Pome

18.42.9 Curvature quadrature domains3CBD30511EFA

The disk comparison above requires the error of the metric weight to be small throughout the disk carrying the norm. A larger range of orders is obtained by absorbing the harmonic part of this error into the holomorphic sections. The transition then takes place on the boundary of the region where the remaining obstacle potential detaches from the weight with first-order contact, and a quadrature identity for the curvature determines that boundary.

Keep $C,x,(L,h),r,z,e$ and $\nu =a(z)\, \mathrm{d}A$ from section 18.42, and assume that the curvature of $(L,h)$ is semipositive on $C$ and strictly positive on $C\setminus \{ x\} $. We now allow integers $q=q_ m\to \infty $ with $q=o(m)$. Write $V_{m,q}=H^0(C,L^ m\otimes \mathfrak m_ x^ q)$, and let $B_{m,q}$ be its density for the norm of section 18.42. Put

\[ \rho =\phi _{z\bar z},\qquad t=q/m,\qquad R=(t/r)^{1/(2r)}. \]

Write $\mathbb D$ for the unit disk.

Lemma 18.42.10.58D1933573E2 Let $b>1$, and let $Q(R,\zeta )$ be a real function, jointly smooth for small real $R$ and $|\zeta |<b$, with $Q(0,\zeta )=|\zeta |^{2r}$. Write $Q_ R=Q(R,\cdot )$, and suppose that every $Q_ R$ is subharmonic and that $Q_{R,\zeta \bar\zeta }>0$ on a fixed neighborhood of the unit circle. For small $R$, let $U_ R$ be the largest subharmonic function on $\{ |\zeta |<b\} $ such that $U_ R\leq Q_ R$ and $U_ R\leq 2r\log |\zeta |+O(1)$ near zero. Then the noncontact set $\Omega _ R=\{ U_ R<Q_ R\} $ is connected, and its boundary is a radial graph near the unit circle. Let $F_ R:\mathbb D\to \Omega _ R$ be the conformal map with $F_ R(0)=0$ and $F_ R'(0)>0$. For every fixed $k$ and all sufficiently small $R$, depending on $k$, the map $F_ R$ and the harmonic function $(U_ R-2r\log |\zeta |)\circ F_ R$ depend smoothly on $R$ in the Hölder spaces $C^{k,\alpha }(\overline{\mathbb D})$, and they differ from their values at $R=0$ by $O(R)$.

The arguments below use only finitely many of these norms, so the dependence of the range of $R$ on $k$ is harmless.

Proof. Fix $k\geq 3$ and $0<\alpha <1$. Let $A^{k,\alpha }$ be the real Banach space of functions that are holomorphic on $\mathbb D$, of class $C^{k,\alpha }$ on its closure, and real at zero. For $h\in A^{k,\alpha }$ near zero set $f(u)=u(1+h(u))$. For small $h$, the map $f$ is univalent, with $f(0)=0$ and $f'(0)>0$. On $|u|=1$ define

\[ K(R,h)=f(u)(Q_ R)_\zeta (f(u))-r, \qquad T(R,h)=(P_-K,\operatorname {Re}[K]_0), \]

where $P_-$ is the projection onto the strictly negative Fourier modes and $[\cdot ]_0$ is the constant Fourier coefficient.

The Cauchy projections are bounded on these Hölder spaces. This follows from the Hilbert kernel on the circle: subtract the value at the evaluation point, split the integral at twice the length of the increment, and use the Hölder bound for differences on the inner part and the bound for the differentiated kernel on the outer part. This gives the $C^{0,\alpha }$ bound, and commuting angular derivatives with the projections gives the $C^{k,\alpha }$ bound. The superposition operator $h\mapsto (Q_ R)_\zeta \circ f$ involves no derivatives of $h$ and is smooth. Thus $T$ is a smooth map into the product of the space of negative-mode functions with $\mathbb R$. At $(R,h)=(0,0)$ we have $T=0$ and

\[ D_ hK(0,0)v=r^2(v+\bar v),\qquad D_ hT(0,0)v=(r^2\overline{v-v(0)},\, 2r^2v(0)). \]

This is a bounded real-linear isomorphism, so the implicit function theorem in Banach spaces gives a unique normalized branch $f_ R$ of solutions of $T(R,h)=0$, depending smoothly on $R$ in the fixed space, with $\| f_ R-\mathrm{id}\| =O(R)$. The branches obtained for larger $k$ agree with this one by uniqueness in the smaller space. For each fixed order, this gives the spatial estimates and the expansions in $R$ needed below, on a parameter interval that depends on the order.

