Theorem 12.8.2E9E119AAA269

Proof. For a maximizing $T$ and any $G\in C(X)$,

Since $B_0=-n\log M$, taking $-n$ times the logarithm and applying Jensen's inequality gives

Hence every probability $\nu _{T,F}$ with $T\in \mathscr M(F)$ belongs to $\mathcal A_0(F)$.

Denote the set of these probabilities by $\mathscr N(F)$; by the preceding lemma it is compact and convex. To prove the reverse inclusion we compute the right derivative $(B_0)'_+(F;h)=\lim _{t\downarrow 0}(B_0(F+th)-B_0(F))/t$ for $h\in C(X)$. Since $\mathfrak D_{F+th}(T)=\int _ X\mathrm{e}^{th/n}\, \mathrm{d}\Gamma _ F(T)$ and the masses of the measures $\Gamma _ F(T)$ are bounded on $\mathscr G_\omega $, a Taylor expansion of the exponential gives, as $t\downarrow 0$,

with the error bounded uniformly in $T$. Testing maximizers at $F$ gives

For the reverse inequality, choose a maximizer $T_ t$ at $F+th$. The multiplicative bounds in the previous proof imply $\mathfrak D_ F(T_ t)\to M(F)$. Every sequence $t\downarrow 0$ has a subsequence along which $T_ t$ converges, and the preceding lemma identifies the limit of $\Gamma _ F(T_ t)$ with the determinant measure of a maximizer at $F$. Since $\mathfrak D_ F(T_ t)\leq M(F)$, the Taylor expansion gives the matching upper bound for the difference quotients of $M$. Since $B_0=-n\log M$, therefore

If $\nu \in \mathcal A_0(F)$, its supporting inequality in direction $th$ implies

If $\nu $ were not in the weakly compact convex set $\mathscr N(F)$, the Hahn–Banach separation theorem in the dual of $C(X)$ would give some $h\in C(X)$ violating this inequality. Hence $\nu \in \mathscr N(F)$, and the two sets are equal. The absolute continuity of their members was proved in theorem 11.34.2. $\square$