THE PRIME NUMBER THEOREM
NIKOLAOS PATTAKOS
1. introduction
In number theory, this Theorem describes the asymptotic distribution of the prime
numbers. The Prime Number Theorem gives a general description of how the primes are
distributed among the positive integers. It formalizes the intuitive idea that primes become
less common as they become larger.
In these notes a proof of the prime number theorem is presented. It is the old classical
proof that uses the Tauberian Theorem of Wiener. It appears in the book “Functional
Analysis” by Dr. Rudin. The Riemann zeta function plays a central role to the proof and
some knowledge of functional analysis is also needed. In the first section we present some
definitions and results that we need for the proof of the Prime Number Theorem. In the
second section we discuss some well known properties of the Riemann zeta function and in
the third section we prove Wiener’s Theorem which implies the Prime Number Theorem.
The letter p will denote a prime number and P will denote the set of all prime numbers.
We will write N = Z+ for the set of all natural numbers excluding zero and throughout
these notes, n, m, k, d will denote natural numbers. We will write d/n everytime the number
d is a divisor of n. Sometimes we will also write x/y to denote the fraction xy . Since we
use the same symbol for two different things the meaning will be clear from the context.
The integral part of a real number r, denoted as [r], is the largest integer that is less than
or equal to r. For x ∈ R+ we define the function
π(x) :=
X
1 = |{p ∈ P : p ≤ x}|,
p≤x
which counts the number of primes that are less than or equal to x. Obviously for x < 2,
π(x) = 0. Our goal is to show the following
Theorem 1. The prime number theorem
lim
x→+∞
π(x) log x
= 1.
x
We will need the following functions
Definition 2. Mangoldt’s function
(
log p,
Λ(n) =
0,
1
n = pk
otherwise
2
NIKOLAOS PATTAKOS
Some values for Λ are Λ(1) = 0, Λ(2) = log 2, Λ(3) = log 3, Λ(4) = log 2, ...
Definition 3. The ψ function defined as
(1)
ψ(x) =
X
Λ(n)
n≤x
and
F (x) =
+∞ X
x
ψ
.
m
m=1
Notice that since ψ(x) = 0 for x < 2, the sum that defines the function F is a finite sum.
Our purpose is to try and bound the fraction that appears in Theorem 1 from above
and below by some other functions that are “easier” to find their behavior at infinity. We
claim that
Lemma 4.
(2)
π(x) log x
1
ψ(x) log x
ψ(x)
,
≤
<
+
x
x
log x x log(x/ log2 x)
for x > e and
F (x) = x log x − x + b(x) log x,
(3)
where b(x) remains bounded as x → +∞.
Observe that if we know
ψ(x)
=1
x
then by inequality (2) we immediately have a proof of Theorem 1. This means that the
problem reduces to proving Lemma 4 and equality (4). We start with the Lemma.
h
i
x
Proof. Let us start by noticing that the natural number log
log p is the number of powers of
p that do not exceed x. Hence,
(4)
lim
x→+∞
ψ(x) =
X h log x i
p≤x
log p
log p,
since all other integers 1 ≤ n ≤ x that are a product of at least two primes give Λ(n) = 0.
Therefore,
ψ(x) ≤
X log x
p≤x
log p
· log p =
X 1 log x = π(x) log x,
p≤x
which proves the left half of inequality (2). Now for the right half if 1 < y < x, then
THE PRIME NUMBER THEOREM
π(x) − π(y) =
X
3
X log p
ψ(x)
≤
,
log y
log y
1≤
y<p≤x
y<p≤x
which together with the trivial observation π(y) < y imply that
ψ(x)
.
log y
( we need this to be less than x, so log2 x > 1 iff x > e )
π(x) < y +
Now for y =
x
log2 x
π(x) <
x
ψ(x)
+
,
2
log x log(x/ log2 x)
and this means we are done.
To finish the proof of the Lemma we need to prove equality (3) for the function F (x).
We first start with
F (n) − F (n − 1) =
+∞ n − 1 X
n
ψ
−ψ
.
m
m
m=1
The m-th summand is 0 except when m divides n in which case it is equal to Λ(n/m).
