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For an annulus $a<r<1$ with $0<a<1$, these factors interpolate the two boundary coefficients of a Fourier harmonic of positive integer order $n$ in a harmonic function. They obey $H_n(1)=1$, $H_n(a)=0$, $K_n(1)=0$, $K_n(a)=1$. Thus a mode with outer coefficient $v$ and inner coefficient $w$ has radial factor $v{H}_n+w{K}_n$.

To derive them, separation of variables in Laplace's equation in polar coordinates gives $r^2{R}^{\prime \prime }+r{R}^{\prime }-n^{2}R=0$, whose solutions are $r^n,r^{-n}$. Imposing the two Dirichlet boundary conditions on $R=\alpha r^n+\beta r^{-n}$ gives $\alpha +\beta =v$ and $a^n\alpha +a^{-n}\beta =w$. Solving yields $\alpha =(v-a^{n}w)/(1-a^{2n})$ and $\beta =a^n(w-a^{n}v)/(1-a^{2n})$, proving the formula. For the zero angular mode, the analogous factor is $v+(w-v)\log r/\log a$, since its radial solutions are $1$ and the logarithm $\log r$.

For boundary data $f(u)$ on the sphere $r=a$, with $u=\cos \theta$, define its Legendre polynomial coefficients by $f_{\ell }=(2\ell +1){\int }_{-1}^{1}f(u)P_{\ell }(u)du/2$. Separation of variables in the Laplace equation gives radial factors $r^{\ell }$ and $r^{-\ell -1}$. Regularity at the origin selects the first for the interior Dirichlet problem; decay at infinity selects the second for the exterior problem. Matching each boundary coefficient gives the displayed series. For continuous boundary data the resulting harmonic functions attain those data, with convergence interpreted through their harmonic extension rather than requiring uniform convergence of the unweighted boundary series. The maximum principle for harmonic functions proves uniqueness in the bounded interior and decaying exterior classes.

A compactness proof of the interior Schauder estimate can magnify a pair of points where the Hölder seminorm of the Hessian matrix is nearly attained. If the proposed estimate fails, normalize that seminorm to one while lower norms and forcing tend to zero. The separation $r_j$ of the chosen pair tends to zero. Subtract the quadratic Taylor polynomial and divide by $r_j^{2+\alpha }$ after rescaling space by $r_j$.

The resulting functions have uniformly bounded Hessian Hölder seminorms on expanding domains and vanishing Taylor data at zero. Compact embedding of Hölder spaces yields an entire harmonic function whose second derivatives have finite global $\alpha$-Hölder seminorms, with $0<\alpha <1$. The polynomial-growth Liouville theorem for harmonic functions makes those derivatives constant, contradicting the normalized nonzero oscillation. The Simon absorption lemma removes the residual small multiple of the larger-ball seminorm.

Although $1/r$ is harmonic away from the origin in three dimensions, it is not a regular harmonic function on a ball containing the origin. Its gradient is $-\hat{r}/r^2$, so the displayed improper volume integral follows by canceling the $r^2$ in the spherical volume element. On the annulus $\epsilon <r<a$, Green's first identity gives outer boundary contribution $-4\pi$ and inner contribution $4\pi \epsilon /a$, recovering the same limit. Its global Laplacian contains a Dirac delta distribution, so the regular-harmonic gradient-pairing theorem is inapplicable.

Two bounded harmonic functions continuous to the quadrant edges with the same Dirichlet boundary conditions are equal. Their difference is bounded and zero on both axes. Odd reflection across each axis gives a bounded harmonic extension away from the corner; the removable singularity for a bounded harmonic function fills in that corner. The harmonic Liouville theorem makes that extension constant, and oddness makes the constant zero. This does not exclude unbounded examples such as the zero-boundary harmonic growth in a quadrant.

For the exterior of the unit disc in two dimensions, with the sign convention ${\Delta }_{x}G=+\delta (x-y)$, this is the Dirichlet Green function that vanishes on the circle and is bounded at infinity. The method of images places an opposite logarithmic source at the inverse point $y/|y{|}^2$. The constant $-(2\pi {)}^{-1}\log |y|$ enforces the zero boundary value. The inversion in a circle identity $|x-y|=|y||x-y/|y{|}^2|$ for $|x|=1$ verifies that condition directly.

