Harmonic function

A harmonic function satisfies Laplace's equation $\Delta u=0$.

If $f:U\to V$ is a holomorphic function between planar open sets and $\psi :V\to \mathbb{R}$ is a harmonic function, then $\psi \circ f$ is harmonic. In fact the chain rule and the Cauchy-Riemann equations, with $f=u+iv$, giveHere the two gradients are orthogonal and have equal length, while $u,v$ are harmonic. The formula holds even where $f^{\prime }=0$. A biholomorphism therefore induces inverse correspondences between harmonic functions by function composition with it and its inverse.

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 $\log r$.

The locally uniform convergence limit of real harmonic functions on a planar domain is harmonic. On a smaller circle, use the Schwarz integral on the unit disk, rescaled to that radius, to construct holomorphic functions whose real parts are the given functions. Uniform convergence of boundary values gives locally uniform convergence of these integrals. Their limit is holomorphic by the Weierstrass convergence theorem, and its real part is the original limit. Applying this construction locally at every point proves the claim without assuming derivative convergence.

On $0<x<a$, $0<y<b$, the bounded harmonic function with value $F$ on the open upper edge and zero on the other three open edges has the sine-hyperbolic series in the title. Separation of variables gives the modes, and the Fourier sine series of a constant selects only odd $n$. The series converges exponentially on compact subsets of the interior, justifying its derivatives in Laplace's equation.

For $F\ne 0$, the upper corners have incompatible one-sided boundary limits, so the solution cannot extend continuously to them. Assigning isolated corner values does not alter the bounded interior Dirichlet problem. For numerical evaluation, the stable mode ratio isThis avoids overflow from separately evaluating large hyperbolic sines.

Suppose the inner and outer boundary values on spheres of radii $0<a<b$ have convergent Legendre polynomial expansions with coefficients $u_n,v_n$. For each degree set $F_n(r)=r^n-a^{2n+1}{r}^{-n-1}$ and $H_n(r)=r^n-b^{2n+1}{r}^{-n-1}$. ThenBoth radial terms solve $(r^2{R}_n^{\prime }{)}^{\prime }=n(n+1)R_n$, so the displayed angular expansion is harmonic wherever the series and required derivatives converge. The two radial factors take endpoint values $(1,0)$ and $(0,1)$, respectively, proving the boundary matching. For continuous data with the usual harmonic-extension convergence, the maximum principle for harmonic functions gives uniqueness. The construction separates the angular orthogonality from the two-boundary radial interpolation.

For a planar harmonic function, $u_{yy}=-u_{xx}$. Its Hessian matrix is therefore $\left(\begin{array}{cc}a& b\\ b& -a\end{array}\right)$, with eigenvalues $\pm \sqrt{a^2+b^2}$. At a stationary point with nonzero Hessian matrix determinant, both signs occur, so the point is a saddle point. A zero determinant requires separate treatment and is not covered by this nondegenerate test. The result is the local quadratic counterpart of the maximum principle for harmonic functions.

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.

A harmonic function on one side of a hyperplane that is continuous up to an open portion of the hyperplane and zero there extends harmonically by odd reflection. One local proof takes a ball centered on the boundary and assigns odd continuous boundary data on its sphere. The Poisson integral on the ball is odd, so it vanishes on the equatorial disk. Uniqueness by the maximum principle for harmonic functions identifies it with the original function on the half-ball. Thus the odd extension is harmonic near each boundary point. A globally bounded zero-boundary harmonic function on a half-space is consequently zero by the Liouville theorem for harmonic functions.

For the upper half-space $H^{d+1}={\mathbb{R}}^d\times (0,\infty )$, this is the space of harmonic functions with finite norm $\Vert u{\Vert }_{h^1}={\sup }_{t>0}{\int }_{{\mathbb{R}}^d}|u(x,t)|dx$. It differs from an analytic Hardy space: no holomorphic function condition is imposed, and its boundary object can be a singular finite Borel measure.

