In principle, this results in an infinite product

4. Infinite products

The basic idea here is to separate the zeros (and poles) of a meromorphic function f(z) as a product component off(z). In principle, this results in an infinite product.

To this end, we first prove

Theorem 4.1. If f(z)is an entire function with no zeros, then there exists another entire function g(z) such that

f(z) =e^g(z).

Proof. Since f(z)6= 0 for all z ∈C, then ^f_f(z)^0(z) is entire. Therefore, f⁰(z)

f(z) = X∞ j=0

ajz^j =a0+a1z+a2z²+· · · .

is a power series representation converging in the wholeC. Consider

h(z) =a₀z+ ¹₂a₁z²+ ¹₃a₃z³+· · ·=z(a₀+ ¹₂a₁z+ ¹₃a₃z²+· · ·). (4.1) Since

lim sup

j→∞

j

r 1

j+ 1|aj|= lim sup

j→∞

1

√j

j+ 1

j

q

|aj|= lim sup

j→∞

j

q

|aj|= 0,

hence the power series (4.1) has radius of convergence = ∞. Therefore, (4.1) determines an entire function. Differentiating term by term, as we may do for a converging power series, we get

h⁰(z) = f⁰(z) f(z). Define now

ϕ(z) :=f(z)e^−h(z), hence

ϕ⁰(z) =f⁰(z)e^−h(z)−f(z)h⁰(z)e^−h(z)=e^−h(z) f⁰(z)−f(z)h⁰(z)

≡0.

Therefore, ϕ(z) is constant, say ϕ(z)≡e^a, a∈C. Note thatϕ(z)6= 0 for allz ∈C. So,

f(z)e^−h(z) =e^a =⇒ f(z) =e^a+h(z). Definingg(z) :=a+h(z), we have the assertion.

Definition 4.2. The infinite product Q_∞

j=1b_j of complex numbers b_j converges, if there exists

nlim→∞

Yn

j=1

bj 6= 0.

15

Remark. Define Pn := Qn

j=1bi. Clearly, Q_∞

j=1bi converges if and only if (Pn) converges and limn→∞Pn 6= 0. Thenbn =Pn/Pn−1 and there exists

nlim→∞bn = limn→∞Pn

lim_n→∞P_n−1 = 1. (4.2)

Therefore, it is customary to use the notation bn = 1 +an; then lim_n→∞a_n= 0 by (4.2).

Theorem 4.3. If a_j ≥0 for all j ∈N, then Q_∞

j=1(1 +a_j) converges if and only if P_∞

j=1a_j converges.

Proof. Observe first that P_n := Qn

j=1(1 +a_j) is a non-decreasing sequence, since a_j ≥0. Therefore, (P_n) either converges to a finite (real) value, or to +∞. Clearly,

a₁+a₂+· · ·+a_n ≤(1 +a₁)(1 +a₂)· · ·(1 +a_n).

On the other hand,

(1 +a1)· · ·(1 +an)≤e^a1· · ·e^an =e^a1+···+an, since e^x ≥1 +x for every x≥0. So, we have

Xn

j=1

aj ≤ Yn

j=1

(1 +aj)≤e

P_n

j=1aj. (4.3)

If (Pn

j=1a_j)_n∈N converges, then (e^Pnj=1aj)_n∈N converges by the continuity of the exponential function. This implies that the increasing sequence Qn

j=1(1 +aj)

n∈N

converges to a non-zero limit by (4.3). If Qn

j=1(1 +a_j)

n∈N converges, then the increasing sequence (Pn

j=1a_j)_n∈N converges, again by (4.3).

Theorem 4.4. If aj ≥ 0, aj 6= 1, for all j ∈ N, then Q_∞

j=1(1−aj) converges if and only if P_∞

j=1aj converges.

Proof. (1) Assume P_∞

j=1aj converges. By the Cauchy criterium, X∞

j=N

aj < ¹₂

for N sufficiently large; then alsoaj <1, j ≥N. Observe that (1−a_N)(1−a_N+1) = 1−a_N −a_N+1+a_Na_N+1

≥1−aN −aN+1 = 1−(aN +aN+1)> ¹₂ . Assume we have proved

(1−a_N)(1−a_N+1)· · ·(1−a_n)≥1−a_N −a_N+1− · · · −a_n. (4.4)

Then

(1−aN)(1−aN+1)· · ·(1−an)(1−an+1)

≥(1−aN −aN+1− · · · −an)(1−an+1)

= 1−a_N −a_N+1− · · · −a_n−a_n+1+ (a_N +· · ·+a_n)a_n+1

≥1−a_N −a_N+1− · · · −a_n+1, and so (4.4) is true for all n≥N. Therefore

(1−aN)(1−aN+1)· · ·(1−an)≥1−(aN +· · ·+an)> ¹₂. This implies that the decreasing sequenceQ_∞

j=N(1−aj) converges to a limitP ≥ ¹2. If N is sufficiently large, then 0<1−aj <1 and so P ≤1. Writing, for n > N,

P_n= Yn

j=1

(1−a_j) =P_N−1· Yn

j=N

(1−a_j),

we get

nlim→∞Pn=PN−1· lim

n→∞

Yn

j=N

(1−aj) =PN−1·P = (1−a1)· · ·(1−aN−1)P 6= 0,

so Q_∞

j=1(1−a_j) converges.