Here we use the fact that
(
h n i hn − 1i
0, n/m ∈
/N
−
=
m
m
1, n/m ∈ N
Thus,
F (n) − F (n − 1) =
(5)
X n X
Λ
=
Λ(d) = log n.
m
m/n
d/n
Let us explain the last equality in equation (5). Any natural number n can be factorized
as a product of powers of prime numbers in a unique way
n=
k
Y
pαi i ,
i=1
where αi ∈ N, for all i ∈ {1, ..., k}. The only terms that contribute to the sum
are exactly the terms p1 , p21 , ..., pα1 1 , ..., pk , p2k , ..., pαk k . Therefore,
X
d/n
Λ(d) =
αi
k X
X
i=1 m=1
Λ(pm
i )
=
αi
k X
X
log pi =
i=1 m=1
k
X
i=1
Since F (1) = 0, we can use induction to show
F (n) = log(n!),
log pαi i
= log
k
Y
i=1
P
d/n Λ(d)
pαi i = log n.
4
NIKOLAOS PATTAKOS
P
for all n ∈ N. By rewriting this last equality as F (n) = nm=1 log m, we see right away
that we should compare F (x) with the integral
ˆ x
J(x) =
log tdt = x log x − x + 1.
1
If n ≤ x ≤ n + 1, then
J(n) < F (n) ≤ F (x) ≤ F (n + 1) < J(n + 2).
The first inequality is true because F (n) is a right Riemann sum of J(n). The second and
third are obvious because our functions are increasing. The forth inequality is true because
F (n + 1) is a left Riemann sum of J(n + 2) which is the same as saying
ˆ
n+1
X ˆ k+1
n+2
J(n + 2) =
log tdt =
1
k=1
log tdt >
k
n+1
X
log k = F (n + 1).
k=1
Thus,
ˆ
n+2
|F (x) − J(x)| ≤ J(n + 2) − J(n) =
log tdt < (n + 2 − n) log(n + 2) ≤ 2 log(x + 2).
n
Observe that F (x) = J(x) + (F (x) − J(x)) = x log x − x + [1 + F (x) − J(x)] and set
1 + F (x) − J(x)
.
log x
Obviously, F (x) = x log x − x + b(x) log x and actually for all x
b(x) =
1
|F (x) − J(x)|
1
2 log(x + 2)
+
<
+
,
log x
log x
log x
log x
which implies that for large values of the variable x, b(x) remains bounded since we have
limx→+∞ ( log1 x + 2 log(x+2)
) = 2.
log x
|b(x)| ≤
2. the zeta function ζ(s)
The zeta function appears naturally in some of the calculations of the next section and
that is why we need to prove some results regarding its zeros on the complex plane. As in
the classical number theory, complex variables will be denoted by s = σ + it, σ, t ∈ R. In
the half plane σ = <(s) > 1 the zeta function is defined by the series
ζ(s) :=
+∞
X
1
,
ns
n=1
which converges absolutely in every compact subset K of the half plane. To see this we
can use the following Lemma with X = K and φn (s) = n1s .
THE PRIME NUMBER THEOREM
5
Lemma 5. Suppose that X is a nonempty set and φn : X → C, n ∈ N, a sequence of
functions. If
+∞
X
sup |φn (x)| < +∞,
n=1 x∈X
then there is a function φ : X → C such that
P+∞
n=1 φn
→ φ uniformly in X.
Indeed, there is positive such that for every s ∈ K, <(s) > 1 + . So,
1
1
1
≤ 1+ ,
sup s =
sup
σ
n
s∈K n
s∈K,σ=<(s) n
which proves that the assumptions of the Lemma are satisfied. We also need the following
classical result of complex analysis.
Theorem 6. Let D be an open subset of C and fn ∈ O(D), for all n ∈ N. Suppose that
f : D → C is a function such that for any compact subset K of D, fn → f uniformly in
K. Then f ∈ O(D) and actually fn0 → f 0 uniformly in every compact subset of D.
This means that ζ(s) is holomorphic in the half plane. Now our purpose is to define
the function ζ(s) in a larger domain so that it remains a holomorphic function. A simple
calculation yields
ˆ
s
N +1
[x]x
−1−s
dx = s
1
N ˆ
X
n+1
x−1−s dx = s
n=1 n
N
X
n=1
N
n·
x−s x=n+1 X
=
n(n−s − (n + 1)−s ).
