5.41 The minimum principle392B79314DF8

Proof. A local smooth correction making $U$ plurisubharmonic can be chosen to depend only on $x$. Hence the restriction of $U$ to $\{ x\} \times S_ I$ is subharmonic or identically $-\infty $. In the first case its distributional Laplacian is the second derivative of $t\mapsto u_ t(x)$, which is therefore nonnegative. The resulting convex representative is finite on $I$, and by the submean inequality it agrees everywhere with the subharmonic function. Finally, $U$ is locally integrable on the product, and Fubini's theorem, applied on products of coordinate balls with a fixed rectangle in $S_ I$, excludes the identically $-\infty $ alternative for almost every $x$. $\square$

Proof. The holomorphic covering map $z=\mathrm{e}^{w/2}$ pulls the function back to $q(\operatorname{Re}w)$, and the argument of lemma 5.41.1 proves necessity. Conversely, a finite convex function $q$ has nonnegative distributional second derivative. Thus $q(\operatorname{Re}w)$ is subharmonic, and subharmonicity descends through the local biholomorphisms of the covering. $\square$

Proof. The classical local Kiselman principle states that if a plurisubharmonic function on the product of a domain with a strip does not depend on the imaginary part of the strip variable, then its infimum over the real part is plurisubharmonic or identically $-\infty $; the interval of the real part may be unbounded. We use this local result from [SGPT].

Fix $x_0\in X$, choose local holomorphic coordinates centered at $x_0$, and let $\kappa $ be their Euclidean form. Fix $\eta >0$. On a sufficiently small coordinate ball the constant form $\beta =\theta (x_0)$ satisfies

Choose a quadratic function $r$ with $\mathrm{dd}^{\mathrm{c}}r=\beta +\eta \kappa $. Then $U+r\circ \pi $ is plurisubharmonic on the product of the ball with $S_ I$. Its infimum over the parameter is $q+r$, so the local principle shows that $q+r$ is plurisubharmonic or identically $-\infty $ on the ball; in the first case

In a fixed atlas, the Euclidean forms are bounded by a fixed multiple of $\omega $, so letting $\eta \downarrow 0$ proves the exact lower bound $\mathrm{dd}^{\mathrm{c}}q\geq -\theta $. Since a plurisubharmonic function cannot be identically $-\infty $ on a nonempty open set, the local alternative propagates across overlapping balls on the connected manifold $X$. Thus either $q\in \mathrm{PSH}(X,\theta )$ or $q\equiv -\infty $. No regularization is needed, because an infimum of upper semicontinuous functions is upper semicontinuous.

Now suppose that $I$ is bounded. By lemma 5.41.1, the function $t\mapsto u_ t(x)$ is finite and convex for almost every $x$, and a supporting affine function at any point of $I$ bounds it from below on the bounded interval $I$. Hence $q(x)>-\infty $ almost everywhere, which excludes the alternative $q\equiv -\infty $. $\square$

Proof. Since $g$ is decreasing and finite somewhere, it is finite on a half-line $(-\infty ,\tau _1)$. Hence the defining supremum is finite for $t>0$: it is bounded above by $ct$ and below by any one of its finite terms. Concavity and upper semicontinuity give $g(c)=\lim _{\tau \uparrow c}g(\tau )$, so including $\tau =c$ does not change the supremum. The Legendre transform $t\mapsto \sup _\tau (g(\tau )+t\tau )$ of $-g$ is infinite for $t<0$. At $t=0$ it equals $m=\sup _{\tau <c}g(\tau )$, and the bounds

show that $f(t)\to m$ as $t\downarrow 0$. The biconjugacy theorem for the closed convex function $-g$ therefore gives the inverse formula; the value at $t=0$ may be omitted without changing the infimum.

Conversely, start with $f$. At each $t>0$, choose the slope $p$ of a supporting line of $f$. The upper bound $f(s)\leq cs$ as $s\to \infty $ forces $p\leq c$. If $p<c$, the supporting line gives $g(p)=f(t)-tp$, so the first formula recovers $f(t)$. If $p=c$, then $f(s)-cs$ is constant for $s\geq t$, because this function is convex and bounded above, hence decreasing, while its supporting line at $t$ is horizontal. Consequently, for every $\tau <c$,

and letting $\tau \uparrow c$ again recovers $f(t)$. In particular, $g$ is finite somewhere. As an infimum of affine functions with slopes $-s<0$, it is upper semicontinuous, concave and decreasing. Letting $s\downarrow 0$ in the bound $f(s)-s\tau \leq s(c-\tau )$ gives $g\leq 0$, and letting $s\to \infty $ gives $g=-\infty $ for $\tau >c$. The endpoint identity now follows from the first part. The case of functions identically equal to $-\infty $ follows directly from the formulas. $\square$

Proof. Fix $\tau <c$ and let $\tau _ j\uparrow \tau $. The decreasing limit of the functions $\Gamma _{\tau _ j}$ dominates $\Gamma _\tau $, so it belongs to $\mathrm{PSH}(X,\theta )$. Choose $r\in (\tau ,c)$. Outside the pole set of $\Gamma _ r$, the concave function $s\mapsto \Gamma _ s(x)$ is finite near $\tau $ and hence continuous there. Thus the decreasing limit equals $\Gamma _\tau $ almost everywhere, and hence everywhere by lemma 5.24.3. This proves left continuity at every $\tau <c$; at $c$ it holds by definition. Since the function is decreasing, left continuity gives upper semicontinuity, also where the value is $-\infty $, and concavity survives the extension. The inverse formula is lemma 5.41.5 at each $x$. $\square$

Proof. Each term $(x,z)\mapsto \Gamma _\tau (x)+\tau \operatorname{Re}z$ of the supremum satisfies the current inequality on the product, since $\operatorname{Re}z$ is pluriharmonic. The bound $\Gamma _\tau \leq 0$ gives $U_ t\leq ct$, while any fixed $\tau _0<c$ gives the minorant $\Gamma _{\tau _0}+t\tau _0$. By the local upper-envelope theorem, as in section 5.40, the upper semicontinuous regularization $V$ of $U$ on the product is quasi-plurisubharmonic, satisfies $\pi ^*\theta +\mathrm{dd}^{\mathrm{c}}V\geq 0$, and agrees with $U$ almost everywhere. It does not depend on $\operatorname {Im}z$ and satisfies $V_ t\leq ct$. The pointwise supremum $U$ is measurable, since by the left continuity in lemma 5.41.8 the parameters may be restricted to rational numbers.

By Fubini's theorem, there is a set $Z\subset X$ of zero smooth volume, containing the pole set of $\Gamma _{\tau _0}$, such that $U_ t(x)=V_ t(x)$ for almost every $t>0$ whenever $x\notin Z$. For such $x$, both sides are finite convex functions of $t$, by lemma 5.41.5, lemma 5.41.1, so by continuity the equality holds for every $t>0$.

For each $\tau <c$, proposition 5.41.3 gives

Off $Z$ it equals $\Gamma _\tau $ by lemma 5.41.8. In particular, $\chi _\tau $ is not identically $-\infty $, and lemma 5.24.3 gives $\chi _\tau =\Gamma _\tau $ everywhere. At each $x$, the convex function $t\mapsto V_ t(x)$ can be recovered from these inverse transforms by lemma 5.41.5, also when it is identically $-\infty $. We obtain

Finally, the minorant $\Gamma _{\tau _0}+t\tau _0$ shows that no slice $U_ t$ is identically $-\infty $, so every slice belongs to $\mathrm{PSH}(X,\theta )$. $\square$