In mathematics , and particularly in the field of complex analysis , the Hadamard factorization theorem asserts that every entire function with finite order can be represented as a product involving its zeroes and an exponential of a polynomial. It is named for Jacques Hadamard .
The theorem may be viewed as an extension of the fundamental theorem of algebra , which asserts that every polynomial may be factored into linear factors, one for each root. It is closely related to Weierstrass factorization theorem , which does not restrict to entire functions with finite orders.
Formal statement [ edit ] Define the Hadamard canonical factors
E n ( z ) := ( 1 − z ) ∏ k = 1 n e z k / k {\displaystyle E_{n}(z):=(1-z)\prod _{k=1}^{n}e^{z^{k}/k}} Entire functions of finite
order ρ {\displaystyle \rho } have
Hadamard 's canonical representation:
[1] f ( z ) = z m e Q ( z ) ∏ n = 1 ∞ E p ( z / a n ) {\displaystyle f(z)=z^{m}e^{Q(z)}\prod _{n=1}^{\infty }E_{p}(z/a_{n})} where
a k {\displaystyle a_{k}} are those
roots of
f {\displaystyle f} that are not zero (
a k ≠ 0 {\displaystyle a_{k}\neq 0} ),
m {\displaystyle m} is the order of the zero of
f {\displaystyle f} at
z = 0 {\displaystyle z=0} (the case
m = 0 {\displaystyle m=0} being taken to mean
f ( 0 ) ≠ 0 {\displaystyle f(0)\neq 0} ),
Q {\displaystyle Q} a polynomial (whose degree we shall call
q {\displaystyle q} ), and
p {\displaystyle p} is the smallest non-negative integer such that the series
∑ n = 1 ∞ 1 | a n | p + 1 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{|a_{n}|^{p+1}}}} converges. The non-negative integer
g = max { p , q } {\displaystyle g=\max\{p,q\}} is called the genus of the entire function
f {\displaystyle f} . In this notation,
g ≤ ρ ≤ g + 1 {\displaystyle g\leq \rho \leq g+1} In other words: If the order
ρ {\displaystyle \rho } is not an integer, then
g = [ ρ ] {\displaystyle g=[\rho ]} is the integer part of
ρ {\displaystyle \rho } . If the order is a positive integer, then there are two possibilities:
g = ρ − 1 {\displaystyle g=\rho -1} or
g = ρ {\displaystyle g=\rho } .
Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent α ≤ ρ ≤ g + 1 {\displaystyle \alpha \leq \rho \leq g+1} .
For example, sin {\displaystyle \sin } , cos {\displaystyle \cos } and exp {\displaystyle \exp } are entire functions of genus g = ρ = 1 {\displaystyle g=\rho =1} .
Critical exponent [ edit ] Define the critical exponent of the roots of f {\displaystyle f} as the following:
α := lim sup r log r N ( f , r ) {\displaystyle \alpha :=\limsup \limits _{r}\log _{r}N(f,r)} where
N ( f , r ) {\displaystyle N(f,r)} is the number of roots with modulus
< r {\displaystyle <r} . In other words, we have an asymptotic bound on the growth behavior of the number of roots of the function:
∀ ϵ > 0 N ( f , r ) ≪ r α + ϵ , and exists a sequence r k such that N ( f , r k ) > k r k α − ϵ {\displaystyle \forall \epsilon >0\quad N(f,r)\ll r^{\alpha +\epsilon },{\text{ and exists a sequence }}r_{k}{\text{ such that }}N(f,r_{k})>kr_{k}^{\alpha -\epsilon }} It's clear that
α ≥ 0 {\displaystyle \alpha \geq 0} .
Theorem: [2] If f {\displaystyle f} is an entire function with infinitely many roots, then
α = inf { β : ∑ k | a k | − β < ∞ } = 1 lim inf k log k | a k | {\displaystyle \alpha =\inf \left\{\beta :\sum _{k}|a_{k}|^{-\beta }<\infty \right\}={\frac {1}{\liminf \limits _{k}\log _{k}|a_{k}|}}} Note: These two equalities are purely about the limit behaviors of a real number sequence
| a 1 | ≤ | a 2 | ≤ ⋯ {\displaystyle |a_{1}|\leq |a_{2}|\leq \cdots } that diverges to infinity. It does not involve complex analysis.
Proposition: α ( f ) ≤ ρ {\displaystyle \alpha (f)\leq \rho } ,[3] by Jensen's formula .
Since f ( z ) / z m {\displaystyle f(z)/z^{m}} is also an entire function with the same order ρ {\displaystyle \rho } and genus, we can wlog assume m = 1 {\displaystyle m=1} .