−s x=n
n=1
Let us recall Abel’s summation formula
N
X
αn (bn+1 − bn ) = (αN +1 bN +1 − α1 b1 ) −
n=1
N
X
bn+1 (αn+1 − αn ),
n=1
which reminds the integration by parts formula, and apply it with αn = n and bn = n−s
to obtain
ˆ
s
N +1
N
h
i
X
[x]x−1−s dx = − (N + 1)(N + 1)−s − 1 −
(n + 1)−s
1
n=1
=
=
N
X
n=1
N
+1
X
n=1
1
+ 1 − (N + 1)(N + 1)−s
(n + 1)s
1
N +1
−
.
s
n
(N + 1)s
6
NIKOLAOS PATTAKOS
But for σ = <(s) > 1 we see that limN →+∞ |(N + 1)/(N + 1)s | = 0. Thus, we have proved
that for s on the half plane <(s) > 1, we have the equality
(6)
ζ(s) =
ˆ +∞
+∞
X
1
=
s
[x]x−1−s dx.
ns
1
n=1
Denote by b(x) = [x] − x and let us have a look at the expression
s
+s
s−1
ˆ
+∞
b(x)x−1−s dx,
1
which makes sense for s ∈ C \ {1}, <(s) > 0. Notice that since b(x) is a bounded function,
the last integral defines a holomorphic function in the half plane <(s) > 0 and that the
s
function s−1
has a simple pole of residue 1 at the point s = 1. In addition, for σ = <(s) > 1
we can split the integral as
s
+s
s−1
and
ˆ
ˆ
+∞
s
+∞
ˆ
[x]x−1−s dx − s
1
since limx→+∞
equality
(7)
x · x−1−s dx,
1
x−s dx = s
1
1/|xs−1 |
+∞
x1−s x=+∞ s
,
=
1 − s x=1
s−1
= limx→+∞ 1/|x|σ−1 = 0. Therefore, for <(s) > 1 we have the
ζ(s) =
s
+s
s−1
ˆ
+∞
b(x)x−1−s dx.
1
This formula allows us to extend the zeta function to the set {s ∈ C \ {1} : <(s) > 0}
as a meromorphic function with a simple pole of residue 1 at the point 1. The following
Theorem is going to play an important role in the proof of Theorem 1 in section (3) and
of course it is a very interesting result on its own.
Theorem 7.
ζ(s) 6= 0,
for all s ∈ {z ∈ C \ {1} : <(s) ≥ 1}.
For the proof we need the following observation.
Lemma 8. For s such that <(s) > 1 we have the representation
ζ(s) =
Y
p∈P
1−
1 −1
.
ps
THE PRIME NUMBER THEOREM
7
Indeed,
1
1
1
1
1 = 1 + ps + p2s + p3s + ...
1 − ps
and since every n has a unique factorization as a product of primes we get that for all s
with <(s) > 1
+∞
X
Y
1
1 −1
=
1
−
.
ns
ps
n=1
p∈P
This connection between the zeta function and prime numbers was discovered by Euler.
Let us also recall the following Lemma.
P
Lemma 9. Suppose that we have a sequence αn ∈ C such that +∞
n=1 |αn | < +∞. Then
Q+∞
the infinite product n=1 (1 + αn ) exists and is not equal to 0.
Now we are ready for the proof of Theorem 7.
Proof. Let {pn }n∈N be an enumeration of P. By applying Lemma 9 for αn = − p1s we get
n
Q
1
that the product +∞
n=1 (1 − ps ) exists and is nonzero. Since,
n
+∞
Y
+∞
1−
n=1
1 Y
1 −1
1
−
= 1,
psn
psn
n=1
we immediately obtain that for all s such that <(s) > 1
+∞
Y
1−
n=1
1 −1
6= 0.