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 $c_d=\Gamma ((d+1)/2)/{\pi }^{(d+1)/2}$.

Away from the origin, a dipole potential $(d\cdot x)/|x{|}^3$ is harmonic. Its gradient is angularly nonuniform, decays as $r^{-3}$, and has zero spherical mean because $⟨n\otimes n⟩=I/3$. That cancellation distinguishes a self-dipole field from the regular incident field that excites a small inclusion.

In three-dimensional Euclidean space, spherical inversion together with the displayed weight maps harmonic functions to harmonic functions on the inverted domain. The Laplacian satisfies $\Delta K_{a}u(\mathbf{x})=(a/r{)}^5(\Delta u)(a^2\mathbf{x}/r^2)$. The formula is for $r>0$; a new singularity may occur at the origin.

For $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$, the chain rule gives ${\nabla }_{\mathbf{r}}=(a^2/r^2)(I-2\hat{\mathbf{r}}{\hat{\mathbf{r}}}^T){\nabla }_{\overline{\mathbf{r}}}$. The radial term is killed by crossing with $\mathbf{r}$, giving the identity. Purely angular vector spherical harmonics are therefore unchanged as angular basis functions. This is distinct from pushing forward a physical vector field by the inversion Jacobian.

For a $C^2$ harmonic function $u$ and Brownian motion stopped before leaving its domain, Itô formula gives $du(B_t)=\nabla u(B_t)\cdot d{B}_t+\frac{1}{2}\Delta u(B_t)dt$. The drift vanishes. On a compact subdomain, the stopped process is a bounded martingale. Optional stopping theorem at an almost-surely finite exit time therefore gives the boundary mean formula. At the centre of a disk, rotational invariance of planar Brownian motion makes the exit law uniform on the boundary circle. This proves the mean value property for harmonic functions probabilistically.

A smooth real function on a complex manifold is pluriharmonic if $\partial \bar{\partial }f=0$. A complex-valued function satisfies the same equation exactly when both real and imaginary parts are pluriharmonic. Restrictions to complex coordinate lines are harmonic functions. On a compact connected complex manifold, choose a maximum of $|f|$ and apply the harmonic maximum principle successively along coordinate lines; the function is constant on a polydisc, then on the manifold by open-and-closed propagation. Without connectedness, constancy is componentwise.

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.

The disk Poisson integral is the harmonic function with boundary data $u$, given byIt is the real part of the Schwarz integral on the unit disk for real data. The nonnegative kernel has integral one; smooth boundary data extend smoothly to the circle. This disk formula should be distinguished from the half-plane formula.

For $F\in L^1(\partial \mathbb{D})$ and a Lebesgue point ${\xi }_0$, its Poisson integral tends to $F({\xi }_0)$ through every fixed Stolz region. If $\delta =1-|z|$, the Poisson kernel there is at most $C_A\delta /({\delta }^2+d(\xi ,{\xi }_0{)}^2)$, where $d$ is shorter angular distance. Dyadic annuli about ${\xi }_0$ bound the near contribution by small local averages of $|F-F({\xi }_0)|$ times a summable sequence $C_A{2}^{-k}$. The contribution outside a fixed small arc is $O(\delta )\Vert F-F({\xi }_0){\Vert }_1$ and vanishes. This is the local kernel argument underlying the Fatou theorem for bounded holomorphic functions.

For $0<r<1$, the geometric series gives $P_r(t)=1+2{\sum }_{n\ge 1}{r}^n\cos nt$. Hence this positive even kernel has normalized integral one, and normalized convolution with it multiplies the $n$th Fourier coefficient by $r^{|n|}$. Its mass outside any fixed neighbourhood of zero tends to zero as $r\uparrow 1$. If an integrable function has finite one-sided limits $A,B$ at zero, evenness gives half the mass to each side and the convolution tends to $(A+B)/2$.

