Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 8 / 4

Use the standard setting of a compact metrizable convex set $K$ in a Hausdorff locally convex real topological vector space, with its metrizable topology. Write $A(K)$ for the continuous affine real functions on $K$. The affine upper envelope of a bounded real function isConstants make this infimum finite, and $f\le \overline{f}$. An infimum of affine continuous majorants is concave and upper semicontinuous, hence Borel. For continuous $f$ it is the upper concave envelope appropriate to barycentric measures; it is not the pointwise maximum of $f$ and a selected affine function.

For a fixed probability $\mu$, define $p(g)={\int }_K\overline{g}d\mu$ on real $C(K)$. Affine majorants showThus $p$ is sublinear and $p(a)=\int ad\mu$ for affine $a$. On the span of $f$, the linear functional taking $f$ to $p(f)$ is dominated by $p$: the negative-scalar condition follows from $-p(f)\le p(-f)$. For $f=0$ start from the zero subspace. The Hahn-Banach theorem extends it to a linear functional $L$ on $C(K)$ with $L\le p$ and $L(f)=p(f)$.

If $g\ge 0$, then $p(-g)\le 0$, so $L(g)\ge 0$. Also $p(1)=1$ and $p(-1)=-1$, forcing $L(1)=1$. Positivity gives $|L(g)|\le \Vert g{\Vert }_{\infty }$, and the Riesz-Markov-Kakutani representation theorem produces a Borel probability $\nu$ with $L(g)=\int gd\nu$. ThereforeFor affine $a$, testing both $a$ and $-a$ gives $\int ad\nu =\int ad\mu$: the two measures have the same barycenter. This proves the requested supporting measure lemma for affine upper envelopes.

Choquet's theorem: every $x\in K$ has a Borel probability measure $\lambda$ concentrated on the extreme points of $K$ whose barycenter is $x$; equivalently,Concentration is a measure-one assertion, not a claim that the topological support must be a closed subset of $\mathrm{Ex}K$.

To prove it, let ${\mathcal{M}}_x$ be the measures satisfying all the displayed affine equalities. It is nonempty because it contains ${\delta }_x$, and it is weakly closed in the compact space $P(K)$, hence compact. Let $h\in C(K)$ be strictly convex, as permitted. Choose $\lambda \in {\mathcal{M}}_x$ maximizing $\int hd\lambda$. Apply the supporting measure lemma with $\mu =\lambda$ and $f=h$. It gives $\nu$ with the same affine integrals, so $\nu \in {\mathcal{M}}_x$, andThus $\overline{h}-h\ge 0$ has integral zero. If $z$ is not extreme, write $z=ty+(1-t)w$ with $0<t<1$ and distinct $y,w\in K$. Strict convexity and every affine majorant giveTherefore the nonnegative gap is strictly positive at every nonextreme point. It is Borel, and $\mathrm{Ex}K$ is Borel by the allowed G-delta set assertion. Its zero integral forces $\lambda (K\setminus \mathrm{Ex}K)=0$, proving Choquet's theorem by strict convexity. No uniqueness is asserted for the representing measure.

For the real $L^{\infty }$ example, give its closed unit ball $K$ the weak-star topology $\sigma (L^{\infty },L^1)$, not the norm topology. The Banach-Alaoglu theorem makes it compact, and separability of $L^1[0,1]$ makes this ball metrizable. Its extreme points are exactly the classes $h$ satisfying $|h|=1$ almost everywhere. Indeed, if $|h|\le 1-\epsilon$ on a positive-measure set, adding and subtracting $\epsilon$ times its indicator decomposes $h$ nontrivially inside the ball. Conversely, if $|h|=1$ almost everywhere and $h=(u+v)/2$ with $|u|,|v|\le 1$, pointwise equality at the endpoints of $[-1,1]$ forces $u=v=h$ almost everywhere. This is the extreme-point criterion for the L-infinity unit ball.

For the hinted case $f=1_A-1_B$, put $C=[0,1]\setminus (A\cup B)$ and $h_{\pm }=1_A-1_B\pm 1_C$. Then $h_{\pm }$ are extreme andhas barycenter $f$. If $C$ is null, the two point masses coincide.

For a general real $f$, choose a measurable representative in $[-1,1]$ and set, for $0\le t\le 1$,Every $H_t$ is extreme. The map into the weak-star compact ball is Borel: for each $u\in L^1$, the function $t\mapsto \int u(s)H_t(s)ds$ is measurable by joint measurability and integration; a countable dense family of such tests generates the ball's topology and Borel sigma-algebra. The measure is independent of changes to $f$ on a null set. For each $s$,The integrand paired with $u$ is dominated by $|u|$, so Fubini's theorem yieldsThese continuous linear tests define the weak-star barycenter, hence that barycenter is $f$. Together with $\nu (\mathrm{Ex}K)=1$, this is the required threshold Choquet representation in L-infinity.

If $L^{\infty }$ is instead taken over complex scalars, its extreme points satisfy the same unit-modulus condition. Write $f(s)=r(s)\zeta (s)$, with $r=|f|$ and $|\zeta |=1$, choosing $\zeta =1$ where $f=0$. Replace the threshold family by $\zeta (s)(21_{\{t\le (1+r(s))/2\}}-1)$. Its members have unit modulus and its average is $f$, so the same pushforward and Fubini argument gives a representing measure in the real locally convex interpretation of the complex ball.