5.41.4 Pointwise Legendre transforms63DEBF0DF568

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 $\mathrm{PSH}(X,\theta )$. $\square$