Let \(\Omega\subseteq \mathbb C\) be open and \(f: \Omega\to \mathbb C\) be holomorphic. Then \(\frac{\partial f}{\partial x}=\frac{1}{i}\frac{\partial f}{\partial y}\), where \(\partial/\partial x\) and \(\partial/\partial y\) denote the usual partial derivatives in the \(x\) and \(y\) variables respectively. If \(f=u+iv\) for some real valued functions \(u, v:\Omega\to \mathbb C\), then we have \[\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \quad \text{ and } \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}. \tag{C-R}\] These relations are called the Cauchy–Riemann equations.
Suppose \(f=u+i v\) is a complex-valued function defined on an open set \(\Omega\). If \(u\) and \(v\) are differentiable in the real sense and satisfy the Cauchy–Riemann equations (C-R) on \(\Omega\), then \(f\) is holomorphic on \(\Omega\).
Let \(\left(x_{0}, y_{0}\right) \in \mathbb{R}^{2}\) and \(u\) be a real-valued function defined in a neighbourhood of \(\left(x_{0}, y_{0}\right)\). Then \(u\) is harmonic at \(\left(x_{0}, y_{0}\right)\) if
\(u\) is continuous at \(\left(x_{0}, y_{0}\right)\).
\(u\) has continuous partial derivatives of the first and the second order at \(\left(x_{0}, y_{0}\right)\) satisfying \[u_{x x}\left(x_{0}, y_{0}\right)+u_{y y}\left(x_{0}, y_{0}\right)=0, \tag{*}\] where \(u_{x y}\left(x_{0}, y_{0}\right)=\frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x}\left(x_{0}, y_{0}\right)\right)\).
The (*) is called the Laplace equation. Further \(u\) is called harmonic in \(\Omega\) if it is harmonic at every point of \(\Omega\).
Remark.
We identify the elements \((x, y)\) of \(\mathbb{R}^{2}\) with \(x+i y\) of \(\mathbb{C}\) and it will be clear from the context whether we are taking \((x, y)\) or \(x+i y\).
For any \(z=x+i y\in\mathbb C\) and any real-valued function \(u=u(x, y)\) we write \[u(z)=u(x, y).\]
If \(u\) is harmonic in \(\Omega\), then \(u+c\) for any constant \(c\) is harmonic in \(\Omega\).
Let \(f \in H(\Omega)\) be given by \[f(z)=u(x, y)+i v(x, y).\] Then \(\operatorname{Re}(f)\) and \(\operatorname{Im}(f)\) are harmonic in \(\Omega\).
Proof: The proof depends on \(f^{\prime} \in H(\Omega)\) and \(f^{\prime \prime} \in H(\Omega)\).
Let \(f(z)=u(x, y)+i v(x, y)\) be given. First, we prove that \(u\) and \(v\) have continuous partial derivatives of orders \(0,1\) and \(2\) at every point of \(\Omega\). We prove the assertion for \(u\) and the proof for \(v\) is similar.
Let \(\left(x_{0}, y_{0}\right) \in \Omega\) and \(z_{0}=x_{0}+i y_{0}\). Then we see that \(u\) is continuous at \(\left(x_{0}, y_{0}\right)\) since \(f\) is continuous at \(z_{0}\).
Further, by the Cauchy–Riemann equations, we have \[f^{\prime}\left(z_{0}\right)=u_{x}\left(x_{0}, y_{0}\right)+i v_{x}\left(x_{0}, y_{0}\right)=v_{y}\left(x_{0}, y_{0}\right)-i u_{y}\left(x_{0}, y_{0}\right).\]
By differentiating this identity, we have \(u_{x}\) and \(u_{y}\) that are continuous at \((x_{0}, y_{0})\) since \(f^{\prime}(z)\) is continuous at \(z_{0}\). Next, we have \[\begin{aligned} f^{\prime \prime}\left(z_{0}\right) & =u_{x x}\left(x_{0}, y_{0}\right)+i v_{x x}\left(x_{0}, y_{0}\right) =v_{y x}\left(x_{0}, y_{0}\right)-i u_{y x}\left(x_{0}, y_{0}\right) \\ & =v_{x y}\left(x_{0}, y_{0}\right)-i u_{x y}\left(x_{0}, y_{0}\right) =-u_{y y}\left(x_{0}, y_{0}\right)-i v_{y y}\left(x_{0}, y_{0}\right) . \end{aligned}\]
This implies \(u\) has continuous partial derivative of order \(2\) at \((x_{0}, y_{0})\), since \(f^{\prime \prime}(z)\) is continuous at \(z_{0}\).
Since \(\left(x_{0}, y_{0}\right)\) is an arbitrary point of \(\Omega\), we conclude that \(u\) has continuous partial derivatives of order \(0,1\) and \(2\) at every point of \(\Omega\).
