§4.4.5 Hadamard Factorization Theorem

4.4.5 Hadamard Factorization Theorem

Theorem 4.4.5.1.

Let

𝑓 (𝑧) = e^{𝜑 (𝑧)} 𝑃 (𝑧)

be the Weierstrass canonical factorization of

𝑓

, where

𝑓

is entire with finite order

𝜌 = 𝜌 (𝑓)

and

𝑓 (0) = 1

. Then

𝜑

is a polynomial of degree

\leq 𝜌

Proof.

By logarithmic differentiation and by taking $𝑝 > 𝜌 - 1$ subsequent derivatives, we have

\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} = 𝜑^{'} (𝑧) + \frac{𝑃^{'} (𝑧)}{𝑃 (𝑧)} ⟹ \frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)}) = 𝜑^{(𝑝 + 1)} (𝑧) + \frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑃^{'} (𝑧)}{𝑃 (𝑧)}) .

By applying Proposition 4.4.4.17 $Proposition 4.4.4.17 Proposition 4.4.4.17$ and Lemma 4.4.4.18 $Lemma 4.4.4.18 Lemma 4.4.4.18$ , we have

- \sum_{𝑘 = 1}^{\infty} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}} = 𝜑^{(𝑝 + 1)} (𝑧) - \sum_{𝑘 = 1}^{\infty} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}} ⟹ 𝜑^{(𝑝 + 1)} \equiv 0 .

Hence,

𝜑

is a polynomial of degree

\leq 𝑝

. Choosing

𝑝 = 1 + ⌊ 𝜌 - 1 ⌋ > 𝜌 - 1

so that

𝑝 \leq 𝜌

, the assertion follows. ⁠

□

Corollary 4.4.5.1.1.

Let

𝑓 (𝑧) = 𝑧^{𝑚} e^{𝜑 (𝑧)} 𝑃 (𝑧)

be the Weierstrass canonical factorization of

𝑓

, where

𝑓

is entire with finite order

𝜌 = 𝜌 (𝑓)

. Then

𝜑

is a polynomial of degree

\leq 𝜌

Proof.

Let $𝑓 (𝑧) = 𝑧^{𝑚} 𝑔 (𝑧)$ , where $𝑓$ and $𝑔$ are entire, $𝑔 (0) \neq 0$ , and $𝑓$ has finite order $𝜌 (𝑓)$ and $𝑀 (𝑟, 𝑓) = 𝑟^{𝑚} 𝑀 (𝑟, 𝑔)$ for $𝑟 > 0$ . For $𝜀 > 0$ , $\exists 𝑟^{'} > 0$ such that $𝑟 > 𝑟^{'}$ implies

𝑀 (𝑟, 𝑓) = 𝑟^{𝑚} 𝑀 (𝑟, 𝑔) \leq e^{𝑟^{𝜌 (𝑓) + 𝜀}} ⟹ 𝑀 (𝑟, 𝑔) \leq e^{𝑟^{𝜌 (𝑓) + 𝜀}} .

Thus, $𝜌 (𝑔) \leq 𝜌 (𝑓)$ by letting $𝜀 \to 0^{+}$ . If $𝜌 (𝑓) = 0$ , this already gives $𝜌 (𝑔) = 𝜌 (𝑓)$ . Otherwise, for any $0 < 𝜀 < \frac{𝜌 (𝑓)}{2}$ , $\forall 𝑟^{'} > 0$ , $\exists 𝑟 > 𝑟^{'}$ such that

𝑀 (𝑟, 𝑓) \geq e^{𝑟^{𝜌 (𝑓) - 𝜀}} ⟹ 𝑀 (𝑟, 𝑔) \geq exp (𝑟^{𝜌 (𝑓) - 2 𝜀} (𝑟^{𝜀} - \frac{𝑚 log 𝑟}{𝑟^{𝜌 (𝑓) - 2 𝜀}})) \geq e^{𝑟^{𝜌 (𝑓) - 2 𝜀}}

because for sufficiently large $𝑟$ ,

𝑟^{𝜀} - \frac{𝑚 log 𝑟}{𝑟^{𝜌 (𝑓) - 2 𝜀}} > 1 .

