Functions of a Complex Variable II Math 562, Spring 2020

Home , Infinite product, Normal convergence

Jens Lorenz

April 27, 2020

Department of Mathematics and Statistics, UNM, Albuquerque, NM 87131

Contents

1 Notations 5

2 The Schwarz Lemma and Aut(D) 6 2.1 Schwarz Lemma ...... 6 2.2 Biholomorphic Maps and Automorphisms ...... 7 2.3 The Automorphism Group of D ...... 7 2.4 The Schwarz–Pick Lemma ...... 9 2.5 Remarks ...... 11

3 Linear Fractional Transformations and the Riemann Sphere 12 3.1 The Riemann Sphere ...... 12 3.2 Linear Fractional Transformations ...... 14 3.3 The Automorphisms of the Riemann Sphere ...... 16 3.4 A Remarkable Geometric Property of Linear Fractional Trans- formations ...... 17 3.4.1 Analytical Description of Circles ...... 18 3.4.2 Circles under Reciprocation ...... 18 3.4.3 Straight Lines under Reciprocation ...... 20 3.5 Example: The Cayley Transform in the Complex Plane . . . . . 21 3.6 The Cayley Transform of a Matrix ...... 22

4 The Riemann Mapping Theorem 24 4.1 Statement and Overview of Proof ...... 24 4.2 The Theorem of Arzela–Ascoli ...... 26 4.3 Montel’s Theorem ...... 29 4.4 Auxiliary Results on Logarithms and Square Roots ...... 32 4.5 Construction of a Map in ...... 33 4.6 A Bound of f 0(P ) for all Ff ...... 35 4.7 Application of| the| Theorems∈ of F Montel and Hurwitz ...... 36 4.8 Proof That f is Onto ...... 37

1 4.9 Examples of Biholomorphic Mappings ...... 39

5 An Introduction to Schwarz–Christoﬀel Formulas 41 5.1 General Power Functions ...... 41 5.2 An Example of a Schwarz–Christoﬀel Map ...... 42

6 Meromorphic Functions with Prescribed Poles 47 6.1 Meromorphic Functions on C ...... 47 6.2 The Mittag–Leﬄer Theorem ...... 47 6.3 An Example ...... 50

7 Infinite Products 52 7.1 Infinite Products of Complex Numbers ...... 52 7.1.1 Two Definitions of Convergence ...... 52 7.1.2 Examples ...... 53 7.2 Infinite Products of Numbers: Convergence Theory ...... 56 7.3 Infinite Products of Functions ...... 61 7.4 Example: The Product Formula for the Sine Function ...... 64 7.5 Euler’s Constant ...... 65 7.6 The Gauss Formula for Γ(z) and Weierstrass’ Product Formula for 1/Γ(z)...... 66 7.6.1 A Formula for (Γ0/Γ)0 ...... 70 7.7 Entire Functions with Prescribed Zeros ...... 71 7.7.1 Construction of f(z); Motivation ...... 72 7.7.2 Weierstrass’ Canonical Factors ...... 72

8 The Bernoulli Numbers and Applications 77 8.1 The Bernoulli Numbers ...... 77 8.2 The Taylor Series for z cot z in Terms of Bernoulli Numbers . . . 79 8.3 The Mittag–Leﬄer Expansion of πz cot(πz)...... 80 8.4 The Values of ζ(2m)...... 81 8.5 Sums of Powers and Bernoulli Numbers ...... 81

9 The Riemann Zeta–Function 84 9.1 Deﬁnition for Re s > 1...... 84 9.2 Simple Bounds of ζ(s) for s > 1...... 84 9.3 Meromorphic Continuation of ζ(z) to Re z > 0 ...... 85 9.4 Analytic Continuation of ζ(z) to C 1 ...... 87 9.5 Euler’s Product Formula for ζ(z)...... \{ } 90 P 1 9.6 The Sum p and the Prime Number Theorem ...... 93 9.7 Auxiliary Results about Fourier Transformation: Poisson’s Sum- mation Formula ...... 94 9.8 A Functional Equation ...... 96 9.9 Growth Estimates for the ζ–Function ...... 98

2 10 Analytic Continuation 101 10.1 Analytic Continuation Using the Cauchy Riemann Equations . . 101 10.2 Exponential Decay of Fourier Coeﬃcients and the Strip of Ana- lyticity ...... 101 10.3 The Schwarz Reﬂection Principle ...... 101 10.4 Examples for Analytic Continuation ...... 101 10.5 Riemann Surfaces: Intuitive Approach ...... 101 10.6 Riemann Surfaces: Germs, Sheafs, and Fibers ...... 101

11 Fourier Series 102 11.1 Convergence Results: Overview ...... 102 11.2 The Dirichlet Kernel ...... 104 11.3 The Fej´erKernel ...... 105 11.4 Convergence of σnf in Maximum Norm ...... 107 11.5 Weierstrass’ Approximation Theorem for Trigonometric Polyno- mials ...... 108 11.6 Convergence of Snf in L2 ...... 108 11.7 The Lebesgue Constant of the Dirichlet Kernel ...... 110 11.7.1 The Lebesgue Constant as Norm of a Functional . . . . . 112 11.8 Divergence of the Fourier Series of a Continuous Function at a Point...... 112 11.9 Isomorphy of L2(0, 1) and l2 ...... 114 11.10Convergence of Snf in Maximum Norm ...... 116 11.11Smoothness of f(x) and Decay of fˆ(k)...... 117 11.12Exponential Decay of fˆ(k) and Analyticity of f ...... 119 11.13Divergence of Snf(0): Explicit Construction of f ...... 121 11.14Fourier Series and the Dirichlet Problem for Laplace’s Equation on the Unit Disk ...... 123

12 Fourier Transformation 124 12.1 Motivation: Application to PDEs ...... 124 12.2 The Fourier Transform of an L1 Function ...... 127 12.3 The Riemann–Lebesgue Lemma ...... 128 12.4 The Fourier Transform on the Schwartz Space ...... 130 12.5 The Fourier Inversion Formula on the SchwartzS Space: Prepara- tions ...... 131 12.6 The Fourier Inversion Formula on the Schwartz Space ...... 134 12.7 Operators ...... 135 12.8 Elementary Theory of Tempered Distributions ...... 136 12.8.1 Ordinary Functions as Tempered Distributions ...... 137 12.9 The Fourier Transform on 0: An Example Using Complex Vari- ables ...... S ...... 137 12.10Decay of the Fourier Transform of f and Analyticity of f .... 138 12.11The Paley–Wiener Theorem ...... 138 12.12The Laplace Transform and Its Inversion ...... 139

3 13 Growth and Zeros of Entire Functions 140 13.1 Jensen’s Formula ...... 140 13.2 The Order of Growth ...... 142 13.3 Zeros and Growth Estimates ...... 143 13.4 Hadamard’s Factorization Theorem ...... 145 13.5 Entire Functions of Non–Integer Order of Growth ...... 148 13.6 Proof of Hadamard’s Factorization Theorem ...... 148

14 The Prime Number Theorem 155 14.1 Functions ...... 155 14.2 Reduction to Asymptotics of ψ1(x)...... 157 14.3 Integral Representation of ψ1(x)...... 161 14.4 Auxiliary Results ...... 161 14.5 The ζ–Function has no Zero on the Line s = 1 + it ...... 163

15 Complex Variables and PDEs 166 15.1 2D Irrotational Euler Flows ...... 166 15.2 Laplace Equation ...... 171 15.2.1 Derivation of the Poisson Kernel via Complex Variables . 171 15.2.2 Derivation of the Poisson Kernel via Separation of Variables175

16 Rearrangement of Series 176 16.1 Rearrangement of Absolutely Convergent Series ...... 176 16.2 Interchanging Double Sums ...... 177 16.3 Rearrangement of a Double Series ...... 180

4 1 Notations

R ﬁeld of real numbers C ﬁeld of complex numbers D(z0,R) z : z z0 < R : open disk of radius R centered at z0 ¯ { | − | } D(z0,R) z : z z0 R : closed disk of radius R centered at z0 ∂D(z ,R) {z : |z − z |= ≤ R} : boundary of disk of radius R centered at z 0 { | − 0| } 0 D D = D(0, 1) = z : z < 1 : open unit disk { | | } H H = z = x + iy : y > 0 : open upper half–plane { } H(U) set of all holomorphic functions f : U C where U C is open → ⊂ C(U) set of all continuous functions f : U C where U C is any set R x ds → ⊂ ln(x) for real positive x: ln(x) = 1 s log(z) complex logarithm

5 2 The Schwarz Lemma and Aut(D) 2.1 Schwarz Lemma

Let D = D(0, 1) denote the open unit disk. In the Schwarz Lemma one considers functions f H(D) with f(0) = 0 and f(D) D¯. The Schwarz Lemma then says that the∈ estimate f(z) 1 can be sharpened.⊂ We recall the maximum| | modulus ≤ theorem, which will be used in the proof of the Schwarz Lemma.

Theorem 2.1 (Maximum Modulus Theorem) Let U C denote an open, connected set and let g H(U). If there exists a point z⊂ U with ∈ 0 ∈ g(z) g(z ) for all z U | | ≤ | 0 | ∈ then g(z) is constant in U. In other words, only constant holomorphic functions attain their maximal value.

This follows from the open mapping theorem: A holomorphic function g(z) on an open, connected set U maps open subsets of U to open sets, unless g(z) is constant.

Theorem 2.2 (Schwarz Lemma) Let f H(D) satisfy ∈ a) f(z) 1 for all z D; b) f| (0) =| ≤ 0. ∈ 0 Then f(z) z for all z D and f (0) 1. In addition, if | | ≤ | | ∈ | | ≤ f(z ) = z | 0 | | 0| 0 for some z0 D 0 or if f (0) = 1, then f is a rotation, i.e., ∈ \{ } | | f(z) = αz for some α C with α = 1. ∈ | | Proof: Set

f(z)/z, 0 < z < 1 g(z) = f 0(0), z =| 0 | Since f(0) = 0 we have

∞ X f(z) = a zj for z < 1 j | | j=1 and obtain that g H(D). Let 0 < ε < 1 and consider g in the closed disk ∈ D¯(0, 1 ε) D. Since f(z) 1 in D one obtains that − ⊂ | | ≤ 1 g(z) for z = 1 ε . | | ≤ 1 ε | | − − By the maximum modulus theorem we conclude that

Terminology: A nonempty, open, connected subset Ω C is called a region. ⊂ Let U, V be regions in C. A map f : U V is called biholomorphic if f is 1 1 and onto and f as well as f −1 are holomorphic.→ It is sufficient to assume that− f : U V is holomorphic, 1 1 and onto. Then f −1 is automatically holomorphic,→ as we have proved earlier.− We also recall that f 0(z) = 0 for all z U if f is biholomorphic. 6 ∈ Definition: Let U be a region in C. If f : U U is 1 1, onto, and holomorphic (thus biholomorphic), then f is called→ an automorphism− of U. The set of all automorphisms of U is denoted by Aut(U). This set forms a group if the product of f, g Aut(U) is defined as the composition, f g. ∈ ◦ Example 1: According to Homework 5, Problem 2, the set Aut(C) consists of all functions f(z) of the form f(z) = az + b where a = 0. 6 2.3 The Automorphism Group of D We first consider some special Möbiustransformations. Let c C and c < 1. Set ∈ | | z c φc(z) = − , z D . 1 cz¯ ∈ − Clearly, φc H(D) for c < 1. Note that ∈ | | φ (z) z for c = 0 . c ≡ We prove:

Lemma 2.1 a) φc(z) < 1 for all z D. | | ∈ b) φc(φ−c(z)) = z for all z D. ∈ c) φc Aut(D) for c < 1. ∈ | |

7 Proof: a ) Clearly, φ (z) < 1 is equivalent to | c | 1 cz¯ 2 > z c 2 , | − | | − | which is equivalent to

(1 cz¯ )(1 cz¯) > (z c)(¯z c¯) , − − − − which is equivalent to

0 < (1 c 2)(1 z 2) = 1 + c 2 z 2 z 2 c 2 − | | − | | | | | | − | | − | | the inequality (2.1) holds. b) With w c z + c φ (w) = − , φ (z) = c 1 cw¯ −c 1 +cz ¯ − we have

z+c c φ (φ (z)) = 1+¯cz − c −c 1 c¯ z+c − 1+¯cz z + c c z c 2 = − − | | 1 +cz ¯ cz¯ c 2 − − | | z(1 c 2) = − | | 1 c 2 = z . − | | c) Since φc(φ−c(z)) = z for all z D, we obtain that φc is onto and φ−c is 1 1. ∈ − Replacing c by c shows that φc is a bijection of D. − Another set of automorphisms of D are the rotations Rα deﬁned by

R (z) = αz, z < 1 , α | | for α = 1. Clearly, R Aut( ) and | | α ∈ D

φc Rα Aut(D) (2.2) ◦ ∈ whenever c < 1, α = 1. The following theorem says that every f Aut(D) has the form| | (2.2).| | ∈

Theorem 2.3 Let f Aut(D) and let b = f(0). Then we have ∈

f = φ− R = R φ− (2.3) b ◦ α α ◦ bα¯ for some α with α = 1. | |

8 Proof: 1) First assume that f(0) = 0 and let g = f −1. By Schwarz Lemma we 0 0 conclude that f (0) 1 and g (0) 1. From f(g(z)) = z for all z D we have | | ≤ | | ≤ ∈

f 0(g(z))g0(z) = 1 , and, in particular,

f 0(0)g0(0) = 1 . It follows that f 0(0) = 1. Another application of Schwarz Lemma yields f(z) = αz. Equation (2.3)| holds| with b = 0. 2) Let b := f(0) and consider h = φ f. We have b ◦

h(0) = φb(b) = 0 . By Part 1) of the proof we conclude that h = R for some α with α = 1. α | | Therefore, f = φ− R . In other words, b ◦ α

f(z) = φ− (αz) = φ− R (z) . b b ◦ α 3) The equations

f(z) = φ−b(αz) αz + b = 1 + ¯bαz z + bα¯ = α 1 + ¯bαz = αφ−bα¯(z) complete the proof of the theorem. 2.4 The Schwarz–Pick Lemma

The automorphisms φc of D, obtained for every c with c < 1, allow to prove an extension of the Schwarz lemma. Note that | |

z c 1 c 2 φ (z) = − , φ0 (z) = − | | . c 1 cz¯ c (1 cz¯ )2 − − Theorem 2.4 (Schwarz–Pick) Let f : D D, f H(D). (It is neither assumed that f is 1 1 nor that f is onto nor→ that f∈(0) = 0.) − 1) For all a D: ∈ 1 f(a) 2 f 0(a) − | | (2.4) | | ≤ 1 a 2 − | | (If a = 0 and f(0) = 0 this reduces to the estimate f 0(0) 1 of the Schwarz lemma.) | | ≤ 2) If a1, a2 D then ∈ 9 f(a ) f(a ) a a | 2 − 1 | | 2 − 1| . (2.5) 1 f(a )f(a ) ≤ 1 a1a2 − 1 2 | | − | Proof: 1) Let b = f(a) and set

F = φ f φ− . b ◦ ◦ a Then we have φ−a(0) = a and

F (0) = φb(f(a))

= φb(b) = 0

The Schwarz lemma is applicable to F and implies that

F 0(0) 1 . | | ≤ Further, by the chain rule,

0 0 0 0 F (0) = φb(b)f (a)φ−a(0) . Here

φ0 (0) = 1 a 2 −a − | | and

1 b 2 1 φ0 (b) = − | | = . b (1 b 2)2 1 b 2 − | | − | | Therefore,

1 a 2 F 0(0) = f 0(a) − | | . 1 b 2 − | | The estimate F 0(0) 1 yields that | | ≤ 1 b 2 f 0(a) − | | . | | ≤ 1 a 2 − | | This proves (2.4). 2) Let b1 = f(a1) and b2 = f(a2). Consider the function

F = φ f φ− . b1 ◦ ◦ a1 As above, we have F (0) = 0 and the Schwarz lemma yields that F (z) z | | ≤ | | for all z D. Setting ∈

w = φ−a1 (z), z = φa1 (w) , the estimate F (z) z becomes | | ≤ | | 10 φ (f(w)) φ (w) . | b1 | ≤ | a1 | Using the deﬁnition of φc this reads: f(w) b w a 1 1 − − . 1 b f(w) ≤ 1 a1w − 1 − If we use this estimate for w = a we obtain (2.5). 2 2.5 Remarks The formula

b a d(a, b) = | − | , a, b D , 1 ab ∈ | − | deﬁnes the so-called pseudo–hyperbolic metric on D. (This is related to Poincar´e geometry on D.) The second part of the Schwarz–Pick Lemma then says that

d(f(a), f(b)) d(a, b) for all a, b D ≤ ∈ if f H(D), f : D D. If f Aut(D) then one can apply the estimate to f and ∈f −1 to obtain → ∈

d(f(a), f(b)) = d(a, b) for all a, b D . (2.6) ∈ In fact, one can prove a converse: If f H(D), f : D D, and if (2.6) holds, ∈ → then f Aut(D). ∈ The introduction of non–Euclidean geometries was historically very important. Recall: Euclid of Alexandria ( 365–300 B.C.) and his Parallel Postulate, the famous ﬁfth Postulate. ∼ Immanuel Kant: The concept of Euclidean space is by no means of empirical origin, but is an inevitable necessity of thought. (Critique of Pure Reason, 1781) Work of Carl Friedrich Gauss (1777–1855) and Janos Bolyai (1802–1860) on non–Euclidean geometry made it clear that Kant’s position was wrong. One has to distinguish between geometry as a mathematical subject and geometry of physical space, which is a subject of observation. Newtonian physics assumes Euclidean space and a time variable independent of space. To create the theory of general relativity, it was necessary to overcome the doctrine of Euclidean space.

11 3 Linear Fractional Transformations and the Rie- mann Sphere

3.1 The Riemann Sphere

It is often useful to compactify the complex plane C by formally adding the point . We use the notation ∞ Cˆ = C . ∪ {∞} One can turn Cˆ into a topological space by using an identiﬁcation (this is nothing but a map which is 1 1 and onto) with the unit sphere in R3, − 3 2 2 2 S = (x1, x2, x3) R : x + x + x = 1 . { ∈ 1 2 3 } This identiﬁcation can be established as follows: Let N = (0, 0, 1) S denote the north pole of S and let ∈

z = x + iy C . ∈ Draw the straight line in R3 from N to (x, y, 0). It intersects S in a unique point, which we call Π(z), the stereographic projection of z to S. The map Π maps C bijectively onto S N : \{ } Π: C S N . → \{ } As z , one obtains that Π(z) N on S. It is therefore natural to extend | | → ∞ → Π as a mapping from Cˆ to S by deﬁning

Π( ) = N = (0, 0, 1) . ∞ In this way we have deﬁned a bijection

Π: Cˆ S. → The set S is a metric space if we use the Euclidean distance in R3 as distance between points in S. We know, then, what it means that a sequence qn S converges to q S. We know what the open and the closed sets in S are∈ etc. Using the map∈ Π we transform the concept of convergence etc. (or the metric) to Cˆ. If the point is not involved, these concepts agree with the standard ∞ concepts in C, but they are now extended in a meaningful way to Cˆ. The space Cˆ is called the one–point compactification of C, or the Riemann sphere. We mention that one can also consider Cˆ as a complex manifold by introducing local coordinates near the point Cˆ. If one has done this, then the ∞ ∈ notion of holomorphy of a function f : Cˆ Cˆ makes sense. In practice, to discuss holomorphy when is involved, one→ uses the mappings z 1/z and w 1/w to map to zero.∞ → → ∞ Holomorphy of functions where is involved: 1) Let us first define what it means∞ that a complex–valued function f(w) is holomorphic at w = . Let Ω Cˆ denote an open set with Ω and let ∞ ⊂ ∞ ∈ 12 1 f :Ω C denote a map. There exists ε > 0 so that z Ω for 0 < z < ε. Define→ ∈ | |

f(1/z) for 0 < z < ε , F (z) = f( ) for z =| 0 |. ∞ Then, by deﬁnition, the function f(w) is holomorphic at w = if and only if the function F (z) is holomorphic at z = 0. If F (z) is holomorphic∞ at z = 0 then

∞ X F (z) = a zj for z < ε j | | j=0 and

P∞ a w−j for 1 < w < , f(w) = j=0 j ε | | ∞ f( ) for w = . ∞ ∞ Here a = F (0) = f( ). 0 ∞ Example 1: Let w f(w) = for w > 3, f( ) = 1 . w + 3 | | ∞ We have

1/z 1 1 F (z) = = for 0 < z < 1/z + 3 1 + 3z | | 3 and F (0) = f( ) = 1. The function F (z) is holomorphic at z = 0. Thus, by deﬁnition, f(w)∞ is holomorphic at w = . ∞ 2) Holomorphy of a function f(w) with f(w0) = for some w0 C: Let ∞ ∈ Ω C denote an open set and let w0 Ω. Let ⊂ ∈

f(w) C for w Ω w0 ∈ ∈ \{ } and let f(w0) = . Assume that f(w) is continuous at w0. Then, for 0 < w w < ε, we have∞ f(w) = 0 and set | − 0| 6 1/f(w) for 0 < w w < ε F (w) = | − 0| 0 for w = w0

If F (w) is holomorphic at w0 then, by deﬁnition, f(w) is holomorphic at w0. Example 2: Let 1 f(w) = for w C 3 , f(3) = . w 3 ∈ \{ } ∞ − We have

F (w) = w 3 for w C 3 ,F (3) = 0 . − ∈ \{ }

13 Clearly, the function F (w) is holomorphic at w0 = 3. Thus, by deﬁnition, f(w) is holomorphic at w0 = 3. 3) Holomorphy of a function f(w) with f( ) = : Let Ω Cˆ denote an open set with Ω and let ∞ ∞ ⊂ ∞ ∈ f(w) C for w Ω , f( ) = . ∈ ∈ \ {∞} ∞ ∞ 1 Assume that f(w) is continuous at w = . Then, for ε < w < , we have f(w) = 0 and set ∞ | | ∞ 6 1 F (z) = for 0 < z < ε, F (0) = 0 . f(1/z) | |

If F (z) is holomorphic at z0 = 0 then, by deﬁnition, f(w) is holomorphic at w = . ∞ Example 3: Let

n X f(w) = a wj where n 1, a = 0 . j ≥ n 6 j=0 Thus f(w) is a non–constant polynomial. Since f(w) as w we set f( ) = . We have | | → ∞ | | → ∞ ∞ ∞

1 n f(1/z) = a + a − z + ... + a z = 0 for 0 < z < ε zn n n 1 0 6 | | and

zn F (z) = for 0 < z < ε, F (0) = 0 . n an + an−1z + ... + a0z | | The function F (z) is holomorphic at z = 0. Thus, by deﬁnition, the polynomial f(w) is holomorphic at w = . ∞ 3.2 Linear Fractional Transformations Let

a b A = , det(A) = 0 , c d 6 × denote a nonsingular matrix in C2 2. Then A determines the linear fractional transformation

az + b φ (z) = , ad bc = 0 . (3.1) A cz + d − 6

(If one would allow det(A) = 0 then the transformation φA(z) would be constant, which is an uninteresting transformation.) The transformations (3.1) are also called M¨obiustransformations. Note that A and qA, q = 0, determine the 6 same transformation, φA = φqA.

14 ˆ It will be convenient to consider φA as a bijection of C onto itself. Then one ˆ ﬁnds that φA is an automorphism of C. Case 1: c = 0. In this case d = 0 and a = 0 and 6 6 a b φ (z) = z + . A d d The transformation φA : C C is 1 1 and onto. As zn , we have φ (z ) . Therefore, we set→ φ ( ) =− and obtain a continuous→ ∞ bijection A n → ∞ A ∞ ∞ of Cˆ. If one considers Cˆ as a one–dimensional complex manifold, then this bijection is holomorphic. Case 2: c = 0. In this case, if z0 = d/c, then φA(z0) is not deﬁned as a complex number.6 If z z = d/c then−φ (z ) . This holds since n → 0 − A n → ∞ 1 a( d/c) + b = (ad bc) = 0 . − − c − 6 Therefore, we set

φ ( d/c) = . (3.2) A − ∞ In this way, φ is continuously extended to z = d/c. Also, if z , then A 0 − n → ∞ φ (z ) a/c . A n → Therefore, we set

φ ( ) = a/c . (3.3) A ∞ ˆ ˆ In this way, φA : C C is deﬁned as a continuous map, which is holomorphic. → We claim that, with the additional deﬁnitions, (3.2) and (3.3), the map ˆ ˆ φA : C C is a bijection and its inverse is φA−1 . This claim follows from the following→ important lemma.

Lemma 3.1 Let

a b α β A = ,B = c d γ δ

× denote nonsingular matrices in C2 2 and let C denote their product, aα + bγ aβ + bδ C = AB = . cα + dγ cβ + dδ Then we have

φ = φ = φ φ . C AB A ◦ B Proof: In all cases where the point is not involved we have ∞ αz + β φ (z) = =: w B γz + δ

15 and

aw + b φ (w) = A cw + d a(αz + β) + b(γz + δ) = c(αz + β) + d(γz + δ) (aα + bγ)z + aβ + bδ = (cα + dγ)z + cβ + dδ

= φC (z)

This proves that

φ (z) = φ φ (z) AB A ◦ B when is not involved. By continuous extension, the equation ∞ ˆ φAB(z) = φA φB(z), z C , ◦ ∈ follows. 3.3 The Automorphisms of the Riemann Sphere

Let M denote the group of all M¨obiustransformations φA(z) where A × ∈ C2 2, det(A) = 0. With GL(2, C) one denotes the group of all nonsingular 6 × matrices in C2 2. The previous lemma says that GL(2, ) M Φ: C (3.4) A 7→ φ (z) → A is a group homomorphism. Clearly, Φ is onto.

Lemma 3.2 We have

M = Aut(Cˆ) , thus every automorphism of Cˆ is a M¨obiustransformation.

Proof: Let f Aut(Cˆ). a) Assume that f( ) = . In this case, the restric- ∈ ∞ ∞ tion of f to C is an automorphism of C. We have shown that

f(z) = az + b, a = 0 . 6 In particular, f M. ∈ b) Let f( ) = b, b C. Consider ∞ ∈ 1 φ(z) = z b − and deﬁne g = φ f. Since g is an automorphism of Cˆ and g( ) = , we have g M and then f◦ = φ−1 g M. ∞ ∞ ∈Thus we have shown that◦ ∈

16 GL(2, ) Aut(ˆ) Φ: C C (3.5) A 7→ φ (z) → A is a group epimorphism. (A homomorphism, which is onto, is called an epimorphism.) By definition, the kernel ker(Φ) of the epimorphism Φ defined in (3.5) consists of all matrices A GL(2, C) for which the Möbiustransformation φA(z) is ∈ ˆ the unit element of the group Aut(C), i.e., A ker(Φ) if and only if φA(z) z. ∈ ≡ By a simple result of group theory, one obtains that Aut(Cˆ) is isomorphic to the quotient group

GL(2, C)/ker(Φ) .

It is not diﬃcult to see that φA(z) z if and only if A = aI for some a C, a = 0. Thus, ≡ ∈ 6

ker(Φ) = aI : a C, a = 0 . { ∈ 6 } If one identiﬁes two matrices A, B GL(2, C) if and only if A = qB for some ∈ q C, then one obtains the group GL(2, C)/ker(Φ). This group is isomorphic ∈ to Aut(Cˆ). With

SL(2, C) = A GL(2, C): det(A) = 1 { ∈ } one denotes the special linear group of all complex 2 2 matrices with deter- minant 1. Then ×

SL(2, ) Aut(ˆ) Φ : C C (3.6) S A 7→ φ (z) → A is an epimorphism and

ker(Φ ) = I, I . S { − } Therefore, Aut(Cˆ) is isomorphic to

SL(2, C)/ I, I . { − } 3.4 A Remarkable Geometric Property of Linear Fractional Transformations Let denote any circle or straight line in the z–plane. We claim that, if C φA(z) is any linear fractional transformation, then the image of under the transformation C

z φ (z) = w → A is a circle or a straight line in the w–plane. Since φA is a composition of transformations of the form

17 1 z az, z z + b, z , → → → z and since the statement is easily shown for transformations z az and z z + b, we consider the transformation → → 1 z = x + iy = w = u + iv . → z

3.4.1 Analytical Description of Circles The circle = (z , r) centered at z with radius r > 0 has the equation C C 0 0 z z 2 = r2 , | − 0| or,

zz¯ zz¯ zz¯ + A = 0 − 0 − 0 with

A = z z¯ r2, r2 = z z¯ A. 0 0 − 0 0 − One obtains:

Lemma 3.3 Let z0 C and let A R. Then the equation ∈ ∈ zz¯ zz¯ zz¯ + A = 0 − 0 − 0 describes a circle if and only if

z z¯ A > 0 . (3.7) 0 0 − Assuming (3.7) to hold, the circle is centered at z0 and has radius r = √z0z¯0 A. The circle passes through the point z = 0 if and only if A = 0. −

3.4.2 Circles under Reciprocation

Let = (z0, r) denote the circle centered at z0 with radius r > 0. Set A = C 2C z0z¯0 r . Case− 1: The circle does not pass through z = 0, i.e., A = 0. We apply the transformation 6 1 T (z) = = w z to the points of . The points w T ( ) satisfy C ∈ C 1 z¯ z 0 0 + A = 0 , ww¯ − w − w¯ or,

1 w¯z¯ wz + Aww¯ = 0 , − 0 − 0 18 or, z z¯ 1 ww¯ w 0 w¯ 0 + = 0 . − A − A A The last equation has the form

ww¯ ww¯ ww¯ + B = 0 − 0 − 0 with z¯ 1 w = 0 ,B = . 0 A A We have

z z¯ 1 w w¯ B = 0 0 0 0 − A2 − A 1 = (z z¯ A) A2 0 0 − > 0 .

