[proofplan] Fixing the pole $y \in \Omega$, we subtract the Green function from the translated fundamental solution and call the difference $H_y$. The bounded $C^2$ hypothesis is used through the existence of the Dirichlet Green function and its [boundary regularity](/theorems/99) away from the pole. The defining distributional equations for $G(\cdot,y)$ and $\Phi(\cdot-y)$ show that their singular Dirac masses cancel, so $H_y$ is harmonic throughout $\Omega$, not merely away from $y$. The zero Dirichlet trace of $G(\cdot,y)$ gives the boundary trace of $H_y$, and the asymptotics follow because harmonic functions are locally bounded while the fundamental solution diverges at its pole. [/proofplan]

[step:Define the regular part by subtracting the Green function from the fundamental solution] Fix $y \in \Omega$. Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure). Define the punctured Euclidean space $\mathbb{R}^m_0 := \mathbb{R}^m \setminus \{0\}$ for each integer $m \geq 1$, and define the open unit ball $B(0,1) := \{z \in \mathbb{R}^n : |z| < 1\}$. Define the geometric constant \begin{align*} \omega_n := \mathcal{L}^n(B(0,1)). \end{align*} Let $\Phi: \mathbb{R}^n_0 \to \mathbb{R}$ be the fundamental solution of $-\Delta$. With this convention, for $n \geq 3$, \begin{align*} \Phi(z)=\frac{1}{n(n-2)\omega_n}|z|^{2-n}, \qquad z \in \mathbb{R}^n_0, \end{align*} and for $n=2$, \begin{align*} \Phi(z)=-\frac{1}{2\pi}\log |z|, \qquad z \in \mathbb{R}^2_0. \end{align*} Define the map $H_y: \Omega \setminus \{y\} \to \mathbb{R}$ by \begin{align*} H_y(x) := \Phi(x-y)-G(x,y). \end{align*} For $x \in \Omega \setminus \{y\}$ this definition is equivalent to \begin{align*} G(x,y)=\Phi(x-y)-H_y(x). \end{align*} The defining property of the Dirichlet Green function for $-\Delta$ is that, in the distributional sense on $\Omega$, \begin{align*} -\Delta_x G(\cdot,y)=\delta_y, \end{align*} where $\delta_y$ is the Dirac measure at $y$. The defining property of the fundamental solution is that \begin{align*} -\Delta_x \Phi(\cdot-y)=\delta_y \end{align*} in the distributional sense on $\Omega$. Since the Dirichlet Green function $G(\cdot,y)$ and the translated fundamental solution $\Phi(\cdot-y)$ are locally integrable on $\Omega$, their difference $H_y$ defines a distribution on all of $\Omega$. Therefore \begin{align*} -\Delta H_y = -\Delta \Phi(\cdot-y)+\Delta G(\cdot,y)=\delta_y-\delta_y=0. \end{align*} as distributions on $\Omega$. By the distributional form of [interior smoothness of harmonic functions](/theorems/36), often called Weyl's lemma for harmonic distributions, $H_y$ has a representative, still denoted $H_y$, satisfying \begin{align*} H_y \in C^\infty(\Omega) \end{align*} and \begin{align*} \Delta H_y=0 \end{align*} pointwise in $\Omega$. On $\Omega \setminus \{y\}$ this smooth representative agrees with the originally defined pointwise function $\Phi(\cdot-y)-G(\cdot,y)$: the two functions define the same distribution there, hence agree $\mathcal{L}^n$-a.e., and both sides are continuous on every [open set](/page/Open%20Set) compactly contained in $\Omega \setminus \{y\}$. Thus $H_y$ is harmonic in $\Omega$ and the identity $G(x,y)=\Phi(x-y)-H_y(x)$ remains valid for every $x \in \Omega \setminus \{y\}$. [/step]

[step:Identify the boundary trace of the regular part]Since a bounded $C^2$ domain is in particular a bounded $C^1$ domain, the trace hypothesis for Dirichlet Green functions recorded in [Green's Representation for Poisson's Equation](/theorems/42) applies to $\Omega$. Hence the Dirichlet Green function has zero continuous boundary trace in the first variable away from the pole. Thus, for every $x_0 \in \partial\Omega$, \begin{align*} G(x,y)\to 0 \end{align*} as $x \to x_0$ from within $\Omega$. Since $y \in \Omega$ and $x_0 \in \partial\Omega$, we have $x_0 \neq y$, and the map \begin{align*} x \mapsto \Phi(x-y) \end{align*} is continuous in a neighbourhood of $\partial\Omega$. Hence \begin{align*} H_y(x)=\Phi(x-y)-G(x,y)\to \Phi(x_0-y). \end{align*} Therefore the continuous boundary trace of $H_y$ is $\Phi(\cdot-y)$ on $\partial\Omega$.[/step]