The Poisson kernel on the circle is nonnegative, has normalized integral one, and its mass outside any fixed arc about zero tends to zero as $r\uparrow 1$. Split the convolution error into a short arc, controlled by the modulus of continuity, and its complement, controlled by the vanishing kernel mass. This proves uniform convergence for every continuous periodic function. Its Fourier multiplier is $r^{|n|}$, so the convolution is the Abel sum of the Fourier series.

The geometric series gives $Q_r(t)=2{\sum }_{n\ge 1}{r}^n\sin nt$. Its convolution multiplier is $-i\mathrm{sgn}(n)r^{|n|}$ for nonzero frequencies and zero at frequency zero. Since the series is uniformly absolutely convergent for $r<1$, these identities hold for every Lebesgue integrable function by termwise integration. If all negative Fourier coefficients vanish, $Q_r\ast f=-i(P_r\ast f-\hat{f}(0))$.

Let an integrable function have finite one-sided limits $A,B$ at zero. Oddness gives $(Q_r\ast f)(0)=(2\pi {)}^{-1}{\int }_0^{\pi }{Q}_r(t)[f(-t)-f(t)]dt$. Also ${\int }_0^{\delta }{Q}_r(t)dt=\log [(1-2r\cos \delta +r^2)/(1-r{)}^2]=2\log (1/(1-r))+O_{\delta }(1)$. Thus $(Q_r\ast f)(0)/\log (1/(1-r))\to (A-B)/\pi$: make the near-zero error in $f(-t)-f(t)$ arbitrarily small and bound the remaining integral uniformly. When negative Fourier coefficients vanish, the conjugate-kernel identity makes this convolution bounded because $P_r\ast f$ tends to $(A+B)/2$. Therefore $A=B$. A removable discontinuity caused solely by altering a point value is not excluded.

For $r<1$, average the real part of the convergent power series for $\log (1+r{e}^{i\theta })$: every nonconstant Fourier mode vanishes. For $r>1$, factor out $r$ and apply the same argument to $r^{-1}$. The boundary singularity at $r=1$ is logarithmically integrable. This identity computes the expected logarithmic distance after an isotropic Gaussian displacement.

For a smooth harmonic function on a manifold without boundary, assume its nonzero superlevel sets are compact. On a regular positive superlevel set $\{U>\epsilon \}$, integration by parts gives $\int |\nabla U{|}^2=-\epsilon {\int }_{\partial \{U>\epsilon \}}|\nabla U|\le 0$. Applying the same argument to $-U$ proves $U=0$. This avoids assuming an unstated decay rate for the boundary flux at infinity.

An element $h\in H_{\mathrm{loc}}^1(U)$ is weakly harmonic on an open set $U$ when ${\int }_U\nabla h\cdot \nabla \varphi =0$ for every test function $\varphi \in C_c^{\infty }(U)$. This is the statement $\Delta h=0$ in distributions. It requires no regularity of the domain boundary. The Weyl lemma, a form of elliptic regularity, identifies such an element with a smooth harmonic function.

A real function $h$ on the vertices of a graph of an unweighted locally finite graph is a discrete harmonic function at $u$ when $\mathrm{deg}(u)h(u)={\sum }_{v\sim u}h(v)$, equivalently $Lh(u)=0$ for the Graph Laplacian. For positive degree this says its value is the average over its graph neighbours; at an isolated vertex of a graph the Laplacian condition is vacuous. On an electrical network the corresponding equation uses conductance weights and the Weighted graph Laplacian. Voltages away from sources and sinks, and hitting probabilities away from their prescribed boundaries, are discrete harmonic functions.

Every bounded harmonic function on a graph is constant if the connected locally finite graph is recurrent. Along its random walk on a graph, the harmonic values form a bounded martingale and converge almost surely. Recurrence visits each vertex infinitely often, making every vertex value a subsequential limit of the same convergent sequence. Thus every value agrees.

