Proposition 10.5.32DF07B2D5AA9

Proof. We treat $\nu (\cdot ,x)$. The proof for $D$ is the same, except that the additivity of Lelong numbers along $D$ is applied at a point of $D$ at which all the functions involved attain their generic Lelong numbers along $D$ (proposition 6.9.7). For $t>0$ we have $\mathrm{PSH}(X,t\theta +\varepsilon \omega )=t\mathrm{PSH}(X,\theta +(\varepsilon /t)\omega )$, hence $V_{t\theta +\varepsilon \omega }=tV_{\theta +(\varepsilon /t)\omega }$, and taking suprema over $\varepsilon $ gives $\nu (t\alpha ,x)=t\nu (\alpha ,x)$. If $\alpha _1,\alpha _2\in \mathcal E$ have representatives $\theta _1,\theta _2$, the sum of the minimal currents with allowance $\varepsilon $ in the two classes is a current $\theta _1+\theta _2+\mathrm{dd}^{\mathrm{c}}(V_{\theta _1+\varepsilon \omega }+V_{\theta _2+\varepsilon \omega })\geq -2\varepsilon \omega $ in $\alpha _1+\alpha _2$. By minimality and additivity of Lelong numbers,

and taking the supremum over $\varepsilon $ gives subadditivity. Together with homogeneity this proves convexity.

Let $\alpha _ j\to \alpha $ in $\mathcal E$, with representatives $\theta _ j$, and fix $\varepsilon >0$. For large $j$, by (10.2.18), the class $\alpha -\alpha _ j$ has a smooth representative $\eta _ j$ with $-\varepsilon \omega \leq \eta _ j\leq \varepsilon \omega $. The current $\theta _ j+\mathrm{dd}^{\mathrm{c}}V_{\theta _ j+\varepsilon \omega }+\eta _ j\in \alpha $ is at least $-2\varepsilon \omega $, and its potential relative to $\theta $ differs from $V_{\theta _ j+\varepsilon \omega }$ by a smooth function. By minimality,

for large $j$, so $\nu (V_{\theta +2\varepsilon \omega },x)\leq \varliminf _ j\nu (\alpha _ j,x)$. Taking the supremum over $\varepsilon $ proves lower semicontinuity.

Finally, a convex function $f$ with finite values on an open convex subset $U$ of a finite-dimensional real vector space is continuous. Indeed, let $Q\subset U$ be a small closed cube; by convexity the function is bounded above on $Q$ by the maximum $M$ of its values at the vertices. If $c$ is the center of $Q$ and $c\pm v\in Q$, then $f(c+v)\geq 2f(c)-f(c-v)\geq 2f(c)-M$, so $f$ is also bounded below on $Q$. If $y,z$ lie in the concentric cube of half the size, extend the segment from $z$ to $y$ beyond $y$ to the point $w$ at distance $s$ from $y$, where $s$ is half the side length of the smaller cube, so that $w\in Q$. The slope inequality for convex functions gives $f(y)-f(z)\leq |y-z|(f(w)-f(y))/|w-y|\leq |y-z|(M-\inf _ Qf)/s$, and by symmetry $f$ is Lipschitz on the smaller cube. Applied to the open convex cone $\mathcal B$, this proves continuity there. $\square$