# Analytic Number Theory/Tools from complex analysis

## Infinite products[edit | edit source]

**Lemma 5.1 (Convergence of real products)**:

Let be such that

converges absolutely. Then if ,

converges.

**Proof**:
Without loss of generality, we assume for all .

Denote

- .

Then we have

- .

We now apply the Taylor formula of first degree with Lagrange remainder to at to obtain for

- , .

Hence, we have for

- , .

Hence, and thus we obtain the (even absolute) convergence of the ; thus, by the continuity of the exponential, also the converge.

**Theorem 5.2 (Comparison test for complex products)**:

Assume that is a *non-negative* real sequence such that

converges. Assume further that is a sequence of complex numbers such that . Then also

converges. Furthermore, for all

- .

**Proof**:

We define

- , . We note that
- .

Without loss of generality we may assume that all the products are nonzero; else we have immediate convergence (to zero).

We now prove that is a Cauchy sequence. Indeed, we have

and furthermore

and therefore

- .

Since , it is a Cauchy sequence, and thus, by the above inequality, so is . The last claim of the theorem follows by taking in the above inequality.

**Theorem 5.3 (Sum test for complex products)**:

Let be a real sequence such that

converges absolutely. Then if ,

converges, where is a complex sequence. Furthermore, for all

- .

**Proof 1**:

We prove the theorem using lemma 5.1 and the comparison test.

Indeed, by lemma 5.1 the product

converges. Hence by theorem 5.2, we obtain convergence and the desired inequality.

**Proof 2 (without the inequality)**:

We prove the theorem except the inequality at the end from lemma 5.1 and by using the Taylor formula on .

We define . Then since every complex number satisfies , we need to prove the convergence of the sequences and .

For the first sequence, we note that the convergence of is equivalent to the convergence of . Now for each

**Theorem 5.4 (Holomorphic products)**:

Let be a sequence of holomorphic functions in a domain such that for each we can find a compact and a sequence such that and

converges absolutely. Then

defines a holomorphic function.

**Proof**:

First, we note that is well-defined for each due to theorem 5.2. In order to prove that the product is holomorphic, we use the fact from complex analysis that if a sequence of functions converging locally uniformly to another function has infinitely many holomorphic members, then the limit is holomorphic as well. Indeed, we note by the inequality in theorem 5.3, that we are given uniform convergence. Hence, the theorem follows.

### Exercises[edit | edit source]

## The Weierstraß factorisation[edit | edit source]

The following lemma is of great importance, since we can deduce three important theorems from it:

- The existence of holomorphic functions with prescribed zeroes
- The
*Weierstraß factorisation theorem*(a way to write any holomorphic function made up from linear factors and the exponential) - The
*Mittag-Leffler theorem*(named after Gösta Mittag-Leffler (one guy))

**Lemma 5.5**:

Let be a sequence of complex numbers such that

and

- .

Then the function

has exactly the zeroes in the correct multiplicity.

**Proof**:

Define for each

- .

Our plan is to prove that converges uniformly in every subcircle of the circle of radius for every . Since the function is holomorphic in a unit ball around zero, it is equal to its Taylor series there, i.e.

- .

Hence, for

- .

Let now be given and be arbitrary. Then we have for , arbitrary

- .

Now summing over , we obtain

for all . Hence, we have uniform convergence in that circle; thus the sum of the logarithms is holomorphic, and so is the original product if we plug everything into the exponential function (note that we *do* have even if is an arbitrary *complex* number).

Note that our method of proof was similar to how we proved lemma 5.1. In spite of this, it is not possible to prove the above lemma directly from theorem 5.4 since the corresponding series does not converge if the are chosen increasing too slowly.

**Theorem 5.6 (Holomorphic functions with given zeroes)**:

Let be a sequence of complex numbers which does not have an accumulation point. Then the function

has zeroes with the right multiplicity, where the sequence are the nonzero elements of the sequence ordered ascendingly with respect to their absolute value and is the number of zeroes within the sequence .