A real function $h$ on a finite connected graph satisfying $Lh=0$ everywhere is constant. At a maximum graph vertex, the identity $\mathrm{deg}(u)h(u)={\sum }_{v\sim u}h(v)$ forces every graph neighbour to have the same maximum. Connectedness propagates it to every graph vertex. This proves uniqueness of voltage up to an additive constant, and identifies sums of oppositely directed occupation voltages.

Every bounded discrete harmonic function on ${\mathbb{Z}}^d$ is constant. After rescaling its range to $[0,1]$, place it in the compact convex set of all such functions; translation invariance and the mean-value identity force every extreme point to be constant, and the Krein-Milman theorem finishes the proof.

A harmonic function on a punctured ball that is bounded near the missing point extends uniquely to a harmonic function on the whole ball. The extension can be defined by the mean value property for harmonic functions; interior estimates then give smoothness across the point.

The harmonic replacement of $u$ in a domain $D$ is the harmonic function $w$ having the same boundary trace as $u$. Thus $u-w\in H_0^1(D)$, and subtracting the weak equations permits local energy estimates for the error $u-w$.

For every closed ball contained in the domain of a harmonic function, the value at its centre equals both its average over the ball and its average over the boundary sphere.

If $u\ge 0$ is a harmonic function on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, use $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$ and the mean value property for harmonic functions. Nonnegativity lets us enlarge the integral, and the ratio of volumes is $3^n$. This compares values without requiring derivative bounds.

For a nonnegative harmonic function on the unit disc, the displayed inequalities are sharp. Apply the Poisson integral on the unit disk on radius $R>|z|$, bound its positive kernel between $(R-|z|)/(R+|z|)$ and its reciprocal, and use the mean value property for harmonic functions. Letting $R\uparrow 1$ proves the assertion. Positive Poisson kernels with one boundary pole give equality in the appropriate radial directions.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.

If a continuous function has the spherical mean value property at every point for some sequence of radii decreasing to zero, then it is harmonic. Comparison on each relatively compact ball with the harmonic function having the same boundary data proves the result by propagating any hypothetical interior extremum along one of the admissible spheres.

The Weyl lemma says that every locally integrable distributional solution of $\Delta u=0$ agrees almost everywhere with a smooth harmonic function. Convolution with a mollifier and the harmonic mean value property provide a standard proof.

If $u$ is harmonic on $B_r(x_0)$, then for every integer $m\ge 1$,The first-order estimate follows by differentiating the ball mean-value formula and applying the divergence theorem; iteration gives higher orders.

If an entire harmonic function on ${\mathbb{R}}^n$ grows like $O(1+|x{|}^{k+\alpha })$ with $k\in \mathbb{N}$ and $0<\alpha <1$, then it is a harmonic polynomial of degree at most $k$. Apply the interior derivative estimate of order $k+1$ on expanding balls.

A harmonic function on all of ${\mathbb{R}}^n$ with a finite global Hölder seminorm of exponent $0<\mu <1$ is constant. Indeed $|w(x)-w(0)|\le C|x{|}^{\mu }$, and the interior derivative estimate for a harmonic function on balls of radius $R$ bounds $|\nabla w(0)|$ by $C{R}^{\mu -1}$. Letting $R\to \infty$ makes this gradient zero, and translating the centre proves the assertion everywhere.

A real-valued function $u$ on a simply connected domain is harmonic exactly whenfor some holomorphic function $f$. One construction observes that $u_x-i{u}_y$ is holomorphic and takes its holomorphic primitive.

A twice differentiable function is strictly subharmonic where $\Delta u>0$. It cannot have an interior local maximum because the Hessian at such a maximum is negative semidefinite.

