§11.1 Consequences of Holomorphy

11.1 Consequences of Holomorphy

Obviously, we will first formally define the concept of holomorphy in higher dimensions.

Definition 11.1.1.

A function

𝑓 : Ω \subseteq ℂ^{𝑛} \to ℂ

is holomorphic if it is holomorphic in each variable when the others are held constant.

If we consider $𝑓$ to be a function of $𝑧_{1}, \overline{𝑧_{1}}, 𝑧_{2}, \overline{𝑧_{2}}, \dots, 𝑧_{𝑛}, \overline{𝑧_{𝑛}}$ , then $𝑓$ is holomorphic iff $\frac{𝜕 𝑓}{𝜕 \overline{𝑧_{𝑘}}} \equiv 0$ for all $1 \leq 𝑘 \leq 𝑛$ and $𝑓$ has all continuous partial derivatives.

Theorem 11.1.2 (Cauchy’s Integral Formula on Polydisks).

Fix $𝒂 = (𝑎_{1}, \dots, 𝑎_{𝑛}) \in ℂ^{𝑛}$ arbitrarily and suppose $𝑟_{1}, 𝑟_{2}, \dots, 𝑟_{𝑛} > 0$ are the radii of the polydisk defined by $Ω = \prod_{𝑘 = 1}^{𝑛} 𝐷 (𝑎_{𝑘}, 𝑟_{𝑘})$ (where the product here is the Cartesian product). Suppose $𝑓 : \overline{Ω} \to ℂ$ is holomorphic. For fixed $𝑘_{1}, 𝑘_{2}, \dots, 𝑘_{𝑛} \in ℤ_{\geq 0}$ , we have that

(\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒛) = \frac{\prod_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}!}{{(2 π i)}^{𝑛}} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \dots \oint_{𝜕 𝐷 (𝑎_{𝑛}, 𝑟_{𝑛})} \frac{𝑓 (𝜁_{1}, \dots, 𝜁_{𝑛}) {d 𝜁}_{𝑛} \dots {d 𝜁}_{1}}{\prod_{𝑗 = 1}^{𝑛} {(𝜁_{𝑗} - 𝑧_{𝑗})}^{𝑘_{𝑗} + 1}}

for any

𝑧 = (𝑧_{1}, 𝑧_{2}, \dots, 𝑧_{𝑛}) \in Ω

Proof.

By Cauchy–Goursat (Theorem 3.2.1 $Theorem 3.2.1 Theorem 3.2.1$ ), we have

\frac{𝜕^{𝑘_{1}}}{𝜕 𝑧_{1}^{𝑘_{1}}} 𝑓 (𝒛) = \frac{𝑘_{1}!}{2 π i} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \frac{𝑓 (𝜁_{1}, 𝑧_{2}, \dots, 𝑧_{𝑛})}{{(𝜁_{1} - 𝑧_{1})}^{𝑘_{1} + 1}} {d 𝜁}_{1}

for $𝑧 \in Ω$ , which is holomorphic. Thus, by the same application on $\frac{𝜕^{𝑘_{1}}}{𝜕 𝑧_{1}^{𝑘_{1}}} 𝑓 (𝒛)$ , we have

\frac{𝜕^{𝑘_{2}}}{𝜕 𝑧_{2}^{𝑘_{2}}} \frac{𝜕^{𝑘_{1}}}{𝜕 𝑧_{1}^{𝑘_{1}}} 𝑓 (𝒛) = \frac{𝑘_{2}! 𝑘_{1}!}{{(2 π i)}^{2}} \oint_{𝜕 𝐷 (𝑎_{2}, 𝑟_{2})} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \frac{𝑓 (𝜁_{1}, 𝜁_{2}, 𝑧_{3}, \dots, 𝑧_{𝑛}) {d 𝜁}_{1} {d 𝜁}_{2}}{{(𝜁_{1} - 𝑧_{1})}^{𝑘_{1} + 1} {(𝜁_{2} - 𝑧_{2})}^{𝑘_{2} + 1}} .

By reiterating

𝑛

times and reversing the order of differentiation and integration, the conclusion follows. ⁠

□

By the boundedness assumption for $𝑓$ , we have:

Corollary 11.1.2.1 (Cauchy’s Estimate on Polydisks).

