Subharmonic Functions and the Dirichlet Problem

For every real $\alpha$, the set $\{z\in \mathrm{\Omega }:u(z)<\alpha \}$ is open because $u$ is upper semicontinuous. Hence the sets $\{u\ge \alpha \}$ are closed, and therefore $u$ is Borel measurable.

The circle $C$ is compact. If $u$ is not identically $-\mathrm{\infty }$ on $C$, upper semicontinuity gives a point of maximum and therefore a finite upper bound $M$ on $C$; if $u\equiv -\mathrm{\infty }$ on $C$, then $-\mathrm{\infty }$ is already an upper bound. So the boundary function on $[0,2\pi ]$ is Borel measurable and bounded above.

The parametrization $t\mapsto a+r{e}^{it}$ is continuous, so composing it with the Borel function from step 1.1 makes $t\mapsto u(a+r{e}^{it})$ Borel measurable on $[0,2\pi ]$.

A Borel measurable function bounded above on a finite interval has an extended-real integral in $[-\mathrm{\infty },\mathrm{\infty })$, so the displayed circle average is well defined.

Assume condition 1. Let $h$ be continuous on $\overline{D(a,r)}$, harmonic on $D(a,r)$, and satisfy $h\ge u$ on $\mathrm{\partial }D(a,r)$. On $D(a,r)$ define $v:=u-h$. For $z\in D(a,r)$ and every $0<\rho <\mathrm{dist}(z,\mathrm{\partial }D(a,r))$, the submean inequality for $u$ and the circle mean-value property for $h$ give [L1, L4, given] $v(z)=u(z)-h(z)\le \frac{1}{2\pi }{\int }_0^{2\pi }(u(z+\rho e^{it})-h(z+\rho e^{it}))dt\mathrm{.}$ Thus $v$ is subharmonic on $D(a,r)$. If some point of $D(a,r)$ satisfied $v>0$, then upper semicontinuity on the compact disc would make $v$ attain a positive interior maximum there, contradicting [L6] because $v\le 0$ on $\mathrm{\partial }D(a,r)$. Hence $v\le 0$ on $D(a,r)$, so $h\ge u$ throughout the disc. This is condition 2.

Assume condition 2. Fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ and write $g(\zeta )=u(\zeta )$ on $\mathrm{\partial }D(a,r)$. By [L2], $g$ is Borel measurable and bounded above. On the compact circle, define [given, L2, construct] ${\varphi }_n(\zeta ):={\sup }_{\eta \in \mathrm{\partial }D(a,r)}(g(\eta )-n\mathrm{\mid }\zeta -\eta \mathrm{\mid })\mathrm{.}$ Each ${\varphi }_n$ is finite and continuous, satisfies ${\varphi }_n\ge g$, and decreases pointwise to $g$ because $g$ is upper semicontinuous.

Transporting the Poisson solution from the unit disc by [L3] and [L4], let $h_n$ be the harmonic function on $D(a,r)$, continuous on $\overline{D(a,r)}$, whose boundary values are ${\varphi }_n$. Since ${\varphi }_n\ge g=u$ on $\mathrm{\partial }D(a,r)$, condition 2 gives $u\le h_n$ on $D(a,r)$. Evaluating at the center and using the Poisson formula at the center of a disc, [step 1.2, L3, L4] $u(a)\le h_n(a)=\frac{1}{2\pi }{\int }_0^{2\pi }{\varphi }_n(a+r{e}^{it})dt\mathrm{.}$

Let $M$ be an upper bound for ${\varphi }_1$ on the circle. Then $M-{\varphi }_n$ is an increasing sequence of nonnegative boundary functions, so [L5] gives [step 1.2, step 2.1, L1, L5] ${\lim }_{n\to \mathrm{\infty }}\frac{1}{2\pi }{\int }_0^{2\pi }{\varphi }_n(a+r{e}^{it})dt=\frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$ Passing to the limit in step 2.1 yields the circle submean inequality at $a$. Since the disc was arbitrary and upper semicontinuity is already part of condition 2, condition 1 follows.

