Lemma 11.34.16FADB0C708F54

Proof. The function $W$ is convex: both one-sided derivatives at zero equal one, and the derivative of $2\chi $ increases from one to two on $[0,\infty )$. Since $\omega _1=\mathrm{dd}^{\mathrm{c}}\chi (x)$ on the affine chart, $W-2\chi $ is $2\omega _1$-plurisubharmonic there. Since $W$ has slope one for $x\leq 0$ and agrees with $2\chi $ for $x\geq 0$, the function $W-2\chi $ extends to a global $2\omega _1$-plurisubharmonic function, with a logarithmic pole along $D$ and equal to zero where $|z|\geq 1$. The asserted bounds follow from $\psi =x-\chi $: the difference $\varphi -\psi $ equals $2\log 2-\chi $ for $x\leq 0$ and $\chi -x$ for $x\geq 0$.

For $C>0$ write $v_ C=P_\theta (\min \{ \psi +C,0\} )$. Rotations of the first factor and the transitive unitary action on the second preserve the whole competitor class, so the envelope is radial and independent of the second factor. The function $V_ C=2\chi +v_ C$ is then convex in $x$ and satisfies $V_ C\leq \chi +x+C$. This upper bound as $x\to -\infty $ forces the slope of $V_ C$ at $-\infty $ to be at least one, and by convexity all its slopes are at least one. For $x<0$,

For $x\geq 0$, the zero obstacle gives $V_ C\leq 2\chi =W$. Consequently $v_ C\leq \varphi $. If $C\geq 2\log 2$, the bounds already proved make $\varphi $ a competitor, so $v_ C=\varphi $. Letting $C\to \infty $ proves $P_\theta [\psi ]=\varphi $. Since $\varphi \sim \psi $ and the ordinary envelope depends only on the singularity type, also $P_\theta [\varphi ]=\varphi $.

Finally

Since $\psi \leq \varphi $, the function $\psi $ is a competitor for $P_{(3/4)\theta }(\varphi )$, which is therefore not identically $-\infty $. $\square$