Differentiating the first (C-R) equation \(u_x=v_y\) with respect to \(x\) and the second \(v_x=-u_y\) with respect to \(y\), we obtain \[u_{x x}\left(x_{0}, y_{0}\right)=v_{y x}\left(x_{0}, y_{0}\right), \quad v_{x y}\left(x_{0}, y_{0}\right)=-u_{y y}\left(x_{0}, y_{0}\right),\] which implies \[u_{xx}\left(x_{0}, y_{0}\right)+v_{yy}\left(x_{0}, y_{0}\right)=0,\] since \(v_{y x}\left(x_{0}, y_{0}\right)=v_{x y}\left(x_{0}, y_{0}\right)\). Hence \(u\) is harmonic in \(\Omega\).$$\tag*{$\blacksquare$}$$
Let \(u\) be harmonic in a region \(\Omega\) and let \(V\) be a non-empty open subset of \(\Omega\) such that \(u=0\) in \(V\). Then \(u=0\) in \(\Omega\).
Proof: Let \(u\) be harmonic in \(\Omega\). For \(z \in \Omega\) with \(z=x+i y\), we consider \[g(z)=u_{x}(x, y)-i u_{y}(x, y).\]
We observe that \(u_{x}\) and \(-u_{y}\) are defined in \(\Omega\) and they satisfy Cauchy–Riemann equations in \(\Omega\) since \(u\) is harmonic in \(\Omega\).
Therefore, \(g\) is holomorphic in \(\Omega\).
Further, \(g=0\) on \(V\), since \(u_{x}\) and \(-u_{y}\) vanish on \(V\). Then \(g=0\) on \(\Omega\) by identity theorem for holomorphic functions.
Then \(u_{x}=u_{y}=0\) in \(\Omega\) which implies that \(u\) is constant in \(\Omega\). $$\tag*{$\blacksquare$}$$
Let \(u\) be harmonic in a region \(\Omega\). Then \(v\) is called a harmonic conjugate of \(u\) in \(\Omega\) if
\(v\) is harmonic in \(\Omega\).
There exists \(f \in H(\Omega)\) such that \[f=u+i v\quad \text { in } \quad \Omega.\]
Remark.
Let \(u\) be harmonic in a region \(\Omega\). Assume that \(v\) and \(v_{1}\) are harmonic conjugates of \(u\) in \(\Omega\). Then there exist \(f \in H(\Omega)\) and \(f_{1} \in H(\Omega)\) such that \[f=u+i v, \quad f_{1}=u+i v_{1} \quad \text { in } \quad \Omega.\]
Then \(v-v_{1}=-i\left(f-f_{1}\right) \in H(\Omega)\) is real valued. Therefore \(v\) and \(v_{1}\) differ by a constant. Why?
Remark.
Let \(f\) be integrable on \([a, b]\) and \[F(x)=\int_{a}^{x} f(t) d t \quad \text { for }\quad a \leq x \leq b.\]
Then \(F(x)\) is continuous in \([a, b]\). If \(f\) is continuous at \(x_{0} \in[a, b]\), then \[F^{\prime}\left(x_{0}\right)=f\left(x_{0}\right).\]
Let \(f\) be integrable on \([a, b]\). If there exists a differentiable function \(F\) on \([a, b]\) such that \(F^{\prime}=f\). Then \[\int_{a}^{b} f(t) d t=F(b)-F(a).\]
Let \(\Omega=D(0, R)\) where \(0<R \leq \infty\). Let \(u\) be harmonic in \(\Omega\). Then there exists a harmonic conjugate of \(u\) in \(\Omega\).
Proof: It suffices to find a real-valued function \(v=v(x, y)\) satisfying:
\(v\) has continuous partial derivatives at every point of \(\Omega\).
\(u\) and \(v\) satisfy the Cauchy–Riemann equations \[u_{x}=v_{y}\quad \text{ and } \quad u_{y}=-v_{x}\] at every point of \(\Omega\).
Then \(f=u+i v \in H(\Omega)\). Now we see from the previous theorem that \(v\) will be a harmonic conjugate of \(u\) in \(\Omega\).
For \((x, t) \in \Omega\), by the first equation in (ii), we have \[u_{x}(x, t)=v_{y}(x, t).\]
We integrate both sides with respect to \(t\) along a vertical line from 0 to \(y\). We have \[\int_{0}^{y} v_{y}(x, t) d t=\int_{0}^{y} u_{x}(x, t) d t.\]
Thus \[v(x, y)-v(x, 0)=\int_{0}^{y} u_{x}(x, t) d t.\]
By putting \(v(x, 0)=h(x)\), we have \[v(x, y)=\int_{0}^{y} u_{x}(x, t) d t+h(x).\]
We determine \(h(x)\) such that the second equation in (ii) is satisfied.