For every holomorphic function and $p>0$, $|f{|}^p$ is subharmonic. Away from its zeros the displayed Laplacian calculation proves this. At the zeros, use the smooth approximations $(|f{|}^2+\epsilon {)}^{p/2}$, whose Laplacians are nonnegative, and let $\epsilon \downarrow 0$ in their mean inequalities. This permits Poisson integral comparison for fractional modulus powers even when a global analytic fractional power does not exist.

Write a holomorphic function as $f=u+iv$. The Cauchy-Riemann equations make $u,v$ harmonic functions and give $u_x=v_y$, $u_y=-v_x$. The product rule therefore gives $\Delta (u^2+v^2)=2(|\nabla u{|}^2+|\nabla v{|}^2)=4|f^{\prime }{|}^2$. Thus its squared modulus is subharmonic and is harmonic only where $f^{\prime }$ vanishes.

A harmonic majorant of a real function $u$ is a real harmonic function $h$ with $h\ge u$ pointwise. A least harmonic majorant lies below every other harmonic majorant. A continuous subharmonic function need not have one: $u(z)=(1-|z{|}^2{)}^{-1}$ on the unit disk has increasing circular means diverging to infinity, whereas any harmonic majorant would have the same finite mean on every centered circle.

For continuous subharmonic function $u$ on the unit disk, let $H_r$ be the Poisson integral of its values on $|z|=r$, evaluated inside that disk. Harmonic comparison gives $H_s\ge H_r\ge u$ for $|z|<r<s<1$. The Harnack inequality for harmonic functions makes their limit either finite and harmonic everywhere or everywhere infinite. In the finite case every harmonic majorant dominates every $H_r$, so their limit is the least harmonic majorant. The supremum must restrict $r>|z|$; the Poisson expression evaluated outside its circle does not have this interpretation.

On a relatively compact coordinate disk $V$ in a Riemann surface, replace a continuous subharmonic function $u$ inside $V$ by the Poisson integral of its boundary values, leaving it unchanged outside. The resulting function $L_{V}u$ is continuous and subharmonic, equals $u$ off $V$, is harmonic on $V$, and satisfies $L_{V}u\ge u$. The maximum principle for subharmonic functions proves the inequality. The pasting assertion follows from harmonic comparison: a harmonic comparison function dominating the pasted function on the boundary of a test domain first dominates $u$, then dominates the disk lift on its part inside $V$.

For a countable family of subharmonic functions, a locally integrable pointwise supremum is subharmonic in the distributional derivative sense. Finite maxima preserve subharmonicity: approximate the maximum by the smooth convex coordinatewise increasing function $(s+t+\sqrt{(s-t{)}^2+{\delta }^2})/2$, apply the chain rule after mollification, then pass to a local L1 norm convergence limit. The maxima $g_N=\max (f_1,\dots ,f_N)$ increase to $g$ and are bounded in absolute value by $|f_1|+|g|$ on compact sets, so the dominated convergence theorem completes the proof. The hypothesis cannot be omitted: $f_j=j$ has infinite supremum. To obtain a pointwise subharmonic function, take the canonical representative of a subharmonic distribution.

A locally integrable real function is superharmonic in the distributional sense if its Laplacian is nonpositive on every nonnegative test function. Equivalently its negative is a subharmonic function. The canonical lower semicontinuous representative satisfies the reverse ball mean inequality. A function both subharmonic and superharmonic has zero distributional Laplacian and a smooth harmonic function representative by the Weyl lemma.

For $u\in C^2(D)\cap C(\overline{D})$ on a bounded domain in ${\mathbb{R}}^n$ with $\Delta u\le 0$, the minimum lies on the boundary. Define $u_{\epsilon }=u-\epsilon |x{|}^2$, so $\Delta u_{\epsilon }\le -2n\epsilon <0$. An interior minimum would have positive semidefinite Hessian matrix and nonnegative trace, a contradiction. Thus $u_{\epsilon }$ attains its minimum on the boundary. Boundedness controls $|u-u_{\epsilon }|$ uniformly by a constant times $\epsilon$; letting $\epsilon \downarrow 0$ proves the weak minimum principle for the original superharmonic function. The perturbation handles the non-strict inequality, including constant functions.

