Example 16.28.409167A00E8F32

Proof. For $t\geq 0$, each competitor is at most $-t$ on the unit circle, so the maximum principle for subharmonic functions bounds it by $-t$ on $D$. This bound persists after regularization on $D$, and the constant $-t$ is a competitor. Hence $F_{\delta ,q_ t}(0)=-t$.

For $t=-b$ with $0\leq b\leq \delta /2$, define

This function is continuous and subharmonic: its Laplacian has density $4b$ inside the disk, density zero outside, and a nonnegative contribution on the circle coming from the jump $\delta -2b\geq 0$ of the radial derivative. It belongs to $\mathcal L_\delta $ and equals $q_{-b}$ on $K$. Hence $F_{\delta ,q_{-b}}(0)\geq u_ b(0)=0$. On $D$ the raw envelope is at most the continuous function $b|z|^2$, so its regularization at zero is at most zero. This proves (16.28.41). In particular, the right and left derivatives in $t$ at $t=0$ are $-1$ and $0$, and corollary 16.28.39, applied in the direction $r=-|z|^2$, gives the final assertion. $\square$