5 Third Approach

5.1 Hadamard factorization

In this file, we prove the Hadamard Factorization theorem for functions of finite order, and prove that the zeta function is such.

5.2 Hoffstein-Lockhart

In this file, we use the Hoffstein-Lockhart construction to prove a zero-free region for zeta.

Hoffstein-Lockhart + Goldfeld-Hoffstein-Liemann

Instead of the “slick” identity \(3+4\cos \theta +\cos 2\theta =2(\cos \theta +1)^2\ge 0\), we use the following more robust identity.

Theorem 5.2.1

For any \(p{\gt}0\) and \(t\in \mathbb {R}\),

\[ 3+p^{2it}+p^{-2it}+2p^{it}+2p^{-it} \ge 0. \]

Proof ▶

This follows immediately from the identity

\[ |1+p^{it}+p^{-it}|^2=1+p^{2it}+p^{-2it}+2p^{it}+2p^{-it}+2. \]

[Note: identities of this type will work in much greater generality, especially for higher degree \(L\)-functions.]

This means that, for fixed \(t\), we define the following alternate function.

Definition 5.2.1

For \(\sigma {\gt}1\) and \(t\in \mathbb {R}\), define

\[ F(\sigma ) := \zeta ^3(\sigma )\zeta ^2(\sigma +it)\zeta ^2(\sigma -it)\zeta (\sigma +2it)\zeta (\sigma -2it). \]

Theorem 5.2.2

Then \(F\) is real-valued, and whence \(F(\sigma )\ge 1\) there.

Proof ▶

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

[Note: I often prefer to avoid taking logs of functions that, even if real-valued, have to be justified as being such. Instead, I like to start with “logF” as a convergent Dirichlet series, show that it is real-valued and non-negative, and then exponentiate...]

From this and Hadamard factorization, we deduce the following.

Theorem 5.2.3

There is a constant \(c{\gt}0\), so that \(\zeta (s)\) does not vanish in the region \(\sigma {\gt}1-\frac{c}{\log t}\), and moreover,

\[ -\frac{\zeta '}{\zeta }(\sigma +it) \ll (\log t)^2 \]

there.

Proof ▶

Use Theorem 5.2.2 and Hadamard factorization.

This allows us to quantify precisely the relationship between \(T\) and \(\delta \) in Theorem 4.4.25....

5.3 Strong PNT

Lemma 5.3.1 AnalyticOn.norm-le-of-norm-le-on-sphere

✓

An application of the Maximum modulus principle.

Proof ▶

This is standard in the literature.

Theorem 5.3.1 borelCaratheodory’

✓

An application of

      Complex.borelCaratheodory_zero.

Proof ▶

This is standard in the literature.

This upstreamed from https://github.com/math-inc/strongpnt/tree/main

Lemma 5.3.2 cauchy-formula-deriv

✓

Let \(f\) be analytic on \(|z|\leq R\). For any \(z\) with \(|z|\leq r\) and any \(r'\) with \(0 {\lt} r {\lt} r' {\lt} R\) we have

\[ f'(z)=\frac{1}{2\pi i}\oint _{|w|=r'}\frac{f(w)}{(w-z)^2}\, dw=\frac{1}{2\pi } \int _0^{2\pi }\frac{r'e^{it}\, f(r'e^{it})}{(r'e^{it}-z)^2}\, dt. \]

Proof ▶

This is just Cauchy’s integral formula for derivatives.

Lemma 5.3.3 DerivativeBound

✓

Let \(R,\, M{\gt}0\) and \(0 {\lt} r {\lt} r' {\lt} R\). Let \(f\) be analytic on \(|z|\leq R\) such that \(f(0)=0\) and suppose \(\Re f(z)\leq M\) for all \(|z|\leq R\). Then we have that

\[ |f'(z)|\leq \frac{2M(r')^2}{(R-r')(r'-r)^2} \]

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

Proof ▶

By Lemma 5.3.2 we know that

\[ f'(z)=\frac{1}{2\pi i}\oint _{|w|=r'}\frac{f(w)}{(w-z)^2}\, dw =\frac{1}{2\pi }\int _0^{2\pi }\frac{r'e^{it}\, f(r'e^{it})}{(r'e^{it}-z)^2}\, dt. \]

