This follows immediately from the identity

That \(\log F(\sigma )\ge 0\) for \(\sigma {\gt}1\) follows from Theorem 5.2.1.

Use Theorem 5.2.2 and Hadamard factorization.

This is standard in the literature.

This is standard in the literature.

This is just Cauchy’s integral formula for derivatives.

By Lemma 5.3.2 we know that

Thus,

Now applying Theorem 5.3.1, and noting that \(r'-r\leq |r'e^{it}-z|\), we have that

Substituting this into Equation (1) and evaluating the integral completes the proof.

Using Lemma 5.3.3 with \(r'=(R+r)/2\), and noting that \(r {\lt} R\), we have that

We let \(J_B(z)=I_{B'/B}(z)\). Then clearly, \(J_B(0)=0\). Now note that

Thus by the fundamental theorem of calculus we have that \(J_B'(z)=B'(z)/B(z)\). Now let \(H(z)=\exp (J_B(z))/B(z)\) and note that

Thus, \(H\) is constant since we know that \(B(z)\, J_B'(z)-B'(z)=0\) from \(J_B'(z)=B'(z)/B(z)\). So since \(H(0)=\exp (J_B(0))/B(0)=1/B(0)\) we know \(H(z)=1/B(0)\) for all \(z\). So we have,

Taking the logarithm of both sides completes the proof.

See above.

Since \(f\) is analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) we know that there exists a series expansion about \(\rho \):

Now if we let \(m\) be the smallest number such that \(a_m\neq 0\), then

Trivially, \(h_\rho (z)\) is analytic at \(\rho \) (we have written down the series expansion); now note that

Look at the definition of \(C_f\) and apply ZeroFactorization.

Expand out \(B_f\) as a product, and observe that each part is analytic on \(\overline{\mathbb {D}_R}\).

Since \(f(0)\neq 0\), we know that \(0\not\in \mathcal{K}_f(r)\). Thus,

Thus, substituting this into Definition 5.3.6,

Applying lemma 5.3.7 we know that

Note that for all \(\rho \in \mathcal{K}_f(r)\) that \(1{\lt}R/|\rho |\) since \(r{\lt}R\). Thus, the result follows.

For \(|z|=R\), we know that \(z\not\in \mathcal{K}_f(r)\). Thus,

Thus, substituting this into Definition 5.3.6,

But note that

So we have that \(|B_f(z)|=|f(z)|\leq B\) when \(|z|=R\). Now by the maximum modulus principle, we know that the maximum of \(|B_f|\) must occur on the boundary where \(|z|=R\). Thus \(|B_f(z)|\leq B\) for all \(|z|\leq R\).

Suppose that \(z\in \mathcal{K}_f(r)\). Then we have that

where \(h_z(z)\neq 0\) according to Lemma 5.3.4. Thus, substituting this into Definition 5.3.6,

Trivially, \(|h_z(z)|\neq 0\). Now note that

However, this is a contradiction because \(z\in \overline{\mathbb {D}_r}\) tells us that \(|z|\leq r {\lt} R\). Similarly, note that

However, this is also a contradiction because \(\rho \in \mathcal{K}_f(r)\) tells us that \(R {\lt} R^2/|\overline{\rho }|=|z|\), but \(z\in \overline{\mathbb {D}_r}\) tells us that \(|z|\leq r {\lt} R\). So, we know that

Applying this to Equation (4) we have that \(|B_f(z)|\neq 0\). So, \(B_f(z)\neq 0\).

Now suppose that \(z\not\in \mathcal{K}_f(r)\). Then we have that

Thus, substituting this into Definition 5.3.6,

We know that \(|f(z)|\neq 0\) since \(z\not\in \mathcal{K}_f(r)\). Now note that

However, this is a contradiction because \(\rho \in \mathcal{K}_f(r)\) tells us that \(R {\lt} R^2/|\overline{\rho }|=|z|\), but \(z\in \overline{\mathbb {D}_r}\) tells us that \(|z|\leq r {\lt} R\). So, we know that

Applying this to Equation (5) we have that \(|B_f(z)|\neq 0\). So, \(B_f(z)\neq 0\).

We have shown that \(B_f(z)\neq 0\) for both \(z\in \mathcal{K}_f(r)\) and \(z\not\in \mathcal{K}_f(r)\), so the result follows.

