Proposition 21.11.12B99B20CD6166

Proof. Let $H$ denote the infimum displayed in the statement, with the infimum of the empty set equal to $+\infty$. It is at most $F_i$ for every $i$: put weight $1$ on that index and weight $0$ on the others, choosing arbitrary points in their nonempty effective domains.

To prove convexity, take admissible representations of $x$ and $y$, multiply their weights by $t$ and $1-t$, and combine them. For each index $i$ with combined weight $\gamma_i>0$, replace its two points by their weighted barycenter. Convexity of $F_i$ makes its new contribution no larger than the sum of the old contributions. If $\gamma_i=0$, choose any point of its domain. This produces a representation of $tx+(1-t)y$. Taking approximations to the infima proves the strict-height convexity criterion proved above, including when an infimum is $-\infty$.

Finally, every convex function $G\leq F_i$ for all $i$ is at most the value of every admissible combination, by convexity applied to finite epigraph heights above the $F_i(x_i)$ and then taking their limits. Thus $G\leq H$. Therefore $H$ is the largest convex minorant of $\inf_iF_i$, which is precisely $F_1\land\cdots\land F_m$.