Theorem 6.2.1.
Suppose \(u\) is harmonic in the region \(G\) and \(\overline D[w,r] \subset G\text{.}\) Then
\begin{equation*}
u(w) \ = \ \frac 1 {2 \pi} \int_0^{2 \pi} u \! \left( w + r \,
e^{it} \right) \diff{t} \, \text{.}
\end{equation*}
Section 6.2 Mean-Value and Maximum/Minimum Principle
We have established an intimate connection between harmonic and holomorphic functions, and so it should come as no surprise that some of the theorems we proved for holomorphic functions have analogues in the world of harmonic functions. Here is such a harmonic analogue of Cauchy’s Integral Formula (Theorem 4.4.1 and Theorem 4.4.5).
Proof.
Exercise 6.3.14 provides \(R\) so that \(\overline D[w,r] \subset
D[w,R]\subset G\text{.}\) The open disk \(D[w,R]\) is simply connected, so by Theorem 6.1.7 there is a function \(f\) holomorphic in \(D[w,R]\) such that \(u = \Re f\) on \(D[w,R]\text{.}\) Now we apply Corollary 4.4.2 to \(f\text{:}\)
\begin{equation*}
f(w) \ = \ \frac 1 {2 \pi} \int_0^{2 \pi} f \! \left( w + r \,
e^{it} \right) \diff{t} \, \text{.}
\end{equation*}
Theorem 6.2.1 follows by taking the real part on both sides.
Corollary 4.4.2 and Theorem 6.2.1 say that holomorphic and harmonic functions have the mean-value property. Our next result is an important consequence of this property to extreme values of a function.
Definition 6.2.2.
Let \(G \subset \C\) be a region. The function \(u : G \to \R\) has a strong relative maximum at \(w \in G\) if there exists a disk \(D[w,r] \subseteq
G\) such that \(u(z) \le u(w)\) for all \(z \in D[w,r]\) and \(u(z_0) \lt u(w)\) for some \(z_0 \in D[w,r]\text{.}\) The definition of a strong relative minimum is analogous.
Theorem 6.2.3.
If \(u\) is harmonic in the region \(G\text{,}\) then it does not have a strong relative maximum or minimum in \(G\text{.}\)
Proof.
Assume, by way of contradiction, that \(w\) is a strong relative maximum. Then there is a disk in \(G\) centered at \(w\) containing a point \(z_0\) with \(u(z_0) \lt u(w)\text{.}\) Let \(r := |z_0 - w|\) and apply Theorem 6.2.1:
\begin{equation*}
u(w) \ = \ \frac 1 {2 \pi} \int_0^{2 \pi} u \! \left( w + r \,
e^{it} \right) \diff{t} \, \text{.}
\end{equation*}
Intuitively, this cannot hold, because some of the function values we’re integrating are smaller than \(u(w)\text{,}\) contradicting the mean-value property. To make this into a thorough argument, suppose that \(z_0 = w + r \, e^{it_0}\) for \(0 \leq t_0
\lt 2\pi\text{.}\) Because \(u(z_0) \lt u(w)\) and \(u\) is continuous, there is a whole interval of parameters \([t_0, t_1] \subseteq
[0, 2 \pi]\) such that \(u( w + r \, e^{it}) \lt u(w)\) for \(t_0 \le t \le t_1\text{.}\) Now we split up the mean-value integral:
\begin{align*}
u(w) \amp \ = \ \frac{1}{2 \pi} \int_0^{2 \pi} u \! \left(
w + r \, e^{it} \right) \diff{t}\\
\amp \ = \ \frac 1 {2 \pi} \left( \int_0^{t_0} u \! \left(
w + r \, e^{it} \right) \diff{t} + \int_{t_0}^{t_1} u \!
\left( w + r \, e^{it} \right) \diff{t} \right.\\
\amp \qquad \qquad + \left. \int_{t_1}^{2 \pi} u
\! \left( w + r \, e^{it} \right) \diff{t} \right) .
\end{align*}
All the integrands can be bounded by \(u(w)\text{;}\) for the middle integral we get a strict inequality. Hence
\begin{equation*}
u(w) \ \lt \ \frac 1 {2 \pi} \left( \int_0^{t_0} u(w) \,
\diff{t} + \int_{t_0}^{t_1} u(w)\, \diff{t} + \int_{t_1}^{2
\pi} u(w) \, \diff{t} \right) \ = \ u(w) \, \text{,}
\end{equation*}
a contradiction.
The same argument works if we assume that \(u\) has a relative minimum. But in this case there’s a shortcut argument: if \(u\) has a strong relative minimum then the harmonic function \(-u\) has a strong relative maximum, which we just showed cannot exist.
So far, harmonic functions have benefited from our knowledge of holomorphic functions. Here is a result where the benefit goes in the opposite direction.
