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Topics in Analytic Number Theory, Lent 2013. Lecture 8: Classical Zero Free Region

This document summarizes a lecture on the classical zero-free region for the Riemann zeta function and Dirichlet L-functions. It states that the zeta function has no zeros with real part greater than 1 - c/log(|γ|+2) for some constant c. Similarly, Dirichlet L-functions have no zeros with real part greater than 1 - c/(log q + log(|γ|+2)). It outlines the proof technique which uses trigonometric identities and estimates for logarithmic derivatives to establish these bounds. Finally, it presents a theorem showing that quadratic Dirichlet L-functions have at most one zero very close to the real line.

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0% found this document useful (0 votes)
38 views4 pages

Topics in Analytic Number Theory, Lent 2013. Lecture 8: Classical Zero Free Region

This document summarizes a lecture on the classical zero-free region for the Riemann zeta function and Dirichlet L-functions. It states that the zeta function has no zeros with real part greater than 1 - c/log(|γ|+2) for some constant c. Similarly, Dirichlet L-functions have no zeros with real part greater than 1 - c/(log q + log(|γ|+2)). It outlines the proof technique which uses trigonometric identities and estimates for logarithmic derivatives to establish these bounds. Finally, it presents a theorem showing that quadratic Dirichlet L-functions have at most one zero very close to the real line.

Uploaded by

Eric Parker
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Topics in analytic number theory, Lent 2013.

Lecture 8: Classical zero free region


Bob Hough
February 6, 2013
Reference for this lecture: Davenport Chaps. 13, 14.

The zero-free region


The classical zero free region for the Riemann zeta function asserts that has
all zeros + i with 1 logc|| for || sufficiently large. This has been
c
improved several times, and it is now known that 1 (log ||)
2/3+ , due to
Vinogradov. For the Dirichlet L-functions, the classical zero-free region is still
the best known.
Theorem 8.1 (Classical zero-free region). Let = + i be a zero of the
Riemann zeta function. There exists a constant c > 0 such that
1

c
.
log(|| + 2)

Suppose that mod q is a primitive complex character with zero . Then


1

c
.
log q + log(|| + 2)

Finally, suppose that mod q is a primitive real character. Let > 0. There
exists a constant c() > 0 such that if is a zero of L(s, ) and || > log q then
1

c()
.
log q + log(|| + 2)

We will see that the proof of each of these statements is essentially the same.
The idea is quite simple. Say ( + i) = 0 with 1. Then we may expect
that (1 +  + i) is small, so that the real part of the logarithm is negative and
large:
X cos(n log p)
< log (1 +  + i) =
<<< 0.
npn(1+)
p,n
This suggests that pi 1 for many p, so that p2i 1 for many p, which
will make (1 +  + 2i) large. Thus we may translate an upper bound for
(1 +  + 2i) into a lower bound for (1 +  + i). This idea is made formal
with the following trigonometric identity.
Lemma 8.2. We have 2(1 + cos )2 = 3 + 4 cos + cos 2 0.
1

Proof of classical zero free region. Consider first with () = 0. We may evidently assume that || > c for some small constant c, since has a pole at 1.
Let > 1. By the trigonometric identity,
3[ 0 /()] + 4<[ 0 /( + i)] + <[ 0 /( + 2i)]

X
log p
=
(3 + 4 cos(n log p) + cos(2n log p) 0.
pn
p,n
Now

1
+ O(1)
1
while we may express the other values in terms of the zeros near (recall

1
< (+i)(+i
0 ) = ()2 +( 0 )2 ):
0 /() =

< 0 /( + i) =

X
(0 )=0
| 0 |<1

+ O(log(|| + 2)).
( )2 + ( 0 )2

The zero is itself included in the sum, and notice that we have used the
restriction || > c. Dropping all but this term, we obtain the inequality
< 0 /( + i)

1
+ O(log(|| + 2)).

Proceeding in the same way, but dropping all zeros in the corresponding sum
for 0 /( + 2i), we obtain
< 0 /( + 2i) O(log(|| + 2)).
Notice that we use here again the restriction || > c.
Combining our estimates, we find
3
4

+ C(log(|| + 2)) 0
1
that is,

4
Choose = 1 +

1
10C log(||+2) ,

3
1

1
.
+ C log(|| + 2)

and write = ( 1) + (1 ) to deduce

1
1
1
1
1
( )
0
.
4
31 40 C log(|| + 2)
C log(|| + 2)
Now consider the case where mod q is primitive and complex. The character 2 may not be primitive, but it is certainly not principal, since it takes
values other than 1. Let 2 be induced by 0 mod q 0 with q 0 |q. Now we let
= + i be a zero of L(s, ), and for > 1 we consider


0
L0
L0
0
< 3 () 4 ( + i, ) ( + 2i, )

L
L
X log p 
X log p 

n in
2 n 2in
3
+
4<((p
)p
)
+
<(
(p
)p
))
+
3 + <(0 (pn )pin )
=
n
n
p
p
p,n
p|q,n

p-q

Both sums are positive, the first by the trigonometric identity, and the second
because the leading 3 dominates the contribution of the character, which is of
size 1.
Again we use
0
1
() =
+ O(1),

1
and now

1
L0
( + i, )
+ O(log q + log(|| + 2))
L

and

L0
( + 2i, 0 ) O(log q 0 ) O(log q).
L
Notice that we no longer require the restriction that || > c since L(s, 0 ) does
not have a pole at 1 (this was a factor only in the estimate for 0 /( + 2it)
above).
From here, the proof proceeds exactly as in the case of , except that error
terms of size O(log(|| + 2)) are replaced with ones of size O(log(|| + 2) + log q).
When is a real character, the above argument can be made, but notice
now that L(s, 0 ) = (s). The positivity of the combination of logarithmic
derivatives is the same, as are the first two estimates. Under the assumption
that || > log q we now have

0
( + 2i) O(log(|| + 2) + 1 log q).

The same argument applies, but now the constant depends on .


The above argument gives no information about zeros of quadratic Dirichlet
L-functions that are very near the real line. The following theorem asserts that
at most one such zero can exist for a given L-function.
Theorem 8.3. There exists a > 0 such that for all q and all primitive
quadratic mod q, L(s, ) has at most one zero in the region <(s) > 1 log q
and |=(s)| < log q . In particular, such a zero is necessarily real.
Proof. The zero would have to be real, because its complex conjugate is also a
zero.
Suppose that L(s, ) has two such zeros, 1 and 2 . Let > 1 and consider
(notice that the logarithmic derivative is already real)
L0
(, ) = O(log q)
L
C log q

X
(,)=0

( )2 + 2

1
2

.
( 1 )2 + 12
( 2 )2 + 22

On the other hand,


X (pn ) log p
X log p
L0
0
1
(, ) =

= () =
+ O(1).
n
n
L
p
p

1
p,n
p,n

It follows that
1
1
2
+ C log q

0.
1
( 1 )2 + 12
( 2 )2 + 22
Choose = 1 + 10 log q . Then i 1 10i , and thus
i
100 1

.
(i i )2 + i2
101 i
On the other hand, i

11
10 (

1), so that

i
10 100 1
1 1

.
(i i )2 + i2
11 101 1
21
We deduce that
C log q

1 1
1
= C log q
log q > 0,
21
20

which gives a contradiction if 1 > 20C.

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