Using the previous lemma, we see that T ( ) is the circle centered at C z¯ w = 0 0 A with radius r R = . A | | Case 2: The circle passes through z = 0, i.e., A = 0. We apply the transformation 1 T (z) = = w z to the points of 0 . The points w = T (z) satisfy C\{ } 1 z¯ z 0 0 = 0 , ww¯ − w − w¯ or,

1 w¯z¯ wz = 0 . − 0 − 0 If

w = x + iy, z0 = a + ib , (with real x, y, a, b) then the above equation reads

1 (x iy)(a ib) (x + iy)(a + ib) = 0 , − − − − or,

19 1 2ax + 2by = 0 . − We see that T ( 0 ) is a straight line. If we let T (0) = , then T ( ) is a straight line togetherC\{ } with w = . ∞ C ∞ 3.4.3 Straight Lines under Reciprocation Let α, β, γ be real numbers, (α, β) = (0, 0). Then the equation 6 αx + βy + γ = 0 describes a straight line . Case 1: The line Ldoes not pass through z = 0, i.e., γ = 0. We rewrite the above equation as L 6

α β 1 + x + y = 0 , γ γ or,

1 2ax + 2by = 0 − with

α β 2a = , 2b = . − γ γ We can also write the equation for as L 1 (x iy)(a ib) (x + iy)(a + ib) = 0 . − − − − Setting

z0 = a + ib we see that the equation for is L 1 z¯z¯ zz = 0 with z = x + iy . − 0 − 0 It is then clear that T ( ) consists of the points w with L z¯ z 1 0 0 = 0 , − w¯ − w or,

ww¯ wz¯ wz¯ = 0 . − 0 − 0 Thus, T ( ) is a circle, centered at z0, that passed through w = 0. CaseL 2: The line passes through z = 0, i.e., γ = 0. We could proceed as above, but here it is simplerL to use the parametric description

iθ z = te , t R , ∈

20 of . The points of T ( ) are L L 1 w = e−iθ . t We see that T ( ) is another straight line through the origin. L 3.5 Example: The Cayley Transform in the Complex Plane Consider the M¨obius transforms

i z z + i 1 i F (z) = − = − ,A = − , i + z z + i 1 i and

1 w iw + i i i G(w) = i − = − ,B = − . 1 + w w + 1 1 1 The map F is called the Cayley transform. We have det(A) = 2i and − 1 i i 1 i i A−1 = − = − . 2i 1 1 2i 1 1 − − − We see that B is a scalar multiple of A−1 and, therefore, G is the inverse of F if F and G are considered as functions on Cˆ. Here we use the extensions

F ( i) = ,F ( ) = 1,G( 1) = ,G( ) = i . − ∞ ∞ − − ∞ ∞ − Let

H = z = x + iy : y > 0 { } denote the open upper half–plane and recall that D denotes the open unit circle.

Lemma 3.4 The transformation i z F (z) = − i + z maps the open upper half–plane H onto the open unit circle D. The map 1 w G(w) = i − 1 + w is the back transform. The boundary of H is the real line,

∂H = z = x : x R . { ∈ } The map F maps ∂H = R bijectively onto ∂D 1 . \ {− }

21 Proof: If z H then the distance of z from i is strictly smaller than the ∈ distance of z from i. Therefore, F (z) < 1 for z H. Similarly, if z = x is real then F (z) =− 1, and if z = |x + iy| with y < ∈0 then F (z) > 1. Since | | | | F ( ) = 1 and F ( i) = and since F : Cˆ Cˆ is a bijection, it follows that ∞ − − ∞ → F : H D is a bijection. Since G inverts F as a function from Cˆ to Cˆ, G also → inverts F : H D. The behavior of F and G on the boundaries of H and D is then clear. → Remark: Let H¯ = H ∂H denote the closed upper half–plane. We have ∪ seen that F maps H¯ bijectively onto D¯ 1 and the inverse map is G. It is \ {− } clear that one cannot obtain a continuous bijection between H¯ and D¯ since D¯ is compact and H¯ is not compact. Any continuous image of D¯ is also compact. However, if we use the topology of Cˆ, then H¯ is a compact compact ∪ {∞} subset of Cˆ and

F : H¯ D¯ ∪ {∞} → is an isomorphism between compact sets.

3.6 The Cayley Transform of a Matrix We have seen that i z F : C i C 1 ,F (z) = − , \ {− } → \ {− } i + z is 1-1 and onto with inverse 1 w G : C 1 C i ,G(w) = i − . \ {− } → \ {− } 1 + w We have also seen that

F (R) = ∂D 1 . (3.8) \ {− } × Deﬁnition: Let A Cn n and assume that i / σ(A). Then ∈ − ∈ F (A) = (iI A)(iI + A)−1 − is called the Cayley transform of A. In analogy to (3.8) we have the following result:

Lemma 3.5 If A = A∗ then V := F (A) is unitary and V = F (A) = I. 6 − Proof: We have

V ∗ = ( iI + A)−1( iI A) = (iI + A)(iI A)−1 , − − − − thus

V ∗V = (iI + A)(iI + A)−1 = I. Suppose that

22 V = F (A) = (iI A)(iI + A)−1 = I. − − Then one obtains that

iI A = iI A, − − − a contradiction. Remark: The Cayley transform

V = F (A) = (iI A)(iI + A)−1 − of an operator A plays a role in functional analysis. One can use it to transform even unbounded Hermitian operators A to unitary operators V = F (A). Unitary operators are obviously bounded, and it is often easier to study them. Then the back transform

V = F (A) A = i(I V )(I + V )−1 → − can give information about the unbounded Hermitian operator A. This is used to study spectral theory of unbounded operators.

23 4 The Riemann Mapping Theorem

4.1 Statement and Overview of Proof

Recall that D denotes the open unit disk. Recall that we know the group Aut(D): First, if c < 1 then the M¨obiustransformation | | z c φ (z) = − , z < 1 , c 1 cz¯ | | − is an automorphism of D with inverse φ−c. Second, if α = 1 then the rotation | | R (z) = αz, z < 1 , α | | is an automorphism of D with inverse Rα¯. Third, if f Aut(D) is arbitrary and b = f(0) then ∈

f = φ− R b ◦ α for some α with α = 1. The Riemann| mapping| theorem is the following remarkable result:

Theorem 4.1 Let U C be open and simply connected and let U = C. Then ⊂ 6 there exists a biholomorphic map f : U D. → A sharper result, containing a uniqueness statement, is formulated in Theo- rem 4.2. If Theorem 4.1 is known, then we can prove Theorem 4.2 rather easily using our knowledge of Aut(D).

Theorem 4.2 Let U C be open and simply connected and let U = C. Let P U be any ﬁxed point⊂ in U. Then there exists a unique biholomorphic6 map ∈ f : U D satisfying → f(P ) = 0, f 0(P ) > 0 . (4.1)

Proof: Existence of f: By Theorem 4.1 there exists a biholomorphic F : U D. Set c = F (P ) and recall that → z c φc(z) = − , z D , 1 cz¯ ∈ − is an automorphism of D with φc(c) = 0. Set

g = φ F. c ◦ Then g : U D is biholomorphic and g(P ) = φc(F (P )) = φc(c) = 0. Write → 0 iγ g (0) = re where r > 0 and γ R , ∈ and set

α = e−iγ .

24 Deﬁne f = R g, thus α ◦ f(z) = αg(z), f 0(z) = αg0(z) . We have

f 0(0) = αg0(0) = ααr¯ = r > 0 . The function

f = Rα φc F : U D ◦ ◦ → is biholomorphic and satisﬁes (4.1). Uniqueness of f: Suppose that f1 and f2 satisfy the conditions of the −1 theorem and set h = f1 f2 . Then h Aut(D) and h(0) = 0. It follows that h is a rotation, h(z) αz,◦ α = 1. From∈ ≡ | | h f = f , f (P ) = f (P ) = 0 , ◦ 2 1 1 2 we obtain

0 0 0 h (0)f2(P ) = f1(P ) . 0 0 Since fj(P ) > 0 and α = 1 it follows that h (0) = α = 1, yielding f1 = f2. Overview of the| Proof| of Theorem 4.1: a) Fix any point P U and let ∈ the set consist of all functions f : U D that have the following properties: 1) fF H(U); → 2) f ∈is 1 1; 3) f(P ) =− 0. We will prove by an explicit construction that is not empty. b) If f then f 0(P ) > 0 since, by assumption,F f is 1 1. Using Cauchy’s inequality,∈ it F is easy| to see| that −

s := sup f 0(P ) : f (4.2) {| | ∈ F} is finite. c) Using Montel’s theorem (see Section 4.3) one obtains the existence of a function f H(U) with ∈ 0 f(U) D, f(P ) = 0, s = f (P ) . ⊂ | | The function f is constructed as the normal limit of a sequence of functions in . F d) Using Hurwitz’s theorem, it follows that the function f obtained in c) is 1 1. Therefore, f . In other words, the supremum defining number s in (4.2)− is in fact a maximum.∈ F e) An argument using the automorphisms of D shows that f maps U onto 0 D. If f would not be onto, then one could construct a map g with g (P ) > f 0(P ) = s, in contradiction to the definition of s. ∈ F | | | |

25 4.2 The Theorem of Arzela–Ascoli The Theorem of Arzela–Ascoli is an important result of real analysis. Montel’s Theorem, which we consider in the next section, is a complex version of the Theorem of Arzela–Ascoli. Let Ω Rs be a compact set. With C(Ω) we denote the linear space of all ⊂ continuous functions u :Ω R. Let →

s Theorem 4.3 Let Ω be a compact subset of R . Let un C(Ω) denote a sequence of functions with the following two properties: ∈ 1) For every ε > 0 there is a δ > 0 so that for all n N: ∈ u (x) u (y) < ε if x y < δ, x, y Ω . | n − n | | − | ∈ (It is important that δ does not depend on n. This property is called equicontinuity of the sequence un.) 2) There is a constant C > 0 with

un ∞ C for all n N . | | ≤ ∈ (This property is called uniform boundedness of the sequence un.)

Under these assumptions, there is a subsequence unj and a function u C(Ω) with ∈

u u ∞ 0 as n . | nj − | → j → ∞ Proof: 1. Using a diagonal sequence argument, we will show that the sequence un has a subsequence uj which is a Cauchy sequence in the Banach space k C(Ω), ∞ . | · | 2. A set Fε Ω is called an ε–net for Ω if for every y Ω there is x Fε with x y < ε⊂. Since Ω is bounded, it is easy to show that∈ for every ε∈ > 0 there| exists− | a finite ε–net for Ω. To see this, cover the set Ω with finitely many boxes of diameter ε and, if a box has an intersection with Ω, choose one point in the intersection as an element in the ε–net. 1 For every n N let F1/n denote a finite –net for Ω and let ∈ n F = F . ∪n 1/n We enumerate the set F by first listing the points in F1, then the points in F1/2, then the points in F1/3 etc:

F = x , x ,... . { 1 2 }

26 We will need the following property of F , which is more special than density of F in Ω: If δ > 0 is given, then there is a ﬁnite integer K = K(δ) so that the points

x , . . . , x F 1 K ∈ form a δ–net for Ω. This is clear from the construction of F . 3. In the following, N1, N2 etc. denote inﬁnite subsets of N. Since un(x1) is bounded, there is N1 so that

un(x1), n N1 , ∈ is a convergent sequence of real numbers,

un(x1) a1, n N1 . → ∈ Since un(x2), n N1, is bounded, there is N2 N1 so that ∈ ⊂

un(x2), n N2 , ∈ is a convergent sequence of real numbers,

un(x2) a2, n N2 . → ∈ Repeat this construction inductively. For every k N obtain a set Nk with ∈

Nk Nk−1 ... N1 N ⊂ ⊂ ⊂ ⊂ so that

un(xk), n Nk , ∈ is a convergent sequence of real numbers,

un(xk) ak, n Nk . → ∈ We now use a diagonal sequence argument: Let

(k) (k) Nk = n < n < . . . { 1 2 } and consider the diagonal sequence

(k) jk := nk , k = 1, 2,...

(k) The diagonal sequence nk :

(1) (1) (1) N1 : n1 n2 n3 ... (2) (2) (2) N2 : n1 n2 n3 ... (3) (3) (3) N3 : n1 n2 n3 ... (4) N4 : n ········· 4

27 4. We claim that

u (x ) a as j , jk ν → ν k → ∞ for every ﬁxed x F . This is clear since the tail of the sequence ν ∈

j1 < j2 < . . . < jν < jν+1 < . . . is a subsequence of Nν and

un(xν) aν, n Nν . → ∈

To summarize, we have shown that the sequence ujk (xν) (where j1 < j2 < j3 < ...) converges for every x F as j . In particular, given any ε > 0 there ν ∈ k → ∞ is N(xν, ε) with 1 u (x ) u (x ) < ε for k, l N(x , ε) . | jk ν − jl ν | 3 ≥ ν

If ε > 0 is given, then there are ﬁnitely many points x1, . . . , xK F so that for every y Ω there is x with 1 ν K and y x < δ. If we∈ set ∈ ν ≤ ≤ | − ν|

N(ε) = max N(xν, ε) 1≤ν≤K then we have for k, l N(ε): ≥

max uj (y) uj (y) < ε . y | k − l | This proves the theorem. Extension: It is clear that the theorem generalizes to sequences of functions k fn :Ω R for any ﬁnite k. →

28 4.3 Montel’s Theorem Montel’s theorem is a result about families of functions f H(U) which are uniformly bounded on compact subsets of U. ∈ We will use the following notation: If K C is a compact set and g C(K), then ⊂ ∈

g = max g(z) : z K | |K {| | ∈ } denotes the maximum norm if g. We recall:

Theorem 4.4 (Morera) Let U C denote an open set and let f C(U). If ⊂ ∈ Z f(z) dz = 0 Γ for every closed triangle Γ in U then f is holomorphic on U.

Morera’s Theorem is often used as follows: Let fn H(U) and let f C(U). If f f 0 as n for every compact subset∈K of U, then f ∈H(U). | n − |K → → ∞ ∈ Theorem 4.5 (Montel) Let U C denote an open set and let denote a set of functions f H(U), i.e., ⊂H(U). Assume that for every compactF subset ∈ F ⊂ K of U there is a constant CK with

f(z) C for all z K and for all f . | | ≤ K ∈ ∈ F Then, if f is a sequence, there is a subsequence f and a function f n ∈ F nj ∈ H(U) so that fnj converges normally to f. (The limit function f may or may not belong to the set .) F We will only need the following simpler version of the theorem.

Theorem 4.6 (Montel’s theorem for a sequence, simple version) Let U C be ⊂ an open set and let fn H(U) be a bounded sequence, i.e., there is a constant M with ∈

fn(z) M for all z U and for all n N . | | ≤ ∈ ∈

Then there is a function f H(U) and a subsequence fnj of fn with fnj f normally in U. This means:∈ If K is any compact subset of U and if ε > 0,→ then there exists an integer J(K, ε) with

max fnj (z) f(z) < ε for j J(K, ε) . z∈K | − | ≥ Essentially, Montel’s theorem follows from the Cauchy inequalities and the Arzela–Ascoli theorem. We now give a detailed proof. See Figure 4.1 for sets referred to in the proof.

29 v

δ L U U c

Figure 4.1: Sets in the proof of Montel’s theorem

1. First, let K be any compact subset of U. We assume that U c = C U is \ not empty. (If U = C then every fn is constant and the claim follows directly from Bolzano–Weierstrass.) Set

δ = inf min z w z∈U c w∈K | − | = min min z w . z∈U c w∈K | − | (Here the inﬁmum equals the minimum since U c is closed, and large z U c play no role in determining δ.) We have δ > 0. ∈ 2. We want to show that the sequence fn is equicontinuous on K. For z C let ∈

dist(z, K) = min z w . w∈K | − | Deﬁne

L = z C : dist(z, K) δ/4 . { ∈ ≤ } Clearly, K L U. If v L then D¯(v, δ/2) U. By Cauchy’s inequality we have for all⊂v ⊂L: ∈ ⊂ ∈ M f 0 (v) =: C . | n | ≤ δ/2 1 If z, w K and z w δ/4 then the line segment ∈ | − | ≤ γ(t) = tz + (1 t)w, 0 t 1 , − ≤ ≤ lies in L since γ(t) w δ/4. We then have for z, w K with z w δ/4: | − | ≤ ∈ | − | ≤

f (z) f (w) = f (γ(1)) f (γ(0)) n − n n − n Z 1 0 0 = fn(γ(t))γ (t) dt 0 Z 1 0 = fn(γ(t)) dt (z w) 0 −

30 Therefore,

δ f (z) f (w) C z w for z, w K if z w . | n − n | ≤ 1| − | ∈ | − | ≤ 4

It is clear that this estimate implies equicontinuity of the sequence fn on K. Thus we can apply the Arzela–Ascoli Theorem to the sequence fn on any compact subset K of U. 3. Given K U as above, there exists a subsequence f and a function ⊂ nj f C(K) with max ∈ f (x) f(x) 0. ∈ x K | nj − | → 4. We now choose a sequence of compact sets Kν in U as follows. For ν = 1, 2,... let n o K = z U : dist(z, U c) 1/ν D¯(0, ν) . ν ∈ ≥ ∩ It is clear that

K K ... U and K = U. 1 ⊂ 2 ⊂ ⊂ ∪ν ν Furthermore, if K is any compact subset of U then there is a suﬃciently large ν with

K K . ⊂ ν The sequence fn has a subsequence fn, n N1, which converges uniformly on K1 to some f C(K ). Then consider the sequence∈ f , n N , on K K . There ∈ 1 n ∈ 1 2 ⊃ 1 is a subsequence fn, n N2 N1, converging uniformly to some g C(K2). ∈ ⊂ ∈ However, g equals f on K1, and we may denote g by f. We repeat the argument and obtain: For every ν there is Nν Nν−1 so that fn, n Nν, converges uniformly on K to f C(K ). Since ⊂ ∈ ν ∈ ν K = U ∪ν ν this process deﬁnes a continuous function f on U. 5. If

(k) (k) Nk = n < n < . . . { 1 1 } then let

(k) jk = nk , k = 1, 2,... In this way we obtain the subsequence

fjk of fn. A tail of fjk is a subsequence of fn, n Nν. Therefore, for every ﬁxed ν, ∈ the subsequence fjk converges uniformly on Kν to f. 6. Finally, if K is an arbitrary compact subset of U then K K for some ⊂ ν large ν. Therefore, fjk converges uniformly on K to f. The limit function f is holomorphic on U by Morera’s theorem. 31 4.4 Auxiliary Results on Logarithms and Square Roots

Lemma 4.1 Let U C be open and simply connected; let f H(U). If f(z) = 0 for all z U⊂then there exists a function h H(U) with∈ 6 ∈ ∈ eh(z) = f(z), z U. ∈ Proof: Motivation: Suppose that eh(z) = f(z). Then we have

f 0(z) = h0(z)f(z) , thus

h0(z) = f 0(z)/f(z) . This motivates to construct h(z) as a function with h0 = f 0/f. Fix z0 U and let Γz denote a curve in U from z0 to z. Fix c C with ∈ ∈ c f(z0) = e and deﬁne

Z f 0(w) h(z) = c + dw, z U. Γz f(w) ∈

0 0 h(z0) c Then we have h (z) = f (z)/f(z) and e = e = f(z0). Consider the function

g(z) = f(z)e−h(z), z U. ∈ We have

g(z0) = 1 and

g0(z) = f 0(z)e−h(z) f(z)e−h(z) h0(z) − = e−h(z) f 0(z) f(z)h0(z) − = 0

It follows that g(z) 1, proving the lemma. ≡ Remark: Since eh(z) = f(z) we may consider the function h(z) as a complex logarithm of f(z),

h(z) = log(f(z)) , where log(w) is any inverse of the exponential function deﬁned on the range of f(z).

32 y

D(b, r)

b • q2 •

x D( b, r) − q1 • •b −

Figure 4.2: The disks D( b, r) and the points q ± 1,2

Lemma 4.2 Let U C be open and simply connected; let f H(U). If f(z) = 0 for all z U⊂then there exists a function g H(U) with∈ 6 ∈ ∈ f(z) = (g(z))2, z U. ∈ Proof: Using the previous lemma, we can write

f(z) = eh(z) and deﬁne

g(z) = eh(z)/2 .

4.5 Construction of a Map in F Lemma 4.3 Let U C be open and simply connected. Further, let U = C. ⊂ 6 Let P U be arbitrary, but ﬁxed. Then there exists f H(U) with f(U) D, f is 1–1,∈ and f(P ) = 0. ∈ ⊂

Proof: If U is bounded then f(z) can be taken as a function of the form f(z) = az + b. We treat this simple case ﬁrst in Case a. In Case b we treat the general case.

33 Case a) Let U be bounded. Choose R > 0 so that U D(P,R) and set ⊂ z P f(z) = − . R Clearly, f is 1 1 and f(P ) = 0. Also, if z U D(P,R) then z P < R, − ∈ ⊂ | − | thus f(z) < 1. This shows that f maps U into D. | | Case b) Let U be an arbitrary open, simply connected subset of C, but U = C. Let Q C U and deﬁne 6 ∈ \ φ(z) = z Q, z U. − ∈ Then φ H(U) is 1 1 and φ(z) = 0 for all z U. By Lemma 4.2 there is h H(U∈) with − 6 ∈ ∈ h2(z) = φ(z), z U. ∈ We will show that there is an open disk D( b, r) which is contained in the complement of h(U): − D( b, r) C h(U) . − ⊂ \ Once this is done, the construction of f will be easy. Clearly, if z , z U are two distinct points, then 1 2 ∈ h(z ) = h(z ) for z = z (4.3) 1 6 − 2 1 6 2 since otherwise

2 2 h (z1) = h (z2) , contradicting φ(z ) = φ(z ). Since h H(U) is not constant, the set h(U) is 1 6 2 ∈ open. There exists b C and r > 0 with ∈ D(b, r) h(U) . ⊂ We may assume that b = 0 and 0 < r < b . We claim that the open disk 6 | | D( b, r) lies in C h(U). Suppose this is not the case. Then set − \ q := h(z ) D( b, r) h(U), z U. 1 1 ∈ − ∩ 1 ∈ Set q := q . From 2 − 1 q + b < r | 1 | obtain that

q b < r , | 2 − | i.e., q2 D(b, r) h(U). Therefore, q2 = h(z2) for some z2 U. Because of 0 < r <∈b the two⊂ disks ∈ | |

34 D( b, r) and D(b, r) − are disjoint. This shows that

q = q = q . 2 − 1 6 1 We then have

h(z ) = q = q = h(z ) , 1 1 6 2 2 thus z = z , but 1 6 2 h(z ) = q = q = h(z ) , 1 1 − 2 − 2 in contradiction to (4.3). This contradiction proves that D( b, r) C h(U) . − ⊂ \ In other words,

h(z) + b r for all z U. | | ≥ ∈ Define r f (z) = , z U. 1 2(h(z) + b) ∈ Then f H(U) is 1 1 and f (z) 1 < 1. Set a := f (P ) and recall that 1 ∈ − | 1 | ≤ 2 1 z a φ (z) = − a 1 az¯ − is an automorphisms of D with φa(a) = 0 . The function f = φa f1 satisfies f(P ) = 0. To summarize, we have constructed a function f H(◦U) which is ∈ 1 1 and satisfies f(U) D and f(P ) = 0. − ⊂ 4.6 A Bound of f 0(P ) for all f | | ∈ F Let f , i.e., f H(U), f(U) D, f(P ) = 0, and f is one–to–one. There is r > 0 with∈ F D¯(P, r)∈ U. Let γ(t)⊂ = P + reit for 0 t 2π. We know that for z D(P, r), ⊂ ≤ ≤ ∈ 1 Z f(w) f(z) = dw . 2πi w z γ − Differentiation yields

1 Z f(w) f 0(z) = dw . 2πi (w z)2 γ − For z = P one obtains the Cauchy inequality

35 1 Z 2π f(P + reit) f 0(P ) r dt | 2 | | | ≤ 2π 0 r 1 . ≤ r Here we have used the bound f(w) 1, which follows from f(U) D. We have shown: | | ≤ ⊂

Lemma 4.4 The number s deﬁned by

s = sup f 0(P ) : f {| | ∈ F} satisﬁes 0 < s 1 if D¯(P, r) U. ≤ r ⊂ 4.7 Application of the Theorems of Montel and Hurwitz There is a sequence of functions f with n ∈ F 1 s f 0 (P ) s , n = 1, 2,... ≥ | n | ≥ − n Since fn(U) D the functions fn are uniformly bounded on U. By Montel’s ⊂ theorem (simple version), there is a subsequence fnj and a function f H(U) so ∈0 that fnj converges normally to f on U. In particular, f(P ) = 0 and f (P ) = s. Also, f(z) 1 for all z U. If f(z) = 1 for some z U then f |is constant,| | | ≤ 0 ∈ | | ∈ contradicting f (P ) = s > 0. Therefore, f(U) D. We now claim| that| f is 1 1. To prove this⊂ we will use Hurwitz Theorem: −

Theorem 4.7 (Hurwitz) Let V C be open and connected. Let hj, h H(V ) ⊂ ∈ and assume that the sequence hj(z) converges normally to h(z) in V . If

h (z) = 0 for all z V and for all j = 1, 2,... j 6 ∈ then either h(z) 0 or ≡ h(z) = 0 for all z V. 6 ∈

Proof: We ﬁrst recall that normal convergence in V of the sequence hj(z) to 0 0 h(z) implies normal convergence in V of hj(z) to h (z). Suppose that P V is an isolated zero of h(z). There exists ε > 0 so that ∈ h(z) = 0 for 0 < z P ε . 6 | − | ≤ If Γ denotes the boundary curve of the disk D(P, ε) then the positive integer

1 Z h0(z) dz 2πi Γ h(z) is the order of the zero P of h(z). On the other hand,

36 1 Z h0 (z) j dz = 0 for all j = 1, 2,... 2πi Γ hj(z) One obtains a contradiction as j . → ∞ Recall that fn(z) denotes a sequence of 1–1 functions with

fn H(U), fn(U) D, fn(P ) = 0 ∈ ⊂ and 1 s f 0 (P ) s , n = 1, 2,... ≥ | n | ≥ − n

By Montel’s Theorem there exists a subsequence fnj and f H(U) so that fnj converges normally to f in U. One obtains that f 0(P ) = s∈ > 0. We claim that f is 1–1 on U. Suppose not.| Then| there are two points z , z U with 1 2 ∈ f(z ) = f(z ), z = z . 1 2 1 6 2 Consider the sequence of functions

h (z) = f (z) f (z ), z V = U z . j nj − nj 2 ∈ \{ 2} The functions h H(V ) converge normally in V to f(z) f(z ). Also, the j ∈ − 2 functions hj(z) have no zero in V since they are 1–1 on U. By Hurwitz’s theorem, the limit function f(z) f(z2) either is identically zero on V or has no zero in V . Since f(z) is not constant,− we conclude that f(z) f(z ) has no − 2 zero in V , contradicting the assumption f(z1) = f(z2).

4.8 Proof That f is Onto We have shown that there is a function f with ∈ F s = f 0(P ) g0(P ) for all g , s > 0 . (4.4) | | ≥ | | ∈ F By deﬁnition of the set we have F f H(U), f(U) D, f is one-to-one, f(P ) = 0 . ∈ ⊂ We claim that if f satisﬁes (4.4) then f : U D is onto. This claim then completes the proof∈ F of the Riemann Mapping Theorem.→ We will show:

Lemma 4.5 Let f and assume that f : U D is not onto D. Then we can construct g ∈with F → ∈ F g0(P ) > f 0(P ) . | | | |

37 Proof: With S : D D we denote the squaring function, i.e., S(v) = v2, v D. Also, we recall that→ for c < 1 the function ∈ | | w c φ (w) = − , w < 1 , c 1 cw¯ | | − is an automorphism of D. Since f : U D is not onto, we can choose → a D f(U) ∈ \ and consider

φ(z) = φ f(z) a ◦ f(z) a = − , z U. 1 af¯ (z) ∈ − Since a / f(U) we have φ(z) = 0 for all z U. Also, φ H(U) is 1 1 and ∈ 6 ∈ ∈ − φ(U) D. By Lemma 4.2 there exists a function ψ H(U) which is 1 1 and satisﬁes⊂ ∈ −

φ(z) = ψ(z)ψ(z) = S ψ(z) for all z U, ψ(U) D . ◦ ∈ ⊂ Deﬁne

g(z) = φ (ψ(z)), z U. ψ(P ) ∈ Then g H(U), g is 1 1, g(U) D, and g(P ) = 0. Therefore, g . Also, ∈ − ⊂ ∈ F if we abbreviate b = ψ(P ), then g = φ−b ψ, thus ψ = φb g. We have − ◦ ◦

f = φ− φ a ◦ = φ− S ψ a ◦ ◦ = φ− S φ g a ◦ ◦ b ◦ We now set

h := φ− S φ . a ◦ ◦ b Then the above equation for f says that

f(z) = h(g(z)), z U. ∈ We have h H(D), h(D) D, and ∈ ⊂ 0 = f(P ) = h(g(P )) = h(0) . By Schwarz Lemma, we conclude that

h0(0) < 1 | |

38 Im z Im z g

z P 0 1 Re z 0 Re z P¯

Figure 4.3: Biholomorphic mapping from upper half–plane onto unit disk

unless h is a rotation, h = Rα. However, if h = Rα, then the squaring function S would be an automorphism of D, which is obviously not the case. It follows that h0(0) < 1. Since | | f(z) = h(g(z)), z U, ∈ we have

f 0(P ) = h0(0)g0(P ) , thus

f 0(P ) < g0(P ) . | | | | This proves the lemma and completes the proof of Riemann’s Mapping Theo- rem. . 4.9 Examples of Biholomorphic Mappings

Recall that D denotes the open unit disk and H denotes the open upper half– plane. Example 1: Let P = a + ib, b > 0, be a point in H. Then P¯ = a ib is the reﬂection of P w.r.t. the real axis. The function −

z P g(z) = − , z = P,¯ z P¯ 6 − maps H onto D, maps the real line onto ∂D 1 , and maps C H¯ onto C D¯. In \{ } \ \ particular, g is a biholomorphic mapping from H onto D with g(P ) = 0. If you 0 want to ﬁnd a biholomorphic mapping f : H D with f(P ) = 0 and f (P ) > 0 then set →

f(z) = eiθg(z) and determine θ so that f 0(P ) > 0.

39 Im z Im z g

z P 0 1 Re z 0 Re z P¯

Figure 4.4: Biholomorphic mapping from upper half–plane onto unit disk

Example 2: Let H denote an open half–plane in C. The boundary of H is a straight line, L = ∂H. Let P H and let Q Hc denote the reﬂection of P w.r.t. L. Then ∈ ∈

z P g(z) = − , z H, z Q ∈ − maps H biholomorphically onto D.

40 5 An Introduction to Schwarz–Christoﬀel Formulas

Schwarz–Christoffel formulas give biholomorphic maps w = w(z) from the open upper half–plane H onto regions enclosed by polygons. The real axis is mapped to the polygon. The maps w(z) are given in terms of integrals. Similar integrals also appear in the theory of elliptic functions. We treat an example of a Schwarz–Christoffel map, which is Example 8, p. 350, from [Hahn, Epstein]. A more comprehensive treatment of Schwarz– Christoffel maps is given in [Hille] and in [Stein, Sharkarchi]. A clear understanding of general power functions z zα is helpful and will be reviewed first. →

5.1 General Power Functions Let

U0 = C z = iy : y 0 \{ ≤ } denote a slit plane. Precisely, U0 is the complex plane C with the negative imaginary axis, including z = 0, removed. On U0 we deﬁne the logarithm log0(z) as a holomorphic function as follows: If z U0 then there is a unique argument ∈ π 3π arg (z) = θ with < θ < 0 − 2 2 so that

z = z eiθ = eln |z|+iθ . | | We then set

log (z) = ln z + iθ, z U , 0 | | ∈ 0 and call log0(z) the principle branch of the logarithm on U0. This function R r log0(z) is holomorphic on U0 and extends the natural logarithm ln r = 1 dx/x, deﬁned for positive real numbers r, to the slit plane U0.

General Powers on U0. Let z U0 and let α C. We then set ∈ ∈ π 3π zα = eα log0(z) = eα ln |z|eiαθ where < θ < . (5.1) − 2 2 If α is real and z is real and positive then the definition of zα by (5.1) agrees with the usual definition of zα in real analysis and zα > 0. For fixed α R, the function z zα given in (5.1) extends the real analysis function xα,∈ defined for real positive→ x, to the slit plane U and this function z zα is holomorphic 0 → on U0. We call the function given in (5.1) the principle branch of the power α function z on U0.

Lemma 5.1 Let z U0 and let α R. Then we have ∈ ∈ zα = z α . | | | | 41 Proof: From

z = z eiθ = eln |z|eiθ | | we obtain

zα = eα ln |z|eiαθ , thus

zα = eα ln |z| = z α . | | | |

The following simple examples show that one has to be careful when com- puting powers of powers. 1) Since 1 = eπi we have − ( 1)(−1) = (eπi)−1 = e−πi = 1 = eπi . − − 2) Therefore,

2/3 ( 1)(−1) = (eπi)2/3 = e2πi/3 . − 3) On the other hand

( 1)−2/3 = e−2πi/3 . − This shows that

2/3 ( 1)(−1) = ( 1)−2/3 . − 6 − We see that, in general,

(z−1)α = z−α . 6 5.2 An Example of a Schwarz–Christoﬀel Map The following example is Example 8, p. 350, from [Hahn, Epstein]. It gives a good introduction to Schwarz–Christoﬀel formulas. For real β let Lβ denote the half–line

L = z = β + iy : y 0 β { ≤ } and let

V := C L−1 L0 L1 , \{ ∪ ∪ } i.e., V is the plane C with three half–lines removed. Consider the holomorphic function g : V C given by → g(ζ) = e2πi/3 (ζ + 1)−5/6 ζ−1/2 (ζ 1)−2/3, ζ V. · · · − ∈

42 g(ζ) | |

1 0 1 ζ −

Figure 5.1: g(ζ) = ζ + 1 −5/6 ζ −1/2 1 ζ −2/3 | | | | | | | − |

Figure 5.1 shows a rough graph of the function

g(ζ) = ζ + 1 −5/6 ζ −1/2 ζ 1 −2/3 | | | | · | | · | − | for real values of ζ. The domain of deﬁnition of g is the plane C with three half–lines removed. By removing the three half–lines, we can use the principle branches of the power functions on U0 for the factors of g(ζ). It is then clear that g H(V ). For the following, note that the domain V is simply connected.∈ If z V ∈ then let Γz denote a curve in V from 0 to z and set Z w(z) = g(ζ) dζ . Γz

Since the integral does not depend on the choice of the curve Γz, but only depends on z, we write

Z z Z = . 0 Γz

(Curve–independence of the integral holds though the starting point of Γz, the point z0 = 0, lies on the boundary of V , but does not lie inside V . Note, however, that the singularity ζ−1/2 of g(ζ) at ζ = 0 is integrable.) Note that the function w(z) is holomorphic on V and continuous on

V 1, 0, 1 . ∪ {− } We will now discuss the function w(z) for z on the real axis. We note that the singularities of the function g(ζ) are integrable and that g(ζ) decays like ζ −2 as ζ . Decay like the power ζ −2 occurs since | | | | | | → ∞ | | 5 1 2 + + = 2 . (5.2) 6 2 3 Considering the arguments of the factors of g(ζ), we have for real ζ:

43  e2πi/3 for ζ > 1   1 for 0 < ζ < 1 g(ζ) = g(ζ) (5.3) | | · e−πi/2 = i for 1 < ζ < 0  − −  e2πi/3 for ζ < 1 − To obtain (5.3) for 0 < ζ < 1 note that ζ + 1 > 0 and ζ > 0, but ζ 1 < 0. Therefore, −

ζ 1 = ζ 1 eπi and (ζ 1)−2/3 = ζ 1 e−2πi/3 . − | − | − | − | For 1 < ζ < 0 the equation (5.3) follows similarly. To obtain (5.3) for ζ < 1 note− that −

−5πi/6 −πi( 1 + 5 ) 2πi/3 ie = e 2 6 = e − since (see (5.2)) 1 5 2 = 2 . −2 − 6 3 − We now consider the straight lines Γ0,..., Γ3 on the real axis of the z–plane (see Figure 5.2) and the image of Γ under the map z w(z) = R z g(ζ) dζ. j → 0 Im z

Γ0 Γ1 Γ2 Γ3 1 0 1 Re z −

Figure 5.2: Path of integration for g(ζ)

We start the consideration by determining w(z) for z Γ . Note that ∈ 2 g(z) = g(z) for z Γ2. The point w(z) moves from w(0) = 0 to the ﬁnite positive| value| w(1) along∈ the positive real axis. See Figure 5.3. Next consider w(z) for z Γ1 and note that g(z) = i g(z) for z Γ1. We have ∈ − | | ∈

Z 0 Z 0 w(z) = g(ζ) dζ = i g(ζ) dζ for 1 z 0 . − z z | | − ≤ ≤ Therefore, if z moves from z = 0 to z = 1 along the real axis, then w(z) moves from w(0) = 0 to a ﬁnite value w( 1) up− the imaginary axis. Next consider z > 1. We have −

Z z Z z w(z) = w(1) + g(ζ) dζ = w(1) + e2πi/3 g(ζ) dζ . 1 1 | |

44 Im w

w( 1) − w(Γ0) w( ) ∞ w(Γ1) w(Γ3)

0 w(Γ2) Re w w(1)

R z Figure 5.3: w(z) = 0 g(ζ)dζ

The point w(z) moves along the hypotenuse of the triangle in Figure 5.3. Note that 2π/3 corresponds to 120 degrees. Therefore, the angle of the triangle at w(1) is 60 degrees. Since g(ζ) decays like ζ −2 as ζ , the point w(z) approaches a ﬁnite limit as z| | along the real| | axis: → ∞ → ∞ Z ∞ 2πi/3 a∞ := lim w(z) = w(1) + e g(ζ) dζ . z→∞ 1 | | Similarly, if z < 1, then − Z −1 Z −1 w(z) = w( 1) g(ζ) dζ = w( 1) e2πi/3 g(ζ) dζ . − − z − − z | | One obtains that w(z) moves along a direction parallel to the direction from 0 to e2πi/3. Again, as z along the real axis, the point w(z) approaches a ﬁnite− limit: → −∞

Z −1 2πi/3 a−∞ := lim w(z) = w( 1) e g(ζ) dζ . z→−∞ − − −∞ | | To complete the picture, it is important to understand that

a−∞ = a∞ . (5.4) To show this, consider the curve from z = R to z = R along the real axis completed by the semi–circle z(θ) = Reiθ, 0 −θ π. Denote this closed curve ≤ ≤ by R and denote the semicircle by ΓR. CBy Cauchy’s theorem,

Z R Z Z g(ζ) dζ + g(ζ) dζ = g(ζ) dζ = 0 . −R ΓR CR (To be completely correct, the straight line along the real axis should avoid the singularities of g(ζ) at ζ = 1, 0, 1 and one should consider a corresponding − curve R,ε. However, since the singularities of g are integrable, one can let ε 0C without diﬃculty.) → 45 Since g(ζ) C ζ −2 for large ζ , the integral along the semicircle is bounded by| C/R| ≤for large| | R. | | We have

Z R Z 0 w(R) = g(ζ) dζ and w( R) = g(ζ) dζ . 0 − − −R Therefore,

Z R Z w(R) w( R) = g(ζ) dζ = g(ζ) dζ . − − −R − ΓR It follows that

w(R) w( R) C/R , | − − | ≤ for large R. As R we obtain that a∞ = a−∞. These considerations→ ∞ prove that the image of the real axis under the map

Z z z w(z) = g(ζ) dζ → 0 is the triangle shown in Figure 5.3 where the point a∞ = a−∞ is removed from the hypotenuse. Another good example of Schwarz–Christoﬀel map is

Z z Z z w(z) = (1 ζ2)−1/2 dζ = (1 + ζ)−1/2(1 ζ)−1/2 dζ . 0 − 0 − Here one should note that

Z 1 dζ p = arcsin(1) arcsin( 1) = π . −1 1 ζ2 − − −

46 6 Meromorphic Functions with Prescribed Poles

6.1 Meromorphic Functions on C Recall that a function f is called meromorphic on C if there is a finite or denumerable set = b1, b2,... C so that D { } ⊂ (a) f is holomorphic on C ; \D (b) every point bk is a pole of f. The set is finite or∈ D denumerable and does not have an accumulation point D in C. (Suppose b is an accumulation point of . If b then the point b is an D ∈ D accumulation point of poles and cannot be a pole itself. If b C then f is holomorphic in b, and poles of f cannot accumulated at b.) ∈ \D If f has infinitely many poles bk then we will order them so that

b b ... and b as k . (6.1) | 1| ≤ | 2| ≤ | k| → ∞ → ∞ If f is meromorphic on C we write f M(C). ∈ 1 Example 1: f(z) = sin(πz) . It is easy to show that the pole set of f is the set Z of all integers j. Also, for the function h(z) = sin(πz) we have h0(j) = π cos(πj) = π( 1)j. This yields that − 1 ( 1)j Res , z = j = − . sin(πz) π

Example 2: Every rational function f(z) = p(z)/q(z) is meromorphic on C with ﬁnitely many poles. Example 3: The function f(z) = e1/z is not meromorphic since z = 0 is not a pole, but an essential singularity.

Let f M(C) and let bk be a pole of f. The Laurent expansion of f about ∈ bk has the form

f(z) = P ((z b )−1) + g (z), 0 < z b < r . k − k k | − k| k Here Pk(w) is a polynomial without constant term and gk H(D(bk, rk)). The function ∈

−1 Pk((z bk) ), z C bk , − ∈ \{ } is called the singular part of f at bk.

6.2 The Mittag–Leﬄer Theorem

Notation: If K C is compact and f C(K) then let ⊂ ∈ f = max f(z) : z K . | |K {| | ∈ }

Deﬁnition: Let U C be open and let fk H(U). The series ⊂ ∈

47 ∞ X fk(z) (6.2) k=1 converges normally in U if for every compact set K U the series ⊂ ∞ X f | k|K k=1 converges. Thus, normal convergence of a series (6.2) implies that the sequence of partial sums,

n X sn(z) = fk(z) k=1 converges locally uniformly on U to a unique limit, f H(U). As usual, the limit is denoted by ∈

∞ X f(z) = fk(z) . k=1 P∞ Also, note that in case of normal convergence of (6.2), the series k=1 fk(z) converges absolutely at every point z U. This implies that the terms of the series can be reordered without changing∈ the value of the limit. In other words, the ordering of the terms fk(z) is not important. We also recall that normal convergence of the series

∞ X f(z) = fk(z) k=1 in U implies normal convergence of

∞ 0 X 0 f (z) = fk(z) k=1 in U.

Theorem 6.1 (Mittag–Leﬄer) Let b1, b2,... denote an inﬁnite sequence of distinct points in C without accumulation point in C and set = b1, b2,... . D { } Further, let Pk(w) denote a sequence of polynomials without constant terms, i.e., Pk(0) = 0 for every k. Then there is a function f M(C) so that: (a) the pole set of f equals ; ∈ (b) the singular part of f atDb equals P ((z b )−1). k k − k

Remark: We cannot simply deﬁne

∞ X f(z) = P ((z b )−1) k − k k=1

48 since, in general, the series does not converge. For example, let bk = k and Pk(w) = w. Then the above series is

∞ X 1 . z k k=1 − This series diverges for every z. Proof of Theorem 6.1: We ﬁrst assume that z = 0 is not an element of . Then we may assume 0 < b1 b2 ... and bk as k . Let DP∞ | | ≤ | | ≤ | | → ∞ → ∞ k=1 ck denote a convergent series of positive numbers, ck > 0. For example, 2 ck = 1/k . The function

z P ((z b )−1) → k − k is singular only at z = bk. Thus we can write

∞ X P ((z b )−1) = a zj, z < b . k − k jk | | | k| j=0 Let

nk X j Qk(z) = ajkz j=0 where the integer nk is chosen so large that

−1 max Pk((z bk) ) Qk(z) ck . (6.3) |z|≤|bk|/2 − − ≤ We claim that the series

∞ X −1 f(z) := Pk((z bk) ) Qk(z) , z C =: U, − − ∈ \D k=1 converges normally in U and the limit f is a meromorphic function with the desired properties. Set

−1 fk(z) = Pk((z bk) ) Qk(z), z C bk . − − ∈ \{ } Let K U be any compact set. There exists R > 0 with ⊂ K D¯(0,R) . ⊂ Since b there exists a positive integer k (R) with | k| → ∞ 0 b 2R for k k (R) . | k| ≥ ≥ 0 If k k (R) then, by (6.3), ≥ 0

f f ¯ c . | k|K ≤ | k|D(0,R) ≤ k 49 Therefore,

∞ X f < . | k|K ∞ k=1 This proves normal convergence of the series

∞ X f(z) := fk(z) in U. k=1 We must show that the pole set of f is precisely and that D P ((z b )−1) k − k is the singular part of f at bk. To this end, ﬁx any R > 0. For z U D(0,R) we have ∈ ∩

k0(R) X X f(z) = fk(z) + fk(z)

k=1 k>k0(R) =: f I (z) + f II (z) .

Note that the decomposition f = f I + f II depends on R. The ﬁrst part, f I (z), is meromorphic in C. In fact, f I is a rational function. In the disk D(0,R) I the function f has poles precisely at those bk which lie in D(0,R) and at each b the singular part of f I is P ((z b )−1). The second function, f II (z), is k k − k holomorphic in D(0,R) because every function fk(z), k > k0(R), is holomorphic in D(0, 2R) and the series deﬁning f II converges uniformly on D¯(0,R). Since R > 0 was arbitrary, it is shown that f has the desired properties. So far we have assumed that b = 0. If b = 0 we just add the term 1 6 1

P1(1/z) to the constructed function. Remark: An important technical point of the proof is that the decomposition, f = f I + f II , depends on R, and we consider it in D(0,R). In this way, for each ﬁnite R, one only considers functions with ﬁnitely poles.

6.3 An Example

As before, let = b1, b2,... denote an inﬁnite set in C without accumulation D { } point in C. For simplicity, let 0 < b1 b2 ... | | ≤ | | ≤ Assume that Pk(w) = w for all k, i.e., we want to construct a meromorphic functions with simple poles and residue 1 at each bk. Assume that X 1 X 1 = , < , 2 bk ∞ bk ∞ k | | k | | where the sum is taken over all b = 0. k 6

50 For b = 0 we expand P ((z b )−1) about z = 0: k 6 k − k 1 1 P ((z b )−1) = = + (z) . k − k z b −b O − k k If we take 1 Qk(z) ≡ −bk we obtain

1 fk(z) = Qk(z) z bk − −1 1 = + z bk bk − z = (z b )b − k k If z R and b 2R then we have | | ≤ | k| ≥ 1 z b b z b , | − k| ≥ | k| − | | ≥ 2 | k| and, therefore,

2R f ¯ . | k|D(0,R) ≤ b 2 | k| It follows that the series

∞ X 1 1 f(z) = + , z C =: U, (6.4) z bk bk ∈ \D k=1 − converges normally in U to a meromorphic function f(z). The function f(z) has a simple pole at each bk with residue 1; the function f(z) has no other poles. We also know that we can diﬀerentiate (6.4) arbitrarily often term by term. Now assume that we want to construct a meromorphic function f(z) with poles precisely at the integers and residue 1 at each integer. According to the above, such a function is

1 X 1 1 f(z) = + + . z z n n n=06 −

51 7 Inﬁnite Products

7.1 Inﬁnite Products of Complex Numbers

Let q1, q2,... denote a sequence of complex numbers. We want to deﬁne what it means that the inﬁnite product

∞ Πj=1 qj converges and what the value of this product is if it converges. Naively, and in analogy to series, one considers the sequence of ﬁnite products,

n pn = Πj=1 qj , for n = 1, 2,... and calls the infinite product convergent to p if pn p. This leads to Definition 7.1 below. Most authors define convergence of→ an infinite product more restrictively. Using the more restrictive definition, the theorems about infinite products become simpler to formulate.

7.1.1 Two Deﬁnitions of Convergence Let

qj, j = 1, 2,... denote a sequence of complex numbers. We can form the ﬁnite products

n pn = Πj=1 qj for n = 1, 2,... In correspondence with infinite series, we define: Definition 7.1: The infinite product

∞ Πj=1 qj converges (in the simple sense) if the sequence of partial products pn converges as n . If p p as n then we write → ∞ n → → ∞ ∞ Πj=1 qj = p (in the simple sense) and call p the value of the infinite product. This definition is used, for example, in [Stein, Sharkarchi]. If one of the factors qj in the infinite product is zero then, using the above definition, the infinite product Πjqj always converges to zero (in the simple sense). Other authors prefer a more restrictive definition of convergence for infinite products. The following definition is used by [Greene, Krantz] and others. Definition 7.2: The infinite product

∞ Πj=1 qj converges if the following holds:

52 1) At most finitely many of the qj are equal to zero. 2) If N > 0 is so large that q = 0 for j > N then 0 j 6 0 N lim Πj=N +1 qj N→∞ 0 exists and is non–zero. If these two conditions are met, then the value of the infinite product is the number ΠN0 q lim ΠN q . j=1 j j=N0+1 j · N→∞ It is easy to show that the value of the product does not depend on the choice of N0. Here we assume that qj = 0 for j > N0. The difference between the two definitions6 is not profound. Like most authors, we prefer here to work with Definition 7.2. Then the formulation of many theorems becomes simpler. For example, later we will consider products of holomorphic functions,

p(z) = Π∞ (1 + a (z)), a H(U) . j=1 j j ∈ Then, using Deﬁnition 7.2, the product p(z) is zero at some z = z0 if and only if at least one factor 1 + aj(z) is zero at z = z0.

7.1.2 Examples Example 1: According to Deﬁnition 7.1, the inﬁnite product

Π∞ j = 0 1 2 ... j=0 · · · converges to zero (in the simple sense). According to Deﬁnition 7.2, this product diverges because

n Πj=1 j = n! does not converge to a finite limit as n . → ∞ Example 2: According to Definition 7.1, the infinite product

1j Π∞ j=1 2 converges to zero (in the simple sense). According to Deﬁnition 7.2, this product diverges because

1j Πn 0 as n , j=1 2 → → ∞ but no factor is zero. Example 3: Let 1 q = 1 , j = 2, 3,... j − j2 Consider the inﬁnite product

53 1 1 1 Π∞ 1 = 1 1 ... (7.1) j=2 − j2 − 22 − 32 Here the partial products are

1 1 1 p = (22 1) (32 1) ... (n2 1) n 22 − 32 − n2 − 1 = (1 3)(2 4) ... ((n 1)(n + 1)) n!n! · · − 1 1 = (n 1)! (n + 1)! n!n! − 2 1 n + 1 = 2 n 1 1 It follows that pn 2 . The infinite product (7.1) converges to 2 , using either definition. → Example 4: Let 1 q = 1 , j = 2, 3,... j − j Consider the infinite product 1 1 1 Π∞ 1 = 1 1 ... (7.2) j=2 − j − 2 − 3 Here the partial products are

1 1 1 p = 1 1 ... 1 n − 2 − 3 − n 1 2 3 n 1 = ... − 2 · 3 · 4 n 1 = . n

It follows that pn 0. According to Definition 7.1, the infinite product (7.2) converges to zero,→ in the simple sense. According to Definition 7.2, the infinite product diverges since p 0, but no factor is zero. n → Example 5: We claim that the infinite product 1 Π∞ 1 + j=1 j also diverges. We have

1 p = Πn 1 + n j=1 j j + 1 = Πn j=1 j 2 3 n + 1 = ... 1 · 2 · · n = n + 1

54 Since p as n the infinite product diverges. n → ∞ → ∞ Example 6: According to [Remmert, Classical Topics in Complex Function Theory], infinite products first appeared in 1579 in the work of F. Vieta. He gave the formula v s u s 2 r1 1 1r1 u1 1 1 1r1 = + t + + ... π 2 · 2 2 2 · 2 2 2 2 2 Let us understand the result. The infinite product can be written as

∞ Πj=2 qj with

r1 q = 2 2 and

r1 1 q = + q for j = 2, 3,... j+1 2 2 j

Thus, the qj obey the recursion

1 1 r1 q2 = + q for j 2 and q = = cos(π/4) . j+1 2 2 j ≥ 2 2 Recall that

cos 2α = cos2 α sin2 α − = 2 cos2 α 1 , − thus 1 1 cos2 α = + cos 2α . 2 2 We can also write this as

β 1 1 cos2 = + cos β . 2 2 2 Since

r1 q = = cos(π/22) 2 2 it follows that

j qj = cos(π/2 ), j = 2, 3,...

This formula for qj makes the determination of the inﬁnite product manageable.

55 Recall that

sin 2α = 2 sin α cos α , thus

1 sin 2α cos α = . 2 sin α We then obtain that

1 sin(π/2j−1) q = cos(π/2j) = , j 2 sin(π/2j) thus

1 sin(π/2) sin(π/22) sin(π/2n−1) q q q = ... 2 3 ··· n 2n−1 · sin(π/22) · sin(π/23) · · sin(π/2n) 1 sin(π/2) = 2n−1 · sin(π/2n) 2 π/2n = π · sin(π/2n) Since t 1 as t 0 sin t → → it follows that the partial products

p = q q q n 2 3 ··· n converge to 2/π as n . Using Deﬁnition 7.1 or 7.2, Vieta’s product converges to 2/π. → ∞

7.2 Inﬁnite Products of Numbers: Convergence Theory

Recall the Cauchy convergence criterion for a sequence pn of complex numbers: The sequence p converges if and only if for every ε > 0 there exists J = J(ε) n ∈ N so that

p p < ε for n > m J(ε) . | n − m| ≥ Lemma 7.1 (Cauchy Criterion) The inﬁnite product

∞ Πj=1qj (7.3) converges in the sense of Deﬁnition 7.2 if and only if for all ε > 0 there is J = J(ε) N so that ∈

Πn q 1 < ε for n > m J(ε) . (7.4) j=m+1 j − ≥

56 Proof: 1) Assume that (7.3) converges. There exists J0 N so that qj = 0 for j J . Set ∈ 6 ≥ 0 p = Πn q for n J . n j=J0 j ≥ 0 By assumption,

p L as n ,L = 0 . n → → ∞ 6 For n > m J we have ≥ 0 pn n = Πj=m+1qj . pm If ε > 0 is given, then there exists K(ε ) J with 0 0 ≥ 0 p L < ε and p L < ε for n > m K(ε ) . | n − | 0 | m − | 0 ≥ 0 Therefore,

p L + η 1 + η /L n = n = n = 1 + η pm L + ηm 1 + ηm/L with

η < ε and η < ε . | n| 0 | m| 0 If ε > 0 is given and ε0 > 0 is chosen small enough, then p n 1 = η < ε . pm − | | This proves (7.4). 2) Conversely, assume that for all ε > 0 there exists J(ε) N so that (7.4) ∈ holds. First, let J0 = J(1/2). Then the estimate (7.4) with ε = 1/2 implies that q = 0 for j > J . Set j 6 0 n pn = Πj=J0+1qj for n > J0 .

We must show that the sequence pn converges to a non–zero limit. We ﬁrst show that the sequence pn is bounded. We have p 1 n 1 for n > m > J0 . pm − ≤ 2 Therefore, 1 p p p for n > m > J . | n − m| ≤ 2| m| 0

If we ﬁx m = J0 + 1 we obtain that pn is bounded, pn C for all n > J0. Let ε > 0 be given and let | | | | ≤ p n n 1 = Πj=m+1qj 1 < ε for n > m J(ε) > J0 . pm − − ≥

57 This estimate implies that p n pn pm = 1 pm < εC for n > m J(ε) > J0 . | − | pm − | | ≥

Therefore, the sequence pn converges, pn L. 1 → The estimate pn pm 2 pm proved above for m = J0 + 1 and all large n implies that L =| 0.− | ≤ | | Using the notation6 of the above proof, we have p j+1 qj+1 1 = 1 < ε for j J(ε) . | − | pj − ≥ This proves the following:

Lemma 7.2 If

∞ Πj=1 qj converges, then

q 1 as j . j → → ∞

The previous lemma suggests to write the factors qj in an inﬁnite product

∞ Πj=1qj in the form

qj = 1 + aj .

If the inﬁnite product Π(1 + aj) converges, then aj 0. The converse is not true, of course, as the example → 1 Π∞ 1 j=2 − j shows. We will need the following simple estimates for the real exponential function.

Lemma 7.3 We have

1 + x ex for x 0 (7.5) ≤ ≥ and

ex/2 1 + x for 0 x 2 . (7.6) ≤ ≤ ≤ Proof: The second assertion follows from e < 3 and convexity of the exponential function. The following theorem relates the convergence of an inﬁnite product to the absolute convergence of a series.

58 P∞ Theorem 7.1 Let aj denote a sequence of complex numbers. Then j=1 aj < if and only if | | ∞ Π∞ (1 + a ) j=1 | j| converges.

Proof: a) First assume that

∞ X S := a < . | j| ∞ j=1 Then let

p = Πn (1 + a ) n j=1 | j| and obtain

p e|a1|+...+|an| eS < . n ≤ ≤ ∞ Because of

1 p p ... p eS < , ≤ 1 ≤ 2 ≤ ≤ n ≤ ∞ convergence

p p, p 1 , n → ≥ follows. b) Assume that

Π∞ (1 + a ) j=1 | j| converges. Since a 0 we have | j| → a 2 for j > J , | j| ≤ thus

e|aj |/2 1 + a for j > J . ≤ | j| Setting

n S1 = lim Π (1 + aj ) n→∞ j=J+1 | | we obtain:

n 1 X exp a Πn (1 + a ) S < 2 | j| ≤ j=J+1 | j| ≤ 1 ∞ j=J+1 for all n J + 1. This yields the bound ≥

59 n X a 2 ln S < , | j| ≤ 1 ∞ j=J+1 which implies convergence of the series P a . | j| It is less clear how to relate convergence of the series X aj to convergence of the product

Π(1 + aj) . However, convergence of the series P a does imply convergence of the product | j| Π(1 + aj). We will formulate this as the next theorem.

P∞ Theorem 7.2 Let aj be a sequence of complex numbers. If j=1 aj < then | | ∞

∞ Πj=1(1 + aj) converges.

The proof uses the following lemma.

Lemma 7.4 Let a1, . . . , an C and set ∈ p = Πn (1 + a ), q = Πn (1 + a ) . n j=1 j n j=1 | j| Then the bound

p 1 q 1 | n − | ≤ n − holds.

Proof: We have

pn = (1 + a1) ... (1 + an) X = 1 + ai1 . . . air where the sum is taken over all indices i1, . . . , ir with

1 i < . . . < i n . ≤ 1 r ≤ Therefore,

X p 1 a . . . a | n − | ≤ | i1 ir | = q 1 n −

60 This proves the lemma. Proof of Theorem 7.2: 1) We have aj = 1 for j J. For simplicity of notation, assume that a = 1 for all j 1.6 Then− deﬁne≥ j 6 − ≥ p = Πn (1 + a ), q = Πn (1 + a ) . n j=1 j n j=1 | j| We have, for n > m 1: ≥

p n pn pm = pm 1 | − | | | pm −

= p Πn (1 + a ) 1 | m| j=m+1 j − q Πn (1 + a ) 1 ≤ | m| j=m+1 | j| − qn = qm 1 | | qm − = q q . | n − m| Therefore, if ε > 0 is given, there exists N(ε) with

p p q q < ε if n > m N(ε) . | n − m| ≤ | n − m| ≥ Here, to estimate qn qm , we use Theorem 7.1. This proves that pn converges to a limit L as n | −. | Using Lemma→ 7.4 ∞ and the Cauchy Criterion Lemma 7.1 one obtains that p q 1 n n 1 1 for n > m N0 . pm − ≤ qm − ≤ 2 ≥ This implies that

p p p | m| | n − m| ≤ 2 for all large n. Letting n yields that → ∞ p L p | m| , | − m| ≤ 2 thus

p L | m| > 0 . | | ≥ 2 7.3 Inﬁnite Products of Functions Looking carefully at the proofs of the previous section, one obtains uniformity of convergence of a product of functions

Π(1 + fj(z)) P if the corresponding series fj(z) converges uniformly. We formulate such a result in the following lemma.| |

61 Lemma 7.5 Let K C denote a compact set and let fj : K C denote a sequence of continuous⊂ functions with →

∞ X fj K < where fj K := max fj(z) . | | ∞ | | z∈K | | j=1 Then the sequence of functions

n pn(z) = Πj=1(1 + fj(z)), n = 1, 2 ... converges uniformly on K.

Proof: Fix z K and set ∈

n pn(z) = Πj=1 (1 + fj(z)) q (z) = Πn (1 + f (z) ) n j=1 | j | Q = Πn (1 + f ) n j=1 | j|K

By Theorem 11.1, the sequence of numbers Qn converges. We have, for n > m 1: ≥

pn(z) pm(z) qn(z) qm(z) | − | ≤ | − | = q (z) Πn (1 + f (z) ) 1 | m | j=m+1 | j | −

Q Πn (1 + f ) 1 ≤ m j=m+1 | j|K − Q n = Qm 1 Qm − = Q Q . | n − m| Therefore, if ε > 0 is given, there is N(ε) with

p (z) p (z) Q Q < ε for n > m N(ε) . | n − m | ≤ | n − m| ≥ Here N(ε) does not depend on z. This proves the lemma. The following result is the workhorse for convergence of inﬁnite products of holomorphic functions.