Thus,

\begin{equation} \label{pickupPoint1} |f'(z)|=\left|\frac{1}{2\pi }\int _0^{2\pi } \frac{r'e^{it}\, f(r'e^{it})}{(r'e^{it}-z)^2}\, dt\right| \leq \frac{1}{2\pi }\int _0^{2\pi } \left|\frac{r'e^{it}\, f(r'e^{it})}{(r'e^{it}-z)^2}\right|\, dt. \end{equation}

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

\[ \left|\frac{r'e^{it}\, f(r'e^{it})}{(r'e^{it}-z)^2}\right| \leq \frac{2M(r')^2}{(R-r')(r'-r)^2}. \]

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

Theorem 5.3.2 BorelCaratheodoryDeriv

✓

Let \(R,\, M{\gt}0\). Let \(f\) be analytic on \(|z|\leq R\) such that \(f(0)=0\) and suppose \(\Re f(z)\leq M\) for all \(|z|\leq R\). Then for any \(0 {\lt} r {\lt} R\),

\[ |f'(z)|\leq \frac{16MR^2}{(R-r)^3} \]

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

Proof ▶

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

\[ |f'(z)|\leq \frac{4M(R+r)^2}{(R-r)^3}\leq \frac{16MR^2}{(R-r)^3}. \]

Definition 5.3.1 TaxicabIntegral

Let \(0 {\lt} R\). Let \(f:\overline{\mathbb {D}_R}\to \mathbb {C}\) be analytic on neighborhoods of points in \(\overline{\mathbb {D}_R}\). Define the functon \(I_f:\mathbb {D}_R\to \mathbb {C}\) by

\[ I_f(z)=z\int _0^1f(tz)\, dt. \]

Theorem 5.3.3 LogOfAnalyticFunction

✓

Let \(0{\lt}r{\lt}R\). Let \(B:\overline{\mathbb {D}_{R}}\to \mathbb {C}\) be analytic on neighborhoods of points in \(\overline{\mathbb {D}_{R}}\) with \(B(z)\neq 0\) for all \(z\in \overline{\mathbb {D}_{R}}\).Then there exists \(J_B:\mathbb {D}_R\to \mathbb {C}\) that is analytic on neighborhoods of points in \(\mathbb {D}_R\) such that

\(J_B(0)=0\)
\(J_B'(z)=B'(z)/B(z)\) for all \(z\in \overline{\mathbb {D}_r}\)
\(\log |B(z)|-\log |B(0)|=\mathfrak {R}J_B(z)\) for all \(z\in \mathbb {D}_R\).

Proof ▶

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

\begin{align*} I_{B'/B}(z)=z\int _0^1(B’/B)(tz)\, dt=\int _0^z(B’/B)(u)\, du. \end{align*}

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

\[ H'(z)=(B(z)\, J_B'(z)-B'(z))\left(\frac{\exp (J_B(z))}{(B(z))^2}\right). \]

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,

\[ \frac{1}{B(0)}=\frac{\exp (J_B(z))}{B(z)}\implies \left|\frac{B(z)}{B(0)}\right| =\exp (\mathfrak {R}J_B(z)). \]

Taking the logarithm of both sides completes the proof.

Theorem 5.3.4 LogOfAnalyticFunction’

✓

A wrapper of the above theorem that will be useful later on.

Proof ▶

See above.

Definition 5.3.2 SetOfZeros

✓

Let \(R{\gt}0\) and \(f:\mathbb {C}\to \mathbb {C}\). Define the set of zeros \(\mathcal{K}_f(R)=\{ \rho \in \mathbb {C}:|\rho |\leq R,\, f(\rho )=0\} \).

Definition 5.3.3 ZeroOrder

Let \(f:\mathbb {C}\to \mathbb {C}\). We define \(m_f(\rho )\) as the order of the zero \(\rho \) w.r.t \(f\).

In LEAN, this corresponds exactly with analyticOrderAt/analyticOrderNatAt.

Definition 5.3.4 ZeroFactor

✓

Let \(f:\mathbb {C}\to \mathbb {C}\) and \(\rho \in \mathbb {C}\). Then there exists \(h_\rho \) such that

\[ f(z)=(z-\rho )^{m_f(\rho )}\, h_\rho (z). \]

In LEAN, this corresponds exactly with (-.analyticOrderAt-ne-top.mp -).choose, but this serves as a wrapper of that with the necessary conditions.