(2) Assume now that P_∞

j=1a_j diverges. If a_j does not converge to zero, then 1−a_j does not converge to one. By the Remark after Definition 4.2,Q_∞

j=1(1−a_j) diverges.

So, we may assume that limj→∞aj = 0. Let N be sufficiently large so that 0≤a_j <1 for j ≥N. Since 1−x≤e^−x for 0 ≤x <1, we have

1−a_j ≤e^−aj, j ≥N.

Therefore,

0≤ Yn

j=N

(1−aj)≤ Yn

j=N

e^−aj =e⁻

P_n

j=Naj, n > N.

Since P_∞

j=Naj diverges, limn→∞Pn

j=N aj = +∞, and so limn→∞e⁻

P_n

j=Naj = 0, implying that

nlim→∞

Yn

j=1

(1−aj) = 0.

By Definition 4.2,Q_∞

j=1(1−a_j) diverges.

Definition 4.5. The infinite product Q_∞

j=1(1 + aj) is absolutely convergent, if Q_∞

j=1(1 +|aj|) converges.

Remark. By Theorem 4.3, this is the case if and only ifP_∞

j=1|a_j| converges.

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Theorem 4.6. An absolutely convergent infinite product is convergent.

Proof. Denote

P_n = Yn

j=1

(1 +a_j) and Q_n :=

Yn

j=1

(1 +|a_j|).

Then

Pn−Pn−1 = Yn

j=1

(1 +aj)−

nY−1

j=1

(1 +aj)

=ⁿY⁻¹

j=1

(1 +aj)

(1 +an−1) =an nY−1

j=1

(1 +aj) and, similarly,

Q_n−Q_n−1 =|a_n|

nY−1

j=1

(1 +|a_j|).

Clearly,

|P_n−P_n−1| ≤Q_n−Q_n−1. Since Q_∞

j=1(1 +|aj|) converges, limn→∞Qn exists. Therefore, P_∞

j=1(Qn−Qn−1) converges, and so by the standard majorant principle,P_∞

j=1(Pn−Pn−1) converges, implying that limn→∞Pn exists.

It remains to show that this limit is non-zero. Since P_∞

j=1|a_j| converges, lim_n→∞a_n= 0, and so lim_n→∞(1 +a_n) = 1. Therefore, P_∞

j=1|1+a^ajj| converges by the majorant principle, since |1 +aj| ≥ ¹2 for j large enough and so |1+a^ajj| ≤2|aj|. Therefore

Y∞ j=1

1− a_j 1 +aj

is absolutely convergent. By the preceding part of the proof, a finite limit

nlim→∞

Yn

j=1

1− aj

1 +a_j

exists. But Yn

j=1

1− a_j 1 +aj

= Yn

j=1

1 1 +aj

= 1

Qn

j=1(1 +aj) = 1 Pn

, and so limn→∞Pn6= 0.

Consider finally a sequence fj(z)

j∈N of analytic functions in a domain G⊂C. Similarly as to Definition 4.2, we say that

Y∞ j=1

1 +f_j(z) converges in G, if

nlim→∞

Yn

j=1

1 +f_j(z)

6

= 0 exists for each z ∈G.

Theorem 4.7. The infinite product Q_∞

j=1 1+fj(z)

is uniformly convergent inG, if the series P_∞

j=1|fj(z)| converges uniformly in G.

Proof. Assume

X∞ j=1

|fj(z)|< M(<∞) for all z ∈G. Then by (4.3),

1 +|f₁(z)|

· · · 1 +|f_n(z)|

≤e^{|f1(z)|+···+|fn(z)|} ≤e^M. Denote

Pn(z) :=

Yn

j=1

1 +|fj(z)| .

Then

Pn(z)−Pn−1(z) =|fn(z)| 1 +|f1(z)|

· · · 1 +|fn(z)|

≤e^M|fn(z)|. Since

X∞ j=2

P_n(z)−P_n−1(z)

≤e^M X∞ j=2

|f_j(z)| ≤e^M X∞ j=1

|f_j(z)|, P_∞

j=2 P_n(z)−P_n−1(z)

converges uniformly, and so (P_n) as well. This means that Q_∞

j=1 1 +f_j(z)

is absolutely (uniformly) convergent, hence (uniformly) convergent by Theorem 4.6.

Exercises.

(1) Show that Y∞ n=1

1− 2

(n+ 1)(n+ 2)

= ¹₃.

(2) Show that Y∞ n=3

n²−4 n²−1 = ¹₄. (3) Show that

Y∞ n=2

n³−1

n³+ 1 converges.

(4) Determine whether or not Y∞ n=0

(1−2⁻ⁿ) is convergent.

(5) Prove that Y∞ k=0

1 + z^k

k!

defines an entire function.

(6) Prove that Y∞ k=0

(1 +z^2k) = 1

1−z for all z in the unit disc |z|<1.

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