12.10.32 A full-mass point pole without a smooth maximizing testB4100C32D89A

Proof. The numerator in the definition of $\psi $ vanishes only at $o$, and locally $\omega +\mathrm{dd}^{\mathrm{c}}\psi =\mathrm{dd}^{\mathrm{c}}\log \sum _{i=1}^ n|z_ i|^2\geq 0$. The function $\chi (t)=1-(1-t)^\alpha $ is increasing and convex on $t\leq 0$, with $0\leq \chi '\leq \alpha $. Applying the chain inequality to bounded truncations of $\psi $ and taking the decreasing limit, we obtain $u=\chi \circ \psi \in \mathrm{PSH}(X,\omega )$, with

The function $u$ is smooth off $o$, and it attains its supremum zero on $\{ z_0=0\} $. Since $u=-(1-\psi )^\alpha +1$, proposition 12.10.24 shows that all the envelopes $P_{\delta \omega }(u)$ are not identically $-\infty $.

To compute the mass, use the chart $z_0=1$ and write $s=\sum _{i=1}^ n|z_ i|^2$. A local potential $F$ for $A$, as a function of $s$, and the function $q=sF'$ are

The eigenvalues of $A$ are $q/s$ in the tangential directions and $q'$ in the radial direction. Polar integration gives the radial top-degree measure $nq^{n-1}q'\, \mathrm{d}s$, where the angular normalization follows from the Fubini–Study case $q=s/(1+s)$, whose radial density is $ns^{n-1}(1+s)^{-n-1}\, \mathrm{d}s$ and whose total mass is one. Consequently, for $0<s_1<s_2$,

Here $q(s)\to 0$ as $s\downarrow 0$ and $q(s)\to 1$ as $s\to \infty $. On each compact annulus the extended measure equals the smooth top power, and it puts no mass on $o$ or on the hyperplane at infinity, since both are pluripolar. Exhausting the affine chart by annuli therefore gives extended mass exactly one. (This computation does not use the vanishing of the Lelong numbers.)

Near $o$ write $L=1-\log s$. Differentiating (12.10.34) gives

The smooth volume form, written in terms of $s$, is comparable to $s^{n-1}\, \mathrm{d}s$, so the contribution of a neighborhood of $o$ to $\int _ Xf\, \mathrm{d}V$ is comparable to $\int ^\infty L^{-1-n(1-\alpha )}\, \mathrm{d}L$, which is finite. For every $p>1$, the corresponding integral of $f^ p$ contains an exponentially growing factor $\exp (n(p-1)L)$ and diverges. The measure is smooth off $o$ and puts no mass on $o$, so it is $f\, \mathrm{d}V$ on all of $X$. Since its total mass is one and $\omega $ is closed, theorem 12.10.23 gives $b_\omega (f)=1$, and hence $\rho _\omega (f^{1/n})=b_\omega (f)^{-1/n}=1$.

Suppose that a smooth semipositive $T\in \mathscr G_\omega $ attained this value. The positive current $A$ satisfies $\int _ XA\wedge T=1$ by Stokes' theorem for currents. On $X\setminus \{ o\} $, the matrix arithmetic–geometric mean inequality gives

The right side has total mass one by attainment, so the nonnegative difference of the two sides vanishes off $o$. Since $A$ is smooth and strictly positive there, the equality case gives $T=tA^{n-1}$ on $X\setminus \{ o\} $, also at points where $T=0$: applying the arithmetic–geometric mean inequality to the positive semidefinite coefficient matrix of $T$ conjugated by $A^{1/2}$ shows that all its eigenvalues agree, possibly at zero. The function $t=(T\wedge A)/A^ n$ is smooth and nonnegative on $X\setminus \{ o\} $.

Since $A$ is closed, the equation $\mathrm{dd}^{\mathrm{c}}T=0$ becomes $\mathrm{dd}^{\mathrm{c}}t\wedge A^{n-1}=0$, a locally strictly elliptic scalar equation on $X\setminus \{ o\} $. All eigenvalues of $A$ tend to infinity at $o$, and since $n\geq 2$, so do all eigenvalues of $A^{n-1}$, which are products of $n-1$ eigenvalues of $A$. Since the coefficient matrix of the smooth form $T$ is bounded, it follows that $t\to 0$ at $o$.

Set $t(o)=0$. The resulting continuous nonnegative function on compact $X$ is not identically zero, by the normalization of $T$, so it attains a positive maximum away from $o$. The strong maximum principle on the connected manifold $X\setminus \{ o\} $ forces it to be constant there, contrary to its zero limit at $o$. This excludes every smooth semipositive maximizing test; no assumption on the degeneracy set of $T$ was needed. $\square$

Proof. Only a neighborhood of $o$ can contribute a divergence. There the preceding estimates give $f\, \mathrm{d}V\asymp L^{-1-n(1-\alpha )}\, \mathrm{d}L$, $\log f\sim nL$ and $-u\asymp L^\alpha $. Hence the entropy integral converges exactly when $n(1-\alpha )>1$, and the weighted integral exactly when $\alpha r<n(1-\alpha )$. In the equality cases both integrals are comparable to $\int ^\infty L^{-1}\, \mathrm{d}L$ and diverge. The negative part of $f\log f$ is bounded, hence integrable for the probability volume, so the entropy criterion is unambiguous. $\square$