Lemma 5.3.4 ZeroFactorization

✓

Let \(f:\mathbb {C}\to \mathbb {C}\) be analytic on \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). For all \(\rho \in \mathcal{K}_f(R)\) with \(R{\lt}1\) there exists \(h_\rho (z)\) such that \(h_\rho (z)\) is analytic at \(\rho \), \(h_\rho (\rho )\neq 0\), and \(f(z)=(z-\rho )^{m_f(\rho )}\, h_\rho (z)\).

Proof ▶

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

\[ f(z)=\sum _{0\leq n}a_n\, (z-\rho )^n. \]

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

\[ f(z)=\sum _{0\leq n}a_n\, (z-\rho )^n=\sum _{m\leq n}a_n\, (z-\rho )^n =(z-\rho )^m\sum _{m\leq n}a_n\, (z-\rho )^{n-m}=(z-\rho )^m\, h_\rho (z). \]

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

\[ h_\rho (\rho )=\sum _{m\leq n}a_n(\rho -\rho )^{n-m}=\sum _{m\leq n}a_n0^{n-m}=a_m\neq 0. \]

Definition 5.3.5 CFunction

✓

Let \(0 {\lt} r {\lt} 1\), and \(f:\mathbb {C}\to \mathbb {C}\) be analytic on \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). We define a function \(C_f:\mathbb {C}\to \mathbb {C}\) as follows. This function is constructed by dividing \(f(z)\) by a polynomial whose roots are the zeros of \(f\) inside \(\overline{\mathbb {D}_r}\).

\[ C_f(z)=\begin{cases} \displaystyle \frac{f(z)}{\prod _{\rho \in \mathcal{K}_f(r)}(z-\rho )^{m_f(\rho )}} \qquad \text{for }z\not\in \mathcal{K}_f(r) \\ \displaystyle \frac{h_z(z)}{\prod _{\rho \in \mathcal{K}_f(r)\setminus \{ z\} } (z-\rho )^{m_f(\rho )}}\qquad \text{for }z\in \mathcal{K}_f(r) \end{cases} \]

where \(h_z(z)\) comes from Lemma 5.3.4.

Lemma 5.3.5 CfAnalytic

✓

If \(f:\mathbb {C}\to \mathbb {C}\) is analytic on \(\overline{\mathbb {D}_1}\) then so too is \(C_f\).

Proof ▶

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

Definition 5.3.6 BlaschkeB

✓

Let \(0 {\lt} r {\lt} R {\lt} 1\), and \(f:\mathbb {C}\to \mathbb {C}\) be analytic \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). We define a function \(B_f:\mathbb {C}\to \mathbb {C}\) as follows.

\[ B_f(z)=C_f(z)\prod _{\rho \in \mathcal{K}_f(r)} \left(R-\frac{z\overline{\rho }}{R}\right)^{m_f(\rho )} \]

Lemma 5.3.6 BlaschkeAnalytic

✓

If \(f:\mathbb {C}\to \mathbb {C}\) is analytic on \(\overline{\mathbb {D}_R}\) then so too is \(B_f\).

Proof ▶

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

Lemma 5.3.7 BlaschkeOfZero

✓

Let \(0 {\lt} r {\lt} R{\lt}1\), and \(f:\mathbb {C}\to \mathbb {C}\) be analytic on \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). Then

\[ |B_f(0)|=|f(0)|\prod _{\rho \in \mathcal{K}_f(r)} \left(\frac{R}{|\rho |}\right)^{m_f(\rho )}. \]

Proof ▶

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

\[ C_f(0)=\frac{f(0)}{\displaystyle \prod _{\rho \in \mathcal{K}_f(r)}(-\rho )^{m_f(\rho )}}. \]

Thus, substituting this into Definition 5.3.6,

\[ |B_f(0)|=|C_f(0)|\prod _{\rho \in \mathcal{K}_f(r)}R^{m_f(\rho )} =|f(0)|\prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{|\rho |}\right)^{m_f(\rho )}. \]