[guided]The boundary statement is not a new singular calculation; it is exactly the Dirichlet condition transferred from $G$ to the correction term. Recall that the regular part is the map $H_y: \Omega \setminus \{y\} \to \mathbb{R}$ defined by \begin{align*} H_y(x) := \Phi(x-y)-G(x,y). \end{align*} The pole $y$ lies in the interior of $\Omega$, so no boundary point equals $y$. Consequently the translated fundamental solution \begin{align*} x \mapsto \Phi(x-y) \end{align*} has no singularity on $\partial\Omega$ and is continuous there. Let $x_0 \in \partial\Omega$. Since $\Omega$ is a bounded $C^2$ domain, it is a bounded $C^1$ domain, so the trace hypothesis for Dirichlet Green functions recorded in [Green's Representation for Poisson's Equation](/theorems/42) applies. Since also $x_0 \neq y$, the Dirichlet Green function has zero continuous boundary trace at $x_0$ in the first variable, so \begin{align*} G(x,y)\to 0 \end{align*} as $x \to x_0$ from inside $\Omega$. Using the definition of $H_y$ on $\Omega \setminus \{y\}$ gives \begin{align*} H_y(x)=\Phi(x-y)-G(x,y). \end{align*} Taking the boundary limit and using continuity of $\Phi(\cdot-y)$ at $x_0$ yields \begin{align*} \lim_{x\to x_0} H_y(x)=\Phi(x_0-y)-0=\Phi(x_0-y). \end{align*} Since $x_0 \in \partial\Omega$ was arbitrary, the boundary trace of $H_y$ is precisely $\Phi(\cdot-y)$ on $\partial\Omega$.[/guided]

[step:Show that the harmonic correction is bounded near the pole] For $a\in\mathbb R^n$ and $r>0$, let $B(a,r):=\{z\in\mathbb R^n:|z-a|<r\}$ and let $\overline{B}(a,r)$ denote its closure. For $a\in\mathbb R^n$ and $E\subset\mathbb R^n$, let $\operatorname{dist}(a,E):=\inf\{|a-w|:w\in E\}$. Choose \begin{align*} r := \frac{1}{2}\operatorname{dist}(y,\partial\Omega)>0. \end{align*} Then $\overline{B}(y,r)\subset \Omega$. Since $H_y$ is harmonic in $\Omega$, it is continuous on $\overline{B}(y,r)$. Define \begin{align*} M_y := \sup_{x \in \overline{B}(y,r)} |H_y(x)|. \end{align*} The compactness of $\overline{B}(y,r)$ and the continuity of $H_y$ imply that $M_y<\infty$. Thus, for all $x \in B(y,r)$, \begin{align*} |H_y(x)|\leq M_y. \end{align*} [/step]

[step:Compare the decomposition with the divergent fundamental solution] Assume first that $n \geq 3$. For $x \in B(y,r)\setminus\{y\}$, \begin{align*} G(x,y)=\frac{1}{n(n-2)\omega_n}|x-y|^{2-n}-H_y(x). \end{align*} Dividing by $\frac{1}{n(n-2)\omega_n}|x-y|^{2-n}$ gives \begin{align*} \frac{G(x,y)}{\frac{1}{n(n-2)\omega_n}|x-y|^{2-n}} = 1-n(n-2)\omega_n H_y(x)|x-y|^{n-2}. \end{align*} Using $|H_y(x)|\leq M_y$ and $n-2>0$, we obtain \begin{align*} \left|n(n-2)\omega_n H_y(x)|x-y|^{n-2}\right| \leq n(n-2)\omega_n M_y |x-y|^{n-2}\to 0 \end{align*} as $x\to y$. Hence \begin{align*} G(x,y) \sim \frac{1}{n(n-2)\omega_n}|x-y|^{2-n}. \end{align*} Assume now that $n=2$. For $x \in B(y,r)\setminus\{y\}$, \begin{align*} G(x,y)=-\frac{1}{2\pi}\log |x-y|-H_y(x). \end{align*} Since $-\log |x-y|\to \infty$ as $x\to y$, division gives \begin{align*} \frac{G(x,y)}{-\frac{1}{2\pi}\log |x-y|} = 1+\frac{2\pi H_y(x)}{\log |x-y|}. \end{align*} The bound $|H_y(x)|\leq M_y$ implies \begin{align*} \left|\frac{2\pi H_y(x)}{\log |x-y|}\right| \leq \frac{2\pi M_y}{|\log |x-y||}\to 0 \end{align*} as $x\to y$. Therefore \begin{align*} G(x,y) \sim -\frac{1}{2\pi}\log |x-y|. \end{align*} [/step]

[step:Conclude the singular decomposition and asymptotic structure] For the fixed point $y \in \Omega$, the function $H_y$ is harmonic in $\Omega$, has boundary trace $\Phi(\cdot-y)$ on $\partial\Omega$, and satisfies \begin{align*} G(x,y)=\Phi(x-y)-H_y(x) \end{align*} for every $x \in \Omega \setminus \{y\}$. The comparison in a ball centred at $y$ shows that the bounded harmonic correction does not affect the leading singular term. Since $y \in \Omega$ was arbitrary, the asserted decomposition and asymptotics hold for every pole $y \in \Omega$. [/step]