Corollary 6.2.4.
If \(f\) is holomorphic and nonzero in the region \(G\text{,}\) then \(|f|\) does not have a strong relative maximum or minimum in \(G\text{.}\)
Proof.
By Exercise 6.3.6, the function \(\ln|f(z)|\) is harmonic on \(G\) and so, by Theorem 6.2.3, does not have a strong relative maximum or minimum in \(G\text{.}\) But then neither does \(|f(z)|\text{,}\) because \(\ln\) is monotonic.
We finish our excursion about harmonic functions with a preview and its consequences. We say a real valued function \(u\) on a region \(G\) has a weak relative maximum at \(w\) if there exists a disk \(D[w,r] \subseteq G\) such that all \(z \in D[w,r]\) satisfy \(u(z) \leq u(w)\text{.}\) We define weak relative minimum similarly. In Chapter 8 we will strengthen Theorem 6.2.3 and Corollary 6.2.4 to Theorem 8.2.4 and Corollary 8.2.7 by replacing strong relative extremum in the hypotheses with weak relative extremum. A special but important case of the maximum/minimum principle for harmonic functions, Corollary 8.2.7, concerns bounded regions. In Chapter 8 we will establish that, if \(u\) is harmonic in a bounded region \(G\) and continuous on its closure, then
1
In particular, we will show that one does not have to assume that \(f\) is nonzero in a region \(G\) to have a strong relative maximum in \(G\text{.}\)
\begin{equation}
\sup_{z \in G} u(z) \ = \ \max_{z \in \partial G} u(z) \qquad
\text{ and } \qquad \inf_{z \in G} u(z) \ = \ \min_{z \in \partial
G} u(z) \tag{6.1}
\end{equation}
where, as usual, \(\partial G\) denotes the boundary of \(G\text{.}\) We will exploit this in the next two corollaries.
Corollary 6.2.5.
Suppose \(u\) is harmonic in the bounded region \(G\) and continuous on its closure. If \(u\) is zero on \(\partial G\) then \(u\) is zero in \(G\text{.}\)
Proof.
By (6.1),
\begin{equation*}
u(z) \ \leq \ \sup_{z \in G} u(z) \ = \ \max_{z \in \partial G}
u(z) \ = \ 0
\end{equation*}
and
\begin{equation*}
u(z) \ \geq \ \inf_{z \in G} u(z) \ = \ \min_{z \in \partial G}
u(z) \ = \ 0 \, \text{,}
\end{equation*}
so \(u\) must be zero in \(G\text{.}\)
Corollary 6.2.6.
Suppose \(u\) and \(v\) are harmonic in the bounded region \(G\) and continuous on its closure. If \(u(z) = v(z)\) for all \(z \in \partial G\) then \(u(z) = v(z)\) for all \(z \in G\text{.}\)
Proof.
\(u-v\) is harmonic in \(G\) (Exercise 6.3.2) and is continuous on the closure \(\overline G\text{,}\) and \(u-v\) is zero on \(\partial G\text{.}\) Now apply Corollary 6.2.5.
Corollary 6.2.6 says that if a function \(u\) is harmonic in a bounded region \(G\) and is continuous on the closure \(\overline G\) then the values of \(u\) at points in \(G\) are completely determined by the values of \(u\) on the boundary of \(G\text{.}\) We should remark, however, that this result is of a completely theoretical nature: it says nothing about how to extend a continuous function \(u\) given on the boundary of a region to be harmonic in the full region. This problem is called the Dirichlet problem, and it has a solution for all bounded simply-connected regions. If the region is the unit disk and \(u\) is a continuous function on the unit circle, define
2
Named after Johann Peter Gustav Dirichlet (1805–1859).
\begin{align*}
\hat u \! \left( e^{ i \phi } \right) \ \amp := u \! \left( e^{ i \phi } \right) \quad \text{ and } \\
\hat u \! \left( r \, e^{ i \phi } \right) \ \amp := \ \frac{ 1 }{ 2 \pi } \int_0^{ 2 \pi } u \! \left( e^{ i t } \right) P_r(\phi - t) \, \diff{t} \quad \text{ for } \quad r\lt 1\, \text{,}
\end{align*}
where \(P_r(\phi)\) is the Poisson kernel which we introduced in Exercise 4.5.31. Then \(\hat u\) is the desired extension: it is continuous on the closed unit disk, harmonic in the open unit disk, and agrees with \(u\) on the unit circle. In simple cases this solution can be converted to solutions in other regions, using a conformal map to the unit disk. All of this is beyond the scope of this book, though Exercise 6.3.13 gives some indication why the above formula does the trick. At any rate, we remark that Corollary 6.2.6 says that the solution to the Dirichlet problem is unique.
