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A Brownian excursion in the upper half-plane from $0$ to infinity is killed planar Brownian motion conditioned to stay in $\mathbb{H}$, with its entrance law at $0$. Precisely, it is the Doob h-transform for $h(z)=\mathrm{Im}z$, obtained by starting at $i\epsilon$ and letting $\epsilon \downarrow 0$. Equivalently,where $W$ is standard Brownian motion and the independent process $R$ is a dimension-three Bessel process started at zero. The transformed generator is $\frac{1}{2}\Delta +y^{-1}{\partial }_y$, which explains this representation.

First compute avoidance from an interior point $z$. Let $\tau$ be the exit time of ordinary planar Brownian motion from $\mathbb{H}\setminus A$. The harmonic correctionhas boundary values $\mathrm{Im}z$ on $A$ and zero on the real axis. The hydrodynamic normalization at infinity gives $g_A(z)=z+\mathrm{hcap}(A)/z+O(|z{|}^{-2})$. Together with the maximum principle for harmonic functions this gives the useful bound $0\le v\le {\sup }_{a\in A}\mathrm{Im}a$. Since ordinary Brownian motion hits the real axis almost surely, the optional stopping theorem applied to this bounded harmonic function givesThe Doob h-transform weights a stopped path by its final height divided by its initial height. Applying this at the first hit of $A$, with bounded-time localization and then a limit, therefore yieldsFor the entrance law the same result follows by conditioning at a small positive time and letting that time decrease to zero: the avoidance function has the boundary limit computed next, and the excursion is initially in a neighbourhood disjoint from $A$.

Because the closure of the hull avoids $0$, the Schwarz reflection principle makes $g_A$ analytic in a neighbourhood of zero, with real $g_A(0)$ and positive real $g_A^{\prime }(0)$. ThusIn fact, the reflected Taylor expansion gives $\mathrm{Im}{g}_A(z)/\mathrm{Im}z=g_A^{\prime }(0)+O(|z|)$ uniformly as $z\to 0$ inside the half-plane, so the same limit applies to the entrance process. The map in the question is ${\psi }_A=g_A-g_A(0)$, so its derivative agrees with $g_A^{\prime }$. We obtain the restriction probability of a Brownian half-plane excursion:

On $\mathbb{H}\setminus (A\cup C)$, the nonnegative harmonic function $h_A+h_C$ dominates the boundary data $\mathrm{Im}z$ on $A\cup C$: at a point of $A$, for example, $h_A(z)=\mathrm{Im}z$ and $h_C(z)\ge 0$. The maximum principle for harmonic functions, or equivalently Brownian motion stopped on the union, givesComparing the coefficients at infinity proves

True. This follows from the maximum principle for harmonic functions. A direct reduction to part (i) is also possible: for $\epsilon >0$, setThen $\Delta {\varphi }_{\epsilon }=4\epsilon >0$, soBecause $\Omega$ is bounded, ${\varphi }_{\epsilon }\to \varphi$ uniformly as $\epsilon \downarrow 0$. Taking the limit yields

Let $\Omega =(-1,1{)}^2$ and use the clamped energy space $V=H_0^2(\Omega )$, whose traces satisfy $u={\partial }_{n}u=0$ on $\partial \Omega$. Two applications of integration by parts giveIf equality holds, then $\Delta u=0$. The maximum principle for harmonic functions and the zero Dirichlet boundary condition imply $u=0$. The biharmonic operator is therefore positive definite. The same identity and conclusion hold for the simply supported conditions $u=\Delta u=0$; either standard interpretation of the paper's phrase “zero boundary conditions” gives the result.

Because $\overline{\Omega }$ is compact, $\varphi$ attains a maximum. If a maximizer lay in $\Omega$, its Hessian matrix would be negative semidefinite, so its trace would satisfy $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. Hence every strictly subharmonic function under the stated hypotheses attains a maximum on $\partial \Omega$.

Statement (i) is true. For $ϵ>0$, the functionhas $\Delta {\varphi }_{ϵ}=4ϵ>0$, soLetting $ϵ\downarrow 0$ givesThis is the maximum principle for harmonic functions.

Choose smooth boundary data ${\phi }_j$ converging uniformly to $\phi$, and let $v_j$ be their harmonic extensions. The maximum principle for harmonic functions givesso $v_j$ converges uniformly on $\overline{B}$ to a continuous function $v$ with boundary value $\phi$. Interior derivative estimates make the convergence smooth on compact subsets of $B$, hence $v$ is harmonic there. Uniqueness follows by applying the maximum principle to the difference of two solutions.

Let $B⋐\Omega$ and let $h$ be the harmonic function whose boundary values are $\max (u_1,u_2)$. The harmonic extensions $(u_i{)}_B$ satisfy $(u_i{)}_B\le h$ by the maximum principle for harmonic functions, while subharmonicity gives $u_i\le (u_i{)}_B$ in $B$. Hence $\max (u_1,u_2)\le h$, proving that the maximum of two subharmonic functions is subharmonic.

Every boundary point of $D$ satisfies the exterior sphere condition. For a point on the inner spherical boundary, use a smaller ball inside the removed ball and tangent at that point; for a point on the cube, use a ball in a supporting exterior half-space. If the exterior ball has centre $y$ and radius $r$, a local positive harmonic barrier isfor $n\ge 3$, while in two dimensions use $w(x)=\log (|x-y|/r)$. Adding a sufficiently large positive multiple of a global superharmonic function extends the local barrier across the bounded domain. Hence every boundary point is regular by part (v), and the Perron method for the Dirichlet problem produces a harmonic function attaining the prescribed continuous boundary data. The maximum principle for harmonic functions gives uniqueness.

For the stated flat metric and orientation,Because the metric coefficients and the coordinate one-forms are constant, the Hodge Laplacian acts coefficientwise:Thus a harmonic one-form $\alpha ={\sum }_i{\alpha }_{i}d{x}^i$ has harmonic coefficient functions. Every harmonic function on the compact connected torus is constant by the maximum principle for harmonic functions. HenceThe Hodge decomposition theorem now gives

No such function exists. Continuity on the compact closed unit disk makes $u$ bounded near the origin. The removable singularity for a bounded harmonic function extends $u$ harmonically across the origin. The extended function is continuous on the closed disk and vanishes on its boundary, so the maximum principle for harmonic functions, applied to both $u$ and $-u$, forces $u=0$ throughout the disk. This contradicts $u(0,0)=1$.

For a spatial initial slice with vanishing extrinsic curvature, the vacuum momentum constraint is automatic and the Hamiltonian constraint requires zero Ricci scalar. In three spatial dimensions, the indicated conformally flat metric has $\mathcal{R}=-8{\psi }^{-5}\Delta \psi$, with a nonzero conformal factor. Thus a harmonic function supplies the data on its domain. A positive example is $\psi =1+a/r$ on a punctured or exterior region. There is no nonconstant globally smooth example on all of Euclidean three-space tending to one at infinity, by the maximum principle for harmonic functions. Time symmetry of an initial slice does not imply that its entire evolved spacetime is static.