Assume first that $u$ is subharmonic. Fix $a\in \mathrm{\Omega }$ and choose $r>0$ with $\overline{D(a,r)}\subseteq \mathrm{\Omega }$. For $0<\rho <r$, Taylor's formula in the direction $e^{it}$ gives [L1, given, algebra] $u(a+\rho e^{it})=u(a)+\rho \mathrm{\nabla }u(a)\text{\hspace{0.17em}⁣}\cdot \text{\hspace{0.17em}⁣}{e}^{it}+\frac{{\rho }^2}{2}{e}^{it}\text{\hspace{0.17em}⁣}{}^T(D^{2}u(a))e^{it}+o({\rho }^2)\mathrm{.}$ Averaging over $t$ kills the linear term and averages the quadratic term to $\frac{{\rho }^2}{4}\mathrm{\Delta }u(a)$, so the submean inequality yields $0\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+\rho e^{it})dt-u(a)=\frac{{\rho }^2}{4}\mathrm{\Delta }u(a)+o({\rho }^2)\mathrm{.}$ Dividing by ${\rho }^2$ and letting $\rho \downarrow 0$ gives $\mathrm{\Delta }u(a)\ge 0$.

Assume now that $\mathrm{\Delta }u\ge 0$ on $\mathrm{\Omega }$. Fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$, and let $h$ be continuous on $\overline{D(a,r)}$, harmonic on $D(a,r)$, and satisfy $h\ge u$ on $\mathrm{\partial }D(a,r)$. Put $w=u-h$. Then $w$ is continuous on $\overline{D(a,r)}$, belongs to $C^2(D(a,r))$, satisfies $\mathrm{\Delta }w=\mathrm{\Delta }u\ge 0$ on $D(a,r)$, and has $w\le 0$ on $\mathrm{\partial }D(a,r)$. If some point $p\in D(a,r)$ had $w(p)>0$, choose $0<\epsilon <w(p)\mathrm{/}{r}^2$ and define $w_{\epsilon }(z):=w(z)+\epsilon \mathrm{\mid }z-a{\mathrm{\mid }}^2$. On the boundary one has $w_{\epsilon }\le \epsilon r^2<w(p)\le w_{\epsilon }(p)$, so $w_{\epsilon }$ attains its maximum at an interior point $q\in D(a,r)$. The second-derivative test there gives $\mathrm{\Delta }{w}_{\epsilon }(q)\le 0$, but $\mathrm{\Delta }{w}_{\epsilon }=\mathrm{\Delta }w+4\epsilon \ge 4\epsilon >0,$ a contradiction. Therefore $w\le 0$ on $D(a,r)$, so $u\le h$ on the whole disc. Since the disc and the harmonic boundary majorant $h$ were arbitrary, [L1] shows that $u$ is subharmonic on $\mathrm{\Omega }$.

Fix a compact disc $\overline{D(a,R)}\subseteq \mathrm{\Omega }$ and a smaller concentric disc $\overline{D(a,\rho )}$ with $0<\rho <R$. The function $u$ is upper semicontinuous on the compact circle $\mathrm{\partial }D(a,R)$, so [L2] gives a finite upper bound there. By the harmonic-comparison theorem [L1], $u$ is therefore bounded above on $\overline{D(a,\rho )}$ by the harmonic majorant obtained from any continuous boundary majorant on $\mathrm{\partial }D(a,R)$.