Let $𝒂 = (𝑎_{1}, 𝑎_{2}, \dots, 𝑎_{𝑛}) \in ℂ^{𝑛}$ be fixed and suppose $𝑟_{1}, 𝑟_{2}, \dots, 𝑟_{𝑛} > 0$ are the radii of the polydisk defined by $Ω = \prod_{𝑘 = 1}^{𝑛} 𝐷 (𝑎_{𝑘}, 𝑟_{𝑘})$ (where the product here is the Cartesian product). Suppose $𝑓 : \overline{Ω} \to ℂ$ is holomorphic. For fixed $𝑘_{1}, 𝑘_{2}, \dots, 𝑘_{𝑛} \in ℤ_{\geq 0}$ , we have that

| (\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒛) | \leq \prod_{𝑗 = 1}^{𝑛} (\frac{𝑘_{𝑗}!}{𝑟_{𝑗}^{𝑘_{𝑗}}}) sup_{𝜻 \in \prod_{𝑗 = 1}^{𝑛} 𝜕 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})} | 𝑓 (𝜻) |

for any

𝒛 = (𝑧_{1}, 𝑧_{2}, \dots, 𝑧_{𝑛}) \in Ω

Proof.

For each $𝑗$ , let $𝜀_{𝑗}$ satisfy $𝐷 (𝑧_{𝑗}, 𝜀_{𝑗}) \subseteq 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})$ . By Cauchy’s Integral Formula (Theorem 11.1.2 $Theorem 11.1.2 Theorem 11.1.2$ ), we have

\begin{aligned} | (\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒛) | \\ \begin{matrix}  \end{matrix} \leq \frac{\prod_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}!}{{(2 π)}^{𝑛}} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \dots \oint_{𝜕 𝐷 (𝑎_{𝑛}, 𝑟_{𝑛})} | \frac{𝑓 (𝜁_{1}, \dots, 𝜁_{𝑛})}{\prod_{𝑗 = 1}^{𝑛} {(𝜁_{𝑗} - 𝑧_{𝑗})}^{𝑘_{𝑗} + 1}} | | {d 𝜁}_{𝑛} | \dots | {d 𝜁}_{1} | . \end{aligned}

For each $𝑗$ , let $𝜁_{𝑗} = 𝑎_{𝑗} + 𝑟_{𝑗} e^{i 𝑡_{𝑗}}$ , and it follows that $d 𝜁_{𝑗} = i 𝑟_{𝑗} e^{i 𝑡_{𝑗}} d 𝑡_{𝑗}$ . Because $| 𝜁_{𝑗} - 𝑧_{𝑗} | > 𝜀_{𝑗}$ , we have, after substitution,

\begin{aligned} | (\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒛) | \leq \frac{\prod_{𝑗 = 1}^{𝑛} (𝑟_{𝑗} 𝑘_{𝑗}!)}{{(2 π)}^{𝑛}} \int_{0}^{2 π} \dots \int_{0}^{2 π} | \frac{𝑓 (𝜁_{1}, \dots, 𝜁_{𝑛})}{\prod_{𝑗 = 1}^{𝑛} 𝜀_{𝑗}^{𝑘_{𝑗} + 1}} | d 𝑡_{𝑛} \dots d 𝑡_{1} \\ \begin{matrix}  \end{matrix} \leq \prod_{𝑗 = 1}^{𝑛} (\frac{𝑟_{𝑗} 𝑘_{𝑗}!}{2 π 𝜀_{𝑗}^{𝑘_{𝑗} + 1}}) sup_{𝜻 \in \prod_{𝑗 = 1}^{𝑛} 𝜕 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})} | 𝑓 (𝜻) | \int \dots \int_{{[0, 2 π]}^{𝑛}} d 𝑡_{𝑛} \dots d 𝑡_{1} \\ \begin{matrix}  \end{matrix} \leq \prod_{𝑗 = 1}^{𝑛} (\frac{𝑘_{𝑗}!}{𝑟_{𝑗}^{𝑘_{𝑗}}}) sup_{𝜻 \in \prod_{𝑗 = 1}^{𝑛} 𝜕 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})} | 𝑓 (𝜻) |, \end{aligned}

since

𝜀_{𝑗} \leq 𝑟_{𝑗}

for all

𝑗

. ⁠

□

Similar to the univariate case, there are Taylor expansions of holomorphic functions in several complex variables.

Theorem 11.1.3.