The outward normal of the exterior domain is $-x$ on the unit circle. Its normal derivative gives the positive exterior Poisson kernelFor bounded harmonic functions this kernel integrates boundary data to the interior value. Boundedness at infinity is essential to uniqueness: without it, a multiple of $\log |y|$ can be added without changing the circle's Dirichlet boundary condition.

If $u$ is a bounded harmonic function on a domain $D$, continuous on its closure, and the Brownian exit time $T_D$ is almost surely finite from $z$, then $u(z)={\mathbb{E}}_z[u(B_{T_D})]$. The Itô formula and compact localization make $u(B_{t\wedge T_D})$ a bounded martingale. bounded convergence theorem at the exit time proves the representation even if the domain is unbounded. Removing boundedness of $u$ can invalidate this conclusion through loss of uniform integrability.

For $d\ge 3$ and $|x|>\epsilon$, stop the harmonic function $u(x)=|x{|}^{2-d}$ at the two boundaries of $\{\epsilon <|x|<R\}$. Bounded optional stopping gives the probability of hitting the inner sphere first as $(u(x)-u(R))/(u(\epsilon )-u(R))$. Letting $R\to \infty$ gives the displayed formula; a finite hitting path is contained in some finite-radius ball.

If a nonconstant complex polynomial $P$ had no zeros, its reciprocal $g=1/P$ would be an entire holomorphic function, bounded and tending to zero at infinity. Its real and imaginary parts are harmonic functions. Thus $g(z+B_t)$, for planar Brownian motion started at zero, is a bounded complex martingale with expectation $g(z)$. For fixed $R$, the Gaussian density bounds $\mathbb{P}(|z+B_t|\le R)$ by $R^2/(2t)$. Outside a sufficiently large disk, $|g|\le \epsilon$. Hence $|g(z)|\le \epsilon +\Vert g{\Vert }_{\infty }{R}^2/(2t)$. Let $t\to \infty$ and then $\epsilon \downarrow 0$ to obtain $g(z)=0$, a contradiction. This proves the fundamental theorem of algebra without the maximum modulus principle.

For standard three-dimensional Brownian motion started at $x$ with $r<|x|<R$,The harmonic function $x\mapsto 1/|x|$ gives a bounded stopped martingale in the annulus, so the optional stopping theorem gives the formula. Taking $R\to \infty$ yields ${\mathbb{P}}_x({\tau }_r<\infty )=r/|x|$ for $|x|>r$. This limit uses path continuity on the finite interval before a hit, rather than assuming transience to prove transience.

Every point in the convex hull of a subset of ${\mathbb{R}}^d$ is a convex combination of at most $d+1$ points of that subset. Thus in the plane three ingredients always suffice, however many samples were supplied. The representation of $p=(2,1)$ in convex hull illustrates this bound. Two ingredients need not suffice: the centroid of a triangle with only its three vertices as input is on none of their three pairwise line segments.

A short proof outline removes redundant ingredients. With more than $d+1$ points, the augmented vectors $(s_i,1)$ are linearly dependent. A dependence lets the weights vary while preserving both their sum and the represented point. Move along it until one positive weight first becomes zero; all weights stay nonnegative. Repeating leaves at most $d+1$ positive weights.

In his 1911 paper, Constantin Carathéodory used convex geometry to investigate the possible Fourier coefficients of positive harmonic functions. Its introduction interprets convex mixtures through the center of mass, connecting finite-dimensional geometry to constraints on a function's coefficients.^[ref]

A locally uniformly bounded sequence of harmonic functions with a pointwise limit converges uniformly on compact subsets, and its limit is harmonic. A radial mollifier of fixed local radius reproduces each harmonic function by the mean value property for harmonic functions. Differentiating that convolution gives uniform interior gradient bounds, hence equicontinuity. A finite covering of a compact set converts pointwise convergence to uniform convergence. Passing to the limit in the radial convolution identity proves smoothness; passing to the limit against the Laplacian of each test function proves harmonicity.

If $u,v$ are harmonic functions on a bounded domain and continuous on its closure, their difference is a harmonic function. Its minimum is attained on the boundary by the maximum principle for harmonic functions applied to its negative. Nonnegative boundary difference therefore implies nonnegative interior difference. Applying the comparison in both directions proves uniqueness for the Dirichlet problem whenever a classical solution exists. This is an order statement and does not itself establish existence.

