Hodge theory: Lecture 4

Single-variable holomorphic functions. We continue our discussion from the previous notes.

Theorem 1. Let $f : U \rightarrow \mathbb{C}$ be a smooth function. The following are equivalent:

• $f$ is holomorphic;
• $f$ is analytic.

Proof. First, assume that $f$ is holomorphic. Fix any $a \in U$ and take an open disk $D$ containing $a$ such that $\overline{D} \subset U.$ For any $z \in D,$ Cauchy's integral formula tells us that $$f(z) = \frac{1}{2\pi i}\int_{\partial D}\frac{f(w)}{w - z} dw.$$ Let's look at the integrand: $$\begin{align*}\frac{f(w)}{w - z} &= \frac{f(w)}{(w - a) - (z - a)} \\ &= \left(\frac{f(w)}{w - a}\right) \frac{1}{1 - (z-a)/(w-a)} \\ &= \left(\frac{f(w)}{w - a}\right) \sum_{n=0}^{\infty} \frac{(z-a)^{n}}{(w-a)^{n}} \\ &= \sum_{n=0}^{\infty}\frac{f(w)}{(w-a)^{n+1}}(z-a)^{n},\end{align*}$$ where $w \in \partial D,$ which ensures that $|z - a| < |w - a|.$ Note that this series is absolutely and uniformly convergent as it is given by the geometric series. Thus, (even a weaker version) Dominate Convergence implies that $$\begin{align*} \frac{1}{2\pi i} \int_{\partial D}\frac{f(w)}{w - z} dw &= \frac{1}{2\pi i} \int_{\partial D} \left(\sum_{n=0}^{\infty}\frac{f(w)}{(w-a)^{n+1}}(z-a)^{n}\right)dw \\ &= \sum_{n=0}^{\infty} \left( \frac{1}{2\pi i} \int_{\partial D}\frac{f(w)}{(w-a)^{n+1}}dw \right) (z-a)^{n} \end{align*}$$ for any $z \in D.$ The uniform convergence and the absolute convergence of the last expression follow from the second expression, but we also may check explicitly as follows. Denote by $r > 0$ the radius of $D.$ If we take a disc $D'$ with radius $r - \epsilon > 0,$ for any $z \in D',$ we have $$\begin{align*}\left|\left( \int_{\partial D}\frac{f(w)}{(w-a)^{n+1}}dw \right) (z - a)^{n}\right| &\leq |z - a|^{n}\int_{\partial D}\frac{|f(w)|}{|w-a|^{n+1}}dw \\ &< r^{-1} (1 - \epsilon/r)^{n}\int_{\partial D}|f(w)|dw \\ &\leq r^{-1}(1 - \epsilon/r)^{n}C,\end{align*}$$ for some constant $C.$

Remark. A more general version for the first direction is available as Theorem 10.7 in Rudin. It is funny to note that $\partial D$ actually means topological boundary. It is equal to the boundary of the $\overline{D}$ in the manifold theory sense, which is really what we are using when we think about Stokes' theorem.

Conversely, let $f$ be analytic. Given $a \in U,$ take a small disc $D$ in $U$ centered at $a$ such that $\overline{D} \subset U$ and $$f(z) = \sum_{n=0}^{\infty}c_{n} (z - a)^{n}$$ for $z \in D,$ where the sum is uniformly and absolute convergent. The partial sum $$P_{n}(z) = \sum_{d=0}^{n}c_{d} (z - a)^{d}$$ is a polynomial, so it is holomorphic. Hence, by Cauchy's integral formula, we have $$P_{n}(z) = \frac{1}{2\pi i}\int_{\partial D} \frac{P_{n}(w)}{w - z}dw.$$ This implies that $$\begin{align*} f(z) &= \lim_{n \rightarrow \infty}P_{n}(z) \\ &= \frac{1}{2\pi i} \lim_{n \rightarrow \infty}\int_{\partial D} \frac{P_{n}(w)}{w - z}dw \\ &= \frac{1}{2\pi i} \int_{\partial D} \frac{f(w)}{w - z}dw\end{align*}$$ by Dominate Convergence. Now an application of Dominate Convergence (e.g., Theorem 2.27 in Folland) lets us have $$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2\pi i} \int_{\partial D} \frac{\partial}{\partial \bar{z}} \frac{f(w)}{w - z}dw = 0,$$ since $1/(w - z)$ is holomorphic in $z \in D,$ for any $w \in \partial D.$ Thus $f$ is holomoprhic, as desired. $\Box$

