TOPICS IN ANALYTIC NUMBER THEORY 1. Arithmetic functions An arithmetic function is a function f

37 Pages·2012·0.37 MB·English

TOPICS IN ANALYTIC NUMBER THEORY
TOM SANDERS
1. Arithmetic functions
An arithmetic function is a function f : N → C; there are many interesting and natural
examples in analytic number theory. To begin with we consider what is perhaps the best
known
(cid:88)
π(n) := 1 (x),
P
x(cid:54)n:
the usual counting function of the primes. Various heuristic arguments suggest that one
shouldexpectxtobeprimewithprobability1/logx, andcoupledwithabodyofnumerical
evidence this prompted Gauss to conjecture that
(cid:90) n 1 n
π(n) ∼ Li(n) := dx ∼ .
logx logn
1
It is the purpose of the first section of the course to prove this result.
As it stands π can be a little difficult to evaluate because the indicator function of the
primes is not very smooth. To deal with this we introduce the von Mangoldt function Λ
defined by
(cid:40)
logp if n = pk
Λ(n) =
0 otherwise.
It will turn out that Λ is smoother than the indicator function of the primes and while
the von Mangoldt function is supported on a larger set (namely all powers of primes), in
applications this difference will be negligible. For now we note that our interest lies in the
sum of the von Mangoldt function, defined to be
(cid:88)
ψ(n) := Λ(x),
x(cid:54)n
and which is closely related to π as the next proposition shows. The key idea in the proof of
thepropositionisatechniquecalledpartialsummationor,sometimes,Abeltransformation.
This is the observation that
(cid:88) (cid:88)
f(x)(g(x)−g(x−1)) = f(n)g(n)− g(x)(f(x+1)−f(x)),
x(cid:54)n x(cid:54)n−1
and may be thought of as a discrete analogue of integration by parts.
Proposition 1.1. ψ(n) ∼ n if and only if π(n) ∼ n/logn.
Last updated: 28th April, 2012.
1
2 TOM SANDERS
Proof. Firstnotethatpowerofprimeslargerthan1makeanegligiblecontributiontoψ(n):
(cid:88) (cid:88) √
ψ(n) = Λ(x) = logp+O( nlogn).
x(cid:54)n p(cid:54)n
On the other hand by partial summation we have
(cid:88) (cid:88)
logp = (π(x)−π(x−1))logx
p(cid:54)n x(cid:54)n
(cid:88)
= π(n)logn+ π(x)(log(x+1)−logx)
x(cid:54)n−1
(cid:88) π(x)
= π(n)logn+ +O(logn),
x
x(cid:54)n−1
whence
(cid:88) π(x) √
(1.1) ψ(n) = π(n)logn+ +O( nlogn).
x
x(cid:54)n
Now, if π(x) ∼ x/logx for all x (cid:54) n, then
(cid:88)
ψ(n) ∼ n+ o(1) ∼ n.
x(cid:54)n
Conversely, if ψ(x) ∼ x for all x (cid:54) n then
1
x (cid:38) (π(x)−π(x1/2))logx1/2 = (π(x)−O(x1/2))logx.
2
It follows that π(x)/x = o(1) which can be inserted into (1.1) again to get that
(cid:88)
n ∼ ψ(n) ∼ π(n)logn+ o(1) ∼ π(n)logn.
x(cid:54)n
It follows that π(n) ∼ n/logn. (cid:3)
In fact this argument leads to explicit error terms – not just asymptotic results – but
this will not concern us since it turns out that the main contribution to the error term
|π(n)−n/logn| comes from approximating Li(n) by n/logn.
There is a natural convolution operation on arithmetic functions. Given two arithmetic
functions f and g we define their convolution f ∗g point-wise by
(cid:88)
f ∗g(x) = f(z)g(y).
zy=x
This convolution has an identity δ and it turns out that 1, the function that takes the
value 1 everywhere, has an inverse: we define the M¨obius µ-function by
(cid:40)
(−1)k if x = p ...p for distinct primes p ,...,p
1 k 1 k
µ(x) :=
0 otherwise,
with the usual convention that 1 has a representation as the empty product.
TOPICS IN ANALYTIC NUMBER THEORY 3
Theorem 1.2 (Mo¨bius inversion). We have the identity
µ∗1 = δ = 1∗µ.
