Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 36 / 3 / a / Solution

For $v\in Z$, write $M(v)={\max }_i{v}_i$. Maximizing a linear functional on the price simplex givesThus $g(v)$ is a nonempty closed convex face of the compact price simplex. Its graph of a set-valued mapping is closed: if $v_n\to v$, $p_n\to p$ and $p_n\in g(v_n)$, then, for every fixed $q\in P$, the inequalities $p_n\cdot v_n\ge q\cdot v_n$ pass to $p\cdot v\ge q\cdot v$.

Each value $z(p)$ is closed in $Z$ by the closed-graph hypothesis, hence compact; it is nonempty and convex by assumption. Therefore $F(p,v)=g(v)\times z(p)$ has nonempty closed convex values in the nonempty compact convex set $K=P\times Z$.

Its graph of a set-valued mapping is also closed. If $(p_n,v_n)\to (p,v)$ and $(q_n,w_n)\to (q,w)$ with $q_n\in g(v_n)$ and $w_n\in z(p_n)$, the two closed-graph properties give $q\in g(v)$ and $w\in z(p)$. Finally, this closed graph implies upper hemicontinuity because the whole output space $K$ is compact. If an open set containing $F(p,v)$ failed to contain the nearby values, select outputs outside it along a convergent input sequence. A subsequence of those outputs converges in $K$; closedness puts its limit in $F(p,v)$, whereas closedness of the open set's complement puts it outside. This contradiction proves upper hemicontinuity.

Consequently all the hypotheses of the Kakutani fixed-point theorem hold: the domain is nonempty, compact and convex, and the set-valued mapping maps it into itself with nonempty closed convex values and upper hemicontinuity.