Theorem 16.28.294044625470CC

Proof. Each defining inequality of $\mathcal J_ a(K)$ is preserved under weak convergence of the probability measures and convergence of the constants, because the test function $f$ is continuous. Since the Borel probability measures on the compact metric space $K$ form a weakly compact set, the pairs in $\mathcal J_ a(K)$ with $c\leq C$ form a compact set for each finite $C$.

Call a pair in $\mathcal J_ a(K)$ active at $\tau $ if it satisfies the equality in (16.28.30). Such pairs exist by theorem 16.28.27, and since $\int _ K\psi \, \mathrm{d}\mu \geq m=\inf _ K\psi $, they satisfy

The set of active pairs is therefore compact, and the minimum of $c$ over it is attained. By theorem 16.28.27 at $\tau +h$, each active pair gives $F_{\tau +h}(a)\leq F_\tau (a)+hc$, hence $D_\tau (a)\leq c$.

Conversely, for $h>0$ choose a pair $(\mu _ h,c_ h)$ attaining the minimum in (16.28.28) at $\tau +h$, with $q=\psi $. The representation at $\tau $ gives

Thus $0\leq c_ h\leq (F_{\tau +h}(a)-F_\tau (a))/h$, and these constants are bounded by (16.28.21). Along a sequence $h\downarrow 0$, compactness gives a limiting pair $(\mu ,c)\in \mathcal J_ a(K)$. The same bound makes $F_{\tau +h}(a)\to F_\tau (a)$ and $hc_ h\to 0$, and continuity of $\psi $ then gives $\int _ K\psi \, \mathrm{d}\mu +\tau c=F_\tau (a)$. Thus the limit pair is active, and it satisfies $c\leq D_\tau (a)$. This proves (16.28.30). Finally, every pair has $c\geq V_ K^*(a)$ by theorem 16.28.27, so the attainment of the minimum gives the stated criterion. $\square$