Proof. Suppose that n is a natural number so that by the fundamental theorem of arith-
metic there are primes p ,...,p and naturals e ,...,e such that n = pe1...pel. Since µ
1 l 1 l 1 l
is only supported on square-free integers, if µ(d) (cid:54)= 0 then d|p ...p . It follows that
1 l
(cid:88) (cid:88) (cid:88)
1∗µ(n) = µ(d) = µ(d) = (−1)|S| = (1−1)l = δ(n).
d|n d|p1...pl S⊂[l]
The results follows by symmetry of convolution. (cid:3)
Convolution has the effect of smoothing or concentrating in Fourier space which is de-
sirable because it means that the function is easier to estimate. As an example of this
we shall estimate the average value of the divisor function τ(n) (sometimes denoted d(n))
defined by
(cid:88)
τ(n) := 1 = 1∗1(n),
d|n
that is the number of divisors of n. Before we begin we recall that the Euler constant is
(cid:90) ∞ {x}
γ := dx,
x(cid:98)x(cid:99)
1
where (cid:98)x(cid:99) is the integer part of x and {x} := x − (cid:98)x(cid:99). There are many open questions
about the Euler constant, although they will not concern us. For our work it is significant
only as the constant in the following elementary proposition.
Proposition 1.3. We have the estimate
(cid:88) 1
= logn+γ +O(1/n)
x
x(cid:54)n
Proof. Given the definition of γ this is essentially immediate:
(cid:88) 1 (cid:88) 1 (cid:90) n+1 1
−logn = − dx+O(1/n)
x x x
x(cid:54)n x(cid:54)n 1
(cid:90) n+1(cid:18) 1 1(cid:19)
= − dx+O(1/n)
(cid:98)x(cid:99) x
1
(cid:90) n+1 {x}
= dx+O(1/n)
x(cid:98)x(cid:99)
1
(cid:90) ∞
= γ + O(1/x2)dx+O(1/n) = γ +O(1/n).
n+1
The result is proved. (cid:3)
We shall now use Dirichlet’s hyperbola method to estimate the average value of the
divisor function.
4 TOM SANDERS
Proposition 1.4. We have the estimate
(cid:88) √
τ(x) = nlogn+(2γ −1)n+O( n).
x(cid:54)n
Proof. An obvious start is to note that
(cid:88) (cid:88) (cid:88) n (cid:88)(cid:16)n (cid:17)
τ(x) = 1 = (cid:98) (cid:99) = +O(1) = nlogn+O(n)
a a
x(cid:54)n ab(cid:54)n a(cid:54)n a(cid:54)n
by Proposition 1.3. The weakness of this argument is that the approximation
n n
(cid:98) (cid:99) = +O(1)
a a
is not a strong statement when a is close to n – the error term is of comparable size to the
main term. However, since ab (cid:54) n we certainly have that at least one of a and b is always
√
at most n. It follows from the inclusion-exclusion principle that
(cid:88) (cid:88) n (cid:88)
1 = 2 (cid:98) (cid:99)− 1.
a
√ √
ab(cid:54)n a(cid:54) n a,b(cid:54) n
This is called the hyperbola method because it is a way of counting lattice points below
the hyperbola xy = n. Now, as before we have that
(cid:88) (cid:88) (cid:16)n (cid:17) √
τ(x) = 2 +O(1) −( n+O(1))2.
a
x(cid:54)n a(cid:54)√n
On the other hand by Proposition 1.3 we have that
(cid:88) n 1
= nlogn+γn+O(1),
a 2
√
a(cid:54) n
and the result follows on rearranging. (cid:3)
Recalling Stirling’s formula (or directly) we have that
(cid:88)
logx = logn! = nlogn−n+O(logn),
x(cid:54)n
thus if we put
(cid:88)
∆(n) := (τ(x)−logx−2γ),
x(cid:54)n
√
then we showed above that ∆(n) = O( n). With additional ideas of a rather different
natureVoronishowedthat∆(n) = O(n1/3logn)andthishassincebeenimprovedtoO(nα)
for some α < 1/3. In the other directly Hardy and Landau showed that ∆(n) = Ω(n1/4),
but the true order of the error term is not known.
We now return to the von Mangoldt function. By the Fundamental Theorem of Arith-
metic if n ∈ N then n = pe1...pel for some primes p ,...,p and naturals e ,...,e . Now,
1 l 1 l 1 l
l
(cid:88) (cid:88) (cid:88)
1∗Λ(n) = Λ(n) = logp = e logp = logn.
i i
d|n pk|n i=1
TOPICS IN ANALYTIC NUMBER THEORY 5
Applying M¨obius inversion then gives us an expression for Λ as a convolution:
Λ = µ∗1∗Λ = µ∗log.