psn
Therefore, by Lemma 8, ζ(s) 6= 0 for all s in the half plane <(s) > 1. The remaining case
is to see what happens on the line 1 + it for t ∈ R \ {0}. Fix such a t. For σ = <(s) > 1
we have
(8)
log |ζ(σ)3 · ζ(σ + it)4 · ζ(σ + 2it)| =
X
m−1 ρ−ms <(3 + 4ρ−imt + ρ−2imt ) ≥ 0,
ρ,m
iθ
iθ
since <(6 + 8eiθ + 2e2iθ ) = (e 2 + e− 2 )4 ≥ 0 for all θ. Hence, by taking exponentials in
inequality (8) we arrive at
3 ζ(σ
+ it) 4
1
(9)
,
|ζ(σ + 2it)| ≥
σ−1
σ−1
for σ > 1. If ζ(1 + it) were 0, the left hand side of inequality (9) would converge to a limit
as σ → 1+ , namely, |ζ 0 (1 + it)|4 |ζ(1 + 2it)|, since (σ − 1)ζ(σ) → 1 (remember that ζ(s) has
a simple pole of residue 1 at the point 1). But the right hand side of inequality (9) goes to
|(σ − 1)ζ(σ)| 8
NIKOLAOS PATTAKOS
infinity. This means that we would have a contradiction. Thus, the zeta function has no
zeros on the line 1 + it, t ∈ R, and the proof is complete.
Before we proceed to the next section let us see what we have done so far. The ζ function
is defined on the half plane <(s) > 1 as a power series and it can be extended to the whole
half plane D = {z ∈ C \ {1} : <(z) > 0} by the use of formula (7). We have showed that it
has no zeros for <(s) > 1 and for s = 1 + it, t ∈ R \ {0}. This means that if ζ(s) has any
zeros in D they must be in the critical strip {z ∈ D : 0 < <(z) < 1}. Riemann’s hypothesis
(RH) claims that for s in the critical strip we have
1
ζ(s) = 0 =⇒ <(s) = .
2
That is all the zeros of ζ lie on the critical line <(s) = 12 . This problem is still open.
3. wiener’s theorem and the prime number theorem
Here some knowledge of standard facts of Functional Analysis is needed. The notion of
the Fourier transform plays a central role and at some point we also use some fundamental
facts of the theory of distributions. All of them are pretty standard. We start with a
Lemma.
Lemma 10. Suppose that f ∈ L1 (Rn ), t ∈ Rn and > 0. Then there is h ∈ L1 (Rn ) with
khk1 < and such that
ĥ(s) = fˆ(t) − fˆ(s)
(10)
for all s in some neighborhood of t.
Proof. Choose g ∈ L1 (Rn ) so that ĝ = 1 in some neighborhood of 0, say B(0, r). For λ > 0
and x ∈ Rn , put
x
gλ (x) = eitx · λ−n · g
λ
ˆ
and define hλ (x) = f (t)gλ (x) − (f ∗ gλ )(x). Since
ˆ
ˆ
eity y −isy
gˆλ (s) =
e
· gλ (y)dy =
e−isy · n · g
dy
λ
λ
Rn
Rn
which is equal to (x = λy )
ˆ
ˆ
y dy
e−iy(s−t) · g
=
e−iλx(s−t) · g(x)dx = ĝ(λ(s − t)).
n
λ
λ
n
n
R
R
This last Fourier transform of g can be chosen to be 1 if λ(s−t) ∈ B(0, r) which is the same
as saying that s − t ∈ B(0, λr ) or s ∈ t + B(0, λr ) =: Vλ . Thus, gˆλ = 1 in some neighborhood
of t. In addition, by the definition of hλ we have
hˆλ (s) = fˆ(t)gˆλ (s) − fˆ(s)gˆλ (s) = fˆ(t) − fˆ(s),
THE PRIME NUMBER THEOREM
9
if s ∈ Vλ which means that (10) is true for hλ in place of h. Next by the definition of hλ
ˆ
hλ (x) =
f (y)[e−ity · gλ (x) − gλ (x − y)]dy
Rn
and observe that the absolute value of the expression in brackets is
x − y 1 x ·
g
−
g
.
n
λ
λ
λ
It follows that
ˆ
ˆ
|g(ξ) − g(ξ − λ−1 y)|dξdy.
|f (y)|
khλ k1 ≤
Rn
Rn
Now the inner integral is at most 2kgk1 and it tends to 0 as λ → +∞ for every fixed
y ∈ Rn . Therefore, khλ k1 → 0 by the Lebesgue Dominated Convergence Theorem.