If f {\displaystyle f} has only finitely many roots, then f ( z ) = e Q ( z ) ∏ k = 1 n E 0 ( z / a k ) {\displaystyle f(z)=e^{Q(z)}\prod _{k=1}^{n}E_{0}(z/a_{k})} with the function e Q ( z ) {\displaystyle e^{Q(z)}} of order ρ {\displaystyle \rho } . Thus by an application of the Borel–Carathéodory theorem , P {\displaystyle P} is a polynomial of degree d e g ( Q ) ≤ floor ( ρ ) {\displaystyle deg(Q)\leq {\text{floor}}(\rho )} , and so we have ρ = d e g ( Q ) = g {\displaystyle \rho =deg(Q)=g} .
Otherwise, f {\displaystyle f} has infinitely many roots. This is the tricky part and requires splitting into two cases. First show that g ≤ floor ( ρ ) {\displaystyle g\leq {\text{floor}}(\rho )} , then show that ρ ≤ g + 1 {\displaystyle \rho \leq g+1} .
Define the function g ( z ) := ∏ k = 1 ∞ E p ( z / a k ) {\displaystyle g(z):=\prod _{k=1}^{\infty }E_{p}(z/a_{k})} where p = floor ( α ) {\displaystyle p={\text{floor}}(\alpha )} . We will study the behavior of f ( z ) / g ( z ) {\displaystyle f(z)/g(z)} .
Bounds on the behaviour of |E p | [ edit ] In the proof, we need four bounds on | E p | {\displaystyle |E_{p}|} :
For any r ∈ ( 0 , 1 ) {\displaystyle r\in (0,1)} , | 1 − E p ( z ) | ≤ O ( | z | p + 1 ) {\displaystyle |1-E_{p}(z)|\leq O(|z|^{p+1})} when | z | ≤ r {\displaystyle |z|\leq r} . For any r ∈ ( 0 , 1 ) {\displaystyle r\in (0,1)} , there exists B > 0 {\displaystyle B>0} such that ln | E p ( z ) | ≥ − B | z | p + 1 {\displaystyle \ln |E_{p}(z)|\geq -B|z|^{p+1}} when | z | ≤ r {\displaystyle |z|\leq r} . For any r ∈ ( 0 , 1 ) {\displaystyle r\in (0,1)} , there exists B > 0 {\displaystyle B>0} such that | E p ( z ) | ≥ | 1 − z | e − B | z | p {\displaystyle |E_{p}(z)|\geq |1-z|e^{-B|z|^{p}}} when | z | ≥ r {\displaystyle |z|\geq r} . ln | E p ( z ) | ≤ O ( | z | p + 1 ) {\displaystyle \ln |E_{p}(z)|\leq O(|z|^{p+1})} for all z {\displaystyle z} , and ln | E p ( z ) | ≤ O ( | z | p ) {\displaystyle \ln |E_{p}(z)|\leq O(|z|^{p})} as | z | → ∞ {\displaystyle |z|\to \infty } . These are essentially proved in the similar way. As an example, we prove the fourth one.
ln | E n ( z ) | = R e ( − z n + 1 ∑ k = 0 ∞ z k k + n + 1 ) ≤ | z | n + 1 | g ( z ) | {\displaystyle \ln |E_{n}(z)|=Re\left(-z^{n+1}\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}\right)\leq |z|^{n+1}|g(z)|} where
g ( z ) := ∑ k = 0 ∞ z k k + n + 1 {\displaystyle g(z):=\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}} is an entire function. Since it is entire, for any
R > 0 {\displaystyle R>0} , it is bounded in
B ( 0 , R ) {\displaystyle B(0,R)} . So
ln | E n ( z ) | = O ( | z | n + 1 ) {\displaystyle \ln |E_{n}(z)|=O(|z|^{n+1})} inside
B ( 0 , R ) {\displaystyle B(0,R)} .
Outside B ( 0 , R ) {\displaystyle B(0,R)} , we have
ln | E n ( z ) | = ln | 1 − z | + R e ( Q ( z ) ) = o ( | z | ) + O ( | z | n ) < O ( | z | n + 1 ) {\displaystyle \ln |E_{n}(z)|=\ln |1-z|+Re(Q(z))=o(|z|)+O(|z|^{n})<O(|z|^{n+1})} g is well-defined[ edit ] Source:[2]
For any R > 0 {\displaystyle R>0} , we show that the sum ∑ k = 1 ∞ ln | E p ( z / a k ) | {\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|} converges uniformly over B ( 0 , R ) {\displaystyle B(0,R)} .