Hence, $𝜌 (𝑔) \geq 𝜌 (𝑓) - 2 𝜀$ . Letting $𝜀 \to 0^{+}$ implies $𝜌 (𝑓) = 𝜌 (𝑔)$ . Let $𝑔 (𝑧) = 𝑐 ℎ (𝑧)$ where $𝑐$ is a constant, so that $ℎ (0) = 1$ . It is also trivial that $𝜌 (𝑔) = 𝜌 (ℎ)$ . Explicitly, we have $ℎ (𝑧) = e^{𝜑 - Log 𝑐} 𝑃 (𝑧)$ .

By Theorem 4.4.5.1 $Theorem 4.4.5.1 Theorem 4.4.5.1$ on $ℎ$ , $𝜑 - Log 𝑐$ is a polynomial of degree $\leq 𝜌$ , and so is $𝜑$ . ⁠ $□$

Proposition 4.4.5.2.

Let $𝜇 \in ℤ_{\geq 0}$ and let ${𝑎_{𝑛}}_{𝑛 \in ℕ} \subset ℂ^{*}$ be a sequence such that $\sum_{𝑛 = 1}^{\infty} {| 𝑎_{𝑛} |}^{- 𝜇 - 1}$ converges. If

𝑃 (𝑧) = \prod_{𝑛 = 1}^{\infty} 𝐸_{𝜇} (\frac{𝑧}{𝑎_{𝑛}}),

then for any $𝑧 \in ℂ$ ,

log | 𝑃 (𝑧) | \leq (2 𝜇 + 1) {| 𝑧 |}^{𝜇 + 1} \sum_{𝑛 = 1}^{\infty} {| 𝑎_{𝑛} |}^{- 𝜇 - 1} .

Proof.

For $| 𝑤 | < 1$ , Taylor’s formula gives

\begin{aligned} log | 𝐸_{𝜇} (𝑤) | & = ℜ𝔢 (log (1 - 𝑤) + \sum_{𝑗 = 1}^{𝜇} \frac{𝑤^{𝑗}}{𝑗}) = ℜ𝔢 (- \sum_{𝑗 = 𝜇 + 1}^{\infty} \frac{𝑤^{𝑗}}{𝑗}) \\ \leq \sum_{𝑗 = 𝜇 + 1}^{\infty} \frac{{| 𝑤 |}^{𝑗}}{𝑗} \leq \frac{{| 𝑤 |}^{𝜇 + 1}}{1 - | 𝑤 |} . \end{aligned}

Multiplying by $1 - | 𝑤 |$ gives

(1 - | 𝑤 |) log | 𝐸_{𝜇} (𝑤) | \leq {| 𝑤 |}^{𝜇 + 1} for | 𝑤 | < 1 .

We now prove by induction that

log | 𝐸_{𝜇} (𝑤) | \leq (2 𝜇 + 1) {| 𝑤 |}^{𝜇 + 1}

for any $𝑤 \in ℂ$ . If $𝜇 = 0$ , then

log | 𝐸_{0} (𝑤) | = log | 1 - 𝑤 | \leq log (1 + | 𝑤 |) \leq | 𝑤 | .

Suppose the assertion holds for $𝜇 - 1$ . Since

𝐸_{𝜇} (𝑤) = 𝐸_{𝜇 - 1} (𝑤) exp (\frac{𝑤^{𝜇}}{𝜇}) ⟹ log | 𝐸_{𝜇} (𝑤) | = log | 𝐸_{𝜇 - 1} (𝑤) | + ℜ𝔢 \frac{𝑤^{𝜇}}{𝜇},

it follows that

log | 𝐸_{𝜇} (𝑤) | \leq (2 𝜇 - 1) {| 𝑤 |}^{𝜇} + | \frac{𝑤^{𝜇}}{𝜇} | \leq (2 𝜇 - 1 + \frac{1}{𝜇}) {| 𝑤 |}^{𝜇} .