Theorem 11. If φ ∈ L∞ (Rn ), Y is a subspace of L1 (Rn ) and f ∗ φ = 0 for every f ∈ Y
then the set
Z(Y ) =:
\
{s ∈ Rn : fˆ(s) = 0}
f ∈Y
contains the support of the tempered distribution φ̂.
Proof. Fix a point t ∈ Rn \ Z(Y ). Then for some f ∈ Y we have that fˆ(t) = 1 (since there
is a function in Y whose Fourier transform is non zero on t and Y is a subspace of L1 (Rn )).
By Lemma (10) there is a function h such that khk1 < 1 and
ĥ(s) = 1 − fˆ(s),
for all s in some small neighborhood V of t. To prove the Theorem we have to show that
φ̂ = 0
in V , or equivalently that φ̂(ψ̂) = 0 for all ψ in the Schwartz class S(Rn ), whose Fourier
transform has its support in V . Fix such ψ. Since φ̂(ψ̂) = φ(ψ̌) = (φ ∗ ψ)(0) it suffices to
show that φ∗ψ = 0. For this purpose define g0 = ψ and for integer m > 0 let gm = h∗gm−1 .
Then kgm k1 ≤ khkm
1 kψk1 and since khk1 < 1 the function
G=
∞
X
gm
m=0
is in
L1 (Rn ).
Since on the support of ψ̂ we have ĥ(s) = 1 − fˆ(s) we obtain
(1 − ĥ(s))ψ̂(s) = ψ̂(s)fˆ(s)
which implies that
10
NIKOLAOS PATTAKOS
ψ̂ =
∞
X
ĥm ψ̂ fˆ = Ĝfˆ.
m=0
Thus, we arrive at the equality ψ = G ∗ f . So by our initial assumption (f ∗ φ = 0) we get
exactly what we needed
ψ ∗ φ = G ∗ f ∗ φ = G ∗ 0 = 0.
The following Theorem is known as Wiener’s Tauberian Theorem.
Theorem 12. If Y is a closed translation invariant subspace of L1 (Rn ) and if Z(Y ) = ∅,
then
Y = L1 (Rn ).
Proof. To say that Y is translation ´invariant we mean that τx f ∈ Y for all f ∈ Y and all
x ∈ Rn . Let φ ∈ L∞ (Rn ) such that f φ̌ = 0 for every f ∈ Y . Observe that
ˆ
f ∗ φ(x) =
(τ−x f )(y)φ̌(y)dy = 0,
Rn
by the translation invariance of Y . By the previous Theorem we have that the support
of the distribution φ̂ lies in Z(Y ) which is an empty set. Hence, φ̂ = 0 and since the
Fourier transform maps S(Rn )0 onto itself in an one to one way we get that φ = 0 as a
distribution. Therefore, φ = 0 in L∞ (Rn ). Thus Y ⊥ = {0} and by the Hahn-Banach
Theorem Y = L1 (Rn ).
Theorem 13. Suppose that K ∈ L1 (Rn ) and let Y be the smallest closed translation
invariant subspace of L1 (Rn ) that contains K. Then
Y = L1 (Rn ) ⇐⇒ K̂(t) 6= 0,
for all t ∈ Rn .
Proof. Note that Z(Y ) = {t ∈ Rn : K̂(t) = 0}. One inclusion is obvious by the definition
of Z(Y ). For the other observe that if K̂(s) = 0 for some value of the variable s then
n
τd
x K(s) = 0 for all x ∈ R and this implies that ĝ(s) = 0 for every linear combination g
of translates of K and since Y is the closed linear span of all translates of K we obtain
fˆ(s) = 0 for all f ∈ Y . Therefore, {t ∈ Rn : K̂(t) = 0} ⊂ Z(Y ).
Now for the proof of the Theorem if K̂(t) 6= 0 for all t ∈ Rn we have Z(Y ) = ∅ and by
the previous Theorem we have Y = L1 (Rn ).
