Reflection Green kernel for the half-space

Since $y_n>0$, we have $y^{†}=(y^{\mathrm{\prime }},-y_n)\mathrm{\notin }\bar{H}$. Thus for $x\in H\setminus \{y\}$ both vectors $x-y$ and $x-y^{†}$ are nonzero, and the formula defines a real function smooth in $x$ off $y$. By [F1] and [F5] the first term is locally integrable, while the second is continuous on all of $H$ because $\mathrm{\mid }x-y^{†}\mathrm{\mid }\ge x_n+y_n>0$. Hence the difference is locally integrable on $H$; write $T$ for its regular distribution, which exists by [F4].

Symmetry. For distinct $x,y\in H$, the vectors $x-y^{†}=(x^{\mathrm{\prime }}-y^{\mathrm{\prime }},x_n+y_n)$ and $y-x^{†}=(y^{\mathrm{\prime }}-x^{\mathrm{\prime }},y_n+x_n)$ have equal Euclidean norms, because their first $n-1$ coordinates differ only by a sign and their last coordinates agree; and $\mathrm{\Phi }$ is even, being a function of $\mathrm{\mid }z\mathrm{\mid }$ only. Hence $\mathrm{\Phi }(x-y^{†})=\mathrm{\Phi }(y-x^{†})$ and $\mathrm{\Phi }(x-y)=\mathrm{\Phi }(y-x)$, so $G_H(x,y)=G_H(y,x)$.

Strict positivity. For distinct $x,y\in H$ the $n-1$ leading coordinates of $x-y^{†}$ and $x-y$ agree, so $\mathrm{\mid }x-y^{†}{\mathrm{\mid }}^2-\mathrm{\mid }x-y{\mathrm{\mid }}^2=(x_n+y_n{)}^2-(x_n-y_n{)}^2=4x_n{y}_n>0$; thus $\mathrm{\mid }x-y^{†}\mathrm{\mid }>\mathrm{\mid }x-y\mathrm{\mid }\ge 0$. Since $n\ge 3$ gives the negative exponent $2-n<0$ and $r\mapsto r^{2-n}$ is strictly decreasing on $(0,\mathrm{\infty })$ (a quotient of positive powers, verified from $r^{2-n}=1\mathrm{/}{r}^{n-2}$), and since the factor $1\mathrm{/}((n-2){\omega }_{n-1})$ of [F1] is positive, we get $\mathrm{\Phi }(x-y^{†})<\mathrm{\Phi }(x-y)$, that is $G_H(x,y)>0$.

Harmonicity in $x$ off the pole. Fix $y\in H$. By [F2] the translate $x\mapsto \mathrm{\Phi }(x-y)$ is smooth and harmonic on ${\mathbb{R}}^n\setminus \{y\}$, hence on $H\setminus \{y\}$; and $x\mapsto \mathrm{\Phi }(x-y^{†})$ is smooth and harmonic on all of $H$, because $H$ is contained in ${\mathbb{R}}^n\setminus \{y^{†}\}$. A difference of harmonic smooth functions is smooth and harmonic, so $x\mapsto G_H(x,y)$ is smooth and harmonic on $H\setminus \{y\}$.

Zero boundary trace. Let $z\in \mathrm{\partial }H=\{x_n=0\}$ and let $x\to z$ with $x\in H$. Then $x-y\to z-y=(z^{\mathrm{\prime }}-y^{\mathrm{\prime }},-y_n)$ and $x-y^{†}\to z-y^{†}=(z^{\mathrm{\prime }}-y^{\mathrm{\prime }},y_n)$, and these two limit vectors have equal norms $\mathrm{\mid }{z}^{\mathrm{\prime }}-y^{\mathrm{\prime }}{\mathrm{\mid }}^2+y_n^2$, a positive number because $y_n>0$; in particular neither limit is the origin. By continuity of $\mathrm{\Phi }$ off the origin, ${\lim }_{x\to z,x\in H}{G}_H(x,y)=\mathrm{\Phi }(z-y)-\mathrm{\Phi }(z-y^{†})=0$. As the formula is continuous on the closed set $\{x:x_n\ge 0,x\mathrm{\ne }y\}$, it extends $G_H(\cdot ,y)$ continuously to $\bar{H}\setminus \{y\}$ with value $0$ on $\mathrm{\partial }H$.

Distributional identity. Let $\varphi \in C_c^{\mathrm{\infty }}(H)$ be a test function and let $\psi$ be its extension by zero to ${\mathbb{R}}^n$, which is smooth and compactly supported. By the derivative rules of [F4], $(\mathrm{\Delta }T)(\varphi )=T(\mathrm{\Delta }\varphi )={\int }_H{G}_H(x,y)\mathrm{\Delta }\varphi (x)dx$, hence $⟨-\mathrm{\Delta }T,\varphi ⟩=-{\int }_H{G}_H(x,y)\mathrm{\Delta }\varphi (x)dx=-{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(x-y)\mathrm{\Delta }\psi (x)dx+{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(x-y^{†})\mathrm{\Delta }\psi (x)dx$, because $\psi =\varphi$ on $H$ and $G_H(x,y)=\mathrm{\Phi }(x-y)-\mathrm{\Phi }(x-y^{†})$ there. By [F3] applied at the poles $y$ and $y^{†}$, $-{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(x-y)\mathrm{\Delta }\psi (x)dx=\psi (y)=\varphi (y)$, while $+{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(x-y^{†})\mathrm{\Delta }\psi (x)dx=-\psi (y^{†})=0$; the last equality holds because $\mathrm{supp}\psi \subseteq H$ avoids $H^c$. Therefore $⟨-\mathrm{\Delta }T,\varphi ⟩=\varphi (y)={\delta }_y(\varphi )$ for every test function, that is $-{\mathrm{\Delta }}_x{T}_{G_H(\cdot ,y)}={\delta }_y$ on $H$ in the sense of [F4].

Steps 2.1, 2.2, 2.3, 2.4 and 2.5 establish that the reflection kernel $G_H(\cdot ,y)$ is symmetric, strictly positive at distinct points of $H$, smooth and harmonic in $x$ off $y$, has zero continuous boundary trace, and represents $-{\mathrm{\Delta }}_x{G}_H(\cdot ,y)={\delta }_y$ distributionally on $H$; it therefore acts as the Green kernel of this unbounded half-space, and no bounded-domain Green definition is applied to $H$ anywhere above.