2.4: Harmonic, Subharmonic, and Plurisubharmonic Functions

Theorem \(\PageIndex{1}\)

Let \(u \colon \partial \mathbb{D} \to \mathbb{R}\) be a continuous function. The function \(Pu \colon \overline{\mathbb{D}} \to \mathbb{R}\), defined by \[Pu(re^{i\theta}) = \int_{-\pi}^\pi u(e^{it}) P_r(\theta-t) \, dt \quad \text{if $r < 1$} \qquad \text{and} \qquad Pu(e^{i\theta}) = u(e^{i\theta}), \] is harmonic in \(\mathbb{D}\) and continuous on \(\overline{\mathbb{D}}\).

Proposition \(\PageIndex{1}\): Mean-Value Property and Sub-Mean-Value Property

Let \(U \subset \mathbb{C}\) be an open set.

1. A continuous function \(f \colon U \to \mathbb{R}\) is harmonic if and only if \[f(a) = \frac{1}{2\pi} \int_0^{2\pi} f(a+re^{i\theta})\, d\theta \qquad \text{whenever} \quad \overline{\Delta_r(a)} \subset U .\]
2. An upper-semicontinuous function \(f \colon U \to \mathbb{R} \cup \{ -\infty \}\) is subharmonic if and only if \[f(a) \leq \frac{1}{2\pi} \int_0^{2\pi} f(a+re^{i\theta})\, d\theta \qquad \text{whenever} \quad \overline{\Delta_r(a)} \subset U .\]

Proposition \(\PageIndex{4}\)

Suppose \(U \subset \mathbb{C}\) is an open set and \(f_\alpha \colon U \to \mathbb{R} \cup \{ -\infty \}\) is a family of subharmonic functions. Let \[\varphi(z) = \sup_\alpha\, f_\alpha(z) . \] If the family is finite, then \(\varphi\) is subharmonic. If the family is infinite, \(\varphi(z) \not= \infty\) for all \(z\), and \(\varphi\) is upper-semicontinuous, then \(\varphi\) is subharmonic.

Proof

Suppose \(\overline{\Delta_r(a)} \subset U\). For any \(\alpha\), \[\frac{1}{2\pi} \int_0^{2\pi} \varphi (a+re^{i\theta})\, d\theta \geq \frac{1}{2\pi} \int_0^{2\pi} f_\alpha (a+re^{i\theta})\, d\theta \geq f_\alpha(a) .\] Taking the supremum on the right over \(\alpha\) obtains the results.

Proposition \(\PageIndex{5}\)

Let \(U \subset \mathbb{C}^n\) be open. A \(C^2\)-smooth \(f \colon U \to \mathbb{R}\) is plurisubharmonic if and only if the complex Hessian matrix \[\left[ \frac{\partial^2 f}{\partial \bar{z}_j \partial z_k} \right]_{jk}\] is positive semidefinite at every point.

Proof

First suppose that the complex Hessian has a negative eigenvalue at a point \(p \in U\). After a translation assume \(p=0\). As \(f\) is real-valued, the complex Hessian \(\left[ \frac{\partial^2 f}{\partial \bar{z}_j \partial z_k} \Big|_0 \right]_{jk}\) is Hermitian. A complex linear change of coordinates acts on the complex Hessian by \(*\)-congruence, and therefore we can diagonalize, using Sylvester’s Law of Inertia again. So assume that \(\left[ \frac{\partial^2 f}{\partial \bar{z}_j \partial z_k} \Big|_0 \right]_{jk}\) is diagonal. If the complex Hessian has a negative eigenvalue, then one of the diagonal entries is negative. Without loss of generality suppose \(\frac{\partial^2 f}{\partial \bar{z}_1 \partial z_1}\Big|_0 < 0\). The function \(z_1 \mapsto f(z_1,0,\ldots,0)\) has a negative Laplacian and therefore is not subharmonic, and thus \(f\) itself is not plurisubharmonic.

