Harmonic Functions and the Poisson Integral

Choose $r>0$ with $D(a,r)\subseteq \mathrm{\Omega }$, and define $g:=u_x-i{u}_y$ on $D(a,r)$. Since $u$ is harmonic and $C^2$, [L1] makes $g$ holomorphic on $D(a,r)$.

The disc $D(a,r)$ is star-shaped, hence homologically simply connected by [L2], so $g$ has a primitive $G$ there with $G^{\mathrm{\prime }}=g$.

Write $G=U+iV$. Because $G^{\mathrm{\prime }}=g=u_x-i{u}_y$, one has $U_x=u_x$ and $U_y=u_y$, so the real-valued function $H:=u-U$ has continuous first partials with $H_x=H_y=0$ on $D(a,r)$. Hence the Cauchy-Riemann equations hold for $H$, [L4] makes $H$ holomorphic there, and [L3] makes it constant.

If $H\equiv c$ on $D(a,r)$, then $F:=G+c$ is holomorphic there and $\mathrm{Re}F=U+c=u$.

Define $g:=u_x-i{u}_y$ on $\mathrm{\Omega }$. Since $u$ is harmonic, [L1] makes $g$ holomorphic on $\mathrm{\Omega }$.

By [L2], the holomorphic function $g$ has a primitive $G$ on $\mathrm{\Omega }$ with $G^{\mathrm{\prime }}=g$.

Write $G=U+iV$. From $G^{\mathrm{\prime }}=g=u_x-i{u}_y$ one gets $U_x=u_x$ and $U_y=u_y$, so $H:=u-U$ has continuous first partials with $H_x=H_y=0$ on $\mathrm{\Omega }$. Hence the Cauchy-Riemann equations hold for $H$, [L4] makes $H$ holomorphic, and [L3] makes it constant.

If $H\equiv c$ on $\mathrm{\Omega }$, then $F:=G+c$ is holomorphic on $\mathrm{\Omega }$ and $\mathrm{Re}F=U+c=u$. Writing $F=u+iv$ defines a harmonic conjugate $v$ of $u$ on $\mathrm{\Omega }$.

By [L1], the functions $F_1:=u+i{v}_1$ and $F_2:=u+i{v}_2$ are holomorphic, so [L2] makes $-i(F_1-F_2)=v_1-v_2$ holomorphic on $\mathrm{\Omega }$.

The function $v_1-v_2$ is real-valued, so [L3] makes it constant on $\mathrm{\Omega }$.

Fix $a\in \mathrm{\Omega }$. By [L1], some disc $D(a,r)\subseteq \mathrm{\Omega }$ and some holomorphic $F=U+iV$ on that disc satisfy $u=U$ there.

By [L2], the coordinate map $(U,V)$ is smooth and real analytic on $D(a,r)$, so its first coordinate $U=u$ is smooth and real analytic there.

Since $a$ was arbitrary, $u$ is smooth and real analytic on all of $\mathrm{\Omega }$.

Because the closed disc $\overline{D(a,r)}$ lies in the open set $\mathrm{\Omega }$, choose $R>r$ with $D(a,R)\subseteq \mathrm{\Omega }$. The restriction of $u$ to this disc is harmonic, so [L1] gives a holomorphic function $F$ on $D(a,R)$ with $\mathrm{Re}F=u$ there.

For every $0<t\le r$, applying [L2] to $F$ on the circle of radius $t$ and taking real parts gives $u(a)=\frac{1}{2\pi }{\int }_0^{2\pi }u(a+t{e}^{i\theta })d\theta \mathrm{.}$

The circle formula of the definition is step 2.1 at $t=r$.

Multiplying the identity of step 2.1 by $2t\mathrm{/}{r}^2$ and integrating from $0$ to $r$ gives the disc formula of the definition, because $\frac{2}{r^2}{\int }_0^{r}tdt=1$.

Suppose $u$ has an interior local maximum at $a\in \mathrm{\Omega }$, with value $M=u(a)$. By [L1], some disc $D\subseteq \mathrm{\Omega }$ and some holomorphic $F$ on $D$ satisfy $\mathrm{Re}F=u$ on $D$; the real part of $F$ has a local maximum at $a$, so [L2] makes $F$ constant on $D$, and therefore $u\equiv M$ on $D$.

