Chapter 8

Infinite Products

As is amply demonstrated by power series expansions, a highly useful technique

in complex analysis is to express an analytic function as an infinite sum of much
simpler functions. Likewise, it can be useful to express an analytic function as
an infinte product of simpler functions. This is especially true in the study of
the zeroes of analytic functions. For example, a polynomial p of degree n can
be written as a product
Yn
p(z) = a (z − zj ),
j=1

where {z1 , z2 , ·, zn } are the zeroes of p. It turns out that product expansions of
a similar type (but with infinitely many factors) are possible for other analytic
functions.
Since the exponential function converts sums to products, we can expect that
the theory of infinite products will be closely related to the theory of infinite
sums.

8.1 Convergence of Infinite Products

In the following discussion, log will be the principal branch of the log function.
Definition 8.1.1. If {uk } is a sequence of complex numbers and
n
Y
pn = uk , (8.1.1)
k=1

then we will say that the infinite product

∞
Y
uk (8.1.2)
k=1

converges to the complex number p if lim pn = p.

n→∞

255
256 CHAPTER 8. INFINITE PRODUCTS

Theorem 8.1.2. If {uk } is a sequence of complex numbers, then the infinite

product (8.1.2) converges to a non-zero number p if and only if the infinite sum
∞
X
log uk (8.1.3)
k=1

converges to a number λ. In this case, p = eλ . Furthermore, if the infinite series

converges absolutely, then the infinite product is unchanged by a rearrangement
of the factors.
Proof. We have to be careful here, because the log function converts products
to sums only up to ±2πi. However, it is true that log uv = log u + log v if u
and v have positive real part since, in this case, log u and log v have imaginary
parts in (−π/2, π/2) and log uv has imaginary part in (−π, π). Thus, log uv and
log u + log v cannot differ by a non-zero multiple of 2πi.
We define the partial products pn as in (8.1.1). If p = lim pn exists and is
n→∞
non-zero, then
lim log(pn /p) = 0,
n→∞

since log is continuous at 1. In particular, there is an N such that

−π/4 < Im (log(pn /p)) < π/4 whenever n ≥ N.

It follows that pn /pm = (pn /p)(pm /p)−1 is in the right half plane for n, m ≥ N .
In particular, un+1 = pn+1 /pn is in the right half plane for n ≥ N . Thus,

log(pn+1 /pN ) = log((pn /pN )un+1 ) = log(pn /pN ) + log un+1

whenever n ≥ N . This equation and an induction argument beginning with

n = N show that
n
X
log(pn /pN ) = log uk
k=N +1

for all n > N . Since the left side of this equality converges as n → ∞ so does
the right side. This implies the convergence of the series (8.1.3).
Conversely, if this series converges and we let
n
X
λn = log un ,
k=1

be its nth partial sum, then the sequence {λn } converges to a number λ. Since

pn = eλn ,

and the exponential function is continuous, the sequence {pn } converges to eλ .

If the series (8.1.3) converges absolutely, then each of its rearrangements
converges to the same number. It follows that each rearrangement of the infinite
product (8.1.2) also converges to the same number.
8.1. CONVERGENCE OF INFINITE PRODUCTS 257

Uniform Convergence of Products

We will be primarily interested in infinite products of analytic functions. In
this situation, whether or not the product converges uniformly is of critical
importance. We say that an the infinite product of a sequence of functions {uk }
converges uniformly on a set S if the sequence pn of partial products converges
uniformly on S.
Theorem 8.1.3. Let uk be a sequence of complex valued functions defined and
bounded on a set S. If the series
∞
X
log uk (z)
k=1

converges uniformly to λ(z) on S, then the infinite product

∞
Y
uk (z)
k=1

converges uniformly to eλ(z) on S.

Proof. Let λn (z) be the nth partial sum of the infinite sum and pn (z) the nth
partial product of the infinite product. Then the uniform convergence of the
series on S implies that λn (z) − λ(z) converges uniformly to 0 on S. Since the
exponential function is continuous at 0, this implies that
pn (z)
= eλn (z)−λ(z)
p(z)
converges uniformly to 1.
The fact that each λn is bounded on S and the convergence is uniform implies
that λ is bounded on S and, hence, that p(z) = e−λ(z) is also bounded on S.
Hence, pn = (pn /p)p converges uniformly to p on S.
Theorem 8.1.4. Let {ak (z)} is a sequence of complex valued functions defined
on a set S. If the series
X∞
|ak (z)| (8.1.4)
k=1

converges uniformly on S, then the infinite product

∞
Y
(1 + ak (z)) (8.1.5)
k=1

converges uniformly on S. Each rearrangement of the infinite product converges

to the same function. If the infinite product converges to p(z), then each zero
of p(z) is a zero, with the same order, of some finite product of the factors
1 + ak (z).
258 CHAPTER 8. INFINITE PRODUCTS

Proof. If |w| < 1/2, then (Exercise 8.1.1)

2
|w| ≤ | log(1 + w)| ≤ 2|w|. (8.1.6)
3
If the series (8.1.4) converges uniformly on S, then there is a K such that
|ak (z)| ≤ 1/2 for k ≥ K and for all z ∈ S. If we use (8.1.6) with w = ak , it
follows that one of the two series
∞
X ∞
X
| log(1 + ak (z))| and |ak (z)|
k=K k=K

converges uniformly on S if and only if the other one does also. Hence, if (8.1.4)
converges uniformly then
X∞
log(1 + ak (z))
k=K

converges uniformly and absolutely. By the previous two theorems, this is im-
plies the uniform convergence of
∞
Y
(1 + ak (z))
k=K

to a function on S with no zeroes. It follows that (8.1.5) converges uniformly

on S, the limit is unaffected by by rearrangements of the factors, and each of
its zeroes is a zero, with the same order, of the product of the factors 1 + ak (z)
for k < K.

Example 8.1.5. Prove that the infinite product

∞
Y
(1 − z 2 /k 2 ) (8.1.7)
k=1

converges uniformly on each bounded subset of C.