For locally integrable real $u$ with nonnegative distributional derivative $\Delta u$, its ball means increase with radius. To see this, mollify: for a smooth subharmonic function the derivative of its spherical mean is the integral of $\Delta u$ over the ball divided by the sphere area. The ball means inherit monotonicity; local $L^1$ convergence passes this to $u$. Their decreasing-radius limit exists, may be minus infinity at exceptional points, and equals $u$ almost everywhere by the Lebesgue differentiation theorem. It is upper semicontinuous as an infimum of continuous ball means and satisfies the mean-value inequality. Arbitrary changes of a representative on a null set can destroy its pointwise maximum principle.

A continuous subharmonic function on the closure of a bounded domain attains its maximum on the boundary. If it is twice differentiable, the strict condition $\Delta u>0$ rules out an interior maximum immediately because the Hessian matrix there would be negative semidefinite.

If the canonical representative of a subharmonic distribution on a connected domain attains a finite global maximum at an interior point, it is constant. The ball mean inequality forces equality almost everywhere on a small ball around the maximum. The canonical representative then equals the maximum everywhere on that ball. The maximum set is open by this argument and closed by upper semicontinuity; connectedness makes it the whole domain.

For a harmonic function $w$ smooth inside a bounded domain and $C^1$ on its closure, $\Delta |Dw{|}^2=2|D^{2}w{|}^2\ge 0$. The weak maximum principle for elliptic operators and continuity of the gradient imply ${\sup }_{\Omega }|Dw|={\sup }_{\partial \Omega }|Dw|$.

A bounded harmonic function on the complex plane is constant. More generally, an entire harmonic function bounded on either side is constant: write it as $\mathrm{Re}f$ by harmonic function as the real part of a holomorphic function, then apply the Liouville theorem to $e^f$ or $e^{-f}$.

A bounded entire harmonic function satisfies $u(x)={\mathbb{E}}_{x}u(B_t)$ because $u(B)$ is a bounded local martingale. The Gaussian heat kernel has a directional derivative with $L^1$ norm $\sqrt{2/\pi }/\sqrt{t}$. Consequently $|u(x)-u(y)|\le \Vert u{\Vert }_{\infty }\sqrt{2/\pi }|x-y|/\sqrt{t}$; letting $t\to \infty$ proves the harmonic Liouville theorem in every dimension.

A harmonic function on ${\mathbb{R}}^2\setminus \{0\}$ bounded on one side is constant. Shift and change sign to make it nonnegative. Its circular mean has the form $a\log r+b$; positivity for every $r>0$ forces $a=0$. For $|x|=r$, the mean value property for harmonic functions bounds its value by the mean integral over $r/2<|z|<3r/2$, giving at most $8b$. A removable singularity for a bounded harmonic function fills the origin, and the harmonic Liouville theorem finishes the proof. The dimensional distinction matters: $|x{|}^{2-n}$ is a positive nonconstant harmonic function on punctured ${\mathbb{R}}^n$ for $n\ge 3$.

For a bounded harmonic function $f$ on ${\mathbb{R}}^d$, the processes $f(B_t^x)$ and $f(B_t^y)$ are bounded martingales. Couple the Brownian motions from $x$ and $y$ so that they coalesce in finite time almost surely. Thenwhich proves that $f$ is constant in every dimension.

Every nonconstant harmonic function $u:\mathbb{C}\to \mathbb{R}$ is unbounded above and below by the harmonic Liouville theorem. The intermediate value theorem then gives $u(\mathbb{C})=\mathbb{R}$.

Let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function forBecause $2\pi i$ is a period, the formulais independent of the branch of the complex logarithm. It is a nonconstant harmonic function onand its other period gives $u(qz)=u(z)$.

A harmonic polynomial is a polynomial annihilated by the Laplacian.

If ${\mathcal{P}}_{\ell }$ is the space of degree-$\ell$ homogeneous polynomials and ${\mathcal{H}}_{\ell }$ its harmonic polynomials, then ${\mathcal{P}}_{\ell }={\mathcal{H}}_{\ell }\oplus |x{|}^2{\mathcal{P}}_{\ell -2}$. Under the Fischer inner product, the second summand is orthogonal to the kernel of the Laplacian. Iteration expresses every polynomial restricted to the unit sphere as a sum of spherical harmonics.

A harmonic conjugate v makes u+iv holomorphic; it exists globally on simply connected domains.

For a holomorphic function $f=u+iv$, the Cauchy-Riemann equations imply $\nabla u\cdot \nabla v=0$ and $|\nabla u|=|\nabla v|=|f^{\prime }|$. At a point where $f^{\prime }\ne 0$, the two regular level curves therefore meet at right angles, since each tangent is perpendicular to its nonzero gradient. The noncritical condition is necessary: for $f(z)=z^2$, the zero level sets are the coordinate axes and the two diagonals, which meet at angles of $\pi /4$ at the critical point zero. Orthogonality at regular points is one geometric expression of conformality.

For a harmonic function $u$, the function $g=u_x-i{u}_y$ is holomorphic. It is the derivative of a single-valued holomorphic function exactly when its integrals around all closed curves vanish. A primitive then has real part $u$ after adjusting a real constant. Simply connected domains satisfy this condition; on an annulus $u=\log |z|$ fails because the derivative has period $2\pi i$.

For $|z|>1$, both logarithms in the display can use their principal logarithms. The real part is $\log (|z+1{|}^2/|z-1{|}^2)$ and the derivative is $2/(z+1)-2/(z-1)$. The two residues cancel on a loop surrounding both points. Equivalently $f=4{\sum }_{m\ge 0}{z}^{-(2m+1)}/(2m+1)$ is globally single-valued outside the unit circle, although individual logarithms of $z+1$ and $z-1$ are not.

// Destination: complex-dynamics.bigb

If $u+iv$ and $p+iq$ are holomorphic functions, with the image of the first inside the domain of the second, then $p(u,v)+iq(u,v)$ is holomorphic by the chain rule. Directly, their Cauchy-Riemann equations imply $[p(u,v){]}_x=[q(u,v){]}_y$ and $[p(u,v){]}_y=-[q(u,v){]}_x$. Differentiating these relations proves harmonicity of both parts. No nonvanishing derivative hypothesis is required.

On a domain carrying a branch of the complex logarithm, the holomorphic function $-i\mathrm{Log}z$ has real part equal to the chosen angle and imaginary part $-\log |z|$. These are harmonic conjugates; their level sets are rays and circles, meeting orthogonally. For the principal real arctangent $\varphi =\arctan (y/x)$, its domain has the two components $x>0$ and $x<0$. The corresponding holomorphic functions are $-i\mathrm{Log}z$ and $-i\mathrm{Log}(-z)$, using the complex logarithm on the right half-plane. This formulation does not assert a globally single-valued angle on the punctured plane.

On a disc, the closed one-form $-u_{y}dx+u_{x}dy$ has a potential $v$ when $\Delta u=0$. The Cauchy--Riemann equations then make $u+iv$ holomorphic.

The function $\log |z|$ is harmonic on the punctured complex plane, but it is not globally the real part of a holomorphic function there. Such a function would have derivative $1/z$, whose integral around the unit circle is $2\pi i$, whereas the integral of a derivative around a closed curve is zero.

Every $C^2$ harmonic function is locally the real part of a holomorphic function and is therefore infinitely differentiable.

Harmonic functions admit locally convergent power-series expansions.

A harmonic function on a connected domain that vanishes on a nonempty open subset vanishes everywhere.