17.17.9 A logarithmic pole on the projective line58086AECC383

Proof. Fix $0<\lambda <1$. For $s=\log |z|^2$, let $q_\lambda (s)=\Gamma _\lambda (z)+\log (1+\mathrm{e}^ s)$. In the first region of (17.17.10) this is the affine function $\lambda s+H(\lambda )$, which is tangent to $\log (1+\mathrm{e}^ s)$ at $s_\lambda =\log (\lambda /(1-\lambda ))$, and in the second region it is $\log (1+\mathrm{e}^ s)$. Its derivative is continuous at the junction and nondecreasing, so $q_\lambda $ is convex. By corollary 5.41.2, $\Gamma _\lambda $ is $\theta $-plurisubharmonic away from zero, and it extends across its logarithmic pole by lemma 5.24.7. It vanishes near infinity and has the ordinary type of $\lambda g$.

To prove $\Gamma _\lambda =P_\theta [\lambda g]$, let $v\leq 0$ be a competitor with $v\preceq \lambda g$. On the punctured disk $0<|z|^2<\lambda /(1-\lambda )$, the function

is subharmonic, since $v+\log (1+|z|^2)$ is subharmonic and $\log |z|^2$ is harmonic there. Since $v\preceq \lambda g$, this difference is bounded above near zero, so it extends subharmonically across zero by lemma 5.24.7. Its upper limits at the boundary circle are at most zero, since $v\leq 0$ and $\Gamma _\lambda =0$ there. The maximum principle gives $v\leq \Gamma _\lambda $ on the disk (at zero both have the value $-\infty $), and outside the disk the same inequality follows from $\Gamma _\lambda =0$. Thus every competitor lies below $\Gamma _\lambda $, which is itself a competitor, so $\Gamma _\lambda =P_\theta [\lambda g]$. Since the envelope depends only on the ordinary type, also $P_\theta [\Gamma _\lambda ]=\Gamma _\lambda $. The zero potential is an envelope fixed point as well.

For a fixed $0<|z|<\infty $, put $p=|z|^2/(1+|z|^2)$. The function $\lambda \mapsto \Gamma _\lambda (z)$ is zero for $\lambda \leq p$. For $p<\lambda <1$ its derivative is

It vanishes at $\lambda =p$ and decreases in $\lambda $, which proves that $\lambda \mapsto \Gamma _\lambda (z)$ is concave and decreasing. At $z=0$, the function is zero for $\lambda \leq 0$ and $-\infty $ for $\lambda >0$; at $z=\infty $, it is zero for $\lambda \leq 1$. These functions are concave and decreasing as well. Finally, $\sup _{\lambda <1}\Gamma _\lambda =0$, and the limit as $\lambda \uparrow 1$ is $g$, so the family is an envelope test curve with cutoff one.

For finite nonzero $z$ and $t>0$, the supremum defining $U_ t(z)$ is attained at a parameter $\lambda >p$, namely at

and substituting this value into the first line of (17.17.10) gives (17.17.12). At $z=0$, only the parameters $\lambda \leq 0$ give finite values, and the supremum is zero; at $z=\infty $, the supremum of $t\lambda $ over $\lambda <1$ is $t$. In the coordinate $w=1/z$, the formula reads $\log ((|w|^2+\mathrm e^ t)/(|w|^2+1))$, so the ray is smooth in both charts. $\square$