Theorem 7.3 (Main Theorem on Convergence of Infinite Products of Holo- morphic Functions) Assume that U is an open subset of C and let fj H(U) denote a sequence of holomorphic functions on U. Assume that the∈ series P∞ j=1 fj K converges for every compact subset K of U. Then the following holds:| | 1) For every z U the infinite product ∈ ∞ Πj=1 1 + fj(z) =: F (z) converges and defines a function F H(U). ∈ 62 2) The sequence of holomorphic functions F (x) = ΠN 1 + f (z) , z U, N j=1 j ∈ converges normally in U to F (z). 3) For every z U the function F (z) has a zero at z if and only if one of 0 ∈ 0 the factors qj(z) = 1 + fj(z) has a zero at z0. Furthermore, the multiplicity of z0 as a zero of F (z) is the finite sum (over j) of the multiplicities of the zero z0 of the factors qj(z). 4) If F (z) = 0 then 6 ∞ 0 F 0(z) X f (z) = j . F (z) 1 + f (z) j=1 j Here the series converges normally in the set

U = z U : F (z) = 0 . 0 { ∈ 6 } Proof: 1) and 2) The convergence of the inﬁnite product for each ﬁxed z U ¯ ∈ 1 follows from Theorem 7.2. Fix z0 U and let K := D(z0, ε) U. Let fj K 2 for j > J. Write ∈ ⊂ | | ≤

F (z) = ΠJ (1 + f (z)) ΠN (1 + f (z)) =: F (z) G (z) . (7.7) N j=1 j · j=J+1 j J · N

Note that the function GN (z) has no zero in K. The ﬁrst ﬁnite product, FJ (z), is a function in H(U). The sequence GN (z) converges uniformly on K to a function G C(K). This follows from the previous lemma. We then have G H(D(z∈, ε)). The sequence F converges uniformly on K to F . It ∈ 0 N follows that F H(D(z0, ε)). Since z0 U was arbitrary, we have shown that F H(U). ∈ ∈ ∈3) In the factorization (7.7) let N to obtain that → ∞ F (z) = F (z) G(z), z D(z , ε) . J · ∈ 0 Here G is holomorphic and nowhere zero in D(z0, ε). This implies 3). 4) For a product of three functions qj(z),

Q = q1q2q3 we have

0 0 0 0 Q = q1q2q3 + q1q2q3 + q1q2q3 . Therefore, if Q(z) = 0, 6 3 0 Q0(z) X q (z) = j . Q(z) q (z) j=1 j It is clear that this generalizes to any ﬁnite sum and we have for F (z) = 0: 6

63 N 0 F 0 (z) X f (z) N = j . F (z) 1 + f (z) N j=1 j Letting N we obtain that → ∞ ∞ 0 F 0(z) X f (z) = j . F (z) 1 + f (z) j=1 j

It is not diﬃcult to prove that the series converges normally in U0. (Homework) 7.4 Example: The Product Formula for the Sine Function We recall: 1

∞ 1 X 2z π cot(πz) = + , z C Z . (7.8) z z2 n2 ∈ \ n=1 − We want to use this to prove:

2 1 ∞ z sin(πz) = z Π 1 , z C . (7.9) π n=1 − n2 ∈ Let

z2 fn(z) = , n N . −n2 ∈ If K is a compact subset of C then K D¯(0,R) for some R > 0. We then have ⊂ R2 f . | n|K ≤ n2 Since P n−2 < we obtain P f < and, by the previous theorem, ∞ | n|K ∞ convergence of the inﬁnite product Πn(1 + fn(z)) follows. The function

z2 P (z) = z Π∞ 1 n=1 − n2 is entire and P (z) has a simple zero at each integer. Also, for z C Z, ∈ \ ∞ P 0(z) 1 X 2z = + = π cot(πz) . P (z) z z2 n2 n=1 − If we set 1 G(z) = sin(πz) π

1 R cot(πζ) Using residue calculus for 2 dζ, z ∈ \ , where γn is a sequence of growing γn (ζ−z) C Z P∞ 1 π rectangles, we have seen that j=−∞ π(j−z)2 = sin2(πz) . Then the formula (7.8) follows by integration.

64 then we also have, from calculus,

G0(z) = π cot(πz), z C Z . G(z) ∈ \ The function P/G has a removable singularity at each integer. We have

P 0 P 0G PG0 P P 0 G0 = − = = 0 . G G2 G · P − G Therefore, P/G = c = const. Considering

P (z) G(z) lim = 1 = lim , z→0 z z→0 z we conclude that c = 1, i.e., P (z) G(z). This proves the product formula (7.9). ≡

7.5 Euler’s Constant We deﬁne here Euler’s constant γ, also called the Euler–Mascheroni constant. It occurs frequently when one discusses the Γ–function. A numerical value is

γ = 0.577 215 66 ... It is unknown whether γ is rational or irrational.

Lemma 7.6 For n = 1, 2,... let

1 1 γ = 1 + + ... + ln (n) n 2 n − 1 1 δ = 1 + + ... + ln (n + 1) n 2 n − We claim that

δn < δn+1 < γn+1 < γn .

The sequences γn and δn converge to the same limit γ, called Euler’s constant.

Proof: We have

1 γ γ = + ln (n) ln (n + 1) n+1 − n n + 1 − 1 Z n+1 dx = . n + 1 − n x For n < x < n + 1 we have 1 1 1 < < , n + 1 x n thus

65 γ γ < 0 . n+1 − n Similarly, 1 δ δ = + ln (n + 1) ln (n + 2) n+1 − n n + 1 − 1 Z n+2 dx = . n + 1 − n+1 x For n + 1 < x < n + 2 we have 1 1 1 < < , n + 2 x n + 1 thus

δ δ > 0 . n+1 − n Also, 1 γ δ = ln (n + 1) ln (n) = ln 1 + > 0 . n − n − n Since γ δ 0 as n , the lemma is proved. n − n → → ∞ 7.6 The Gauss Formula for Γ(z) and Weierstrass’ Product For- mula for 1/Γ(z) Recall that Z ∞ Γ(z) = tz−1e−t dt, Re z > 0 . 0 Splitting the integral as

Z ∞ Z 1 Z ∞ = ... + ... =: g(z) + h(z) 0 0 1 we obtain h H(C) and use the series ∈ ∞ X ( 1)j e−t = − tj j! j=0 to obtain

∞ X ( 1)j Z 1 g(z) = − tz−1+j dt j! j=0 0 ∞ X ( 1)j 1 = − . j! z + j j=0 In this way we obtain a formula for the Γ–function valid for

66 z C 0, 1, 2,... =: U, ∈ \{ − − } namely,

∞ X ( 1)j 1 Z ∞ Γ(z) = − + tz−1e−t dt, z U. j! z + j ∈ j=0 1 It is clear that Γ H(U). The function Γ has a simple pole at each n 0, 1, 2,... and ∈ ∈ { − − } ( 1)j Res(Γ, j) = − for j = 0, 1, 2,... − j! We also recall the reﬂection formula π Γ(z)Γ(1 z) = , z C Z , − sin(πz) ∈ \ which implies that Γ has no zero. Therefore, the function 1 ∆(z) := Γ(z) is entire. The function ∆(z) has a simple zero at each n 0, 1, 2,... and has no other zeros. The zeros of ∆(z) clearly show up in∈ the { following− − product} formula.

Theorem 7.4 (Weierstrass) We have

1 γz ∞ z −z/j ∆(z) = = e z Π 1 + e , z C , (7.10) Γ(z) · · j=1 j ∈ where γ is Euler’s constant,

γ = lim γn = 0.577 21 ... n→∞ with 1 1 γ = 1 + + ... + ln (n) . n 2 n − Convergence of the inﬁnite product in (7.10): We have z e−z/j = 1 + η (z), η (z) C(R)j−2 for z R. − j j | j | ≤ | | ≤ We then have

z z2 1 + e−z/j = 1 +η ˜ (z) j − j2 j with

η˜ (z) C˜(R)j−2 for z R. | j | ≤ | | ≤

67 It is then clear that the inﬁnite product in (7.10) converges uniformly for z R, where R > 0 is arbitrary. Therefore, | | ≤ z Q(z) := eγz z Π∞ 1 + e−z/j (7.11) · · j=1 j is an entire function. This function has a simple zero at each n 0, 1, 2,... and has no other zeros. ∈ { − − } We want to prove that Q(z) = ∆(z). To do this, we will prove the equality for 0 < x = z 1 and then apply the identity theorem. Recall that≤φ(x) = ln (Γ(x)), x > 0, satisﬁes φ00(x) > 0 for x > 0. (See Math 561.)

00 Lemma 7.7 Let φ : (0, ) R be a C2 function with φ (x) > 0 for all x > 0. If 0 < a < b < c then we∞ have→

φ(b) φ(a) φ(c) φ(b) − − b a ≤ c b − − and φ(b) φ(a) φ(c) φ(a) − − . b a ≤ c a − − Proof: The ﬁrst estimate follows from the mean value theorem. To prove the second estimate, consider the function

φ(s) φ(a) h(s) = − for s > a . s a − Diﬀerentiation yields

1 φ(s) φ(a) h0(s) = φ0(s) − . s a − s a − − Then, using the mean value theorem, we ﬁnd that h0(s) 0 for s > a. Application: We let φ(x) = ln Γ(x), x > 0. Let n ≥be an integer,n 2, and let 0 < x 1. We have ≥ ≤ 0 < n 1 < n < n + x n + 1 − ≤ and obtain

φ(n) φ(n 1) φ(n + x) φ(n) φ(n + 1) φ(n) − − − − . n (n 1) ≤ n + x n ≤ n + 1 n − − − − Since

Γ(n) φ(n) φ(n 1) = ln = ln(n 1) − − Γ(n 1) − − and

Γ(n + x) φ(n + x) φ(n) = ln − Γ(n)

68 we obtain that

Γ(n + x) x ln(n 1) ln x ln n . − ≤ Γ(n) ≤ Therefore,

Γ(n + x) (n 1)x nx , − ≤ Γ(n) ≤ thus

(n 1)! (n 1)x Γ(n + x) (n 1)! nx . − − ≤ ≤ − From the functional equation Γ(z + 1) = zΓ(z) we have

Γ(n + x) = (x + n 1)(x + n 2) (x + 1)x Γ(x) . − − ··· This yields the bounds

(n 1)! (n 1)x (n 1)! nx − − Γ(x) − . x(x + 1) ... (x + n 1) ≤ ≤ x(x + 1) ... (x + n 1) − − In the lower bound, we may replace n 1 by n. Then we have − n! nx n! nx x + n Γ(x) . x(x + 1) ... (x + n) ≤ ≤ x(x + 1) ... (x + n) · n Denote

n! nx A = . n x(x + 1) ... (x + n) We have shown that x + n A Γ(x) A . n ≤ ≤ n · n Equivalently, n Γ(x) A Γ(x) . n + x ≤ n ≤ We obtain for n : → ∞ Lemma 7.8 Let 0 < x 1. Then we have ≤ n! nx Γ(x) = lim . n→∞ x(x + 1) ... (x + n)

Deﬁne, for z C, ∈

z(z + 1) ... (z + n) h (z) = n n! nz z z z = e−(ln n) z z 1 + 1 + ... 1 + · · 1 2 n z z z = eγnz z 1 + e−z/1 1 + e−z/2 ... 1 + e−z/n · · 1 2 n

69 The above lemma implies that, for 0 < x 1: ≤ 1 Γ(x) = lim . (7.12) n→∞ hn(x) Also,

h (z) Q(z) n → normally in C where Q(z) is deﬁned in (7.11). Therefore, 1 1

hn(z) → Q(z) normally in U. By (7.12) we obtain that 1 Γ(x) = for 0 < x 1 . Q(x) ≤ The identity theorem then yields that 1 Γ(z) = for z U. Q(z) ∈ This proves the Weierstrass’ product formula for ∆(z) = 1/Γ(z). The following result is easy to show:

Lemma 7.9 Let K C be compact. Let an, a C(K and assume that an ⊂ ∈ | − a K 0 as n . If a(z) = 0 for all z K then | → → ∞ 6 ∈ 1 1

0 as n . an − a K → → ∞ One obtains the following Gauss’ formula for Γ(z):

Theorem 7.5 (Gauss) We have

n! nz Γ(z) = lim n→∞ z(z + 1) (z + n) ··· with normal convergence in U = C 0, 1, 2,... . \{ − − } 7.6.1 A Formula for (Γ0/Γ)0

Let U = C 0, 1, 2,... and set \{ − − } f (z) = n!e(ln n)zΠn (z + j)−1, z U. n j=0 ∈ By Gauss’ formula we have f (z) Γ(z) normally in U. We have n → n f 0 (z) X n = ln n (z + j)−1 f (z) − n j=0 and

70 n d f 0 (z) X n = (z + j)−2 . dz f (z) n j=0 For n obtain: → ∞ ∞ d Γ0(z) X = (z + j)−2 . dz Γ(z) j=0 Recall that we have used the inequality φ00(x) > 0 for 0 < x < where φ(x) = ln Γ(x). We now have shown that ∞

∞ X 1 φ00(x) = , 0 < x < . (x + j)2 ∞ j=0

7.7 Entire Functions with Prescribed Zeros

If a1, . . . , an C then the polynomial ∈ p(z) = (a z) (a z) 1 − ··· n − vanishes precisely at a1, . . . , an. Now let a1, a2,... denote an inﬁnite sequence in C without accumulation point in C, i.e., an . We want to construct an entire function f(z) with f(z) = 0 if and| only| → if z ∞ a , a ,... . We allow ∈ { 1 2 } repetitions in the aj. Then, if a number a appears q times among the aj, we want f(z) to have a zero precisely of order q at a. We will show below how to construct such a function f(z). Remarks on Non–Uniqueness. If f(z) vanishes precisely at the aj (to the correct order) then any function

h(z) g(z) = f(z)e , h H(C) , ∈ has the same property. We show:

Lemma 7.10 Let aj be a sequence in C without accumulation point in C. Let f(z) and g(z) be entire functions that vanish precisely at the aj. If a appears q times in the sequence, then let f(z) and g(z) have a zero of order q at a. Under these assumptions, there is h H(C) with ∈ h(z) g(z) = f(z)e , z C . ∈ Proof: The quotient

f(z) Q(z) = , z C a1, a2,... , g(z) ∈ \{ } is bounded near every aj. By Riemann’s removability theorem, Q(z) extends to an entire function. Also, Q(z) = 0 for all z. Therefore, Q(z) has the form Q(z) = eh(z) where h(z) is entire. 6

71 7.7.1 Construction of f(z); Motivation One might try to construct f(z) as the inﬁnite product

(a z)(a z) 1 − 2 − ··· However, since aj , this product never converges. Somewhat smarter| | → ∞ is the following: Let us assume that

a = a = ... = a = 0 and a = 0 for j > m . 1 2 m j 6 Then try the inﬁnite product z z zm 1 1 (7.13) − am+1 − am+2 ··· This works if

X 1 < . a ∞ j>m | j| Under this assumption the above infinite product defines an entire function with the desired property. For example, it suffices that

a c j1+ε, j J, | j| ≥ ≥ for some c > 0 and ε > 0. However, if

a j or a pj , | j| ∼ | j| ∼ the inﬁnite product will typically not converge. Weierstrass’ idea was to use extra factors in the inﬁnite product (7.13), which ensure convergence, but do not introduce new zeros.

7.7.2 Weierstrass’ Canonical Factors Let us assume a = 0 for all j, for simplicity. Instead of the inﬁnite product j 6 z z 1 1 (7.14) − a1 − a2 ··· try z z E E (7.15) a1 a2 ··· where E(z) is an entire function of the form

E(z) = (1 z)φ(z) . − Here φ(z) should be suitably chosen, with φ(z) = 0 for all z. Then one obtains 6 z z z z z z E E = 1 φ 1 φ (7.16) a1 a2 ··· − a1 a1 − a2 a2 ···

72 The factors φ(z/aj) should make the product converge without introducing new zeros. Note that, as aj , the argument z/aj of φ converges to zero. We then want to construct| φ|( →z) ∞ with

(1 z)φ(z) 1 for z < ε − ∼ | | to enhance convergence of the product. However, if we would require (1 z)φ(z) = 1, then φ(z) would become singular at z = 1, and we do not obtain− an entire function. We want to construct φ(z) so that a) (1 z)φ(z) = 1 + (zk+1) for z 0; b) φ is− entire. O ∼ c) φ(z) = 0 for all z C. To ensure6 that φ(z)∈ never vanishes, let us construct φ(z) in the form

φ(z) = eh(z) . We then have the requirement, for small z , | | 1 eh(z) = φ(z) = + (zk+1) , 1 z O or, − 1 h(z) = log + (zk+1) = log(1 z) + (zk+1) . 1 z O − − O − The second equation holds since log 1 + ε = log(1 z) + (ε) for z 0. 1−z − − O ∼ Set r(z) = log(1 z). Then we have r(0) = 0 and − − 1 r0(z) = = 1 + z + z2 + ... 1 z − Therefore,

z2 z3 r(z) = log(1 z) = z + + + .... − − 2 3 This leads to the following requirement for h(z):

z2 z3 h(z) = h (z) = z + + + ... + (zk+1) . k 2 3 O Therefore, we set

z2 z3 zk h (z) = z + + + ... + . k 2 3 k This leads to the following deﬁnition of the Weierstrass’ canonical factors.

Deﬁnition: For k = 0, 1,... deﬁne Ek(z) by

z2 zk E (z) = 1 z, E (z) = (1 z) exp z + + ... + for k 1 . 0 − k − 2 k ≥

73 Clearly, Ek(z) vanishes only at z = 1 and has a simple zero at z = 1. Therefore, the function z E (z/a ) vanishes to ﬁrst order at z = a and → k j j vanishes nowhere else. We will show now that 1 Ek(z) vanishes to order k + 1 at z = 0. In this sense, E (z) 1 to order |k +− 1 at z |= 0. k ∼ Lemma 7.11 There is a constant C > 0, independent of k and z, with 1 1 E (z) C z k+1 for z . | − k | ≤ | | | | ≤ 2 One can choose C = 2(e 1). − Proof: In the following let z 1 . Let | | ≤ 2 z2 zk h (z) = z + + ... + . k 2 k We have

Ek(z) = (1 z) exp(hk(z)) − = exp log(1 z) + h (z) − k = ew with

w = log(1 z) + h (z) . − k Since

∞ X zj log(1 z) = − − j j=1 we have

∞ X zj w = . − j j=k+1 Therefore,

∞ X 1 w z k+1 z j−k−1 | | ≤ | | j | | j=k+1 1 1 z k+1 1 + + + ... ≤ | | 2 4 = 2 z k+1 . | | In particular, w 1 for z 1 . Thus, | | ≤ | | ≤ 2

74 w 1 Ek(z) = 1 e | − | | − | 2 3 = w + w /2! + w /3! + ... w 1 + 1/2! + 1/3! + ... ≤ | | = w (e 1) | | − 2(e 1) z k+1 ≤ − | |

Remark: It will be useful below that the constant C in the previous lemma does not depend on k. This is not completely obvious from the construction, which only yields 1 E (z) = (zk+1). − k O

Theorem 7.6 Let aj denote a sequence of complex numbers with aj as j . Assume that | | → ∞ → ∞ a = ... = a = 0 < a a ... 1 m | m+1| ≤ | m+2| ≤ Then the inﬁnite product

m ∞ f(z) = z Π Ej(z/aj), z C , j=m+1 ∈ deﬁnes an entire function f(z) which has zeros precisely at the aj. The multiplicity of a zero an of f(z) equals the number of occurrences of an in the sequence aj.

Proof: Assume m = 0. Fix R > 0 and note that aj 2R for j j0(R). If z R then | | ≥ ≥ | | ≤ z 1

for j j0(R) . aj ≤ 2 ≥ Therefore, by Lemma 7.11,

z j+1 1 E (z/a ) C | | C 2−(j+1) . | − j j | ≤ a j+1 ≤ | j| If one writes

Ej(z/aj) = 1 + fj(z) then

∞ X f ¯ < . | j|∞,D(0,R) ∞ j=1 The claim now follows from Theorem 7.3. In the previous theorem, the index j of Ej increases unboundedly. It is of interest to give a condition under which this index can be chosen constant.

75 Theorem 7.7 Let aj denote a sequence of complex numbers with aj as j . Assume that | | → ∞ → ∞ a = ... = a = 0 < a a ... 1 m | m+1| ≤ | m+2| ≤ Also, assume that

∞ X 1 < a s ∞ j=m+1 | j| for some s > 0. If k is an integer with s k + 1 then the inﬁnite product ≤ m ∞ f(z) = z Π Ek(z/aj), z C , j=m+1 ∈ deﬁnes an entire function f(z) which has zeros precisely at the aj. The multiplicity of a zero a of f(z) equals the number of occurances of a in the sequence aj.

Proof: Assume m = 0. Fix R > 0 and note that aj 2R for j j0(R). If z R then | | ≥ ≥ | | ≤ z 1

for j j0(R) . aj ≤ 2 ≥ Therefore, by Lemma 7.11,

z k+1 1 1 E (z/a ) C | | C(R, k) . | − k j | ≤ a k+1 ≤ a k+1 | j| | j| The claim now follows from

∞ X 1 < a k+1 ∞ j=m+1 | j| and Theorem 7.3.

76 8 The Bernoulli Numbers and Applications

8.1 The Bernoulli Numbers The function g(z) deﬁned by

g(z) = z/(ez 1) for 0 < z < 2π, g(0) = 1 , − | | is holomorphic in D(0, 2π). We write its Taylor series as

∞ X B g(z) = ν zν, z < 2π , (8.1) ν! | | ν=0 where the numbers Bν are, by deﬁnition, the Bernoulli numbers. Since

1 g(z) = 1 1 2 1 + 2 z + 6 z + ... 1 = 1 z + ... − 2 it follows that 1 B = 1,B = . 0 1 −2 Lemma 8.1 The function z h(z) = g(z) + 2 is even. Consequently,

B = 0 for ν 3, ν odd . ν ≥ Proof: We must show that z z g( z) = g(z) + , − − 2 2 i.e.,

g( z) g(z) = z . − − With a = ez we have

z z g( z) g(z) = − − − e−z 1 − ez 1 −1 1− = z − 1 1 − a 1 a − − and

77 a 1 1 a − = − = 1 . 1 a − a 1 1 a − − − One can compute the Bernoulli numbers easily using a recursion. We claim:

Lemma 8.2 For n 1 we have ≥ n X n + 1 B = 0 . ν ν ν=0

Proof: We have, for 0 < z < 2π: | |

ez 1 z 1 = − z · ez 1 ∞ − ∞ X zµ X B = ν zν (µ + 1)! · ν! µ=0 ν=0 ∞ X B = ν zµ+ν ν!(µ + 1)! µ.ν=0 ∞ n X X B = ν zn ν!(n + 1 ν)! n=0 ν=0 − Since

n + 1 (n + 1)! = ν ν!(n + 1 ν)! − the lemma is proved. Using Pascal’s triangle, we can compute the binomial coeﬃcients. Then, using the previous lemma and B0 = 1 we obtain: For n = 1: 1 B + 2B = 0, thus B = . 0 1 1 −2 For n = 2: 1 B + 3B + 3B = 0, thus B = . 0 1 2 2 6 For n = 3:

B0 + 4B1 + 6B2 + 4B3 = 0, thus B3 = 0 .

For n = 4: 1 B + 5B + 10B + 10B + 5B = 0, thus B = . 0 1 2 3 4 4 −30 Continuing this process, one obtains the following non–zero Bernoulli numbers:

78 1 B = 6 42 1 B = 8 −30 5 B = 10 66 691 B = 12 −2730 7 B = 14 6 etc. Remark: The sequence B2ν is unbounded since otherwise the series (8.1) would have a ﬁnite radius of| convergence.| We will see below that ( 1)ν+1B > − 2ν 0. Thus, the sign pattern observed for B2 to B14 continuous.

8.2 The Taylor Series for z cot z in Terms of Bernoulli Numbers Recall that

∞ w X B g(w) = = ν wν . ew 1 ν! − ν=0 We now express the Taylor series for z cot z about z = 0 in terms of Bernoulli numbers. Note that

1 cos z = (eiz + e−iz) 2 1 sin z = (eiz e−iz) 2i − eiz + e−iz cot z = i eiz e−iz − 1 + e−2iz = i 1 e−2iz − 1 e−2iz + 2e−2iz = i − 1 e−2iz − 2 = i 1 + e2iz 1 − Therefore, 1 2iz cot z = i + , z · e2iz 1 − thus

79 z cot z = iz + g(2iz) ∞ 1 X B = iz + 1 (2iz) + ν (2iz)ν − 2 ν! ν=2 ∞ X 4ν = 1 + ( 1)ν B z2ν . − (2ν)! 2ν ν=1 We substitute πz for z and summarize this:

Lemma 8.3 If Bν denotes the sequence of the Bernoulli numbers, then we have for z < 1: | | ∞ X (2π)2ν πz cot(πz) = 1 + ( 1)ν B z2ν . (8.2) − (2ν)! 2ν ν=1

8.3 The Mittag–Leﬄer Expansion of πz cot(πz) One can derive the following Mittag–Leﬄer expansion (see, for example, [Hahn, Epstein, p. 214]). One can also use residue calculus; see Math 561. Note that the function on the left side and the function on the right side of the following formula both have a simple pole at each integer.

∞ 1 X 1 π cot(πz) = + 2z . z z2 n2 n=1 − Therefore,

∞ X z2 πz cot(πz) = 1 2 . − n2 z2 n=1 − Here, for z < 1: | | z2 (z/n)2 = n2 z2 1 (z/n)2 − ∞− X z 2m = n m=1 Therefore,

∞ ∞ X X z 2m πz cot(πz) = 1 2 (8.3) − n n=1 m=1 ∞ ∞ X X z 2m = 1 2 (8.4) − n m=1 n=1 ∞ X = 1 2 ζ(2m) z2m (8.5) − m=1

80 8.4 The Values of ζ(2m) Comparing the expressions (8.5) and (8.2), we obtain the following result about the values of the Riemann ζ–function at even integers. This result was already known to Euler in 1734.

Theorem 8.1 For m = 1, 2,...:

∞ X 1 1 (2π)2m ζ(2m) = = ( 1)m+1 B . (8.6) n2m 2 − (2m)! 2m n=1

m+1 Remark: Since, clearly, ζ(2m) > 0 we obtain that ( 1) B2m > 0. Examples: − 1 For m = 1 we have B2 = 6 , thus (2π)2 1 π2 ζ(2) = = . 2 2 · 6 6 · For m = 2 we have B = 1 , thus 4 − 30 (2π)4 1 π4 ζ(4) = = . 2 4! · 30 90 · 1 For m = 3 we have B6 = 42 , thus (2π)6 1 π6 ζ(6) = = . 2 6! · 42 945 · For m = 4 one obtains

π8 ζ(8) = . 9450 Remark: According to [Temme, p. 6], no nice formula for ζ(2m + 1) seems to be known if m is an integer.

8.5 Sums of Powers and Bernoulli Numbers It is not diﬃcult to show the following formulae by induction in n:

n−1 X 1 1 j = n2 n 2 − 2 j=1 n−1 X 1 1 1 j2 = n3 n2 + n 3 − 2 6 j=1 n−1 X 1 1 1 j3 = n4 n3 + n2 4 − 2 4 j=1

The formulae follow the pattern:

81 n−1 X 1 1 jk = nk+1 nk + ... + B n , k + 1 − 2 k j=1 but it is not obvious how the general formula should read. Deﬁne the sum

n−1 X S (n 1) = jk k − j=0 where k = 0, 1, 2, 3,... and n = 1, 2, 3,.... We claim that, for every ﬁxed integer k 0, the sum S (n 1) is a polynomial ≥ k −

Φk(n) of degree k + 1 in the variable n and that the coeﬃcients of Φk(n) can be obtained in terms of Bernoulli numbers. Precisely:

Theorem 8.2 For every integer k 0, let Φ denote the polynomial ≥ k k 1 X k + 1 Φ (n) = B nk+1−µ . k k + 1 µ µ µ=0 Then we have

S (n 1) = Φ (n) for all n = 1, 2, . k − k ···

Remark: Writing out a few terms of Φk(n), the theorem says that

1 1 1 k + 1 S (n 1) = nk+1 nk + B nk−1 + + B n . k − k + 1 − 2 k + 1 2 2 ··· k Proof of Theorem: The trick is to write the ﬁnite geometric sum

E (w) = 1 + ew + e2w + + e(n−1)w n ··· in two ways and then to compare coeﬃcients. We have

n−1 X jw En(w) = e j=0 n−1 ∞ X X jk = wk k! j=0 k=0 ∞ n−1 XX 1 = jk wk k! k=0 j=0 ∞ X 1 = S (n 1) wk k! k − k=0

82 (Here we have used the convention 00 = 1.) On the other hand, we have

enw 1 E (w) = − n ew 1 w− enw 1 = − ew 1 · w ∞− ∞ X B X nλ+1 = µ wµ wλ µ! · (λ + 1)! µ=0 λ=0 ∞ X X B = µ nλ+1 wk µ!(λ + 1)! k=0 µ+λ=k

Comparison yields that

X k! S (n 1) = B nλ+1 k − µ!(λ + 1)! µ µ+λ=k k 1 X (k + 1)! = B nk+1−µ k + 1 µ!(k + 1 µ)! µ µ=0 − This proves the claim since

k + 1 (k + 1)! = . µ µ!(k + 1 µ)! −

83 9 The Riemann Zeta–Function

9.1 Deﬁnition for Re s > 1 Let s = σ + it. If σ > 1 then the formula

∞ X ζ(s) = n−s (9.1) n=1 deﬁnes ζ(s). For every n = 1, 2,... the function

s n−s = e−s ln n → is entire. Therefore, for N = 1, 2,..., the ﬁnite sum

N X −s SN (s) = n n=1 is an entire function. The sequence SN (s) converges uniformly for

Re s 1 + δ > 1 . ≥ Thus, (9.1) deﬁnes a holomorphic function in the half–plane

U = s : Re s > 1 . 1 { } 9.2 Simple Bounds of ζ(s) for s > 1 Let n be a positive integer and let s > 1. Then we have

(n + 1)−s x−s n−s for n x n + 1 . ≤ ≤ ≤ ≤ Therefore,

∞ ∞ X Z ∞ dx X (n + 1)−s n−s . ≤ xs ≤ n=1 1 n=1 This says that 1 ζ(s) 1 ζ(s) for s > 1 . − ≤ s 1 ≤ − In other words, 1 s ζ(s) for s > 1 . s 1 ≤ ≤ s 1 − − The lower bound ζ(s) > 1 for s > 1 is also obvious.

84 9.3 Meromorphic Continuation of ζ(z) to Re z > 0 Let

U = z : Re z > 0 . 0 { } We will show that ζ(z) can be continued as a holomorphic function deﬁned in U0 1 . Also, ζ(z) has a simple pole at z = 1 with residue equal to 1. It turns out\{ that} ζ(z) 1/(z 1) is in fact an entire function, but we will not show this here since our− analytic− continuation is only valid for Re z > 0. Notation: If x R then there is a unique integer n with n x < n + 1. We then write ∈ ≤

x = x n, x = n + x = [x] + x { } − { } { } and call x the fractional part and [x] = n the integer part of x. We use{ } the function [x] to derive an integral representation for ζ(z). The following holds for all z C: ∈

N X n n−z (n + 1)−z = 1 2−z + 2 2−z 3−z + ... + N N −z (N + 1)−z − − − − n=1 = 1 + 2−z + 3−z + ... + N −z N(N + 1)−z − Here, for all z C: ∈

x=n+1 n−z (n + 1)−z = x−z − − x=n Z n+1 = z x−z−1 dx n Multiply the last equation by n and sum over n from 1 to N to obtain:

N N X Z n+1 X z [x]x−z−1 dx = n n−z (n + 1)−z − n=1 n n=1 N X = n−z N(N + 1)−z − n=1 Thus we have for all z C and all N = 1, 2,...: ∈ N Z N+1 X z [x]x−z−1 dx = n−z N(N + 1)−z . − 1 n=1 In this equation we now assume Re z > 1 and let N . This yields the integral representation → ∞ Z ∞ z [x]x−z−1 dx = ζ(z), Re z > 1 . 1

85 If we replace the integer part of x, denoted by [x], by x itself, then we can evaluate the integral. For Re z > 0:

Z ∞ Z ∞ z z x x−z−1 dx = z x−z dx = . · z 1 1 1 − Taking the diﬀerence between the last two equations and noting that x [x] = x we have − { } Z ∞ z z (x [x])x−z−1 dx = ζ(z) , − z 1 − 1 − thus

z Z ∞ ζ(z) = z x x−z−1 dx, Re z > 1 . (9.2) z 1 − { } − 1 Here the function z/(z 1) is holomorphic in C 1 . In (9.2), we have identiﬁed the singular behavior of− ζ(z) at z = 1. \{ } Since x is bounded, the integral { } Z ∞ H(z) = x x−z−1 dx 1 { } deﬁnes a holomorphic function of z for Re z > 0. We obtain that

z Z ∞ z x x−z−1 dx (9.3) z 1 − { } − 1 is holomorphic in U0 1 and, by (9.2), agrees with ζ(z) for Re z > 1. Thus, the holomorphic extension\{ } of ζ(z) to U 1 exists and is given by 0 \{ }

z Z ∞ ζ(z) = z x x−z−1 dx z 1 − 1 { } −1 = + F (z) z 1 − with Z ∞ F (z) = 1 z x x−z−1 dx . (9.4) − 1 { } Lemma 9.1 The function ζ(z) is meromorphic in U = z : Re z > 0 . It 0 { } has a simple pole at z = 1 with residue equal to 1. It has no other poles in U0.