Let $𝑓 : ℂ^{𝑛} \to ℂ$ be holomorphic on (a neighborhood of) the closure $\overline{Ω}$ of a polydisk $Ω = \prod_{𝑘 = 1}^{𝑛} 𝐷 (𝑎_{𝑘}, 𝑟_{𝑘})$ centered at $𝒂 = (𝑎_{1}, 𝑎_{2}, \dots, 𝑎_{𝑛}) \in ℂ^{𝑛}$ . Then, for any $𝒛 = (𝑧_{1}, 𝑧_{2}, \dots, 𝑧_{𝑛}) \in Ω$ , we have the expansion

𝑓 (𝒛) = \sum_{𝑘_{1} = 0}^{\infty} \dots \sum_{𝑘_{𝑛} = 0}^{\infty} 𝑎_{𝑘_{1}, \dots, 𝑘_{𝑛}} {(𝑧_{1} - 𝑎_{1})}^{𝑘_{1}} \dots {(𝑧_{𝑛} - 𝑎_{𝑛})}^{𝑘_{𝑛}}, (11.1 . 1)

where $\forall 𝑘_{1}, \dots, 𝑘_{𝑛} \in ℤ_{\geq 0}$ ,

𝑎_{𝑘_{1}, \dots, 𝑘_{𝑛}} = \frac{1}{\prod_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}!} (\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒂) .

The series converges absolutely and uniformly on

Ω

Proof.

By Theorem 11.1.2 $Theorem 11.1.2 Theorem 11.1.2$ we have

𝑓 (𝒛) = \frac{1}{{(2 π i)}^{𝑛}} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \dots \oint_{𝜕 𝐷 (𝑎_{𝑛}, 𝑟_{𝑛})} \frac{𝑓 (𝜻)}{(𝜁_{1} - 𝑧_{1}) \dots (𝜁_{𝑛} - 𝑧_{𝑛})} {d 𝜁}_{𝑛} \dots {d 𝜁}_{1} .

For each $𝑗$ , since $| 𝑧_{𝑗} - 𝑎_{𝑗} | < 𝑟_{𝑗} = | 𝜁_{𝑗} - 𝑎_{𝑗} |$ on $𝜕 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})$ , the geometric series expansion holds:

\frac{1}{𝜁_{𝑗} - 𝑧_{𝑗}} = \frac{1}{𝜁_{𝑗} - 𝑎_{𝑗}} \cdot \frac{1}{1 - \frac{𝑧_{𝑗} - 𝑎_{𝑗}}{𝜁_{𝑗} - 𝑎_{𝑗}}} = \sum_{𝑘_{𝑗} = 0}^{\infty} \frac{{(𝑧_{𝑗} - 𝑎_{𝑗})}^{𝑘_{𝑗}}}{{(𝜁_{𝑗} - 𝑎_{𝑗})}^{𝑘_{𝑗} + 1}},

which converges uniformly in $𝜁_{𝑗}$ on $𝜕 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})$ . Hence, we have

𝑓 (𝒛) = \frac{1}{{(2 π i)}^{𝑛}} \sum_{𝑘_{1} = 0}^{\infty} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \dots \oint_{𝜕 𝐷 (𝑎_{𝑛}, 𝑟_{𝑛})} \frac{𝑓 (𝜻) {(𝑧_{1} - 𝑎_{1})}^{𝑘_{1}} {d 𝜁}_{𝑛} \dots {d 𝜁}_{1}}{{(𝜁_{1} - 𝑎_{1})}^{𝑘_{1} + 1} (𝜁_{2} - 𝑧_{2}) \dots (𝜁_{𝑛} - 𝑧_{𝑛})}

where uniform convergence has allowed the interchange of summation and integration. Reiteration of this process gives

𝑓 (𝒛) = \sum_{𝑘_{1} = 0}^{\infty} \dots \sum_{𝑘_{𝑛} = 0}^{\infty} \frac{\prod_{𝑗 = 1}^{𝑛} {(𝑧_{𝑗} - 𝑎_{𝑗})}^{𝑘_{𝑗}}}{{(2 π i)}^{𝑛}} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \dots \oint_{𝜕 𝐷 (𝑎_{𝑛}, 𝑟_{𝑛})} \frac{𝑓 (𝜻) {d 𝜁}_{𝑛} \dots {d 𝜁}_{1}}{\prod_{𝑗 = 1}^{𝑛} {(𝜁_{𝑗} - 𝑎_{𝑗})}^{𝑘_{𝑗} + 1}} .

By the Cauchy Integral Formula (Theorem 11.1.2 $Theorem 11.1.2 Theorem 11.1.2$ ), we have