Let $S:=\{z\in \mathrm{\Omega }:u\equiv M\text{ on some neighbourhood of }z\}$. Step 1.1 gives $a\in S$, and $S$ is open by definition. If $b\in \bar{S}$, choose a disc $D_b\subseteq \mathrm{\Omega }$ and a holomorphic $G$ on $D_b$ with $\mathrm{Re}G=u$ there by [L1]; since $S\cap D_b$ contains a nonempty open set on which $\mathrm{Re}G=M$, both $\mathrm{Re}(G-M)$ and $\mathrm{Re}(M-G)$ have local maxima there, so [L2] makes $G$ constant on $D_b$, hence $u\equiv M$ on $D_b$ and $b\in S$. Thus $S$ is closed in $\mathrm{\Omega }$.

Because $\mathrm{\Omega }$ is connected by [L3], the nonempty set $S$ that is open and closed in $\mathrm{\Omega }$ must equal $\mathrm{\Omega }$. Thus a local interior maximum forces $u$ to be constant on $\mathrm{\Omega }$. Applying the same argument to $-u$, which is harmonic because $(-u{)}_{xx}+(-u{)}_{yy}=-(u_{xx}+u_{yy})=0$, gives the local minimum statement as well.

Now assume $\mathrm{\Omega }$ is bounded and $u$ is continuous on $\bar{\mathrm{\Omega }}$. The closure $\bar{\mathrm{\Omega }}$ is nonempty, closed and bounded in ${\mathbb{R}}^2$, hence compact by [L4], so [L5] says that the continuous extension of $u$ attains a maximum and a minimum there. If either extremum were attained at an interior point and $u$ were nonconstant, step 3.1 would force $u$ to be constant. Therefore both extremal values are realized on $\mathrm{\partial }\mathrm{\Omega }$, and the displayed equalities follow.

Let $w:=u-v$. Then $w$ is continuous on $\bar{\mathrm{\Omega }}$, harmonic on $\mathrm{\Omega }$, and satisfies $w=0$ on $\mathrm{\partial }\mathrm{\Omega }$.

Applying [L1] to $w$ gives ${\sup }_{\bar{\mathrm{\Omega }}}w={\sup }_{\mathrm{\partial }\mathrm{\Omega }}w=0$, so $w\le 0$ on $\bar{\mathrm{\Omega }}$; applying [L1] to $-w$ gives ${\sup }_{\bar{\mathrm{\Omega }}}(-w)=0$, so $w\ge 0$ on $\bar{\mathrm{\Omega }}$. Hence $w=0$ everywhere.

Therefore $u=v$ on $\bar{\mathrm{\Omega }}$.

Suppose first that $u$ is bounded above. By [L1], choose a harmonic conjugate $v$ and set $F:=u+iv$, a holomorphic function on $\mathbb{C}$. Then $\exp (F)$ is entire by [L3], and its modulus is $e^u$, which is bounded because $u$ is. So [L2] makes $\exp (F)$ constant.

Differentiating the identity $\exp (F)\equiv c$ gives $0=(\exp (F){)}^{\mathrm{\prime }}=\exp (F)F^{\mathrm{\prime }}$. Since an exponential value is never $0$, [L3] gives $F^{\mathrm{\prime }}=0$, and [L3] again makes $F$ constant. Therefore $u=\mathrm{Re}F$ is constant.

If instead $u$ is bounded below, then $-u$ is harmonic and bounded above, so step 2.1 applied to $-u$ makes $-u$, and therefore $u$, constant.

Let $S:=\{z\in \mathrm{\Omega }:u=0\text{ on some neighbourhood of }z\}$. Then $U\subseteq S$, so $S$ is nonempty, and $S$ is open by definition.

Let $a\in \bar{S}$. By [L1], choose a disc $D\subseteq \mathrm{\Omega }$ around $a$ and a holomorphic function $F$ on $D$ with $\mathrm{Re}F=u$ there. Since $S\cap D$ contains a nonempty open set on which $\mathrm{Re}F=0$, both $\mathrm{Re}F$ and $\mathrm{Re}(-F)$ have interior local maxima $0$ on $D$; [L2] therefore makes both $F$ and $-F$ constant, so $u=\mathrm{Re}F$ vanishes on all of $D$. Hence $a\in S$.

Thus $S$ is closed in $\mathrm{\Omega }$, and [L3] makes the nonempty clopen set $S$ equal to all of $\mathrm{\Omega }$. Therefore $u=0$ everywhere on $\mathrm{\Omega }$.