Lemma 5.3.8 norm-fOfZero-le-norm-BlaschkeOfZero

✓

Let \(0 {\lt} r {\lt} R{\lt}1\), and \(f:\mathbb {C}\to \mathbb {C}\) be analytic on \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). Then

\[ |f(0)|\leq |B_f(0)|. \]

Proof ▶

Applying lemma 5.3.7 we know that

\[ |B_f(0)|=|f(0)|\prod _{\rho \in \mathcal{K}_f(r)} \left(\frac{R}{|\rho |}\right)^{m_f(\rho )}. \]

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

Lemma 5.3.9 DiskBound

✓

Let \(0 {\lt} r {\lt} R{\lt}1\). If \(f:\mathbb {C}\to \mathbb {C}\) is a function analytic on \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\) such that \(|f(z)|\leq B\) for \(|z|\leq R\), then \(|B_f(z)|\leq B\) for \(|z|\leq R\) also.

Proof ▶

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

\[ C_f(z)=\frac{f(z)}{\displaystyle \prod _{\rho \in \mathcal{K}_f(r)}(z-\rho )^{m_f(\rho )}}. \]

Thus, substituting this into Definition 5.3.6,

\[ |B_f(z)|=|f(z)|\prod _{\rho \in \mathcal{K}_f(r)} \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|^{m_f(\rho )}. \]

But note that

\[ \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right| =\frac{|R^2-z\overline{\rho }|/R}{|z-\rho |} =\frac{|z|\cdot |\overline{z-\rho }|/R}{|z-\rho |}=1. \]

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 \(|z|\leq R\).

Lemma 5.3.10 BlaschkeNonZero

✓

Let \(0 {\lt} r {\lt} R{\lt}1\) and \(f:\overline{\mathbb {D}_1}\to \mathbb {C}\) be analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). Then \(B_f(z)\neq 0\) for all \(z\in \overline{\mathbb {D}_r}\).

Proof ▶

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

\[ C_f(z)=\frac{h_z(z)}{\displaystyle \prod _{\rho \in \mathcal{K}_f(r)\setminus \{ z\} } (z-\rho )^{m_f(\rho )}}. \]

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

\begin{equation} \label{pickupPoint2} |B_f(z)|=|h_z(z)|\cdot \left|R-\frac{|z|^2}{R}\right|^{m_f(z)} \prod _{\rho \in \mathcal{K}_f(r)\setminus \{ z\} } \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|^{m_f(\rho )}. \end{equation}

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

\[ \left|R-\frac{|z|^2}{R}\right|=0\implies |z|=R. \]

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

\[ \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|=0\implies |z|=\frac{R^2}{|\overline{\rho }|}. \]

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

\[ \left|R-\frac{|z|^2}{R}\right|\neq 0\qquad \text{and}\qquad \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|\neq 0 \quad \text{for all}\quad \rho \in \mathcal{K}_f(r)\setminus \{ z\} . \]

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

\[ C_f(z)=\frac{f(z)}{\displaystyle \prod _{\rho \in \mathcal{K}_f(r)}(z-\rho )^{m_f(\rho )}}. \]

Thus, substituting this into Definition 5.3.6,

\begin{equation} \label{pickupPoint3} |B_f(z)|=|f(z)|\prod _{\rho \in \mathcal{K}_f(r)} \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|^{m_f(\rho )}. \end{equation}

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

\[ \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|=0\implies |z|=\frac{R^2}{|\overline{\rho }|}. \]

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

\[ \left|\frac{R-z\overline{\rho }/R}{z-\rho }\right|\neq 0 \quad \text{for all}\quad \rho \in \mathcal{K}_f(r). \]

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.

Theorem 5.3.5 ZerosBound

✓

Let \(0{\lt} r {\lt} R{\lt}1\). If \(f:\mathbb {C}\to \mathbb {C}\) is a function analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(f(0)=1\) and \(|f(z)|\leq B\) for \(|z|\leq R\), then

\[ \sum _{\rho \in \mathcal{K}_f(r)}m_f(\rho )\leq \frac{\log B}{\log (R/r)}. \]

Proof ▶

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

\[ |B_f(0)| =|f(0)|\prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{|\rho |}\right)^{m_f(\rho )} =\prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{|\rho |}\right)^{m_f(\rho )}. \]

Thus, substituting this into Definition 5.3.6,

\[ (R/r)^{\sum _{\rho \in \mathcal{K}_f(r)}m_f(\rho )} =\prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{r}\right)^{m_f(\rho )} \leq \prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{|\rho |}\right)^{m_f(\rho )} =|B_f(0)|\leq B \]

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.