(\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒂) = \frac{\prod_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}!}{{(2 π i)}^{𝑛}} \oint_{𝜕 𝐷 (𝑎_{1}, 𝑟_{1})} \dots \oint_{𝜕 𝐷 (𝑎_{𝑛}, 𝑟_{𝑛})} \frac{𝑓 (𝜻) {d 𝜁}_{𝑛} \dots {d 𝜁}_{1}}{\prod_{𝑗 = 1}^{𝑛} {(𝜁_{𝑗} - 𝑎_{𝑗})}^{𝑘_{𝑗} + 1}},

and hence if we let

𝑎_{𝑘_{1}, \dots, 𝑘_{𝑛}} = \frac{1}{\prod_{𝑗 = 1}^{𝑛} 𝑘_{𝑗}!} (\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓 (𝒂),

then (11.1.1) $(11.1.1) (11.1.1)$ follows. Cauchy’s Estimate (Corollary 11.1.2.1 $Corollary 11.1.2.1 Corollary 11.1.2.1$ ) gives that

| 𝑎_{𝑘_{1}, \dots, 𝑘_{𝑛}} | \leq 𝑀 \prod_{𝑗 = 1}^{𝑛} (\frac{1}{𝜌_{𝑗}^{𝑘_{𝑗}}}),

where $𝑀 = {sup}_{𝜻 \in \prod_{𝑗 = 1}^{𝑛} 𝜕 𝐷 (𝑎_{𝑗}, 𝑟_{𝑗})} | 𝑓 (𝜻) |$ for some $𝜌_{𝑗} > 𝑟_{𝑗}$ for all $𝑗$ . Hence,

\begin{aligned} | \sum_{𝑘_{1} = 0}^{\infty} \dots \sum_{𝑘_{𝑛} = 0}^{\infty} 𝑎_{𝑘_{1}, \dots, 𝑘_{𝑛}} \prod_{𝑗 = 1}^{𝑛} {(𝑧_{𝑗} - 𝑎_{𝑗})}^{𝑘_{𝑗}} | & \leq \sum_{𝑘_{1} = 0}^{\infty} \dots \sum_{𝑘_{𝑛} = 0}^{\infty} | 𝑎_{𝑘_{1}, \dots, 𝑘_{𝑛}} | \prod_{𝑗 = 1}^{𝑛} {| 𝑧_{𝑗} - 𝑎_{𝑗} |}^{𝑘_{𝑗}} \\ \leq 𝑀 \sum_{𝑘_{1} = 0}^{\infty} \dots \sum_{𝑘_{𝑛} = 0}^{\infty} \prod_{𝑗 = 1}^{𝑛} {| \frac{𝑟_{𝑗}}{𝜌_{𝑗}} |}^{𝑘_{𝑗}} \\ = 𝑀 \prod_{𝑗 = 1}^{𝑛} \sum_{𝑘_{𝑗} = 0}^{\infty} {| \frac{𝑟_{𝑗}}{𝜌_{𝑗}} |}^{𝑘_{𝑗}} < \infty . \end{aligned}

By the Weierstrass $𝑀$ –Test (Theorem 2.3.3 $Theorem 2.3.3 Theorem 2.3.3$ ), the series converges absolutely and uniformly on $Ω$ . ⁠ $□$

Theorem 11.1.4 (Identity).

Let

𝑓

be a holomorphic function on

Ω \subseteq ℂ^{𝑛}

. If the set

{𝑧 \in Ω : 𝑓 (𝑧) = 0}

has an accumulation point in

Ω

, then

𝑓 \equiv 0

Ω

Theorem 11.1.5 (Maximum Modulus Principle).

Let $Ω \subset ℂ^{𝑛}$ be a open bounded region, and suppose that $𝑓 : Ω \to ℂ$ is holomorphic. If

𝑀 = sup_{𝜁 \in 𝜕 Ω} lim_{\begin{matrix} 𝑧 \to 𝜁 \\ 𝑧 \in Ω \end{matrix}} | 𝑓 (𝑧) |,

then

| 𝑓 (𝑧) | < 𝑀

for all

𝑧 \in Ω

, unless

𝑓

is constant.

Theorem 11.1.6 (Weierstrass).

Suppose that $Ω \subseteq ℂ^{𝑛}$ is a region and that ${𝑓_{𝑘}}_{𝑘 \in ℕ}$ is a sequence of holomorphic functions $Ω \to ℂ$ . If ${𝑓_{𝑘}}_{𝑘 \in ℕ}$ converges locally uniformly to $𝑓$ on $Ω$ , then $𝑓$ is holomorphic on $Ω$ . Moreover, $\forall 𝑘_{1}, \dots, 𝑘_{𝑛} \in ℤ_{\geq 0}$ ,

(\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓_{𝑘} ⇉ (\prod_{𝑗 = 1}^{𝑛} \frac{𝜕^{𝑘_{𝑗}}}{𝜕 𝑧_{𝑗}^{𝑘_{𝑗}}}) 𝑓

on compact subsets of

Ω

Theorem 11.1.7 (Montel).

A family

ℱ︀

of holomorphic functions on some region

Ω \subseteq ℂ^{𝑛}

is normal iff it is locally uniformly bounded on

Ω