Lemma 9.2 Let F (z) be deﬁned in (9.4). Then we have

F (1) = γ

Pn 1 where γ = lim →∞ ln n is Euler’s constant. n j=1 j −

86 Proof: We have F (1) = 1 lim →∞ J with − N N

Z N+1 −2 JN = x x dx 1 { } N X Z n+1 = (x n)x−2 dx − n=1 n N Z N+1 dx X Z n+1 dx = n x − x2 1 n=1 n N X 1 = ln(N + 1) − n + 1 n=1 N+1 X 1 = ln(N + 1) − n n=2 N+1 X 1 = ln(N + 1) + 1 − n n=1 Therefore,

N+1 X 1 F (1) = lim ln(N + 1) = γ . N→∞ n − n=1

This proves for the Laurant expansion of ζ(s) centered at s = 1:

∞ 1 X ζ(s) = + γ + a (s 1)j . s 1 j − − j=1

9.4 Analytic Continuation of ζ(z) to C 1 \{ } We claim that

Z ∞ xz−1 Γ(z)ζ(z) = dx for Re z > 1 . ex 1 0 − This follows from the geometric sum formula

∞ 1 X = e−nx ex 1 − n=1 and

87 ∞ X 1 Z ∞ Γ(z)ζ(z) = e−ttz−1 dt (t = nx) nz n=1 0 ∞ X Z ∞ = e−nxxz−1 dx n=1 0 ∞ Z ∞ X = e−nxxz−1 dx 0 n=1 Z ∞ xz−1 = dx ex 1 0 − For Re z > 1 we can write

Z 1 xz−1 Z ∞ xz−1 Γ(z)ζ(z) = dx + dx =: I (z) + I (z) . ex 1 ex 1 1 2 0 − 1 − Here the second integral, I2(z), deﬁnes an entire function of z. To discuss the ﬁrst integral, recall that

∞ x X B = m xm, x < 2π , ex 1 m! | | − m=0 2 where B0,B1,... is the sequence of Bernoulli numbers. For Re z > 1 we have

Z 1 1 xm+z−2 dx = , m = 0, 1,... m + z 1 0 − and one obtains that

Z 1 xz−1 I (z) = dx 1 x 0 e 1 ∞ − X B Z 1 = m xm+z−2 dz m! · m=0 0 ∞ X B 1 = m , Re z > 1 . m! · z + m 1 m=0 − For m = 0, 1, 2,... the functions 1 z (1 m) − − have poles at 1, 0, 1, 2, 3,... − − − 2 Recall that Bm = 0 if m is odd and m ≥ 3. Also, by Hadamard’s formula, 1/m 1 lim supm→∞ |Bm/m!| = 2π .

88 However, since Bm = 0 for m = 3, 5, 7,... the functions

Bm 1 m! · z (1 m) − − have poles at 1, 0, 1, 3, 5,... − − − Since

B 1 | m| for m m , m! ≤ 2m ≥ 0 it follows that

∞ X Bm 1 I1(z) = , z C 1, 0, 1, 3, 5,... m! · z (1 m) ∈ \{ − − − } m=0 − − is an analytic function with simple poles at

1, 0, 1, 3, 5,... − − − We have 1 1 ζ(z) = I (z) + I (z) for Re z > 1 Γ(z) 1 Γ(z) 2 and, since 1 ∆(z) = Γ(z) is zero at

0, 1, 2, 3,... − − − the term 1 I (z) = ∆(z)I (z) Γ(z) 1 1 is an analytic function in C 1 . One obtains that \{ }

ζ(z) = ∆(z)I1(z) + ∆(z)I2(z), z C 1 , ∈ \{ } with

1 ∆(z) = , ∆(z) entire Γ(z) ∞ X Bm 1 I1(z) = , z C 1, 0, 1, 3, 5,... m! z + m 1 ∈ \{ − − − } m=0 − Z ∞ xz−1 I (z) = dx, I (z) entire . 2 ex 1 2 1 − 89 The function ∆(z)I2(z) is the product of two entire functions. The function ∆(z)I1(z) is analytic in C 1 with a simple pole at z = 1. Recall that ∆(z) = 1/Γ(z) has a simple\{ zero} at

n = 0, 1, 2, 3, 4,... − − − − and I1(z) has a simple pole at

n = 1, 0, 1, 3, 5 ... − − − Therefore ζ(z) = ∆(z) I1(z) + I2(z) has a simple zero at

n = 2, 4, 6,... − − − These zeros are called the trivial zeros of the zeta–function.

9.5 Euler’s Product Formula for ζ(z)

Let pj, j = 1, 2,... denote the sequence of prime numbers, i.e.,

p1 = 2, p2 = 3, p3 = 5,... Euler’s product formula for the ζ–functions says that 1 ζ(z) = Π∞ for Re z > 1 . j=1 1 p−z − j Let us ﬁrst prove convergence of the inﬁnite product.

Lemma 9.3 The inﬁnite product 1 P (z) = Π∞ for Re z > 1 j=1 1 p−z − j converges and deﬁnes a holomorphic function for Re z > 1. The function P (z) has no zero in U = z : Re z > 1 . 1 { } Proof: Since p 2 we have j ≥ 1 p−z for Re z > 1 . | j | ≤ 2 For ε 1 we have | | ≤ 2 1 = 1 + ε(1 + ε + ε2 + ...) = 1 + R(ε) 1 ε − with 1 R(ε) 2 ε for ε . | | ≤ | | | | ≤ 2

90 Therefore, if Re z = 1 + η > 1, then 1 = 1 + f (z) 1 p−z j − j with 2 2 f (z) . | j | ≤ p 1+η ≤ j1+η | j| Using Theorem 7.3, we obtain convergence of the inﬁnite product 1 P (z) = Π∞ for Re z > 1 , j=1 1 p−z − j the function P (z) is holomorphic for Re z > 1 and

P (z) = 0 for Re z > 1 . 6

An intuitive, but somewhat imprecise argument for the equation 1 ζ(z) = Π∞ for Re z > 1 j=1 1 p−z − j is the following: We have 1 = 1 + p−z + p−2z + ... 1 p−z j j − j and

1 1 = (1 + p−z + p−2z + ...) (1 + p−z + p−2z + ...) 1 p−z · 1 p−z ··· 1 1 · 2 2 ··· − 1 − 2 X −z = pk1 pkJ 1 ··· J where the sum is taken over all J = 1, 2,... and nonnegative integers k1, . . . , kJ . By the Fundamental Theorem of Arithmetic, every n = 1, 2,... has a unique representation as a product of prime powers,

n = pk1 pkJ 1 ··· J and, therefore, the above sum equals

∞ X n−z = ζ(z), Re z > 1 . n=1 The argument is not rigorous since we multiplied inﬁnitely many series somewhat carelessly. We now give a rigorous argument and ﬁx

J, K N, s > 1 . ∈ 91 We have, for every j = 1, 2,...

1 = 1 + p−s + p−2s + ... 1 p−s j j − j K X p−ks . ≥ j k=0 Taking the ﬁnite product over j = 1,...,J yields

K 1 X ΠJ ΠJ p−ks j=1 1 p−s ≥ j=1 j − j k=0 X −s = pk1 pkJ 1 ··· J 0≤kν ≤K X = n−s n∈S(J,K)

Here S(J, K) is the set of all positive integers n of the form

n = pk1 pkJ with 0 k K for ν = 1,...,J. 1 ··· J ≤ ν ≤ Since the above product is ﬁnite, we can let K and obtain → ∞ 1 X ΠJ = n−s j=1 1 p−s − j n∈S(J) where S(J) consists of all positive integers n of the form

n = pk1 pkJ with k 0, 1,... for ν = 1,...,J. 1 ··· J ν ∈ { } If 1 n J then n S(J). Therefore, ≤ ≤ ∈ J X X n−s n−s ζ(s), s > 1 . ≤ ≤ n=1 n∈S(J) Taking the limit as J and recalling that → ∞ 1 X ΠJ = n−s j=1 1 p−s − j n∈S(J) we have shown that

∞ 1 X Π∞ = n−s = ζ(s), s > 1 . j=1 1 p−s − j n=1 Since P (z) and ζ(z) are holomorphic for Re z > 1 we have shown:

92 Theorem 9.1 For all z with Re z > 1 we have

∞ 1 X Π∞ = n−z = ζ(z) . j=1 1 p−z − j n=1 The function ζ(z) is zero–free in Re z > 1.

P 1 9.6 The Sum p and the Prime Number Theorem

Theorem 9.2 (Euler) If p1, p2,... denotes the sequence of primes, then

∞ X 1 = . p ∞ j=1 j

Proof: Suppose that ∞ X 1 = L < . p ∞ j=1 j Then let s > 1 and recall that

1 −s 2 −s = 1 + fj(s) with 0 < fj(s) 2pj . 1 p ≤ ≤ pj − j Therefore, for any ﬁnite J,

J 1 J 2 Πj=1 −s Πj=1 1 + 1 p ≤ pj − j ΠJ e2/pj ≤ j=1 1 1 = exp 2( + ... + ) p1 pJ e2L ≤ In this estimate we can let J and obtain → ∞ ζ(s) e2L for s > 1 . ≤ However, we know that 1 ζ(s) for s > 1 s 1 ≤ − and obtain a contradiction as s 1+. → It is clear that pj >> j for all large j. One can ask how fast, in comparison with j, the sequence pj grows as j . As a consequence of the previous → ∞ 1+ε theorem, the following Corollary says that pj cannot grow as fast as j if ε > 0. Corollary: If ε > 0 then numbers cε > 0 and J N with ∈ c j1+ε p for j J ε ≤ j ≥ 93 do not exist. Proof: Otherwise one would obtain 1 1 c for j J, ε 1+ε pj ≤ j ≥ in contradiction to the previous theorem. 1+ε Instead of comparing pj with j , it is common to introduce the function

π(x) = # p : p x , x 1 , { j j ≤ } ≥ and to compare the growth of π(x) with the growth of x as x . Here π(x) is the number of primes x. → ∞ ≤ Roughly speaking, Theorem 9.2 says that the sequence pj of primes grows rather slowly, slower than the sequence

j1+ε, j = 1, 2,... for any ε > 0 since

∞ X 1 < . j1+ε ∞ j=1

Since the sequence pj grows only ’a little’ faster than the sequence j (there are ’many’ primes), the function π(x) goes to inﬁnity only ’a little’ slower than x. Thus, one expects that

π(x) x goes to zero as x , but only quite slowly. In fact, one can prove that → ∞ π(x) 1 as x . x ∼ ln x → ∞ Precisely:

Theorem 9.3 (The Prime Number Theorem)

ln(x)π(x) lim = 1 . x→∞ x

9.7 Auxiliary Results about Fourier Transformation: Poisson’s Summation Formula ∞ The Schwartz space S(R) consists of all C functions f : R C for which all derivatives decay rapidly. More precisely, for all k = 0, →1, 2,... and all j = 0, 1, 2,... there exists a constant Cjk so that

(k) j f (x)x Cjk for all x R . | | ≤ ∈ For f S(R) the Fourier transform is ∈

94 Z ∞ −2πiξx fˆ(ξ) = f(x)e dx, ξ R . −∞ ∈ The Fourier inversion formula holds: Z ∞ 2πiξx f(x) = fˆ(ξ)e dξ, x R . −∞ ∈ Let f S(R). The function ∈ ∞ X F (x) := f(x + n), x R , ∈ n=−∞ is called the periodization of the function f(x). Clearly, F (x + 1) F (x). Another way to obtain a 1–periodic function from f(x) is to consider≡

∞ X 2πinx F˜(x) := fˆ(n)e , x R . ∈ n=−∞ The following lemma says that F (x) F˜(x). ≡ Lemma 9.4 For all f S(R) we have ∈ ∞ ∞ X X 2πinx F (x) := f(x + n) = fˆ(n)e , x R . (9.5) ∈ n=−∞ n=−∞

Proof: For all m Z: ∈

∞ Z 1 X Z 1 F (x)e2πimx dx = f(x + n)e−2πim(x+n) dx 0 n=−∞ 0 ∞ X Z n+1 = f(y)e−2πimy dy n=−∞ n Z ∞ = f(y)e−2πimy dy −∞ = fˆ(m)

For the function F˜(x) we also have

Z 1 F˜(x)e−2πimx dx = fˆ(m) 0 since

Z 1 2π(n−m)x e dx = δmn . 0 The lemma follows.

95 If one sets x = 0 in formula (9.5) one obtains Poisson’s Summation Formula for f S(R): ∈ ∞ ∞ X X f(n) = fˆ(n) (9.6) n=−∞ n=−∞

Fourier Transform of a Gaussian: Consider the Gaussian f(x) = e−πx2 with Fourier transform Z ∞ −πx2 −2πiξx fˆ(ξ) = e e dx, ξ R . −∞ ∈ We know that fˆ(0) = 1. Also, using integration by parts,

∞ Z 2 fˆ0(ξ) = e−πx ( 2πix)e−2πiξx dx −∞ − ∞ Z d 2 = i e−πx e−2πiξx dx −∞ dx = 2πξfˆ(ξ) − Thus, the function fˆ(ξ) satisﬁes the initial values problem

fˆ0(ξ) = 2πξfˆ(ξ), fˆ(0) = 1 . − Since the Gaussian h(ξ) = e−πξ2 satisﬁes the same initial value problem and since the solution of the initial value pfoblem is unique, it follows that

2 fˆ(ξ) = e−πξ . Thus, the Gaussian f(x) = e−πx2 has the Fourier transform fˆ(ξ) = e−πξ2 .

9.8 A Functional Equation This section needs details about the theta function. For t > 0 one deﬁnes the theta function by

∞ X 2 θ(t) = e−πn t . n=−∞ Using the Poisson summation formula one ﬁnds that

θ(t) = t−1/2θ(1/t), t > 0 . (9.7)

Lemma 9.5 For Re s > 1 we have Z ∞ −s/2 1 s −1 π Γ(s)ζ(s) = t 2 (θ(t) 1) dt . 2 0 −

96 Proof: We have

∞ 1 X 2 (θ(t) 1) = e−πn t . 2 − n=1 s −1 Multiply by t 2 and integrate:

Z ∞ ∞ Z ∞ 1 s −1 X s −1 −πn2t t 2 (θ(t) 1) dt = t 2 e dt . 2 − 0 n=1 0 Using the substitution

2 2 −1 s −1 2 1− s s −1 πn t = q, dt = (πn ) dq, t 2 = (πn ) 2 q 2 , one ﬁnds that

Z ∞ Z ∞ s −1 −πn2t 2 −s/2 s −1 −q t 2 e dt = (πn ) q 2 e dq 0 0 = (πn2)−s/2Γ(s/2) = π−s/2n−sΓ(s/2) Summation over n proves the lemma. The function

ξ(s) = π−s/2Γ(s/2)ζ(s) is meromorphic in C with simple poles at s = 1 (from ζ(s)) and at s = 0 (from Γ(s/2)). The simple poles of Γ(s/2) at s = 2, 4, 6,... are compensated by simple zeros of ζ(s) at these poles. Thus, ξ(−s) has− no− poles except at s = 0 and s = 1. Theorem 9.4 The function ξ(s) satisﬁes

ξ(s) = ξ(s 1) for s C 0, 1 . − ∈ \{ } Proof: See [Stein, Shakarchi, Theorem 2.3]. The proof is based on the previous lemma and the functional equation (9.7). One obtains:

s− 1 ζ(s) = π 2 ∆(s/2)Γ((1 s)/2)ζ(1 s) − − with 1 ∆(s/2) = . Γ(s/2) For Re s < 0 we know that ζ(1 s) = 0 and Γ((1 s)/2) = 0. Also, ∆(s/2) has a zero precisely at s = 2, −4, 66 ,... if Re s < −0. 6 We have shown: − − − Theorem 9.5 a) If Re s > 1 then ζ(s) = 0 by Euler’s product formula. b) If Re s < 0 then ζ(s) = 0 precisely6 for s = 2, 4, 6,... These so–called trivial zeros of ζ are all simple. − − −

97 9.9 Growth Estimates for the ζ–Function We use the notation

s = σ + it where σ R and t R . ∈ ∈ For n = 1, 2,... we deﬁne the entire function

Z n+1 Z n+1 −s −s −s −s δn(s) = (n x ) dx = n x dx . n − − n Clearly,

Z n+1 −s −s n = x dx + δn(s) . n If σ = Re s > 1 we sum over all n = 1, 2,... and obtain

∞ X ζ(s) = n−s n=1 Z ∞ ∞ −s X = x dx + δn(s) 1 n=1 ∞ 1 X = + δ (s) s 1 n − n=1

The following estimates for the functions δn(s) will imply that the equation

∞ 1 X ζ(s) = + δ (s) (9.8) s 1 n − n=1 holds for all s = 1 with σ = Re s > 0. 6 Lemma 9.6 If s = σ + it and σ > 0 then 2 δ (s) (9.9) | n | ≤ nσ and

s δ (s) | | . (9.10) | n | ≤ n1+σ Proof: a) We have n−s = n−σ and | | x−s = x−σ n−σ for x n . | | ≤ ≥ The estimate (9.9) follows. b) Set f(x) = x−s and note that

98 Z n+1 δn(s) = f(n) f(x) dx − n Z n+1 = (f(n) f(x)) dx n − We have f 0(x) = sx−s−1 and − s f 0(x) = s x−σ−1 | | for x n . | | | | ≤ nσ+1 ≥ The estimate

s f(n) f(x) | | for x n | − | ≤ nσ+1 ≥ follows, which yields (9.10). Lemma 9.7 For every 0 < ε 1 there is a constant C > 0 so that ≤ ε ζ(s) C t ε for 1 σ 2, t 1 . | | ≤ ε| | ≤ ≤ | | ≥ Here s = σ + it.

Proof: The formula (9.8) yields

∞ 1 X ζ(s) + δ (s) | | ≤ s 1 | n | | − | n=1 where

s 1 2 = (σ 1)2 + t2 1 , | − | − ≥ 1 thus |s−1| 1. Using the estimates for δn(s) from the previous lemma, we obtain ≤ | |

δ (s) = δ (s) ε δ (s) 1−ε | n | | n | | n | s ε 2 | | ≤ nε(σ+1) · n(1−ε)σ 2 s ε = | | nσ+ε 2 s ε | | (since σ 1) ≤ n1+ε ≥ Here

s 2 = σ2 + t2 4 + t2 5t2 , | | ≤ ≤ thus

s ε √5 t ε . | | ≤ | | 99 We obtain:

Proof: Proceeding as in the proof of the previous lemma, we have

δ (s) = δ (s) ε δ (s) 1−ε | n | | n | | n | 2 s ε | | ≤ nσ+ε 2 s ε | | (since σ 1 ε/2) ≤ n1+ε/2 ≥ − The claim follows as in the proof of the previous lemma. We can now bound the growth of ζ0(1 + it) for t using a Cauchy inequality. | | → ∞ Recall: If f H(D(z ,R)) and 0 < r < R then ∈ 0 1 f 0(z ) max f(z) : z z = r . | 0 | ≤ r {| | | − 0| } Apply this estimate to the ζ–function with z0 = 1 + it and r = ε. Using the estimates of the two previous lemmas, one obtains that 1 ζ0(1 + it) C t ε . | | ≤ ε ε| | Lemma 9.9 For every 0 < ε 1 there is a constant C > 0 so that ≤ 2 ε ζ0(1 + it) C t ε for t 2 . | | ≤ ε| | | | ≥ We also need a bound for 1/ζ(1 + it) as t . | | → ∞

100 10 Analytic Continuation

10.1 Analytic Continuation Using the Cauchy Riemann Equa- tions Discussion of the 2π–periodic Cauchy problem for the Cauchy–Riemann equations. It is ill–posed.

10.2 Exponential Decay of Fourier Coeﬃcients and the Strip of Analyticity 10.3 The Schwarz Reﬂection Principle 10.4 Examples for Analytic Continuation 10.5 Riemann Surfaces: Intuitive Approach 10.6 Riemann Surfaces: Germs, Sheafs, and Fibers

101 11 Fourier Series

11.1 Convergence Results: Overview

Let X denote the linear space of all continuous functions f : [0, 1] C with f(0) = f(1). (Some of the results that we discuss hold for more general→ functions than functions in X.) For f, g X we use the L –inner product, ∈ 2 Z 1 ¯ (f, g)L2 = f(x)g(x) dx , 0 the L2 norm,

1/2 f L = (f, f) , k k 2 L2 and the maximum norm,

f ∞ = max f(x) . | | x | | The sequence of functions

2πikx φk(x) = e , k Z , ∈ is orthonormal in X, i.e.,

(φj, φk)L2 = δjk . Let f X be given by a series, ∈ X f(x) = akφk(x) . k If the series converges uniformly, then one obtains that

ak = (φk, f)L2 . This motivates the deﬁnition of the Fourier coeﬃcients: If f L (0, 1) then set ∈ 1 ˆ f(k) = (φk, f)L2 , k Z . ∈ The series

X ˆ f(k)φk(x) k is called the Fourier series of f. Basic questions are: Under what assumptions on f does the Fourier series converge? In which sense does it converge (e.g., pointwise, or uniformly, or in L2–norm)? If the series converges, does it converge to f? With

102 n X ˆ Snf(x) = f(k)φk(x) k=−n we denote the n–th partial sum of the Fourier series of f. A reasonable question is: Given f X, is it true that ∈ S f(x) f(x) as n n → → ∞ for all x [0, 1]? In 1873, du Bois–Raymond proved that the answer is No, in general.∈ It is diﬃcult to construct an explicit example. However, using the Principle of Uniform Boundedness of functional analysis, one can show rather easily that a function f X exists for which the sequence of numbers Snf(0) is unbounded. ∈ Let f X. The sequence of arithmetic means, ∈ n 1 X σ f(x) = S f(x) , n n + 1 k k=0 can be shown to converge uniformly to f. These means are the so–called Cesaro means of the partial sums Snf(x).

Theorem 11.1 Let f X. Then we have ∈

f σ f ∞ 0 as n . | − n | → → ∞

We will prove this important result below. Let = span φ : n n n and let Tn { k − ≤ ≤ } = . T ∪nTn The functions in are called trigonometric polynomials. Using TheoremT 11.1 it is easy to prove:

Theorem 11.2 (Weierstrass) The space of trigonometric polynomials is T dense in X with respect to ∞. | · | From Weierstrass’ theorem it follows rather easily that the Fourier series S f converges to f w.r.t. : n k · kL2

Theorem 11.3 Let f X (or, more generally, let f L2(0, 1)). Then we have ∈ ∈

f S f 0 as n . k − n kL2 → → ∞

103 11.2 The Dirichlet Kernel Let f X. We have ∈ Z 1 fˆ(k) = e−2πikyf(y) dy 0 and

n X ˆ 2πikx Snf(x) = f(k)e k=−n n Z 1 X = e2πik(x−y) f(y) dy 0 k=−n Z 1 = Dn(x y)f(y) dy 0 − where

n X 2πikt Dn(t) = e , t R , (11.1) ∈ k=−n is called the Dirichlet kernel.

Lemma 11.1 The Dirichlet kernel (11.1) has the following properties:

1. ∞ Dn C (R); ∈ 2. D (t) D (t + 1) ; n ≡ n 3. Z 1 Dn(t) dt = 1 ; 0 4. sin(π(2n + 1)t) D (t) = ; n sin(πt) 5. Z 1 4 L := D (t) dt = ln n + (1) as n . n n 2 0 | | π O → ∞

Proof: We prove 4. using the geometric sum formula. Let q = e2πit = 1. We have 6

104 Dn(t)

2n + 1

0 t

Figure 11.1: Dirichlet Kernel

2n −n X j Dn(t) = q q j=0 q2n+1 1 = q−n − q 1 − qn+1/2 q−n−1/2 = − q1/2 q−1/2 − sin(π(2n + 1)t) = sin(πt) Property 5 will be shown below in Lemma 11.4. Note that, because of Property 4,

D (t) 2n + 1 for t 0 . n ∼ ∼ 1 1 A plot of Dn(t) shows that Dn(t), 2 t 2 , is concentrated near t = 0 for large n. Together with Property 3,− this≤ makes≤ it plausible that

Z 1 Snf(x) = Dn(x y)f(y) dy f(x) 0 − ∼ for large n. In fact, convergence S f(x) f(x) holds under suitable assump- n → tions on f. However, the oscillatory nature of Dn(t) makes a convergence analysis as n diﬃcult. In fact, in Section 11.8 we will use Property 5 to show existence→ of ∞ a function f X for which S f(0) diverges. ∈ n 11.3 The Fej´erKernel Let f X. We have ∈ Z 1 σnf(x) = Fn(x y)f(y) dy 0 −

105 where Fn(t) is the Fej´erkernel:

n 1 X F (t) = D (t) . n n + 1 k k=0

The following result shows that Fn(t) 0. This crucial inequality will allow us to prove Theorem 11.1. ≥

Lemma 11.2 We have

1 sin2(π(n + 1)t) F (t) = . n n + 1 sin2(πt)

Proof: Using the formula

sin(π(2k + 1)t) D (t) = k sin(πt) we must show that

n X sin2(π(n + 1)t) sin(π(2k + 1)t) = . (11.2) sin(πt) k=0 We will prove this equation using again the geometric sum formula. Let

r = eπit and q = r2 . We have

2i sin(π(2k + 1)t) = eiπ(2k+1)t e−iπ(2k+1)t 1 − = rqk q−k − r Therefore,

n X qn+1 1 1 q−n−1 1 2i sin(π(2k + 1)t) = r − − ( recall that q = r2) q 1 − r q−1 1 k=0 − − qn+1 1 q−n−1 1 = − − r r−1 − r−1 r − − r2(n+1) 2 + r−2(n+1) = − r r−1 − Here the numerator is (recall that r = eπit):

2 N = rn+1 r−n−1 = (2i)2 sin2(π(n + 1)t) − and we obtain

106 Fn(t) n

0 t

Figure 11.2: Fej´erKernel

n X sin2(π(n + 1)t) sin(π(2k + 1)t) = 2i . · r r−1 k=0 − Since

r r−1 = 2i sin(πt) − the equation (11.2) follows and the lemma is proved.

11.4 Convergence of σnf in Maximum Norm Let f X. We extend f as a 1–periodic continuous function deﬁned for all ∈ x R. ∈Below we will use that

Z 1 Fn(t) dt = 1 . 0 Furthermore, if 0 < δ 1 , then we have ≤ 2 1 sin(πt) δ > 0 for δ t . ≥ ≤ ≤ 2 Therefore, 1 1 1 F (t) for δ t . (11.3) n ≤ n + 1 δ2 ≤ ≤ 2 We have

Z 1 σnf(x) = Fn(x y)f(y) dy 0 − Z 1 = Fn(t)f(x t) dt 0 − Z 1/2 = Fn(t)f(x t) dt −1/2 −

107 and

Z 1/2 f(x) σnf(x) = Fn(t)(f(x) f(x t)) dt . − −1/2 − − First ﬁx an arbitrary (small) δ > 0. We have

Z Z f(x) σnf(x) Fn(t) f(x) f(x t) dt + Fn(t) f(x) f(x t) dt | − | ≤ |t|≤δ | − − | δ≤|t|≤1/2 | − − | Z 1/2 Mδ + 4 f ∞ Fn(t) dt (11.4) ≤ | | δ with

Mδ = max f(x) f(x t) : x R, t δ . {| − − | ∈ | | ≤ } Note that Mδ 0 as δ 0 since f is uniformly continuous. The integral in (11.4) is bounded→ by 1/(→δ2(n + 1)). If ε > 0 is given, we can ﬁnd δ > 0 with M ε/2 and then have δ ≤ Z 1/2 ε 4 f ∞ Fn(t) dt for n N(ε) . | | δ ≤ 2 ≥ This proves Theorem 11.1. 11.5 Weierstrass’ Approximation Theorem for Trigonometric Polynomials Since σ f it is clear that Theorem 11.2 follows from Theorem 11.1. n ∈ Tn

11.6 Convergence of Snf in L2 Let f X. Recall that ∈ n X ˆ 2πikx Snf(x) = f(k)e k=−n denotes the n–th partial sum of the Fourier series of f. It is clear that Snf n. The following result says that S f is the unique best approximation to f ∈in T n Tn w.r.t. the L2–norm.

Theorem 11.4 Let f X. If g is arbitrary and g = S f, then we have ∈ ∈ Tn 6 n f S f < f g . k − n kL2 k − kL2 Proof: We have

0 = (e2πijx, f S f) for j n . − n L2 | | ≤ Therefore,

108 0 = (h, f S f) for all h . − n L2 ∈ Tn If h is arbitrary, h = 0, then ∈ Tn 6

f S f h 2 = f S f 2 + h 2 k − n − kL2 k − n kL2 k kL2 > f S f 2 k − n k which proves the theorem. To prove Theorem 11.3 for f L (0, 1) we will need the following result: ∈ 2

Lemma 11.3 a) The space X is dense in L2(0, 1) with respect to the L2–norm. b) If g L (0, 1) is arbitrary, then S g g . ∈ 2 k n kL2 ≤ k kL2

Proof: a) Step functions are dense in L2 (integration theory). Then, every step function can be approximated in L2–norm by an element in X by ‘rounding the corners’ and enforcing periodicity. b) Let g L (0, 1) be arbitrary. By construction of S g we have ∈ 2 n (φ , g S g) = 0 for j n . j − n L2 | | ≤ This says that the approximation error, η := g S g, is orthogonal to the n − n space of trigonometric polynomials of degree n, i.e., to n. In particular, η = g S g is orthogonal to S g. Therefore, ≤ T n − n n

g 2 = (g S g + S g, g S g + S g) k kL2 − n n − n n L2 = (ηn + Sng, ηn + Sng)L2 = η 2 + S g 2 . k nkL2 k n kL2

Proof of Theorem 11.3: First let f X. By Theorem 11.1, given any ε > 0, there is g with ∈ ∈ T

f g ∞ < ε . | − | If g then we have for all n N: ∈ TN ≥

f S f f g f g ∞ < ε . k − n kL2 ≤ k − kL2 ≤ | − |

This proves that f Snf L2 0 as n if f X. Second, let f k L−(0, 1).k Given→ ε > 0→ there ∞ is f∈ε X with ∈ 2 ∈ f f ε < ε . k − kL2 We have

f S f f f ε + f ε S f ε + S f ε S f . k − n kL2 ≤ k − kL2 k − n kL2 k n − n kL2

109 The last term is

S (f ε f) f ε f < ε . k n − kL2 ≤ k − kL2 It follows that

f S f 2ε + f ε S f ε . k − n kL2 ≤ k − n kL2 Therefore, if n N(ε), ≥ f S f 3ε . k − n kL2 ≤ 11.7 The Lebesgue Constant of the Dirichlet Kernel Recall that

sin(π(2n + 1)t D (t) = n sin(πt) denotes the Dirichlet kernel.