Theorem 2 ($\bar{\partial}$-lemma in single variable). Let $D$ be a nonempty open disk in an open subset $U \subset \mathbb{C}$ such that $\overline{D} \subset U.$ Then for any smooth function $g : U \rightarrow \mathbb{C},$ there is a function $f : D \rightarrow \mathbb{C}$ such that $$g = \frac{\partial f}{\partial \bar{z}}.$$ More specifically, we can construct such an $f$ as follows: $$f(z) = \frac{1}{2\pi i} \int_{\overline{D}}g(w) \frac{dw \wedge d\bar{w}}{w - z}$$ for $z \in D.$

Remark. From a change of variable argument into the polar coordinate from the previous notes, we note that the integral above is well-defined.

Proof. Fix any $z_{0} \in D.$ Take any open discs $D_{1}, D_{2}$ centered at $z_{0}$ in $D$ such that $$\overline{D_{1}} \subsetneq \overline{D_{2}} \subsetneq D.$$ Then $\mathbb{C} \smallsetminus D_{1}$ and $D_{2}$ cover $\mathbb{C} = \mathbb{R}^{2},$ so (e.g., by Proposition 13.6 in Tu's book) we may find smooth (partition of unity) $\rho_{1}, \rho_{2} : \mathbb{C} \rightarrow \mathbb{R}$ such that

• $\rho_{1} + \rho_{2} = 1$ on $\mathbb{C},$
• $\rho_{1} = 0$ on $D_{1},$ and
• $\rho_{2} = 0$ on $\mathbb{C} \setminus D_{2}.$

Thus, we have $g = g_{1} + g_{2}$ with $g_{i} := \rho_{i}g$ smooth functions $U \rightarrow \mathbb{C}$ such that 

• $g_{1} = 0$ on $\overline{D_{1}},$ and
• $g_{2} = 0$ on $\mathbb{C} \setminus D_{2}.$

Hence, we may write $f = f_{1} + f_{2}$ with $$f_{j}(z) := \frac{1}{2 \pi i}\int_{\overline{D}} g_{j}(w) \frac{dw \wedge d\bar{w}}{w - z}.$$ We have $$f_{1}(z) = \frac{1}{2 \pi i}\int_{D \setminus \overline{D_{1}}} g_{1}(w) \frac{dw \wedge d\bar{w}}{w - z},$$ so for $z \in D_{1},$ we have $$\frac{\partial f_{1}}{\partial \bar{z}} = \frac{1}{2\pi i}\int_{D \setminus \overline{D_{1}}}\frac{\partial}{\partial \bar{z}} \left(\frac{g_{1}(w)}{w - z}\right) dw \wedge d\bar{w} = 0$$ because $$\frac{\partial}{\partial \bar{z}} \left(\frac{g_{1}(w)}{w - z}\right) = 0$$ when $w \in D \smallsetminus \overline{D_{1}}.$ (Again, we are using a consequence of Dominated Convergence, say Theorem 2.27 in Folland.) Note that $$f_{2}(z) = \frac{1}{2 \pi i}\int_{\mathbb{C}} g_{2}(w) \frac{dw \wedge d\bar{w}}{w - z} = -\frac{1}{\pi} \int_{0}^{2\pi}e^{-i\theta}\int_{0}^{\infty}g_{2}(z + re^{i\theta})drd\theta,$$ which can be computed with $w = z + re^{i \theta}$ where $r \in (0, \infty)$ and $\theta \in (0, 2\pi).$ Since $g_{2} = 0$ outside $D_{2},$ its derivative is also $0$ outside $\overline{D_{2}}.$ Thus, we have $$\begin{align*}\frac{\partial f_{2}}{\partial \bar{z}} &= -\frac{1}{\pi} \int_{0}^{2\pi}e^{-i\theta}\int_{0}^{\infty}\frac{\partial g_{2}(z + re^{i\theta})}{\partial \bar{z}}drd\theta \\ &= \frac{1}{2 \pi i}\int_{\mathbb{C}} \frac{\partial g_{2}}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z} \\ &= \frac{1}{2 \pi i}\int_{D} \frac{\partial g_{2}}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z}, \end{align*}$$ because $g_{2} = 0$ outside $D.$ Now, Cauchy's formula gives us $$g_{2}(z) = \frac{1}{2\pi i} \int_{\partial D} \frac{g_{2}(w)}{w - z}dw + \frac{1}{2\pi i} \int_{D} \frac{\partial g_{2}}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z}dw.$$ But the first integral vanishes because $g_{2} = 0$ on $\partial D,$ as it is outside $D_{2}.$ Therefore, it follows that $$\frac{\partial f}{\partial \bar{z}} = \frac{\partial f_{2}}{\partial \bar{z}} = g$$ on $D_{1}.$ Since $D_{1} \subset D$ is an arbitrary open disk, this shows that the above equality is true on $D.$ $\Box$