Solution: We have | − z 2 /k 2 | ≤ R2 /k 2 for all z in the disc DR (0). Since
the positive termed series
∞
X R2
k2
k=1

converges, it follows that the series

∞
X |z 2 |
k2
k=1

converges uniformly on DR (0). Hence, by the previous theorem, the infinite

product (8.1.7) also converges uniformly on DR (0) for each R and, hence, on
each bounded subset of C.
8.1. CONVERGENCE OF INFINITE PRODUCTS 259

Logarithmic Derivative of a Product

If an analytic function f on an open set U has an analytic logarithm g on U –
that is, if f = eg on U with g analytic – then g ′ = f ′ /f . The expression f ′ /f
is independent of which logarithm is chosen for f . Furthermore, as long as f
is not identically zero on any component of U , f ′ /f exists (as a meromorphic
function on U ) even if f does not have an analytic logarithm on U . Note that f
cannot have an analytic or even a meromorphic logarithm in any neighborhood
of a point where it has the value zero (Exercise 8.1.4).
Definition 8.1.6. Let f be an analytic function on an open set U and suppose
that f is not identically 0 on any component of U . Then the meromorphic
function f ′ /f is called the logarithmic derivative of f on U .
Logarithmic derivative is quite a well behaved notion. The logarithmic
derivative of the product of two functions is the sum of there logrithmic deriva-
tives (Exercise 8.1.5). Furthermore, the following theorem states that logarith-
mic derivative is preserved by uniform limits. The proof is left to the exercises
(Exercise 8.1.6).
Theorem 8.1.7. Let {fn } be a sequence of analytic functions on a connected
open set U . If this sequence converges uniformly to f on U then the sequence
{fn′ /fn } converges uniformly to f ′ /f on compact subsets of U \ S, where S is
the set of zeroes of f .
When applied to infinite products, this immediately implies the following
corollary.
Corollary 8.1.8. Let {uk } be a sequence of analytic functions on a connected
open set U . If the product
∞
Y
f (z) = uk (z)
k=1
converges uniformly on compact subsets of U to a function f which is not iden-
tically 0, then the infinite sum
∞
X u′ (z)
k
uk (z)
k=1
′
converges uniformly to f /f on compact subsets of U \ S, where S is the set of
zeroes of f .
Example 8.1.9. Show that the function
∞ 
z2
Y 
f (z) = πz 1− 2
k
k=1

has a logarithmic derivative which can be written as

∞
f ′ (z) 1 X 2z
= + (8.1.8)
f (z) z z 2 − k2
k=1
260 CHAPTER 8. INFINITE PRODUCTS

or as
n
f ′ (z) X 1
= lim (8.1.9)
f (z) n→∞ z−k
k=−n

Solution Note that the infinite product in the expression for f converges
uniformly on each compact disc in the plane by Example 8.1.5.
By the previous theorem,
∞ ∞
f ′ (z) 1 X −2z/k 2 1 X 2z
= + = + .
f (z) z 1 − z 2 /k 2 z z 2 − k2
k=1 k=1

This proves (8.1.8). Since

2z 1 1
= + ,
z2 −k 2 z−k z+k
the nth partial sum of the series (8.1.8) can be re-written as the sum that
appears in (8.1.9).

The logarithmic derivative of f , as computed in the above example, will be

used in the problem set to prove that f (z) = sin πz. That is,
∞ 
z2
Y 
sin(πz) = πz 1− 2 . (8.1.10)
k
k=1

Exercise Set 8.1

1. Prove that if w is a complex number with |w| ≤ 1/2, then
2
|w| ≤ | log(1 + w)| ≤ 2|w|.
3

2. Does the infinite product

∞  
Y 1
1+
k
k=1

converge? How about the product

∞  
Y 1
1 + 3/2 ?
k
k=1

3. Show that the infinite product

∞ 
Y z  z/k
1− e
k
k=1

converges uniformly on compact subsets of the plane.

8.2. WEIERSTRASS PRODUCTS 261

4. Prove that if f is analytic on U and has a zero at z0 ∈ U , then there is

no meromorphic function g defined in a neighborhood V of z0 such that
f = eg on V .
5. Prove that the logarithmic derivative of the product f g of two analytic
functions is the sum of the logarithmic derivative of f and the logarithmic
derivative of g. Also prove the analogous statement for the quotient f /g.
6. Prove Theorem 8.1.7. Hint: first prove that it is true on any disc in U on
which f has no zeroes.
7. Prove that the logarithmic derivative of a meromorphic function f on C
is also a meromorphic function on C and is odd (even) if f is odd (even).
8. If f is the function defined in Example 8.1.9, prove that the logarithmic
derivative of f is an odd meromorphic function which is periodic of pe-
riod 1. Observe that the logarithmic derivative of sin πz has the same
properties.
9. Prove that if f is the function of the previous exercise, and we set
sin(πz)
g(z) = ,
f (z)
then g is an entire function with no zeroes and, hence, has a logarithm h
which is entire. Then, sin(πz) = f (z)eh(z) .
10. Prove that if f , g and h are the functions of the previous exercise, then
the logarithmic derivative of g is h′ (z) = π cot πz − f ′ (z)/f (z).
11. With h as above, prove that h′ is bounded on the strip 0 ≤ Re (z) ≤ 1
(use (8.1.8)). Show that this implies it is bounded on the entire plane and,
hence, is constant.
12. With h as above, prove that h′ (0) = 0 and, hence, that h′ is identically 0
and h is a constant. Then use the fact that limz→0 z −1 sin z = 1 to show
that this constant is 0. Conclude that
∞ 
z2
Y 
sin(πz) = πz 1− 2 .
k
k=1

8.2 Weierstrass Products

In this section we will show that, given any sequence of points of an open set
U ⊂ C, with no limit point in U , there is an analytic function on U with exactly
the points of this sequence as its zeroes, with each zero having order equal to
the number of times it appears in the sequence. The analytic function will be
constructed as an infinite product of certain simple functions, each of which has
exactly one zero. These simple functions are constructed as follows.
262 CHAPTER 8. INFINITE PRODUCTS

For p = 0, 1, 2, · · · we define entire functions Ep (z) by E0 (z) = 1 − z and

2
/2+···+z p /p
Ep (z) = (1 − z)ez+z for p > 0.

