Poisson kernel and bounded Dirichlet problem on a half-space

Work under [F9] and put $R(x,z):=\mathrm{\mid }x-(z,0)\mathrm{\mid }=(\mathrm{\mid }{x}^{\mathrm{\prime }}-z{\mathrm{\mid }}^2+t^2{)}^{1\mathrm{/}2}>0$ for $x=(x^{\mathrm{\prime }},t)\in H$. Write $y=(z,0)$.

Derivative of the fundamental kernel: since $\mathrm{\Phi }(w)=\mathrm{\mid }w{\mathrm{\mid }}^{2-n}\mathrm{/}((n-2){\omega }_{n-1})$ on $w\mathrm{\ne }0$, the chain rule and real-power rule [F8] give $\mathrm{\nabla }\mathrm{\Phi }(w)=\mathrm{\mid }w{\mathrm{\mid }}^{-n}w\cdot (2-n)\mathrm{/}((n-2){\omega }_{n-1})=-w\mathrm{/}({\omega }_{n-1}\mathrm{\mid }w{\mathrm{\mid }}^n)$ for every $w\mathrm{\ne }0$.

Normalisation, first reduction. By [F6] applied to the nonnegative measurable function $z\mapsto P_H(x,z)$ and the $C^1$ diffeomorphism $z=x^{\mathrm{\prime }}+tw$ with Jacobian $t^{n-1}$, ${\int }_{{\mathbb{R}}^{n-1}}{P}_H(x,z)dz=\frac{2t}{{\omega }_{n-1}}{\int }_{{\mathbb{R}}^{n-1}}\frac{dz}{(\mathrm{\mid }{x}^{\mathrm{\prime }}-z{\mathrm{\mid }}^2+t^2{)}^{n\mathrm{/}2}}=\frac{2t{t}^{n-1}}{{\omega }_{n-1}{t}^n}{\int }_{{\mathbb{R}}^{n-1}}\frac{dw}{(1+\mathrm{\mid }w{\mathrm{\mid }}^2{)}^{n\mathrm{/}2}}=\frac{2}{{\omega }_{n-1}}I,I:={\int }_{{\mathbb{R}}^{n-1}}\frac{dw}{(1+\mathrm{\mid }w{\mathrm{\mid }}^2{)}^{n\mathrm{/}2}}\mathrm{.}$

The boundary derivative. Fix $x=(x^{\mathrm{\prime }},t)\in H$ and a boundary coordinate $z\in {\mathbb{R}}^{n-1}$. For $0<s<t$, the formula in [F1] gives $G_H(x,(z,s))=\mathrm{\Phi }(x^{\mathrm{\prime }}-z,t-s)-\mathrm{\Phi }(x^{\mathrm{\prime }}-z,t+s)$. Both arguments stay nonzero through $s=0$, so this explicit expression extends smoothly to the boundary pole $(z,0)$. By step 1.2, differentiating in $s$ at $0$ gives ${\mathrm{\partial }}_s{G}_H(x,(z,s)){\mathrm{\mid }}_{s=0}=2t\mathrm{/}({\omega }_{n-1}(\mathrm{\mid }{x}^{\mathrm{\prime }}-z{\mathrm{\mid }}^2+t^2{)}^{n\mathrm{/}2})$. Since the outward normal is $\nu =-e_n$, the negative outward derivative is $-{\mathrm{\partial }}_{\nu }{G}_H=+{\mathrm{\partial }}_s{G}_H$, which is the displayed positive kernel. This calculation uses the explicit reflected formula and its smooth boundary extension; it does not apply the interior-pole statement of [F1] at a boundary pole.