Since \(f(0)=1\), by Lemma 5.3.7 we know that

Thus, substituting this into Definition 5.3.6,

whereby Lemma 5.3.9 we know that \(|B_f(z)|\leq B\) for all \(|z|\leq R\). Taking the logarithm of both sides and rearranging gives the desired result.

By Lemma 5.3.9 we immediately know that \(|B_f(z)|\leq B\) for all \(|z|\leq R\). Now since \(L_f=J_{B_f}\) by Definition 5.3.7, by Theorem 5.3.3 we know that

for all \(|z|\leq r\). Note that in the above

because of Lemma 5.3.8. So by Theorem 5.3.2, it follows that

for all \(|z|\leq r'\).

Since \(z\in \overline{\mathbb {D}_{r'}}\setminus \mathcal{K}_f(R')\) we know that \(z\not\in \mathcal{K}_f(R')\); thus, by Definition 5.3.5 we know that

Substituting this into Definition 5.3.6 we have that

Taking the complex logarithm of both sides we have that

Taking the derivative of both sides we have that

By Definition 5.3.7 and Theorem 5.3.3, since \(L_f(z)=J_{B_f}(z)\) we have \(L_f'(z)=J'_{B_f}(z)=(B_f'/B_f)(z)\). Thus,

Now since \(z\in \overline{\mathbb {D}_{r'}}\subseteq \overline{\mathbb {D}_{R'}}\) and \(\rho \in \mathcal{K}_f(r)\subseteq \mathcal{K}_f(R')\), we know that \(R^2/R'-R'\leq |z-R^2/\overline{\rho }|\). Thus by the triangle inequality we have

Now by Theorem 5.3.5 and 5.3.6 we get our desired result with a little algebraic manipulation.

From the Euler product expansion of \(\zeta \), we have that for \(\Re s{\gt}1\)

Thus, we have that

Now note that \(|1-p^{-(3/2+it)}|\leq 1+|p^{-(3/2+it)}|=1+p^{-3/2}\). Thus,

for all \(t\in \mathbb {R}\) as desired.

Note that

With minor simplifications we have

The first integral evaluates to \(1/s\) (when \(0{\lt}\mathfrak {R}s\)), so this term when multiplied by the \(s/2\) cancels with the \(-1/2\). This exactly gives \(\zeta _1\).

This immediately follows from Lemmas 5.3.9 and 4.4.9.

For the sake of clearer proof writing let \(z=s+3/2+it\). Since \(|s|\leq 1\) we know that \(1/2\leq \mathfrak {R}z\); additionally, as \(|t|\geq 2\), we know \(1\leq |\mathfrak {I}z|\). So, \(z\in S\). Thus, from Lemma 5.3.10 we know that

by applying the triangle inequality. Now note that \(|z-1|\geq 1\). Likewise,

Thus we have that,

Expand out \(\zeta \) as a summation (we can do this as we are in the region of convergence) and ignore the norm as every term is positive and real. Comparing term by term proves the result.

Let \(g(z)=\zeta (z+3/2+it)/\zeta (3/2+it)\). Note that \(g(0)=1\) and for \(|z|\leq R\)

by Theorems 5.3.8 and 5.3.11. Thus by Theorem 5.3.7 we have that

Now note that \(f'/f=g'/g\), \(\mathcal{K}_f(r)=\mathcal{K}_g(r)\), and \(m_g(\rho )=m_f(\rho )\) for all \(\rho \in \mathcal{K}_f(r)\). Thus we have that,

where the implied constant \(C\) is taken to be

If we suppose the opposite, i.e. that there are an infinite number of zeros in this region, then \(f\equiv 0\) by the identity theorem. This is a contradiction, so the statement must be true.

