Theorem 11.34.286A7D85266AA

Proof. The function $C_\varphi =-B_\varphi $ is convex and Lipschitz. Fix $F\in C(X)$. For a direction $h\in C(X)$, the limit

exists and is finite: convexity makes the difference quotients monotone in $t>0$, and their absolute values are at most $\| h\| _\infty $. Clearly $p$ is positively homogeneous. Given another direction $k\in C(X)$, apply convexity at the midpoint of $F+2th$ and $F+2tk$ and let $t\downarrow 0$; this gives $p(h+k)\leq p(h)+p(k)$. Thus $p$ is sublinear.

By the Hahn–Banach theorem, the zero functional on $\{ 0\} $ extends to a linear functional $\ell \leq p$. Since $p(h)\leq \| h\| _\infty $, applying this bound to $h$ and $-h$ shows that $\ell $ is continuous. Monotonicity of the convex difference quotients gives

which is (11.34.1) once $\ell $ is represented by a probability. Testing the translation rule with positive and negative constants gives $\ell (1)=1$. If $h\geq 0$, monotonicity of $B_\varphi $ gives $C_\varphi (F-th)\leq C_\varphi (F)$ for $t>0$, whereas the inequality just proved gives the lower bound $C_\varphi (F)-t\ell (h)$. Hence $\ell (h)\geq 0$. By the Riesz representation theorem, $\ell (h)=\int _ Xh\, \mathrm{d}\nu $ for a probability measure $\nu $, and $\nu \in \mathcal A_\varphi (F)$. Each inequality in (11.34.1) defines a weakly closed half-space in the compact space of probability measures, so their intersection is closed and convex, hence weakly compact.

It remains to prove absolute continuity. Let $K$ be compact with $\mathrm{d}V(K)=0$, and choose continuous functions $0\leq \chi _ j\leq 1$, equal to one on $K$, whose supports have volumes tending to zero. For any $\nu \in \mathcal A_\varphi (F)$, the supporting inequality with $G=F+\chi _ j$ gives

The right-hand side tends to zero by theorem 11.33.5, so $\nu (K)=0$. By inner regularity of finite Borel measures on $X$, the same holds for every Borel set of zero volume. The Radon–Nikodym theorem now gives $\mathrm{d}\nu =h_\nu \mathrm{d}V$ with $h_\nu \geq 0$ and $\int _ Xh_\nu \mathrm{d}V=1$. $\square$