By substituting \(v(x, y)\) in the second equation in (ii), we have \[\begin{aligned} u_{y}(x, y) & =-\frac{\partial}{\partial x} \int_{0}^{y} u_{x}(x, t) d t-h^{\prime}(x) =-\int_{0}^{y} u_{x x}(x, t) d t-h^{\prime}(x) \\ & =\int_{0}^{y} u_{y y}(x, t) d t-h^{\prime}(x) =u_{y}(x, y)-u_{y}(x, 0)-h^{\prime}(x). \end{aligned}\]
Therefore, \(h^{\prime}(x)=-u_{y}(x, 0)\), which is satisfied if \[h(x)=-\int_{0}^{x} u_{y}(s, 0) d s+C,\] where \(C\) is any constant. Then \[v(x, y)=\int_{0}^{y} u_{x}(x, t) d t-\int_{0}^{x} u_{y}(s, 0) d s+C.\]
We check that \(v\) satisfies (i) and (ii) and hence \(v\) is a harmonic conjugate of \(u\) in \(\Omega\).$$\tag*{$\blacksquare$}$$
A region \(\Omega\) is simply connected if and only if every harmonic function in \(\Omega\) has a harmonic conjugate in \(\Omega\).
Let \(u=u(x, y)\) and \(v=v(x, y)\) be harmonic function in a region \(\Omega\). For \((x, y) \in \Omega\), let \[R=R(x, y)=\frac{1}{2} \log \left((u(x, y))^{2}+(v(x, y))^{2}\right).\] Then \(R\) is harmonic in \(\Omega\).
Proof: It is clear that \(R\) is continuous and it has continuous partial derivatives of orders \(1\) and \(2\) at every point of \(\Omega\).
We show that \(R\) satisfies the Laplace equation at every point of \(\Omega\).
At \((x, y) \in \Omega\), we have \[R_{x}=\frac{u u_{x}+v v_{x}}{u^{2}+v^{2}}, \quad R_{y}=\frac{u u_{y}+v v_{y}}{u^{2}+v^{2}},\] and \[\begin{aligned} \left(u^{2}+v^{2}\right)^{2}\left(R_{x x}+R_{y y}\right) =&\left(u^{2}+v^{2}\right)\left(u_{x}^{2}+v_{x}^{2}+u_{y}^{2}+v_{y}^{2}\right)\\ &-2\left(u u_{x}+v v_{x}\right)^{2}-2\left(u u_{y}+v v_{y}\right)^{2}, \end{aligned}\] by using \(u_{x x}+u_{y y}=0\) and \(v_{x x}+v_{y y}=0\).
Simplifying, we obtain \[\begin{aligned} \left(u^{2}+v^{2}\right)^{2}\left(R_{x x}+R_{y y}\right)= & u^{2} v_{x}^{2}+u^{2} v_{y}^{2}+v^{2} u_{x}^{2}+v^{2} u_{y}^{2}\\ &-\left(u^{2} u_{y}^{2}+u^{2} u_{x}^{2}+v^{2} v_{y}^{2}+v^{2} u_{x}^{2}\right) \\ & -2 u v u_{x} u_{y}-2 u v v_{x} v_{y} = 0 \end{aligned}\] by using the Cauchy-Riemann equations.$$\tag*{$\blacksquare$}$$
Let \(\Omega=\mathbb{C} \backslash\{0\}\). For \(z \in \Omega\) with \(z=x+i y\), let \[u(x, y)=\log |z|=\frac{1}{2} \log \left(x^{2}+y^{2}\right).\] Then \(u\) is harmonic in \(\Omega\).
Proof: We observe that \(u\) is continuous in \(\Omega\) where it has continuous partial derivatives of orders \(1\) and \(2\), since \[u_{x}=\frac{x}{x^{2}+y^{2}}, \quad u_{y}=\frac{y}{x^{2}+y^{2}}\] and \[u_{x x}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}, \quad u_{y y}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}} .\] The latter equation implies that \(u\) is harmonic in \(\Omega\).$$\tag*{$\blacksquare$}$$
Let \(D_{1}\) and \(\Omega_{1}\) be open discs. Let \(F\) be holomorphic function from \(D_{1}\) into \(\Omega_{1}\) and \(u\) be harmonic in \(\Omega_{1}\). Then \(u \circ F\) is harmonic in \(D_{1}\).
Proof: Let \(F(z)=A(x, y)+i B(x, y)\) for \(z=x+i y \in D_{1}\).
Since \(u\) is harmonic in \(\Omega_{1}\) and \(D_{1}\) is a disc, then there exists \(G \in H\left(\Omega_{1}\right)\) such that \[G(z)=\phi(x, y)+i \psi(x, y)\quad \text { for } \quad z=x+i y \in \Omega_{1},\] where \(\phi(x, y)=u(x, y)\). Then, for \(z=x+i y \in D_{1}\), we have \[\begin{aligned} G \circ F(z) & =G(A(x, y)+i B(x, y)) \\ & =\phi(A(x, y), B(x, y))+i \psi(A(x, y), B(x, y)) \\ & =u(A(x, y), B(x, y))+i \psi(A(x, y), B(x, y)), \end{aligned}\] and \(\operatorname{Re}(G \circ F(z))=u(A(x, y), B(x, y))=u \circ F(z)\). Now we conclude that \(u \circ F\) is harmonic in \(D_{1}\) and we are done.$$\tag*{$\blacksquare$}$$
A region \(\Omega\) is simply connected if and only if every harmonic function in \(\Omega\) has a harmonic conjugate in \(\Omega\).
