Lemma 11.44.19.0E0F3A0EC129 The functional $\mathcal F_{\lambda ,x}$ is finite, concave, decreasing and $\lambda ^{-1}$-Lipschitz for the uniform norm on $C(X)$. It satisfies
\[ \mathcal F_{\lambda ,x}(0)=w_\lambda (x),\qquad \mathcal F_{\lambda ,x}(r+b) =\mathcal F_{\lambda ,x}(r)-b/\lambda . \]
There is a probability measure $\nu _{\lambda ,x}$ such that, for every $r\in C(X)$ and every $s$ admissible for $r$ in (11.44.18),
D8630F60DEBD
11.44.20
\begin{equation} \lambda s(x)+\int _ Xr\, \mathrm{d}\nu _{\lambda ,x} \leq \lambda w_\lambda (x). \end{equation}
Proof.
By part (2) of corollary 15.13.3, $w_\lambda $ is the greatest admissible potential for $r=0$. If $m=\min _ Xr$ and $M=\max _ Xr$, then $w_\lambda -M/\lambda $ is admissible for $r$, and translating any potential admissible for $r$ by $m/\lambda $ makes it admissible for $r=0$. Hence
\[ w_\lambda (x)-M/\lambda \leq \mathcal F_{\lambda ,x}(r) \leq w_\lambda (x)-m/\lambda . \]
The mixed density inequality on $\Omega $ and the arithmetic–geometric mean inequality show that a convex combination of potentials admissible for $r_0$ and $r_1$ is admissible for the same convex combination of the forcing terms. Approximating the two suprema at $x$ proves concavity. Inclusion of the admissible classes proves monotonicity, and translation by constants proves the translation rule and the Lipschitz bound.
Now apply the Hahn–Banach construction in the proof of theorem 11.34.2 to $-\lambda \mathcal F_{\lambda ,x}$ at zero. This functional is finite, continuous, convex and increasing, and it satisfies $-\lambda \mathcal F_{\lambda ,x}(r+b)=-\lambda \mathcal F_{\lambda ,x}(r)+b$ for constants $b$. The same tests with nonnegative functions and constants show that its subgradient is a probability measure $\nu _{\lambda ,x}$. The resulting supporting inequality, combined with $s(x)\leq \mathcal F_{\lambda ,x}(r)$, gives (11.44.20); it is not necessary that the supremum be attained.
$\square$
of theorem 11.44.2.
By proposition 11.44.9 and the moment bound following it, $\mathcal P_ f$ has the stated properties, and lemma 11.44.15 gives precompactness of the centered potentials. It remains to identify every cluster potential.
Step 1: limits of the forcing probabilities. Fix $x\in \Omega $. For $r\in C(X)$, choose a critical potential $\psi _ r$ for $\mathrm{e}^ r f$ with supremum zero. By the definition of $\kappa $,
\[ \mathrm{MA}_\theta (\psi _ r) =c\mathrm{e}^{-\kappa (r)+r}f\, \mathrm{d}V. \]
Since $\psi _ r\leq 0$, the potential $\psi _ r-\kappa (r)/\lambda $ is admissible in (11.44.18). Thus
AABB829EFF5F
11.44.21
\begin{equation} \int _ Xr\, \mathrm{d}\nu _{\lambda ,x} \leq \kappa (r)+\lambda \bigl (w_\lambda (x)-\psi _ r(x)\bigr ). \end{equation}
For fixed $x$ and $r$, the values $w_\lambda (x)$ are uniformly bounded by lemma 11.44.15, and $\psi _ r(x)$ is finite. Hence every weak cluster point of $\nu _{\lambda ,x}$ as $\lambda \downarrow 0$ belongs to $\mathcal P_ f$.
Step 2: a pointwise comparison. Let $s$ satisfy (11.44.3). The function $-\lambda s$ is lower semicontinuous and bounded below on $X$, with the value $+\infty $ allowed, so metric inf-convolution and truncation give continuous functions $r_ j\uparrow -\lambda s$. Since $r_ j\leq -\lambda s$, the potential $s$ is admissible for every $r_ j$, and (11.44.20) bounds the integrals of the $r_ j$ from above. Monotone convergence, after subtracting a common lower bound, shows that $s$ is integrable against $\nu _{\lambda ,x}$ and gives
E3635828F0E0
11.44.22
\begin{equation} w_\lambda (x)\geq s(x)-\int _ Xs\, \mathrm{d}\nu _{\lambda ,x}. \end{equation}
Step 3: cluster potentials belong to the class (11.44.3). Take a sequence $\lambda _ j\downarrow 0$ with $w_{\lambda _ j}\to z$ locally uniformly on $\Omega $ and in $L^1(X)$. The bounds (11.44.16) show that $z$ has minimal type, and $z$ is continuous on $\Omega $. Since $c\mathrm{e}^{\lambda _ j w_{\lambda _ j}}f\to cf$ in $L^ p$, by dominated convergence and the common bound $w_\lambda \leq K$, continuity of local bounded products gives $\mathrm{MA}_\theta (z)=cf\, \mathrm{d}V$ on $\Omega $, and then on $X$ because neither side charges $X\setminus \Omega $.
For each $\nu \in \mathcal P_ f$, the bound $|w_\lambda |\leq |\psi |+K$ has a $\nu $-integrable right-hand side, and $\nu $ charges only $\Omega $. Dominated convergence in $\int _ Xw_\lambda \, \mathrm{d}\nu \leq 0$ therefore gives $\int _ Xz\, \mathrm{d}\nu \leq 0$. Hence $z$ belongs to the class (11.44.3).
Step 4: $z$ is the greatest element of the class. For any $s$ in the class and fixed $x\in \Omega $, pass to a further subsequence with $\nu _{\lambda _ j,x}\rightharpoonup \nu $, so that $\nu \in \mathcal P_ f$ by Step 1. The potential $s$ is upper semicontinuous and bounded above on $X$, so
\[ \limsup _ j\int _ Xs\, \mathrm{d}\nu _{\lambda _ j,x} \leq \int _ Xs\, \mathrm{d}\nu \leq 0. \]
For an unbounded $s$, this follows from the portmanteau theorem applied to $\max (s,-M)$, followed by $M\to \infty $; the final integral is finite by (11.44.11). Passing to the limit in (11.44.22) gives $z(x)\geq s(x)$. This holds for every $x\in \Omega $, and hence on $X$ by lemma 5.24.3. Thus every cluster potential is the greatest element $w$ of the class, and precompactness gives convergence of the whole family.
Step 5: the normalizations. The maximum of $\int _ Xw\, \mathrm{d}\nu $ over $\mathcal P_ f$ is attained, because under (11.44.10) it is the maximum of the weakly continuous functional $\eta \mapsto A^{-1}\int _ Xaw\eta \, \mathrm{d}V$ on $\mathcal H_ a$; note that $aw\in L^\ell $. If the maximum were negative, a small positive constant translate of $w$ would still satisfy (11.44.3), contrary to greatestness. Thus it is zero.
Finally, $k_\lambda \to \sup _ Xw$. Indeed, from any subsequence choose a further one with $k_\lambda \to k$. Then $v_\lambda =w_\lambda -k_\lambda $ converges in $L^1$ to $w-k$, and theorem 5.25.2 gives $\sup _ X(w-k)=0$. Thus $k=\sup _ Xw$ for every cluster point of $(k_\lambda )$. Subtracting the constants proves the convergence of $u_\lambda -\sup _ Xu_\lambda $. If $\theta $ is Hermitian, we may take $\rho =0$ and $\Omega =X$, so that local uniform convergence is uniform convergence.
$\square$
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