Polar evaluation of $I$. With $m=n-1\ge 2$ and $w=r\theta$, [F6] gives $I={\omega }_{n-2}{\int }_0^{\mathrm{\infty }}\frac{r^{n-2}dr}{(1+r^2{)}^{n\mathrm{/}2}}$; substituting $s=r^2$, $r^{n-2}dr=\frac{1}{2}{s}^{(n-1)\mathrm{/}2-1}ds$, this is $I=\frac{{\omega }_{n-2}}{2}{\int }_0^{\mathrm{\infty }}{s}^{(n-1)\mathrm{/}2-1}(1+s{)}^{-n\mathrm{/}2}ds$. The further substitution $t=\frac{s}{1+s}$ turns the last integral into ${\int }_0^1{t}^{(n-1)\mathrm{/}2-1}(1-t{)}^{1\mathrm{/}2-1}dt=B(\frac{n-1}{2},\frac{1}{2})$ by [F6]; hence $I=\frac{{\omega }_{n-2}}{2}B(\frac{n-1}{2},\frac{1}{2})<+\mathrm{\infty }$. [step 1.3, F6, algebra] 3.1 Positivity: for $x\in H$ we have $t>0$ and ${\omega }_{n-1}>0$ by [F6], while the denominator is a positive real number; hence $P_H(x,z)>0$ for every $z\in {\mathbb{R}}^{n-1}$.

Evaluation of the constants. By [F6], $\mathrm{\mid }{B}_1^n\mathrm{\mid }={\pi }^{n\mathrm{/}2}\mathrm{/}\mathrm{\Gamma }(n\mathrm{/}2+1)$ and $\mathrm{\mid }{B}_1^n\mathrm{\mid }={\omega }_{n-1}\mathrm{/}n$, so ${\omega }_{n-1}=n{\pi }^{n\mathrm{/}2}\mathrm{/}\mathrm{\Gamma }(n\mathrm{/}2+1)=2{\pi }^{n\mathrm{/}2}\mathrm{/}\mathrm{\Gamma }(n\mathrm{/}2)$; replacing $n$ by $n-1$ gives ${\omega }_{n-2}=2{\pi }^{(n-1)\mathrm{/}2}\mathrm{/}\mathrm{\Gamma }((n-1)\mathrm{/}2)$. By [F6] again, $B(\frac{n-1}{2},\frac{1}{2})=\mathrm{\Gamma }\left(\frac{n-1}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{/}\mathrm{\Gamma }\left(\frac{n}{2}\right)=\mathrm{\Gamma }\left(\frac{n-1}{2}\right)\sqrt{\pi }\mathrm{/}\mathrm{\Gamma }\left(\frac{n}{2}\right)$. Substituting into steps 1.3 and 2.2, ${\int }_{{\mathbb{R}}^{n-1}}{P}_H(x,z)dz=\frac{2}{{\omega }_{n-1}}\cdot \frac{{\omega }_{n-2}}{2}\cdot \frac{\mathrm{\Gamma }\left(\frac{n-1}{2}\right)\sqrt{\pi }}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}=\frac{2{\pi }^{(n-1)\mathrm{/}2}}{2{\pi }^{n\mathrm{/}2}}\cdot \sqrt{\pi }=1$.

Boundary convergence. Fix $z_0\in \mathrm{\partial }H$ and $\eta >0$; by continuity of $g$ at $z_0$ choose $\delta >0$ with $\mathrm{\mid }g(z)-g(z_0)\mathrm{\mid }<\eta$ for $\mathrm{\mid }z-z_0\mathrm{\mid }<\delta$. Let $C=\{z:\mathrm{\mid }z-z_0\mathrm{\mid }<\delta \}$ and $D={\mathbb{R}}^{n-1}\setminus C$. For $z\in C$, $\mathrm{\mid }g(z)-g(z_0)\mathrm{\mid }\le \eta$; for $z\in D$ we use $\mathrm{\mid }g(z)-g(z_0)\mathrm{\mid }\le 2\parallel g{\parallel }_{\mathrm{\infty }}$, and the mass of $D$ is small: if $\mathrm{\mid }{x}^{\mathrm{\prime }}-z_0\mathrm{\mid }<\delta \mathrm{/}2$ and $z\in D$, then $\mathrm{\mid }{x}^{\mathrm{\prime }}-z\mathrm{\mid }\ge \mathrm{\mid }z-z_0\mathrm{\mid }-\mathrm{\mid }{x}^{\mathrm{\prime }}-z_0\mathrm{\mid }>\delta \mathrm{/}2$, so by step 1.3 ${\int }_D{P}_H(x,z)dz\le \frac{2}{{\omega }_{n-1}}{\int }_{\mathrm{\mid }w\mathrm{\mid }>\delta \mathrm{/}(2t)}\frac{dw}{(1+\mathrm{\mid }w{\mathrm{\mid }}^2{)}^{n\mathrm{/}2}}$, and this tail tends to $0$ as $t\downarrow 0$ by [F7] and the finiteness in step 2.2, since the integrands are dominated by the integrable function $(1+\mathrm{\mid }w{\mathrm{\mid }}^2{)}^{-n\mathrm{/}2}$ and vanish pointwise on the shrinking domain. Hence $\mathrm{\mid }{U}_g(x)-g(z_0)\mathrm{\mid }\le \eta +2\parallel g{\parallel }_{\mathrm{\infty }}{\int }_D{P}_H(x,z)dz$, so ${\mathrm{lim\hspace{0.17em}sup}}_{x\to (z_0,0)}\mathrm{\mid }{U}_g(x)-g(z_0)\mathrm{\mid }\le \eta +2\parallel g{\parallel }_{\mathrm{\infty }}\cdot 0=\eta$, and $\eta >0$ was arbitrary; so $U_g(x)\to g(z_0)$ as $x\to (z_0,0)$ from inside $H$.