Note that z + z 2 /2 + · · · + z p /p is the pth partial sum for the power seriies
expansion of − log(1 − z) about z = 0 and so, although Ep (1) = 0, the sequence
Ep (z) will converge uniformly to (1 − z)(1 − z)−1 = 1 on each disc of radius less
than 1 centered at 0. More precisely:
Theorem 8.2.1. Each Ep (z) is an entire function with the following properties:
(a) the only zero of Ep (z) occurs at z = 1;
(b) if |z| ≤ 1, then |Ep (z) − 1| < |z|p+1 .
Proof. Part (a) is obvious. To prove Part (b), we note that the derivative of
1 − Ep (z) is (Exercise 8.2.1)
2
/2+···+z p /p
(1 − Ep (z))′ = −Ep′ (z) = z p ez+z . (8.2.1)

Since this has a zero of order p at z = 0, the function 1 − Ep (z) has a zero of
order p + 1 at z = 0.
The function (8.2.1) has a power series expansion about 0 with all of its coef-
ficients non-negative real numbers, since this is true of the exponential function
and the function z + z 2 /2 + · · · + z p /p. It follows that the function
1 − Ep (z)
h(z) = ,
z p+1
also has non-negative real numbers as coefficients for its power series expansion
about 0. This implies that the maximum value achieved by |h(z)| for |z| ≤ 1 is
h(1) = 1. That is,  
 1 − Ep (z) 
 z p+1  ≤ 1 for |z| ≤ 1.
 

Part (b) follows from this.

If f is an analytic function on U , then we will say that a sequence {zk } ⊂ U
is a list of the zeroes of f counting multiplicity if each zk is a zero of f , and if
each zero w of f occurs m(w) times in this sequence, where m(w) is the order
of the zero w.
Let {zk } be a sequence of non-zero complex numbers converging to ∞. The
next theorem show how to use scaled versions of the functions Ep to construct
an entire function with this sequence as a list of its zeroes counting multiplicity.
The resulting product is called a Weierstrass product.
Theorem 8.2.2. Let A be a subset of C. If {zk } is a sequence of non-zero
complex numbers and {pk } is a sequence of integers such that
∞ 
 r pk +1

X
 
 zk  < ∞ for all r > 0, (8.2.2)
k=1
8.2. WEIERSTRASS PRODUCTS 263

then the Weierstrass product

∞
Y
f (z) = Epk (z/zk ), (8.2.3)
k=1

converges uniformly on compact subsets of C to an entire function which has

{zk } as a list of its zeroes counting multiplicity.

Proof. By part (b) of the previous theorem, we have

 pk +1
z
|Epk (z/zk ) − 1| ≤   if |z| ≤ |zk |.
zk

The condition |z| ≤ |zk | must be satisfied for all sufficiently large k if the series
(8.2.2) converges. The theorem follows by applying Theorem 8.1.4 with ak (z) =
Epk (z/zk ) − 1.

The Weierstrass Theorem

Theorem 8.2.3. If {zk } is any sequence of complex numbers converging to
infinity, then there is an entire function with {zk } as a list of its zeroes counting
multiplicity.

Proof. Suppose m of the zk are equal to 0 (m might be 0). We may as well

assume these are the first m terms of the sequence. Then {zk }∞ k=m+1 is a
sequence of non-zero complex numbers.
If R > 0, then, since zk → ∞, there is a K > m such that |zk | > 2R for all
k ≥ K. Then the series
X  z k
 
 
 zk 
k=m+1

converges uniformly on |z| ≤ R by comparison with the geometric series with

ratio 1/2. Thus, the hypotheses of the previous theorem are satisfied if we
choose pk = k − 1 for each k. The resulting Weierstrass product
∞
Y
Ep (z/zk )
k=m+1

converges uniformly on compact subsets of C to an entire function which has

{zk }∞
k=m+1 as a list of its zeroes counting multiplicity. Then

∞
Y
f (z) = z m Ep (z/zk )
k=m+1

is an entire function with {zk }∞

k=1 as a list of its zeroes counting multiplicity.
264 CHAPTER 8. INFINITE PRODUCTS

Example 8.2.4. Find an entire function which has a zero of order k at each
positive integer k.
Solution: We construct a sequence

1, 2, 2, , 3, 3, 3, 4, 4, 4, 4, · · ·

in which each n appears k times and the terms are arranged in increasing order.
Then, for this sequence {zk } and a given positive integer p, we have
∞ ∞ ∞
X 1 X 1 X 1
p+1
= k p+1 =
|zk | k kp
k=1 k=1 k=1

If we choose p = 2, then the right side is the convergent series

∞
X 1
.
k2
k=1

The Weierstrass product for the sequences {zk } and {pk = 2} is

∞  k
Y 2 2
(1 − z/k)ez/k+z /(2k ) .
k=1

By Theorem 8.2.2 this infinite product converges to an entire function with the
required zeroes.

Weierstrass Factorization
The Weierstrass Theorem for the plane leads immediately to the Weierstrass
Factorization Theorem for entire functions:
Theorem 8.2.5. Let f be an entire function which is not identically zero. Let
m be the order of the zero of f at 0, and let {zk } be a list of the non-zero zeroes
of f counting multiplicity. Then there exists non-negative integers p1 , p2 , · · ·
and an entire function h such that
∞
Y
f (z) = eh(z) z m Epk (z/zk ).
k=1

The sequence {pk } may be chosen in any way which satisfies (8.2.2).
Proof. The product
∞
Y
g(z) = z m Epk (z/zk )
k=1

converges uniformly on compact sets if {pk } is chosen such that (8.2.2) holds
(pk = k − 1 is one choice which always works, but there may be better choices
for a given f ). Furthermore, the resulting function g has the same zeroes as f
8.2. WEIERSTRASS PRODUCTS 265

with the same multiplicities. Thus, f g −1 is an entire function with no zeroes

(after removable singularities are removed). It follows that

f g −1 = eh

for some entire function h. The theorem follows from this.

