Proposition 17.16.309355B610E2C3

Proof. In the chart $z_0=1$, write $s=\log |z'|^2$, $L=\log (1+\mathrm e^ s)$, $P=\mathrm e^ s/(1+\mathrm e^ s)$, and $T=1+L-s\geq 1$. The local potential $\Phi =L+1-T^ a$ has derivative

Both $A$ and $A'$ are positive, so the radial complex Hessian is positive. Its bounded-above local potential extends plurisubharmonically across $o$ by lemma 5.24.7. The function extends smoothly across the hyperplane at infinity. Its bounded plurisubharmonic truncations decrease to a nontrivial canonical potential: near $o$, $|v_ a|=O((1-s)^ a)$ is integrable against radial smooth volume $O(\mathrm e^{ds})\, \mathrm ds$.

The radial mass flux is $Q(s)=A(s)^ d$. Its sphere coefficient is one, since $\mathrm d^{\mathrm c}\log |z'|^2$ integrates to one on the Hopf circle and $\mathrm{dd}^{\mathrm{c}}\log |z'|^2$ induces mass-one Fubini–Study on $\mathbb P^{d-1}$. Stokes on an annulus gives mass $Q(s_2)-Q(s_1)$. Since $Q(-\infty )=0$ and $Q(+\infty )=1$, finite-level locality and annular exhaustion give non-pluripolar mass one. The pole has no atom, and smoothness gives no mass on the hyperplane at infinity.

As $s\to -\infty $,

The squared energy therefore converges exactly when $2a-d(1-a)<0$. At equality the integrand has order $T^{-1}$, giving logarithmic divergence. All other compact regions have smooth potentials and measures. Fiber integration gives the pullback assertions. Finally, $v_ a-v_ b=T^ b-T^ a\to +\infty $ at $o$, so mutual bounded comparison fails. Both points lie in the completed metric space and are distinct, giving finite positive distance. $\square$