Proof: Assume that \(\Omega\) is simply connected and let \(u\) be harmonic in \(\Omega\). We show that \(u\) has a harmonic conjugate in \(\Omega\).
We may assume that \(\Omega \neq \mathbb{C}\) otherwise the assertion follows from the previous theorem.
Then, by the Riemann mapping theorem, there exists an analytic homeomorphism \(F\) from \(D\) onto \(\Omega\).
In the previous lemma, we take \(D_{1}=D\left(z_{0}, s\right), \Omega_{1}=D\left(F\left(z_{0}\right), r\right)\) and \(F\) is holomorphic function from \(D_{1}\) into \(\Omega_{1}\). Since \(\Omega_{1} \subseteq \Omega\), we see that \(u\) is harmonic in \(\Omega_{1}\). Let \(u \circ F=u_{1}\).
Let \(z_{0} \in D\). Then \(F\left(z_{0}\right) \in \Omega\) and there exist \(0<s<r<1\) such that \[F\left(D\left(z_{0}, s\right)\right) \subseteq D\left(F\left(z_{0}\right), r\right) \subseteq \Omega.\]
Then \(u_{1}\) is harmonic in \(D_{1}\) by the previous lemma. In particular, \(u_{1}\) is harmonic at \(z_{0}\).
Since \(z_{0}\) is an arbitrary point of \(D\), we see that \(u_{1}\) is harmonic in \(D\). Hence, there exist \(v_{1}\) harmonic in \(D\) and \(f_{1} \in H(D)\) such that \[f_{1}=u_{1}+i v_{1} \quad \text { in }\quad D.\]
Then \[f_{1} \circ F^{-1}=u+i v_{1} \circ F^{-1} \quad \text { in }\quad \Omega,\] and \(f_{1} \circ F^{-1} \in H(\Omega)\). Hence, we conclude that \(v_{1} \circ F^{-1}\) is harmonic conjugate of \(u\) in \(\Omega\).
Now, let \(\Omega\) be a region and assume that every harmonic function in \(\Omega\) has a harmonic conjugate in \(\Omega\). We show that \(\Omega\) is simply connected.
We may assume that \(\Omega \neq \mathbb{C}\) otherwise the assertion follows since \(\mathbb{C}\) is simply connected.
It suffices to show that for every \(f \in H(\Omega)\) with \(\frac{1}{f} \in H(\Omega)\), there exists \(g \in H(\Omega)\) such that \(f(z)=g^{2}(z)\) for \(z \in \Omega\). Then \(\Omega\) is conformally equivalent to \(D\). Hence \(\Omega\) is simply connected as desired.
Let \(f \in H(\Omega)\) with \(\frac{1}{f} \in H(\Omega)\). We set \[\operatorname{Re}(f)=u, \quad \operatorname{Im}(f)=v.\]
Then \(u\) and \(v\) are harmonic in \(\Omega\). For \(x+i y \in \Omega\), we set \[R(x, y)=\log |f(x+i y)|=\frac{1}{2} \log \left((u(x, y))^{2}+(v(x, y))^{2}\right),\] which is defined since \(f(z) \neq 0\) for \(z \in \Omega\).
\(R(x, y)\) is harmonic in \(\Omega\) as it was shown above.
Then, by our assumption, there exists a harmonic function \(S\) in \(\Omega\) and \(g_{1} \in H(\Omega)\) such that \[g_{1}=R+i S \quad \text { in } \quad \Omega.\] Let \[h(z)=e^{g_{1}(z)} \quad \text { for }\quad z \in \Omega.\]
Then \(\frac{f(z)}{h(z)} \in H(\Omega)\) and \[\left|\frac{f(z)}{h(z)}\right|=1 \quad \text { for } \quad z \in \Omega.\]
Therefore \(\frac{f(z)}{h(z)}\) is constant in \(\Omega\) by the open mapping theorem.
Then \(f(z)=c e^{g_{1}(z)}=e^{g_{1}(z)+c_{1}}\), where \(c\) and \(c_{1}\) are constants.
By putting \[g(z)=e^{\frac{g_{1}(z)+c_{1}}{2}},\] we see that \(g(z) \in H(\Omega)\) and \(f(z)=(g(z))^{2}\) for \(z \in \Omega\).$$\tag*{$\blacksquare$}$$
Let \(u\) be real-valued continuous function in a region \(\Omega\). Then \(u\) has mean value property (MVP) in \(\Omega\) if for every \(a \in \Omega\), we have \[u(a)=\frac{1}{2 \pi} \int_{0}^{2 \pi} u\left(a+r e^{i \theta}\right) d \theta,\] whenever \(\overline{D}(a, r) \subseteq \Omega\).
