Several Complex Variables (II): Holomorphic functions

04 Sep 2018

• several complex variables

In this post, we look at holomorphic, analytic, and complex differentiable functions in several complex variables. As in the one variable case, we show that many of these notions are equivalent.

Recap: one variable

Let $f:U\to\bb C$, where $U\subseteq\bb C$ is open and connected. Recall the following definitions:

1. $f$ satisfies the Cauchy-Riemann equations if $\frac{\del f}{\del\overline z}=0$ holds in $U$.

2. $f$ has ‘Cauchy integral representation’ (non-standard term) if for any disc $\Delta=\Delta(z_0,r)$ with $\overline\Delta\subseteq U$, $f$ can be written as

   $$f(z)=\frac1{2\pi i}\int_{\del\Delta}\!\frac{f(\zeta)}{\zeta-z}d\zeta$$

   for all $z\in\Delta$.

3. $f$ is analytic if for any $z_0\in U$, $f$ has a power series expansion

   $$f(z)=c_0+c_1(z-z_0)+c_2(z-z_0)^2+\cdots,$$

   with $c_k\in\bb C$, which converges for all $z$ in some nonempty open disc $\Delta(z_0,r)$.

4. $f$ is holomorphic, or complex differentiable, if the limit

   $$f'(z_0)\coloneqq\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$$

   exists for all $z_0\in U$.

It is a fundamental result in classical complex analysis that these four definitions are equivalent. Here we give an outline of the proof.

$(1)\Rightarrow(2)$: For $z\in\Delta$, check that $g(\zeta)=\frac{f(\zeta)-f(z)}{\zeta-z}$ satisfies $\frac{\del g}{\del\overline\zeta}=0$. By Green’s formula (or the generalised Cauchy integral formula) on the region $\Delta\backslash\overline\Delta(z,\eps)$, we get

$$\int_{\del\Delta}\!g(\zeta)\,d\zeta=\int_{\del\Delta(z,\eps)}g(\zeta)\,d\zeta.$$

As $\eps\to0^+$, the RHS is the integral of a bounded function over a curve of length $\to0$, so it goes to 0. Hence

$$\int_{\del\Delta}\!\frac{f(\zeta)}{\zeta-z}d\zeta=\int_{\del\Delta}\!\frac{f(z)}{\zeta-z}d\zeta=2\pi if(z).$$

$(2)\Rightarrow(3)$: For $z\in\Delta=\Delta(z_0,r)$, we have

$$\begin{aligned} f(z)&=\frac1{2\pi i}\int_{\del\Delta}\!\frac{f(\zeta)}{\zeta-z}d\zeta\\ &=\frac1{2\pi i}\int_{\del\Delta}\!\frac{f(\zeta)}{\zeta-z_0}\frac1{1-\frac{z-z_0}{\zeta-z_0}}d\zeta\\ &=\frac1{2\pi i}\sum_{n=0}^\infty\left(\int_{\del\Delta}\!\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta\right)(z-z_0)^n, \end{aligned}$$

where the geometric series converges normally (ie. uniformly on compact subsets), justifying the exchange of sum and integral in the last line. Hence we may take

$$c_n=\frac1{2\pi i}\int_{\del\Delta}\!\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}d\zeta.$$

$(3)\Rightarrow(4)$: We expect that the derivative $f’$ is given by termwise differentiation. This is a theorem from real analysis about power series, whose proof carries over to the complex case.

$(4)\Rightarrow(1)$: We proved this in the previous post. In summary, taking the limit for the derivative in two different directions gives the usual formulation of the Cauchy-Riemann equations

$$\frac{\del u}{\del x}=\frac{\del v}{\del y},\qquad\frac{\del v}{\del x}=-\frac{\del u}{\del y},$$

which can be verified to be equivalent to $\frac{\del f}{\del\overline z}=0$ by explicit computation.

Several variables

We want to extend the above results to the case of several complex variables, which we write as $z=(z_1,z_2,\ldots,z_n)\in\bb C^n$. Analogously to the one variable setting, we can define $z_j=x_j+iy_j$,

$$\begin{aligned} dz_j&=dx_j+i\,dy_j\\ d\overline{z_j}&=dx_j-i\,dy_j \end{aligned}\qquad \begin{aligned} \frac\del{\del z_j}&=\frac12\left(\frac\del{\del x_j}-i\frac\del{\del y_j}\right)\\ \frac\del{\del\overline z_j}&=\frac12\left(\frac\del{\del x_j}+i\frac\del{\del y_j}\right). \end{aligned}$$

Let $f:U\to\bb C$, where $U\subseteq\bb C^n$ is open and connected. We make the following definitions:

1. $f$ satisfies the Cauchy-Riemann equations if $\frac{\del f}{\del\overline z_j}=0$ holds in $U$ for all $1\leq j\leq n$.