In many cases, the sequence {pk } can be chosen to be constant.

Example 8.2.6. Find a Weierstrass factorization for sin(πz).

Solution: This function has a zero of order 1 at each integer and has no
other zeroes. Since
∞
X 1
< ∞,
k2
k=1

the condition (8.2.2) holds if we choose pk = 1 for every k. Then the above
theorem tells us that
Y Y
sin(πz) = eh(z) z E1 (z/k) = eh(z) z (1 − z/k)ez/k ,
k6=0 k6=0

where the product is over all non-zero integers k. Note that if the factors for k
and −k in this product are paired, the result is

(1 − z/k)ez/k (1 + z/k)e−z/k = 1 − z 2 /k 2 .

We conclude from Exercise 8.1.12 that eh(z) = π, that

Y
sin(πz) = πz (1 − z/k)ez/k
k6=0

is a Weierstrass factorization of sin(πz), and that this factorization is equivalent

to the factorization
Y∞
sin(πz) = πz (1 − z 2 /k 2 ).
k=1

The General Weierstrass Theorem

If C is replaced by an arbitrary non-empty, proper open subset of S 2 , the ana-
logue of Theorem 8.2.3 holds with only a slightly more complicated proof.

Theorem 8.2.7. Let U be a non-empty, proper open subset of S 2 . If {zk } is

any sequence of points of U with no limit points in U , then there is an analytic
function f on U with {zk } as a list of its zeroes counting mulltiplicity.

Proof. Either U or its image under some linear fractional transformation will
contain ∞. Thus, we may as well assume ∞ ∈ U . Then the complement of U
in S 2 is a compact subset K of the plane.
266 CHAPTER 8. INFINITE PRODUCTS

Since {zk } has no limit point in U , the distance between zk and K must
approach 0 as k → ∞. It follows that we may choose a sequence {wk } of points
of K such that lim |zk − wk | = 0.
We set
∞  
Y zk − wk
f (z) = Ek .
n=1
z − wk
The product converges uniformly on compact subsets of U , since lim |zk −wk | = 0
implies the uniform convergence on compact subsets of U of
∞ 
 zk − wk k+1

X

 z − wk 
 .
k=1

The function f is analytic in U as has {zk } as a list of its zeroes counting

multiplicity.

Meromorphic Functions
On a connected open set U , the set of analytic functions forms an integral
domain – that is, it is a commutative ring with the property that the product of
two elements is zero if and only if one of them is zero. The set of meromorphic
functions of U forms a field – that is, a commutative ring in which every non-zero
element has an inverse. The next theorem shows that the field of meromorphic
function is actually the quotient field of the ring of analytic functions. That is,
every meromorphic function is the quotient f /g of two analytic functions.
Theorem 8.2.8. If U is a connected open subset of C, then each meromorphic
function on U has the form f /g, where f and g are analytic on U and g is not
identically zero.
Proof. Let h be a meromorphic function on U and let {zk } be a sequence con-
sisting of the poles of h, with each zk listed as many times as the order of the
pole at zk . By Theorem 8.2.7 there is an analytic function g on U with {zk }
as a list of its zeroes counting multiplicity. Then, after removing removable
singularities, f = gh is an analytic function on U . Thus, h = f /g with f and g
analytic on U .

The Mittag-Leffler Theorem

The Weierstrass Theorem (Theorem 8.2.7) gives the existence of an analytic
function with a specified list of zeroes counting multiplicity. The Mittag-Leffler
Theorem is a companion theorem. It gives the existence of a meromorphic
function with a specified list of poles and principal parts. We prove it only for
discs, although it is true for general open sets.
Theorem 8.2.9. Let R be a positive number or ∞. Let S be a discrete set of
points of DR (0) and {hw : w ∈ S} a set of polynomials with no constant terms.
Then there exists a meromorphic function f with a pole at w with principal part
hk ((z − w)−1 ) for each w ∈ S and with no other poles.
8.2. WEIERSTRASS PRODUCTS 267

Proof. We choose an increasing sequence of radii {rn } with rn → R and we let

S1 be the subset of S which lies in Dr1 (0) and, for n > 1, and let

Sn = {w ∈ S : rn−1 < |w| ≤ rn .

Then, for each n, X

gn (z) = hk ((z − w)−1 )
w∈Sn

is a meromorphic function on the plane with a pole at w with the required

principal part for each w ∈ Sn and with no other poles.
We might hope to construct the function we are after by simply taking the
infinite sum of the functions gn . Unfortunately, there is no reason to think this
sequence should converge on DR (0). However, we can modify each gn , without
changing its poles and pricipal parts, in such a way as to end up with an infinite
series which does converge.
For each n > 1,the function gn is analytic on an open set containing the
closed disc Drn−1 (0). Hence, it is the uniform limit on this closed disc of its
power series at 0. It follows that there is a polynomial pn such that

|gn (z) − pn (z)| < 2−n for |z| ≤ rn−1 .

If we set f1 = g1 and fn = gn − pn for n > 1, then, for each m > 1, the series
∞
X
fn (z)
n=m+1

converges uniformly to an analytic function on Drm (0). This means that

∞
X
f (z) = fn (z)
n=1

is defined as a meromorphic function on Drm (0) and has the required poles and
principal parts at those points of S which lie in this disc. Since this is true for
each m, and lim rm = R, f is meromorphic on all of DR (0) and has the required
poles and principal parts.

Exercise Set 8.2

2
/2+···+z p /p
1. Show that the derivative of Ep (z) is z p ez+z .

