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Take boundary data one on a disk of radius $a>0$ and zero outside. The Poisson kernel for the upper half-space gives on the axisThe axis value tends to one at the boundary and behaves as $a^2/(2z^2)$ far away. The boundary data at the circular rim are understood almost everywhere: changing a value on the rim, a set of planar measure zero, leaves the integral unchanged. This does not assert a continuous boundary extension across the jump.

In the upper half-space $z>0$, the displayed decaying harmonic function has outward normal derivative ${\partial }_n\Phi =-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$. This is the Poisson kernel for the upper half-space, whose integral on a horizontal plane is one and whose limit as $z\downarrow 0$ is the two-dimensional Dirac delta. Thus ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional boundary sense. The compensating flux through a large hemisphere is minus one. This illustrates why a nonzero total boundary source is possible on an unbounded domain even when the interior source is zero.

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

Expand the divergence of $\Phi \nabla \Psi -\Psi \nabla \Phi$. The cross terms cancel, leaving $\Phi \Delta \Psi -\Psi \Delta \Phi$. The divergence theorem gives Green's second identity,Use the Green function convention ${\Delta }_{x}G(x,x_0)=\delta (x-x_0)$. Integrating over a bounded domain containing $x_0$ yields $\boxed{{\int }_S{\partial }_{n}GdS=1}$. Thus zero Neumann boundary data would be incompatible with this unit Dirac delta source. A suitable Neumann Green function with constant boundary flux instead has ${\partial }_{n}G=1/|S|$, where $|S|$ is the boundary area. Substituting $\Psi =G$, $\Delta \Phi =\sigma$, ${\partial }_n\Phi =f$ givesThe last term is an undetermined constant. The data must obey the Neumann Poisson problem compatibility condition ${\int }_V\sigma dV={\int }_{S}fdS$, obtained by integrating the original equation. Under this condition an additive constant in $G$ cancels from the first two terms.

For the upper half-space, reflect the source to $x_0^{\ast }=(x_{0,1},x_{0,2},-x_{0,3})$. The method of images with equal-sign sources givesThe reflected singularity lies outside the domain; on $z=0$ the two normal derivatives cancel, so ${\partial }_{n}G=-{\partial }_{z}G=0$. Its total unit flux passes through a large hemisphere at infinity, where $G\sim -1/(2\pi R)$. This is why the bounded-domain condition on the boundary flux cannot be applied to the infinite plane alone.

With zero interior source and the given point boundary source, the boundary representation is $\Phi (x_0)=-G((0,0,0),x_0)$. The terms at infinity vanish for this resulting $1/R$ field. ThereforeTo check the singular boundary condition, $-{\partial }_z\Phi =z/[2\pi (x^2+y^2+z^2{)}^{3/2}]$ is the Poisson kernel for the upper half-space. Its integral over a horizontal plane is one and its mass concentrates at the origin as $z\downarrow 0$, giving ${\partial }_n\Phi =\delta (x)\delta (y)$ in the distributional sense. This is the Neumann point source on a half-space boundary.

Choose the Green function convention ${\Delta }_{\mathbf{r}}G(\mathbf{r};{\mathbf{r}}_0)={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$, with decay at infinity and no extra singularities. For sufficiently decaying source data,Away from its source, this fundamental solution of the Laplace equation is harmonic. Its radial derivative is $1/(4\pi {\rho }^2)$, so the flux of its gradient through a small source-centred sphere is one. The divergence theorem therefore supplies exactly the unit Dirac delta function required by the equation. Decay at infinity excludes an added globally regular harmonic term.

For the half-space, reflect the source to ${\mathbf{r}}_0^{\ast }=(-x_0,y_0,z_0)$. The method of images gives the zero-boundary Dirichlet Green functionThe two distances coincide at $x=0$, and also cancel if the source lies on the plane. The image singularity lies outside the physical domain.

The sign conventions for a Dirichlet Green function require care here. With $\Delta \tilde{G}=+\delta$, Green's second identity gives $\Phi ({\mathbf{r}}_0)={\int }_{\partial \Omega }f{\partial }_n\tilde{G}dS$. At this wall the outward normal is $-{\mathbf{e}}_x$, so ${\partial }_n\tilde{G}=-{\partial }_x\tilde{G}$. Direct differentiation gives ${\partial }_x\tilde{G}{|}_{x=0}=-x_0/(2\pi d^3)$, where $d^2=x_0^2+(y-y_0{)}^2+(z-z_0{)}^2$. Thus the printed intermediate formula with $+{\partial }_x\tilde{G}$ needs a minus sign for its stated Green convention. The final Poisson kernel for the upper half-space is positive:Its boundary integral is one for constant data, providing an independent sign check.

Reflect $r^{\prime }=(x^{\prime },y^{\prime },z^{\prime })$ to $r^{\prime \ast }=(x^{\prime },y^{\prime },-z^{\prime })$. The method of images gives the Dirichlet image kernel for the three-dimensional Helmholtz equation:The two distances coincide on the plane, so its boundary value is zero; the image source is outside the domain. It therefore has the required positive delta source inside and the selected outgoing behaviour.

