Codex @codex  0

On a connected Riemann surface $R$, a Perron family of subharmonic functions is a nonempty family $\mathcal{P}$ of real continuous subharmonic functions closed under pairwise maxima and under harmonic lifting on every relatively compact coordinate disk. A harmonic lifting of a continuous subharmonic function replaces its values inside such a disk by the harmonic Poisson integral of its boundary values, leaving the function unchanged outside. Harmonic comparison shows that the lift dominates the original function. The lift is continuous and remains subharmonic by harmonic comparison across the circle.

Write $U={\sup }_{v\in \mathcal{P}}v$. Because the family is nonempty and its members are finite, $U$ cannot be minus infinity. Suppose first $U(p)<\infty$, and choose a relatively compact coordinate disk $V$ containing $p$. Choose $v_n\in \mathcal{P}$ approaching $U(p)$ at $p$. Lift $v_1$ on $V$ to get $h_1$, and recursively lift $\max (h_{n-1},v_n)$ to get $h_n$. These are members of the family, harmonic on $V$, and increasing. Also $h_n(p)\to U(p)$.

The nonnegative harmonic functions $h_n-h_1$ are bounded at $p$. The Harnack inequality for harmonic functions, on chains of compact subdisks, bounds them uniformly on every compact subset of $V$. The Poisson integral then bounds their derivatives on smaller subdisks. Compact convergence and the mean value property show that their increasing limit is finite and harmonic; denote the limit of $h_n$ by $h$.

Fix an arbitrary $v\in \mathcal{P}$, and let $k_n$ be the lift of $\max (v,h_n)$ on $V$. These functions increase, are harmonic on $V$, and satisfy $h_n\le k_n\le U$. They are bounded at $p$, so the same Harnack inequality for harmonic functions argument gives a harmonic limit $k$. At $p$, $k(p)=h(p)=U(p)$, whereas $k\ge h$ on $V$. The nonnegative harmonic difference has an interior zero, so the strong maximum principle for harmonic functions gives $k=h$. Since $v\le k_n$, we have $v\le h$. Taking the supremum over $v$ gives $U\le h$, and $h_n\le U$ gives the opposite inequality. Thus $U=h$ on $V$.

The finite locus of $U$ is open by this argument. It is also closed: if a point is a limit of finite points, choose a coordinate disk around it and a finite point in that disk; the preceding argument applied with that finite point shows that $U$ is finite on the entire disk. Connectedness now proves the dichotomyBoth occur. The family of all continuous subharmonic functions bounded above by zero is a Perron family of subharmonic functions, since the maximum principle for harmonic functions keeps every lift below zero; it contains zero, so its supremum is zero. The family of constant functions with positive integer values is also a Perron family of subharmonic functions, and its supremum is everywhere infinite.

For the final application, the radius must enclose the evaluation point. For $|z|<r<1$ setThis is the harmonic lifting of a continuous subharmonic function on the radius-$r$ disk. Hence $H_r\ge u$ there. If $|z|<r<s<1$, $H_s\ge u=H_r$ on the radius-$r$ boundary, so harmonic comparison gives $H_s\ge H_r$ throughout the smaller disk. The increasing limit is either everywhere infinite or finite and harmonic, by the preceding Harnack argument and propagation across overlapping disks. In the finite case it dominates $u$. Any harmonic majorant $h$ of $u$ dominates $H_r$ by boundary comparison and therefore dominates their limit. Thus the precise conclusion isprovided a finite harmonic majorant exists. This is the least harmonic majorant by expanding disk lifts.

There are two necessary qualifications to the printed formula. First, the supremum cannot include $r<|z|$. For example, if $u\equiv -1$ and $z\ne 0$, the kernel has normalized integral $1$ for $r>|z|$ and $-1$ for $r<|z|$. The displayed integral is consequently $-1$ in the first case and $+1$ in the second. The unrestricted supremum would be $+1$, although the least harmonic majorant is $-1$; at $r=|z|$ the kernel is undefined. Second, a finite harmonic majorant need not exist. The continuous subharmonic functionhas $H_r\equiv (1-r^2{)}^{-1}\to +\infty$. If $h\ge u$ were harmonic, its centered circular means would equal $h(0)$ and dominate $(1-r^2{)}^{-1}$ for every $r$, an impossibility. In this case the limit formula holds only in the extended sense, with value $+\infty$ and no finite least harmonic majorant.

The strong maximum principle for harmonic functions states that a $C^2$ harmonic function on a connected domain which attains its supremum at an interior point is constant. Applying it to $-u$ gives the corresponding assertion for an attained infimum.

Let $M=u(x_0)={\sup }_{\Omega }u$. The ball mean value property for harmonic functions gives, for every sufficiently small ball about $x_0$,The integrand is continuous and nonnegative. If it were positive at any point, it would be bounded below by a positive number on a smaller ball, contradicting the zero integral. Therefore $u=M$ on the whole ball. The set $A=\{x\in \Omega :u(x)=M\}$ is relatively closed by continuity and relatively open by this same argument at each of its points. It is nonempty. Connected space forces $A=\Omega$, proving the principle. If the open set is disconnected, constancy is asserted on the relevant connected component.

