Application of Wirtinger Derivatives to Harmonic Functions
This article is inspired by Exercise 1 in Chapter 1 of Krantz’s Several Complex Variables. We first recall the definition of Wirtinger derivatives:
\[\begin{align*} \frac{\partial}{\partial z}&=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial }{\partial y})\\ \frac{\partial}{\partial \bar{z}}&=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial }{\partial y}) \end{align*}\]They are useful for and closely related to harmonic functions because of the following well-known identity:
\[\begin{align*} \Delta=4\frac{\partial^2}{\partial z\partial \bar{z}} \end{align*}\]Using this identity and chain rules of Wirtinger derivatives, we can prove some interesting theorems about (complex) harmonic functions (and at the same time we can see that harmonicity is a very nice property). Let’s start with the following question. Let $f:\Omega\subset\mathbb{C}\rightarrow \mathbb{C}$ be a nonvanishing complex-valued harmonic function. Suppose $1/f$ is also harmonic. Prove that $f$ is either holomorphic or anti-holomorphic. In particular, if $f$ is real-valued, then $f$ must be constant.
Proof. We will use the calculus of Wirtinger derivative:
\[\begin{align*} \frac{\partial}{\partial \bar{z}}\frac{\partial f}{\partial z}&=0\text{ (because $f$ is harmonic)}\\ \frac{\partial}{\partial \bar{z}}\frac{\partial (1/f)}{\partial z}&=0\text{ (because $1/f$ is harmonic)}\\ \frac{\partial}{\partial \bar{z}}[(-\frac{1}{f^2})\frac{\partial f}{\partial z}]&=0\text{ (Chain Rule)}\\ \frac{\partial (-1/f^2)}{\partial \bar{z}}\frac{\partial f}{\partial z}-\frac{1}{f^2}\frac{\partial}{\partial \bar{z}}\frac{\partial f}{\partial z}&=0\text{ (Product Rule)}\\ \frac{\partial (-1/f^2)}{\partial \bar{z}}\frac{\partial f}{\partial z}&=0\\ \frac{2}{f^3}\frac{\partial f}{\partial \bar{z}}\frac{\partial f}{\partial z}&=0\\ \frac{\partial f}{\partial \bar{z}}\frac{\partial f}{\partial z}&=0 \end{align*}\]This is the product of two complex-valued harmonic functions, so one of them must be identically equal to $0$. $\frac{\partial f}{\partial z}\equiv 0$ implies that $f$ is anti-holomorphic and $\frac{\partial f}{\partial \bar{z}}\equiv 0$ implies that $f$ is holomorphic. $\square$
Here are some similar questions. Let $f:\Omega\subset\mathbb{C}\rightarrow \mathbb{C}$ be a complex-valued harmonic function. Suppose $f^2$ is also harmonic. Prove that $f$ is either holomorphic or anti-holomorphic. In particular, if $f$ is real-valued, then $f$ must be constant.
Proof. We repeat the calculations:
\[\begin{align*} \frac{\partial}{\partial\bar{z}}\frac{\partial (f^2)}{\partial z}&=0\\ \frac{\partial}{\partial\bar{z}}[2f\frac{\partial f}{\partial z}]&=0\\ 2\frac{\partial f}{\partial\bar{z}}\frac{\partial f}{\partial z}+2f\frac{\partial}{\partial\bar{z}}\frac{\partial f}{\partial z}&=0\\ \frac{\partial f}{\partial\bar{z}}\frac{\partial f}{\partial z}&=0 \end{align*}\]We obtain the same conclusion. $\square$
Let $f:\Omega\subset\mathbb{C}\rightarrow \mathbb{C}-[0,\infty)$ be a complex-valued harmonic function. Suppose $\log f$ is also harmonic. Prove that $f$ is either holomorphic or anti-holomorphic. In particular, if $f$ is real-valued, then $f$ must be constant.
Proof.
\[\begin{align*} \frac{\partial}{\partial\bar{z}}\frac{\partial (\log f)}{\partial z}&=0\\ \frac{\partial}{\partial\bar{z}}[\frac{1}{f}\frac{\partial f}{\partial z}]&=0\\ \frac{\partial (1/f)}{\partial\bar{z}}\frac{\partial f}{\partial z}+\frac{1}{f}\frac{\partial}{\partial\bar{z}}\frac{\partial f}{\partial z}&=0\\ (-\frac{1}{f^2})\frac{\partial f}{\partial\bar{z}}\frac{\partial f}{\partial z}&=0\\ \frac{\partial f}{\partial\bar{z}}\frac{\partial f}{\partial z}&=0 \end{align*}\]which completes the proof. $\square$
Notice that the above arguments can be generalized to complex valued harmonic functions defined on $\mathbb{C}^n$.