2. $f$ has ‘Cauchy integral representation’ if for any polydisc $\Delta=\Delta(a,r)$ with $\overline\Delta\subseteq U$, $f$ can be written as

   $$f(z)=\frac1{(2\pi i)^n}\int_{\del\Delta(a_n,r_n)}\!\!\!\!\cdots\int_{\del\Delta(a_1,r_1)}\!\frac{f(\zeta_1,\ldots,\zeta_n)\,d\zeta_1\cdots d\zeta_n}{(\zeta_1-z_1)\cdots(\zeta_n-z_n)}$$

   for all $z\in\Delta$.

3. $f$ is analytic if for any $a\in U$, $f$ has a power series expansion

   $$f(z)=\sum_{k_1,\ldots,k_n\geq0}c_{k_1,\ldots,k_n}(z_1-a_1)^{k_1}\cdots(z_n-a_n)^{k_n},$$

   with $c_{k_1,\ldots,k_n}\in\bb C$, which converges for all $z$ in some nonempty open disc $\Delta(a,r)$.

4. $f$ is holomorphic, or complex differentiable, if for each $a\in U$ there exists a linear functional $Df(a):\bb C^n\to\bb C$ such that

   $$\lim_{z\to a}\frac{|f(z)-f(a)-Df(a)(z-a)|}{|z-a|}=0.$$

We can also consider functions that are analytic separately in each variable:

0. $f$ is separately analytic if for any $a\in U$, the functions $$f_j:\Delta(0,\eps)\subseteq\bb C\to\bb C,\qquad w\mapsto f(a+we_j)$$ are analytic at $w=0$ for each $1\leq j\leq n$. Here $\{e_1,\ldots,e_n\}$ is the basis of $\bb C^n$ dual to coordinates $\{z_1,\ldots,z_n\}$.

Note that this definition depends on the choice of coordinates.

Now we can investigate the relationship between these notions.

$(0)\Leftrightarrow(1)$: Clear.

$(0)\Rightarrow(2)$: This follows directly from $n$ applications of the Cauchy integral formula.

$(2)\Rightarrow(3)$: This is known as Osgood’s Lemma, when we add an assumption that $f$ is bounded on compact subsets of $U$. In particular, this holds when $f$ is continuous.

We proceed as in the one variable case:

$$\begin{aligned} f(z)&=\frac1{(2\pi i)^n}\int\!\!\cdots\!\!\int\!\frac{f(\zeta_1,\ldots,\zeta_n)\,d\zeta_1\cdots d\zeta_n}{(\zeta_1-z_1)\cdots(\zeta_n-z_n)}\\ &=\frac1{(2\pi i)^n}\int\!\!\cdots\!\!\int\!\frac{f(\zeta_1,\ldots,\zeta_n)\,d\zeta_1\cdots d\zeta_n}{(\zeta_1-a_1)\cdots(\zeta_n-a_n)}\frac1{1-\frac{z_1-a_1}{\zeta_1-a_1}}\cdots\frac1{1-\frac{z_n-a_n}{\zeta_n-a_n}}\\ &=\sum_{k_1,\ldots,k_n\geq0}c_{k_1,\ldots,k_n}(z_1-a_1)^{k_1}\cdots(z_n-a_n)^{k_n}, \end{aligned}$$

where

$$c_{k_1,\ldots,k_n}=\frac1{(2\pi i)^n}\int\!\!\cdots\!\!\int\!\frac{f(\zeta_1,\ldots,\zeta_n)\,d\zeta_1\cdots d\zeta_n}{(\zeta_1-a_1)^{k_1+1}\cdots(\zeta_n-a_n)^{k_n+1}}.$$

We can exchange sum and integral in the last line since the infinite sum converges normally by the boundedness assumption.

In general, $(2)\Rightarrow(3)$ still holds without the boundedness assumption; this is known as Hartogs’s theorem. We omit the proof because it requires a number of technical results. In practice, it is usually easy to show the continuity of $f$, so Osgood’s lemma is sufficient to deduce analyticity from separate analyticity.

$(3)\Rightarrow(4)$: Again, we expect that the derivative is given by termwise differentiation.

From the one variable case, we know that as a power series in $z_1$, $f$ can be differentiated termwise to give

$$\begin{aligned} \frac\del{\del z_1}f(z)&=\sum_{k_1\geq1}C_{k_1}(z_1-a_1)^{k_1-1},\\ \text{where }C_{k_1}&=k_1\sum_{\mathclap{k_2,\ldots,k_n\geq0}}c_{k_1,\ldots,k_n}(z_2-a_2)^{k_2}\cdots(z_n-a_n)^{k_n}, \end{aligned}$$

which is a convergent power series in $\Delta=\Delta(a,r)$; in particular, $\frac{\del f}{\del z_1}$ is continuous in $\Delta$.

Similarly, all partial derivatives of $f$ are continuous in $\Delta$; hence the total derivative $Df$ exists.

$(4)\Rightarrow(0)$: Taking $z=a+we_j$ gives

$$\begin{aligned} \lim_{w\to0}\frac{|f(a+we_j)-f(a)-wDf(a)(e_j)|}{|w|}&=0\\ \implies\lim_{w\to0}\frac{f(a+we_j)-f(a)}w&=Df(a)(e_j). \end{aligned}$$

In particular, $f$ is holomorphic (hence analytic) separately in each $z_j$.