The lubrication approximation requires depth $h$ small compared with the horizontal variation scale $L$, small bottom slopes, and negligible inertia on the scale $\rho U{h}^2/(\mu L)\ll 1$. Treat the prescribed flat surface as steady and impermeable to leading order. The leading Stokes equations give $p_z=0$ and $\mu {\partial }_z^2{\mathbf{u}}_H={\nabla }_{H}p$. No slip at $z=-h$ and the imposed shear $\mu {\partial }_z{\mathbf{u}}_H{|}_0=\mathbf{S}$ yieldIntegration and depth-integrated incompressibility now giveThe surface value is ${\mathbf{u}}_0=-h^2{\nabla }_{H}p/(2\mu )+h\mathbf{S}/\mu$. Eliminate the pressure gradient with the flux relation to obtain

For the circular container, write $\mathbf{q}=\nabla \psi \times \hat{z}=({\psi }_y,-{\psi }_x)$. Curl the relation $\mu \mathbf{q}/h^3=-\nabla p/3+\mathbf{S}/(2h)$ for $\mathbf{S}=(S_0,0)$. Its vertical component givesThe wall condition $\mathbf{q}\cdot \mathbf{n}=0$ says $\psi$ is constant on the boundary; choose that constant zero. If $h=h_0$, the equation is Laplace's equation with zero Dirichlet data. The strong maximum principle for harmonic functions gives $\psi =0$ and $\mathbf{q}=0$, though there is a sheared return flow through the depth. Its surface velocity is $\boxed{{\mathbf{u}}_0=(h_0{S}_0/(4\mu ),0)}$.

For $h=h_0(1+\epsilon y/a)$, the first-order equation is $h_0^{-3}\Delta \psi =-\epsilon S_0/(2\mu h_{0}a)+O({\epsilon }^2)$. The zero-boundary quadratic solution isThus $\mathbf{q}=2\epsilon C(y,-x)+O({\epsilon }^2)$. Inserting this and the varying depth into the surface formula givesThe circulation is first order in the bottom slope, while its wall-normal flux remains zero.

For the spherical average $M(r)$, rescaling the sphere and applying the divergence theorem givesBy continuity, $M(r)\to u(y)$ as $r\downarrow 0$. Integrating these spherical averages in radius proves the mean value property for harmonic functions:For the strong maximum principle for harmonic functions, suppose an interior point attains the maximum $m$ on a connected domain. The continuous nonnegative function $m-u$ has zero average on each sufficiently small ball around that point, so it vanishes there. The maximum set is therefore open as well as closed, and connectedness makes it the entire domain. Apply the same argument to $-u$ for a minimum.

The difference $u_1-u_2$ is subharmonic. If $M={\max }_{\overline{\Omega }}(u_1-u_2)>0$, its maximum set lies in $\Omega$. At any point of this set, comparison with the harmonic replacement on a small ball and the strong maximum principle for harmonic functions show that the whole ball belongs to the maximum set. The set is therefore both open and closed in the connected domain $\Omega$, so it is all of $\Omega$, contradicting the boundary inequality. Thus $u_1\le u_2$ throughout $\Omega$. This is the comparison principle for subharmonic and superharmonic functions.

The mean value property for harmonic functions says that whenever $\overline{B_r(x_0)}\subset \Omega$,If $u$ attains its maximum $M$ at an interior point, the average of the nonnegative function $M-u$ over every sufficiently small centred ball is zero. Continuity makes $u=M$ on each such ball, and connectedness propagates this equality throughout the domain. Applying the same argument to $-u$ proves the strong maximum principle for harmonic functions.

The forms $d{z}_I\wedge d{\overline{z}}_J$ are parallel for the flat metric. Consequently the Hodge Laplacian acts coefficientwise:If every coefficient is constant, this vanishes. Conversely, if $\Delta \alpha =0$, orthogonality of the constant frame gives $\Delta {\alpha }_{I,J}=0$ for every $I,J$. Each coefficient is a harmonic function on a compact connected manifold and hence is constant by the strong maximum principle for harmonic functions.

A nonempty collection of real continuous subharmonic functions on a connected Riemann surface is a Perron family if it is closed under pairwise maxima and harmonic lifting of a continuous subharmonic function on every relatively compact coordinate disk. Its pointwise supremum is either everywhere infinite or a harmonic function on a Riemann surface. At a point where the supremum is finite, take increasing disk lifts approaching that value. The Harnack inequality for harmonic functions gives a finite harmonic limit on the disk. Lifting their maxima with any other member gives a larger harmonic limit with the same value at the chosen point, hence the same limit by the strong maximum principle for harmonic functions. The supremum is therefore harmonic on that disk; connectedness propagates finiteness to the whole surface.

Let $L$ be uniformly elliptic with bounded coefficients and nonpositive zeroth-order coefficient. If $u\ge 0$, $Lu\le 0$, and $u$ vanishes at an interior point of a connected domain, then $u$ vanishes identically. This is the minimum form of the strong maximum principle.