Lemma 11.43.2832F0D3BF7D6

Proof. The Legendre transform of $L$ is $L^*=\ell $ on $\Sigma $ and $+\infty $ outside it. Since $-\log (n+1)\leq \ell \leq 0$,

A finite convex function of the logarithmic moduli is plurisubharmonic: for a smooth convex function the complex Hessian is positive, and convolution gives the general case. On $T\cap \{ Z_ j\ne 0\} $, add the local Fubini–Study potential $\log \sum _ i|Z_ i/Z_ j|^2$; the resulting functions are again convex functions of the logarithmic moduli of the affine coordinates, and they are bounded above by that smooth potential because $h_ K,g_ K\leq L$. Lemma 5.24.7 therefore gives the extensions across the coordinate hyperplanes and the stated bounds. If $p\in \operatorname {int}K$, then $p\in \operatorname {int}\Sigma $, and at $x_ i=\log (p_ i/(1-\sum _ k p_ k))$ the supremum defining $L=\ell ^*$ is attained at $p\in K$, so $g_ K(x)=L(x)$. This proves $\sup \Phi _ K=0$.

For $C\geq \log (n+1)$, let $v_ C=P_{\omega _{\mathrm{FS}}}(\psi _ K+C,0)$, the envelope of the obstacle $\min (\psi _ K+C,0)$. Since $\Phi _ K$ is a competitor, $\Phi _ K\leq v_ C\leq 0$. The obstacle is upper semicontinuous, so $v_ C\leq \psi _ K+C$ everywhere. The compact torus acting by rotations of the coordinates preserves the obstacle, so $v_ C$ is invariant, and on $T$ the function $f=v_ C+L$ is a finite convex function of $x$ with $f\leq L$ and $f\leq h_ K+C$. The second inequality shows that the Legendre transform $f^*$ is $+\infty $ outside $K$, and the first gives $f^*\geq \ell $. Thus

Hence $v_ C\leq \Phi _ K$ on $T$, and since both functions are quasi-plurisubharmonic and $T$ is the complement of an analytic set, $v_ C=\Phi _ K$ on $\mathbb P^ n$. Letting $C\to \infty $ gives $P_{\omega _{\mathrm{FS}}}[\psi _ K]=\Phi _ K$. Since $\psi _ K\leq \Phi _ K\leq \psi _ K+\log (n+1)$, the envelopes of $\min (\Phi _ K+C,0)$ and $\min (\psi _ K+C,0)$ interlace as $C\to \infty $, so also $P_{\omega _{\mathrm{FS}}}[\Phi _ K]=\Phi _ K$.

For the final assertion put $\varepsilon =1-\delta $, $b=\sum _ i p_{*,i}$ and $\chi =p_*\cdot x-\varepsilon L$ on $T$. The inclusion $p_*+\delta \Sigma \subset \Sigma $ implies $p_{*,i}\geq 0$ and $b\leq \varepsilon $, and the extension of $\chi $ to $\mathbb P^ n$ satisfies

Since $h_ K\geq p_*\cdot x+\delta h_\Sigma $ and $L-h_\Sigma \leq \log (n+1)$, one has $\chi -\delta \log (n+1)\leq \Phi _ K$. Thus $\chi -\delta \log (n+1)$ is an $\varepsilon \omega _{\mathrm{FS}}$-plurisubharmonic competitor for $P_{\varepsilon \omega _{\mathrm{FS}}}(\Phi _ K)$, which proves the claim. $\square$