2. Compute the logarithmic derivative of Ep (z).

3. Find an entire√function (given by a Weierstrass product) that has a zero

of order 1 at n for n = 1, 2, 3, · · · and no other zeroes.

4. Find an entire√function (given by a Weierstrass product) that has a zero

of order 2 at n for n = 1, 2, 3, · · · and has no other zeroes.
268 CHAPTER 8. INFINITE PRODUCTS

5. Find an entire function (given by a Weierstrass product) that has a zero

of order n at n2 for n = 1, 2, 3, · · · and has no other zeroes.
6. If f is an entire function, show that f = g n for some entire function g if
and only if the order of each zero of f is divisible by n.
7. Suppose f is an entire function such that {zk } is a list of its non-zero
zeroes counting multiplicity and suppose that
∞
X 1
< ∞.
|zk |
k=1

Describe the simplest Weierstrass factorization of f .

8. Suppose f is an odd entire function, the order of the zero at 0 is m, and
{zk } is a list of the other zeroes of f counting multiplicity. Show that m
is positive and odd. If
∞
X 1
< ∞,
|zk |2
k=1
prove that f has a factorization of the form
∞ 
z2
Y 
m h(z)
f (z) = z e 1− 2
zk
k=1

where h is an entire function.

9. Show that, if U is any non-empty open subset of the plane, then there is
an analytic function on U which cannot be extended to be analytic on any
larger open set. Hint: Use the general Weierstrass Theorem to construct
an analytic function on U with a lot of zeroes.
10. Prove that, given a sequence {zk } of complex numbers converging to in-
finity and a sequence {nk } of integers, there is an entire function f with
given values for f and its derivatives up to order nk at zk for each k. Hint:
Use the Mittag-Leffler and Weierstrass Theorems together.
11. Prove that if f1 and f2 are two entire functions with no common zeroes,
then there exist entire functions g1 and g2 such that

g1 f1 + g2 f2 = 1.

Hint: Use the Mittag-Leffler Theorem to show that an entire function g2

can be chosen so that, at each zero of f1 , the function 1 − g2 f2 has a zero
of order at least as large.
12. Let f1 , f2 , · · · fn be entire functions. Show that there are entire functions
h1 , h2 , ·, hn , and u such that fj = uhj for j = 1, · · · , n and the func-
tions h1 , h2 , · · · , hn have no common zeroes. Hint: Use the Weierstrass
Theorem.
8.3. ENTIRE FUNCTIONS OF FINITE ORDER 269

13. Let f1 , f2 , · · · fn be entire functions with no common zero. Use induction

and the preceding two exercises to show that there are entire functions
g1 , g2 , · · · , gn such that

g1 f1 + g2 f2 + · · · + gn fn = 1.

14. Those who are familiar with commutative ring theory may want to do this
exercise. Let E be the ring of entire functions. Show that the following ring
theoretic properties of E are consequences of the preceding two exercises:
(a) every finitely generated ideal of E is a principle ideal;
(b) every finitely generated maximal ideal of E is of the form

Mw = {f ∈ E : f (w) = 0} for some w ∈ C.

15. The conclusions of the last four exercises actually hold for the ring of
analytic functions on any open subset of the plane. However, to prove
them all in this generality would require a stronger form of the Mittag-
Leffler Theorem than the one proved here. Prove these results for the
largest class of open sets that you can using the machinery developed in
this text.

8.3 Entire Functions of Finite Order

Definition 8.3.1. An entire function f is said to be of finite order if there is
a number t such that t
|f (z)| ≤ e|z|
for all z with |z| sufficiently large. The infimum of all such numbers t is called
the order of f .
p
For each non-negative integer p, the function ez is an entire function of
finite order p. More generally:
Example 8.3.2. Show that eh(z) is an entire function of finite order p if h is a
polynomial of degree p.
Solution: If t > p, then limz→∞ |z|−t |h(z)| = 0. This implies that there is
an R > 0 such that
|h(z)| < |z|t for |z| > R.
Then t
|eh(z) | ≤ e|z| for |z| > R. (8.3.1)
h(z)
Since such a statement is true for all t > p, by definition e has finite order
at most p.
On the other hand, if t < p, then limz→∞ |z|−t |h(z)| = +∞. Hence, there is
no R for which (8.3.1) holds. We conclude that the order of f is at least p and,
hence, is equal to p.
270 CHAPTER 8. INFINITE PRODUCTS

Non-vanishing Entire Functions of Finite Order

It turns out that the functions eh(z) of the preceding example are the only entire
functions of finite order which are non-vanishing.
To prove this, we will need the following theorem of Borel-Carathéodory
relating the growth of the real part of an analytic function to the growth of the
the absolute value of the function.
Theorem 8.3.3. Suppose 0 < r < R and let g be a function analytic on an
open set containing DR (0). Then
2r R+r
|g(z)| ≤ sup{Re (g(w)) : |w| = R} + |g(0)| if |z| ≤ r.
R−r R−r
Proof. We suppose first that g(0) = 0. We set
m = sup{Re (g(w)) : |w| = R}.
Note that the Mean Value Theorem for harmonic functions implies that m ≥ 0.
If |w| = R and u = Re (g(w)), then u ≤ m and
u − 2m ≤ u ≤ 2m − u.
Thus, |u| ≤ |2m − u|, from which it follows that
|g(w)| ≤ |2m − g(w)|,
since the numbers g(w) and 2m − g(w) have the same imaginary parts and have
real parts u and 2m − u, respectively.
We conclude from the above, that the function
g(z)
h(z) =
z(2m − g(z))
satisfies the inequality
1
|h(w)| ≤ for |w| = R.
R
Since the analytic function h has a removable singularity at 0, this inequality
holds throughout the disc DR (0) by the Maximum Modulus Theorem. Thus,
|g(z)| 1
≤ whenever |z| = r,
r|2m − g(z)| R
which implies
r
|g(z)| ≤ (2m + |g(z)|).
R
If we collect terms involving |g(z)| on the left and divide by 1 − r/R, the result
is
2r
|g(z)| ≤ m for |z| ≤ r.
R−r
This concludes the proof in the case where g(0) = 0. This general case follows
from applying this result to the function g0 (z) = g(z) − g(0). The details are
left to the exercises.
8.3. ENTIRE FUNCTIONS OF FINITE ORDER 271

