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follows because n(2|z|) ≤ c|z|s by Theorem 2.1. Since we restricted s to satisfy s > ρ0, we can take an initial s suﬃciently close to ρ0, so that the assertion of the lemma is established (with s being replaced by s′). Corollary 5.4 There exists a sequence of radii, r1, r2, . . ., with rm →∞, such that∣∣∣∣∣ ∞∏ n=1 Ek(z/an) ∣∣∣∣∣ ≥ e−c|z|s for |z| = rm. Proof. Since ∑ |an|−k−1 <∞, there exists an integer N so that ∞∑ n=N |an|−k−1 < 1/10. Therefore, given any two consecutive large integers L and L + 1, we can ﬁnd a positive number r with L ≤ r ≤ L+ 1, such that the circle of radius r centered at the origin does not intersect the forbidden discs of Lemma 5.3. For otherwise, the union of the intervals In = [ |an| − 1|an|k+1 , |an|+ 1 |an|k+1 ] 5. Hadamard’s factorization theorem 151 (which are of length 2|an|−k−1) would cover all the interval [L,L+ 1]. (See Figure 1.) This would imply 2 ∑∞ n=N |an|−k−1 ≥ 1, which is a con- tradiction. We can then apply the previous lemma with |z| = r to con- clude the proof of the corollary. I2 I3 a1 I1 a2 In a3 an Figure 1. The intervals In Proof of Hadamard’s theorem Let E(z) = zm ∞∏ n=1 Ek(z/an). To prove that E is entire, we repeat the argument in the proof of Theo- rem 4.1; we take into account that by Lemma 4.2 |1−Ek(z/an)| ≤ c ∣∣∣∣ zan ∣∣∣∣k+1 for all large n, and that the series ∑ |an|−k−1 converges. (Recall ρ0 < s < k + 1.) More- over, E has the zeros of f , therefore f/E is holomorphic and nowhere vanishing. Hence f(z) E(z) = eg(z) 152 Chapter 5. ENTIRE FUNCTIONS for some entire function g. By the fact that f has growth order ρ0, and because of the estimate from below for E obtained in Corollary 5.4, we have eRe(g(z)) = ∣∣∣∣ f(z)E(z) ∣∣∣∣ ≤ c′ec|z|s for |z| = rm. This proves that Re(g(z)) ≤ C|z|s, for |z| = rm. The proof of Hadamard’s theorem is therefore complete if we can estab- lish the following ﬁnal lemma. Lemma 5.5 Suppose g is entire and u = Re(g) satisﬁes u(z) ≤ Crs whenever |z| = r for a sequence of positive real numbers r that tends to inﬁnity. Then g is a polynomial of degree ≤ s. Proof. We can expand g in a power series centered at the origin g(z) = ∞∑ n=0 anz n. We have already proved in the last section of Chapter 3 (as a simple application of Cauchy’s integral formulas) that (6) 1 2π ∫ 2π 0 g(reiθ)e−inθ dθ = { anr n if n ≥ 0 0 if n < 0. By taking complex conjugates we ﬁnd that (7) 1 2π ∫ 2π 0 g(reiθ)e−inθ dθ = 0 whenever n > 0, and since 2u = g + g we add equations (6) and (7) to obtain anr n = 1 π ∫ 2π 0 u(reiθ)e−inθ dθ, whenever n > 0. For n = 0 we can simply take real parts of both sides of (6) to ﬁnd that 2Re(a0) = 1 π ∫ 2π 0 u(reiθ) dθ. 6. Exercises 153 Now we recall the simple fact that whenever n �= 0, the integral of e−inθ over any circle centered at the origin vanishes. Therefore an = 1 πrn ∫ 2π 0 [u(reiθ)− Crs]e−inθ dθ when n > 0, hence |an| ≤ 1 πrn ∫ 2π 0 [Crs − u(reiθ)] dθ ≤ 2Crs−n − 2Re(a0)r−n. Letting r tend to inﬁnity along the sequence given in the hypothesis of the lemma proves that an = 0 for n > s. This completes the proof of the lemma and of Hadamard’s theorem. 6 Exercises 1. Give another proof of Jensen’s formula in the unit disc using the functions (called Blaschke factors) ψα(z) = α− z 1− αz . [Hint: The function f/(ψz1 · · ·ψzN ) is nowhere vanishing.] 2. Find the order of growth of the following entire functions: (a) p(z) where p is a polynomial. (b) ebz n for b = 0. (c) ee z . 