In this post, we will discuss the product representation of entire functions. In order to do so, we first discuss the convergence of infinite products.
1. Infinite products
Given a sequence of complex numbers , we say the product converges if the limit exists.
Theorem 1 If , then converges. Moreover, the product converges to 0 if and only if one of its factors is 0.
Proof: WLOG, we may assume that for all . Thus we may use power series to define logarithm so that . Then we obtain the following partial products
For each , we make the estimate
Using continuity of , we obtain the first result. To prove the second assertion, we note that if for all , then clearly the products converges to which is nonzero. Similarly we have convergence theorem for functions.
Theorem 2 Suppose is a sequence of holomorphic functions in an open set . If for all we have
then converges to a holomorphic function. Moreover, if does not vanish in for any , then
2. Weisterass product
Theorem 3 Given any sequence of complex numbers with as , there exists an entire function that vanishes at all and nowhere else. Any other entire function is of the form where is entire.
In general the function may not converge, so we need to insert extra factors to make it converge without adding any zeros. For each integer , we define and . Then we prove the following estimate:
Lemma 4 If , then for some constant .
Proof: Since , we obtain the taylor expansion . Thus we have
Clearly . In particular, , thus
This provides a desired bound. Proof: Define . Fix , we will show that converges uniformly on . Decompose and we only need to consider the latter factor. Clearly for all and all and by previous lemma we see that the product converges uniformly in . Moreover, they never vanish in which implies the limit does not vanish also. Hence the function provides the desired function.
3. Hadamard product
In order to refine the result, we need some elementary results. First tool is Jensen’s formula.
Theorem 5 Let be an open set contains the closure of a disk and suppose is holomorphic in , , and never vanishes on the circle . If denote the zeros of inside the disc, then
Proof: First note that if the theorem holds true for and then it holds true for . Consider which is holomorphic function in that vanishes nowhere. So we only need to show that the theorem is true for and . Since has no zeros we need to show . This is clearly true by mean-value property of the harmonic function . It remains to show
This boils down to show for any . Making change of variables shows this is equivalent to . Consider the function which is holomorphic and vanishes nowhere in the unit disk. Thus applying mean-value property to we have
Next we claim that if and does not vanish on the circle , then
This is immediate if we prove
Lemma 6 If are the zeros of inside then
Proof: We start from the right-hand side.
4. Functions of finite order
Let be an entire function. We call has order of growth if
In particular, the order of is .
Theorem 7 If is an entire function that has order of growth , then for sufficiently large . Moreover if are zeros of , then converges for any .
Proof: We shall use Jensen’s formula. Note that is an increasing function of . Now choose , then clearly we have
By Jensen’s formula we have
This proves the first part. Then we directly estimate
5. Hadamard products
Theorem 8 Suppose is entire and has order of growth . Let be the integer such that . If are zeros of , then
where is a polynomial of degree , is the order of zero at .
Proof: The convergence of the canonical products is now easily solved. Using Lemma 2.2, we have . Thus the convergence of implies the convergence of the product. In order to show that is a polynomial of degree , we need more efforts.
Lemma 9 We have the following estimates
Proof of Lemma. If , we directly have
If , then
Lemma 10 For any , we have
except possibly for .
Proof of Lemma. Again we use usual decomposition
We first estimate the second product. Note that for all . By previous lemma, we have
Note that for some constant as . Thus we have
because . To estimate the first prouct we use previous lemma to obtain
Using a similar trick we have . Thus
It remains to estimate . But using the property that , we have
Now we consider
This completes the proof of the lemma.
Lemma 11 There exists a sequence of complex numbers with such that
Proof of Lemma. First we note that converges so we may take . For each positive integer we may find such that the the disk does not intersect the forbidden areas in previous lemma. Suppose it is not the case, then there is a particular such that all balls of radius with will intersect the forbidden area. In particular, by rotating the those balls on the real line so that the diameter will lie in the positive real axis, we see that . This implies which is a contradiction.
Now we prove the Hadamard’s theorem. Let . Previously we’ve shown that converges to a holomorphic which has zeros as does and does not vanish everywhere else. Thus we have and in particular
This implies on . Finally we show that this is enough to guarantee that is a polynomial of degree less than .
Lemma 12 Suppose is entire and satisfies
for a sequence of positive real numbers that tends to infinity. Then is a polynomial of degree .
Proof of Lemma. First we write . Let denote the circle centerer at the origin with radius . By Cauchy’s integral formula, we have whenver . This implies
which further implies
Recall that whenever . Thus for , we have
Letting shows that if . Hence is a polynomial of degree .