This expression will allow us to relate ψ to the function M defined by
(cid:88)
M(n) := µ(x).
x(cid:54)n
The M¨obius µ-function takes the values −1 and 1 (and 0) and so we have the trivial upper
bound |M(n)| (cid:54) n. Since µ does not display any obvious additive patterns we might hope
that µ takes the values 1 and −1 in a random way. If it did it would follow from the central
limit theorem that we would have square-root cancellation:
M(n) = O (n1/2+ε) for all ε > 0.
ε
The Riemann hypothesis is the conjecture that this is the case and it is far from being
know. On the other hand we shall be able to show that there is some cancellation so that
M(n) = o(n) and this already implies the prime number theorem.
Proposition 1.5. If M(n) = o(n) then ψ(n) ∼ n.
Proof. We use the hyperbola method again: let B be a parameter to be optimised later,
and then note that
(cid:88)
ψ(n)−n+2γ = (Λ(x)−1(x)+2γδ(x))
x(cid:54)n
(cid:88)
= µ(a)(logb−τ(b)+2γ)
ab(cid:54)n
(cid:88)
= M(n/b)(logb−τ(b)+2γ)
b(cid:54)B
(cid:88) (cid:88)
+ µ(a) (logb−τ(b)+2γ).
a(cid:54)n/B B<b(cid:54)n/a
By hypothesis we have that
(cid:88)
M(n/b)(logb−τ(b)+2γ) = o(n).O(B.(logB +τ(B)+2γ)) = o(B2n)
b(cid:54)B
which covers the first term. For the second we note that
(cid:88) (cid:88) (cid:88)
| µ(a) (logb−τ(b)+2γ)| (cid:54) |∆(n/a)−∆(B)|
a(cid:54)n/B B<b(cid:54)n/a a(cid:54)n/B
√
(cid:88) (cid:112)
= O( n/a+ B)
a(cid:54)n/B
by the bound for ∆ given by Proposition 1.4. On the other hand by integral comparison
we have
(cid:88) √ (cid:90) n/B 1 (cid:112)
1/ a (cid:54) √ da = O( n/B),
a
a(cid:54)n/B 0
6 TOM SANDERS
whence
√ √
(cid:88) (cid:112)
O( n/a+ B) = O(n/ B).
a(cid:54)n/B
Combining what we have shown we see that
√
ψ(n)−n+2γ = o(B2n)+O(n/ B).
It remains to choose B tending to infinity sufficiently slowly which gives the result. (cid:3)
2. The Fourier transform
In this section we shall develop some of the basic ideas of Fourier analysis. There are
many references for this material, with the classic being Rudin’s book ‘Fourier analysis on
groups’.
TheFouriertransformonR, Z, Tandfinitegroupsiswellknownandinthe20thCentury
some efforts were made to unify the theories and it was discovered that the natural setting
was that of locally compact abelian groups; to begin with let G be such. Those unfamiliar
with the theory of topological groups may just as well think of G as just being one of the
aforementioned examples.
Fourier analysis is important because of its relationship with convolution: suppose that
µ and ν are (complex Borel) measures on G. Then the convolution µ ∗ ν is the measure
defined by
(cid:90)
µ∗ν(A) = 1 (x+y)dµ(x)dν(y).
A
This form of convolution is a generalisation of convolution of arithmetic functions as we
shall see later. We norm the space of complex Borel measures in the usual way:
(cid:90) (cid:90)
(cid:107)µ(cid:107) := d|µ| := sup{ fdµ : f ∈ C (G)},
0
and write M(G) for the space of complex Borel measures on G with finite norm. Convo-
lution on this space interacts well with the norm in that we have Young’s inequality:
(cid:107)µ∗ν(cid:107) (cid:54) (cid:107)µ(cid:107)(cid:107)ν(cid:107) for all µ,ν ∈ M(G).
We shall write G(cid:98) for the dual group of G, that is the locally compact abelian group of
continuous homomorphisms γ : G → S1, where S1 := {z ∈ C : |z| = 1}, and are now in a
position to record the Fourier transform. Given µ ∈ M(G) we define its Fourier transform
in (cid:96)∞(G(cid:98)) by
(cid:90)
µ(γ) := γdµ for all γ ∈ G(cid:98).
(cid:98)
It is easy to check that
µ(cid:91)∗ν(γ) = µ(γ)ν(γ).