Derivative bounds and integrability. Every partial derivative $D_x^{\alpha }{P}_H(x,z)$ is continuous on $H\times {\mathbb{R}}^{n-1}$ and, on each compact $K\subset H$, satisfies $\mathrm{\mid }{D}_x^{\alpha }{P}_H(x,z)\mathrm{\mid }\le C_{\alpha ,K}(1+\mathrm{\mid }z\mathrm{\mid }{)}^{-n}$. Indeed, writing $\lambda =\mathrm{\mid }x\mathrm{\prime }-z{\mathrm{\mid }}^2+t^2$, on $K$ the height $t$ is bounded away from $0$ and both $t$ and $\mathrm{\mid }x\mathrm{\prime }\mathrm{\mid }$ are bounded above; for large $\mathrm{\mid }z\mathrm{\mid }$, $\lambda$ is comparable to $\mathrm{\mid }z{\mathrm{\mid }}^2$. Each horizontal derivative of ${\lambda }^{-n\mathrm{/}2}$ contributes a factor $O(\mathrm{\mid }z\mathrm{\mid })$ and one extra factor ${\lambda }^{-1}$, gaining decay; each vertical derivative either differentiates the numerator $t$, leaving the base decay $O(\mathrm{\mid }z{\mathrm{\mid }}^{-n})$, or differentiates a denominator factor and gains decay with bounded factors of $t$. Repeating these rules shows that no derivative decays more slowly than $\mathrm{\mid }z{\mathrm{\mid }}^{-n}$; bounded $z$ are covered by compactness and smoothness on $K$. Since $n>n-1$, this majorant is integrable over ${\mathbb{R}}^{n-1}$. Also $\mathrm{\mid }{U}_g(x)\mathrm{\mid }\le \parallel g{\parallel }_{\mathrm{\infty }}{\int }_{{\mathbb{R}}^{n-1}}{P}_H(x,z)dz=\parallel g{\parallel }_{\mathrm{\infty }}$ by steps 3.1 and 3.2; in particular $U_g$ is absolutely convergent and bounded on $H$.

