Step 1. Analytic Properties of the Riemann Zeta Function

Home , Cauchy sequence, Uniform convergence

Step 1 . Analytic Properties of the Riemann zeta function [2 lectures ]

The Riemann zeta function is the infinite sum of terms 1 /n s, n ≥ 1. For each n, the 1 /n s is a continuous function of s, i.e. 1 1 lim = , s s0 s→s0 n n for all s0 ∈ C, and is differentiable, i.e. d 1 d log n = e−s log n = − (log n) e−s log n = − , ds ns ds ns for all s ∈ C. We know from second year analysis that we can add a finite number of continuous (differentiable) functions to get a continuous (differentiable) function, but this is not necessarily true for an infinite sum. We need more than the sum converges.

C Deﬁnition 4.3 A sequence {xn}n≥1 in is a Cauchy Sequence if, and only if, ∀ε > 0, ∃N ≥ 1 : ∀n, m ≥ N, |xn − xm| < ε.

Depending on how you have constructed C you may deﬁne that a series converges if, and only if, it is a Cauchy Sequence. Unfortunately this was not how convergence of a sequence was deﬁned in MATH20101 say.

C Theorem 4.4 A sequence {xn}n≥1 converges in if, and only if, {xn}n≥1 is a Cauchy Sequence.

Proof (⇒) Assume that {xn}n≥1 converges to x say. Let ε > 0 be given. Then by the ε − N deﬁnition of convergence, ∃N ≥ 1 : |xn − x| < ε/ 2. Let m, n ≥ N then

|xn − xm| = |xn − x + x − xm| ≤ | xn − x| + |x − xm| < ε, and we have verified the definition of a Cauchy Sequence. (⇐) This follows as a defining property of C, basically that C is complete.

7 Deﬁnition 4.5 Let {Fn (z)}n≥1 be a sequence of functions deﬁned on a set D ⊆ C.

The sequence {Fn}n≥1 converges to F on D iﬀ

∀ε > 0, ∀z ∈ D , ∃N = N (ε, z ) : ∀n ≥ N, |Fn (z) − F (z)| < ε.

The sequence {Fn}n≥1 converges uniformly to F on D iﬀ

∀ε > 0, ∃N = N (ε) : ∀z ∈ D , ∀n ≥ N, |Fn (z) − F (z)| < ε.

Be careful to understand the difference in these definitions, for convergence the value of N depends on the point z, whereas for uniform convergence the N will work simultaneously for all z in the domain D. Equivalently the definition of uniform convergence can be given in terms of a Cauchy sequence as

∀ε > 0, ∃N = N (ε) : ∀z ∈ D , ∀n, m ≥ N, |Fn (z) − Fm (z)| < ε. All deﬁnitions for sequences translate into deﬁnitions for series, as in

∞ Deﬁnition 4.6 A series j=1 fj (z) converges uniformly on D if, and n only if, the sequence of partialP sums Fn (z) = j=1 fj (z) converges uniformly on D. That is, P

n ∀ε > 0, ∃N = N (ε) : ∀z ∈ D , ∀n, m ≥ N, fj (z) < ε.

j=Xm+1

This leads to Weierstrass’s M-Test for Uniform Convergence of a Series : If there exists a sequence {Mi}i≥1 of non-negative numbers for which |fj (z)| < M j ∞ ∞ for all z ∈ D and all i ≥ 1 and j=1 Mj converges then j=1 fj (z) converges uniformly on D. P P Veriﬁcation In fact, assuming k > ℓ , the partial sums satisfy

|Fk (z) − Fℓ (z)| = |fk (z) + fk−1 (z) + ... + fℓ+1 (z)|

≤ | fk (z)| + |fk−1 (z)| + ... + |fℓ+1 (z)|

< M k + Mk−1 + ... + Mℓ+1 . On account of the convergence of the series the last sum can be made smaller than a given ε > 0. Hence the sequence of partial sums Fm (z) converges uniformly.

