Subharmonic Functions and the Dirichlet Problem — Examples

Writing $z=x+iy$, one has $u(x+iy)=x^2+y^2$, so [L1, given, algebra] $\mathrm{\Delta }u=u_{xx}+u_{yy}=2+2=4\ge 0.$ By [L1], $u$ is subharmonic on $\mathbb{C}$.

The identity map $f(z)=z$ is holomorphic on $\mathbb{C}$, and [L2, given] $\log \mathrm{\mid }f(z)\mathrm{\mid }=\log \mathrm{\mid }z\mathrm{\mid }$ with value $-\mathrm{\infty }$ at the zero $z=0$. Therefore [L2] makes $v$ subharmonic on $\mathbb{C}$. Away from $0$, the function $v$ is the real part of the holomorphic logarithm and so is harmonic there.

On the circle $\mathrm{\mid }z\mathrm{\mid }=r$, the boundary values of $u$ are the constant $r^2$. Hence the constant function $h(z)=r^2$ is harmonic on $D(0,r)$ and has exactly the boundary values required by the Poisson modification.

By [L1], the modified function agrees with $u$ outside $D(0,r)$ and is harmonic inside. Since step 1.1 gives a harmonic candidate with the correct boundary data on the inner disc, [L2] forces the inside harmonic piece to be exactly $h(z)=r^2$. This gives the displayed formula.

Every point of the two boundary circles of $A_{r,R}$ admits an exterior disc, so [L1] makes the annulus regular and hence ensures existence of a Perron solution.

The function [given, algebra] $h(z)=\alpha +\frac{\beta -\alpha }{\log (R\mathrm{/}r)}\log \frac{\mathrm{\mid }z\mathrm{\mid }}{r}$ is harmonic on $A_{r,R}$ because $\log \mathrm{\mid }z\mathrm{\mid }$ is harmonic away from $0$, and it satisfies $h=\alpha$ on $\mathrm{\mid }z\mathrm{\mid }=r$ and $h=\beta$ on $\mathrm{\mid }z\mathrm{\mid }=R$.

By step 1.1, the Perron solution exists; by step 1.2, $h$ is a continuous harmonic function on the closure of the annulus with the required boundary data. The uniqueness statement [L2] therefore forces the Perron solution to equal $h$.

On the first quadrant one may choose the holomorphic branch of $z^{2\mathrm{/}3}$. If $z=r{e}^{it}$ with $0<t<\pi \mathrm{/}2$, then [given, algebra] $z^{2\mathrm{/}3}=r^{2\mathrm{/}3}{e}^{2it\mathrm{/}3}$ has argument in $(0,\pi \mathrm{/}3)$, so $\mathrm{Re}(z^{2\mathrm{/}3})>0$. Therefore $b(z)=-\mathrm{Re}(z^{2\mathrm{/}3})<0$ on $Q$ near the corner, and $b(z)\to 0$ as $z\to 0$.

On any set in the square that stays a positive distance from $0$, the quantity $\mathrm{Re}(z^{2\mathrm{/}3})$ has a positive minimum, so $b$ stays uniformly below a negative constant there. Thus $b$ has exactly the shape required in [L2], and it is the concrete barrier predicted abstractly by [L1].

Suppose $u$ were a continuous harmonic solution of this boundary-value problem on $\mathrm{\Omega }$. Then $u$ is bounded on every punctured neighbourhood of $0$ because it extends continuously to the puncture with value $1$. By [L1], $u$ extends to a harmonic function $U$ on the full unit disc.

The extension $U$ still has boundary value $0$ on the unit circle, so [L2] forces $U\equiv 0$ on the closed unit disc. But then $U(0)=0$, contradicting the prescribed puncture value $1$. Therefore no such harmonic solution exists.

Since one continuous boundary datum is not solvable on $\mathrm{\Omega }$, [L3] shows that $\mathrm{\Omega }$ cannot have all boundary points regular. In particular, the puncture is an irregular boundary point.

The punctured unit disc is a bounded plane domain, and [L1] supplies a continuous boundary datum on it that is not attained by any harmonic function.

That single bounded counterexample contradicts the universal claim, so the statement is false.

Let $H$ be the regularized Perron envelope for the punctured-disc datum from [L2]. By [L1], $H$ is harmonic on the punctured disc.

If the universal claim were true, then $H$ would also attain the prescribed boundary values at the puncture and on the outer circle, so it would be a harmonic solution of exactly the boundary-value problem ruled out in [L2]. Therefore the claim is false.

If the claim were true, some nonconstant subharmonic function would attain a finite interior maximum.

But [L1] says that any subharmonic function with a finite interior maximum must be constant on the connected component where that maximum occurs. This contradicts step 1.1.

The function $\mathrm{Re}z$ is harmonic on the unit disc, so $n\mathrm{Re}z$ is harmonic for each $n$. By [L1], each [L1, given, algebra] $u_n(z)=\max \{n\mathrm{Re}z,-1\}$ is subharmonic.

Their pointwise supremum is [step 1.1, algebra] $u(z)={\sup }_n{u}_n(z)=\{\begin{array}{ll}\mathrm{Re}z,& \mathrm{Re}z<0,\\ 0,& \mathrm{Re}z=0,\\ +\mathrm{\infty },& \mathrm{Re}z>0.\end{array}$ This function is not even finite-valued on the right half-disc, so it cannot be subharmonic in the page's convention.

Therefore the arbitrary-supremum claim is false. Step 2.1 is also exactly why [L2] insists on local boundedness above and upper-semicontinuous regularization.