Example: the product formula for the sine function

Before proceeding with the general theory of Weierstrass products, we consider the key example of the product formula for the sine function:

(3) sin πz

This identity will in turn be derived from the sum formula for the cotan-gent function (cot πz = cos πz/ sin πz): The first formula holds for all complex numbers z, and the second when-ever z is not an integer. The sum_∞

n=−∞1/(z + n) needs to be properly understood, because the separate halves corresponding to positive and negative values of n do not converge. Only when interpreted symmetri-cally, as lim_N→∞

|n|≤N1/(z + n), does the cancellation of terms lead to a convergent series as in (4) above.

3. Infinite products 143 We prove (4) by showing that both π cot πz and the series have the same structural properties. In fact, observe that if F (z) = π cot πz, then F has the following three properties:

(i) F (z + 1) = F (z) whenever z is not an integer.

(ii) F (z) = 1

z + F₀(z), where F₀ is analytic near 0.

(iii) F (z) has simple poles at the integers, and no other singularities.

Then, we note that the function

∞

also satisfies these same three properties. In fact, property (i) is simply the observation that the passage from z to z + 1 merely shifts the terms in the infinite sum. To be precise,



Letting N tend to infinity proves the assertion. Properties (ii) and (iii) are evident from the representation _z¹+_∞

n=1 2z

z²−n² of the sum.

Therefore, the function defined by

∆(z) = F (z)−

∞ n=−∞

1 z + n

is periodic in the sense that ∆(z + 1) = ∆(z), and by (ii) the singularity of ∆ at the origin is removable, and hence by periodicity the singularities at all the integers are also removable; this implies that ∆ is entire.

To prove our formula, it will suffice to show that the function ∆ is bounded in the complex plane. By the periodicity above, it is enough to do so in the strip |Re(z)| ≤ 1/2. This is because every z^∈ C is of the form z^= z + k, where z is in the strip and k is an integer. Since ∆ is holomorphic, it is bounded in the rectangle|Im(z)| ≤ 1, and we need only control the behavior of that function for |Im(z)| > 1. If Im(z) > 1 and z = x + iy, then

cot πz = i e^iπz+ e^−iπz

e^iπz− e^−iπz = i e^−2πy+ e^−2πix e^−2πy− e^−2πix,

144 Chapter 5. ENTIRE FUNCTIONS

and in absolute value this quantity is bounded. Also 1 therefore if y > 1, we have

 Now the sum on the right-hand side is majorized by

_∞

0

y y²+ x² dx ,

because the function y/(y²+ x²) is decreasing in x; moreover, as the change of variables x → yx shows, the integral is independent of y, and hence bounded. By a similar argument ∆ is bounded in the strip where Im(z) <−1, hence is bounded throughout the whole strip |Re(z)| ≤ 1/2.

Therefore ∆ is bounded in C, and by Liouville’s theorem, ∆(z) is con-stant. The observation that ∆ is odd shows that this constant must be 0, and concludes the proof of formula (4).

To prove (3), we now let

Proposition 3.2 and the fact that

1/n²<∞ guarantee that the prod-uct P (z) converges, and that away from the integers we have

P^(z)

and so P (z) = cG(z) for some constant c. Dividing this identity by z, and taking the limit as z→ 0, we find c = 1.

Remark. Other proofs of (4) and (3) can be given by integrating analogous identities for π²/(sin πz)² derived in Exercise 12, Chapter 3, and Exercise 7, Chapter 4. Still other proofs using Fourier series can be found in the exercises of Chapters 3 and 5 of Book I.

4. Weierstrass infinite products 145 4 Weierstrass infinite products

We now turn to Weierstrass’s construction of an entire function with prescribed zeros.

Theorem 4.1 Given any sequence {an} of complex numbers with

|an| → ∞ as n → ∞, there exists an entire function f that vanishes at all z = a_n and nowhere else. Any other such entire function is of the form f (z)e^g(z), where g is entire.

Recall that if a holomorphic function f vanishes at z = a, then the multiplicity of the zero a is the integer m so that

f (z) = (z− a)^mg(z),

where g is holomorphic and nowhere vanishing in a neighborhood of a.

Alternatively, m is the first non-zero power of z− a in the power series expansion of f at a. Since, as before, we allow for repetitions in the sequence {an}, the theorem actually guarantees the existence of entire functions with prescribed zeros and with desired multiplicities.