Lemma 11.4 There is a constant c > 0 with

Z 1/2 Ln := Dn(t) dt c ln n . −1/2 | | ≥ More precisely, 4 L = ln n + (1) . n π2 O

The number Ln is called the Lebesgue constant of the Dirichlet kernel Dn(t). Proof: 1) For 0 < t ε we have ≤ 0 < sin(πt) = πt(1 + (t2)) , O thus 1 1 0 < = (1 + (t2)) . sin πt πt O This yields that 1 1 1 = + (1) for 0 < t . sin πt πt O ≤ 2 2) We have

110 Z 1/2 Ln = 2 Dn(t) dt 0 | | Z 1/2 sin(πt(2n + 1)) = 2 | | dt + (1) (set t(2n + 1) = s) 0 πt O n+ 1 2 Z 2 1 = sin(πs) ds + (1) π 0 s| | O n−1 2 X Z k+1 1 = sin(πs) ds + (1) π s| | O k=0 k n−1 2 Z 1 X 1 = sin(πs) ds + (1) π | | s + k O 0 k=1 For 0 s 1 and k 1 we have ≤ ≤ ≥ 1 1 1 , 1 + k ≤ s + k ≤ k thus

n−1 n−1 X 1 X 1 ln n + ln n = 1 + k − 1 + k k=1 k=1 n−1 X 1 ≤ s + k k=1 n−1 X 1 ≤ k k=1 n−1 X 1 = ln n + ln n . k − k=1 This yields 3

n−1 X 1 ln n C ln n + C. − ≤ s + k ≤ k=1 Since

Z 1 2 sin(πs) ds = 0 π the claim, 4 L = ln n + (1) , n π2 O follows. 3 1 1 Recall that γn = 1 + 2 + ... + n − ln n → γ = 0.57721 ...

111 11.7.1 The Lebesgue Constant as Norm of a Functional

As above, let X denote the linear space of all continuous functions f : [0, 1] C with f(0) = f(1). → On X we use the maximum norm, ∞. Then (X, ∞) is a Banach space. | · | | · | Deﬁne the linear functional An : X C by → Z 1 Anf = Snf(0) = Dn(y)f(y) dy . 0 One deﬁnes the norm of An by

A = sup A f : f X, f ∞ = 1 . k nk {| n | ∈ | | } It is not diﬃcult to show that An : X C is a bounded linear functional → with norm An = Ln. In fact, the estimate An Ln is easy to show. To see that it isk sharp,k set 4 k k ≤

h(y) = sgn(Dn(y)) 5 and let hε X denote a smooth approximation of h with hε ∞ = 1. We then have for ε ∈ 0: | | → Z 1 Z 1 Anhε = Dn(y)hε(y) dy Dn(y) dy = Ln . 0 → 0 | | The main point is that, by the previous lemma, we obtain that

A as n . k nk → ∞ → ∞ 11.8 Divergence of the Fourier Series of a Continuous Function at a Point In 1873, du Bois–Raymond constructed a continuous function whose Fourier series diverges at a point. We use the Uniform–Boundedness Principle of functional analysis. (This is also called the Banach–Steinhaus Theorem or the Resonance Principle.) First recall some simple concepts from functional analysis. If X,Y are normed spaces, then a linear operator

T : X Y → is called bounded if there is a constant C 0 with ≥ T f C f for all f X. (11.5) k kY ≤ k kX ∈ If T : X Y is a bounded linear operator, then its operator norm is deﬁned by →

4Here sgn denotes the sign–function, sgn(x) = 1 for x > 0, sgn(x) = −1 for x < 0, and sgn(0) = 0. 5 R 1 The functions hε(y) must satisfy 0 |h(y) − hε(y)|dy → 0 as ε → 0. One can choose hε(y) as a continuous, piecewise linear function.

112 T = min C 0 : (11.5) holds k k { ≥ } = sup T x : x X, x = 1 {k kY ∈ k kX } The linear space L(X,Y ) of all bounded linear operators from X to Y , together with the operator norm, is a normed space. The Uniform–Boundedness Principle is the following remarkable result:

Theorem 11.5 Let X be a Banach space and Y be a normed space. Let L(X,Y ). Assume that for every f X there is a constant C with T ⊂ ∈ f T f C for all T . k kY ≤ f ∈ T Then there is a constant C with

T C for all T . k k ≤ ∈ T The following is a reformulation:

Theorem 11.6 Let X be a Banach space and Y be a normed space. Let L(X,Y ). T ⊂ Assume that

sup T : T = . (11.6) {k k ∈ T } ∞ Then there is f X so that ∈ sup T f : T = . (11.7) {k kY ∈ T } ∞ This reformulation is called Resonance Principle. Roughly, if (11.6) holds, then there exists f X which resonates with the family of operators, and (11.7) holds. ∈ T Application: Let (X, ∞) denote the Banach space of all continuous functions | · |

f : [0, 1] C → with f(0) = f(1) equipped with the maximum norm. Let Y = C with z Y = z . For n = 1, 2,... let k k | |

An : X C → be deﬁned by

Z 1 Anf = Snf(0) = Dn(y)f(y) dy, f X. 0 ∈

As stated above, the operator norm of An is Ln and Ln by Lemma 11.4. We apply the resonance principle to the family → ∞

113 = A : n = 1, 2,... . T { n } Since An as n the resonance principle implies existence of a functionk fk →X for ∞ which→ ∞ ∈

Anf = Snf(0) is unbounded.

11.9 Isomorphy of L2(0, 1) and l2 So far, we have shown that

f σ f ∞ 0 as n for f X | − n | → → ∞ ∈ and

f S f 0 as n for f L (0, 1) . (11.8) k − n kL2 → → ∞ ∈ 2 Also, if f L2(0, 1), then the construction of Snf shows that the approximation error ∈

f S f − n is orthogonal to . Therefore, Tn f 2 = f S f 2 + S f 2 for f L (0, 1) . k kL2 k − n kL2 k n k ∈ 2 Since

n X S f 2 = fˆ(k) 2 k n kL2 | | k=−n one obtains Parseval’s relation:

∞ X f 2 = fˆ(k) 2 . (11.9) k kL2 | | k=−∞

Deﬁnition: Let l2 denote the linear space of all sequences

a = (ak)k∈ , ak C , Z ∈ with X a 2 < . | k| ∞ k For a, b l deﬁne the l inner product by ∈ 2 2 X (a, b)l2 = a¯kbk . k

114 As usual, the corresponding norm is deﬁned by

a 2 = (a, a) . k kl2 l2

It is not diﬃcult to show:

Theorem 11.7 The sequence space l2 with the above inner product is complete, i.e., it is a Hilbert space.

Theorem 11.8 The mapping F : L (0, 1) l deﬁned by 2 → 2 ˆ F (f) = (f(k)) ∈ , f L (0, 1) , k Z ∈ 2 is a Hilbert space isomorphism.

Proof: Parseval’s relation (11.9) shows that F maps L2(0, 1) into l2 and is norm preserving. The linearity of F is clear. Since F f = f it is also k kl2 k kL2 clear that F is one–to–one. To prove that F is onto, let a l2 be given and deﬁne ∈

n X fn(x) = akφk(x) . k=−n Then f is a Cauchy sequence in L (0, 1), thus there exists f L (0, 1) with n 2 ∈ 2 f f 0 as n . k − nkL2 → → ∞ We have

(φk, f)L = lim (φk, fn)L = ak . 2 n→∞ 2 Therefore, F f = a, showing that F is onto. It remains to prove that F preserves the inner product:

(F f, F g)l2 = (f, g)L2 . (11.10) The polarization equality,

4(f, g) = f + g 2 f g 2 + i if + g 2 i if g 2 , k k − k − k k k − k − k which is valid in any inner product space over C, shows that one can express the inner product in terms of the norm. Therefore, (11.10) follows from (11.9).

115 11.10 Convergence of Snf in Maximum Norm Lemma 11.5 Let f : R C, f(x + 1) f(x). If f Cr then → ≡ ∈ ˆ −r f(k) Cr k , k Z, k = 0 , | | ≤ | | ∈ 6 with

Z 1 −r (r) Cr = (2π) f (x) dx . 0 | | Consequently,

f S f ∞ 0 as n | − n | → → ∞ if r 2. ≥ Proof: Through integration by parts one ﬁnds that 1 fˆ(k) = (f 0)ˆ(k) . 2πik Applying this result r times and noting that

Z 1 (f (r))ˆ(k) = e−2πikxf (r)(x) dx 0 the estimate of fˆ(k) follows. Therefore, if r 2, we have for n > m N(ε): | | ≥ ≥ X S f S f ∞ fˆ(k) < ε . | n − m | ≤ | | m<|k|≤n

It follows that S f is a Cauchy sequence in (X, ∞), and there exists g X n | · | ∈ with g S f ∞ 0. This implies g S f 0. We have shown that | − n | → k − n kL2 → f Snf L2 0, and conclude that g = f. k −One cank relax→ the assumption f C2 slightly. ∈ Theorem 11.9 Let f : R C, f(x + 1) f(x), f C1. Then the Fourier sums → ≡ ∈

n X ˆ 2πikx Snf(x) = f(k)e k=−n converge uniformly to f(x),

f S f ∞ 0 as n . | − n | → → ∞ Proof: From 1 fˆ(k) = (f 0)ˆ(k) 2πik and

116 ∞ X (f 0)ˆ(k) 2 = f 0 2 | | k kL2 k=−∞ we conclude that

∞ X k 2 fˆ(k) 2 =: Q < . | | | | ∞ k=−∞ Using the Cauchy–Schwarz inequality,

X S f S f ∞ fˆ(k) | n − m | ≤ | | m<|k|≤n X 1 = k fˆ(k) k | || | m<|k|≤n | | X 1 1/2 Q1/2 . ≤ k 2 m<|k|≤n | |

As in the proof of the previous lemma, it follows that Snf is a Cauchy sequence in (X, ∞), and the claim follows. | · | 11.11 Smoothness of f(x) and Decay of fˆ(k)

Let f : R C, f(x) f(x + 1), f C. Since → ≡ ∈ X f 2 = fˆ(k) 2 k kL2 | | k we know that

fˆ(k) 0 as k . → | | → ∞ We want to show that the smoothness of f is related to the decay rate of fˆ(k). Smoothness implies decay:

− Lemma 11.6 Let r 1, 2,... . Let f Cr 1(R) have period one, and assume that Drf = f (r) L ∈. {Then we} have ∈ ∈ 1 fˆ(k) C k −r for k = 0 | | ≤ r| | 6 with

Z 1 −r r Cr = (2π) D f(x) dx . 0 | | Proof: This follows through integration by parts,

117 Z 1 fˆ(k) = e−2πikxf(x) dx 0 1 Z 1 = e−2πikxDf(x) dx 2πik 0 1 Z 1 = e−2πikxDrf(x) dx r (2πikx) 0

Decay implies smoothness:

Lemma 11.7 Let r 0, 1, 2,... . Let f : R C have period one and let f L (0, 1). Assume∈ that { there exists} A > 0, δ >→0 with ∈ 1 fˆ(k) A k −(r+1+δ) for all k = 0 . | | ≤ | | 6 Then f Cr. ∈ Proof: We assume that the above estimate holds for r = 1 and must show that f C1. We have ∈

n X ˆ 2πikx Snf(x) = f(k)e k=−n n 0 X ˆ 2πikx (Snf) (x) = 2πikf(k)e k=−n For n > m:

0 0 X (S f) (S f) ∞ 2π k fˆ(k) | n − m | ≤ | || | m<|k|≤n X 2πA k −(1+δ) ≤ | | m<|k|≤n ε ≤ for n > m N(ε). Therefore, theer exists g X with ≥ ∈ 0 g (S f) ∞ 0 as n . | − n | → → ∞ Also,

f S f ∞ 0 as n . | − n | → → ∞ Let n in the equation → ∞ Z x 0 (Snf)(x) (Snf)(0) = (Snf) (y) dy − 0

118 to obtain that

Z x f(x) f(0) = g(y) dy . − 0 It follows that f C1 and f 0 = g. The proof for other values of r is similar. ∈ 11.12 Exponential Decay of fˆ(k) and Analyticity of f

For α > 0 let Sα denote the horizontal strip of width 2α along the real axis:

Sα = z = x + iy : x R, y < α . { ∈ | | } Lemma 11.8 Let f H(S ), f(z) f(z + 1). Then, for any 0 β < α, there ∈ α ≡ ≤ is a constant Cβ = Cβ(f) with

ˆ −2π|k|β f(k) Cβe for all k Z . | | ≤ ∈ Proof: For deﬁniteness, let k < 0. Consider the rectangle

Γ1 + Γ2 + Γ3 + Γ4 where

Γ : z(x) = x, 0 x 1 , 1 ≤ ≤ Γ : z(y) = 1 + iy, 0 y β , 2 ≤ ≤ Γ : z(x) = x + iβ, 0 x 1 , − 3 ≤ ≤ Γ : z(y) = iy, 0 y β . − 4 ≤ ≤ Let

g(z) = f(z)e−2πikz . We then have Z fˆ(k) = g(z) dz Γ1 and, by Cauchy’s theorem,

4 X Z g(z) dz = 0 . j=1 Γj The periodicity of f also implies that Z Z g(z) dz + g(z) dz = 0 . Γ2 Γ4 Consequently,

119 Z fˆ(k) = g(z) dz . −Γ3

We have justiﬁed to move the integration path Γ1 upwards by β. If z Γ3 then z = x + iβ and ∈

e−2πikz = e2πkβ = e−2π|k|β for k < 0 . | | It follows that

fˆ(k) C e−2π|k|β | | ≤ β with

Cβ = max f(x + iβ) . 0≤x≤1 | | If k > 0 we can argue similarly, moving the path of integration down by β. The previous lemma says that the Fourier coeﬃcients fˆ(k) of a periodic function f(x) decay exponentially as k if f(x) can be continued analytically into a strip along the real axis.| The| → following ∞ lemma shows a converse.

Lemma 11.9 Let γ > 0 and let a = (a ) l denote a sequence with k ∈ 2 −2π|k|γ ak Ce for all k Z . | | ≤ ∈ Then there is a unique function f H(S ) with f(z) f(z + 1) and ∈ γ ≡ ˆ f(k) = ak, k Z . ∈ Proof: Set

a e2πikz Ce−2π|k|(γ−β) where γ β > 0 . | k|| | ≤ − ¯ If K Sγ is any compact set, then K Sβ for some β with 0 < β < γ, and the above⊂ estimate implies that the series⊂

X 2πikz ake k converges normally in S to an analytic function f H(S ). It is clear that γ ∈ γ fˆ(k) = a and that f is unique. k

120 11.13 Divergence of Snf(0): Explicit Construction of f The following is close to [Koerner]. We show existence of a function f X for which ∈

Snf(0) is unbounded. Recall that

Z 1/2 Snf(y) = Dn(y x)f(x) dx , −1/2 − thus

Z 1/2 Z 1/2 Snf(0) = Dn( x)f(x) dx = Dn(x)f(x) dx . −1/2 − −1/2 Here

sin(π(2n + 1)x) D (x) = n sin(πx) is the Dirichlet kernel. We have shown that the Lebesgue constants diverge,

Z 1/2 Ln = Dn(x) dx as n . −1/2 | | → ∞ → ∞ Let

hn(x) = sgn Dn(x) . We then have

S h (0) = L as n . n n n → ∞ → ∞ The functions hn are not continuous, but we can approximate hn by gn,ε X so that ∈

Z 1/2 hn(x) gn,ε dx < ε and gn,ε ∞ = 1 . −1/2 | − | | | Choosing εε(n) > 0 small enough, one ﬁnds that

S g (0) L 1, g ∞ = 1 . n n,ε ≥ n − | n,ε| Since the trigonometric polynomials are dense in (X, ∞), there exists fn 0 | · | ∈ T with g f ∞ < ε . One then ﬁnds that | n,ε − n|

S f (0) L 2, f ∞ 2 . n n ≥ n − | n| ≤ 1 Dividing by 2, we have obtained a trigonometric polynomial 2 fn with 1 1 1 S ( f )(0) (L 2), f ∞ 1 . n 2 n ≥ 2 n − |2 n| ≤ Since L as n , the following lemma holds: n → ∞ → ∞ 121 Lemma 11.10 Given any constant A > 0 there exists H and N N with ∈ T ∈

S H(0) A, H ∞ 1 . | N | ≥ | | ≤ The remaining part of the proof may be called piling up of bad functions. 2k Take A = Ak = 2 for k = 1, 2,... By the previous lemma we obtain sequences Hk and n(k) N with ∈ T ∈ 2k S H (0) 2 , H ∞ 1 . | n(k) k | ≥ | k| ≤ We write

X ˆ 2πijx Hk(x) = Hk(j)e (11.11) |j|≤q(k) and may assume that

q(k) n(k) . ≥ Set

k X p(k) = (2q(j) + 1) j=1 and deﬁne the trigonometric polynomials

n X −k 2πip(k)x fn(x) = 2 Hk(x)e , n = 1, 2,... (11.12) k=1 If n > m then

n X −k −m f f ∞ 2 2 . | n − m| ≤ ≤ k=m+1

This shows that fn is a Cauchy sequence in X. Since (X, ∞) is complete there exists f X with | · | ∈

f f ∞ 0 . | − n| → We will show that the sequence Snf(0) is unbounded. Convergence f f ∞ 0 implies that | − n| → ˆ ˆ fn(r) f(r) as n for all r Z . → → ∞ ∈ The main point of the construction is the following: The integers p(k) are so large that the terms in the sum (11.12) have no common frequencies. This makes the following analysis possible. Substituting the right side of (11.11) for Hk(x) in (11.12) yields

n X −k X ˆ fn(x) = 2 Hk(j) exp(2πi(p(k) + j)x) k=1 |j|≤q(k)

122 and we obtain

fˆ (p(k) + j) = 2−kHˆ (j) if n k and j q(k) . n k ≥ | | ≤ Letting n we see that → ∞ fˆ(p(k) + j) = 2−kHˆ (j) if j q(k) . k | | ≤ Now consider

X S f(0) S f(0) = fˆ(p(k) + j) p(k)+n(k) − p(k)−n(k)−1 |j|≤n(k) −k X ˆ = 2 Hk(j) |j|≤n(k) −k = 2 Sn(k)Hk(0)

(In the second equation we have used that n(k) q(k).) We now take absolute values in the above equation and recall the estimate≤

S H (0) 22k | n(k) k | ≥ to obtain

S f(0) S f(0) 2k, k = 1, 2,... | p(k)+n(k) − p(k)−n(k)−1 | ≥ This implies that the sequence Snf(0) is unbounded.

11.14 Fourier Series and the Dirichlet Problem for Laplace’s Equation on the Unit Disk

The Poisson kernel Pr(α) is the Abel sum of

∞ 1 X eikα . 2π k=−∞ In other words,

∞ ∞ 1 X X P (α) = (reiα)k + (re−iα)k for 0 r < 1 . r 2π ≤ k=0 k=1

123 12 Fourier Transformation

12.1 Motivation: Application to PDEs Example 1: Consider the PDE

ut + aux = 0 for a function u(x, t) where a R. Give an initial condition ∈ ikx u(x, 0) = e , x R . ∈ The initial function

eikx = cos(kx) + i sin(kx) is a wave with wavenumber k (assumed to be real). Roughly, the wave eikx has k waves in the interval 0 x 2π and the wavelength is 2π/ k . To obtain a | | ≤ ≤ | | ikx solution of the PDE ut + aux = 0 with initial condition u(x, 0) = e try the ansatz

u(x, t) = α(t)eikx . One obtains that

α0(t) + iakα(t) = 0, α(0) = 1 , thus

α(t) = e−iakt, u(x, t) = eik(x−at) . The solution u(x, t) describes that the initial wave eikx moves undistorted at speed a. If a > 0 the wave moves to the right, if a < 0 it moves to the left. Example 2: Consider the heat equation

ut = uxx for x R, t 0 , ∈ ≥ with initial condition

ikx u(x, 0) = e , x R . ∈ The ansatz

u(x, t) = α(t)eikx leads to

α0(t) = k2α(t), α(0) = 1 , − with solution

2 α(t) = e−k t .

124 ikx The solution of the PDE ut = uxx with initial condition u(x, 0) = e is

2 u(x, t) = e−k teikx . We see that the amplitude decays exponentially as t increases if k = 0. The smaller the wavelength of the initial function u(x, 0) = eikx, the more6 rapid is the decay in time. We expect, therefore, that the solution of the heat equation will become smooth if we start with rough initial data u(x, 0) = f(x). Example 3: Consider the linearized Korteweg–de Vries equation

ut = uxxx for x R, t 0 , ∈ ≥ with initial condition

ikx u(x, 0) = e , x R . ∈ The ansatz

u(x, t) = α(t)eikx leads to

α0(t) = ik3α(t), α(0) = 1 , − with solution

3 α(t) = e−ik t . ikx The solution of the PDE ut = uxxx with initial condition u(x, 0) = e is

2 u(x, t) = eik(x−k t) . We see that the amplitude α(t) satisfies α(t) = 1, thus there is neither decay nor growth. Here the wave speed is k2,| i.e.,| the wave speed depends on the wave length of the initial function eikx. We have the effect of dispersion. Example 4: Consider the free–space Schrödingerequation

ut = iuxx for x R, t 0 , ∈ ≥ with initial condition

ikx u(x, 0) = e , x R . ∈ The ansatz

u(x, t) = α(t)eikx leads to

α0(t) = ik2α(t), α(0) = 1 , − with solution

125 2 α(t) = e−ik t .

ikx The solution of the PDE ut = iuxx with initial condition u(x, 0) = e is

u(x, t) = eik(x−kt) . We see that the amplitude α(t) satisfies α(t) = 1, thus there is neither decay nor growth. Here the wave speed is k, i.e.,| the| wave speed depends on the wave length of the initial function eikx. We have the effect of dispersion. More General Equation: Consider the constant–coefficient equation

m X j ut = Lu ajD u for x R, t 0 , ≡ ∈ ≥ j=0 where

∂ aj C,D = . ∈ ∂x Given an initial condition

ikx u(x, 0) = e , x R , ∈ the ansatz u(x, t) = α(t)eikx leads to

α0(t) = Lˆ(ik)α(t), α(0) = 1 , where

m ˆ X j L(ik) = aj(ik) j=0 is the so–called symbol of L. The solution of the amplitude equation is

α(t) = eLˆ(ik)t

ikx and the solution of the PDE ut = Lu with initial condition u(x, 0) = e is

u(x, t) = eLˆ(ik)teikx . If one wants to solve the initial value problem

ut = Lu, u(x, 0) = f(x) , with more general functions f(x) then one might try to write f(x) as a super- position of the simple waves eikx. The tool is the Fourier transform.

126 12.2 The Fourier Transform of an L1 Function

For f L1 = L1(R, C) one deﬁnes its Fourier transform by ∈ Z ∞ −2πixξ fˆ(ξ) = f(x)e dx, ξ R . (12.1) −∞ ∈ (We follow the conventions in [Stein, Shakarchi]. Other forms, like Z ∞ 1 −ixξ fˆ(ξ) = f(x)e dx, ξ R , √2π −∞ ∈ are also used.) Remark: The condition f L1 is necessary if one wants to use the standard meaning of the integral in (12.1).∈ However, using the notions of distribution theory, one can deﬁne a Fourier transform for functions and distributions that are not in L1.

Lemma 12.1 If f L then fˆ is uniformly continuous. ∈ 1

Z ˆ ˆ −2πixξ1 −2πixξ2 f(ξ1) f(ξ2) f(x) e e dx | − | ≤ |x|>R | || − | Z + f(x) e−2πixξ1 e−2πixξ2 dx |x|≤R | || − | ε −2πixξ1 −2πixξ2 2 + f L1 max e e . ≤ 4 · k k |x|≤R | − |

Here Z ∞

f L1 = f(x) dx . k k −∞ | | The maximum in the above formula equals

−2πxξ M = max 1 e where ξ = ξ2 ξ1 . |x|≤R | − | − Since the function

g(x, ξ) = 1 e−2πxξ | − | is uniformly continuous on the set

127 x R, ξ 1 | | ≤ | | ≤ and since

g(x, 0) = 0 , there is δ > 0 so that

g(x, ξ) < ε0 for ξ < δ, x R. | | | | ≤ Choosing ε 1 ε0 = 2 f k kL1 we obtain fˆ(ξ ) fˆ(ξ ) < ε for ξ ξ < δ. | 1 − 2 | | 1 − 2| What can be said about the decay of fˆ(ξ) as ξ ? Through integration by parts one obtains: | | | | → ∞

Lemma 12.2 Let f C1, i.e., f is a C1 function with compact support. Then ∈ 0 1 fˆ(ξ) f 0 , ξ = 0 . | | ≤ 2π ξ k kL1 6 | | Proof: We have

Z ∞ fˆ(ξ) = f(x)e−2πixξ dx −∞ 1 1 Z ∞ = f(x) e−2πixξ x=∞ + f 0(x)e−2πixξ dx 2πiξ |x=−∞ 2πiξ − −∞ The estimate follows. 12.3 The Riemann–Lebesgue Lemma The following result is known as the Riemann–Lebesgue Lemma:

Lemma 12.3 If f L then fˆ(ξ) 0 as ξ . ∈ 1 | | → | | → ∞ To prove the Riemann–Lebesgue Lemma, we need the following auxiliary result:

∞ ∞ Lemma 12.4 The set C0 of all C functions with compact support is dense in (L , ). 1 k · kL1 Proof: Let f L and let ε > 0 be given. Choose R > 0 so large that ∈ 1 Z ε f(x) dx < . |x|>R | | 2

128 We apply a cut–oﬀ to the function f and set

f(x) x R g(x) = . 0 |x|> ≤ R | | ε We then have f g L1 2 and g has compact support. We now mollifyk − gk. This≤ process is as follows: Choose a C∞ function Φ : R R with → Φ(y) > 0 for 1 < y < 1, Φ(y) = 0 for y 1 − | | ≥ and Z ∞ Φ(y) dy = 1 . −∞ Note that

Z ∞ 1 Φ(y/ε) dy = 1 for ε > 0 −∞ ε and the function y Φ(y/ε) is supported in y ε. Deﬁne → | | ≤ Z ∞ 1 gε(x) = Φ((x y)/ε)g(y) dy . −∞ ε − It is not diﬃcult to prove that g C∞ and g g 0 as ε 0. ε ∈ 0 k − εkL1 → → Proof of Riemann–Lebesgue Lemma: Let f L1 and let ε > 0 be given. There exists f C∞ so that ∈ ε ∈ 0 ε f f < . k − εkL1 2 This implies that

ˆ ˆ ε f(ξ) fε(ξ) < for ξ R . | − | 2 ∈ Since

C fˆ (ξ) ε for ξ = 0 | ε | ≤ ξ 6 | | we obtain the estimate

ε C fˆ(ξ) + ε < ε for ξ R . | | ≤ 2 ξ | | ≥ ε | | Example 1: Let 1 x 1 f(x) = . 0 |x|> ≤ 1 | | We have

129 Z 1 fˆ(ξ) = e−2πixξ dx −1 1 = e−2πixξ x=1 2πiξ |x=−1 − sin(2πξ) = . πξ Since

Z R fˆ(ξ) dξ as R 0 | | → ∞ → ∞ ˆ we conclude that f / L1. The example shows that, in general, the assumption f L does not imply∈ fˆ L . ∈ 1 ∈ 1 12.4 The Fourier Transform on the Schwartz Space S A function f : R C is called rapidly decreasing if → n x f(x) , x R , | | ∈ is bounded for all n N. The Schwartz space = (R, C) consists of all f ∞ ∈ S S ∈ C (R, C) for which all derivatives f (k)(x), k = 0, 1, 2,... are rapidly decreasing. For f the Fourier transform ∈ S Z ∞ −2πixξ fˆ(ξ) = f(x)e dx, ξ R , −∞ ∈ is well–deﬁned.

Theorem 12.1 a) If f then fˆ . b) If f then ∈ S ∈ S ∈ S (f 0)ˆ(ξ) = 2πiξfˆ(ξ) . c) If f then 2πixf(x) and ∈ S − ∈ S (fˆ)0(ξ) = 2πixf(x) ˆ(ξ) . −

Proof: We know that fˆ C and fˆ is bounded. Further, f 0 and, through integration by parts, ∈ ∈ S

Z (f 0)ˆ(ξ) = f 0(x)e−2πixξ dx Z = f(x)(2πiξ)e−2πixξ dx

= 2πiξfˆ(ξ)

130 This shows that ξfˆ(ξ) is bounded and b) holds. By further diﬀerentiations, we see that ξnfˆ(ξ) is bounded for every n. We now| show| that fˆ C1 and ∈ (fˆ)0(ξ) =g ˆ(ξ) with g(x) = 2πixf(x) . − To this end, let ξ R be ﬁxed and let h be a real number, h = 0. Consider ∈ 6

1 Q := (fˆ(ξ + h) fˆ(ξ)) gˆ(ξ) h h − − Z 1 = f(x)e−2πixξ (e−2πixh 1) + 2πix dx . h − We must show that for any ε > 0 there is h > 0 so that Q < ε for 0 < h 0 | h| | | ≤ h0. It is clear that for real α:

1 2/α for α 1 (e−2πiα 1) | | ≥ α − ≤ C for 0 < α 1 | | ≤ Therefore, the function (e−2πiα 1)/α is bounded for real α = 0. It follows that − 6 1 (e−2πixh 1) + 2πix C x h − ≤ | | for xh = 0. For any R > 0 we have 6

Z Z 1 −2πixh Qh C f(x) x dx + f(x) (e 1) + 2πix dx . | | ≤ |x|≥R | || | |x|≤R | | h −

We choose R > 0 so that the ﬁrst term is bounded by ε/2. Then, for x R, we have | | ≤ 1 (e−2πixh 1) + 2πix C h . h − ≤ R| |

It is then clear that Qh < ε if h is small enough. This proves c). With the same arguments as above,| | it follows| | that ξn(fˆ)0(ξ) is bounded for every n. The argument can be repeated for every derivative| of| fˆ(ξ), and the theorem is proved. 12.5 The Fourier Inversion Formula on the Schwartz Space: Preparations The following is an important deﬁnite integral:

Lemma 12.5 ∞ Z 2 J := e−πx dx = 1 . −∞

131 Proof: Using Fubini’s theorem and transformation to polar coordinates, we have

∞ ∞ Z Z 2 2 J 2 = e−π(x +y ) dxdy −∞ −∞ ∞ Z 2 = 2π re−πr dr 0 Z ∞ = e−ρ dρ 0 = 1 .

We next show that the Fourier transform of a Gaussian is a Gaussian:

Lemma 12.6 The Fourier transform of

2 f(x) = e−πx is

2 fˆ(ξ) = e−πξ .