3. Show that if τ is ﬁxed with Im(τ ) > 0, then the Jacobi theta function Θ(z|τ ) = ∞∑ n=−∞ eπin 2τe2πinz is of order 2 as a function of z. Further properties of Θ will be studied in Chap- ter 10. [Hint: −n2t + 2n|z| ≤ −n2t/2 when t > 0 and n ≥ 4|z|/t.] 4. Let t > 0 be given and ﬁxed, and deﬁne F (z) by F (z) = ∞∏ n=1 (1− e−2πnte2πiz). Note that the product deﬁnes an entire function of z. 154 Chapter 5. ENTIRE FUNCTIONS (a) Show that |F (z)| ≤ Aea|z|2 , hence F is of order 2. (b) F vanishes exactly when z = −int + m for n ≥ 1 and n,m integers. Thus, if zn is an enumeration of these zeros we have∑ 1 |zn|2 =∞ but ∑ 1 |zn|2+� <∞. [Hint: To prove (a), write F (z) = F1(z)F2(z) where F1(z) = N∏ n=1 (1− e−2πnte2πiz) and F2(z) = ∞∏ n=N+1 (1− e−2πnte2πiz). Choose N ≈ c|z| with c appropriately large. Then, since( ∞∑ N+1 e−2πnt ) e2π|z| ≤ 1 , one has |F2(z)| ≤ A. However, |1− e−2πnte2πiz| ≤ 1 + e2π|z| ≤ 2e2π|z|. Thus |F1(z)| ≤ 2Ne2πN|z| ≤ ec′|z|2 . Note that a simple variant of the function F arises as a factor in the triple product formula for the Jacobi theta function Θ, taken up in Chapter 10.] 5. Show that if α > 1, then Fα(z) = ∫ ∞ −∞ e−|t| α e2πizt dt is an entire function of growth order α/(α− 1). [Hint: Show that −|t| α 2 + 2π|z||t| ≤ c|z|α/(α−1) by considering the two cases |t|α−1 ≤ A|z| and |t|α−1 ≥ A|z|, for an appropriate constant A.] 6. Prove Wallis’s product formula π 2 = 2 · 2 1 · 3 · 4 · 4 3 · 5 · · · 2m · 2m (2m− 1) · (2m + 1) · · · . [Hint: Use the product formula for sin z at z = π/2.] 7. Establish the following properties of inﬁnite products. 6. Exercises 155 (a) Show that if ∑ |an|2 converges, then the product ∏(1 + an) converges to a non-zero limit if and only if ∑ an converges. (b) Find an example of a sequence of complex numbers {an} such that ∑ an converges but ∏ (1 + an) diverges. (c) Also ﬁnd an example such that ∏ (1 + an) converges and ∑ an diverges. 8. Prove that for every z the product below converges, and cos(z/2) cos(z/4) cos(z/8) · · · = ∞∏ k=1 cos(z/2k) = sin z z . [Hint: Use the fact that sin 2z = 2 sin z cos z.] 9. Prove that if |z| < 1, then (1 + z)(1 + z2)(1 + z4)(1 + z8) · · · = ∞∏ k=0 (1 + z2 k ) = 1 1− z . 10. Find the Hadamard products for: (a) ez − 1; (b) cos πz. [Hint: The answers are ez/2z ∏∞ n=1(1 + z 2/4n2π2) and ∏∞ n=0(1− 4z2/(2n + 1)2), respectively.] 11. Show that if f is an entire function of ﬁnite order that omits two values, then f is constant. This result remains true for any entire function and is known as Picard’s little theorem. [Hint: If f misses a, then f(z)− a is of the form ep(z) where p is a polynomial.] 12. Suppose f is entire and never vanishes, and that none of the higher derivatives of f ever vanish. Prove that if f is also of ﬁnite order, then f(z) = eaz+b for some constants a and b. 13. Show that the equation ez − z = 0 has inﬁnitely many solutions in C. [Hint: Apply Hadamard’s theorem.] 14. Deduce from Hadamard’s theorem that if F is entire and of growth order ρ that is non-integral, then F has inﬁnitely many zeros. 15. Prove that every meromorphic function in C is the quotient of two entire functions. Also, if {an} and {bn} are two disjoint sequences having no ﬁnite limit 156 Chapter 5. ENTIRE FUNCTIONS points, then there exists a meromorphic function in the whole complex plane that vanishes exactly at {an} and has poles exactly at {bn}. 16. Suppose that Qn(z) = Nn∑ k=1 cnk z k are given polynomials for n = 1, 2, . . .. Suppose also that we are given a sequence of complex numbers {an} without limit points. Prove that there exists a meromorphic function f(z) whose only poles are at {an}, and so that for each n, the diﬀerence f(z)−Qn ( 1 z − an ) is holomorphic near an. In other words, f has a prescribed poles and principal parts at each of these poles. This result is due to Mittag-Leﬄer. 17. Given two countably inﬁnite sequences of complex numbers {ak}∞k=0 and {bk}∞k=0, with limk→∞ |ak| =∞, it is always possible to ﬁnd an entire function