For the other direction, suppose the complex Hessian is positive semidefinite at all points. Let \(p \in U\). After an affine change of coordinates assume that the line \(\xi \mapsto a+b\xi\) is simply setting all but the first variable to zero, that is, \(a=0\) and \(b=(1,0,\ldots,0)\). As the complex Hessian is positive semidefinite, \(\frac{\partial^2 f}{\partial \bar{z}_1 \partial z_1} \geq 0\) for all points \((z_1,0,\ldots,0)\). We proved above that \(\nabla^2 g \geq 0\) implies \(g\) is subharmonic, and we are done.

Theorem \(\PageIndex{2}\)

That is, \(f\) is a limit of smooth plurisubharmonic functions. The idea of the proof is important and useful in many other contexts.

Proof

We smooth \(f\) out by convolving with so-called mollifiers, or approximate delta functions. Many different mollifiers work, but let us use a specific one for concreteness. For \(\epsilon > 0\), define \[g(z) = \begin{cases} C e^{-1/(1-||z||^2)} & \text{ if $||z|| < 1$,} \\ 0 & \text{ if $||z|| \geq 1$,} \end{cases} \qquad \text{and} \qquad g_\epsilon(z) = \frac{1}{\epsilon^{2n}} g(z/\epsilon) .\] It is left as an exercise that \(g\), and so \(g_\epsilon\), is smooth. The function \(g\) has compact support as it is only nonzero inside the unit ball. The support of \(g_\epsilon\) is the \(\epsilon\)-ball. Both are nonnegative. Choose \(C\) so that \[\int_{\mathbb{C}^n} g\, dV = 1 , \qquad \text{ and therefore } \qquad \int_{\mathbb{C}^n} g_\epsilon\, dV = 1 .\] Here \(dV\) is the volume measure. The function \(g\) only depends on \(||z||\). To get an idea of how these functions work, see Figure \(\PageIndex{3}\).

Figure \(\PageIndex{3}\)

Compare the graphs to the graphs of the Poisson kernel as a function of \(\theta\), which is also a type of mollifier. In fact, the idea of integrating against the right approximate delta function with the desired properties is similar to the solution of the Dirichlet problem using the Poisson kernel.

The function \(f\) is bounded above on compact sets as it is upper semicontinuous. If \(f\) is not bounded below, we replace \(f\) with \(\max \bigl\{ f , \frac{-1}{\epsilon} \bigr\}\), which is still plurisubharmonic. Therefore, without loss of generality we assume that \(f\) is locally bounded.

For \(z \in U_\epsilon\), we define \(f_\epsilon\) as the convolution with \(g_\epsilon\): \[f_\epsilon(z) = (f * g_\epsilon)(z) = \int_{\mathbb{C}^n} f(w) g_\epsilon (z-w) \, dV(w) = \int_{\mathbb{C}^n} f(z-w) g_\epsilon (w) \, dV(w) .\] The two forms of the integral follow easily via change of variables. We are perhaps abusing notation a bit as \(f\) is only defined on \(U\), but it is not a problem as long as \(z \in U_\epsilon\), because \(g_\epsilon\) is then zero when \(f\) is undefined. By differentiating the first form under the integral, we find that \(f_\epsilon\) is smooth.

Let us show that \(f_\epsilon\) is plurisubharmonic. We restrict to a line \(\xi \mapsto a+b\xi\). We wish to test subharmonicity by the sub-mean-value property using a circle of radius \(r\) around \(\xi = 0\): \[\begin{align}\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} f_\epsilon(a+bre^{i\theta})\, d\theta & = \frac{1}{2\pi} \int_0^{2\pi} \int_{\mathbb{C}^n} f\bigl(a+bre^{i\theta}-w\bigr) g_\epsilon (w) \, dV(w) \,d\theta \\ & = \int_{\mathbb{C}^n} \left( \frac{1}{2\pi} \int_0^{2\pi} f\bigl(a-w+bre^{i\theta}\bigr) \, d\theta \right) g_\epsilon (w) \, dV(w) \\ & \geq \int_{\mathbb{C}^n} f(a-w) g_\epsilon (w) \, dV(w) = f_\epsilon(a). \end{aligned}\end{align}\] For the inequality we used \(g_\epsilon \geq 0\). So \(f_\epsilon\) is plurisubharmonic.