8 The Question then arises what properties of a uniformly converging sequence are inherited by its limit?

We have Weierstrass’s Theorem for Series (see Background Notes 0.7) that states Theorem 4.7 Weierstrass’s Theorem for Series . Assume

f1 (z) , f 2 (z) , f 3 (z) , ... ∞ are holomorphic in an open set D, and i=1 fi (z) converge uniformly on every closed and bounded subset of D. ThenP ∞ (i) The F (z) = i=1 fi (z) is holomorphic on D, P ∞ (k) (ii) For all k ≥ 1, the series i=1 fi (z) converges on D, and converge uniformly on every closed and boundedP subset of D with limit F (k) (z). (So the series can be diﬀerentiated term-by-term.) Our ﬁrst example of use of this theorem is the Riemann zeta function.

The Riemann zeta function is a particular example of a Dirichlet Series, ∞ −s C n=1 ann . For a given s ∈ such a series may converge or not and if it Pconverges it may converge absolutely or not. Results on how such regions are connected for general Dirichlet Series are not give in this course. We will, ∞ though, make use of the fact that if a series n=1 an converges absolutely ∞ ∞ then we have an “ infinite” triangle inequality P| n=1 an| ≤ n=1 |an| . P P For Re s ≥ 1 + δ the Riemann zeta function converges absolutely and to ∞ s see this use the Comparison Test on n=1 |1/n |, for each term satisfies s 1+ δ ∞ 1+ δ |1/n | ≤ 1/n and n=1 1/n converges.P P To go further we need to show that the series defining ζ (s) converges uniformly on some domain. Theorem 4.8 Assume δ > 0. For Re s ≥ 1 + δ the Riemann zeta function ∞ 1 ζ (s) = ns Xn=1 converges uniformly, is holomorphic in Rs > 1, with derivative ∞ log n ζ′ (s) = − ns Xn=1 for Re s > 1.

9 Proof We have to show only that the series deﬁning ζ (s) converges uniformly 1+ δ Re s ≥ 1+ δ. We could simply apply Weierstrass’s M-test with Mn = 1 /n or do it directly as

N ∞ ∞ 1 1 1 ζ (s) − = ≤ ns ns ns Xn=1 n=XN+1 n=XN+1

by “inﬁnite” triangle inequality

∞ 1 ≤ since Re s ≥ 1 + δ n1+ δ n=XN+1 ∞ du 1 ≤ 1+ δ = δ . ZN u δN Given any ε > 0, then 1 /δN δ is less than ε when N is suﬃciently large, independent of s, showing uniform convergence. We can apply Weierstrass’s Theorem since any closed and bounded subset of Re s > 1 is contained in Re s ≥ 1 + δ for some δ > 0. Each term in the series is holomorphic on C with derivative d 1 d log n = e−s log n = − . ds ns ds ns The stated result then follows from Weierstrass’s Theorem.

In Theorem 1.8 we saw 1 −1 ζ (s) = 1 − , ps Yp for Re s > 1. This with Theorem 4.8 shows that the Euler product is continuous and diﬀerentiable. This was not obvious. For each prime p, the (1 − 1/p s)−1 is a continuous function of s, i.e.

1 −1 1 −1 lim 1 − = 1 − , s s0 s→s0 p p for all s0 ∈ C, and is diﬀerentiable, i.e.

d 1 −1 1 −2 d 1 1 −2 d 1 1 − = − 1 − 1 − = 1 − ds ps ps ds ps ps ds ps log p 1 −2 = − 1 − , ps ps

10 for all s ∈ C. We know from second year analysis that we can multiply a finite number of continuous (differentiable) functions to get a continuous (differentiable) function, but this is not necessarily true for an infinite product.

There is a Weierstrass’s Theorem for Inﬁnite Products but it is not given here.

Earlier in the course we looked at replacing sums by integrals. In the same vein we can use Partial Summation to replace the Dirichlet Series deﬁning the Riemann zeta-function by an integral.

Theorem 4.9 For s =6 1 we have 1 1 N 1−s N du = 1 + + − s {u} (5) ns s − 1 1 − s Z us+1 1≤Xn≤N 1 where {u} is the fractional part of u (so {u} = u − [u]).