Theorem 8.3.4. An entire function f with no zeroes has finite order p if and
only if p is a non-negative integer and f has the form
f (z) = eh(z) ,
where h is a polynomial of degree p.
Proof. In view of Example 8.3.2, we need only show that every non-vanishing
entire function f of finite order p has the above form.
Since f has no zeroes and the plane is simply connected, there is an entire
function h such that
f (z) = eh(z) for all z ∈ C.
Since f has finite order p, for each t > p there is an M > 0 such that
t
eRe(h(z)) = |f (z)| ≤ e|z| for |z| ≥ M.
This implies
Re (h(z)) ≤ |z|t for |z| ≥ M.
We apply the previous theorem with r > M and R = 2r to conclude
|h(z)| ≤ 2|z|t + 3|h(0)| if |z| = r.
Since this is true for all r > M , Exercise 3.3.9 implies that h must be a poly-
nomial of degree at most t. Since t was an arbitrary number greater than p, we
conclude that h is a polynomial of degree at most p. If it were a polynomial of
degree less than p, then f would have order less than p. Hence, the degree of
the polynomial h is exactly p. This, of course, implies that p is a non-negative
integer.

Canonical Products
Given a sequence {zk }, we let µ be the inf of the numbers t such that
∞
X 1
< ∞. (8.3.2)
|zk |t
k=1

If there is no such t, then we set µ = ∞. The number µ is called the exponent

of convergence for the sequence {zk }.
If {zk } has finite exponent of convergence µ, then we can write down a
convergent Weierstrass product (8.2.3), using {zk }, in which the sequence {pk }
is a constant p. We choose p to be the smallest integer such that µ < p + 1.
Then the condition
∞
X 1
< ∞, (8.3.3)
|zk |p+1
k=1
is satisfied. Hence, by Theorem 8.2.2, the Weierstrass product
∞
Y
f (z) = Ep (z/zk ) (8.3.4)
k=1
272 CHAPTER 8. INFINITE PRODUCTS

converges. This is called the canonical product for the sequence {zk }.
The significance of the choice of p made for the canonical product is that,
with this choice, the resulting product is an entire function with order λ equal
to the exponent of convergence µ of the sequence {zk }. The next theorem yields
part of what is needed to prove this. The remainder of the proof will come in
the next section.

Theorem 8.3.5. The canonical product for a sequence {zk }, with finite expo-
nent of convergence µ, is an entire function of finite order λ ≤ µ.

Proof. We choose p to be the smallest integer such that µ < p + 1, and let t be
any number in the range µ < t < p + 1.
We claim that there is a positive constant A such that
t
|Ep (z)| ≤ eA|z| (8.3.5)

for all z.
If |z| ≤ 1/2, this follows from (8.1.6) with w = Ep (z) − 1 and Theorem 8.2.1.
These combine to show that

| log Ep (z)| ≤ 2|z|p+1 ≤ 2|z|t ,

and this implies (8.3.5) holds with A = 2.

If |z| > 1/2, then |z|k ≤ 2t−k |z|t , and so
p
Re z k


X
log |Ep (z)| = log |1 − z| +
k
k=1
p
X
≤ |z| + |z|k ≤ (p + 1)2t |z|t .
k=1

Thus, (8.3.5) holds with A = (p + 1)2t in this case.

To prove the theorem, we note that, if f is given by the canonical product
(8.3.4), then by (8.3.5),
∞
Y t t
|f (z)| ≤ eA|z/zk | = eB|z| ,
k=1

where
∞
X
B=A 1/|zk |t .
k=1

The series in this expression converges because t is larger than the exponent of
convergence µ.
Since for any s > t, we have B|z|t ≤ |z|s for |z| sufficiently large, it follows
that f has finite order at most t. Since t was an arbitrary number strictly
between µ and p + 1, we conclude that f has order at most µ.
8.3. ENTIRE FUNCTIONS OF FINITE ORDER 273

One might guess, based on Theorem 8.3.4 that the order of an entire function
of finite order must be a non-negative integer. This is not the case, as is shown
by the following example.
Example 8.3.6. Find an entire function with finite order 1/2.
Solution: The function
∞ 
z2

sin πz Y
= 1− 2
πz k
k=1

has order 1 (Exercise 8.3.3). It seems reasonable that if we replace z 2 by z in

this product that the result would be an entire function of order 1/2. In fact,
the resulting function has a zero of order 1 at k 2 for each positive integer k and
∞
X 1
<∞
(k 2 )t
k=1

for every t > 1/2 and for no smaller values of t. Hence, the sequence {1/k 2}
has exponent of convergence 1/2. Since 0 is the smallest integer p such that
1/2 < p + 1, the preceding theorem implies that the canonical product
∞ 
Y z 
f (z) = 1−
k2
k=1

is an entire function of finite order at most 1/2.

In fact, it is easy to directly compute the order of f if we note that
√
sin π z
f (z) = √ .
π z
This expression on the right is entire and is independent of the choice of the
square root function because the function (πz)−1 sin πz is an even function. It
is easy to see from this that, since (πz)−1 sin πz has order 1, f has order 1/2
(Exercise 8.3.5).