Since only finitely many | a k | < K R {\displaystyle |a_{k}|<KR} , we can split the sum to a finite bulk and an infinite tail:
∑ k = 1 ∞ ln | E p ( z / a k ) | = ∑ | a k | < K R ln | E p ( z / a k ) | + ∑ | a k | ≥ K R ln | E p ( z / a k ) | {\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|=\sum _{|a_{k}|<KR}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq KR}\ln |E_{p}(z/a_{k})|} The bulk term is a finite sum, so it converges uniformly. It remains to bound the tail term.
By bound (1) on | E p | {\displaystyle |E_{p}|} , | 1 − E p ( z / a k ) | ≤ B | z / a k | p + 1 ≤ B / K p + 1 {\displaystyle |1-E_{p}(z/a_{k})|\leq B|z/a_{k}|^{p+1}\leq B/K^{p+1}} . So if K {\displaystyle K} is large enough, for some B ′ > 0 {\displaystyle B'>0} ,[nb 1]
∑ | a k | ≥ K R ln | 1 − ( 1 − E p ( z / a k ) ) | ≤ ∑ | a k | ≥ K R B ′ B | z / a k | p + 1 < | z | p + 1 B ′ B ∑ k 1 | a k | p + 1 {\displaystyle \sum _{|a_{k}|\geq KR}\ln |1-(1-E_{p}(z/a_{k}))|\leq \sum _{|a_{k}|\geq KR}B'B|z/a_{k}|^{p+1}<|z|^{p+1}B'B\sum _{k}{\frac {1}{|a_{k}|^{p+1}}}} Since
p + 1 = floor ( α ) + 1 > α {\displaystyle p+1={\text{floor}}(\alpha )+1>\alpha } , the last sum is finite.
g ≤ floor(ρ ) [ edit ] As usual in analysis, we fix some small ϵ > 0 {\displaystyle \epsilon >0} .
Then the goal is to show that f ( z ) / g ( z ) {\displaystyle f(z)/g(z)} is of order ≤ ρ + ϵ {\displaystyle \leq \rho +\epsilon } . This does not exactly work, however, due to bad behavior of f ( z ) / g ( z ) {\displaystyle f(z)/g(z)} near a k {\displaystyle a_{k}} . Consequently, we need to pepper the complex plane with "forbidden disks", one around each a k {\displaystyle a_{k}} , each with radius 1 | a k | ρ + ϵ {\displaystyle {\frac {1}{|a_{k}|^{\rho +\epsilon }}}} . Then since ∑ k 1 | a k | ρ + ϵ < ∞ {\displaystyle \sum _{k}{\frac {1}{|a_{k}|^{\rho +\epsilon }}}<\infty } by the previous result on α {\displaystyle \alpha } , we can pick an increasing sequence of radii R 1 < R 2 < ⋯ {\displaystyle R_{1}<R_{2}<\cdots } that diverge to infinity, such that each circle ∂ B ( 0 , R n ) {\displaystyle \partial B(0,R_{n})} avoids all these forbidden disks.
Thus, if we can prove a bound of form ln | f ( z ) g ( z ) | = O ( | z | ρ + ϵ ) = o ( | z | ρ + 2 ϵ ) {\displaystyle \ln \left|{\frac {f(z)}{g(z)}}\right|=O(|z|^{\rho +\epsilon })=o(|z|^{\rho +2\epsilon })} for all large z {\displaystyle z} [nb 2] that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem, d e g ( Q ) ≤ floor ( ρ + 2 ϵ ) {\displaystyle deg(Q)\leq {\text{floor}}(\rho +2\epsilon )} for any ϵ > 0 {\displaystyle \epsilon >0} , and so as we take ϵ → 0 {\displaystyle \epsilon \to 0} , we obtain g ≤ floor ( ρ ) {\displaystyle g\leq {\text{floor}}(\rho )} .
Since ln | f ( z ) | = o ( | z | ρ + ϵ ) {\displaystyle \ln \left|f(z)\right|=o(|z|^{\rho +\epsilon })} by the definition of ρ {\displaystyle \rho } , it remains to show that − ln | g ( z ) | = O ( | z | ρ + ϵ ) {\displaystyle -\ln \left|g(z)\right|=O(|z|^{\rho +\epsilon })} , that is, there exists some constant B > 0 {\displaystyle B>0} such that
∑ k = 1 ∞ ln | E p ( z / a k ) | ≥ − B | z | ρ + ϵ {\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|\geq -B|z|^{\rho +\epsilon }} for all large
z {\displaystyle z} that avoids these forbidden disks.
As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many a k {\displaystyle a_{k}} with modulus < | z k | / 2 {\displaystyle <|z_{k}|/2} and infinitely many a k {\displaystyle a_{k}} with modulus ≥ | z k | / 2 {\displaystyle \geq |z_{k}|/2} . So we have to bound:
∑ | a k | < | z k | / 2 ln | E p ( z / a k ) | + ∑ | a k | ≥ | z k | / 2 ln | E p ( z / a k ) | {\displaystyle \sum _{|a_{k}|<|z_{k}|/2}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq |z_{k}|/2}\ln |E_{p}(z/a_{k})|} The upper-bounding can be accomplished by the bounds (2), (3) on
| E p | {\displaystyle |E_{p}|} , and the assumption that
z {\displaystyle z} is outside every forbidden disk. Details are found in.