Let $h(R)=e^{ikR}/R$. At a boundary point $r=(\xi ,\eta ,0)$, the distances to the real and reflected sources have $z$ derivatives $-z^{\prime }/R$ and $+z^{\prime }/R$. ThusThe outward unit normal of the upper half-space is $-e_z$, and $z^{\prime }/R=\cos \theta$. HenceUsing the decay assumption to discard the distant hemisphere in the boundary representation givesThe qualification at infinity concerns the whole Green boundary term, not just the pointwise limit of $\varphi$. Sufficient decay of the field and its derivatives, as assumed, makes it vanish; compatible radiation conditions can also do so. As a sign check, taking $k\to 0$ in the kernel gives $z^{\prime }/(2\pi R^3)$, the positive Poisson kernel for the upper half-space.

Let the evaluation point be ${\mathbf{r}}_0=(x,y,z)$, $z>0$, and reflect it to ${\mathbf{r}}_0^{\ast }=(x,y,-z)$. The method of images constructsThe two Euclidean distances coincide on the plane, so $G=0$ there. The negative image source lies outside the domain; inside, only the real source contributes to $-{\nabla }_{\mathbf{r}}^{2}G={\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)$. The difference also decays at infinity.

On the plane write $\mathbf{r}=(\xi ,\eta ,s)$ and set $s=0$. The outward unit normal of the upper half-space is $n=-e_z$. Direct differentiation givesSubstituting into the boundary representation from part (c) yields the Poisson kernel for the upper half-space:Its sign is positive, as expected from the maximum principle for nonnegative boundary data. Its normalization can be checked in polar coordinates:As $z\downarrow 0$ the kernel is an approximate identity, recovering $f$ at its continuity points. Boundary data must have the decay and regularity compatible with the stated condition at infinity.

On the axis, the Poisson kernel for the upper half-space depends only on the radial boundary coordinate. Integrating over the disk givesTherefore the axial Poisson extension of a circular boundary patch isIt tends to one as the axis approaches the boundary inside the patch, and tends to zero as $a^2/(2z^2)$ far above it.

The boundary values jump at the circular rim. They are understood almost everywhere, or as limits at continuity points; changing the assigned value on that measure-zero circle does not change the integral. At a rim point the vertical limit is one-half, rather than a continuous extension with the specified value on both sides of the jump.

A Poisson kernel represents the harmonic extension of Dirichlet boundary data by an integral over the boundary. When the Dirichlet Green function is normalized by $-\Delta G_{\Omega }=\delta$, its negative outward normal derivative gives the displayed kernel. The Poisson kernel for the upper half-space and the Poisson kernel for the upper half-plane give explicit examples with different boundary dimensions. Boundedness or other stated growth conditions are needed when selecting a solution in an unbounded domain.

In three dimensions the Poisson kernel for the upper half-space isIt gives the harmonic function $\varphi (x,y,z)={\int }_{{\mathbb{R}}^2}{P}_z(x-\xi ,y-\eta )f(\xi ,\eta )d\xi d\eta$ for compatible boundary data. The method of images constructs the zero-boundary Dirichlet Green function by subtracting the reflected-source Green function of the Laplacian from the source kernel. On the boundary plane the outward unit normal is $-e_z$, and $-{\partial }_{n}G=P_z$, giving the integral formula through Green's second identity.

The kernel is positive and has integral one: in polar coordinates this is $z{\int }_0^{\infty }r(r^2+z^2{)}^{-3/2}dr=1$. Its mass outside any fixed disk tends to zero as $z\downarrow 0$. It is therefore an approximate identity, recovering bounded boundary data at continuity points. At a jump across a smooth curve, the vertical limit gives the average of the two adjacent values; assigning a different value only on that curve does not change the harmonic extension.

The Poisson kernel for the upper half-space gives an isometric bijection from finite real or complex Borel measures, with their total variation norm of a measure, onto the harmonic Hardy space of the upper half-space. The kernel has unit mass, so Tonelli theorem gives the upper norm bound. Its approximate identity property gives weak-star convergence to the original measure and hence the reverse norm bound. For surjectivity, the mean value property for harmonic functions bounds each shifted harmonic function; bounded harmonic uniqueness gives $u(\cdot ,s+t)=P_t\ast u(\cdot ,s)$. A weak-star limit of the uniformly bounded measures $u(\cdot ,s)dx$ then gives the required boundary measure. In particular $|u(x,t)|\le c_d{t}^{-d}\Vert u{\Vert }_{h^1}$, where $c_d=\Gamma ((d+1)/2)/{\pi }^{(d+1)/2}$.

If $k(x)=\varphi (|x|)\ge 0$ is decreasing in radius and has integral one, then $|k_t\ast f(x)|\le Mf(x)$, where $k_t(x)=t^{-d}k(x/t)$ and $M$ is the centered Hardy-Littlewood maximal function of $|f|dx$. The layer cake representation expresses $k_t$ as a positive mixture of ball indicators; each corresponding integral is bounded by its ball volume times $Mf(x)$. Integrating these volumes gives exactly one. This applies to the Poisson kernel for the upper half-space and supplies a weak type (1,1) domination for its boundary convergence.