Observation (for later). The proof can be modified so that if $$g = g(z_{1}, \dots, z_{n}) : U_{1} \times \cdots \times U_{n} \rightarrow \mathbb{C}$$ is a smooth function where each $U_{j}$ is open and holomorphic in variables $z_{2}, \dots, z_{n}$ (separably), then so is (modified) $f.$

Holomorphic functions of several variables. Let $U \subset \mathbb{C}^{n}$ be a nonempty open subset. The coordinate functions $(z_{1}, \dots, z_{n}),$ where we write $z_{j} = x_{j} + iy_{j} = (x_{j}, y_{j}) \in \mathbb{R}^{2}$ as usual.

A smooth function $f : U \rightarrow \mathbb{C}$ is said to be holomorphic if it is holomorphic in each $z_{j},$ which in other words, we have $$\frac{\partial f}{\partial \bar{z_{j}}} = 0$$ on $U.$ We say $f$ is analytic if for every $a \in U,$ we have a polydisk $$D(a) = D_{r_{1}}(a_{1}) \times \cdots \times D_{r_{n}}(a_{n})$$ such that $$f(z) = \sum_{k_{1}=0}^{\infty} \cdots \sum_{k_{n}=0}^{\infty}c_{k_{1},\dots,k_{n}}(z_{1} - a_{1})^{k_{1}} \cdots (z_{n} - a_{n})^{k_{n}}$$ for $z \in D,$ absolutely and uniformly.

Theorem. Given a smooth function $f : U \rightarrow \mathbb{C},$ the following are equivalent:

(1) $f$ is holomorphic;
(2) $f$ is analytic;
(3) For every polydisc $D_{1} \times \cdots \times D_{n} \subset U,$ we have $$f(z) = \left(\frac{1}{2 \pi i}\right)^{n} \int_{\prod_{j=1}^{n}\partial D_{j}} \frac{f(w)}{(w_{1} - z_{1}) \cdots (w_{n} - z_{n})} dw_{1} \wedge \cdots \wedge dw_{n},$$ where the integral is over the product of circles with product orientation.

Proof. Assume (2). Then $f$ is analytic in each variable, so it must be holomorphic in each variable. This implies (1).

To see that (1) implies (3), we use Cauchy's integral formula for single variable $n$ times, and apply Fubini (e.g., 2.37 b in Folland) as $f$ is continuous.

We can apply the same argument as in the single variable case to ensure that (3) implies (2). This finishes the proof. $\Box$