A half-disk small-energy estimate with a nonlinear normal derivative bound

Reflect the density evenly, $\widehat{e}(s,t)=e(s,\mathrm{\mid }t\mathrm{\mid })$, on the full disk. It is continuous and piecewise $C^2$. Distributional integration by parts in the upper and lower half-planes gives $\mathrm{\Delta }\widehat{e}=(\mathrm{\Delta }e{)}^{\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n}}+2e_t(s,0+){\delta }_{\{t=0\}},$ where ${\delta }_{\{t=0\}}$ denotes integration along the flat line. Indeed the jump of the first $t$ derivative is $e_t(s,0+)-(-e_t(s,0+))=2e_t(s,0+)$. Thus the atom is bounded below by $-2(ce(s,0{)}^{3\mathrm{/}2}+de(s,0)){\delta }_{\{t=0\}}$. Its sign is not discarded. The reflected integral is ${\int }_{B_r}\widehat{e}=2E$.

We shall use [F1] also for continuous distributionally subharmonic functions. Here is the needed extension. On any relatively compact disk convolve the continuous function with a nonnegative smooth unit-integral bump supported in a radius-$\epsilon$ disk. Such bumps are obtained by normalizing $\exp (-1\mathrm{/}(1-\mathrm{\mid }x{\mathrm{\mid }}^2))$ on the unit disk and scaling. Positivity of the distributional Laplacian gives nonnegative ordinary Laplacian after convolution, since its value is the distribution evaluated on a translated nonnegative bump. Apply [F1] on disks whose closed $\epsilon$ neighborhood remains in the domain. Uniform continuity on compact subsets makes the convolutions converge uniformly to the original continuous function, so their mean-value inequalities pass to the limit. This proves the stated extension without assuming that a reflected function is $C^2$.

If $e(0,0)=0$, the conclusion follows from $E\ge 0$. Otherwise maximize $(r-\mathrm{\mid }z\mathrm{\mid }{)}^2\widehat{e}(z)$ on the full closed disk, and let $z_0=(s_0,t_0)$ be a point of positive maximum. Put $M=\widehat{e}(z_0)>0$ and $\mathrm{\ell }=(r-\mathrm{\mid }{z}_0\mathrm{\mid })\mathrm{/}2$. Maximality gives $\widehat{e}\le 4M$ on ${\bar{B}}_{\mathrm{\ell }}(z_0)$ and $r^{2}e(0,0)\le 4{\mathrm{\ell }}^{2}M$. On this smaller disk the bulk Laplacian is at least $-K$, with $K=16A{M}^2+4BM$, and its line atom is at least $-2L{\delta }_{\{t=0\}}$, with $L=8c{M}^{3\mathrm{/}2}+4dM$. Consequently $F(z)=\widehat{e}(z)+K\mathrm{\mid }z-z_0{\mathrm{\mid }}^2\mathrm{/}4+L\mathrm{\mid }t\mathrm{\mid }$ is distributionally subharmonic there, since $\mathrm{\Delta }\mathrm{\mid }t\mathrm{\mid }=2{\delta }_{\{t=0\}}$.

Apply step 1.2 to $F$ on a radius $0<\rho \le \mathrm{\ell }$ about $z_0$; the endpoint radius follows by continuity if needed. The average of $\mathrm{\mid }z-z_0{\mathrm{\mid }}^2$ is ${\rho }^2\mathrm{/}2$, and $\mid \mathrm{\mid }t\mathrm{\mid }-\mathrm{\mid }{t}_0\mathrm{\mid }\mid \le \rho$ there. Hence $M\le \frac{1}{\pi {\rho }^2}{\int }_{B_{\rho }(z_0)}\widehat{e}+M(2AM+B\mathrm{/}2){\rho }^2+M(8c\sqrt{M}+4d)\rho \mathrm{.}$ Choose $\rho =\min \{\mathrm{\ell },\frac{1}{4\sqrt{AM}},\frac{1}{64c\sqrt{M}}\},$ interpreting a term with zero denominator as $+\mathrm{\infty }$. The two nonlinear error terms are at most $M\mathrm{/}8$ each. The remaining terms satisfy $MB{\rho }^2\mathrm{/}2\le MB{r}^2\mathrm{/}8\le M\mathrm{/}32$ and $4Md\rho \le 2Mdr\le M\mathrm{/}8$, because $\mathrm{\ell }\le r\mathrm{/}2$. Their sum is $13M\mathrm{/}32<M\mathrm{/}2$. Thus ${\int }_{B_{\rho }(z_0)}\widehat{e}\ge \frac{\pi }{2}M{\rho }^2\mathrm{.}$

If $\rho <\mathrm{\ell }$, one of the nonlinear terms determines the minimum, so $M{\rho }^2$ equals $1\mathrm{/}(16A)$ or $1\mathrm{/}(4096c^2)$, or their common minimum when both are present. In all cases $M{\rho }^2\ge 1\mathrm{/}H$. Step 3.1 would give $2E\ge \pi \mathrm{/}(2H)$, contradicting $E<\pi \mathrm{/}(8H)$. Therefore $\rho =\mathrm{\ell }$. Step 3.1 then gives $M{\mathrm{\ell }}^2\le 4E\mathrm{/}\pi$, and the center comparison in step 2.1 proves $e(0,0)\le 16E\mathrm{/}(\pi r^2)$. If both nonlinear coefficients vanish, the minimum defining $\rho$ is already $\mathrm{\ell }$; the same argument applies. The proof never uses a normal derivative at the curved edge, where the maximizing weight vanishes.