We apply Theorem 5.3.13 where \(r'=2/3\), \(r=3/4\), \(R'=4/5\), and \(R=5/6\). Thus, for all \(z\in \overline{\mathbb {D}_{2/3}}\setminus \mathcal{K}_f(4/5)\) we have that

where \(f(z)=\zeta (z+3/2+it)\) for \(t\in \mathbb {R}\) with \(|t|\geq 2\). Now if we let \(z=-1/2+\delta \), then \(z\in (-1/2,1/2)\subseteq \overline{\mathbb {D}_{2/3}}\). Additionally, \(f(z)=\zeta (1+\delta +it)\), where \(1+\delta +it\) lies in the zero-free region where \(\sigma {\gt}1\). Thus, \(z\not\in \mathcal{K}_f(4/5)\). So,

But now note that if \(\rho \in \mathcal{K}_f(3/4)\), then \(\zeta (\rho +3/2+it)=0\) and \(|\rho |\leq 3/4\). Thus, \(\rho +3/2+it\in \mathcal{Z}_t\) (the argument works in reverse as well). Additionally, note that \(m_f(\rho )=m_\zeta (\rho +3/2+it)\). So changing variables using these facts gives us that

Note that, for \(\rho \in \mathcal{Z}_{2t}\)

Now since \(\rho \in \mathcal{Z}_{2t}\), we have that \(|\rho -(3/2+2it)|\leq 3/4\). So, we have \(\Re \rho \in [3/4,9/4]\) and \(\mathfrak {I}\rho \in [2t-3/4,2t+3/4]\). Additionally, we know that \(\zeta (\rho )=0\). This implies the stronger condition that \(\Re \rho \in [3/4,1]\). Thus,

Which implies that

Note that, from Lemma 5.3.15, we have

Since \(m_\zeta (\rho )\geq 0\) for all \(\rho \in \mathcal{Z}_{2t}\), the inequality from Equation (6) tells us that by subtracting the sum from both sides we have

Noting that \(\log |2t|=\log (2)+\log |t|\leq 2\log |t|\) completes the proof.

Note that for \(\rho '\in \mathcal{Z}_t\)

Now since \(\rho '\in \mathcal{Z}_t\), we have that \(|\rho '-(3/2+it)|\leq 3/4\). So, we have \(\mathfrak {R}\rho '\in [3/4,9/4]\) and \(\mathfrak {I}\rho '\in [t-3/4,t+3/4]\). Additionally, we know that \(\zeta (\rho ')=0\). This implies the stronger condition that \(\mathfrak {R}\rho '\in [3/4,1]\). Thus,

Which implies that

Note that, from Lemma 5.3.15, we have

Now we proceed via cases. If \(\rho \in \mathcal{Z}_t\), then \(\mathcal{Z}_t=\rho \sqcup (\mathcal{Z}_t\setminus \rho )\) and the inequality from Equation (7) tells us that by subtracting the sum over all \(\rho '\in \mathcal{Z}_t\setminus \{ \rho \} \) from both sides we have

But of course we have that \(\mathfrak {R}(1+\delta +it-\rho )=1+\delta -\sigma \). So subtracting this term from both sides and recalling the implied constant we have

We have that \(\sigma \leq 1\) since \(\zeta \) is zero free on the right half plane \(\sigma {\gt}1\). Thus, \(0{\lt}1+\delta -\sigma \). Noting this in combination with the fact that \(1\leq m_\zeta (\rho )\) gives us that

Now if \(\rho \not\in \mathcal{Z}_t\), then \(|\sigma -3/2|=|\rho -(3/2+it)|{\gt}3/4\) since \(\zeta (\rho )=0\). Additionally, \(\sigma \leq 1\) since \(\zeta (\rho )=0\), thus \(3/2-\sigma =|\sigma -3/2|{\gt}3/4\). So it follows that

Thus, by subtracting the sum over \(\rho '\in \mathcal{Z}_t\) from both sides we have

From Theorem 4.4.10 we know that

Changing variables \(s\mapsto 1+\delta \) and applying the triangle inequality we have that

We know that \(\Lambda (n)\) is \(O(n^\varepsilon )\) since it is bounded above by \(\log (n)\). Thus,

Since \(\varepsilon \) can be arbitrarily small, we know the left hand side converges by the \(p\)-test.

If we suppose the opposite, i.e. that there are an infinite number of zeta zeros in this region, then \(\zeta \equiv 0\) by the identity theorem. This is a contradiction, so the statement must be true.

From Theorem 4.5.1 when \(\Re s{\gt}1\) we have

Thus,

Now applying Euler’s identity

Since

we know that the series on the right hand side is bounded below by \(0\), and by Lemmas 5.3.16, 5.3.17, and 5.3.18 we have an upper bound on the left hand side. So,

where \(A\), \(B\), and \(C\) are the implied constants coming from Lemmas 5.3.18, 5.3.17, and 5.3.16 respectively. By choosing \(D\geq 5A+4B+C\geq 3A/\log 2+4B+C\) we have

by some manipulation. Now if we choose \(\delta =(2D\log |t|)^{-1}\) then we have

So with some manipulation we have that

This is exactly the desired result with the constant \(E=(14D)^{-1}\). Since \(D\geq 5A+4B+C\geq 1\) we have that \(E\in (0,1/14)\).