Let \(u\) be harmonic in a region \(\Omega\). Then \(u\) satisfies MVP in \(\Omega\).
Proof Let \(a \in \Omega\) with \(\overline{D}(a, r) \subseteq \Omega\). There exists an open disc \(E\) such that \[\overline{D}(a, r) \subseteq E \subseteq \Omega\] and \(u\) has a harmonic conjugate in \(E\).
Therefore there exists \(f \in H(E)\) such that \(u=\operatorname{Re}(f)\). Now \[f(a)=\frac{1}{2 \pi i} \int_{|z-a|=r} \frac{f(z)}{z-a} d z\] by the Cauchy integral formula.
By putting \(z-a=r e^{i \theta}\) with \(0 \leq \theta \leq 2 \pi\), we have \[f(a)=\frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{f\left(a+r e^{i \theta}\right) i r e^{i \theta}}{r e^{i \theta}} d \theta=\frac{1}{2 \pi} \int_{0}^{2 \pi} f\left(a+r e^{i \theta}\right) d \theta.\]
By comparing the real parts on both the sides, we obtain \[u(a)=\frac{1}{2 \pi} \int_{0}^{2 \pi} u\left(a+r e^{i \theta}\right) d \theta.\]
This holds for every \(a \in \Omega\) whenever \(\overline{D}(a, r) \subseteq \Omega\).$$\tag*{$\blacksquare$}$$
Let \(u\) be real-valued continuous function in a region \(\Omega\) and assume that \(u\) has MVP in \(\Omega\). Suppose that there exists \(a \in \Omega\) such that \[u(z) \leq u(a) \quad \text { for all }\quad z \in \Omega.\] Then \(u\) is constant in \(\Omega\).
Proof: We assume that \(u\) is not constant in \(\Omega\). Let \(u\) be continuous in a region satisfying \(M V P\) in \(\Omega\) and there exists \(a \in \Omega\) such that \(u(z) \leq u(a)\) for \(z \in \Omega\). We consider \[A=\{z \in \Omega : u(z)=u(a)\} .\]
We may assume that \(A \neq \varnothing\), since \(a \in A\). It suffices to show that \(A\) is both open and closed.
Then \(A=\Omega\), since \(\Omega\) is connected and hence \(u\) is constant in \(\Omega\).
Let \(z \in \overline{A}\). Then there exists a sequence \((z_{n})_{n\in\mathbb N}\subseteq A\) such that \(\lim _{n \rightarrow \infty} z_{n}=z\). Since \(u\) is continuous, we have \(\lim _{n \rightarrow \infty} u\left(z_{n}\right)=u(z)\). But \(u\left(z_{n}\right)=a\) for \(n \geq 1\) since \(z_{n} \in A\). Therefore \(u(z)=u(a)\) which implies that \(z \in A\). Thus \(\overline{A} \subseteq A\) and hence \(A\) is closed.
Now we show that \(A\) is open. Let \(z_{0} \in A\) and there exists \(r>0\) with \(D\left(z_{0}, r\right) \subseteq \Omega\) such that \(D\left(z_{0}, r\right)\) is not contained in \(A\).
Then there exists \(b \in D\left(z_{0}, r\right)\) and \(b \notin A\). Thus \[u(b)<u(a)=u\left(z_{0}\right)\]
Since \(u\) is continuous, there exists \(s>0\) such that \[u(z)<u(a) \quad \text { for }\quad z \in D(b, s) .\]
Let \(\left|b-z_{0}\right|=\rho<r\). Then there exists an arc on the circle \(\left|z-z_{0}\right|=\rho\) containing \(b\) of positive length where \(u(z)<u\left(z_{0}\right)\) and \(u(z) \leq u(a)=u\left(z_{0}\right)\) elsewhere on the circle.
Therefore \[\frac{1}{2 \pi} \int_{0}^{2 \pi} u\left(z_{0}+\rho e^{i \theta}\right) d \theta<u\left(z_{0}\right).\]
On the other hand, we have \[\frac{1}{2 \pi} \int_{0}^{2 \pi} u\left(z_{0}+\rho e^{i \theta}\right) d \theta=u\left(z_{0}\right),\] since \(u\) satisfies \(M V P\) by the assumption. This is a contradiction.$$\tag*{$\blacksquare$}$$
Let \(\Omega\) be a bounded region. Assume that \(u\) is a non-constant real-valued continuous function defined on \(\overline{\Omega}\) and \(u\) has MVP in \(\Omega\). Then there exists \(a \in \partial \Omega\) such that \[u(z)<u(a) \quad \text { for }\quad z \in \Omega.\]
Proof: Prove it!
For \(0 \leq r<1\) and \(0 \leq \theta \leq 2 \pi\), the function \[P_{r}(\theta)=\sum_{n=-\infty}^{\infty} r^{|n|} e^{i n \theta} \tag{*}\] is called the Poisson kernel.
We understand that \(0^{0}=1\) in the sum on the right-hand side of (*) so that \(P_{r}(\theta)=1\) if \(r=0\).