Let us show that \(f_\epsilon(z) \geq f(z)\) for all \(z \in U_\epsilon\). As \(g_\epsilon(w)\) only depends on \(|w_1|,\ldots,|w_n|\), we notice that \(g_\epsilon(w_1,\ldots,w_n) = g_\epsilon(|w_1|,\ldots,|w_n|)\). Without loss of generality we consider \(z=0\), and we use polar coordinates for the integral.

\[\begin{align}\begin{aligned} f_{\epsilon}(0)&=\int_{\mathbb{C}^{n}}f(-w)g_{\epsilon}(|w_{1}|,\cdots ,|w_{n}|)dV(w) \\ &=\int_{0}^{\epsilon}\cdots\int_{0}^{\epsilon}\left(\int_{0}^{2\pi}\cdots\int_{0}^{2\pi}f(-r_{1}e^{1\theta_{i}}, \cdots ,-r_{n}e^{i\theta_{n}})d\theta_{1}\cdots d\theta_{n}\right) \\ & \qquad\qquad\qquad\qquad\qquad\qquad g_{\epsilon}(r_{1},\cdots ,r_{n})r_{1}\cdots r_{n}dr_{1}\cdots dr_{n} \\ &\geq\int_{0}^{\epsilon}\cdots\int_{0}^{\epsilon}\left(\int_{0}^{2\pi}\cdots\int_{0}^{2\pi}(2\pi )f(0, -r_{2}e^{i\theta_{2}},\cdots ,-r_{n}e^{-\theta _{n}})d\theta_{2}\cdots d\theta_{n}\right) \\ &\qquad\qquad\qquad\qquad\qquad\qquad g_{\epsilon}(r_{1},\cdots ,r_{n})r_{1}\cdots r_{n}dr_{1}\cdots dr_{n} \\ &\geq f(0)\int_{0}^{\epsilon}\cdots\int_{0}^{\epsilon}(2\pi )^{n}g_{\epsilon}(r_{1},\cdots r_{n})r_{1}\cdots r_{n}dr_{1}\cdots dr_{n} \\ &=f(0)\int_{\mathbb{C}^{n}}g_{\epsilon}(w)dV(w)=f(0).\end{aligned}\end{align}\]

The second equality above follows as \(g_\epsilon\) is zero outside the polydisc of radius \(\epsilon\). For the inequalities, we again needed that \(g_\epsilon \geq 0\). The penultimate equality follows from the fact that \(2\pi = \int_0^{2\pi}d \theta\).

Finally, we show \(\lim_{\epsilon \to 0} f_\epsilon (z) = f(z)\). For subharmonic, and so for plurisubharmonic, functions, \(\limsup_{\zeta\to z} f(\zeta) = f(z)\). So given \(\delta >0\) find an \(\epsilon >0\) such that \(f(\zeta)-f(z) \leq \delta\) for all \(\zeta \in B_\epsilon(z)\). \[\begin{align}\begin{aligned} f_\epsilon(z) - f(z) & = \int_{B_\epsilon(0)} f(z-w) g_\epsilon (w) \, dV(w) - f(z) \int_{B_\epsilon(0)} g_\epsilon (w) \, dV(w) \\ & = \int_{B_\epsilon(0)} \bigl(f(z-w)-f(z)\bigr)\, g_\epsilon (w) \, dV(w) \\ & \leq \delta \int_{B_\epsilon(0)} g_\epsilon (w) \, dV(w) = \delta . \end{aligned}\end{align}\] Again we used that \(g_\epsilon \geq 0\). We find \(0 \leq f_\epsilon - f(z) \leq \delta\), and so \(f_\epsilon(z) \to f(z)\).