Uniqueness. Let $w$ be bounded and harmonic on $H$, continuous on $\bar{H}$, with $w=0$ on $\mathrm{\partial }H$; it suffices to show $w\equiv 0$. If $w$ is complex-valued, apply the argument below separately to its real and imaginary parts, so assume $w$ is real-valued. Fix $p\in \mathrm{\partial }H$ and a ball $\mathbb{B}:=B_{\rho }(p)$ with $\rho >0$. Define $g_{\rho }$ on $\mathrm{\partial }\mathbb{B}$ by $g_{\rho }(y)=w(y)$ for $y_n\ge 0$ and $g_{\rho }(y)=-w(y^{\mathrm{\prime }},-y_n)$ for $y_n<0$; this is continuous on $\mathrm{\partial }\mathbb{B}$ because $w$ is continuous on $\bar{H}$ and $w=0$ on the plane, where the two clauses agree. By [F3] let $W$ be the harmonic function on $\mathbb{B}$ with trace $g_{\rho }$; since $g_{\rho }$ is odd under the reflection $\sigma (y)=(y^{\mathrm{\prime }},-y_n)$, the function $y\mapsto -W(\sigma (y))$ is harmonic on $\mathbb{B}$ with the same trace $g_{\rho }$ (because $g_{\rho }\circ \sigma =-g_{\rho }$), so [F3] gives $W(\sigma (y))=-W(y)$: $W$ is odd. In particular $W=0$ on the flat part $\mathrm{\partial }\mathbb{B}\cap \mathrm{\partial }H$, and on the upper half ball ${\mathbb{B}}^{+}:=\mathbb{B}\cap H$ both $W$ and $w$ are harmonic, continuous on the closure of ${\mathbb{B}}^{+}$, and agree on its boundary (the upper hemisphere carries $g_{\rho }=w$, and the flat part carries $w=0=W$); the weak maximum principle [F4] applied to $W-w$ and to $w-W$ gives $W=w$ on ${\mathbb{B}}^{+}$. Therefore the odd extension $\tilde{w}$ of $w$ (namely $\tilde{w}(y)=w(y)$ for $y_n>0$ and $\tilde{w}(y)=-w(y^{\mathrm{\prime }},-y_n)$ for $y_n<0$) coincides with the harmonic function $W$ on $\mathbb{B}$, hence is harmonic on a neighbourhood of $p$; as $p$ was arbitrary and $\tilde{w}$ is harmonic off the plane, $\tilde{w}$ is harmonic on all of ${\mathbb{R}}^n$. It is bounded by $\parallel w{\parallel }_{\mathrm{\infty }}$, so [F5] makes it constant, and its value at the plane is $0$; hence $\tilde{w}\equiv 0$ and $w\equiv 0$.

Smoothness and harmonicity. By step 4.1 the domination hypothesis of [F7] holds on every compact $K\subset H$ and all admissible derivatives, so induction over the coordinate directions as in [F7] gives $U_g\in C^{\mathrm{\infty }}(H)$ with $D^{\alpha }{U}_g(x)={\int }_{{\mathbb{R}}^{n-1}}{D}_x^{\alpha }{P}_H(x,z)g(z)dz$. Moreover ${\mathrm{\Delta }}_x(w_n\mathrm{\mid }w{\mathrm{\mid }}^{-n})=0$ for $w\mathrm{\ne }0$: by [F8] and step 1.2, $\mathrm{\Delta }{w}_n=0$, $\mathrm{\nabla }(w_n)=e_n$, $\mathrm{\nabla }\mathrm{\mid }w{\mathrm{\mid }}^{-n}=-n\mathrm{\mid }w{\mathrm{\mid }}^{-n-2}w$ and $\mathrm{\Delta }\mathrm{\mid }w{\mathrm{\mid }}^{-n}=2n\mathrm{\mid }w{\mathrm{\mid }}^{-n-2}$, so the product rule gives $\mathrm{\Delta }(w_n\mathrm{\mid }w{\mathrm{\mid }}^{-n})=2e_n\cdot (-n\mathrm{\mid }w{\mathrm{\mid }}^{-n-2}w)+w_n\cdot 2n\mathrm{\mid }w{\mathrm{\mid }}^{-n-2}=0$. Since $P_H(x,z)=\frac{2}{{\omega }_{n-1}}\phi (x-(z,0))$ with $\phi (w)=w_n\mathrm{\mid }w{\mathrm{\mid }}^{-n}$ and $x-(z,0)$ never vanishes for $x\in H$, the chain rule gives ${\mathrm{\Delta }}_x{P}_H(x,z)=0$ for all $x\in H$, $z\in {\mathbb{R}}^{n-1}$, and therefore $\mathrm{\Delta }{U}_g(x)={\int }_{{\mathbb{R}}^{n-1}}{\mathrm{\Delta }}_x{P}_H(x,z)g(z)dz=0$.

If $v$ is any bounded harmonic function on $H$, continuous on $\bar{H}$, with trace $g$, then $w:=v-U_g$ is bounded, harmonic by step 5.1, continuous on $\bar{H}$ and zero on the plane by step 3.3, so step 4.2 gives $w\equiv 0$ and $v=U_g$; for complex data both $U_g$ and the difference are complex, and the maximum-principle and Liouville steps were applied to the real and imaginary parts. Together with steps 2.1, 3.1, 3.2, 4.1, 5.1 and 3.3 this proves every clause of the statement.