The set $A=\{z\in \mathrm{\Omega }:u(z)=-\mathrm{\infty }\}$ has empty interior. Suppose instead that $D(c_0,R)\subseteq A$ for some $R>0$, and fix any point $x$ in the connected component of $\mathrm{\Omega }$ containing $c_0$. By [L3], choose a polygonal path $\gamma$ in that component from $c_0$ to $x$. Because $\gamma ([0,1])$ is compact and lies in the open set $\mathrm{\Omega }$, choose $r>0$ with $3r<R$ and $\overline{D(y,4r)}\subseteq \mathrm{\Omega }$ for every $y\in \gamma ([0,1])$. Subdivide the path by points $c_0,c_1,\dots ,c_N=x$ on $\gamma$ with $\mathrm{\mid }{c}_{j+1}-c_j\mathrm{\mid }<r\mathrm{/}2$ for every $j$. We claim inductively that $D(c_j,3r)\subseteq A$ for all $j$. The case $j=0$ holds because $3r<R$. If $D(c_j,3r)\subseteq A$ and $b\in D(c_j,7r\mathrm{/}2)$, choose $\rho$ with $\max (0,\mathrm{\mid }b-c_j\mathrm{\mid }-3r)<\rho <r\mathrm{/}2$. Then the circle $\mathrm{\mid }z-b\mathrm{\mid }=\rho$ lies in $D(c_j,4r)\subseteq \mathrm{\Omega }$ and meets $D(c_j,3r)$ in an open arc, so $u=-\mathrm{\infty }$ on a set of positive arc-length measure on that circle. By [L2], the circle values are Borel measurable and bounded above, hence the circle average is $-\mathrm{\infty }$. The submean inequality therefore gives $u(b)=-\mathrm{\infty }$, proving $D(c_j,7r\mathrm{/}2)\subseteq A$. Since $\mathrm{\mid }{c}_{j+1}-c_j\mathrm{\mid }<r\mathrm{/}2$, one has $D(c_{j+1},3r)\subseteq D(c_j,7r\mathrm{/}2)\subseteq A$, completing the induction. In particular $x\in A$. Because $x$ was arbitrary in the component, this contradicts the subharmonic convention that $u$ is not identically $-\mathrm{\infty }$ there. Thus $A$ has empty interior, so every open subdisc contains a point where $u$ is finite.

Fix $a\in \mathrm{\Omega }$ and choose $r>0$ with $\overline{D(a,3r)}\subseteq \mathrm{\Omega }$. Step 1.1, applied with outer radius $3r$ and inner radius $2r$, gives a finite upper bound $M$ for $u$ on $D(a,2r)$. By step 1.2 choose $c\in D(a,r\mathrm{/}2)$ with $u(c)>-\mathrm{\infty }$, and then choose $r+\mathrm{\mid }c-a\mathrm{\mid }<s<2r-\mathrm{\mid }c-a\mathrm{\mid }\mathrm{.}$ These inequalities give $D(a,r)\subseteq D(c,s)\subseteq D(a,2r)$.

For every $0<t<s$, the submean inequality at $c$ gives ${\int }_0^{2\pi }(M-u(c+t{e}^{i\theta }))d\theta \le 2\pi (M-u(c))\mathrm{.}$ The integrand is nonnegative and Borel by [L2]. Multiply by $t$, integrate from $0$ to $s$, and apply [L4] to obtain ${\int }_{D(c,s)}(M-u)dA\le \pi s^2(M-u(c))<\mathrm{\infty }\mathrm{.}$ Thus the negative part of $u$ is integrable on $D(c,s)$, while its positive part is bounded there by $M$. Hence $u\in L^1(D(c,s))$, and therefore $u\in L^1(D(a,r))$.

Every point $a\in \mathrm{\Omega }$ admits such a disc $D(a,r)$, so $u\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$.

Let $D\subseteq \mathrm{\Omega }$ be a disc on which $f$ has no zeros. By [L3], there is a holomorphic function $L$ on $D$ with $\exp L=f$. Writing $L=\alpha +i\beta$, one has $\alpha =\log \mathrm{\mid }f\mathrm{\mid }$ on $D$. Since $L$ is holomorphic and smooth by [L4], the Cauchy-Riemann equations imply $\mathrm{\Delta }\alpha =0$, so [L1] makes $\log \mathrm{\mid }f\mathrm{\mid }$ subharmonic on every zero-free disc.

