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An open set satisfies an exterior cone condition at $z$ if a solid cone with vertex $z$ lies outside the set in some neighbourhood of $z$. The stronger global version requires the closures of the cone and the set to meet only at $z$. In two dimensions, a power barrier for an exterior cone makes such a point a regular boundary point for the Dirichlet problem. Local cone conditions suffice for local boundary regularity; a global cone is an additional geometric assumption.

Fix $x_0\in \partial \Omega$, $\epsilon >0$, and the power barrier for an exterior cone $g$ from part (i). Write $M=\Vert f{\Vert }_{\infty }$. By continuity of the boundary data, there is $\rho >0$ such that $|\phi (x)-\phi (x_0)|<\epsilon$ on $\partial \Omega \cap B_{\rho }(x_0)$. On the rest of the boundary, $-g\ge m{\rho }^{\nu }$. The boundary is compact, so $\phi$ is bounded. Choose $A>0$ large enough that $A\delta \ge M$ andThen the Perron subfunction for the Poisson equation and its superfunction counterpartsatisfy $\Delta v_{-}\ge M\ge f$, $\Delta v_{+}\le -M\le f$, and $v_{-}\le \phi \le v_{+}$ on the whole boundary. The near-boundary ordering uses the choice of $\rho$; the remaining ordering uses the choice of $A$. In particular the Perron family is nonempty and bounded above.

The Perron method for the Dirichlet problem and the permitted comparison give $v_{-}\le u\le v_{+}$. Since $g(x)\to 0$ as $x\to x_0$, we obtainLetting $\epsilon \downarrow 0$ provesThe interior Perron theorem gives continuity inside $\Omega$, and the displayed limit together with continuity of $\phi$ proves the continuous extension $u\in C^0(\overline{\Omega })$ with the required boundary trace. Thus every boundary point is a regular boundary point.

For the bounded-domain correction, let $\mathcal{S}$ be the family of Perron subfunctions for the Poisson equation $v\in C^0(\overline{\Omega })$ with $v\le \phi$ on $\partial \Omega$. The subfunction condition is ball comparison: whenever $\overline{B}\subset \Omega$ and a classical solution $H$ of $\Delta H=f$ on $B$ satisfies $v\le H$ on $\partial B$, it also satisfies $v\le H$ throughout $B$. For a $C^2$ function this is the sign condition $\Delta v\ge f$. A superfunction reverses both inequalities.

The Perron solution is the pointwise upper envelope of the subfunctions:This is the Perron method for the Dirichlet problem with the Poisson equation replacing the homogeneous Laplace equation. The power barrier for an exterior cone supplies a global member of $\mathcal{S}$ and a global superfunction, as constructed below. The permitted subfunction-superfunction comparison therefore makes this envelope finite. The standard interior Perron theorem, available in the question, gives $u\in C^2(\Omega )$ and $\Delta u=f$; neither result needs to be proved in this part.

A bounded-domain hypothesis is needed for the requested global construction and uniqueness. The printed question does not include it. For example, the upper half-plane satisfies the stated exterior cone condition, but both $u=0$ and $u(x_1,x_2)=x_2$ are harmonic functions with zero Dirichlet boundary condition. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the exterior cone condition and all other data remain as printed.

The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi )$, varying $r$ on a fixed ray forces $\nu =2$ if $r^{\nu -2}(h^{\prime \prime }+{\nu }^{2}h)\ge \delta >0$ for every $r>0$. We would then have $h^{\prime \prime }+4h\ge \delta$ and $h<0$. Put $a=\pi /4$ and $b=3\pi /4$. Twice integration by parts givescontradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).

Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$ and setOn the complement of the cone, unwrap the angle in polar coordinates as $\vartheta \in (\alpha ,2\pi -\alpha )$, measured from the cone axis. Then $\psi =\vartheta -\pi$ is the angle from the opposite axis and $|\psi |\le \pi -\alpha$ on $\overline{\Omega }\setminus \{x_0\}$. Define the power barrier for an exterior cone byThis has the requested separated form $r^{\nu }h(\vartheta )$. In the printed signed-angle convention, the same function is $-r^{\nu }\cos (\beta (\pi -|\theta |))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos (\beta \psi )$.

Since $\beta (\pi -\alpha )<\pi /2$, put $m=\cos (\beta (\pi -\alpha ))>0$. The cosine is at least $m$, so $g\le -m{r}^{\nu }<0$ away from $x_0$, and $g$ is continuous at $x_0$. The Laplacian in polar coordinates givesLet $D={\sup }_{x\in \Omega }|x-x_0|<\infty$. Since $\nu <2$, the barrier for the Dirichlet problem satisfiesBoth the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.

For the corrected bounded-domain problem, the interior theorem for the Perron method for the Dirichlet problem gives a classical solution of the Poisson equation, and the power barrier for an exterior cone argument gives its continuous extension with the required Dirichlet boundary condition. Hence existence holds in $C^2(\Omega )\cap C^0(\overline{\Omega })$.

If $u_1,u_2$ are two such classical solutions, then $w=u_1-u_2$ is a harmonic function with zero boundary trace. The weak maximum principle for elliptic operators, applied to $w$ and $-w$ on the bounded domain, gives $w\le 0$ and $-w\le 0$. ConsequentlyThere is exactly one solution after adding boundedness of $\Omega$. For the printed unbounded-domain version, the half-plane counterexample in part (i) disproves uniqueness; an additional condition at infinity or a suitable restricted solution class would instead have to be specified.