**Proof**:

We order increasingly according to the modulus and the standard greater or equal order on the real numbers. We go on to observe that then , since if it were to remain bounded, there would be an accumulation point according to the Heine–Borel theorem. Also, the sequence is zero only finitely many often (otherwise zero would be an accumulation point). After eliminating the zeroes from the sequence we call the remaining sequence . Let the number of zeroes in . Then due to lemma 5.5, the function

has the required properties.

**Theorem 5.7 (Weierstraß factorisation theorem)**:

Let be holomorphic and not the constant zero function with zeroes , let are the nonzero elements of the sequence ordered ascendingly with respect to their absolute value, and if , let be the order of the zero of . Then there exists a holomorphic function such that

- .

**Proof**:

First, we note that does not have an accumulation point, since otherwise would be the constant zero function by the identity theorem from complex analysis. From theorem 5.6, we obtain that the function has exactly the zeroes with the right multiplicity, where the sequence are the nonzero elements of the sequence ordered ascendingly with respect to their absolute value and is the number of zeroes within the sequence . We have that has no zeroes and is bounded and hence holomorphic due to Riemann's theorem on resolvable singularities. For, if were unbounded, it would have a singularity at a zero of . This singularity can not be essential since dividing by finitely many linear factors would eliminate that singularity. Hence we have a pole, and this would be resolvable by multiplying linear factors to . But then has a zero of the order of that pole, which is not possible since we may eliminate all the zeroes of by writing , holomorphic and nonzero at , where is the order of the zero of at .

Hence, has a holomorphic logarithm on , which we shall denote by . This satisfies

- .

**Corollary 5.8 (Mittag-Leffler's theorem)**:

Let be a sequence of complex numbers which does not have an accumulation point. Then there exists a meromorphic function which has exactly the poles , where the pole has order .

**Proof**:

From theorem 5.7 we obtain a function with zeroes in the right multiplicity. Set .

### Exercises[edit | edit source]

## The Hadamard factorisation[edit | edit source]

In this subsection, we strive to factor certain holomorphic functions in a way that makes them even easier to deal with than the Weierstraß factorisation. This is the Hadamard factorisation. It only works for functions satisfying a certain growth estimate, but in fact, many important functions occuring in analytic number theory do satisfy this estimate, and thus that factorisation will give us ways to prove certain theorems about those functions.

In order to prove that we may carry out a Hadamard factorisation, we need some estimates for holomorphic functions as well as some preparatory lemmata.

### Estimates for holomorphic functions[edit | edit source]

**Theorem 5.9**:

Let be a holomorphic function such that , and let be the sequence of zeroes of that function ordered ascendingly by absolute value. Let . If we denote the number of zeroes of inside by , then

- .

**Proof**:

Set and define the function by

- ,

where the latter limit exists by developing into a power series at and observing that the constant coefficient vanishes. By Riemann's theorem on removable singularities, is holomorphic. We now have

- ,

and if further , then and hence we may multiply that number without change to anything to obtain for

- .

Now writing and , we obtain on the one hand

and on the other hand

- .

Hence,

- ,

which is why both and have the same distance to , since lies on the real axis.

Hence, due to the maximum principle, we have

- .

**Theorem 5.10**:

Let , let be holomorphic within and define . Then

- .

**Proof**:

First, we consider the case and . We may write in its power series form

- ,

where . If we write and , we obtain by Euler's formula

and thus

- .

Since the latter sum is majorised by the sum

- ,

it converges absolutely and uniformly in . Hence, by exchanging the order of integration and summation, we obtain

due to

and further for all

due to

- ,

as can be seen using integration by parts twice and . By monotonicity of the integral, we now have

- .

This proves the theorem in the case . For the general case, we define

- .

Then , hence by the case we already proved

- .

**Theorem 5.11**:

### Further preparations[edit | edit source]

**Definition 5.12 (exponent of convergence)**:

Let be a sequence of complex numbers not containing zero such that

converges for a . Then

is called the **exponent of convergence** of the sequence .

**Definition 5.13 (holomorphic functions of finite order)**:

Let be holomorphic, and define for

- .

If there exists such that

for a suitable , then is said to be **of finite order**. In this case,

is called the **order** of .

**Lemma 5.14**: