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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.

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.

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 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.

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.

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.

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 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.

The Kähler Laplacian identity ${\Delta }_d^{\mathrm{H}}=2{\Delta }_{\bar{\partial }}$ implies that every holomorphic function on a Kähler manifold is a harmonic function. On functions, $\bar{\partial }u=0$ and ${\bar{\partial }}^{\ast }u=0$ by degree, so both terms of the Dolbeault Laplacian vanish. Compactness is unnecessary. Changing the sign convention for the scalar Laplacian does not change this conclusion.

Suppose an unbounded domain contains the upper strip $(-1,1)\times (1,\infty )$ with its two vertical sides on the boundary. A bounded harmonic function with constant values $a$ on the right side and $b$ on the left side, and with valid bounded harmonic representation with almost sure Brownian exit, satisfies $u(x,y)\to a(1+x)/2+b(1-x)/2$ as $y\to \infty$. The horizontal Brownian exit time is finite; on that finite random interval the vertical Brownian path has a finite minimum. Consequently reaching the bottom before a side has probability tending to zero. The remaining two exit probabilities are the one-dimensional interval probabilities.

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 $e^{-f}$.

For a harmonic function in the quadrant with boundary derivatives $u_y(0,y)=g_1(y)$ and $u_x(x,0)=g_2(x)$, integrating the data to equal corner values and applying the Schwarz integral formula after the conformal map $w=z^2$ givesThese are tangential boundary derivatives, so uniqueness requires an additional normalization or growth restriction. A homogeneous contribution has the form $ih(z^2)/z$ with $h$ holomorphic in the complex upper half-plane and real on its boundary away from zero. For example, $u=xy$ has zero prescribed derivatives on both edges.

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 $u-w$.

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.

On a complete nonsingular spatial quotient with no inner boundary and standard asymptotic flatness, a bounded harmonic function tending to zero on all ends is zero. The static vacuum conformal spatial metric is then Ricci flat; three-dimensional curvature from the Ricci tensor makes it flat. Completeness and an ordinary Euclidean asymptotic end exclude nontrivial flat quotients, yielding Minkowski spacetime with the standard global time coordinate.

Laplace's equation isIts solutions are harmonic functions.

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 random set is local for a Gaussian free field if, conditionally on the set and its associated harmonic function, the remaining field is an independent zero-boundary field in its complement. This extends the deterministic Domain Markov property of the Gaussian free field to appropriate random revealed domains. Arbitrary field-dependent random sets need not have this property.

If $B(y,R)$ is an exterior ball tangent to a planar domain at $x_0$, thenis a harmonic function in the domain, vanishes at $x_0$, and is positive at every other point of its closure. It is therefore a barrier for the Dirichlet problem and proves that $x_0$ is a regular boundary point.

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.