PomeExample 12.10.41.2A68BBAE5086 For every $n\geq 2$, there is a smooth nonnegative density $f$ on $\mathbb P^ n$, vanishing precisely on the coordinate hyperplanes, such that $b_\omega (f)=1$ and $D_{f^{1/n}}$ has a smooth semipositive maximizer strictly positive off those hyperplanes. The threshold solution is smooth, but $f^{1/n}$ is not differentiable at a general point of any coordinate hyperplane. There is no smooth strictly positive maximizing form. The normalized greatest exponential solutions converge uniformly to the normalized threshold solution.
Example 12.10.412A68BBAE5086
Proof. Define the holomorphic map $F:\mathbb P^ n\to \mathbb P^ n$ and the semipositive form $A$ by
\[ F[Z_0:\cdots :Z_ n]=[Z_0^2:\cdots :Z_ n^2],\qquad A=\tfrac 12F^*\omega . \]
The homogeneous formula
\[ u=\tfrac 12\log \! \left(\sum _{i=0}^ n|Z_ i|^4\right) -\log \! \left(\sum _{i=0}^ n|Z_ i|^2\right) \]
defines a smooth function on projective space, and $A=\omega +\mathrm{dd}^{\mathrm{c}}u$. The form $A$ is semipositive, and it is strictly positive exactly where every homogeneous coordinate is nonzero. Define
\[ f=A^ n/\omega ^ n,\qquad T=A^{n-1}. \]
Since $\omega $ and $A$ are closed and cohomologous, Stokes' theorem gives $\int _ X f\omega ^ n=\int _ X\omega \wedge T=1$, and $\mathrm{dd}^{\mathrm{c}}T=0$; thus $T\in \mathscr G_\omega $. Every $v\in \mathrm{PSH}(X,\omega )$ has extended mass at most one: each bounded truncation $\max (v,-k)$ has mass one, and restricting its Monge–Ampère measure to $\{ v>-k\} $ can only decrease that mass. Consequently an inequality $\mathrm{MA}^{\mathrm{ext}}_\omega (v)\geq c f\omega ^ n$ forces $c\leq 1$. The smooth solution $u$ gives the reverse bound, so $b_\omega (f)=1$.
The determinant formula for smooth semipositive forms gives
\[ k_ T=f^{(n-1)/n},\qquad D_{f^{1/n}}(T)=\int _{\mathbb P^ n}f\omega ^ n=1. \]
By (12.10.2) and $\rho _\omega (f^{1/n})=b_\omega (f)^{-1/n}=1$, the form $T$ is a maximizer. It is strictly positive away from the coordinate hyperplanes because $A$ is; on these hyperplanes $A$ has deficient rank, and hence so does the coefficient matrix of $T=A^{n-1}$.
For the density root, use the chart $Z_0=1$ with coordinates $z_1,\ldots ,z_ n$. Pulling back the Fubini–Study volume by $(z_1,\ldots ,z_ n)\mapsto (z_1^2,\ldots ,z_ n^2)$ and including the factor $2^{-n}$ gives
\[ f(z)=2^ n\left(\prod _{i=1}^ n|z_ i|^2\right) \frac{(1+\sum _ i|z_ i|^2)^{n+1}}{(1+\sum _ i|z_ i|^4)^{n+1}}. \]
Near a point of $\{ z_1=0\} $ at which all the other coordinates are nonzero, $f^{1/n}$ is $|z_1|^{2/n}$ times a smooth strictly positive function. Restricting to a real line transverse to that hyperplane proves nondifferentiability, also in the case $n=2$, where $|z_1|^{2/n}=|z_1|$. By symmetry of the coordinates, the same holds on every coordinate hyperplane. The absence of a smooth strictly positive maximizer follows from corollary 12.10.14, since $f$ has zeros. Finally, $f$ is bounded, so corollary 12.10.31 gives the uniform convergence of the normalized exponential solutions. $\square$
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