Theorem \(\PageIndex{3}\): Radó

Let \(U \subset \mathbb{C}^n\) be open and \(f \colon U \to \mathbb{C}\) a continuous function that is holomorphic on the set \[U' = \bigl\{ z \in U : f(z) \not= 0 \bigr\} . \] Then \(f \in \mathcal{O}(U)\).

Proof

First assume \(n=1\). As the theorem is local, it is enough to prove it for a small disc \(\Delta\) such that \(f\) is continuous on the closure \(\overline{\Delta}\), let \(\Delta'\) be the part of the disc where \(f\) is nonzero. If \(\Delta'\) is empty, then we are done as \(f\) is just identically zero and hence holomorphic.

Let \(u\) be the real part of \(f\). On \(\Delta'\), \(u\) is a harmonic function. Let \(Pu\) be the Poisson integral of \(u\) on \(\Delta\). Hence \(Pu\) equals \(u\) on \(\partial \Delta\), and \(Pu\) is harmonic in all of \(\Delta\). Consider the function \(Pu(z) - u(z)\) on \(\overline{\Delta}\). The function is zero on \(\partial \Delta\) and it is harmonic on \(\Delta'\). By rescaling \(f\) we can without loss of generality assume that \(|f(z)| < 1\) for all \(z \in \overline{\Delta}\). For any \(t >0\), the function \(z \mapsto t \log |f(z)|\) is subharmonic on \(\Delta'\) and upper-semicontinuous on \(\overline{\Delta'}\). Further, it is negative on \(\partial \Delta\). The function \(z \mapsto -t \log |f(z)|\) is superharmonic (minus a subharmonic function) on \(\Delta'\), lower-semicontinuous on \(\overline{\Delta'}\) and positive on \(\partial \Delta\). On the set \(\Delta \setminus \Delta'\) where \(f\) is zero, the two functions are \(-\infty\) and \(\infty\) respectively. See Figure \(\PageIndex{4}\). Therefore, for all \(t > 0\) and \(z \in \partial \Delta \cup (\Delta \setminus \Delta')\), we have \[\label{eq:3} t \log |f(z)| \leq Pu(z)-u(z) \leq -t \log |f(z)| .\] Applying the maximum principle to the subharmonic functions \(z \mapsto t \log |f(z)| - \bigl(Pu(z)-u(z)\bigr)\) and \(z \mapsto t \log |f(z)| - \bigl(u(z)-Pu(z)\bigr)\) shows that \(\eqref{eq:3}\) holds for all \(z \in \Delta'\) and all \(t > 0\).

Figure \(\PageIndex{4}\)

Taking the limit \(t \to 0\) shows that \(Pu = u\) on \(\Delta'\). Let \(W = \Delta \setminus \overline{\Delta'}\). On \(W\), \(u=0\) and so \(Pu-u\) is harmonic on \(W\) and continuous on \(\overline{W}\). Furthermore, \(Pu-u=0\) on \(\overline{\Delta'} \cup \partial \Delta\), and so \(Pu-u=0\) on \(\partial W\). By the maximum principle, \(Pu=u\) on \(W\) and therefore on all of \(\overline{\Delta}\). Similarly, if \(v\) is the imaginary part of \(f\), then \(Pv = v\) on \(\overline{\Delta}\). In other words, \(u\) and \(v\) are harmonic on \(\Delta\). As \(\Delta\) is simply connected, let \(\tilde{v}\) be the harmonic conjugate of \(u\) that equals \(v\) at some point of \(\Delta'\). As \(f\) is holomorphic on \(\Delta'\), the harmonic functions \(\tilde{v}\) and \(v\) are equal the nonempty open subset \(\Delta'\) of \(\Delta\) and so they are equal everywhere. Consequently, \(f = u +iv\) is holomorphic on \(\Delta\).

The extension of the proof to several variables is left as an exercise.