Theorem 16.28.146D209F2CFC94

Proof. Let $m=\inf _ K\psi $. Inclusion of the growth classes and scaling of competitors give, for $0<\alpha <\beta $,

Indeed, if $u$ is a competitor for $V_\beta $, then $(\alpha /\beta )(u-m)+m$ is a competitor for $V_\alpha $, since $\psi \geq m$. The constant $m$ is also a competitor, so all terms are finite and the right-hand side is nonnegative. The estimate shows that $V_\delta (z)$ depends continuously on $\delta >0$ at every point $z$. Since $V_\delta \leq \psi $ on $K$ and $V_\delta $ increases with $\delta $, the set $\Lambda $ is closed upwards, and by continuity it is relatively closed in $(0,\infty )$. This proves the three possible forms of $\Lambda $.

For $0\leq t\leq 1$, convex combinations of competitors give

By the preceding proposition, the same inequality holds for the regularized envelopes almost everywhere, and hence everywhere by lemma 5.24.3. Thus $F_\delta (z)$ is a concave nondecreasing function of $\delta $ at every point $z$.

For $a\in K$ and $\delta \in \Lambda $, the defect $d_\delta =F_\delta (a)-\psi (a)$ is nonnegative, concave and nondecreasing in $\delta $. Suppose $d_\alpha =0$, and that there is $\eta \in \Lambda $ with $\eta <\alpha $. For every $\beta >\alpha $, write $\alpha $ as a convex combination of $\eta $ and $\beta $ with positive weights. Concavity then forces both nonnegative defects $d_\eta $ and $d_\beta $ to vanish, and monotonicity gives $d_\delta =0$ for all $\delta \in \Lambda $ with $\delta <\alpha $. Hence a zero of the defect at a parameter other than the smallest element of $\Lambda $ forces every defect to vanish. This proves the alternatives for each $\Lambda _{\mathrm{reg}}(a)$, and intersecting them gives the alternatives for $\bigcap _{a\in K}\Lambda _{\mathrm{reg}}(a)$. $\square$