For \(0 \leq r<1\) and \(0 \leq \theta \leq 2 \pi\), we have \[P_{r}(\theta)=\operatorname{Re}\left(\frac{1+r e^{i \theta}}{1-r e^{i \theta}}\right)=\frac{1-r^{2}}{1-2 r \cos \theta+r^{2}}. \tag{**}\]
Proof: For \(0 \leq|z|<1\), we have \[\begin{aligned} \frac{1+z}{1-z}&=(1+z)(1-z)^{-1}\\ &=(1+z)\left(1+z+z^{2}+\cdots\right)\\ &=1+2 \sum_{n=1}^{\infty} z^{n}. \end{aligned}\] Here the rearrangement of terms of the series is permissible since the series is absolutely convergent.
By putting \(z=r e^{i \theta}\) with \(0 \leq r<1\) above, we have \[\frac{1+r e^{i \theta}}{1-r e^{i \theta}}=1+2 \sum_{n=1}^{\infty} r^{n} e^{i n \theta}.\]
Now \[\begin{aligned} \operatorname{Re}\left(\frac{1+r e^{i \theta}}{1-r e^{i \theta}}\right) & =1+2 \sum_{n=1}^{\infty} r^{n} \cos n \theta\\ &=1+\sum_{n=1}^{\infty} r^{n}\left(e^{i n \theta}+e^{-i n \theta}\right) \\ & =1+\sum_{n=1}^{\infty} r^{n} e^{i n \theta}+\sum_{n=-\infty}^{-1} r^{|n|} e^{i n \theta}\\ &=1+\sum_{\substack{n=-\infty \\ n \neq 0}}^{\infty} r^{|n|} e^{i n \theta}=P_{r}(\theta). \end{aligned}\]
Further \[\frac{1+r e^{i \theta}}{1-r e^{i \theta}}=\frac{\left(1+r e^{i \theta}\right)\left(1-r e^{-i \theta}\right)}{\left|1-r e^{i \theta}\right|^{2}}=\frac{1-r^{2}+2 i r \sin \theta}{\left|1-r e^{i \theta}\right|^{2}}\] and \[\left|1-r e^{i \theta}\right|^{2}=1-2 r \cos \theta+r^{2} .\]
Therefore \[P_{r}(\theta)=\operatorname{Re}\left(\frac{1+r e^{i \theta}}{1-r e^{i \theta}}\right)=\frac{1-r^{2}}{1-2 r \cos \theta+r^{2}}.\]
For \(0 \leq r<1\), we have \(P_{r}(\theta)>0\) for \(0 \leq \theta \leq 2 \pi\) and \(P_{r}(\theta)\) is periodic with period \(2 \pi\). Further \[\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}(\theta) d \theta=1.\]
Let \(\delta>0\). Then \[\lim _{r \rightarrow 1^{-}} P_{r}(\theta)=0\] uniformly in \(\theta\) with \(\delta \leq|\theta| \leq \pi\).
Proof of (a): It is clear that \(P_{r}(\theta)>0\) for \(0 \leq \theta \leq 2 \pi\) and periodic with period \(2 \pi\), since \[P_{r}(\theta)=\operatorname{Re}\left(\frac{1+r e^{i \theta}}{1-r e^{i \theta}}\right)=\frac{1-r^{2}}{1-2 r \cos \theta+r^{2}}.\]
By integrating both sides of the previous identity, we obtain \[\begin{aligned} \int_{-\pi}^{\pi} P_{r}(\theta) & =\int_{-\pi}^{\pi} \sum_{n=-\infty}^{\infty} r^{|n|} e^{i n \theta} d \theta =\sum_{n=-\infty}^{\infty} r^{|n|} \int_{-\pi}^{\pi} e^{i n \theta} d \theta=2 \pi, \end{aligned}\] since the series converges uniformly in \(\theta\) and \[\int_{-\pi}^{\pi} e^{i n \theta}= \begin{cases} 2 \pi & \text{ if } n=0,\\ 0 & \text{ if } n\neq 0. \qquad \tag*{$\blacksquare$} \end{cases}\]
Proof of (b): Let \(\delta>0\) and \(0<r<1\).