Note that \(\delta _t=E/\log |t|\) where \(E\) is the implied constant from Lemma 5.3.21. But we know that \(E=(14D)^{-1}\) where \(D\geq 5A+4B+C\) where \(A\), \(B\), and \(C\) are the constants coming from Lemmas 5.3.18, 5.3.17, and 5.3.16 respectively. Thus,

But note that \(A\geq 1\), \(B{\gt}0\), and \(C{\gt}0\) by Lemmas 5.3.18, 5.3.17, and 5.3.16 respectively. So, by a very lazy estimate we have \(E\leq 1/14\). Thus,

By Lemma 5.3.22 we have that

We apply Theorem 5.3.13 where \(r'=2/3\), \(r=3/4\), \(R'=4/5\), and \(R=5/6\). Thus for all \(z\in \overline{\mathbb {D}_{2/3}}\setminus \mathcal{K}_f(4/5)\) we have that

where \(f(z)=\zeta (z+3/2+it)\) for \(t\in \mathbb {R}\) with \(|t|\geq 2\). Now if we let \(z=\sigma -3/2\), then \(z\in (-29/54,0)\subseteq \overline{\mathbb {D}_{2/3}}\). Additionally, \(f(z)=\zeta (\sigma +it)\), where \(\sigma +it\) lies in the zero free region given by Lemma 5.3.21 since \(\sigma \geq 1-\delta _t/3\geq 1-\delta _t\). Thus, \(z\not\in \mathcal{K}_f(4/5)\). So,

But now note that if \(\rho \in \mathcal{K}_f(3/4)\), then \(\zeta (\rho +3/2+it)=0\) and \(|\rho |\leq 3/4\) (and the argument works in reverse). Additionally, note that \(m_f(\rho )=m_\zeta (\rho +3/2+it)\). So changing variables using these facts gives us that

Let \(\rho =\sigma '+it'\) and note that since \(\rho \in \mathcal{Z}_t\), we have \(t'\in (t-3/4,t+3/4)\). Thus, if \(t{\gt}1\) we have

And otherwise if \(t{\lt}-1\) we have

So by taking reciprocals and multiplying through by a constant we have that \(\delta _t\leq 2\delta _{t'}\). Now note that since \(\rho \in \mathcal{Z}_t\) we know that \(\sigma '\leq 1-\delta _{t'}\) by Theorem 5.3.21 (here we use the fact that \(|t|\geq 3\) to give us that \(|t'|\geq 2\)). Thus,

Take \(F=E/3\) where \(E\) comes from Theorem 5.3.21. Then we have that \(\sigma \geq 1-\delta _t/3\). So, we apply Lemma 5.3.23, which gives us that

Using the reverse triangle inequality and rearranging, we have that

where \(C\) is the implied constant in Lemma 5.3.23. Now applying Lemma 5.3.24 we have that

Now let \(f(z)=\zeta (z+3/2+it)/\zeta (3/2+it)\) with \(\rho =\rho '+3/2+it\). Then if \(\rho \in \mathcal{Z}_t\) we have that

with the same multiplicity of zero, that is \(m_\zeta (\rho )=m_f(\rho ')\). And also if \(\rho \in \mathcal{Z}_t\) then

Thus we change variables to have that

Now note that \(f(0)=1\) and for \(|z|\leq 5/6\) we have

by Theorems 5.3.8 and 5.3.11. Thus by Theorem 5.3.5 we have that

where \(D\) is taken to be sufficiently large. Recall, by definition that, \(\delta _t=E/\log |t|\) with \(E\) coming from Theorem 5.3.21. By using this fact and the above, we have that

where the implied constant is taken to be bigger than \(6D/E+C\). We know that the RHS is bounded above by \(\ll \log ^2|t|\); so the result follows.

Note that

by 5.3.19.

Put the previous two Lemmas together.

This follows as a corollary of Theorems 4.6.1 and 5.3.27.