Definition 5.3.7 JBlaschke

✓

Let \(0 {\lt} r {\lt} R{\lt}1\). If \(f:\mathbb {C}\to \mathbb {C}\) is a function analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(f(0)=1\), define \(L_f(z)=J_{B_f}(z)\) where \(J\) is from Theorem 5.3.3 and \(B_f\) is from Definition 5.3.6.

Theorem 5.3.6 JBlaschkeDerivBound

✓

Let \(B{\gt}1\) and \(0 {\lt} r' {\lt} r {\lt} R{\lt}1\). If \(f:\mathbb {C}\to \mathbb {C}\) is a function analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(f(0)=1\) and \(|f(z)|\leq B\) for all \(|z|\leq R\), then for all \(|z|\leq r'\)

\[ |L_f'(z)|\leq \frac{16\log (B)\, r^2}{(r-r')^3}. \]

Proof ▶

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

\[ L_f(0)=0\qquad \text{and}\qquad \Re L_f(z)=\log |B_f(z)|-\log |B_f(0)|\leq \log |B_f(z)|\leq \log B \]

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

\[ 0=\log |f(0)|\leq \log |B_f(0)| \]

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

\[ |L_f'(z)|\leq \frac{16\log (B)\, r^2}{(r-r')^3} \]

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

Theorem 5.3.7 FinalBound

✓

Let \(B{\gt}1\) and \(0 {\lt} r' {\lt} r {\lt} R' {\lt} R{\lt}1\). If \(f:\mathbb {C}\to \mathbb {C}\) is a function analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(f(0)=1\) and \(|f(z)|\leq B\) for all \(|z|\leq R\), then for all \(z\in \overline{\mathbb {D}_{r'}}\setminus \mathcal{K}_f(R')\) we have

\[ \left|\frac{f'}{f}(z)-\sum _{\rho \in \mathcal{K}_f(r)}\frac{m_f(\rho )}{z-\rho }\right| \leq \left(\frac{16r^2}{(r-r')^3}+\frac{1}{(R^2/R'-R')\, \log (R/R')}\right)\log B. \]

Proof ▶

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

\[ C_f(z)=\frac{f(z)}{\displaystyle \prod _{\rho \in \mathcal{K}_f(r)}(z-\rho )^{m_f(\rho )}}. \]

Substituting this into Definition 5.3.6 we have that

\[ B_f(z)=f(z)\prod _{\rho \in \mathcal{K}_f(r)} \left(\frac{R-z\overline{\rho }/R}{z-\rho }\right)^{m_f(\rho )}. \]

Taking the complex logarithm of both sides we have that

\[ \mathrm{Log}\, B_f(z)=\mathrm{Log}\, f(z) +\sum _{\rho \in \mathcal{K}_f(r)}m_f(\rho )\, \mathrm{Log}(R-z\overline{\rho }/R) -\sum _{\rho \in \mathcal{K}_f(r)}m_f(\rho )\, \mathrm{Log}(z-\rho ). \]

Taking the derivative of both sides we have that

\[ \frac{B_f'}{B_f}(z)=\frac{f'}{f}(z) +\sum _{\rho \in \mathcal{K}_f(r)}\frac{m_f(\rho )}{z-R^2/\overline{\rho }} -\sum _{\rho \in \mathcal{K}_f(r)}\frac{m_f(\rho )}{z-\rho }. \]

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,

\[ \frac{f'}{f}(z)-\sum _{\rho \in \mathcal{K}_f(r)}\frac{m_f(\rho )}{z-\rho } =L_f'(z)-\sum _{\rho \in \mathcal{K}_f(r)}\frac{m_f(\rho )}{z-R^2/\overline{\rho }}. \]

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

\[ \left|\frac{f'}{f}(z)-\sum _{\rho \in \mathcal{K}_f(r)}\frac{m_f(\rho )}{z-\rho }\right| \leq |L_f'(z)|+\left(\frac{1}{R^2/R'-R'}\right)\sum _{\rho \in \mathcal{K}_f(r)}m_f(\rho ). \]