Fix a zero $a$ of $f$, and let $m={\mathrm{ord}}_a(f)$. By [L2], on a small disc about $a$ one has $f(z)=(z-a{)}^{m}g(z)$ with $g(a)\mathrm{\ne }0$. Shrinking if necessary, $g$ has no zeros there, so step 1.1 makes $\log \mathrm{\mid }g\mathrm{\mid }$ harmonic and hence subharmonic on that disc.

On the punctured disc around $a$, [step 2.1, algebra] $u(z)=m\log \mathrm{\mid }z-a\mathrm{\mid }+\log \mathrm{\mid }g(z)\mathrm{\mid }\mathrm{.}$ The function $\log \mathrm{\mid }z-a\mathrm{\mid }$ is harmonic on the punctured disc, and at the center $a$ its value is $-\mathrm{\infty }$ while every circle average is finite; hence it is subharmonic there. Therefore the right-hand side is subharmonic on the whole disc, agreeing with $u$ away from $a$ and with $u(a)=-\mathrm{\infty }$ at the center.

Every point of $\mathrm{\Omega }$ lies either on a zero-free disc covered by step 1.1 or on a zero-containing disc covered by step 3.1. So $u$ is subharmonic throughout $\mathrm{\Omega }$.

Put $u=\log \mathrm{\mid }f\mathrm{\mid }$. By [L1], for every closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$, [L1, algebra] $u(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$ Exponentiating and using Jensen's inequality for the convex increasing map $x\mapsto e^{px}$ gives $\mathrm{\mid }f(a){\mathrm{\mid }}^p=e^{pu(a)}\le \frac{1}{2\pi }{\int }_0^{2\pi }{e}^{pu(a+r{e}^{it})}dt=\frac{1}{2\pi }{\int }_0^{2\pi }\mathrm{\mid }f(a+r{e}^{it}){\mathrm{\mid }}^{p}dt\mathrm{.}$

The function $\mathrm{\mid }f{\mathrm{\mid }}^p$ is continuous, hence upper semicontinuous, and is not identically zero on a connected component because $f$ is not identically zero there. Step 1.1 is exactly the submean inequality, so $\mathrm{\mid }f{\mathrm{\mid }}^p$ is subharmonic.

Let $I=\{j:{\alpha }_j>0\}$. If $I$ is empty, the combination is the harmonic zero function. Otherwise it means the well-defined extended-real sum ${\sum }_{j\in I}{\alpha }_j{u}_j$; no product $0\cdot (-\mathrm{\infty })$ occurs. Finite sums with positive coefficients preserve upper semicontinuity, and [L3] shows that all summands are finite simultaneously almost everywhere, so their sum is not identically $-\mathrm{\infty }$. On any closed disc, multiply the submean inequality for $u_j$ by ${\alpha }_j>0$ and sum over $I$ to obtain the submean inequality for the combination.

For the finite maximum, upper semicontinuity is preserved by finite maxima. Let $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ and let $h$ be continuous on the closure, harmonic on the disc, and satisfy $h\ge \max (u_1,\dots ,u_m)$ on the boundary. Then $h\ge u_j$ on the boundary for every $j$, so [L1] gives $h\ge u_j$ throughout the disc for every $j$. Therefore $h\ge \max (u_1,\dots ,u_m)$ on the disc. Another use of [L1] shows that the maximum is subharmonic.

Steps 1.1 and 1.2 prove the two closure properties.

A decreasing limit of upper semicontinuous functions is upper semicontinuous, so $u$ is upper semicontinuous on $\mathrm{\Omega }$. If $u\equiv -\mathrm{\infty }$, the first alternative of the statement holds and there is nothing more to prove. Assume from now on that $u$ is finite at least at one point.