For the other direction, if Y = L1 (Rn ) then Z(Y ) = ∅ since there is no s ∈ Rn such that
2
fˆ(s) = 0 for every f ∈ L1 (Rn ). For example, take f (x) = e−π|x| whose Fourier transform
has no zeros.
THE PRIME NUMBER THEOREM
11
Theorem 14. Suppose φ ∈ L∞ (Rn ), K ∈ L1 (Rn ), K̂(t) 6= 0 for all t ∈ Rn and there is a
constant α such that
lim (K ∗ φ)(x) = α · K̂(0).
|x|→∞
Then for every function f ∈ L1 (Rn ) we get
lim (f ∗ φ)(x) = α · fˆ(0).
|x|→∞
Proof. Put ψ = φ − α and define
Y =: {f ∈ L1 (Rn ) : lim (f ∗ ψ)(x) = 0}.
|x|→∞
L1 (Rn ).
It is clear that Y is a subspace of
We claim that Y is closed. To see this let fn ∈ Y
and fn → f in L1 (Rn ) as n → ∞. Since,
kf ∗ ψ − fn ∗ ψk∞ ≤ kf − fn k1 kψk∞
we get that fn ∗ ψ → f ∗ ψ uniformly on Rn . Thus, lim|x|→∞ (f ∗ ψ)(x) = 0. Also, Y is
translation invariant because we have the equalities
((τy f ) ∗ ψ)(x) = [τy (f ∗ ψ)](x) = (f ∗ ψ)(x − y).
Finally, since by our assumption the function K ∈ Y and since by the previous Theorem
the smallest closed translation invariant subspace of L1 (Rn ) that contains K is the whole
space we arrive at the desired result Y = L1 (Rn ).
The next and final Theorem is known as Ingham’s Tauberian Theorem.
Theorem 15. Suppose g is a real nondecreasing function on (0, ∞), g(x) = 0 for x < 1,
and let
G(x) =
∞ X
x
.
g
n
n=1
If G(x) = αx log(x) + bx + x(x), where α, b are constants and (x) → 0 as x → ∞, then
lim
x→∞
Proof. First we show that
g(x) − g(x/2) ≤
∞
X
g(x)
x
g(x)
= α.
x
is bounded. Since g is nondecreasing we have
(−1)n+1 g(x/n) = G(x) − 2G(x/2) = x · (α log 2 + (x) − (x/2)),
n=1
which is less than Ax where A is some positive constant. Since we have the equality
12
NIKOLAOS PATTAKOS
g(x) = g(x) − g(x/2) + g(x/2) − g(x/4) + . . .
it follows
x
x
+ A · + . . . = 2Ax.
2
P4
Now for x ∈ R define h(x) = g(ex ) and H(x) = ∞
n=1 h(x − log n). Then h(x) = 0 if x < 0,
H(x) = G(ex ) = α · ex log(ex ) + b · ex + ex (ex ) = ex (α · x + b + 1 (x)), where 1 (x) → 0 as
x → ∞.
Now let φ(x) = h(x)/ex . Obviously, φ is bounded by 2A and we have to prove that
limx→∞ φ(x) = α. Put k(x) = [ex ]e−x and let λ ∈ R+ \ Q. Define
g(x) ≤ Ax + A ·
K(x) = 2k(x) − k(x − 1) − k(x − λ).
(11)
Using the definition of the integer part of a real number it is easy to see that we have
ex K(x) ∈ L∞ (R) which means that K(x) ∈ L1 (R).
If we let s = σ + it, for σ > 0 we get
ˆ
ˆ
ˆ ∞
−xs
x −x(s+1)
k(x)e dx = [e ]e
dx =
[ex ]e−x(s+1) dx,
R
0
R
since for x < 0 the quantity ex < 1 and therefore, [ex ] = 0. It is time to make the change
of variables y = ex to get
ˆ
ˆ ∞
ζ(1 + s)
−xs
k(x)e dx =
[y]y −2−s dy =
,
1+s
R
1
and this is because for σ1 > 1 we have the equality (proved in the previous section)
ˆ ∞
[x]x−1−σ1 −it dx.
