Proof.
The function $p\mapsto \sum _ i\sqrt{p_ i}$ is concave, so $P_ a$ is convex. Set
\[ b=\frac{n+1}{2n},\qquad p_{*,i}=\frac bn, \qquad \delta =\frac{n-1}{4n}. \]
Then $b+\delta <1$ and $\sum _ i\sqrt{p_{*,i}}=\sqrt{(n+1)/2}>1$. Adding a point of $\delta \Sigma $ preserves the last inequality and keeps the sum of the coordinates below one, so $p_*+\delta \Sigma $ lies in every $P_ a$.
All the standard basis vectors $e_ i$ belong to $P_ a$, and therefore
\[ \psi _{P_ a}\geq \log \frac{\max _{1\leq i\leq n}|Z_ i|^2}{\sum _{i=0}^ n|Z_ i|^2}. \]
Hence the reference is locally bounded away from $o$. The Cauchy–Schwarz inequality gives $\sum _ ip_ i\geq a/n$ on $P_ a$. When all $x_ i<0$, this yields $h_{P_ a}(x)\leq (a/n)\max _ i x_ i$, which tends to $-\infty $ as $z\to o$, and lemma 11.43.2 gives the pole at $o$. Local boundedness outside one point implies that all generic Lelong numbers along prime divisors vanish, and pulling back to the blowup at a different point introduces no pole.
To test analyticity, fix positive $\lambda _1,\ldots ,\lambda _ n$ and approach $o$ along $|z_ i|^2=\mathrm{e}^{-R\lambda _ i}$, $R\to \infty $. Writing $y_ i=\sqrt{p_ i}$, the Cauchy–Schwarz inequality gives
6654DDDC658F
11.43.12
\begin{equation} \min _{p\in P_ a}\sum _ i\lambda _ i p_ i =\frac a{\sum _ i\lambda _ i^{-1}}=:d_ a(\lambda ),\qquad y_ i=\frac{\sqrt a}{\lambda _ i\sum _ k\lambda _ k^{-1}}, \end{equation}
where the minimizer satisfies $\sum _ ip_ i\leq a\leq 1$, so that it lies in $P_ a$. Since $L(-R\lambda )\to 0$ and $\Phi _{P_ a}-\psi _{P_ a}$ is bounded, $-\Phi _{P_ a}/R\to d_ a(\lambda )$ along these curves.
Suppose that the type were that of $c\log \sum _ k|f_ k|^2$ near $o$, with $c>0$ and holomorphic germs $f_ k=\sum _\alpha c_{k,\alpha }z^\alpha $. Exponentiating the two inequalities expressing this equivalence and averaging over the arguments of the $z_ i$ gives the positive series
\[ \sum _{k,\alpha }|c_{k,\alpha }|^2\mathrm{e}^{-R\alpha \cdot \lambda }, \]
whose exponential decay rate in $R$ must be $d_ a(\lambda )/c$. That rate is the minimum of $\alpha \cdot \lambda $ over the monomials with nonzero coefficients: a minimizing term gives one bound, and convergence at a fixed $R_0$ bounds the series above by a constant times $\mathrm{e}^{-(R-R_0)\min \alpha \cdot \lambda }$. Near any $\lambda $ with positive coordinates, only finitely many monomials can attain the minimum, because these coordinates have a common positive lower bound and any one nonzero coefficient bounds the degree of the minimizing monomials. The rate is therefore locally piecewise linear in $\lambda $. But
\[ d_ a(s,1,\ldots ,1)=\frac{as}{1+(n-1)s},\qquad \frac{\mathrm{d}^2}{\mathrm{d}s^2}d_ a(s,1,\ldots ,1) =-\frac{2a(n-1)}{(1+(n-1)s)^3}<0. \]
This contradiction proves nonanalyticity. The assertion persists on a blowup at a point different from $o$, which does not change a neighborhood of the pole.
The removed corner $C_ a=\{ p\in \mathbb R^ n:p_ i\geq 0,\ \sum _ i\sqrt{p_ i}<\sqrt a\} =aC_1$ lies in $\Sigma $. The substitution $p_ i=y_ i^2$ gives
\[ \operatorname {Vol}_ n(C_1) =2^ n\int _{y_ i\geq 0,\, \sum y_ i<1}\prod _ i y_ i\, \mathrm{d}y =\frac{2^ n}{(2n)!}. \]
The last equality follows by iterated integration of monomials over a simplex. Removing the corner from $\Sigma $, whose volume is $1/n!$, and applying (11.43.4) gives (11.43.11). Since $n\geq 2$, one has $\kappa _ n<1$.
$\square$
Pome
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