Exercise Set 8.3

1. Finish the proof of Theorem 8.3.3 by showing that, if it is true in the case
where g(0) = 0, then it is true in general.
2. Show that a polynomial has finite order 0.
3. Show that sin z, z −1 sin z, and cos z all have finite order 1.
4. If f is an entire function of order λ(f ), k is a non-negative integer, and
g(z) = f (z k ), then prove that λ(g) = kλ(f ), where λ(g) is the order of g.
5. Prove
√ that if g(z) is an even entire function of finite order λ and f (z) =
g( z), then
√ f is an entire function of finite order λ/2. In particular, show
that cos z has order 1/2.
274 CHAPTER 8. INFINITE PRODUCTS

6. Prove that the order of the sum or product of two entire functions is less
than or equal to the maximum of the orders of the two functions.
7. What is the order of the entire function esin z ?
8. Suppose f is an entire function which satisfies the inequality |f (z)| ≤ |z||z|
for |z| sufficiently large. Prove that f has finite order at most 1.
9. Find the exponent of convergence of the following sequences: {2k }, {k r }
(r > 0), {log k}.
10. Given an arbitrary non-negative real number µ, show that there is a se-
quence of complex numbers {zk } with exponent of convergence µ.
11. Does the order of an entire function necessarily have to be the same as the
exponent of convergence of its sequence of zeroes? Justify your answer.

8.4 Hadamard’s Factorization Theorem

Our goal in this section is to complete the characterization of entire functions of
finite order λ. We will prove a theorem of Hadamard which asserts that every
such function factors as a power of z times a canonical product of order at most
λ times the exponential of a polynomial of degree at most λ. The key ingredient
in the proof is Jensen’s Formula relating the density of the zeroes of an entire
function to the rate of growth at infinity of the function.

Jensen’s Formula
Theorem 8.4.1. If f is analytic in an open set containing the disc Dr (0), f
has no zeroes on the boundary of this disc, f (0) 6= 0, and z1 , z2 , · · · zn are the
zeroes, counting multiplicity, of f in Dr (0), then
Z 2π
|f (0)|rn
 
1
log = log(|f (rei θ)|) dθ.
|z1 | · |z2 | · · · · |zn | 2π 0
Proof. We first prove this in the case where r = 1. We divide f by a product of
linear fractional transformations which preserve the unit circle and have zeroes
at the points zi . This yields a function
1 − z1z 1 − z2 z 1 − znz
g(z) = f (z) ··· .
z − z1 z − z2 z − zn
This function is analytic and non-vanishing in an open set containing the closed
unit disc D, and has the same modulus on the unit circle as does f . Thus, g
has an analytic logarithm in an open set containing D. Then log |g(x)| is the
real part of an analytic function in this set and, hence, is harmonic. The Mean
Value Theorem for harmonic functions implies that
  Z 2π
|f (0)| 1
log = log |g(0)| = log |f (eiθ )| dθ. (8.4.1)
|z1 | · |z2 | · · · · |zn | 2π 0
8.4. HADAMARD’S FACTORIZATION THEOREM 275

To prove the theorem for general r, it suffices to apply (8.4.1) with f replaced
by the function f (rz). If f has zeroes at z1 , z2 , · · · , zn in the disc Dr (0), then
f (rz) has zeroes z1 /r, z2 /r, · · · , zn /r in the unit disc D. Thus, the equation of
the theorem follows directly from (8.4.1) applied to f (rz).
This leads to the following estimate on the number of zeroes of an entire
function inside a disc Dr (0).
Theorem 8.4.2. If f is an entire function with |f (0)| = 1, n(r) is the number
of zeroes of f inside a disc Dr (0), and M (2r) is the supremum of |f (z)| on the
boundary of D2r (0), then
log M (2r)
n(r) ≤ .
log 2
Proof. Let n = n(r) and m = n(2r), and let z1 , z2 , · · · zm be the zeroes of f
inside the disc D2r (0) ordered so that |zj | ≤ |zk | for j ≤ k. Then Jensen’s
Theorem with r replaced by 2r implies that
 
 2r 2r 2r 2r 
log f (0)
 ··· ··· ≤ log M (2r).
z1 z2 zn zm 
Since
2r 2r
2< if j≤n and 1 < if j > n,
|zj | |zj |
this implies that
log |f (0)2n | ≤ log M (2r),
or
log |f (0)| + n log 2 ≤ log M (2r).
The theorem follows from this, since log |f (0)| = 0.

Zeroes of Functions of Finite Order

The preceding theorem has the following consequence for entire functions of
finite order.
Theorem 8.4.3. Let f be an entire function of finite order λ and with f (0) 6= 0.
Let {zk } be a list of the zeroes of f , counted according to multiplicity and indexed
in order of increasing modulus, and let µ be the exponent of convergence of {zk },
then µ ≤ λ.
Proof. We claim that, for each t > λ, there are constants N , C > 0 and q > 1
such that
|zk |t ≥ Ck q for all k ≥ N. (8.4.2)
Assuming this, we conclude that the series
∞
X 1
|zk |t
k=1
276 CHAPTER 8. INFINITE PRODUCTS

converges for all t > λ, by comparison with the series

∞
X 1
,
kq
k=1

which converges for q > 1. This, in turn, implies the exponent of convergence
µ is at most λ.
To complete the proof, we must verify the claim concerning (8.4.2). In
doing this, we may as well assume that |f (0)| = 1, since, if this is not so, we
may make it so by replacing f by f divided by a constant times a power of z.
Such a replacement will have no effect on whether the above claim is true.
Let rk = |zk |. Since the zeroes are indexed in such a way that the modulus
is a non-decreasing function of k, there are at least k zeroes of f with modulus
less than or equal to rk . By Theorem 8.4.2,

log M (2rk )
k≤ ,
log 2

where M (2rk ) is the sup of |f (z)| on the circle |z| = 2rk .

We choose s with λ < s < t. Since f has order λ, there is an R such that
rk ≥ R implies
s
M (2rk ) ≤ e(2rk ) .
Hence, for rk ≥ R,
(2rk )s
k≤ .
log 2
This implies
(log 2)t/s t/s
rkt ≥ k = Ck q ,
2t
where
(log 2t/s ) t
C= and q = > 1.
2t s
This is true provided rk = |zk | > R. However, since lim zk = ∞, there is an N
such that k > N implies |zk | > R. This completes the proof

The above theorem, when combined with Theorem 8.3.5, yields the following
corollary.

Corollary 8.4.4. The canonical product for a sequence {zk } with exponent of
convergence µ has finite order λ = µ.

