PomeLemma 16.28.18.50B3044AA630 For $0<\alpha <\beta $, one has on $\mathbb C^ n$
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16.28.19
\begin{equation} (\beta -\alpha )V_ K^*\leq F_\beta -F_\alpha \leq \frac{\beta -\alpha }{\alpha }(F_\alpha -m). \end{equation}
16.28.16 Growth slopes and supporting probabilitiesFBAA435459A9
By concavity, the rate at which $F_\delta $ grows when the growth allowance $\delta $ is increased is at least its rate of growth for large $\delta $, and the latter is the classical extremal function of $K$, without weight. For continuous weights, supporting probabilities express the initial rate of growth through tests by functions of the fixed class $\mathcal L_1$.
When the weight varies, we write $V_{\delta ,q}$ and $F_{\delta ,q}=V_{\delta ,q}^*$ for the raw and regularized envelopes defined as above, with a bounded obstacle $q$ on the fixed compact set $K$ in place of $\psi $. Thus $F_\delta =F_{\delta ,\psi }$. Set
\[ V_ K=V_{1,0},\qquad V_ K^*=F_{1,0},\qquad m=\inf _ K\psi ,\qquad M=\sup _ K\psi . \]
Thus $V_ K^*$ is the extremal function of $K$. Since the constant $0$ is a competitor for $V_ K$, and since competitors can be scaled and translated by constants, we have $V_ K^*\geq 0$ and
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16.28.17
\begin{equation} \delta V_ K^*+m\leq F_\delta \leq \delta V_ K^*+M. \end{equation}
Proof. If $u$ is a competitor for $V_\alpha =V_{\alpha ,\psi }$ and $v$ is a competitor for $V_ K=V_{1,0}$, then $u+(\beta -\alpha )v$ is a competitor for $V_\beta $. Taking the suprema over $u$ and $v$ independently gives
\[ V_\beta \geq V_\alpha +(\beta -\alpha )V_ K. \]
By proposition 16.28.10, the raw envelopes agree almost everywhere with their regularizations, so this inequality holds almost everywhere for the regularizations, and hence everywhere by lemma 5.24.3. This is the first inequality. The same argument applied to (16.28.15) gives the second. $\square$
Proposition 16.28.20.22685BEFF668 For every $a\in \mathbb C^ n$ and $\tau >0$, the right derivative
\[ D_\tau (a)=\lim _{h\downarrow 0}\frac{F_{\tau +h}(a)-F_\tau (a)}{h} \]
exists and is finite. The difference quotients decrease as $h>0$ increases, and
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16.28.21
\begin{equation} V_ K^*(a)\leq \frac{F_{\tau +h}(a)-F_\tau (a)}h \leq D_\tau (a)\leq \frac{F_\tau (a)-m}{\tau }. \end{equation}
As $h\to \infty $, they converge to $V_ K^*(a)$, with the error bound
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16.28.22
\begin{equation} 0\leq \frac{F_{\tau +h}(a)-F_\tau (a)}h-V_ K^*(a) \leq \frac{M-m}h. \end{equation}
Moreover, $D_\tau (a)=V_ K^*(a)$ if and only if
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16.28.23
\begin{equation} F_\delta (a)=F_\tau (a)+(\delta -\tau )V_ K^*(a) \qquad (\delta \geq \tau ). \end{equation}
Proof. For $0<h_1<h_2$, concavity of $F_\delta (a)$ gives
\[ F_{\tau +h_1}(a)-F_\tau (a) \geq \frac{h_1}{h_2}\bigl (F_{\tau +h_2}(a)-F_\tau (a)\bigr ). \]
Thus the difference quotients decrease with $h$. Lemma 16.28.18 bounds them between $V_ K^*(a)$ and $(F_\tau (a)-m)/\tau $, which proves the existence and finiteness of $D_\tau (a)$ and (16.28.21). Combining the lower bound for the increments with (16.28.17) gives (16.28.22), with a bound independent of $a$. If $D_\tau (a)=V_ K^*(a)$, then by (16.28.21) every difference quotient equals $V_ K^*(a)$, which gives (16.28.23). Conversely, that formula gives $D_\tau (a)=V_ K^*(a)$. $\square$
Corollary 16.28.24.0D01C895AF55 Let $a\in K$ and $\tau \in \Lambda _{\mathrm{reg}}(a)$. Regular contact at $a$ holds for every $\delta >\tau $ if and only if $D_\tau (a)=0$. If $D_\tau (a)>0$, regular contact at $a$ fails for every $\delta >\tau $, and $\tau $ is the smallest element of $\Lambda $.