Proof: Set

∞ Z 2 F (ξ) = e−πx e−2πixξ dx . −∞ We then have F (0) = 1 and, using integration by parts,

Z 2 F 0(ξ) = e−πx ( 2πix)e−2πixξ dx − Z d 2 = i (e−πx ) e−2πixξ dx dx = 2πξF (ξ) − Thus we have shown that the function fˆ(ξ) = F (ξ) satisﬁes

F 0(ξ) = 2πξF (ξ),F (0) = 1 . − Unique solubility of this initial value problem proves that

2 fˆ(ξ) = F (ξ) = e−πξ .

By linear scalings (x/√δ = y) one obtains:

132 Lemma 12.7 For δ > 0 let

1 −πx2/δ −πξ2δ Kδ(x) = e ,Gδ(ξ) = e . √δ Then we have

ˆ ˆ Kδ = Gδ and Gδ = Kδ .

Further properties of the family of functions Kδ(x) are:

Lemma 12.8 a) Z ∞ Kδ(x) dx = 1, δ > 0 . −∞ b) If η > 0 is ﬁxed, then Z Kδ(x) dx 0 as δ 0 . |x|≥η → → c) Z ∞ x Kδ(x) dx 0 as δ 0 . −∞ | | → → d) If f then ∈ S Z ∞ f(x)Kδ(x) dx f(0) as δ 0 . −∞ → →

Proof: a) follows from Z ∞ Kδ(x) dx = Gδ(0) = 1 . −∞ Using the substitution x/√δ = y we have

Z ∞ Z ∞ √ 1 −π(x/ δ)2 Kδ(x) dx = e dx η √δ η Z ∞ −πy2 = √ e dy η/ δ and b) follows. To show c) we compute

Z ∞ Z ∞ √ 1 −π(x/ δ)2 xKδ(x) dx = xe dx 0 √δ 0 ∞ Z 2 = √δ e−πy dy . 0 Finally,

133 Z ∞ Z ∞ f(x)Kδ(x) dx = f(0) + (f(x) f(0))Kδ(x) dx . −∞ −∞ − Here f(x) f(0) C x , and the claim follows from c). | − | ≤ | | We next prove the following multiplication rule:

Theorem 12.2 a) If f, g then fg . Therefore, fgˆ . b) If f, g then ∈ S ∈ S ∈ S ∈ S Z ∞ Z ∞ f(x)ˆg(x) dx = fˆ(y)g(y) dy . −∞ −∞ Proof: If f, g then Leibniz’ rule yields fg . Furthermore, using Fubini’s theorem, ∈ S ∈ S

Z ∞ Z ∞ Z ∞ f(x)ˆg(x) dx = f(x)g(y)e−2πixy dydx −∞ −∞ −∞ Z ∞ Z ∞ = f(x)g(y)e−2πixy dxdy −∞ −∞ Z ∞ = fˆ(y)g(y) dy −∞ This proves the multiplication rule. 12.6 The Fourier Inversion Formula on the Schwartz Space Any function f can be written as a Fourier integral. The formula (12.2) is called the Fourier∈ S representation of f(x) or the Fourier inversion formula.

Theorem 12.3 If f then we have ∈ S Z ∞ 2πixξ f(x) = fˆ(ξ)e dξ, x R . (12.2) −∞ ∈

Proof: First, since the Fourier transform of Gδ(ξ) is Kδ(x), we have for any δ > 0:

∞ ∞ Z 1 2 Z 2 f(x) e−πx /δ dx = fˆ(ξ)e−πξ δ dξ . (12.3) −∞ √δ −∞ In this equation we take the limit δ 0 and obtain → Z ∞ f(0) = fˆ(ξ) dξ . (12.4) −∞ This proves the Fourier inversion formula for x = 0. Next, ﬁx x and set

F (y) = f(x + y), y R . ∈ We have

134 Z ∞ Fˆ(ξ) = f(x + y)e−2πiyξ dy −∞ Z ∞ = f(z)e−2πizξe2πixξ dz −∞ = fˆ(ξ) e2πixξ .

Using (12.4) with f replaced by F yields

f(x) = F (0) Z ∞ = Fˆ(ξ) dξ −∞ Z ∞ = fˆ(ξ)e2πixξ dξ . −∞ This proves the inversion formula. 12.7 Operators We can summarize the results by deﬁning the following operators from into itself: S

Z ( f)(ξ) = f(x)e−2πixξ dx F Z ( g)(x) = g(ξ)e2πixξ dξ G (Sf)(x) = f( x) − (Bf)(x) = f¯(x)

The Fourier Inversion Theorem says that

= idS . GF We also see that

= S = B B. G F F Since SS = BB = idS one obtains that

4 = S, = idS . FF F In particular, : is linear, one–to–one, and onto. The followingF S result → S is Plancherel’s formula.

Theorem 12.4 For f, g deﬁne the L inner product by ∈ S 2 Z ¯ (f, g)L2 = f(x)g(x) dx .

135 Then we have

ˆ (f, g)L2 = (f, gˆ)L2 .

Proof: Deﬁne

φ = Bf = B f . G F Then, using the multiplication formula,

Z ¯ (f, g)L2 = fg dx Z = φgˆ dx Z = φgˆ dξ Z = (Bfˆ)ˆgdξ ˆ = (f, gˆ)L2

Corollary For all f we have ∈ S f = fˆ . (12.5) k k k k 12.8 Elementary Theory of Tempered Distributions On the Schwartz space one deﬁnes a convergence concept as follows: If g S n ∈ S then gn 0 in means that for all j, k N we have → S ∈ j (k) sup x gn (x) 0 as n . x | | → → ∞

If gn, g then gn g in means that g gn 0 in . This convergence concept∈ is S very restrictive.→ S − → S A linear functional φ : C is called continuous if gn 0 in implies S → → S that φ(gn) 0 in C. Since the convergence concept in is very restrictive, the continuity requirement→ of a linear functional is very mild.S The linear space of all continuous linear functionals on is called the dual space of , denote by 0. Every element of 0 is called a temperedS distribution. Notation:S If φ S 0 and g we oftenS write ∈ S ∈ S

φ(g) = (φ, g)S0S to denote the value of φ at g. This notation emphasizes the linearity of the expression φ(g) in φ and in g.

136 12.8.1 Ordinary Functions as Tempered Distributions If f L satisﬁes an estimate ∈ 1,loc f(x) K x k for x K , (12.6) | | ≤ | | | | ≥ 1 for some constants K,K1 and k, then f determines the tempered distribution φf deﬁned by Z ∞ φf (g) = f(x)g(x) dx, g . −∞ ∈ S We also denote the expression by

(f, g)S0S and identify φf with f. In particular, since every f satisﬁes an estimate (12.6), we obtain the inclusion 0. ∈ S S ⊂ S 12.9 The Fourier Transform on 0: An Example Using Complex Variables S For f, g we have ∈ S Z ∞ Z ∞ ( f)(x)g(x) dx = f(x)( g)(x) dx . −∞ F −∞ F Therefore, if φ 0 we deﬁne φ 0 by ∈ S F ∈ S

( φ, g)S0S = (φ, g)S0S for all g . F F ∈ S We want to show by an example how the theory of complex variables is used in determining the Fourier transform of the distribution (determined by)

−πx2is f(x) = e , x R , ∈ 0 where s R, s = 0, is ﬁxed. Note that f L∞, so f . However, f / L1, so the Fourier∈ 6 transform of f cannot be∈ obtained directly∈ S by evaluating∈ the Fourier integral formula. Recall the deﬁnitions

−πx2δ 1 −πξ2/δ Gδ(x) = e ,Kδ(ξ) = e √δ where x and ξ are real and where δ > 0. The functions Gδ and Kδ belong to and we have S

Gˆ = K in δ δ S for every δ > 0. Formally setting δ = is suggests that the Fourier transform of f(x) is

1 2 F (ξ) = e−πξ /is √is

137 in the sense of distributions. Here √z is obtained by analytically continuing the function √δ, δ > 0, to C ( , 0]. \ −∞ 0 Note that f and F both belong to L∞, thus f and F both belong to . We claim that (f) = F in 0. This means that we have for all g : S F S ∈ S ∞ ∞ Z 2 1 Z 2 e−πx isgˆ(x) dx = e−πξ /isg(ξ) dξ . −∞ √is −∞ To show this, ﬁx g and introduce the functions ∈ S

∞ Z 2 L(z) = e−πx zgˆ(x) dx −∞ ∞ 1 Z 2 R(z) = e−πξ /zg(ξ) dξ √z −∞ for Re z 0, z = 0. Then we know that ≥ 6 L(δ) = R(δ) for δ > 0 . Furthermore, L(z) and R(z) are analytic functions in the open right half–plane

H = z : Re z > 0 . r { } Therefore, L(z) = R(z) in Hr. Now ﬁx s R, s = 0. The point z0 = is is a ∈ 6 boundary point of the half–plane Hr. We claim that

lim L(z) = L(is) , z→is,z∈Hr and a similar relation hold for R(z). To see this, let z H and let ε > 0 be ∈ r given. For suﬃciently large x0 we have

∞ Z 2 2 L(is) L(z) e−πx is e−πx z gˆ(x) dx | − | ≤ −∞ | − || |

Z Z 2 2 2 gˆ(x) dx + e−πx is e−πx z gˆ(x) dx ≤ |x|≥x0 | | |x|≤x0 | − || | −πx2is −πx2z ε + gˆ L1 max e e . ≤ k k |x|≤x0 | − |

This estimate shows that L(is) L(z) 2ε if z H is suﬃciently close to | − | ≤ ∈ r is. Since a similar result holds for R(z) and since L(z) = R(z) for z Hr, the claim fˆ = F follows. ∈

12.10 Decay of the Fourier Transform of f and Analyticity of f 12.11 The Paley–Wiener Theorem Application to the ﬁnite speed of propagation for the wave equation and the Klein–Gordon equation.

138 12.12 The Laplace Transform and Its Inversion Relation between the Laplace and the Fourier transform: Let f : [0, ) C denote a continuous function satisfying the estimate ∞ → f(t) Ceαt, t 0 , | | ≤ ≥ for some C > 0 and real α. For complex s with Re s > α the Laplace transform of f is Z ∞ ( f)(s) = e−stf(t) dt . L 0 Using certain conventions, the Fourier transform of a function g : R C is → Z ∞ −iyt ( g)(y) = e g(t) dt, y R . F −∞ ∈ Denote the Heaviside function by H(t) and let s = x + iy. We have

Z ∞ ( f(t))(x + iy) = H(t)f(t)e−xte−iyt dt L −∞ = (H(t)f(t)e−xt) (y) F Assuming the Fourier inversion formula to be valid for the function t H(t)f(t)e−xt one obtains that →

1 Z ∞ H(t)f(t)e−xt = eiyt (f(t))(x + iy) dy . 2π −∞ L This yields

1 Z ∞ H(t)f(t) = e(x+iy)t (f(t))(x + iy) dy . 2π −∞ L In terms of a line integral one obtains that

1 Z H(t)f(t) = est( f)(s) ds . 2πi Γx L

Here Γx is the straight line parameterized by

s = x + iy, < y < . −∞ ∞ For applications, it is important that the function ( f)(s) is often analytic in L a region larger than the half–plane Re s > α and that the curve Γx may be deformed in the region of analyticity.

139 13 Growth and Zeros of Entire Functions

13.1 Jensen’s Formula Theorem 13.1 Let D = D(0,R + ε) and let f H(D). Assume that ∈ f(0) = 0 and f(z) = 0 for z = R. 6 6 | | Let a1, . . . , aN denote the zeros of f with aj < R, repeated according to their multiplicity. Then the following formula holds:| |

N X a 1 Z 2π ln f(0) = ln | j| + ln f(Reit) dt . (13.1) | | R 2π | | j=1 0

Before proving the formula, we recall Cauchy’s integral formula: If f ∈ D(z0,R + ε) then 1 Z f(z) f(z ) = dz . 0 2πi z z Γ − 0 Here Γ has the parameterization

z(t) = z + Reit, 0 t 2π . 0 ≤ ≤ We can also write the formula in mean–value form:

Z 2π 1 it f(z0) = f(z0 + Re ) dt . 2π 0 Here we have used that

dz = Rieit dt = (z z )i dt , − 0 thus

dz = dt . i(z z ) − 0 If u : D R is harmonic, then u is the real–part of a function f H(D). Therefore, → ∈

Z 2π 1 it u(z0) = u(z0 + Re ) dt . (13.2) 2π 0

Lemma 13.1 Let a C, a < 1. Then we have ∈ | | Z 2π ln eit a dt = 0 . 0 | − | Proof: Set f(z) = 1 za. Then f has no zero in D(0, 1 + ε) =: D. We have f(z) = eφ(z) for some −φ H(D). Therefore, f(z) = eRe φ(z) and ∈ | | u(z) := ln f(z) = Re φ(z) | | 140 is harmonic in D. Also, u(0) = ln 1 = 0. Formula (13.2) with R = 1 and u0 = 0 yields

1 Z 2π 1 Z 2π 0 = u(0) = u(eit) dt = ln 1 aeit dt , 2π 0 2π 0 | − | thus Z 2π 0 = ln 1 aeit dt . (13.3) 0 | − | (Note that (13.3) is the special case of (13.1) obtained for R = 1, f(z) = 1 az.) In (13.3) substitute t = s to obtain − −

Z −2π 0 = ln 1 ae−is ds − 0 | − | Z 0 = ln 1 ae−is ds −2π | − | Z 2π = ln 1 ae−is ds . 0 | − | In the last equation we have used periodicity of the function which is integrated. Note that

1 ae−is = e−is(eis a) = eis a . | − | | − | | − | This proves the lemma. Proof of Theorem 13.1: a) Assume that f has no zero aj with aj < R. Then we can write | |

f(z) = eφ(z), φ H(D) . ∈ The function

u(z) = ln f(z) = Re φ(z) | | is harmonic in D. Formula (13.2) yields

1 Z 2π ln f(0) = ln f(Reit) dt . | | 2π 0 | | This shows formula (13.1) if f has no zero. b) Fix w C with 0 < w < R and consider the function f(z) = z w. Then formula∈ (13.1) claims that| | −

ln Reit w = ln R + ln eit a . | − | | − | Lemma 13.1 yields that

141 Z 2π ln eit a dt = 0 . 0 | − | Formula (13.4) follows. This shows that (13.1) holds for a function of the form f(z) = z w. − c) Assume two functions f1(z) and f2(z) satisfy the assumptions of the theorem and assume formula (13.1) holds for f1 and f2. Then the product f(z) = f1(z)f2(z) also satisﬁes the assumptions of the theorem and formula (13.1) holds for f. d) Let f H(D) denote any function satisfying the assumptions of the theorem. We∈ can write

f(z) = g(z)(z a )(z a ) ... (z a ) − 1 − 2 − N where g H(D) has no zeros in D. Using a), b), and c), the formula (13.1) follows for∈ f. 13.2 The Order of Growth Deﬁnition: Let f denote an entire function. One says that f has ﬁnite growth order if there exist positive constants A, B, ρ so that

B|z|ρ f(z) Ae for all z C . (13.5) | | ≤ ∈ If f is an entire function of finite growth order one defines its order of growth, denote by ρ0 = ρ0(f), as the infimum of all positive ρ for which the estimate (13.5) holds for some constants A = Aρ and B = Bρ. Examples: 1. Every polynomial has order of growth ρ0 = 0. z 2. The function f(z) = e has order of growth ρ0 = 1. 3. The function f(z) = cos z has order of growth ρ0 = 1. 4. Recall that

∞ j X w2 cos w = ( 1)j − (2j)! j=0 and deﬁne

∞ X zj f(z) = cos √z = ( 1)j . − (2j)! j=0 It is then clear that

f(z2) = cos z . 1 The order of growth of f is ρ0 = 2 . 5. One can show that 1 AeB|z| ln |z| for z 1 . Γ(z) ≤ | | ≥ | | 142 The function 1/Γ(z) has order of growth ρ0 = 1. [Stein, Shakarchi], p. 165 (The estimate should also follow from the inﬁnite product representation.) 6. It can be shown that the function (s 1)ζ(s) has order of growth ρ0 = 1. [Stein, Shakarchi], p. 202 −

13.3 Zeros and Growth Estimates

Let f denote an entire functions which has a sequence of zeros aj. We order the zeros so that

a a ... | 1| ≤ | 2| ≤ Each zero is listed according to its multiplicity. For 0 r < let n(r) denote ≤ ∞ the number of zeros aj with aj r. Clearly, n(r) is a piecewise constant function which increases. We will| | show ≤ that a growth estimate for f(z) implies a growth estimate for n(r) as r . | | → ∞ Theorem 13.2 Assume that the entire function f(z) satisﬁes a growth estimate

B|z|ρ f(z) Ae for all z C (13.6) | | ≤ ∈ with positive constants A, B, ρ. Then there are positive constants C and r0 so that

n(r) Crρ for r r . (13.7) ≤ ≥ 0 Proof: a) Using the Heaviside function

1, x 0 H(x) = 0, x <≥ 0 we have 1, aj r H(r aj ) = | | ≤ − | | 0, aj > r | | and can write

∞ X n(r) = H(r a ) . − | j| j=1 For each ﬁxed r the sum is ﬁnite. b) Assume f(0) = 0. Then we can write 6

143 N Z R dr X Z R dr n(r) = H(r a ) r − | j| r 0 j=1 0 N X Z R dr = r j=1 |aj | N X = ln(R/ a ) . | j| j=1

Here the zeros a1, . . . , aN are the zeros of f with aj R. Using Theorem 13.1 we obtain the following result: | | ≤

Lemma 13.2 As above, let f denote an entire function and assume

f(0) = 0 and f(z) = 0 for z = R. (13.8) 6 6 | | Then we have

Z R dr 1 Z 2π n(r) = ln f(Reit) dt ln f(0) . 0 r 2π 0 | | − | | c) We continue the proof of Theorem 13.2 and let f denote an entire function with (13.8) satisfying the growth estimate (13.6). We let R = 2r and have

ln f(Reit) ln A + BRρ | | ≤ C rρ for r r . ≤ 1 ≥ 0 Now the estimate of the previous lemma yields

Z R ds ρ n(s) C2r for r r0 . 0 s ≤ ≥ Furthermore,

Z 2r ds n(r) ln 2 = n(r) r s Z 2r ds n(s) ≤ r s C rρ ≤ 2 for r r0. This≥ proves (13.7) as long as f satisﬁes (13.8) for R = 2r. If f has a zero aj with aj = R = 2r, we simply choose 0 < ε r0 so that f has no zero on the circle| with| radius 2r + 2ε. We then have ≤

144 n(r) n(r + ε) C (r + ε)ρ C rρ for r r . ≤ ≤ 2 ≤ 3 ≥ 0 d) Assume that f has a zero of order m at z = 0. We set F (z) = f(z)/zm. The entire function F (z) satisﬁes the same growth estimate as f does. Since n (r) = n (r) + m the claim n (r) Crρ follows. f F f ≤ Theorem 13.3 Let f H(C) satisfy the growth estimate of the previous the- ∈ orem and let aj denote the non–zero zeros of f. If s > ρ then we have X a −s < . | j| ∞ j

Proof: For any positive integer l the number of zeros aj of f with

2l < a 2l+1 | j| ≤ is less than or equal to n(2l+1). Therefore, for any large l,

X −s −ls l+1 aj 2 n(2 ) l l+1 | | ≤ 2 <|aj |≤2 C2−ls 2(l+1)ρ ≤ C 2−l(s−ρ) ≤ 1 Since 2−(s−ρ) < 1 the geometric series X 2−l(s−ρ), s > ρ , l converges, and the claim follows. 13.4 Hadamard’s Factorization Theorem

Let f denote an entire function with growth order ρ0 and let k = [ρ0] denote the integer part of ρ0. We choose s and ρ with

k = [ρ ] ρ < ρ < s k + 1 . 0 ≤ 0 ≤ Then f satisﬁes a growth estimate

B|z|ρ f(z) Ae for all z C . | | ≤ ∈ We assume that f has a sequence of zeros aj which we order so that

a = ... = a = 0 < a a ... 1 m | m+1| ≤ | m+2yy| ≤ As above, n(r) denotes the number of zeros aj with aj r. We know that n(r) Crρ for large r and | | ≤ ≤ ∞ X a −s < . (13.9) | j| ∞ j=m+1

145 Recall the Weierstrass’ canonical factors,

w2 wk E (w) = (1 w) exp w + + ... + . k − 2 k Convergence of the series

∞ X a −k−1 < (13.10) | j| ∞ j=m+1 implies that

m ∞ z F (z) = z Πj=m+1Ek (13.11) aj is an entire function which has precisely the same zeros as f(z). Consequently, one can write the given function f(z) in the form

f(z) = eg(z)F z where g(z) is an entire function. Hadamard’s Factorization Theorem is the following remarkable result. It gives more information about the function g(z).

Theorem 13.4 Let f(z) denote an entire function with growth order ρ0 and let k = [ρ0] denote the integer part of f. Assume that f(z) has a sequence of zeros aj with

a = ... = a = 0 < a a ... 1 m | m+1| ≤ | m+2yy| ≤ Then f(z) has the form

g(z) m ∞ z f(z) = e z Πj=m+1Ek (13.12) aj where g(z) is a polynomial of degree k. The function g(z) is unique, except ≤ that one can add a constant 2πim, m Z. ∈ An idea of the proof is to turn the growth estimate for f(z) into an estimate for Re g(z). This requires to show lower bounds for the product F (z) and is rather technical. Once these lower bounds are derived, the following| lemma| will complete the proof of Hadamard’s Factorization Theorem.

Lemma 13.3 Let g(z) denote an entire function. Let k 0, 1,... and let k s < k + 1. Assume that there is a constant C and a∈ sequence { of} positive numbers≤ r with r so that Re g(z) satisﬁes the following upper bound: ν ν → ∞ Re g(z) Crs if z = r , ν = 1, 2,... (13.13) ≤ ν | | ν Under these assumptions, the function g(z) is a polynomial of degree k. ≤

146 Proof: First ﬁx any r > 0 and any integer n 0. With Γr we denote the circle with parameterization z = reit, 0 t 2π. Let≥ ≤ ≤ ∞ X n g(z) = gnz . n=0 We obtain

g(z) g = ... + n + ... zn+1 z thus

Z g(z) Z 2π 2πig = dz = i g(reit)r−ne−int dt . n n+1 Γr z 0 This shows that

Z 2π n it −int 2πgnr = g(re )e dt for n 0 . (13.14) 0 ≥ Also, by Cauchy’s theorem,

Z 2π g(reit)eint dt = 0 for n 1 , 0 ≥ thus

Z 2π g¯(reit)e−int dt = 0 for n 1 . 0 ≥ 1 Therefore, if we set u(z) = Re g(z) = 2 (g(z) +g ¯(z)), then we ﬁnd that Z 2π n it −int πgnr = u(re )e , dt for n 1 . 0 ≥ Furthermore, setting n = 0 in (13.14) and taking real–parts yields

Z 2π it 2π Re g0 = u(re ) dt . 0 We then have, for n 1, ≥

1 Z 2π πg = u(reit)e−int dt n n r 0 1 Z 2π = u(reit) Crs e−int dt n r 0 − Here the term in ... is 0i for r = r , by assumption. Taking absolute ≤ ν values we obtain

147 1 Z 2π π g Crs u(reit) dt n n | | ≤ r 0 − = 2πCrs−n 2π(Re g )r−n − 0

Now recall the assumption s < k+1 and assume n k+1. Letting r = rν in the above estimate yields that g = 0 for n > k≥. This proves the lemma.→ ∞ n 13.5 Entire Functions of Non–Integer Order of Growth The previous lemma has the following important implication.

Theorem 13.5 Let f(z) denote an entire function with order of growth ρ0. If ρ0 is not an integer, then f has inﬁnitely many zeros.

Proof: Suppose f has only ﬁnitely many zeros a1, . . . , aN . We may assume that N 1 and let ≥ p(z) = (z a ) ... (z a ) − 1 − N and write

f(z) = eg(z)p(z) where g(z) is an entire function. (If f(z) has no zeros, then take p(z) 1.) If ≡ k < ρ0 < s < k + 1 with integer k then f(z) satisﬁes an estimate

s f(z) AeB|z| . | | ≤ Since p(z) 1 for all large z we obtain | | ≥ | |

eRe g(z) = eg(z) | | s AeB|z| ≤ if z = r is large. This yields | | Re g(z) Crs for z = r ≤ | | if r is large. The previous lemma then yields that g(z) is a polynomial and the representation f(z) = eg(z)p(z) implies the order of growth of f(z) to be an integer. This contradiction proves the theorem. 13.6 Proof of Hadamard’s Factorization Theorem The main technical lemma is the following lower bound for the product

∞ z E(z) = Πj=1Ek . (13.15) aj

148 Lemma 13.4 Let 0 k ρ < s < k + 1 where k is an integer. Let aj denote a sequence of complex≤ numbers≤ with

0 < a a ... | 1| ≤ | 2| ρ and assume n(r) Cr for r r0, where n(r) is the number of aj with aj r. Consider the entire≤ function E≥(z) deﬁned in (13.15). There is a constant| c| >≤ 0, independent of z, and there is a sequence rν of positive radii with rν so that → ∞

s E(z) e−c|z| if z = r , ν = 1, 2,... | | ≥ | | ν For later reference, we recall that the assumption n(r) Crρ implies that ≤ ∞ X a −s < for s > ρ . | j| ∞ j=1 Before we prove the main lemma, Lemma 13.4, we show two simple auxiliary results for the Weierstrass function Ek(z).

Lemma 13.5 For small z we have | | k+1 1 E (z) e−2|z| if z . | k | ≥ | | ≤ 2

w Proof: We have Ek(z) = e with

w = log(1 z) + h (z) − k zk+1 zk+2 = + + ... − k + 1 k + 2 This yields the bound w 2 z k+1 if z 1 . Therefore, | | ≤ | | | | ≤ 2 k+1 E (z) = ew e−|w| e−2|z| . | k | | | ≥ ≥

If z is bounded away from zero, we have the following lower bound for E (z)|:| | k |

Lemma 13.6 There is a constant c = ck > 0 with

k 1 E (z) 1 z e−c|z| if z . | k | ≥ | − | | | ≥ 2

Proof: The deﬁnition of Ek(z) yields

E (z) = 1 z ehk(z) | k | | − || | 1 z e−|hk(z)| ≥ | − |

149 Here z2 zk 1 h (z) z + + ... + c z k if z . | k | ≤ 2 k ≤ k| | | | ≥ 2

The estimate of the next lemma is central for the proof of the main lemma, Lemma 13.4.

Lemma 13.7 Under the assumptions of Lemma 13.4, deﬁne the following union of open discs:

Ω = ∞ D(a , a −k−1) . ∪j=1 j | j| There are positive constants c and R so that

−c|z|s E(z) e if z R and z C Ω . | | ≥ | | ≥ ∈ \

Proof: a) Write E(z) = P1P2 with

P1 = Π|aj |≤2|z|Ek(z/aj),P2 = Π|aj |>2|z|Ek(z/aj) . We ﬁrst estimate P from below: If a > 2 z then, by Lemma 13.5, 2 | j| | | k+1 E (z/a ) e−2|z/aj | . | k j | ≥ Therefore,

k+1 X P e−2|z| Q with Q = a −k−1 . 2 ≥ | j| |aj |>2|z| Since z −1 > 2 a −1 we have | | | j| a −k−1 = a −s a s−k−1 C a −s z s−k−1 . | j| | j| | j| ≤ | j| | | This yields the upper bound

s P e−c|z| 2 ≥ follows. b) Next consider the ﬁnite product

P1 = Π|aj |≤2|z| Ek(z/aj) . Since z/a 1 we can apply Lemma 13.6 and obtain | j| ≥ 2

150 z k −c|z/ak| Ek(z/aj) 1 e . | | ≥ − aj Therefore,

P P P | 1| ≥ 3 4 with z P = Π 1 3 |aj |≤2|z| − aj and

k −c|z/aj | P4 = Π|aj |≤2|z| e .

c) To estimate P4 from below we write

k −c|z| Q1 P4 = e with X Q = a −k . 1 | j| |aj |≤2|z| Since

a −k = a −s a s−k C a −s z s−k | j| | j| | j| ≤ | j| | | we have X Q C z s−k a −s = C z s−k . 1 ≤ | | | j| 1| | j This shows that

s P e−c1|z| , c = cC . 4 ≥ 1 1 d) It remains to estimate z P = Π 1 3 |aj |≤2|z| − aj from below. Here it is important that z does not lie in any of the discs D(a , a −k−1), i.e., j | j| a z a −k−1 . | j − | ≥ | j| One obtains that

z a z 1 = | j − | a −k−2 − a a ≥ | j| j | j| and

151 P Π a −k−2 . 3 ≥ |aj |≤2|z| | j| Therefore, X ln P (k + 2) ln a . 3 ≥ − | j| |aj |≤2|z| Finally,

X ln a n(2 z ) ln(2 z ) | j| ≤ | | | | |aj |≤2|z| C z ρ ln(2 z ) ≤ | | | | C z s ≤ 1| | for all suﬃciently large z . The lower bound | | ln P c z s, c > 0 , 3 ≥ − | | follows and the inequality

s P e−c|z| 3 ≥ is shown. e) To summarize, we have for z R, z C Ω: | | ≥ ∈ \

E(z) = P1P2 s P e−c|z| 2 ≥ P1 P3P4 | | ≥ −c|z|s P3 e ≥ s P e−c|z| 4 ≥ where c > 0 is a constant. This implies that

s E(z) e−3c|z| . | | ≥ The lemma is proved. Recall that

0 < a a ... | 1 ≤ | 2| ≤ and k ρ < k + 1. Consequently, ≤ ∞ X a −k−1 < . | j| ∞ j=1 Also, recall the deﬁnition

Ω = ∞ D(a , a −k−1) . ∪j=1 j | j| 152 Lemma 13.8 Under the above assumptions, there is a sequence of positive radii r , r ,... with r so that none of the circles 1 2 ν → ∞ z = r , ν = 1, 2,... | | ν intersects the union of discs Ω.

Proof: Choose J so large that

∞ X 1 a −k−1 . | j| ≤ 10 j=J The set

Ω = J−1D(a , a −k−1) J ∪j=1 j | j| is bounded and we choose an integer n so large that ΩJ D(0, n). We claim that the interval ⊂

n r n + 1 ≤ ≤ contains a number r so that the circle z = r does not intersect the set Ω. In | | fact, for any r n, the circle z = r does not intersect ΩJ since ΩJ D(0, r) and it remains≥ to show that we| | can choose r in the interval n r ⊂n + 1 so that the circle z = r does not intersect the union ≤ ≤ | | ∞ D(a , a −k−1) . ∪j=J j | j| To see this, consider the closed interval h i I = a a −k−1, a + a −k−1 , j J, j | j| − | j| | j| | j| ≥ of length

length(I ) = 2 a −k−1 . j | j| We have

X 1 length(I ) . j ≤ 5 j≥J Therefore, there exists a number r in the interval n r n + 1 which does not ≤ ≤ lie in the union j≥J Ij. It is then clear∪ that the circle z = r does not intersect Ω. This proves Lemma 13.8. The| main| technical lemma, Lemma 13.4, follows as a consequence of Lemma 13.7 and Lemma 13.8. We summarize:

153 Theorem 13.6 (Hadamard) Let f(z) denote an entire function of ﬁnite order of growth ρ0 and let k = [ρ0] denote the integer part of ρ0. Case 1: f(z) has only ﬁnitely many zeros. Then f(z) has the form

f(z) = eg(z) p(z) where p(z) and g(z) are polynomials. The polynomial g(z) has degree k. The order of growth of f(z) equals k = ρ0. Case 2: f(z) has inﬁnitely many zeros aj with

a = ... = a = 0 < a a ... 1 m | m+1| ≤ | m+2| ≤ Then f(z) has the form

g(z) m ∞ f(z) = e z Πj=m+1Ek(z/aj) where g(z) is a polynomial of degree k. In particular, Case 1 cannot occur≤ unless the order of growth of f(z) is an integer. If the order of growth of f(z) is not an integer, then f(z) has inﬁnitely many zeros.