ζ(σ1 + it) = (σ1 + it)
1
Repeat this with k(x −1) and k(x−λ) in place of k(x) and let σ → 0 (Lebesgue Dominated
Convergence Theorem) to arrive at
ˆ
ζ(1 + it)
K(x)e−itx dx = (2 − e−it − e−iλt ) ·
.
1 + it
R
Since ζ(1 + it) 6= 0 and since λ ∈ R+ \ Q we have K̂(t) 6= 0 for all t 6= 0. Taking into
account that ζ has a simple pole of residue 1 at s = 1 the RHS of the previous equality
(with the use of Del Hopital’s rule) tends to
h
ζ(1 + it) · it i
ie−it + iλe−iλt
lim (2 − e−it − e−iλt ) ·
= lim
= 1 + λ 6= 0,
t→0
t→0
it · (1 + it)
i − 2t
which implies K̂(0) 6= 0.
Our next goal is to apply Wiener’s Tauberian Theorem. Let u(x) = [ex ], v = χ[0,∞) and
µ be the measure that assigns mass 1 to each point of the set
THE PRIME NUMBER THEOREM
13
X := {log n : n ∈ Z+ },
and whose support is this set. Observe that
H(x) =
∞
X
ˆ
h(x − log n) =
h(x − log n)dµ(n) = h ∗ µ(x).
X
n=1
Also,
ˆ
v ∗ µ(x) =
v(x − log n)dµ(n) =
X
∞
X
v(x − log n) =
n=1
∞
X
χ[0,∞) (x − log n) =
X
1,
n≤ex
n=1
where in this last sum the only terms that contribute are those n ≤ ex and every time we
find one we add 1. But this is exactly the integer part of ex which is by definition u(x).
Hence,
ˆ x
h ∗ u(x) = (h ∗ v ∗ µ)(x) = H ∗ v(x) =
H(y)dy,
0
where here we have used the fact that g(x) = 0 for 0 < x < 1. Now
ˆ
ˆ
y−x
y −y
−x
−x
φ ∗ k(x) =
e h(x − y)[e ]e dy = e (h ∗ u)(x) = e
x
H(y)dy,
0
R
and this last integral is the same as
ˆ
e−x
x
ˆ
h
α · yey + b · ey + 1 (y)y · ey dy = α · x − α + b + e−x α − b +
x
i
1 (y)ey dy .
0
0
By this and equality (11) we immediately obtain
lim (K ∗ φ)(x) = (1 + λ)α = αK̂(0).
x→∞
By Wiener’s Tauberian Theorem we get that the same equality is true for all functions
f ∈ L1 (R) in place of K.
For the last step of the proof assume that f1 , f2 are positive functions whose integral is 1
and such that the first one is supported on [0, ] and the second one on [−, 0], where is a
fixed positive number. Since ex φ(x) = h(x) = g(ex ) the function ex φ(x) in nondecreasing.
Thus, if x − ≤ y ≤ x we have
φ(y) ≤ e φ(x)
(observe ey φ(y) ≤ ex φ(x) =⇒ φ(y) ≤ ex−y φ(x) ≤ e φ(x)) and if x ≤ y ≤ x + we have
φ(y) ≥ e− φ(x).
Consequently,
14
NIKOLAOS PATTAKOS
e− (f1 ∗ φ)(x) ≤ φ(x) ≤ e (f2 ∗ φ)(x).
(12)
For instance, let us prove the second inequality as follows
ˆ x+
ˆ x+
f2 ∗ φ(x) =
f2 (x − y)φ(y)dy ≥
f2 (x − y)e− φ(x)dy = e− φ(x).
x
x
Finally, we let x → ∞ in (12) to derive that
α · e− ≤ lim inf φ(x) ≤ lim sup φ(x) ≤ α · e ,
x→∞
x→∞
for all positive which is the same as saying that
lim φ(x) = α.
x→∞
The Prime Number Theorem follows now if we take g(x) = ψ(x) where ψ is the function
defined in formula (1) and use Lemma (4) and Theorem (15).
References
[1] Walter Rudin, Functional Analysis. International Series in Pure and Applied Mathematics, Second
Edition.
N. Pattakos: School of Mathematics, University of Birmingham, Edgbaston, England.
e-mail address: nikolaos.pattakos@gmail.com