For the holomorphic case, fix $a\in U$. By [L1], choose a neighbourhood $W$ of $\varphi (a)$ and a holomorphic function $F$ on $W$ with $\mathrm{Re}F=u$ there. Shrinking around $a$ if necessary, $\varphi$ maps that neighbourhood into $W$, so [L2] makes $F\circ \varphi$ holomorphic and $\mathrm{Re}(F\circ \varphi )=u\circ \varphi$ there. Thus $u\circ \varphi$ is harmonic near $a$, and since $a$ was arbitrary it is harmonic on $U$.

For the antiholomorphic case, fix $a\in U$ and choose $W$ and $F$ as in step 1.1 around $\psi (a)$. By [L3], the map $\tilde{\psi }(z):=\overline{\psi (z)}$ is holomorphic on $U$, and the map $\tilde{F}(w):=\overline{F(\stackrel{‾}{w})}$ is holomorphic on $\bar{W}:=\{\bar{\zeta }:\zeta \in W\}$. Therefore [L2] makes $\tilde{F}\circ \tilde{\psi }$ holomorphic on a neighbourhood of $a$, and its real part is $\mathrm{Re}\tilde{F}(\tilde{\psi }(z))=\mathrm{Re}\overline{F(\psi (z))}=\mathrm{Re}F(\psi (z))=u(\psi (z))\mathrm{.}$ Hence $u\circ \psi$ is harmonic near $a$, and therefore on $U$.

Steps 1.1 and 2.1 prove the two invariance statements.

Since $1-r^2>0$ and $1-2r\cos \theta +r^2=(1-r{)}^2+2r(1-\cos \theta )>0,$ the quotient $P_r(\theta )$ is positive for every $\theta$.

By [L2], $1=\frac{1}{2\pi }{\int }_0^{2\pi }\frac{1+r{e}^{i\theta }}{1-r{e}^{i\theta }}d\theta \mathrm{.}$ Taking real parts and using [L1] gives $\frac{1}{2\pi }{\int }_0^{2\pi }{P}_r(\theta )d\theta =1$.

If $\delta \le \mathrm{\mid }\theta \mathrm{\mid }\le \pi$, then $\cos \theta \le \cos \delta$, so $0<P_r(\theta )\le \frac{1-r^2}{1-2r\cos \delta +r^2}\mathrm{.}$ The denominator tends to $2(1-\cos \delta )>0$ as $r\to 1^{-}$, while the numerator tends to $0$, so the right-hand side tends to $0$, proving the uniform concentration estimate on representatives in $[-\pi ,\pi ]$. Periodicity gives the equivalent formulation using circular distance from $0$.

By [L1], the parameter integral $F(z)=\frac{1}{2\pi }{\int }_0^{2\pi }\frac{e^{it}+z}{e^{it}-z}\phi (e^{it})dt$ is holomorphic on $\mathbb{D}$.

Taking real parts under the integral and using [L2] gives $\mathrm{Re}F(z)=\frac{1}{2\pi }{\int }_0^{2\pi }P(z,e^{it})\phi (e^{it})dt=P[\phi ](z)\mathrm{.}$

By [L3], the real part of the holomorphic function $F$ is harmonic. Since step 2.1 identifies that real part with $P[\phi ]$, the Poisson integral is harmonic on $\mathbb{D}$.

Let $\epsilon >0$. By uniform continuity of $\phi$ on the compact unit circle, choose $\delta \in (0,\pi ]$ such that $\mathrm{\mid }\phi (e^{it})-\phi (e^{i\alpha })\mathrm{\mid }<\epsilon \mathrm{/}2$ whenever the circular distance from $t$ to $\alpha$ is less than $\delta$.

Writing $z=r{e}^{i\alpha }$ and subtracting $\phi (e^{i\alpha })$ inside the Poisson integral, positivity and total mass one from [L1] give $\mathrm{\mid }P[\phi ](z)-\phi (e^{i\alpha })\mathrm{\mid }\le \frac{1}{2\pi }{\int }_{\mathrm{\mid }t-\alpha \mathrm{\mid }<\delta }{P}_r(\alpha -t)\frac{\epsilon }{2}dt+\frac{1}{2\pi }{\int }_{\mathrm{\mid }t-\alpha \mathrm{\mid }\ge \delta }{P}_r(\alpha -t)2\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}dt\mathrm{.}$

The first integral in step 2.1 is at most $\epsilon \mathrm{/}2$ because the kernel mass is $1$, and the second is at most $2\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$ times the far-arc mass from [L1], which is $<\epsilon \mathrm{/}2$ for all $\alpha$ once $r$ is close enough to $1$. Therefore $\mathrm{\mid }P[\phi ](r{e}^{i\alpha })-\phi (e^{i\alpha })\mathrm{\mid }<\epsilon$ uniformly in $\alpha$.