Let $𝐴 = 2 𝜇 - 1 + \frac{1}{𝜇}$ . If $| 𝑤 | \geq \frac{𝐴}{2 𝜇 + 1}$ , then

𝐴 {| 𝑤 |}^{𝜇} \leq (2 𝜇 + 1) {| 𝑤 |}^{𝜇 + 1} .

If $| 𝑤 | < \frac{𝐴}{2 𝜇 + 1}$ , then $𝐴 \leq 2 𝜇$ implies $| 𝑤 | < \frac{2 𝜇}{2 𝜇 + 1} < 1$ , giving $\frac{1}{1 - | 𝑤 |} < 2 𝜇 + 1$ , so from the first estimate, we have

log | 𝐸_{𝜇} (𝑤) | \leq \frac{{| 𝑤 |}^{𝜇 + 1}}{1 - | 𝑤 |} \leq (2 𝜇 + 1) {| 𝑤 |}^{𝜇 + 1} .

Therefore,

log | 𝑃 (𝑧) | \leq \sum_{𝑛 = 1}^{\infty} log | 𝐸_{𝜇} (\frac{𝑧}{𝑎_{𝑛}}) | \leq (2 𝜇 + 1) {| 𝑧 |}^{𝜇 + 1} \sum_{𝑛 = 1}^{\infty} {| 𝑎_{𝑛} |}^{- 𝜇 - 1} . □

Then the results of Corollary 4.4.5.1.1 $Corollary 4.4.5.1.1 Corollary 4.4.5.1.1$ , Theorem 4.4.4.12 $Theorem 4.4.4.12 Theorem 4.4.4.12$ , and Proposition 4.4.5.2 $Proposition 4.4.5.2 Proposition 4.4.5.2$ may be consolidated into a single statement:

Theorem 4.4.5.3 (Hadamard Factorization Theorem).

Let

𝜇

be the genus of

𝑓

and let

𝜌

be the order of

𝑓

, where

𝑓

is entire with finite order. Then

𝜇 \leq 𝜌 \leq 𝜇 + 1

Proof.

Let $𝑝$ be the rank of $𝑓$ and let $𝑞$ be the degree of the polynomial $𝜑$ in its canonical factorization (Definition 4.4.4.14 $Definition 4.4.4.14 Definition 4.4.4.14$ ). By Theorem 4.4.4.12 $Theorem 4.4.4.12 Theorem 4.4.4.12$ and Corollary 4.4.5.1.1 $Corollary 4.4.5.1.1 Corollary 4.4.5.1.1$ , $𝑝 \leq ⌊ 𝜌 ⌋$ and $𝑞 \leq 𝜌$ . Therefore, $𝜇 = max {𝑝, 𝑞} \leq 𝜌$ .

By Proposition 4.4.5.2 $Proposition 4.4.5.2 Proposition 4.4.5.2$ ,

log | 𝑃 (𝑧) | \leq (2 𝑝 + 1) {| 𝑧 |}^{𝑝 + 1} \sum_{𝑛 = 1}^{\infty} {| 𝑎_{𝑛} |}^{- 𝑝 - 1} .

Also, ${max}_{| 𝑧 | = 𝑟} ℜ𝔢 𝜑 (𝑧) \leq 𝐶 𝑟^{𝑞}$ for sufficiently large $𝑟$ and some $𝐶 > 0$ . Hence,

log 𝑀 (𝑟, 𝑓) \leq 𝑚 log 𝑟 + 𝐶 𝑟^{𝑞} + (2 𝑝 + 1) 𝑟^{𝑝 + 1} \sum_{𝑛 = 1}^{\infty} {| 𝑎_{𝑛} |}^{- 𝑝 - 1} \leq 𝐶^{'} 𝑟^{𝜇 + 1}

for sufficiently large $𝑟$ and some $𝐶^{'} > 0$ . Therefore, $𝜌 \leq 𝜇 + 1$ . ⁠ $□$

Theorem 4.4.5.4.

