Subharmonic Functions: A First Glimpse
Proving a Theorem Using the Knowledge of Subharmonic Functions
Theorem. Suppose \(X\) is a Riemann surface and \(f:X\to \mathbb{C}\) is a nonconstant complex-valued harmonic function. Then \(|f|\) does not attain its maximum.
Proof. A one-line proof: it's evident because the function \(|f|\) is subharmonic and nonconstant.
The essential reason of \(|f|\) being subharmonic is as follows, based on the various propositions in the next section.
Let \(f(x+iy)=u(x,y)+iv(x,y)\) be a nonconstant harmonic function on a open subset \(U\) of \(\mathbb{C}\). By direct computation, on \(\{z\in U:f(z)\neq 0\}\) there holds
and thus \(\Delta|f|\ge 0\) by Cauchy-Schwartz inequality. Hence \(|f|\) is subharmonic on \(\{z\in U:f(z)\neq 0\}\). As the subharmonicity of \(|f|\) at the zeros of \(f\) is immediate, we conclude that \(|f|\) is subharmonic on \(U\). \(\blacksquare\)
(Continuous) Subharmonic Functions
Let's recall subharmonic functions in an informal but detailed way.
The standard definition of subharmonic functions considers upper semi-continuous functions instead of (the more restricted) continuous functions. But we'll restrict our attention to the space of continous functions for a quicker acquaintance of subharmonicity.
A real-valued continuous function \(v\) on an open subset \(U\) of \(\mathbb{C}\) is said to be subharmonic, if for any harmonic function \(u\) on an arbitrary connected open subset \(V\) of \(U\), \(v-u\) does not attain its maximum as a function on \(V\), unless it's constant. By the maximum principle of harmonic functions, we see that any harmonic function is subharmonic.
Subharmonicity is a local property: \(v\) is subharmonic iff \(v\) is subharmonic at each point of the domain. By \(v\) is subharmonic at \(z\in U\), we mean there exists an open neighborhood \(U_z\subset U\) such that \(v|_{U_z}\) is subharmonic.
The only if part is obvious. For the if part, suppose \(v-u\) attains its maximum at \(z_0\in V\), then \(v-u\) is constant when restricted to \(U_{z_0}\cap V\). Consider the set
which is clearly a nonempty closed subset of \(V\). Suppose \(z_1\in E\), i.e., \(v-u\) attains its maximum at \(z_1\), then \(v-u\) is constant on an open neighborhood \(U_{z_1}\cap V\) of \(z_1\), and so \(U_{z_1}\cap V\subset E\), whence \(E\) is an open subset of \(V\). Since \(V\) is connected, we conclude that \(E=V\), i.e., \(v-u\) is constant on \(V\). Therefore, \(v\) is subharmonic and the if part is proved.
In the general setting, we show that \(v\) is subharmonic iff
\[ \begin{equation*} \text{$v(z_0)\le \dfrac{1}{2\pi}\displaystyle\int_{0}^{2\pi} v(z_0+re^{i\theta}) \,\mathrm{d}\theta\quad$ whenever $\quad \overline{D(z_0,r)}\subset U$} \end{equation*} \]
The only if part is immediate from the Possion integral (which gives the solution to the Dirichlet problem of the Laplace equation) and the mean value property of harmonic functions. The if part is again a consequence of the mean value property of harmonic functions. Let \(z_0\in U\) and \(D(z_0,r)\subset U\). Suppose \(u\) is a harmonic function on \(D(z_0,r)\) such that \(v-u\) attains its maximum at \(z_1\in D(z_0,r)\). Assume without loss of generality that \((v-u)(z_1)=0\), for otherwise we may replace \(u\) by the function \(\widetilde{u}=u+(v-u)(z_1)\). Take \(0<r'<r-|z_1-z_0|\) so that \(\overline{D(z_1,r')}\subset D(z_0,r)\). We note that when restricted to \(\overline{D(z_1,r')}\), \(u-v\) attains its maximum at \(z_1\) and in particular, \(v(z)\le u(z)\) for all \(z\in \partial D(z_1,r')\). Thus, by the mean value property of harmonic functions, we have
But \((v-u)(z_1)=0\), so both the inequalities above are equalities. Therefore, by the continuity of \(v\) and \(u\), we must have \(u(z)=v(z)=v(z_1)\) for all \(z\in \partial D(z_1,r')\). Since \(0<r'<r-|z_1-z_0|\) is arbitrary, we conclude that \(v-u\equiv 0\) on \(D(z_1,r-|z_1-z_0|)\). Consequently, we see that
is a nonempty open and closed subset of \(D(z_0,r)\), and thus \(E=D(z_0,r)\) by connectedness, i.e., \(v-u\) is identically zero. Hence \(v\) is subharmonic at any given point \(z_0\in U\), which proves the if part.
It's immediate from the above characterizations that \(v\) is subharmonic on \(U\) iff for any \(\overline{D(z,r)}\subset U\) and for any function \(u\) harmonic on \(D(z,r)\) and continuous on \(\overline{D(z,r)}\), we have
\[ \begin{equation*} \text{$v\le u$ on $\partial D(z,r)$}\implies \text{$v\le u$ on $D(z,r)$} \end{equation*} \]
This is an analogy to the 1-dimensional case where we consider lines and convex functions. A more technical result is the following.
If \(v\in C^2(U)\), then \(v\) is subharmonic iff \(\Delta v\ge 0\) on \(U\).
To prove this, we first establish a general formula, which should be familiar to readers knowing Green functions. Take \(\overline{D(z_0,r)}\subset U\). For any \(0<\varepsilon<r\) and for any \(u\in C^2(\overline{D(z_0,r)}-D(z_0,\varepsilon))\), by Green's second identity, we have
where \(\mathbf{n}\) denotes the outward unit normal vector. Consider the function \(u(z)=\dfrac{1}{2\pi}\ln |z-z_0|\), which is harmonic on \(\mathbb{C}-\{z_0\}\). On \(\partial D(z_0,r)\), we have \(u\equiv \dfrac{1}{2\pi}\ln r\) and \(\dfrac{\partial u}{\partial \mathbf{n}}\equiv \dfrac{1}{2\pi r}\). Similar results hold for \(\partial D(z_0,\varepsilon)\). Therefore,
Letting \(\varepsilon\to 0^+\), we obtain
Note that \(\displaystyle\int_{\partial D(z_0,r)} \dfrac{\partial v}{\partial \mathbf{n}} = \displaystyle\int_{D(z_0,r)} \Delta v\), we finally get
from which it's clear that \(\Delta v\ge 0\) on \(U\) iff
So we're done.