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

Theorem 5.3.8 ZetaFixedLowerBound

For all \(t\in \mathbb {R}\) one has

\[ |\zeta (3/2+it)|\geq \frac{\zeta (3)}{\zeta (3/2)}. \]

Proof ▶

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

\[ \zeta (s)=\prod _p\frac{1}{1-p^{-s}}. \]

Thus, we have that

\[ \frac{\zeta (2s)}{\zeta (s)}=\prod _p\frac{1-p^{-s}}{1-p^{-2s}}=\prod _p\frac{1}{1+p^{-s}}. \]

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

\[ |\zeta (3/2+it)|=\prod _p\frac{1}{|1-p^{-(3/2+it)}|} \geq \prod _p\frac{1}{1+p^{-3/2}}=\frac{\zeta (3)}{\zeta (3/2)} \]

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

Lemma 5.3.11 ZetaAltFormula

Let

\[ \zeta _0(s)=1+\frac{1}{s-1}-s\int _1^\infty \{ x\} \, x^{-s}\, \frac{dx}{x}. \]

We have that \(\zeta (s)=\zeta _0(s)\) for \(\sigma {\gt}1\).

Proof ▶

Note that for \(\sigma {\gt}1\) we have

\[ \zeta (s)=\sum _{n=1}^\infty \frac{1}{n^s} =\sum _{n=1}^\infty \frac{n}{n^s}-\sum _{n=1}^\infty \frac{n-1}{n^s} =\sum _{n=1}^\infty \frac{n}{n^s}-\sum _{n=0}^\infty \frac{n}{(n+1)^s} =\sum _{n=1}^\infty \frac{n}{n^s}-\sum _{n=1}^\infty \frac{n}{(n+1)^s}. \]

Thus

\[ \zeta (s)=\sum _{n=1}^\infty n\, (n^{-s}-(n+1)^{-s}). \]

Now we note that

\[ s\int _n^{n+1}x^{-s}\, \frac{dx}{x} =s\left(-\frac{1}{s}\, x^{-s}\right)_n^{n+1}=n^{-s}-(n+1)^{-s}. \]

So, substituting this we have

\[ \zeta (s)=\sum _{n=1}^\infty n\, (n^{-s}-(n+1)^{-s}) =s\sum _{n=1}^\infty n\int _n^{n+1}x^{-s}\, \frac{dx}{x} =s\int _1^\infty \lfloor x\rfloor \, x^{-s}\, \frac{dx}{x}. \]

But noting that \(\lfloor x\rfloor =x-\{ x\} \) we have that

\[ \zeta (s)=s\int _1^\infty \lfloor x\rfloor \, x^{-s}\, \frac{dx}{x} =s\int _1^\infty x^{-s}\, dx-s\int _1^\infty \{ x\} \, x^{-s}\, \frac{dx}{x}. \]

Evaluating the first integral completes the result.

Lemma 5.3.12 ZetaAltFormulaAnalytic

We have that \(\zeta _0(s)\) is analytic for all \(s\in S\) where \(S=\{ s\in \mathbb {C}:\Re s{\gt}0,\, s\neq 1\} \).

Proof ▶

Note that we have

\[ \left|\int _1^\infty \{ x\} \, x^{-s}\, \frac{dx}{x}\right| \leq \int _1^\infty |\{ x\} \, x^{-s-1}|\, dx \leq \int _1^\infty x^{-\sigma -1}\, dx=\frac{1}{\sigma }. \]

So this integral converges uniformly on compact subsets of \(S\), which tells us that it is analytic on \(S\). So it immediately follows that \(\zeta _0(s)\) is analytic on \(S\) as well, since \(S\) avoids the pole at \(s=1\) coming from the \((s-1)^{-1}\) term.

Lemma 5.3.13 ZetaExtend

We have that

\[ \zeta (s)=1+\frac{1}{s-1}-s\int _1^\infty \{ x\} \, x^{-s}\, \frac{dx}{x} \]

for all \(s\in S\).

Proof ▶

This is an immediate consequence of the identity theorem.