(cid:98) (cid:98)
TOPICS IN ANALYTIC NUMBER THEORY 7
There is a special measure on G called Haar measure which is the unique (up to scaling)
translation invariant measure on G which we shall denote µ . If f ∈ L1(µ ) we may thus
G G
define the Fourier transform of f by
(cid:90)
f(cid:98)(γ) := γfdµ .
G
The advantage of this is that we have an inverse theorem.
Theorem 2.1 (TheFourierinversiontheorem). Suppose that G is a locally compact abelian
group, f ∈ L1(µ ) and f(cid:98)∈ L1(µ ). Then
G G(cid:98)
(cid:90)
f(x) = f(cid:98)(γ)γ(x)dµ (γ) for a.e. x ∈ G.
G(cid:98)
WehavebeendeliberatelyvagueaboutscalingtheHaarmeasureonG; inallapplications
the scaling will be clear.
It may be instructive to consider an example. Suppose that G = Z/pZ for some prime
p. Then the characters on G are just the maps
x (cid:55)→ exp(2πirx/p),
and so G(cid:98) is isomorphic to Z/pZ – the natural measure on each of the groups is different
however. We shall think of G as endowed with normalised counting measure and G(cid:98) as
endowed with counting measure so that
µ (A) := |A|/|G| and µ (Γ) := |Γ|.
G G(cid:98)
Given a function f : G → C, we may think of it as a sum of canonical basis vectors (δ )
x x∈G
where
(cid:40)
|G| if y = x;
δ (y) :=
x
0 otherwise.
We then have the decomposition
(cid:90)
f(x) = f(y)δ (x)dµ (y);
y G
the Fourier transform changes this basis to the set of characters. Indeed
(cid:90)
f(cid:98)(γ) = fγdµ = (cid:104)f,γ(cid:105)
G
and the inversion formula just says that
(cid:90)
(cid:88) (cid:88)
f = (cid:104)f,γ(cid:105)γ = f(cid:98)(γ)γ = f(cid:98)(γ)γdµ (γ).
G(cid:98)
γ∈G(cid:98) γ∈G(cid:98)
Now, given a function f it also induces a convolution operator
M : L2(µ ) → L2(µ );g (cid:55)→ f ∗g
f G G
8 TOM SANDERS
and the Fourier transform serves to diagonalize this operator. Indeed if we decompose g
in the Fourier basis:
(cid:88)
g = (cid:104)g,γ(cid:105)γ,
γ∈G(cid:98)
then it is easy to check that
(cid:88)
M g = f(cid:98)(γ)(cid:104)g,γ(cid:105)γ.
f
γ∈G(cid:98)
Note, of course, that f(cid:98)(γ) = (cid:104)f,γ(cid:105) too but we use the above notation to make things more
suggestive.
Many problems in mathematics involve convolution: in probability theory the law of the
sum of two independent random variables is the convolution of their laws. For enumerative
problems 1 ∗1 (x) is the (scaled) number of ways of writing x as a sum a+b where a ∈ A
A B
and b ∈ B. Indeed, before the end of the course we shall find ourselves in a position where
we want to examine
1 ∗1 ∗1 (x)
A A A
for A the set of primes less than or equal to N and so we shall be pleased to have access
to a basis which diagonalizes this expression.
Returning now to the more immediate concern of arithmetic functions we suppose that
G is the reals under addition and f is an arithmetic function. Then we write m for the
f
atomic measure defined by
(cid:88)
m (A) := f(n)1 (logn).
f A
n∈N
Shortly we shall damp our measures so that they are better behaved, but before that it is
easy to see that if f and g are arithmetic functions then
m ∗m = m .
f g f∗g
An important additional class of measures is the class of damped versions of the above.
Indeed, suppose that σ > 1, and write m for the exponentially damped measure defined
f,σ
by
(cid:88)
m (A) := f(n)1 (logn)exp(−σlogn).
f,σ A
n∈N
Usefully (and not entirely coincidentally) we have the identity
m ∗m = m
f,σ g,σ f∗g,σ
for all arithmetic functions f and g.
If|f(n)| = logO(1)nthenweseethatm isfinitesothat,forexample, m ,m ,m ∈
f,σ 1,σ µ,σ Λ,σ
M(R); in particular we may take the Fourier transform and, in fact,
m (t) = ζ(σ +it) for all t ∈ R,
(cid:100)1,σ
is the classical ζ-function.