Since $\epsilon$ was arbitrary, the convergence is uniform as $r\to 1^{-}$.

By [L1], the function $P[\phi ]$ is harmonic on $\mathbb{D}$.

For $z=r{e}^{i\alpha }$ with $0\le r<1$, [L2] gives $P[\phi ](r{e}^{i\alpha })\to \phi (e^{i\alpha })$ uniformly in $\alpha$ as $r\to 1^{-}$. Therefore defining the boundary values of $P[\phi ]$ by $\phi$ produces a continuous extension to $\bar{\mathbb{D}}$.

If $u$ is any other continuous harmonic function on $\bar{\mathbb{D}}$ with $u=\phi$ on $\mathrm{\partial }\mathbb{D}$, then [L3] applied to $u$ and the extended Poisson integral forces $u=P[\phi ]$ on $\bar{\mathbb{D}}$.

Define $v(w):=u(a+Rw)$ on $\bar{\mathbb{D}}$. Then $v$ is continuous on $\bar{\mathbb{D}}$, harmonic on $\mathbb{D}$, and has boundary values $v(e^{it})=u(a+R{e}^{it})$.

By [L1], the Poisson integral of the boundary function $t\mapsto u(a+R{e}^{it})$ is a continuous harmonic function on $\bar{\mathbb{D}}$ with the same boundary values as $v$; [L2] therefore makes it equal to $v$ throughout $\bar{\mathbb{D}}$.

Evaluating step 2.1 at $w=(z-a)\mathrm{/}R=(\rho \mathrm{/}R)e^{i\varphi }$ gives exactly the displayed formula, because the unit-disc kernel there is $\frac{1-(\rho \mathrm{/}R{)}^2}{1-2(\rho \mathrm{/}R)\cos (\varphi -t)+(\rho \mathrm{/}R{)}^2}=\frac{R^2-{\rho }^2}{R^2-2R\rho \cos (\varphi -t)+{\rho }^2}\mathrm{.}$

Fix a closed disc $\overline{D(a,R)}\subseteq \mathrm{\Omega }$. By [L1], the boundary values $t\mapsto u(a+R{e}^{it})$ have a unique continuous harmonic Poisson extension $v$ to $\overline{D(a,R)}$, and [L2] makes $v$ satisfy the same mean-value property there. Put $w:=u-v$. Then $w$ is continuous on $\overline{D(a,R)}$, has the local mean-value property on $D(a,R)$, and vanishes on the boundary circle.

By [L3], the closed disc is compact, so $w$ attains a maximum $M$ and a minimum $m$ on $\overline{D(a,R)}$. If $M>0$, then the boundary values being $0$ force the maximum to occur at some interior point $b$. For a small circle centered at $b$, the circle mean-value property gives $M=w(b)$ as the average of values all bounded above by $M$, so every value on that circle is also $M$. Repeating this argument on overlapping small circles shows that the set $\{w=M\}$ is both open and closed in the connected disc, hence all of $D(a,R)$; this contradicts the boundary value $0$. Therefore $M\le 0$.

Applying the same argument to $-w$ gives ${\max }_{\overline{D(a,R)}}(-w)\le 0$, so $-w\le 0$ and therefore $w\ge 0$ on the closed disc. Together with step 2.1 this gives $w=0$ on $\overline{D(a,R)}$, so $u=v$ there.

Since the closed disc was arbitrary and $v$ is harmonic on its interior, $u$ is harmonic on every point of $\mathrm{\Omega }$, hence on all of $\mathrm{\Omega }$.

Fix $z_0$ with $0<\mathrm{\mid }{z}_0-a\mathrm{\mid }<R\mathrm{/}2$, and let $\rho :=\mathrm{\mid }{z}_0-a\mathrm{\mid }\mathrm{/}4$. Then the disc $D(z_0,2\rho )$ lies in $0<\mathrm{\mid }z-a\mathrm{\mid }<R$. By [L1], choose a holomorphic function $F$ on $D(z_0,2\rho )$ with $\mathrm{Re}F=u$ there.