154 14 The Prime Number Theorem

For 1 x < let π(x) denote the number of primes p less than or equal to x. The Prime≤ Number∞ Theorem states that

π(x) ln(x) lim = 1 . x→∞ x If one deﬁnes a remainder R(x) by x π(x) = (1 + R(x)) ln x then R(x) 0 as x . Sharp estimates of the remainder are related to the zeros of the→ Riemann→ζ–function ∞ in the critical strip 0 < Re s < 1.

14.1 Functions The following functions will be used: If n x < n + 1, where n is an integer, then let ≤ x = n + x = [x] + x { } { } where [x] is the integer part and x is the fractional part of the real number x. { } Von Mangoldt function:

Λ(n) = ln p if n = pm, Λ(n) = 0 otherwise

Here p is prime and m N. n Λ(n) ∈ 1 0 2 ln 2 3 ln 3 4 ln 2 5 ln 5 6 0 7 ln 7 8 ln 2 9 ln 3

155 π(x) = number of primes x X ≤ ψ(y) = Λ(n) 1≤n≤y X = ln p (Chebyshev’s ψ function) − pm≤y X hln y i = ln p ln p p≤y X ψ(y) = Λ(n)H(y n) (with Heaviside function) − 1≤n≤y X = Λ(n)H(y n) − n≥1 Here we use the Heaviside function

1 for ξ 0 H(ξ) = 0 for ξ≥ < 0 An auxiliary integral:

Lemma 14.1 Let Γc denote the straight line with parametrization

Γ : s(t) = c + it, < t < . c −∞ ∞ For c > 0 we have

1 Z as 0 for 0 < a 1 ds = ≤ 2πi s(s + 1) 1 1 for a > 1 Γc − a Proof: See [Stein, Sharkarchi], Chapter 7, Lemma 2.4. We will show below that

∞ ζ0(s) X = Λ(n)n−s for Re s > 1 . − ζ(s) n=1 −s −1 (This follows from the product formula ζ(s) = Πp(1 p ) .) Multiply the above equation by −

1 xs+1 2πi s(s + 1) to obtain:

∞ 1 xs+1 ζ0(s) x X (x/n)s = Λ(n) . (14.1) 2πi s(s + 1) − ζ(s) 2πi s(s + 1) n=1

Now let c > 1 and integrate the equation along Γc. We will justify below that we may interchange summation and integration. We have, by the previous lemma:

156 1 Z (x/n)s 0 for 1 x < n ds = ≤ 2πi s(s + 1) 1 n for 1 n x Γc − x ≤ ≤ Therefore, equation (14.1) yields that

1 Z xs+1 ζ0(s) X n ds = x Λ(n) 1 . 2πi s(s + 1) − ζ(s) − x Γc 1≤n≤x In the following we will use that

Z x x n for x n H(y n) dy = − ≥ − 0 for x n 1 ≤ The integrated Chebyshev function and its integral representation:

Z x ψ1(x) = ψ(y) dy 1 X Z x = Λ(n) H(y n) dy − n≥1 1 X = Λ(n)(x n) − 1≤n≤x X n = x Λ(n) 1 − x 1≤n≤x x Z xs ζ0(s) = ds 2πi Γc s(s + 1) − ζ(s) Riemann’s ζ–function:

∞ X ζ(s) = n−s n=1 −1 −s = Π ∈ 1 p , Re s > 1 p P − ∞ d X log ζ(s) = Λ(n)n−s −ds n=1 1 Z ∞ x ζ(s) = + 1 s { } dx, Re s > 0 s 1 − xs−1 − 1 ξ(s) = π−s/2Γ(s/2)ζ(s)

14.2 Reduction to Asymptotics of ψ1(x) Let g : [1, ) R denote a function. We will use the following notation: ∞ → lim inf = lim inf g(y) . x→∞ x→∞ y≥x Write

157 x π(x) = (1 + R(x)) ln x ψ(x) = x(1 + r(x)) x2 ψ (x) = (1 + r (x)) 1 2 1 Theorem 14.1 a) If r(x) 0 as x then R(x) 0 as x . b) If r (x) 0 as x → then r(→x) ∞ 0 as x →. → ∞ 1 → → ∞ → → ∞

Remark: One should quantify precisely how estimates for r1(x) yield estimates for R(x). Proof: Part a1) lower bound for

π(x) ln x x For x 1 we have ≥

X hln xi ψ(x) = ln p ln p p≤x X ln x ln p ≤ ln p p≤x X = ln x p≤x = π(x) ln x

Therefore,

ψ(x) π(x) ln x for all x 1 . x ≤ x ≥ Therefore, if

ψ(x) 1 as x x → → ∞ then

π(x) ln x lim inf 1 . x→∞ x ≥ Part a2) upper bound for

π(x) ln x x Fix 0 < α < 1. Note that if x 1 and xα < p then ln xα < ln p. We have ≥

158 X hln xi ψ(x) = ln p ln p p≤x X ln p ≥ p≤x X ln p ≥ xα

ψ(x) + απ(xα) ln x απ(x) ln x ≥ and

ψ(x) π(xα) ln x π(x) ln x + α α . x x ≥ x It is clear that

π(xα) xα , ≤ thus

π(xα) ln x 0 as x . x → → ∞ Therefore, if

ψ(x) 1 as x x → → ∞ then

π(x) ln x 1 α lim sup . ≥ x→∞ x Since 0 < α < 1 was arbitray, we obtain that

π(x) ln x 1 lim sup . ≥ x→∞ x The upper and lower bounds imply that

π(x) ln x lim = 1 . x→∞ x Part b) We claim that if

ψ (x) 1 1 as x x2/2 → → ∞

159 then

ψ(x) 1 as x . x → → ∞ The proof essentially only uses that the function ψ(x) increases monotonically. Let β > 1. We have

1 Z βx ψ(x) ψ(y) dy ≤ (β 1)x x −1 = ψ (βx) ψ (x) (β 1)x 1 − 1 − Therefore,

ψ(x) 1 ψ (βx) ψ (x) 1 β2 1 . x ≤ β 1 (βx)2 − x2 − As x we obtain that → ∞ ψ(x) 1 β2 1 β + 1 lim sup = . x→∞ x ≤ β 1 2 − 2 2 − Since β > 1 was arbitray we obtain that

ψ(x) lim sup 1 . x→∞ x ≤ Similarly, ﬁx 0 < α < 1 and obtain that

1 Z x ψ(x) ψ(y) dy ≥ (1 α)x αx −1 = ψ (x) ψ (αx) (1 α)x 1 − 1 − Therefore,

ψ(x) 1 ψ (x) ψ (αx) 1 α2 1 . x ≥ 1 α x2 − (αx)2 − As x we obtain that → ∞ ψ(x) 1 1 α2 1 + α lim inf − = . x→∞ x ≥ 1 α 2 2 − Since α < 1 was arbitray we obtain that

ψ(x) lim inf 1 . x→∞ x ≥

160 14.3 Integral Representation of ψ1(x)

For c R let Γc denote the verticle line with parametrization ∈ Γ : s(t) = c + it, < t < . c −∞ ∞

Theorem 14.2 For c > 1 and x 1 the function ψ1(x) has the integral representation ≥

1 Z xs+1 ζ0(s) ψ1(x) = ds . 2πi Γc s(s + 1) − ζ(s) The proof is based on auxiliary results given below. Remark: Set

xs+1 ζ0(s) F (x, s) = , s(s + 1) − ζ(s) i.e., F (x, s) is the integrand in the above representation of ψ1(x). We know that 1 ζ(s) = + h(s), ζ0(s) = (s 1)−2 + h0(s) s 1 − − − where h(s) is holomorphic near s = 1. Therefore, Res ζ(s), s = 1 = 1 .

It follows that

x2 Res F (x, s), s = 1 = . 2

The claim is that this term gives the asymptotic behaviour of ψ1(x) to leading order. If denotes the circle of radius ε centered at s = 1, then we have Cε 1 Z x2 F (x, s) ds = . 2πi Cε 2

14.4 Auxiliary Results Lemma 14.2 For c > 0 and a > 0 we have

1 Z as 0 if 0 < a 1 ds = 1 ≤ 2πi c s(s + 1) 1 if a 1 Γ − a ≥ Lemma 14.3 For ε < 1 we have | | ∞ X 1 log(1 ε) = εj . − − j j=1

161 Proof: Set

f(ε) = log(1 ε) for ε < 1 . − − | | Then f(0) = 0 and

∞ 1 X f 0(ε) = = εj . 1 ε − j=0 The claim follows by integration. Lemma 14.4 For Re s > 1 we have

∞ ζ0(s) X = Λ(n)n−s . − ζ(s) n=1 Proof: From the previous lemma,

∞ X 1 log(1 ε) = εj for ε < 1 . − − j | | j=1 Also, for Re s > 1:

ζ(s) = Π (1 p−s)−1 , p −

X log ζ(s) = log(1 p−s) − − p ∞ X X 1 = p−js j p j=1 ∞ X −s = cnn n=1 where

1 if n = pj c = j . n 0 if p is not a prime power Therefore,

∞ X −s ln n log ζ(s) = cne n=1 and diﬀerentiation yields that

ζ0(s) d = log ζ(s) ζ(s) ds ∞ X = c n−s ln n . − n n=1

162 If n = pj then 1 c ln n = ln(pj) = ln p = Λ(n) . n j If n is not a prime power, then

cn ln n = 0 = Λ(n) . This proves the formula

∞ ζ0(s) X = Λ(n)n−s . − ζ(s) n=1 14.5 The ζ–Function has no Zero on the Line s = 1 + it Let

s = σ + it, σ > 1, t R . ∈ We have

−s −1 ζ(s) = Πp(1 p ) X− log ζ(s) = log(1 p−s) − − p ∞ X X 1 = (pj)−s j p j=1 ∞ X −s = cnn n=1 where

1 if n = pj c = j . n 0 if p is not a prime power We have

n−s = n−σn−it = n−σe−it ln n = n−σ cos(t ln n) i sin(t ln n) − thus

Re n−s = n−σ cos(t ln n) . Therefore,

163 ∞ X −σ Re log ζ(s) = cnn cos(t ln n) . n=1 From

ζ(s) = elog ζ(s) we obtain that

ζ(s) = eRe log ζ(s) | | and

∞ X ln ζ(s) = Re log ζ(s) = c n−σ cos(t ln n) . | | n n=1 Now consider

ln ζ3(σ)ζ4(σ + it)ζ(σ + 2it) = 3 ln ζ(σ) + 4 ln ζ(σ + it) + ln ζ(σ + 2it) | | | | | | ∞ X −σ = cnn 3 + 4 cos(t ln n) + cos(2t ln n) n=1 The term in brackets turns out to be non–negative. We know that

cos(2α) = cos2(α) sin2(α) = 2 cos2(α) 1 − − thus

3 + 4 cos(α) + cos(2α) = 2(1 + 2 cos(α) + cos2(α)) = 2(1 + cos(α))2 and obtain that

ln ζ3(σ)ζ4(σ + it)ζ(σ + 2it) 0 . ≥ This yields the lower bound

ζ3(σ)ζ4(σ + it)ζ(σ + 2it) 1 . (14.2) ≥ for all

σ > 1 and t R . ∈ Now suppose that

ζ(1 + it) = 0

164 for some real t = 0. Then consider the function 6

f(σ) = ζ3(σ)ζ4(σ + it)ζ(σ + 2it) for 1 < σ 2 . ≤ We obtain for some constant C. > 0 the estimates

ζ(σ + it) C(σ 1) | | ≤ − ζ(σ + 2it) C | | ≤ ζ(σ) 2(σ 1)−1 | | ≤ − for 1 < σ 2. Therefore, ≤ f(σ) K(σ 1)−3(σ 1)4 = K(σ 1) for 1 < σ 2 . ≤ − − − ≤ This estimate contradicts the lower bound which was established in (14.2). The contradiction shows that the ζ–function has no zero on the line

s = 1 + it, t R . ∈

165 15 Complex Variables and PDEs

15.1 2D Irrotational Euler Flows The Euler equations for 2D incompressible ﬂows read

ut + uux + vuy + px = 0

vt + uvx + vvy + py = 0

ux + vy = 0

Here the vector

2 u(x, y, t), v(x, y, t) R ∈ is the velocity and p(x, y, t) is the pressure. The equation ux + vy = 0 is the incompressibility condition. The corresponding 3D velocity ﬁeld

(u, v, 0) has the vorticity

ω = curl(u, v, 0) = (u, v, 0) ∇ × = (0, 0, v u ) x − y A velocity ﬁeld is called irrotational if its vorticity is zero. For 2D incompressible, irrotational ﬂows one obtains the conditions

u + v = 0, v u = 0 x y x − y which can also be written as

u = v , u = ( v ) . x − y y − − x These are the Cauchy–Riemann equations for the pair

(u, v) . − Lemma 15.1 Let U C denote an open set and let f H(U). Deﬁne u, v : ⊂ ∈ U R by → f(x + iy) = u(x, y) iv(x, y) − and set 1 p(x, y) = u2(x, y) + v2(x, y) . −2 Then

166 (u, v, p) is an irrotational solution of the stationary incompressible Euler equations.

Proof: We know that (u, v) satisfy the Cauchy–Riemann equations, thus − u + v = 0, v u = 0 . x y x − y Also,

1 2 2 (u + v ) = uux + vvx 2 x = uux + vuy = p − x and, similarly,

1 2 2 (u + v ) = uuy + vvy 2 y = uvx + vvy = p − y

Remark: For the constructed solution (u, v, p) we have 1 u2 + v2 + p = 0 . 2 Roughly, this implies that the pressure is low (negative, large in absolute value) where flow speed is high and, conversely, the pressure is high (negative, small in absolute value) where the flow speed is low. This is a simple form of Bernoulli’s law relating the pressure to the flow speed. The Velocity Field has a Potential Let F H(U) and let F 0 = f. As above, define u and v by ∈

f(x + iy) = u(x, y) iv(x, y) . − Deﬁne φ and ψ by

F (x + iy) = φ(x, y) + iψ(x, y) . We have

u iv = f − = F 0

= φx + iψx = φ iφ x − y 167 thus

u = φx and v = φy . This says that the function φ(x, y) is a potential of the velocity ﬁeld u(x, y), v(x, y) = φ(x, y) . ∇ Flow Lines From the Cauchy–Riemann equations

φ = ψ , φ = ψ x y y − x one obtains that

φxψx + φyψy = 0 . If Γ is a line described by

ψ(x, y) = const and if P = (x, y) Γ, then the vector ∈ ψx(P ), ψy(P ) is orthogonal to the tangent to Γ at P . At P = (x, y) we have

0 = φxψx + φyψy

= uψx + vψy which implies that the velocity vector u(P ), v(P ) is tangent to Γ at P . In other words, the lines

ψ(x, y) = const are ﬂow lines. Example 1: F (z) = z = x + iy We have

φ = x, ψ = y and

u = 1, v = 0 . The velocity ﬁeld is u(x, y), v(x, y) = (1, 0)

168 is a uniform ﬂow in the direction of the x–axis. The pressure is constant. The value of the constant is irrelevant since only px and py occur in the Euler equations.

Example 2: F (z) = az = (a1 + ia2)(x + iy) We have

f(z) = a = a + ia = u iv . 1 2 − The velocity vector u(x, y), v(x, y) = (a , a ) 1 − 2 is constant in space. The velocity field describes a uniform flow. z2 Example 3: F (z) = 2 (flow around a corner in the first quadrant) Here

f(z) = z = x + iy , thus

u = x, v = y . − We have

(x + iy)2 1 F (z) = = (x2 y2) + ixy . 2 2 − Therefore,

ψ(x, y) = xy . The ﬂow lines are hyperbolas c y = . x On the y–axis the velocity is

(u(0, y), v(0, y)) = (0, y) , − i.e., the ﬂow is parallel to the y–axis. On the x–axis the velocity is

(u(x, 0), v(x, 0)) = (x, 0) i.e., the ﬂow is parallel to the x–axis. The ﬁrst quadrant is

U = Q = (x, y) : 0 < x, y < . 1 { ∞} On the boundary of U the velocity is tangent to the boundary. This is the usual boundary condition for Euler ﬂow.

169 1 Example 4: F (z) = z + z (ﬂow around a cylinder) We have x iy F (z) = x + iy + − x2 + y2 thus y ψ(x, y) = y . − x2 + y2 The stream lines are given by the equation y y = c = const . − x2 + y2 If we take c = 0 we see that the circle

x2 + y2 = 1 and the y–axis are flow lines. We can interpret the flow as a flow around the unit–circle. We have

1 2x2 u = φ = 1 + x x2 + y2 − (x2 + y2)2 and 2xy v = φ = y −(x2 + y2)2 This yields that

u(x, y) 1 and v(x, y) 0 → → for x + y . | | | | → ∞ Remarks: The Navier–Stokes equations for 2D incompressible ﬂows read

ut + uux + vuy + px = ν∆u

vt + uvx + vvy + py = ν∆v

ux + vy = 0 where ν > 0 is the kinematic viscosity of the ﬂuid. If u and v are constructed as above, then

∆u = ∆v = 0 , thus

(u, v, p) is a solution of the stationary Navier–Stokes equations. However, for the Navier–Stokes equations one typically requires the boundary condition

170 u = v = 0 on every wall of the domain. For the Euler equations one only requires that (u, v) is tangential to the wall. The solutions of the Euler equations which we have constructed will not satisfy the boundary conditions u = v = 0 at a wall unless the solution is trivial. In regions away from walls the Euler equations may still be useful since often one has that 0 < ν << 1. The viscosity terms are important, however, in boundary layers near walls.

15.2 Laplace Equation We will to derive the Poisson kernel for the Dirichlet problem for Laplace equation in the unit disc. Dirichlet Problem: Let D = D(0, 1) denote the unit disc with boundary curve Γ = ∂D. Let g :Γ R denote a continuous function. Find → 2 u : D¯ R with u C (D) C(D¯) → ∈ ∩ so that

∆u = 0 in D, u = g on Γ .

15.2.1 Derivation of the Poisson Kernel via Complex Variables Let f H(D(0, 1 + ε)). By Cauchy’s integral formula: ∈ 1 Z f(ζ) f(z) = dζ for z D. 2πi ζ z ∈ Γ − For z D, z = 0, deﬁne the reﬂected (w.r.t. Γ) point ∈ 6 z 1 z = = . 1 z 2 z¯ | | Since z > 1 we have that | 1| 1 Z f(ζ) 0 = dζ for z D. 2πi ζ z ∈ Γ − 1 Subtracting this equation from the equation for f(z) we obtain

1 Z ζ ζ dζ f(z) = f(ζ) (15.1) 2π ζ z − ζ z iζ Γ − − 1 for 0 < z < 1. We have| | for 0 < z < 1 and ζ Γ: | | ∈

171 ζ ζ ζ ζ = ¯ ζ z − ζ z1 ζ z − ζ ζζ − − − − z¯ ζ z¯ = ζ z − ζ¯ z¯ − − ζ 2 z 2 = | | − | | ζ z 2 | − | 1 z 2 = − | | ζ z 2 | − | The kernel

1 1 z 2 K(z, ζ) = − | | where ζ Γ and z D (15.2) 2π ζ z 2 ∈ ∈ | − | is called the Poisson kernel for the unit disc D. If f H(D(0, 1 + ε)) then we have ∈

Z dζ f(z) = K(z, ζ)f(ζ) for z D. (15.3) Γ iζ ∈

(In (15.1) we had to exclude z = 0 since z1 is undeﬁned for z = 0. However, both sides of (15.1) extend continuously to z = 0 and one obtains the validity of (15.3) also for z = 0.) Properties of K(z, ζ): 1. K(z, ζ) > 0 for z D, ζ Γ. 2. ∈ ∈

Z dζ K(z, ζ) = 1 for z D Γ iζ ∈ (Take f 1 in (15.3).) 3. The≡ function z K(z, ζ) is harmonic in D for every ζ Γ. This follows since→ the function ∈ 1 z , z D, → ζ z ∈ − is holomorphic and the function 1 z , z D 0 , → ζ z ∈ \{ } − 1 is the complex conjugate of a holomorphic function. Therefore, 1 1 z Re → ζ z − ζ z − − 1 is harmonic in D 0 , and this function extends smoothly to z = 0. \{ } Other notation: In formula (15.2) let

172 z = reiθ and ζ = eiφ . One obtains that

K(z, ζ) = K(reiθ, eiφ) 1 1 r2 = − 2π eiφ reiθ 2 | − | Here

eiφ reiθ 2 = (cos φ r cos θ)2 + (sin φ r sin θ)2 | − | − − = 1 2r cos(φ θ) + r2 − − One deﬁnes

1 1 r2 P (r, α) = − . 2π 1 2r cos α + r2 − The kernel P (r, α) is also called the Poisson kernel for the unit disc. We have

K(reiθ, eiφ) = P (r, φ θ) . − Formula (15.3) becomes

Z 2π f(reiθ) = P (r, φ θ)f(eiφ) dφ for 0 r < 1 . 0 − ≤ This result suggests the following:

Theorem 15.1 The solution of the Dirichlet problem is

Z 2π u(reiθ) = P (r, φ θ)g(eiφ) dφ for 0 r < 1 (15.4) 0 − ≤ and

u(reiθ) = g(eiθ) for r = 1 .

Proof: Since the Poisson kernel K(z, ζ) is a harmonic function of z D for every ﬁxed ζ Γ it follows that u(z) is harmonic in D. The diﬃculty∈ is to prove that u attains∈ its boundary values continuously. Fix ζ = eiθ0 Γ. We must prove that 0 ∈

lim u(z) = g(ζ0) . z→ζ0,z∈D Using the notation

iφ iθ0 ζ = e , ζ0 = e

173 we have for all z D: ∈

Z dζ u(z) g(ζ0) = K(z, ζ) g(ζ) g(ζ0) − Γ − iζ Z 2π = K(z.eiφ) g(eiφ) g(eiθ0 ) dφ 0 −

Let ε > 0 be given. There exists δ1 > 0 so that ε g(eiφ) g(eiθ0 ) for φ θ 2δ . | − | ≤ 2 | − 0| ≤ 1 Therefore, Z ε iφ u(z) g(ζ0) + 2 g ∞ K(z, e ) dφ . | − | ≤ 2 | | |φ−θ0|≥2δ1 Let

z = reiθ and θ θ δ . | − 0| ≤ 1 We must bound

K(z, eiφ) = K(reiθ, eiφ) for

φ θ 2δ . | − 0| ≥ 1 Note that

φ θ 2δ and θ θ δ | − 0| ≥ 1 | − 0| ≤ 1 implies that

φ θ δ . | − | ≥ 1 We have

K(z, eiφ) = K(reiθ, eiφ) 1 1 r2 = − 2π eiφ reiθ 2 | − | Here

eiφ reiθ 2 = eiα r 2 with α = φ θ δ . | − | | − | | | | − | ≥ 1 Set

c1 := sin δ1 > 0 . Then we have that

174 eiα r 2 = (cos α r)2 + sin2 α c2 . | − | − ≥ 1 This estimate yields the bound

2 iφ iθ iφ 1 1 r K(z, e ) = K(re , e ) −2 ≤ 2π c1 for

φ θ 2δ , θ θ δ , 0 r < 1 . | − 0| ≥ 1 | − 0| ≤ 1 ≤ Therefore, there exists δ2 > 0 so that Z iφ ε 2 g ∞ K(z, e ) dφ | | |φ−θ0| ≤ 2 if

θ θ δ and 1 δ r < 1 . | − 0| ≤ 1 − 2 ≤ We have proved the limit relation

lim u(z) = g(ζ0) , z→ζ0,z∈D which, together with the continuity of g, implies that u C(D¯). ∈ Remark: In the circle D(0,R) of radius R > 0 the solution formula (15.4) is replaced by

Z 2π R2 r2 u(reiθ) = − g(eiφ) dφ for 0 r < R . (15.5) R2 + r2 2rR cos(φ θ) ≤ 0 − − 15.2.2 Derivation of the Poisson Kernel via Separation of Variables

175 16 Rearrangement of Series

16.1 Rearrangement of Absolutely Convergent Series Theorem 16.1 Assume

∞ X a < | j| ∞ j=1 and let

∞ X aj =: A. j=1 If β : N N is bijective then → ∞ X aβ(j) = A. j=1

Proof: For ε > 0 let νε N be chosen so that ∈ X a ε . (16.1) | j| ≤ j>νε We then have

ν Xε a A ε . (16.2) j − ≤ j=1 Set F := β−1 1, 2, . . . , ν ; ε { ε} clearly,

β(F ) = 1, 2, . . . , ν . ε { ε} Set

nε := max Fε . We then have for all n n : ≥ ε F 1, 2, . . . , n 1, 2, . . . , n , ε ⊂ { ε} ⊂ { } thus β(F ) = 1, 2, . . . , ν β 1, 2, . . . , n . ε { ε} ⊂ { } For all n n obtain the following: ≥ ε

176 n X X X a A = a A + a β(j) − β(ν) − β(ν) j=1 ν∈Fε ν∈{1,2,...,n}\Fε X X a A + a ≤ β(ν) − | β(ν)| ν∈Fε ν∈N\Fε ν Xε X = a A + a j − | j| j=1 j>νε 2ε ≤ by (16.1) and (16.2). 16.2 Interchanging Double Sums

Let akl C for kl = (k, l) N N. Assume ∈ ∈ × ∞ ∞ X X a =: S < . | kl| ∞ k=1 l=1 Then, for all K,L N: ∈ L K K L X X X X a = a S < . | kl| | kl| ≤ ∞ l=1 k=1 k=1 l=1 This implies that for all L N: ∈ L ∞ X X a S < , | kl| ≤ ∞ l=1 k=1 and

∞ ∞ X X S˜ := a S < . | kl| ≤ ∞ l=1 k=1 With a similar argument one obtains that S S˜, thus ≤ ∞ ∞ ∞ ∞ X X X X a = a . | kl| | kl| k=1 l=1 l=1 k=1 The following result about interchanging double sums is more diﬃcult to prove:

Theorem 16.2 Assume that

∞ ∞ X X a < . | kl| ∞ k=1 l=1 Then

177 ∞ ∞ ∞ ∞ X X X X akl = akl . k=1 l=1 l=1 k=1

Proof: For K,L N set ∈

K L X X A(K,L) = akl k=1 l=1 L K X X B(K,L) = akl l=1 k=1 These are ﬁnite sums, and it is clear that

A(K,L) = B(K,L) for all K,L N . ∈ The proof of the theorem will be based on the following lemma.

Lemma 16.1 Assume that

∞ ∞ X X a < | kl| ∞ k=1 l=1 and set

∞ ∞ X X A := akl . k=1 l=1

Then, for all ε > 0, there exists Kε,Lε N with ∈ A A(K,L) ε for K K ,L L . | − | ≤ ≥ ε ≥ ε Proof: Set

∞ ∞ X X S = a , S = S. k | kl| k l=1 k=1

There exists Kε with X S ε , (16.3) k ≤ k>Kε thus

∞ X X X a = S ε . (16.4) | kl| k ≤ k>Kε l=1 k>Kε Since

178 ∞ X a < | kl| ∞ l=1 for all k N there exists L(k, ε) N with ∈ ∈ X ε a . (16.5) | kl| ≤ k2 l>L(k,ε) Set

Lε := max L(k, ε) 1≤k≤Kε where Kε is determined so that (16.3) holds. In the following, let K K and L L . We have: ≥ ε ≥ ε

∞ ∞ X X A = akl k=1 l=1 K ∞ ∞ X X X X = akl + akl k=1 l=1 k>K l=1 K L K ∞ X X X X X X = akl + akl + akl k=1 l=1 k=1 l>L k>K l=1 Thus,

K ∞ X X X X A A(K,L) a + a . (16.6) | − | ≤ | kl| | kl| k=1 l>L k>K l=1 The second term on the right–hand side is ε by (16.4). For the ﬁrst term on the right–hand side of (16.6) we have, using≤ (16.5):

K K X X Xε X X X a = a + a | kl| | kl| | kl| k=1 l>L k=1 l>L KεL K Xε ε X + S ≤ k2 k k=1 k>Kε π2 ε + ε ≤ 6 This proves the lemma. We can complete the proof of Theorem 16.2. Recall that for all ﬁnite K,L we have

L K K L X X X X B(K,L) = akl = akl = A(K,L) . l=1 k=1 k=1 l=1

179 We set

∞ ∞ X X B := akl . l=1 k=1 As in Lemma 16.1, it follows that

B B(K,L) ε | − | ≤ if K and L are suﬃciently large. Therefore, since B(K,L) = A(K,L), the equation B = A follows. This completes the proof of Theorem 16.2. 16.3 Rearrangement of a Double Series

Theorem 16.3 Let akl C for kl = (k, l) N N and assume that ∈ ∈ × ∞ ∞ X X a < . (16.7) | kl| ∞ k=1 l=1 Let

∞ ∞ X X A := akl k=1 l=1 and let β : N N N be bijective. Then → × ∞ X aβ(j) = A. j=1

Proof: For K,L N let ∈

R(K,L) = kl = (k, l) N N : 1 k K and 1 l L , { ∈ × ≤ ≤ ≤ ≤ } thus R(K,L) denotes the sets of all points in N N which lie in the rectangle [1,K] [1,L]. × × Let ε > 0 be given. Because of (16.7) there exist Kε,Lε N so that ∈ X a ε (16.8) | kl| ≤ kl∈N×N\Rε where

Rε = R(Kε,Lε) . We then have X a A ε . (16.9) kl − ≤ kl∈Rε Set

180 −1 Fε := β (Rε) and let

nε := max Fε . For n n we have ≥ ε F 1, 2, . . . , n 1, 2, . . . , n , ε ⊂ { ε} ⊂ { } thus R = β(F ) β 1, 2, . . . , n . ε ε ⊂ { } For all n n obtain the following: ≥ ε

n X X X a A = a A + a β(j) − β(ν) − β(ν) j=1 ν∈Fε ν∈{1,2,...,n}\Fε X X a A + a ≤ β(ν) − | β(ν)| ν∈Fε ν∈N\Fε X X = a A + a kl − | kl| kl∈Rε kl∈N×N\Rε 2ε ≤ by (16.8) and (16.9).

181