Proof. At $\tau $ one has $F_\tau (a)=\psi (a)$, and raw contact holds for all $\delta \geq \tau $ by theorem 16.28.14. If $D_\tau (a)=0$, then by (16.28.21) every difference quotient at $\tau $ vanishes, so $F_\delta (a)=\psi (a)$ for all $\delta >\tau $. If $D_\tau (a)>0$, then $F_\delta (a)>\psi (a)$ for all $\delta $ slightly larger than $\tau $, and by monotonicity for all $\delta >\tau $. In this case $\Lambda _{\mathrm{reg}}(a)$ contains $\tau $ but no larger parameter, so by theorem 16.28.14 it equals $\{ \lambda \} $, where $\lambda =\tau $ is the smallest element of $\Lambda $. $\square$
Assume now that the weights are continuous. If the obstacle is imposed only quasi-everywhere on $K$, the regularized envelope itself becomes a competitor, so the supremum needs no regularization; a constraint imposed at every point would miss the values created by regularization. Indeed, for $q\in C(K,\mathbb R)$,
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16.28.25
\begin{equation} F_{\delta ,q}(z)=\sup \{ u(z):u\in \mathcal L_\delta , \ u\leq q\text{ q.e. on }K\} . \end{equation}
The envelope $F_{\delta ,q}$ itself is a competitor on the right, since it agrees with the raw envelope off a pluripolar set by proposition 16.28.10. Conversely, the argument of corollary 16.28.13 shows that every competitor on the right lies below $F_{\delta ,q}$ everywhere.
For $a\in \mathbb C^ n$, let $\mathcal J_ a(K)$ be the set of supporting pairs $(\mu ,c)$, where $\mu $ is a Borel probability measure on $K$ and $c\geq 0$, such that
whenever $u\in \mathcal L_1$, $f\in C(K,\mathbb R)$ and $u\leq f$ quasi-everywhere on $K$. We call $\mu $ a supporting probability at $a$.
Theorem 16.28.27.865AACD2F96A For every $a\in \mathbb C^ n$, $q\in C(K,\mathbb R)$ and $\delta >0$,
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16.28.28
\begin{equation} F_{\delta ,q}(a)=\min _{(\mu ,c)\in \mathcal J_ a(K)} \left(\int _ K q\, \mathrm{d}\mu +\delta c\right). \end{equation}
In particular, $\mathcal J_ a(K)$ is nonempty and
\[ \min _{(\mu ,c)\in \mathcal J_ a(K)}c=V_ K^*(a). \]
Proof. On the real Banach space $C(K)$, put $T(f)=F_{1,f}(a)$. By (16.28.25), $T$ is concave, since convex combinations of competitors are competitors. Comparison of obstacles and translation by constants show that $T$ is nondecreasing and satisfies
\[ T(f+t)=T(f)+t,\qquad |T(f)-T(g)|\leq \| f-g\| _\infty . \]
Fix $f_0\in C(K)$. The directional derivative of the convex Lipschitz function $-T$ at $f_0$ is finite and sublinear. By the Hahn–Banach theorem, there is a continuous linear functional dominated by it, and since difference quotients of a convex function are monotone, this functional supports $-T$ at $f_0$. Changing the sign, we obtain a continuous linear functional $\ell $ with
\[ T(f)\leq T(f_0)+\ell (f-f_0)\qquad (f\in C(K)). \]
Testing $f=f_0+th$ with $h\geq 0$ and $t>0$, and using monotonicity, gives $\ell (h)\geq 0$. Testing with $f=f_0\pm t$ gives $\ell (1)=1$. By the Riesz representation theorem, $\ell (f)=\int _ K f\, \mathrm{d}\mu $ for a Borel probability measure $\mu $ on $K$. Set $c=T(f_0)-\int _ K f_0\, \mathrm{d}\mu $. The supporting inequality at $f=0$ gives $c\geq T(0)=V_ K^*(a)\geq 0$. Together with (16.28.25), it shows that $(\mu ,c)\in \mathcal J_ a(K)$ and
\[ T(f_0)=\int _ K f_0\, \mathrm{d}\mu +c. \]
Conversely, applying (16.28.26) to $u=F_{1,f_0}$, which lies below $f_0$ quasi-everywhere on $K$, shows that $T(f_0)\leq \int _ Kf_0\, \mathrm{d}\mu +c$ for every pair in $\mathcal J_ a(K)$. This proves (16.28.28) for $\delta =1$, with the minimum attained. The identity $F_{\delta ,q}=\delta F_{1,q/\delta }$ gives the general case, and the zero obstacle gives the last assertion. $\square$
Theorem 16.28.29.4044625470CC If $\psi \in C(K,\mathbb R)$, then for every $a\in \mathbb C^ n$ and $\tau >0$,
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16.28.30
\begin{equation} D_\tau (a)=\min \left\{ c:(\mu ,c)\in \mathcal J_ a(K), \ \int _ K\psi \, \mathrm{d}\mu +\tau c=F_\tau (a)\right\} . \end{equation}
Consequently, $D_\tau (a)=V_ K^*(a)$ if and only if there is a Borel probability measure $\mu $ on $K$ such that
\[ (\mu ,V_ K^*(a))\in \mathcal J_ a(K),\qquad \int _ K\psi \, \mathrm{d}\mu =F_\tau (a)-\tau V_ K^*(a). \]
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
\[ 0\leq c\leq \frac{F_\tau (a)-m}{\tau }. \]
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
\[ F_\tau (a)\leq \int _ K\psi \, \mathrm{d}\mu _ h+\tau c_ h =F_{\tau +h}(a)-hc_ h. \]
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. $\square$
At a point $a$ of regular contact, $F_\tau (a)=\psi (a)$, so the criterion involves only the original weight and tests by functions in $\mathcal L_1$. If the defining tests instead required $u\leq f$ at every point of $K$, the pair $(\delta _ a,0)$, where $\delta _ a$ is the unit point mass at $a\in K$, would always be allowed; the resulting infimum would be at most $\psi (a)$ and would not detect a defect created by regularization.
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