We may assume that \(|\theta| \leq \frac{\pi}{2}\) otherwise the assertion follows immediately from the formula \[P_{r}(\theta)=\operatorname{Re}\left(\frac{1+r e^{i \theta}}{1-r e^{i \theta}}\right)=\frac{1-r^{2}}{1-2 r \cos \theta+r^{2}}.\]
By differentiating both sides with respect to \(\theta\) in the previous formula and setting \(\theta=t\), we have \[P_{r}^{\prime}(t)=\frac{-\left(1-r^{2}\right) 2 r \sin t}{\left(1-2 r \cos t+r^{2}\right)^{2}}.\]
Then \[P_{r}^{\prime}(t)<0 \quad \text { for }\quad \delta \leq t \leq \frac{\pi}{2}.\]
Thus \[P_{r}(\theta) \leq P_{r}(\delta) \quad \text { for }\quad \delta \leq \theta \leq \frac{\pi}{2} .\]
Since \(P_{r}(\theta)=P_{r}(-\theta)\), we obtain \[P_{r}(\theta) \leq P_{r}(\delta) \quad \text { for }\quad \delta \leq|\theta| \leq \frac{\pi}{2}.\]
Since \(\lim _{r \rightarrow 1^{-}} P_{r}(\delta)=0\), we derive that \(\lim _{r \rightarrow 1^{-}} P_{r}(\theta)=0\) uniformly in \(\delta \leq|\theta| \leq \frac{\pi}{2}\). This completes the proof. $$\tag*{$\blacksquare$}$$
Let \(a\in \mathbb{C}\), \(\rho>0\) and \(f\) be real-valued continuous function defined on the circle \(C(a, \rho)\). Then there exists unique real-valued continuous function \(u\) in \(\overline{D}(a, \rho)\) such that \(u\) is harmonic in \(D(a, \rho)\) and \[u(z)=f(z)\quad \text { for } \quad z\in C(a, \rho).\]
Proof: We claim that there is no loss of generality in assuming that \(a=0\) and \(\rho=1\). Indeed, suppose that the assertion of the theorem is valid with \(a=0\) and \(\rho=1\). Let \(f\) be real-valued continuous function on \(C(a, \rho)\).
Then we consider \[g(z)=f(a+\rho z)\quad \text { for } \quad |z|=1.\]
We note that \(g\) is continuous on \(|z|=1\). Then there is a real-valued continuous function \(v(z)\) in \(\overline{D}\) and harmonic in \(D\) such that \[v(z)=g(z) \quad \text { for } \quad |z|=1.\]
Let \[u(z)=v\left(\frac{z-a}{\rho}\right) \quad \text { for }\quad z \in \overline{D}(a, \rho).\]
Then the conclusion follows, since \(u\) is a real-valued continuous function in \(\overline{D}(a, \rho)\) and harmonic in \(D(a, \rho)\) and such that \(u(z)=f(z)\) for \(|z-a|=\rho\).
Let \(M=\max \left\{\left|f\left(e^{i \phi}\right)\right|: |\phi| \leq 2 \pi\right\}\). We prove the theorem with \[u\big(r e^{i \theta}\big)= \begin{cases}\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}(\theta-\phi) f\left(e^{i \phi}\right) d \phi & \text { if } 0 \leq r<1,0 \leq \theta \leq 2 \pi, \\ f\left(e^{i \theta}\right) & \text { if } r=1,0 \leq \theta \leq 2 \pi . \end{cases}\]
Let \(0 \leq r<1\). We show that \(u\) is real part of an analytic function and then it is harmonic in \(D\).
By the formula \(P_{r}(\theta)=\operatorname{Re}\left(\frac{1+r e^{i \theta}}{1-r e^{i \theta}}\right)=\frac{1-r^{2}}{1-2 r \cos \theta+r^{2}},\) we have \[u\big(r e^{i \theta}\big)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f\big(e^{i \phi}\big) \operatorname{Re}\left(\frac{1+r e^{i(\theta-\phi)}}{1-r e^{i(\theta-\phi)}}\right) d \phi.\]
We observe that \[u\big(r e^{i \theta}\big)=\operatorname{Re}(g(z)) \quad \text { with }\quad z=r e^{i \theta},\] where \[g(z)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f\big(e^{i \phi}\big)\left(\frac{e^{i \phi}+z}{e^{i \phi}-z}\right) d z,\] which is analytic in \(D\).
Therefore \(u\) is harmonic in \(D\), hence it is continuous in \(D\). Further \(u\left(e^{i \theta}\right)=f\left(e^{i \theta}\right)\) for \(0 \leq \theta \leq 2 \pi\).
Now we show that \(u\) is continuous on \(|z|=1\).
We have \(\big|u\big(e^{i \theta}\big)\big|=\big|f\big(e^{i \theta}\big)\big| \leq M\) for \(0 \leq \theta \leq 2 \pi\).
Further \(f\left(e^{i \theta}\right)\) with \(0 \leq \theta \leq 2 \pi\) is uniformly continuous. Therefore for \(\varepsilon>0\), there exists \(\delta>0\) such that \[\big|u\big(e^{i \theta}\big)-u\big(e^{i \phi}\big)\big|=\big|f\big(e^{i \theta}\big)-f\big(e^{i \phi}\big)\big|<\varepsilon,\] whenever \(|\theta-\phi| \leq \delta\). Let \(A\) be an arc of the circle \(|z|=1\) with \(e^{i \theta}\) as the centre of the arc and subtending an angle \(\delta\) at the origin. Then \(|\theta-\phi| \leq \delta\) whenever \(e^{i \phi} \in A\).