Proof By Partial Summation (and this argument, for s = 1, has been seen before in the previous Chapter) 1 1 1 1 = − − ns N s N s ns 1≤Xn≤N 1≤Xn≤N

N N du = − (−s) N s Z us+1 1≤Xn≤N n

N N du = + s 1 N s Z us+1 1 1≤Xn≤u

N N du = s + s [u] s+1 N Z1 u N N du N du = s + s u s+1 − s (u − [u]) s+1 N Z1 u Z1 u N s N du = + N 1−s − 1 − s {u} . (6) N s 1 − s Z us+1 1

Let N → ∞ to get

11 Theorem 4.10 For Re s > 1, 1 ∞ {u} ζ (s) = 1 + − s 1+ s du. (7) s − 1 Z1 u Further, the right hand side of (7) deﬁnes a function holomorphic in Re s > 0, except for a simple pole, residue 1 at s = 1 .

This is an example of an analytic continuation of ζ (s). The right hand side of (7) defines a function holomorphic on a region containing Re s > 1, where it agrees with the Dirichlet series definition of ζ (s). It is in fact possible to continue analytically ζ (s) to all of C, though we do not do so in this course. For the remainder of the course when we talk of the Riemann zeta function we will be thinking of the function given by the right hand side of (7) for Re s > 0. When Re s > 1 we will know that it has an alternative definition as a Dirichlet Series. Note On Problem Sheet 3 you are asked to show that

∞ (−1) n 1 = 1 − ζ (s) , ns 2s−1 Xn=1 for Re s > 1. It can be shown that the Dirichlet Series on the left is, in fact, holomorphic in Re s > 0 and so this equality gives another analytic continuation of ζ (s) to Re s > 0. It can be shown though that if a function has an analytic continuation, then that continuation is unique.

Proof Remember that Re s = σ > 1. Let N → ∞, then

N 1−s = N 1−σ → 0 while {u} 1 ≤ (8) u1+ s u1+ σ

and so the integral in (6) converges. Hence 1 ∞ {u} ζ (s) = 1 + − s 1+ s du (9) s − 1 Z1 u as required. In fact, from (8), we see that the integral in (9) converges absolutely for Re s > 0. There is a theory for uniform convergence of integrals ; see Weierstrass’s The- orem for Inﬁnite Integrals in Background Notes 0.7. But this is not directly

12 applicable here since the integrand {u} u−1−s is not a continuous function of u for fixed s. Instead, consider the infinite integral in (9) as an infinite sum of integrals over intervals of length 1: ∞ ∞ ∞ {u} n+1 {u} du du = = fn (s) . Z u1+ s Z u1+ s 1 Xn=1 n Xn=1 We can show (but will only do so for level 4) that for any δ > 0, ∞ (i) the series n=1 fn (s) converges uniformly on Re s > δ,

(ii) each fn (sP) is holomorphic on Re s > δ, with derivative n+1 df n (s) {u} = 1+ s (− log u) du. ds Zn u This is suﬃcient, by Weierstrass’s Theorem for Series, part i of Theorem 4.7 above, to imply that the integral in (9) is holomorphic on Re s > 0.

Diffferentiate (5) w.r.t s, allowable since the sum and integral are finite. This is harder to justify is we differentiated (7). In this way we get Theorem 4.11 for integral N ≥ 1 and s =6 1 we have

N log n 1 − (1 − s)1−s log N + N 1−s − = − + ns 2 2 Xn=1 (s − 1) (1 − s) N {u} N {u} log u − 1+ s du + s 1+ s du. Z1 u Z1 u Let N → ∞ to derive Corollary 4.12 For Re s > 0, ∞ ∞ ′ 1 {u} {u} log u ζ (s) = − 2 − s+1 du + s s+1 du. (10) (s − 1) Z1 u Z1 u As in the proof of Theorem 4.10 this only follows for Re s > 1, when N 1−s log N N 1−s → 0 and → 0 1 − s (1 − s)2 as N → ∞. But once derived, the right hand side of (10) deﬁnes a holomorphic function for Re s > 0, s =6 1 .