TOPICS IN ANALYTIC NUMBER THEORY 9
3. The prime number theorem
We now turn to proving the prime number theorem. We should like to examine the inner
product
(cid:90)
(cid:88)
µ(x) = I dm
σ µ,σ
x(cid:54)n
where
(cid:40)
exp(σx) if x (cid:54) logn
I (x) =
σ
0 otherwise.
Unfortunately the function I is not smooth enough and we are not able to control the
σ
Fourier transform of m well enough.
µ,σ
We shall estimate the Fourier transform of m through m via the usual convolution
µ,σ 1,σ
identity:
m (t) = 1/m (t).
(cid:100)µ,σ (cid:100)1,σ
First we note a simple estimate – we shall make further use of the basic method in Lemma
3.5 so it is worth becoming familiar now.
Lemma 3.1. We have the estimate
1
m (t) = +O(log(1+|t|)).
(cid:100)1,σ
σ −1+it
Proof. Naturally we proceed by integral comparison: let T > 1 be a parameter to be
optimised later.
(cid:90) ∞ (cid:90) ∞
m (t) = x−σ−itdx+ ((cid:98)x(cid:99)−σ−it −x−σ−it)dx
(cid:100)1,σ
1 1
1 (cid:90) T (cid:90) ∞
= + O(x−σ)dx+ ((cid:98)x(cid:99)−σ−it −x−σ−it)dx
σ −1+it
1 T
1 (cid:90) ∞
= +O(logT)+ ((cid:98)x(cid:99)−σ−it −x−σ−it)dx.
σ −1+it
T
To estimate the second integral we just note that
(cid:90) ∞ (cid:88)∞ (cid:90) 1 ((1+θ/n)σ+it −1)
((cid:98)x(cid:99)−σ−it −x−σ−it)dx = dθ
(n+θ)σ+it
T 0
n=T
∞
(cid:88)
= O(|t|/nσ+1) = O(|t|/T).
n=T
The result follows on setting T = 1+|t|. (cid:3)
We encode the Fundamental Theorem of Arithmetic in an analytic way as follows.
Lemma 3.2 (The Euler Product formula). For σ > 1, t ∈ R we have the equivalence
(cid:89)
m (t) = (1−p−σ−it)−1.
(cid:100)1,σ
p
10 TOM SANDERS
Proof. Write
(cid:89) (cid:89)
P := (1−p−σ−it)−1 = (1+p−1(σ+it) +p−2(σ+it) +...).
N
p(cid:54)N p(cid:54)N
By the Fundamental Theorem of Arithmetic if n (cid:54) N then there are powers e ,...,e and
1 r
primes p ,...,p (cid:54) N such that n = pe1...per whence (multiplying out P ) we have
1 r 1 r N
N ∞
(cid:88) (cid:88)
|P − n−σ−it| (cid:54) n−σ.
N
n=1 n=N+1
On the other hand
N ∞
(cid:88) (cid:88)
|m (t)− n−σ−it| (cid:54) n−σ,
(cid:100)1,σ
n=1 n=N+1
so
∞
(cid:88)
|P −m (t)| (cid:54) 2 n−σ = O((σ −1)−1N1−σ)
N (cid:100)1,σ
n=N+1
by the triangle inequality and integral comparison. The lemma is proved on letting N →
∞. (cid:3)
We now record the number theoretic content of this section.
Lemma 3.3. There is an absolute constant c > 0 such that if σ ∈ (1,1+c) then we have
the estimates
(cid:107)m (cid:107) = (σ −1)−1 +O(1)
µ,σ
and
|m (t)| = O((σ −1)−3/4log1/2(2+|t|)).
(cid:100)µ,σ
Proof. First we note that
(cid:107)m (cid:107) = (cid:107)m (cid:107) = (σ −1)−1 +O(1)
µ,σ 1,σ
by integral comparison. The second is where we have the main idea: begin by noting that
1 (cid:54) (1+(1+p−it +pit)2p−σ)
= 1+3p−σ +2p−it−σ +2pit−σ +p−2it−σ +p2it−σ
= (1+O(p−2σ))(1−p−σ)−3|1−p−it−σ|−4|1−p−2it−σ|−2.
It should be remarked that this is, perhaps, the most mysterious part of the proof of the
prime number theorem and has defied good explanation.
Thus
(cid:89)
1 = O( (1−p−σ)−3|1−p−it−σ|−4|1−p−2it−σ|−2)
p
= O((cid:107)m (cid:107)3|m (t)|4|m (2t)|2),
1,σ 1,σ 1,σ