To begin the proof, note first that if f₁and f₂ are two entire functions that vanish at all z = a_n and nowhere else, then f₁/f₂ has removable singularities at all the points a_n. Hence f₁/f₂ is entire and vanishes nowhere, so that there exists an entire function g with f1(z)/f2(z) = e^g(z), as we showed in Section 6 of Chapter 3. Therefore f₁(z) = f₂(z)e^g(z) and the last statement of the theorem is verified.

Hence we are left with the task of constructing a function that vanishes at all the points of the sequence{an} and nowhere else. A naive guess, suggested by the product formula for sin πz, is the product

n(1− z/an).

The problem is that this product converges only for suitable sequences {an}, so we correct this by inserting exponential factors. These factors will make the product converge without adding new zeros.

For each integer k≥ 0 we define canonical factors by

E₀(z) = 1− z and E_k(z) = (1− z)e^{z+z2/2+···+zk/k}, for k≥ 1.

The integer k is called the degree of the canonical factor.

Lemma 4.2 If|z| ≤ 1/2, then |1 − Ek(z)| ≤ c|z|^k+1 for some c > 0.

Proof. If |z| ≤ 1/2, then with the logarithm defined in terms of the power series, we have 1− z = e^log(1−z), and therefore

E_k(z) = elog(1−z)+z+z²/2+···+z^k/k = e^w,

146 Chapter 5. ENTIRE FUNCTIONS

where w =−_∞

n=k+1zⁿ/n. Observe that since|z| ≤ 1/2 we have

|w| ≤ |z|^k+1

∞ n=k+1

|z|^n−k−1/n≤ |z|^k+1

∞ j=0

2^−j ≤ 2|z|^k+1.

In particular, we have|w| ≤ 1 and this implies that

|1 − Ek(z)| = |1 − e^w| ≤ c^|w| ≤ c|z|^k+1.

Remark. An important technical point is that the constant c in the statement of the lemma can be chosen to be independent of k. In fact, an examination of the proof shows that we may take c^= e and then c = 2e.

Suppose that we are given a zero of order m at the origin, and that a₁, a₂. . . are all non-zero. Then we define the Weierstrass product by

f (z) = z^m

∞ n=1

E_n(z/a_n).

We claim that this function has the required properties; that is, f is entire with a zero of order m at the origin, zeros at each point of the sequence{an}, and f vanishes nowhere else.

Fix R > 0, and suppose that z belongs to the disc |z| < R. We shall prove that f has all the desired properties in this disc, and since R is arbitrary, this will prove the theorem.

We can consider two types of factors in the formula defining f , with the choice depending on whether|an| ≤ 2R or |an| > 2R. There are only finitely many terms of the first kind (since|an| → ∞), and we see that the finite product vanishes at all z = a_n with|an| < R. If |an| ≥ 2R, we have|z/an| ≤ 1/2, hence the previous lemma implies

|1 − En(z/an)| ≤ c

_a^z_n

^n+1≤ ₂^{n+1c .}

Note that by the above remark, c does not depend on n. Therefore, the product



|an|≥2R

E_n(z/a_n)

defines a holomorphic function when |z| < R, and does not vanish in that disc by the propositions in Section 3. This shows that the function

5. Hadamard’s factorization theorem 147 f has the desired properties, and the proof of Weierstrass’s theorem is complete.

5 Hadamard’s factorization theorem

The theorem of this section combines the results relating the growth of a function to the number of zeros it possesses, and the above product theorem. Weierstrass’s theorem states that a function that vanishes at the points a₁, a₂, . . . takes the form

e^g(z)z^m

∞ n=1

E_n(z/a_n).

Hadamard refined this result by showing that in the case of functions of finite order, the degree of the canonical factors can be taken to be constant, and g is then a polynomial.

Recall that an entire function has an order of growth≤ ρ if

|f(z)| ≤ Ae^B|z|ρ,

and that the order of growth ρ₀ of f is the infimum of all such ρ’s.

A basic result we proved earlier was that if f has order of growth≤ ρ, then

n(r) ≤ Cr^ρ, for all large r,

and if a₁, a₂, . . . are the non-zero zeros of f , and s > ρ, then

|an|^−s<∞.

Theorem 5.1 Suppose f is entire and has growth order ρ0. Let k be the integer so that k≤ ρ0 < k + 1. If a₁, a₂, . . . denote the (non-zero) zeros of f , then

f (z) = e^{P (z)}z^m

∞ n=1

Ek(z/an),

where P is a polynomial of degree≤ k, and m is the order of the zero of f at z = 0.

Main lemmas

Here we gather a few lemmas needed in the proof of Hadamard’s theorem.

148 Chapter 5. ENTIRE FUNCTIONS