A vector field in Euclidean space is componentwise harmonic when each Cartesian component is a harmonic function, equivalently its componentwise Laplacian vanishes. Constant and linear vector fields are examples, as are constant linear combinations of derivatives of $1/r$ away from the origin in three dimensions. Such fields supply the vector potentials in the Papkovich–Neuber representation. Componentwise harmonicity does not by itself require zero divergence or zero curl.

For a conformal map extending to the relevant boundary sets, compose the target harmonic measure with the map. The planar Laplacian transforms by the positive factor $|f^{\prime }{|}^2$, so the composition is a harmonic function with the correct boundary values. Uniqueness of the Dirichlet problem gives the displayed identity on regular domains, including the smooth-circle domains in the harmonic measure of an annulus.

The principal square root maps the upper $Z$ half-plane bijectively onto $\Re z>0$ with the ray $[1,\infty )$ removed. The prevertex zero is the slit tip, and $\pm 1$ represent the two ends at infinity. The inverse is the unique root $Z^2=1-z^{-2}$ with positive imaginary part. Pulling back ${\pi }^{-1}\arg ((Z-1)/(Z+1))$ gives the bounded harmonic function with value one on the slit banks and zero on the imaginary axis. The harmonic function $\Im [Z/(1-Z^2)]$ has zero finite boundary values but grows at infinity, explaining the role of boundedness.

Let a continuous function satisfy both viscosity inequalities for the Laplace operator. On any relatively compact ball, take its harmonic replacement $h$ with matching boundary values. If $u-h$ has a positive maximum, the function $h+\epsilon (R^2-|x-a{|}^2)$ still lies below $u$ somewhere for sufficiently small positive $\epsilon$, while matching it on the boundary. At an interior maximum of their difference it is a valid upper test, but its Laplacian is $-2n\epsilon <0$, a contradiction. The opposite perturbation excludes a negative minimum. Thus $u=h$ throughout the ball and is a smooth harmonic function. A locally defined test is extended near its touching point using a smooth cutoff function when global tests are required.

With the convention ${\Delta }_z{G}_D=\delta (z-\zeta )$, the displayed Dirichlet Green function vanishes on the unit circle. Indeed $|z-\zeta |=|1-z\bar{\zeta }|$ when $|z|=1$. For $\zeta \ne 0$, the second logarithm corresponds to an opposite image source at $\zeta /|\zeta {|}^2$, outside the unit disc; it is a harmonic function throughout the interior. At $\zeta =0$ the formula becomes $(2\pi {)}^{-1}\log |z|$. Its outward normal derivative in the boundary variable is $(2\pi {)}^{-1}(1-|\zeta {|}^2)/|z-\zeta {|}^2$, the Poisson kernel divided by $2\pi$. Green's second identity therefore gives the Poisson integral on the unit disk.

The conformal map $w=z^2$ sends $x,y>0$ to the complex upper half-plane. For bounded continuous boundary values $g_2$ on the horizontal axis and $g_1$ on the vertical axis with equal corner values, set $f(s)=g_2(\sqrt{s})$ for $s>0$ and $f(s)=g_1(\sqrt{-s})$ for $s<0$. The bounded harmonic function with those Dirichlet boundary conditions isEquivalently, if $z^2=a+ic$ with $c>0$, this is $\frac{2c}{\pi }{\int }_0^{\infty }r\left[g_2(r)/((r^2-a{)}^2+c^2)+g_1(r)/((r^2+a{)}^2+c^2)\right]dr$. The Poisson kernel is a positive approximate identity, so the prescribed edge values are attained. Boundedness is essential for uniqueness on this unbounded domain.

For a harmonic function between $\tau =a$ and $\tau =b$, $0<a<b$, in bipolar coordinates, suppose $\Phi (\sigma ,a)=0$ and $\Phi (\sigma ,b)=F(\sigma )$. Write the periodic boundary values as a Fourier seriesSeparation of variables givesEach radial factor solves $M^{\prime \prime }-n^{2}M=0$ and has endpoint values zero and one; the mean mode solves $M^{\prime \prime }=0$. Sufficiently regular boundary data give the required Fourier series solution. Uniqueness follows from the maximum principle for a harmonic function with zero Dirichlet boundary conditions on both circles. The formula follows the coordinate values; $\tau =b$ is the geometrically smaller circle.