Hadamard’s Theorem
In the proof of the next theorem, we will need the following estimates on the
size of the inverse Ep−1 (z) of the function Ep (z).
8.4. HADAMARD’S FACTORIZATION THEOREM 277

Lemma 8.4.5. If p is a non-negative integer, p ≤ t ≤ p + 1, and z ∈ C, then

there is a constant A such that
1 t
≤ eA|z| (8.4.3)
|Ep (z)|

if |z| ≥ 2 or |z| ≤ 1/2.

Proof. If |z| ≥ 2, then |1 − z| ≥ 1 and |z k /k| ≤ |z|t for k ≤ p. Hence,

2
/2−···−z p /p t
|Ep−1 (z)| = |1 − z|−1 |e−z−z | ≤ ep|z|

and so (8.4.3) holds with A = p in this case.

On the other hand, if |z| ≤ 1/2, then
p
X ∞
X
log Ep (z) = log(1 − z) + z k /k = − z k /k,
k=1 k=p+1

and so
∞
X
| log Ep (z)| ≤ |z|p+1 |z|k ≤ 2|z|p+1 ≤ 2|z|t
j=0

Thus, (8.4.3) holds with A = 2 in this case. If we choose A = max{2, p}, then
(8.4.3) holds in both cases.

We are now in a position to prove Hadamard’s Theorem characterizing entire

functions of finite order. This will be used in the proof of the Prime Number
Theorem in the next chapter.

Theorem 8.4.6. If f is an entire function of order λ, and p is the smallest

integer such that p + 1 > λ, then f factors as
∞
Y
f (z) = z m eh(z) Ep (z/zk ), (8.4.4)
k=1

where m is the order of the zero of f at 0, {zk } is a list of the other zeroes of f
counting multiplicity, and h(z) is a polynomial of degree at most p.

Proof. According to the Weierstrass Factorization Theorem (Theorem 8.2.5) f

has a factorization of the form (8.4.4), where h is an entire function. Thus, the
only thing to be proved is that h is a polynomial of degree at most p. This will
follow from Theorem 8.3.4 if we can show that the function
f (z)
g(z) = eh(z) = Q∞
zm j=1 Ep (z/zk )

has finite order at most λ.

278 CHAPTER 8. INFINITE PRODUCTS

Let t be any number with λ < t ≤ p + 1 and let r ≥ 1 be any radius which
is not one of the numbers |zk |. We factor g(z) as g(z) = g1 (z)g2 (z), where
Y
g1 (z) = f (z)z −m Ep−1 (z/zk ), (8.4.5)
|zk |≤2r

and Y
g2 (z) = Ep−1 (z/zk ). (8.4.6)
|zk |>2r

Suppose |z| = 4r = R. Then |z/zk | ≥ 2 for all k with |zk | ≤ 2r. By the
previous lemma, there is a positive constant A1 such that
Y t
|g1 (z)| ≤ |f (z)| eA1 |z/zk | .
|zk |≤2r

Since f has finite order λ, for sufficiently large r we have

t
|f (z)| ≤ e|z|

and, hence,
t
|g1 (z)| ≤ eB1 r (8.4.7)
where
∞
!
t
X 1
B1 = 4 1 + A1 .
|zk |t
k=1

The infinite series in this expression converges by Theorem 8.4.3. Since g1 (z) is
an entire function (once the removable singularities at the zk with |zk | < 2r are
removed), if the inequality (8.4.7) holds for |z| = 4r = R it must hold for all
z in the disc |z| ≤ R, by the maximum modulus principle. In particular, this
inequality holds for all z with |z| = r.
Also if |z| = r, then |z/zk | < 1/2 if |zk | > 2r, and the previous lemma
implies that there is a constant A2 such that
Y t t
|g2 (z)| ≤ eA2 |z/zk | ≤ eB2 r , (8.4.8)
|zk |>2r

where
∞
X 1
B2 = A2 .
|zk |
k=1

If we set B = B1 + B2 and combine (8.4.7) and (8.4.8), we obtain

t
|g(z)| ≤ eB|z| .

Since t is an arbitrary number larger than λ and less than or equal to p + 1, g

has order at most λ. This completes the proof.
8.4. HADAMARD’S FACTORIZATION THEOREM 279

Exercise Set 8.4

1. What does Hadamard’s Factorization Theorem say about an entire func-
tion of order λ < 1?
2. If√a non-constant entire function of finite order λ has zeroes at the points
i n what are the possible values for λ?
3. Show that an even entire function of order 1 has the form
Y z2

Cz m 1− 2 ,
zk
k

where m is even, C is a non-zero constant, and the sequence

{z1 , −z1 , z2 , −z2 , · · · , zk , −zk , · · · }
is a list of the zeroes of f counting multiplicity.
4. What is the exponent of convergence for the sequence of zeroes in the
preceding exercise.
5. State and prove the analogues of the previous two exercises for odd entire
functions of order 1.
6. Prove that if f is an entire function of order λ and λ is not an integer,
then f has infinitely many zeroes.
7. Under the hypotheses of the preceding exercise, prove that f takes on
every complex value infinitely many times.
8. Prove that if f and g are entire functions of finite order λ and if f (zk ) =
g(zk ) on a sequence which satisfies
X 1
=∞
|zk |t
k=1

for some t > λ, then f (z) = g(z) identically.

9. Use the previous exercise to prove that if two functions of finite order agree
at the points of the sequence {log n}∞ n=1 }, then they agree identically.
Thus, ez is the only entire function of finite order which has the value n
at the point log n for n = 1, 2, · · · .
10. Find an entire function which has zeroes at the points of the sequence
{log n}∞
n=1 . Does it have finite order?

11. Suppose f is an entire function of finite order λ and µ is the exponent of

convergence of the list of zeroes of f . Prove that if µ < λ, then λ is an
integer.
12. Is there an entire function of order 3/2 which has the integers as its list
of zeroes, counting multiplicity?