[2] ρ ≤ g + 1 [ edit ] This is a corollary of the following:
If f ( z ) = e Q ( z ) z m ∏ k E p ( z / a k ) {\displaystyle f(z)=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})} has genus g {\displaystyle g} , then ln | f ( z ) | = O ( | z | g + 1 ) {\displaystyle \ln |f(z)|=O(|z|^{g+1})} .
Split the sum to three parts:
ln | f ( z ) | = R e ( Q ( z ) ) + m R e ( ln z ) + ∑ k ln | E p ( z / a k ) | . {\displaystyle \ln |f(z)|=Re(Q(z))+m\,Re(\ln z)+\sum _{k}\ln |E_{p}(z/a_{k})|.} The first two terms are
O ( | z | p ) = o ( | z | g + 1 ) {\displaystyle O(|z|^{p})=o(|z|^{g+1})} . The third term is bounded by bound (4) of
| E p | {\displaystyle |E_{p}|} :
∑ k ln | E p ( z / a k ) | ≤ B | z | q + 1 ∑ k 1 | a k | q + 1 . {\displaystyle \sum _{k}\ln |E_{p}(z/a_{k})|\leq B|z|^{q+1}\sum _{k}{\frac {1}{|a_{k}|^{q+1}}}.} By assumption,
p = floor ( α ) {\displaystyle p={\text{floor}}(\alpha )} , so
∑ k 1 | a k | q + 1 < ∞ {\displaystyle \sum _{k}{\frac {1}{|a_{k}|^{q+1}}}<\infty } . Hence the above sum is
O ( | z | q + 1 ) = O ( | z | g + 1 ) . {\displaystyle O(|z|^{q+1})=O(|z|^{g+1}).} Applications [ edit ] With Hadamard factorization we can prove some special cases of Picard's little theorem .
Theorem:[4] If f {\displaystyle f} is entire, nonconstant, and has finite order, then it assumes either the whole complex plane or the plane minus a single point.
Proof: If f {\displaystyle f} does not assume value z 0 {\displaystyle z_{0}} , then by Hadamard factorization, f ( z ) − z 0 = e Q ( z ) {\displaystyle f(z)-z_{0}=e^{Q(z)}} for a nonconstant polynomial Q {\displaystyle Q} . By the fundamental theorem of algebra , Q {\displaystyle Q} assumes all values, so f ( z ) − z 0 {\displaystyle f(z)-z_{0}} assumes all nonzero values.
Theorem:[4] If f {\displaystyle f} is entire, nonconstant, and has finite, non-integer order ρ {\displaystyle \rho } , then it assumes the whole complex plane infinitely many times.
Proof: For any w ∈ C {\displaystyle w\in \mathbb {C} } , it suffices to prove f ( z ) − w {\displaystyle f(z)-w} has infinitely many roots. Expand f ( z ) − w {\displaystyle f(z)-w} to its Hadamard representation f ( z ) − w = e Q ( z ) z m ∏ k E p ( z / a k ) {\displaystyle f(z)-w=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})} . If the product is finite, then ρ = g {\displaystyle \rho =g} is an integer.
References [ edit ] ^ Conway, J. B. (1995), Functions of One Complex Variable I, 2nd ed. , springer.com: Springer, ISBN 0-387-90328-3 ^ a b c Dupuy, Taylor. "Hadamard's Theorem and Entire Functions of Finite Order — For Math 331" (PDF) . ^ Kupers, Alexander (April 30, 2020). "Lectures on complex analysis" (PDF) . Lecture notes for Math 113 . , Theorem 12.3.4.ii. ^ a b Conway, John B. (1978). Functions of One Complex Variable I . Graduate Texts in Mathematics. Vol. 11. New York, NY: Springer New York. doi :10.1007/978-1-4612-6313-5 . ISBN 978-0-387-94234-6 . Chapter 11, Theorems 3.6, 3.7. ^ so that B / K p + 1 < 1 / 2 {\displaystyle B/K^{p+1}<1/2} , then we can use the bound ln | 1 − w | ≤ B ′ | w | ∀ | w | < 1 / 2 {\displaystyle \ln |1-w|\leq B'|w|\forall |w|<1/2} to get ^ That is, we fix some yet-to-be-determined constant R {\displaystyle R} , and use " z {\displaystyle z} is large" to mean | z | > R {\displaystyle |z|>R} .