Thus it suffices to show that for any \(e^{i \theta}\) with \(0 \leq \theta \leq 2 \pi\), we have \[\big|u\big(r e^{i \theta}\big)-u\big(e^{i \theta}\big)\big|<2 \varepsilon\quad \text { whenever }\quad r \rightarrow 1^{-}.\]
We also have \[u\big(r e^{i \theta}\big)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}\left(\theta-\gamma\right) u\left(e^{i \gamma}\right) d \gamma \quad \text { for }\quad 0 \leq r<1.\]
By setting \(\theta-\gamma=\phi\), we obtain for \(0 \leq r<1\) that \[u\big(r e^{i \theta}\big)=\frac{1}{2 \pi} \int_{-\pi+\theta}^{\pi+\theta} P_{r}(\phi) u\big(e^{i(\theta-\phi)}\big) d \phi=\frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}(\phi) u\big(e^{i(\theta-\phi)}\big) d \phi,\] since the integrand is periodic with period \(2 \pi\).
We further observe that \[\begin{aligned} u\big(r e^{i \theta}\big)-u\big(e^{i \theta}\big) = & \frac{1}{2 \pi} \int_{-\pi}^{\pi} P_{r}(\phi)\big(u\big(e^{i(\theta-\phi)}\big)-u\big(e^{i \theta}\big)\big) d \phi \\ = & \frac{1}{2 \pi} \int_{|\phi|<\delta} P_{r}(\phi)\big(u\big(e^{i(\theta-\phi)}\big)-u\big(e^{i \theta}\big)\big) d \phi \\ & +\frac{1}{2 \pi} \int_{\pi \geq|\phi| \geq \delta} P_{r}(\phi)\big(u\big(e^{i(\theta-\phi)}\big)-u\big(e^{i \theta}\big)\big) d \phi. \end{aligned}\]
Since \(P_{r}(\phi)>0\), the absolute value of the first integral is at most \[\frac{\varepsilon}{2 \pi} \int_{-\pi}^{\pi} P_{r}(\phi) d \phi=\varepsilon.\]
The absolute value of the second integral is at most \[2 M \max _{\delta \leq|\phi| \leq \pi} P_{r}(\phi)<2 M \frac{\varepsilon}{2 M}\] when \(r \rightarrow 1^{-}\), hence \[\big|u\big(r e^{i \theta}\big)-u\big(e^{i \theta}\big)\big|<2 \varepsilon\quad \text { whenever }\quad r \rightarrow 1^{-}.\]
It remains to show that \(u\) is unique satisfying the assertion of the theorem. Let \(v\) be a continuous function in \(\overline{D}\) such that \(v\) is harmonic in \(D\) and \(v(z)=f(z)\) for \(|z|=1\).
Now we consider the function \(w=u-v\). Then \(w\) is harmonic in \(D\), and therefore it has (MVP) in \(D\). Since \(w=0\) on \(|z|=1\), we conclude by the maximum principle, that \(w=0\) in \(\overline{D}\). Why?
Hence \(v=u\). The proof is completed. $$\tag*{$\blacksquare$}$$
Let \(u\) be a real-valued continuous function with (MVP) in a region. Then \(u\) is harmonic in \(\Omega\).
Proof: Let \(u\) be a real-valued continuous function with (MVP) in \(\Omega\).
Let \(a \in \Omega\). Since \(\Omega\) is open, there exists \(\rho>0\) such that \(\overline{D}(a, \rho) \subseteq \Omega\). It suffices to show that \(u\) is harmonic in \(D(a, \rho)\). Then \(u\) is harmonic at \(a\) and the assertion follows since \(a\) is an arbitrary point in \(\Omega\).
Since \(\overline{D}(a, \rho) \subseteq \Omega\), we see that \(u\) is continuous in \(\overline{D}(a, \rho)\) and it has (MVP) in \(D(a, \rho)\). By the Dirichlet problem, there exists a real-valued continuous function \(v\) in \(\overline{D}(a, \rho)\) such that \(v\) is harmonic in \(D(a, \rho)\) and such that \[u(z)=v(z) \quad \text { if }\quad |z-a|=\rho.\]
Further \(v\) has (MVP) in \(\Omega\), since \(v\) is harmonic. Next we consider \[g=u-v \quad \text { in }\quad \overline{D}(a, \rho).\]
We observe that \(g\) is real-valued continuous function in \(\overline{D}(a, \rho)\) and it has MVP in \(D(a, \rho)\).
Further \[g(z)=0 \quad \text { if }\quad |z-a|=\rho.\]
Assume that \(g\) is not a constant function. Then \(g(z)<0\) in \(D(a, \rho)\) by the maximum principle and \(g(z)>0\) in \(D(a, \rho)\) by the minimum principle. This is a contradiction.
Therefore \(g\) is a constant function \(c\) in \(D(a, \rho)\). In fact \(c=0\) since \(g\) is continuous in \(\overline{D}(a, \rho)\) and zero on \(|z-a|=\rho\).
Hence \(u=v\) is harmonic in \(D(a, \rho)\) as desired. $$\tag*{$\blacksquare$}$$