Theorem 5.3.9 GlobalBound

For all \(s\in \mathbb {C}\) with \(|s|\leq 1\) and \(t\in \mathbb {R}\) with \(|t|\geq 2\), we have that

\[ |\zeta (s+3/2+it)|\leq 7+2\, |t|. \]

Proof ▶

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 5.3.13 we know that

\[ |\zeta (z)|\leq 1+\frac{1}{|z-1|} +|z|\cdot \left|\int _1^\infty \{ x\} \, x^{-z}\, \frac{dx}{x}\right| \]

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

\[ |z|\cdot \left|\int _1^\infty \{ x\} \, x^{-z}\, \frac{dx}{x}\right| \leq |z|\int _1^\infty |\{ x\} \, x^{-z-1}|\, dx \leq |z|\int _1^\infty x^{-\Re z-1}\, dx=\frac{|z|}{\Re z}\leq 2\, |z|. \]

Thus we have that,

\[ |\zeta (s+3/2+it)|=|\zeta (z)|\leq 1+1+2\, |z|=2+2\, |s+3/2+it| \leq 2+2\, |s|+3+2\, |it|\leq 7+2\, |t|. \]

Theorem 5.3.10 LogDerivZetaFinalBound

Let \(t\in \mathbb {R}\) with \(|t|\geq 2\) and \(0 {\lt} r' {\lt} r {\lt} R' {\lt} R{\lt}1\). If \(f(z)=\zeta (z+3/2+it)\), then for all \(z\in \overline{\mathbb {D}_{r'}}\setminus \mathcal{K}_f(R')\) we have that

\[ \left|\frac{f'}{f}(z)-\sum _{\rho \in \mathcal{K}_f(R')}\frac{m_f(\rho )}{z-\rho }\right| \ll \left(\frac{16r^2}{(r-r')^3}+\frac{1}{(R^2/R'-R')\, \log (R/R')}\right)\log |t|. \]

Proof ▶

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

\[ |g(z)|=\frac{|\zeta (z+3/2+it)|}{|\zeta (3/2+it)|} \leq \frac{\zeta (3/2)}{\zeta (3)}\cdot (7+2\, |t|)\leq \frac{13\, \zeta (3/2)}{3\, \zeta (3)}\, |t| \]

by Theorems 5.3.8 and 5.3.9. Thus by Theorem 5.3.7 we have that

\[ \left|\frac{g'}{g}(z)-\sum _{\rho \in \mathcal{K}_g(R')}\frac{m_g(\rho )}{z-\rho }\right| \leq \left(\frac{16r^2}{(r-r')^3}+\frac{1}{(R^2/R'-R')\, \log (R/R')}\right) \left(\log |t|+\log \left(\frac{13\, \zeta (3/2)}{3\, \zeta (3)}\right)\right). \]

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,

\[ \left|\frac{f'}{f}(z)-\sum _{\rho \in \mathcal{K}_f(R')}\frac{m_f(\rho )}{z-\rho }\right| \ll \left(\frac{16r^2}{(r-r')^3}+\frac{1}{(R^2/R'-R')\, \log (R/R')}\right)\log |t| \]

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

\[ C\geq 1+\frac{\log ((13\, \zeta (3/2))/(3\, \zeta (3)))}{\log 2}. \]

Definition 5.3.8 ZeroWindows

Let \(\mathcal{Z}_t=\{ \rho \in \mathbb {C}:\zeta (\rho )=0,\, |\rho -(3/2+it)|\leq 5/6\} \).

Lemma 5.3.14 SumBoundI

For all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 2\) we have

\[ \left|\frac{\zeta '}{\zeta }(1+\delta +it) -\sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{1+\delta +it-\rho }\right|\ll \log |t|. \]

Proof ▶

We apply Theorem 5.3.10 where \(r'=2/3\), \(r=3/4\), \(R'=5/6\), and \(R=8/9\). Thus, for all \(z\in \overline{\mathbb {D}_{2/3}}\setminus \mathcal{K}_f(5/6)\) we have that

\[ \left|\frac{\zeta '}{\zeta }(z+3/2+it) -\sum _{\rho \in \mathcal{K}_f(5/6)}\frac{m_f(\rho )}{z-\rho }\right|\ll \log |t| \]

where \(f(z)=\zeta (z+3/2+it)\) for \(t\in \mathbb {R}\) with \(|t|\geq 3\). 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, \(z\not\in \mathcal{K}_f(5/6)\). So,

\[ \left|\frac{\zeta '}{\zeta }(1+\delta +it) -\sum _{\rho \in \mathcal{K}_f(5/6)}\frac{m_f(\rho )}{-1/2+\delta -\rho }\right| \ll \log |t|. \]

