Lemma 18.42.1058D1933573E2

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

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

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

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

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

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 $R=0$,

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 $F_ R=f_ R$. $\square$