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For a compact H-hull, $v(z)=\mathrm{Im}{g}_K(z)$ is harmonic and satisfies $0<v(z)\le \mathrm{Im}z$. Local boundedness and the interior continuity of $g_K^{-1}$ show that $v$ tends to zero at every finite boundary point. The maximum principle for harmonic functions applied to $v-\mathrm{Im}z$, with an exhaustion controlling infinity through the Laurent expansion, proves the inequality without boundary smoothness.

The inner circle has centre $2/5$ and radius $2/5$. The pole $z=1/2$ of the Möbius transformation lies inside that removed disc, so the transformation is holomorphic on $D$, with derivative $3/(2z-1{)}^2\ne 0$.

For $z=x+iy$,On the outer boundary this is one. On the inner boundary, $|z{|}^2=4x/5$, and the ratio is four. More directly, inside the outer circle the numerator exceeds the denominator by $3(1-|z{|}^2)>0$, whereas outside the inner circle four times the denominator exceeds the numerator by $3(5|z{|}^2-4x)>0$. Thus the image is $1<|w|<2$. The inverse transformation is $z=(w-2)/(2w-1)$, so this is a bijective conformal map onto that annulus.

A radial harmonic function on an annulus has form $A+B\log |w|$. The two boundary values give $A=1$, $B=1/\log 2$. By conformal invariance of harmonicity,It has the required boundary values, and the maximum principle for harmonic functions proves uniqueness among solutions continuous on the closed region.

Set $v(z)=\mathrm{Im}{g}_K(z)$, a positive harmonic function on $D$. First justify its boundary behavior. For $z_n\to p\in \partial D$ with $p$ finite, part (c) makes $g_K(z_n)$ bounded. Any subsequential limit $w$ lies in $\overline{\mathbb{H}}$. If $\mathrm{Im}w>0$, continuity of the inverse inside $\mathbb{H}$ would give $p=g_K^{-1}(w)\in D$, a contradiction. ThusThis boundary degeneration under a mapping-out function is valid without a smooth or locally connected hull boundary.

The harmonic function $u(z)=v(z)-\mathrm{Im}z$ consequently has nonpositive finite-boundary values. Its value tends uniformly to zero at infinity, by the Laurent series. On the bounded domain $D\cap \{|z|<R\}$, the maximum principle for harmonic functions bounds $u$ by $\max (0,{\epsilon }_R)$, where ${\epsilon }_R={\sup }_{D\cap \{|z|=R\}}|g_K(z)-z|\to 0$. Let $R\to \infty$ with $z$ fixed. We obtainThis is the height contraction of a hydrodynamically normalized mapping-out function. The exhaustion controls infinity explicitly, which is necessary when using a maximum principle on an unbounded domain.

On the punctured region $r>0$, choose a positive constant $a$ and setFor a radial function in three dimensions, $\Delta \psi =r^{-2}{\partial }_r(r^2{\partial }_r\psi )$. Since $r^2{\partial }_r\psi =-a$, this harmonic function has $\Delta \psi =0$ at every $r>0$, is nonconstant and positive, and approaches one at infinity. It defines the requested time-symmetric conformally flat vacuum initial data on that punctured or exterior region. One often writes $a=M/2$, giving the time-symmetric spatial metric of the Schwarzschild metric in isotropic coordinates.

A puncture or inner boundary is necessary if global regularity was intended. A smooth harmonic function on all of ${\mathbb{R}}^3$ that tends to one at infinity must equal one everywhere: applying the maximum principle for harmonic functions to larger and larger balls bounds $|\psi -1|$ by its arbitrarily small boundary values. The nonconstant example is therefore not a smooth solution through $r=0$, and no such globally regular example exists under those stronger assumptions.

The construction is well defined. Since $D$ is contained in a ball, its Brownian exit time is finite almost surely: at successive integer times there is a fixed positive probability that the next independent unit-time Brownian motion increment has length greater than the ball's diameter, forcing an exit. The survival probability is therefore bounded by a geometric sequence. Path continuity gives $B_{{\tau }_D}\in \partial D$. Thus the bounded Dirichlet boundary data give a bounded Borel function $u$ on $D$.