Fix a closed disc $\overline{D(a,r)}\subseteq \mathrm{\Omega }$. For every $n$, [L1] gives [L1, L2, L3, choose] $u_n(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }{u}_n(a+r{e}^{it})dt\mathrm{.}$ By [L2], the boundary functions are measurable and bounded above. Choose a constant $M$ larger than $u_1$ on the circle. Then $M-u_n(a+r{e}^{it})$ is an increasing sequence of nonnegative measurable functions of $t$, so [L3] yields ${\lim }_{n\to \mathrm{\infty }}\frac{1}{2\pi }{\int }_0^{2\pi }{u}_n(a+r{e}^{it})dt=\frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$

Passing to the limit in the inequalities of step 1.2 gives [step 1.1, step 1.2, L1] $u(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+r{e}^{it})dt\mathrm{.}$ Since the disc was arbitrary and step 1.1 supplied upper semicontinuity, [L1] makes $u$ subharmonic.

For every compact set $K\subseteq \mathrm{\Omega }$, the hypothesis gives a real number $M_K$ with $u\le M_K$ on $K$, so both $u$ and $U=u^{\ast }$ are locally bounded above and never take the value $+\mathrm{\infty }$. Because $\mathcal{F}$ is nonempty and every $v\in \mathcal{F}$ satisfies $v\le u\le U$, the function $U$ is not identically $-\mathrm{\infty }$ on any connected component. By [L3], $U$ is upper semicontinuous and satisfies $u\le U$.

Let $\overline{D(a,r)}\subseteq \mathrm{\Omega }$ and let $h$ be continuous on the closure, harmonic on the disc, and satisfy $h\ge U$ on $\mathrm{\partial }D(a,r)$. Because $u\le U$, one also has $h\ge u$ on the boundary.

Fix any $v\in \mathcal{F}$. Since $h\ge v$ on $\mathrm{\partial }D(a,r)$, [L2] gives $h\ge v$ on $D(a,r)$. The same is therefore true for every finite maximum of members of $\mathcal{F}$, and [L1] keeps those maxima subharmonic. Taking the supremum over all $v\in \mathcal{F}$ yields $h\ge u$ on $D(a,r)$.

Since $h$ is continuous and dominates $u$, it also dominates the least upper-semicontinuous majorant $U$ by [L3]. Thus $h\ge U$ on $D(a,r)$. Another use of [L2] shows that $U$ is subharmonic on $\mathrm{\Omega }$.

Let [given] $S=\{z\in \mathrm{\Omega }:u(z)=M\}\mathrm{.}$ Because $u$ is upper semicontinuous, $S$ is closed in $\mathrm{\Omega }$, and it is nonempty because $a\in S$.

Choose $r>0$ with $\overline{D(a,r)}\subseteq \mathrm{\Omega }$. For every $0<\rho <r$, [L1] gives [L1, given] $M=u(a)\le \frac{1}{2\pi }{\int }_0^{2\pi }u(a+\rho e^{it})dt\le M,$ so the average equals $M$. Since the integrand never exceeds $M$, it equals $M$ almost everywhere on the circle $\mathrm{\mid }z-a\mathrm{\mid }=\rho$. If some point of that circle had value $<M$, upper semicontinuity would make the value $<M$ on a short arc, forcing the average below $M$. Hence $u\equiv M$ on every circle $\mathrm{\mid }z-a\mathrm{\mid }=\rho$ with $0<\rho <r$.

Step 1.2 shows that every point of $D(a,r)$ lies in $S$, so $S$ is open in $\mathrm{\Omega }$. Since $\mathrm{\Omega }$ is connected and $S$ is nonempty, closed, and open, one has $S=\mathrm{\Omega }$. Therefore $u\equiv M$ on $\mathrm{\Omega }$.

On $D$, the function $w=\max \{u,v\}$ is subharmonic by [L2]. Away from $\mathrm{\partial }D$ its upper semicontinuity is therefore clear. At $\zeta \in \mathrm{\partial }D\cap \mathrm{\Omega }$, the inside limsup is at most $u(\zeta )$ by the hypothesis and upper semicontinuity of $u$, while the outside limsup is at most $u(\zeta )=w(\zeta )$. Thus $w$ is upper semicontinuous on $\mathrm{\Omega }$.

Let $\bar{B}\subseteq \mathrm{\Omega }$ be a closed disc and let $h$ be continuous on $\bar{B}$, harmonic on $B$, and satisfy $h\ge w$ on $\mathrm{\partial }B$. Because $w\ge u$ there, [L1] first gives $h\ge u$ throughout $B$.

