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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.

The PDF specifies decay as $y\to \infty$; the local TeX drops this limit from its displayed boundary data. Take $a=4\pi$, so the conformal map is $w=\sin (z/4)$. Its components areFor $-2\pi <x<2\pi$ and $y>0$, $\eta >0$. The bottom edge maps monotonically to $(-1,1)$, and the two vertical sides map to the real rays $(-\infty ,-1)$ and $(1,\infty )$. The derivative $\cos (z/4)/4$ never vanishes in the interior. Injectivity follows from the identities for equal sines: the alternatives $z_1/4=z_2/4+2\pi n$ and $z_1/4=\pi -z_2/4+2\pi n$ give only the identical point within this strip. Surjectivity can be seen directly: for fixed $\xi$ and $v=y/4>0$, solve $\sin (x/4)=\xi /\cosh v$; thenOn $v>\max (0,\mathrm{arcosh}|\xi |)$, with the lower endpoint read as $0$ when $|\xi |\le 1$, the right side increases strictly from zero to infinity. Thus every point of the upper half-plane has a unique preimage.

By conformal invariance of harmonicity, write the transformed harmonic function as $\Phi (\xi ,\eta )$. Its real-axis boundary data areThe inverse sine on $(-1,1)$ is its real principal branch. For bounded integrable data, take the Fourier transform convention $\hat{\Phi }(k,\eta )={\int }_{\mathbb{R}}\Phi (\xi ,\eta )e^{-ik\xi }d\xi$. The Laplace equation becomes ${\hat{\Phi }}_{\eta \eta }-k^2\hat{\Phi }=0$, with the bounded decaying solution $\hat{\Phi }=\hat{F}(k)e^{-|k|\eta }$. Inverting the Fourier transform givesConsequently the Poisson kernel for the upper half-plane yieldsThe decay is uniform across the strip: $|w{|}^2={\sin }^2(x/4)+{\sinh }^2(y/4)\ge {\sinh }^2(y/4)$ and, for $|w|>1$, $|\Phi |\le \Vert F{\Vert }_{L^1}|w|/(\pi (|w|-1{)}^2)\to 0$. This verifies the printed decay condition. Boundary limits give the prescribed values at continuity points away from the corners.

For the given sine data,Values assigned at $s=\pm 1$ do not alter the integral. There is no solution continuous on the entire closed half-strip for these data: the bottom edge has corner limits $\pm 1$, but the side edges have corner limits zero. The formula is the bounded Dirichlet problem solution with the specified limits away from those two corners. This is the usual interpretation of the printed problem; boundedness excludes additional boundary-singular harmonic functions, while decay alone without a regularity or boundedness condition does not establish uniqueness.

The canonical Dirichlet Green function on the unit disc with its pole at zero is $-\log |w|$. Conformal invariance of the planar Green kernel therefore givesComparing with $G_D(z,y)=-\log |z-y|-G_z(y)$ yieldsThis also explains the boundary data: when $|w|=1$, the correction is $-\log |\varphi (w)-z|$. The composition with ${\varphi }^{-1}$ is harmonic by conformal invariance of harmonicity. Its apparent singularity at $z$ is removable because the quotient defining $\psi$ extends to ${\varphi }^{\prime }(0)$.

Evaluating there proves the regular part of the planar Dirichlet Green function formulaOn an unbounded domain, boundary values alone need not specify a unique harmonic function. Here $G_z$ is the correction belonging to the canonical Dirichlet Green kernel, which fixes that ambiguity.

Lift $u$ through the complex exponential by settingThe conformal invariance of harmonicity makes $v$ harmonic. Since $e^{w+2\pi i}=e^w$ and $u(2z)=u(z)$,Thus $v$ is bounded on the compact rectangleand its two periods make it bounded on the whole plane. The harmonic Liouville theorem makes $v$ constant. Since the exponential map is surjective onto ${\mathbb{C}}^{\ast }$, $u$ is constant.

For the requested counterexample after deleting a countable set, takeand let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function for the latticeDefineChanging a branch of the logarithm adds $2\pi i$, a period of ${\wp }_{\Lambda }$, so $u$ is well defined. Its poles project precisely to $S$, and the conformal invariance of harmonicity makes $u$ harmonic away from them. The other period givesMoreover, $z\notin S$ implies $2z\notin S$. Finally, $u$ is nonconstant because as positive real $z\to 1$,which is real and unbounded. This is a scale-periodic harmonic function from an elliptic function.