The equation $T=0$ prescribes only the real part of $[K]_0$; we show that the imaginary part vanishes as well. Since $P_-K=0$, the function $K$ extends holomorphically to $\mathbb D$, and $K(0)=[K]_0$ is purely imaginary. On the other hand, Stokes' formula gives

\[ \frac1{2\pi \mathrm{i}}\int _{|u|=1}(Q_ R)_\zeta (f_ R(u))f_ R'(u)\, \mathrm{d}u =\frac1\pi \int _{f_ R(\mathbb D)}(Q_ R)_{\zeta \bar\zeta }\, \mathrm{d}A\in \mathbb R. \]

On the circle the integrand equals $(r+K)f_ R'/f_ R$, whose residue at zero is $r+K(0)$. Hence $K(0)$ is real as well, and therefore zero. Consequently

\[ S_ R(\zeta )=\frac{r+K(f_ R^{-1}(\zeta ))}{\zeta } =r/\zeta +H_ R(\zeta ) \]

with $H_ R$ holomorphic on $\Omega _ R=f_ R(\mathbb D)$, and $S_ R=(Q_ R)_\zeta $ on the boundary of $\Omega _ R$. Let

\[ U_ R=2r\log |\zeta |+2\operatorname {Re}\int _0^\zeta H_ R(w)\, \mathrm{d}w+C_ R. \]

Its complex gradient agrees with that of $Q_ R$ on the boundary, so the difference $Q_ R-U_ R$ is constant there; choose the constant $C_ R$ to make it zero. Then the difference and its gradient vanish on the boundary. Since the tangential second derivative of the difference also vanishes there, its inward normal second derivative equals $\Delta Q_ R>0$. Hence $Q_ R-U_ R>0$ in an inner collar of uniform width. At $R=0$,

\[ U_0=1+2r\log |\zeta |,\qquad |\zeta |^{2r}-1-2r\log |\zeta |>0\quad (0<|\zeta |<1). \]

By uniform convergence, this gap keeps a positive minimum on the remaining compact subsets of the interior for small $R$, and the logarithmic term controls a neighborhood of zero.

Set $U_ R=Q_ R$ outside $\Omega _ R$. Since values and gradients match across the boundary, the Laplacian of the patched function carries no measure on the boundary. Its curvature is $r\delta _0$ inside $\Omega _ R$ and $\mathrm{dd}^{\mathrm{c}}Q_ R\geq 0$ outside, so it is an admissible minorant. If $v$ is any competitor, then $v-U_ R$ is subharmonic on $\Omega _ R\setminus \{ 0\} $, bounded above near zero and nonpositive on the boundary. Removability of the singularity at zero and the maximum principle give $v\leq U_ R$ in $\Omega _ R$, and outside $\Omega _ R$ this follows from $v\leq Q_ R$. This identifies the envelope and shows that there are no further noncontact components. The radial graph property and the estimates for the harmonic part follow from the normalized map and its dependence on the parameter. We henceforth write $F_ R=f_ R$. $\square$

We apply the lemma to the given weight. By Taylor's formula with integral remainder, for small real $R$,

\[ Q_ R(\zeta )=R^{-2r}\phi (R\zeta ) =|\zeta |^{2r}+R E(R,\zeta ) \]

with a jointly smooth function $E$. Choose the holomorphic primitive with $G_ R(0)$ real, and write the envelope in $\Omega _ R$ as

\[ U_ R=2r\log |\zeta |+2\operatorname {Re} G_ R(\zeta ),\qquad G_ R=1/2+O(R). \]

Set $D_ t=R\Omega _ R$. In the original coordinate $z$, the obstacle potential is $\phi _ t(z)=R^{2r}U_ R(z/R)$ inside $D_ t$ and $\phi $ outside. Replacing $\phi $ by $\phi _ t$ in this chart and keeping $h$ elsewhere defines a global singular metric on $L$. Its local potential is $C^1$ across the boundary, bounded above by $\phi $, and has the pole $2t\log |z|+O(1)$. On the boundary, $\phi _ z=t/z+g_ t'$ for a holomorphic function $g_ t$. Green's formula therefore gives the quadrature identity

8F00DDA42676 18.42.11
\begin{equation} \int _{D_ t}\rho (z)f(z)\, \mathrm dA=\pi t f(0) \end{equation}

for every function $f$ holomorphic near $\overline{D_ t}$. The curvature of the singular metric is $t\delta _ x$ inside $D_ t$ and the original curvature outside. Finally, we write the boundary as $\Gamma _ t=\{ R c_ R(\theta )\mathrm{e}^{\mathrm{i}\theta }\} $ with a positive periodic function $c_ R$, where $c_ R-1=O(R)$ in each fixed norm used below.

Cite this page

Comments

Corrections, questions, and mathematical discussion are welcome. Sign in with GitHub to comment. Comments are public.