(x)$. For $t\ge 0$ this equals $p(tx)$; for $t<0$ it is nonpositive and therefore at most $p(tx)$. Extend it to $T\le p$ on $V$. ThenApplying domination to $v$ and $-v$ also givesThus $T$ is a continuous linear functional, andThis proves separation from an absorbing convex set with no closedness assumption on $E$. The proof also gives the norm-preserving real extension theorem used earlier: take $p(v)=C\Vert v\Vert$ when the original functional has norm at most $C$.

If $F$ is empty, the zero linear functional satisfies the conclusion vacuously. Otherwise choose one $f_0\in F$, put $\delta =ϵ/2$, and form the convex setIt contains $B(0,\delta )$ because $f_0\in F$. The point $x=-f_0$ is outside $E$: membership would imply $0=f+u$ for some $f\in F$ with $\Vert u\Vert <\delta$, contradicting the assumed absence of points of $F$ in $B(0,ϵ)$.

Apply part (i). There is a continuous linear functional $T$ with $T(-f_0)\ge 1$ and $T(f+u-f_0)\le 1$ for every $f\in F$, $\Vert u\Vert <\delta$. Thus $T\ne 0$ andFor fixed $f$, take the supremum of $Tu$ over this open ball; it is $\delta \Vert T\Vert$. Consequently $Tf\le -\delta \Vert T\Vert$ for every $f\in F$. The normalizationis a continuous linear functional satisfying $Sf\ge 1$ for every $f\in F$. No compactness, closedness or attainment of the supremum on the open ball is needed.

Let $F$ be the convex hull of all terms of the sequence, meaning the set of their finite convex combinations. Suppose $B(0,ϵ)\cap F=\varnothing$. Part (ii) then gives a continuous linear functional $S$ with $Sf\ge 1$ for every $f\in F$. In particular $S{f}_n\ge 1$ for every $n$, contradicting $S{f}_n\to 0$. Therefore the ball meets $F$.

A point of that intersection is a finite combination ${\sum }_{k=1}^m{\alpha }_k{f}_{n_k}$ with nonnegative coefficients summing to one. Set $N={\max }_k{n}_k$, combine repeated indices and insert zero coefficients for omitted indices. This givesThus zero lies in the norm closure of the convex hull. The same argument applies to every tail of the sequence and, by choosing errors tending to zero, also gives the usual Mazur lemma. Completeness of $V$ is not required for this conclusion.

The hypothesis needed is precisely convergence to zero against every continuous linear functional, namely weak convergence. The PDF says only “continuous” for the testing maps. If that is interpreted as including arbitrary nonlinear continuous maps, the constant map $T\equiv 1$ makes the hypothesis impossible. The proof above establishes the substantive conclusion under the intended, weaker linear-testing hypothesis, and therefore also the implication under the literal stronger wording.