The factorization

sin 𝑧 = 𝑧 \prod_{𝑘 = 1}^{\infty} (1 - \frac{𝑧^{2}}{π^{2} 𝑘^{2}})

defines an entire function and uniformly converges on any compact disk

\overline{𝐷 (0, 𝑟)}

Proof.

The zeros of $sin$ are simple at each of $ℤ$ . Aside from the simple zero at $𝑧 = 0$ , let

𝑎_{𝑘} = {\begin{cases} - π 𝑘 / 2 & if 𝑘 \in 2 ℕ, \\ π (𝑘 + 1) / 2 & if 𝑘 \in ℕ ∖ 2 ℕ \end{cases}

enumerate the zeros of $sin$ . By Example 4.4.4.9 $Example 4.4.4.9 Example 4.4.4.9$ , and the Hadamard Factorization Theorem (Theorem 4.4.5.3 $Theorem 4.4.5.3 Theorem 4.4.5.3$ ), the order of $sin$ is 1, the genus does not exceed 1, and

sin 𝑧 = 𝑧 e^{𝜑 (𝑧)} \prod_{𝑘 = 1}^{\infty} 𝐸_{1} (\frac{𝑧}{𝑎_{𝑘}}) = 𝑧 e^{𝜑 (𝑧)} \prod_{𝑘 = 1}^{\infty} (1 - \frac{𝑧}{𝑎_{𝑘}}) exp (\frac{𝑧}{𝑎_{𝑘}}),

where $𝜑 (𝑧) = 𝑎 𝑧 + 𝑏$ is a polynomial (and where the product locally uniformly converges in $ℂ$ ). Since the partial products ${𝑃_{𝑛}}_{𝑛 \in ℕ}$ , where

𝑃_{𝑛} = \prod_{𝑘 = 1}^{𝑛} (1 - \frac{𝑧}{𝑎_{𝑘}}) exp (\frac{𝑧}{𝑎_{𝑘}}),

have a single accumulation point, the subsequence ${𝑃_{2 𝑛}}_{𝑛 \in ℕ}$ converges to the same point. Since

\begin{aligned} 𝑃_{2 𝑛} & = \prod_{𝑘 = 1}^{2 𝑛} (1 - \frac{𝑧}{𝑎_{𝑘}}) exp (\frac{𝑧}{𝑎_{𝑘}}) \\ = \prod_{𝑘 = 1}^{𝑛} [(1 - \frac{𝑧}{π 𝑘}) exp (\frac{𝑧}{π 𝑘}) (1 + \frac{𝑧}{π 𝑘}) exp (- \frac{𝑧}{π 𝑘})], \end{aligned}

we have

sin 𝑧 = 𝑧 e^{𝑎 𝑧 + 𝑏} \prod_{𝑘 = 1}^{\infty} (1 - \frac{𝑧^{2}}{π^{2} 𝑘^{2}}) .

Then from $sin 𝑧 = - sin (- 𝑧)$ we have

𝑧 e^{𝑎 𝑧 + 𝑏} \prod_{𝑘 = 1}^{\infty} (1 - \frac{𝑧^{2}}{π^{2} 𝑘^{2}}) \equiv 𝑧 e^{- 𝑎 𝑧 + 𝑏} \prod_{𝑘 = 1}^{\infty} (1 - \frac{𝑧^{2}}{π^{2} 𝑘^{2}}) ⟹ e^{2 𝑎 𝑧} \equiv 1 ⟹ 𝑎 = 0 .

Since ${lim}_{𝜁 \to 0} \frac{sin 𝜁}{𝜁} = 1$ , we have

lim_{𝑧 \to 0} e^{𝑏} \prod_{𝑘 = 1}^{\infty} (1 - \frac{𝑧^{2}}{π^{2} 𝑘^{2}}) = 1 ⟹ 𝑏 = 0 . □