Let $C$ be a connected component of $B\cap D$. On $\mathrm{\partial }C\cap \mathrm{\partial }B$ one has $h\ge w\ge v$. At a boundary point of $C$ inside $B$, the seam hypothesis and step 1.2 give ${\mathrm{lim\hspace{0.17em}sup}}_C(v-h)\le u-h\le 0$. Thus the subharmonic function $v-h$ has boundary limsup at most $0$ on the bounded domain $C$. If it were positive somewhere, upper semicontinuity and the boundary bound would make it attain a positive interior maximum, contradicting [L3]. Hence $h\ge v$ on every such component.

Steps 1.2 and 2.1 give $h\ge u$ on $B$ and $h\ge v$ on $B\cap D$, hence $h\ge w$ throughout $B$. Every harmonic boundary majorant therefore majorizes $w$, so [L1] and step 1.1 make $w$ subharmonic on $\mathrm{\Omega }$.

Choose a boundary approximation $({\varphi }_n)$ for $u{\mathrm{\mid }}_{\mathrm{\partial }D}$, and let $h_n$ be the associated harmonic functions from [L1]. Because ${\varphi }_n\ge u$ on $\mathrm{\partial }D$, [L2] gives $h_n\ge u$ on $D$ for every $n$. The sequence $(h_n)$ is decreasing because the boundary data are decreasing.

Let $M$ be an upper bound for ${\varphi }_1$ on $\mathrm{\partial }D$. Then each $M-h_n$ is a nonnegative harmonic function on $D$, and the sequence $(M-h_n)$ is increasing. By [L5], choose $z_0\in D$ with $u(z_0)>-\mathrm{\infty }$. Step 1.1 gives $0\le M-h_n(z_0)\le M-u(z_0)$, so [L4] applied to $(M-h_n)$ yields a harmonic limit $H$ on $D$. Consequently $h:=M-H={\inf }_n{h}_n$ is harmonic on $D$.

The inside function $h$ is independent of the chosen boundary approximation. Indeed, if $k$ is obtained from another approximation $({\psi }_m)$, then $h$ is harmonic on $D$ and its boundary limsup satisfies [step 1.1, step 2.1, L2] ${\mathrm{lim\hspace{0.17em}sup}}_{\begin{array}{c}z\to \zeta \\ z\in D\end{array}}h(z)\le {\varphi }_n(\zeta )(\zeta \in \mathrm{\partial }D)$ for every fixed $n$, hence $\mathrm{lim\hspace{0.17em}sup}h\le u\le {\psi }_m$ on the boundary. Applying [L2] to the harmonic function $k_m$ extending ${\psi }_m$ gives $h\le k_m$ on $D$ for every $m$, so $h\le k$. Symmetry gives $k\le h$.

By step 1.1, $h\ge u$ on $D$, and by step 3.1 this harmonic function is intrinsic. Moreover, for every $\zeta \in \mathrm{\partial }D$ and every fixed $n$, one has $h\le h_n$ on $D$ and $h_n(\zeta )={\varphi }_n(\zeta )$, so [step 1.1, step 2.1] ${\mathrm{lim\hspace{0.17em}sup}}_{\begin{array}{c}z\to \zeta \\ z\in D\end{array}}h(z)\le {\varphi }_n(\zeta )\mathrm{.}$ Letting $n\to \mathrm{\infty }$ yields ${\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta ,z\in D}h(z)\le u(\zeta )$.

The Poisson modification equals $h$ on $D$ and $u$ on $\mathrm{\Omega }\setminus D$. Step 4.1 provides the seam inequality, so [L3] shows that $P_{D}u$ is subharmonic on $\mathrm{\Omega }$. It is harmonic on $D$ by step 2.1, equals $u$ outside $D$ by definition, and majorizes $u$ on $D$ by step 4.1. Hence $P_{D}u\ge u$ on all of $\mathrm{\Omega }$.