Hunting for Zero-Free Regions of the Zeta Function (Part II)
The Riemann Hypothesis asserts that all nontrivial zeros of
the Riemann zeta function lie on the critical line 
. However, before we can approach the line itself, we must
first establish that no zeros exist outside of it—specifically near the
boundary 
.
Historically, Charles de la Vallée Poussin mapped the first
classic "zero-free" zone in 1899 (see Part I).
Today, we will take a different approach.
Recall that previously
we proved the following:
Lemma: Let 
be analytic in the disk 
, and suppose that 
for 
throughout the disk. If 
ranges over the zeros of 
satisfying 
, then
for 
and an absolute constant 
.
Example
1: A polynomial with
finitely many zeros
Take
and let 
, 
, with 
.
Then
Hence for all zeros in 
,
Thus,
which satisfies the conclusion.
This is the extreme case where every contribution to 
comes from listed zeros.
Example 2: An exponential
Take
There are no zeros.
Hence
The maximum-growth condition on 
is
so we may take 
. The lemma predicts
but so 
. The estimate is sharp up to the
absolute constant. This example shows where the 
scale comes from.
Example 3: One nearby zero
Let
Then
We have
Therefore, one may take
and apply the lemma to get
However, if 
is very small the estimate is very far from
equality. The bound has lots of room to spare.
Example 4: Many zeros but harmless analytic
factor
Take
Then
and
Moreover,
Let
If 
then
Choose
and apply the lemma to get
If 
is tiny then the bound is far from equality
Example 5: If you want to achieve a natural fit
to the lemma, it is better to place the zeros away from the center.
Consider
with 
is fixed.
Left as an Exercise! ■
Now the Lemma has a couple of corollaries.
Suppose that is analytic in the disk , satisfying the growth condition for throughout the disk.
Let 
denote the zeros of satisfying , with 
. That is, has no zeros in the right-hand half of the
disk .
Then of course
for so for 
implying
For each zero in the sum, let 
. By hypothesis, .
Compute the real part
of 
:
Since every term 
in the sum 
is non-negative, removing
the sum can only increase the right-hand side. Hence
Corollary 1: Suppose that is analytic in the disk , satisfying the growth condition for throughout the disk.
Let denote the zeros of , all satisfying and assume that for every such zero.
Then
■
Now suppose that one zero 
lies on the segment joining 
and 
. Then 
, 
.
Then
Since contributes one of the
terms in this sum and writing 
,
and
Thus,
■
Let 
and 
the Riemann Zeta function.
Take
Because 
for 
such that 
and 
(
, in the disk we estimate
Using 
and (see here) 
Thus,
Apply Corollary 1:
if a root lies on the segment joining 
and 
then 
, 
and
and if is very close to 
, 
is very small, hence a zero extremely close to
forces
to become very negative. However, as shown with the 3-4-1 inequality,
we have the De laVallée Poussin Lower Bound
Hence
(
and 
, 
)
So
Therefore, the argument gives that there are no zeros of 
such that 
, 
and
Thus, we have a
zero-free region of the shape
Even more striking is
the fact that mathematicians have proved a zero-free region far stronger than
one might initially expect: for some constant 
, the Riemann zeta function has no zeros with 
large satisfying
This is the celebrated Korobov-Vinogradov zero-free region.
Our argument yields a zero-free region of width
It is therefore natural
to compare it with the best-known result, which is substantially stronger.
■
We began with the simple3-4-1 inequality, which led us to regions where zeta cannot vanish. Along
the way, a lemma deduced from Borel-Carathéodory Theorem revealed how growth and the location of
zeros shape an analytic function’s behavior.
The lesson is worth remembering:
sometimes, understanding where zeros cannot be is as revealing as finding where
they are.
Let’s wrap up today by putting that Lemma to work
one more time.
Let satisfy the conditions of the lemma, that is, is analytic in the disk , satisfying the growth condition for throughout the disk.
Take its zeros in 
and suppose that 
for 
, where 
. Then of course
So
We have for , hence in that region each term of
the sum
is positive.
Now let 
satisfy . Since has no zeros in this
half-plane, every zero 
occurring in the sum must
satisfy
Therefore
Consequently,
Thus, every term in the
sum is positive, and hence
Returning to the previous inequality, we obtain
Now recall
Borel-Carathéodory Theorem: Let a
function be analytic on a closed disk of radius 
centered at the origin. Suppose that 
. Then, we have the
following inequality
where the norm on the left-hand side
denotes the maximum value of in the closed disc:
(the last equality is due to the Maximum
Modulus Principle).
Apply the theorem to 
on the disk 
(i.e., 
) with 
.
Therefore, for 
Also assume that
Then
Corollary
2: Let be analytic in the disk , satisfying the growth condition for throughout the disk.
Take its zeros in and suppose that for , where .
Then
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