For interior harmonicity, fix a ball with closure in $D$, centred at $x$, and let $\sigma$ be its Brownian exit time. The Strong Markov property, followed by the tower property of conditional expectation, givesThe orthogonal invariance of Brownian motion makes $B_{\sigma }$ uniform on the boundary sphere. Hence $u$ has the spherical mean value property for harmonic functions for every such ball. A locally bounded Borel function with this property is smooth and harmonic: integrating the spherical averages against any smooth radial mollifier gives $u=u\ast \rho$ locally, which first proves smoothness; the mean value property for harmonic functions then implies $\Delta u=0$.

For the boundary limit at $\xi$, choose the harmonic barrier for the Dirichlet problem $h_{\xi }$ from (i). The Itô formula, localized inside $D$, makes $h_{\xi }(B_{t\wedge {\tau }_D})$ a bounded martingale. Compact localization and the bounded convergence theorem, first to the exit and then as $t\to \infty$, yieldFor $\epsilon >0$, if $\partial D\setminus B(\xi ,\epsilon )$ is nonempty, compactness and barrier positivity giveIf that boundary subset is empty, the probability is already zero. ThereforeFirst let $x\to \xi$ and then $\epsilon \downarrow 0$. The continuity of the Dirichlet boundary data proves $u(x)\to f(\xi )$.

Finally, the difference of two continuous solutions is a harmonic function vanishing on the boundary. The maximum principle for harmonic functions on the bounded domain gives that it is zero. Thus the Kakutani solution of the Dirichlet problem exists, attains every prescribed boundary value, and is unique.

Write the pole as $(x_0,y_0)$ with $x_0,y_0>0$. The method of images uses three reflected poles with signs $-,-,+$. ThusThe logarithms cancel in pairs on each axis. Only the original pole is in the interior, so this is the required Dirichlet Green function in a quadrant.

On the horizontal boundary $(t,0)$ the outward normal is $-{\mathbf{e}}_y$. Differentiating the four logarithms givesMultiply by the horizontal Dirichlet boundary condition $e^{-t^2}$ and integrate to obtainBy exchanging the axes, the vertical boundary contributes $F(y_0,x_0)$. The representation in part (a) consequently givesFor the unbounded-domain condition, the same expression is the Poisson integral after the conformal change of variables $w=(x+iy{)}^2$ to the upper half-plane, with real-axis data $e^{-|s|}$. This continuous boundary function tends to zero at both ends. Its Poisson integral tends to zero as $|w|\to \infty$: split the data into a compact part, whose kernel integral tends to zero, and a tail uniformly smaller than any prescribed positive number. Hence the constructed function has the required decay, and its two boundary traces agree at the corner. The maximum principle for harmonic functions on expanding quarter-discs proves uniqueness among decaying solutions. The corner value is obtained by continuity as $1$.

Let $u=v+\psi$, so $v$ has zero Dirichlet boundary condition. Using the supplied form of the minimal surface equation for a graph, setFor $\Vert \psi {\Vert }_{C^{2,\alpha }}\le \beta \le 1$ and $\Vert v\Vert ,\Vert w\Vert \le 1$, the difference bound impliesandThe shifted arguments stay in the permitted radius-two ball, and $\Vert f_{\psi }{\Vert }_{C^{0,\alpha }}\le C{\beta }^2+C_n\beta$. The Laplacian with zero Dirichlet boundary condition has trivial null space by the maximum principle for harmonic functions. Choose $\beta >0$ small enough that $2C\beta \le \delta$ and $C{\beta }^2+C_n\beta \le {\epsilon }_0$, for the constants in part (c) with $L=\Delta$. Its small-data existence for a nonlinear elliptic Dirichlet problem supplies $v$, and $\boxed{u=v+\psi }$ solves the minimal surface equation for a graph with the required boundary values.