But now note that if \(\rho \in \mathcal{K}_f(5/6)\), then \(\zeta (\rho +3/2+it)=0\) and \(|\rho |\leq 5/6\). Thus, \(\rho +3/2+it\in \mathcal{Z}_t\). Additionally, note that \(m_f(\rho )=m_\zeta (\rho +3/2+it)\). So changing variables using these facts gives us that

\[ \left|\frac{\zeta '}{\zeta }(1+\delta +it) -\sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{1+\delta +it-\rho }\right| \ll \log |t|. \]

Lemma 5.3.15 ShiftTwoBound

For all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 2\) we have

\[ -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +2it)\right)\ll \log |t|. \]

Proof ▶

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

\begin{align*} \Re \left(\frac{1}{1+\delta +2it-\rho }\right) & =\Re \left(\frac{1+\delta -2it-\overline{\rho }}{(1+\delta +2it-\rho )(1+\delta -2it-\overline{\rho })}\right) \\ & =\frac{\Re (1+\delta -2it-\overline{\rho })}{|1+\delta +2it-\rho |^2} =\frac{1+\delta -\Re \rho }{(1+\delta -\Re \rho )^2+(2t-\mathfrak {I}\rho )^2}. \end{align*}

Now since \(\rho \in \mathcal{Z}_{2t}\), we have that \(|\rho -(3/2+2it)|\leq 5/6\). So, we have \(\Re \rho \in (2/3,7/3)\) and \(\mathfrak {I}\rho \in (2t-5/6,2t+5/6)\). Thus, we have that

\[ 1/3{\lt}1+\delta -\Re \rho \qquad \text{and}\qquad (1+\delta -\Re \rho )^2+(2t-\mathfrak {I}\rho )^2{\lt}16/9+25/36=89/36. \]

Which implies that

\begin{equation} \label{pickupPoint4} 0\leq \frac{12}{89} {\lt}\frac{1+\delta -\Re \rho }{(1+\delta -\Re \rho )^2+(2t-\mathfrak {I}\rho )^2} =\Re \left(\frac{1}{1+\delta +2it-\rho }\right). \end{equation}

Note that, from Lemma 5.3.14, we have

\[ \sum _{\rho \in \mathcal{Z}_{2t}}m_\zeta (\rho )\, \Re \left(\frac{1}{1+\delta +2it-\rho }\right) -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +2it)\right) \leq \left|\frac{\zeta '}{\zeta }(1+\delta +2it) -\sum _{\rho \in \mathcal{Z}_{2t}}\frac{m_\zeta (\rho )}{1+\delta +2it-\rho }\right| \ll \log |2t|. \]

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

\[ -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +2it)\right)\ll \log |2t|. \]

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

Lemma 5.3.16 ShiftOneBound

There exists \(C{\gt}0\) such that for all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\); if \(\zeta (\rho )=0\) with \(\rho =\sigma +it\), then

\[ -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +it)\right) \leq -\frac{1}{1+\delta -\sigma }+C\log |t|. \]

Proof ▶

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

\begin{align*} \Re \left(\frac{1}{1+\delta +it-\rho '}\right) & =\Re \left(\frac{1+\delta -it-\overline{\rho '}}{(1+\delta +it-\rho ')(1+\delta -it-\overline{\rho '})}\right) \\ & =\frac{\Re (1+\delta -it-\overline{\rho '})}{|1+\delta +it-\rho '|^2} =\frac{1+\delta -\Re \rho '}{(1+\delta -\Re \rho ')^2+(t-\mathfrak {I}\rho ')^2}. \end{align*}

Now since \(\rho '\in \mathcal{Z}_t\), we have that \(|\rho -(3/2+it)|\leq 5/6\). So, we have \(\Re \rho '\in (2/3,7/3)\) and \(\mathfrak {I}\rho '\in (t-5/6,t+5/6)\). Thus we have that

\[ 1/3{\lt}1+\delta -\Re \rho '\qquad \text{and}\qquad (1+\delta -\Re \rho ')^2+(t-\mathfrak {I}\rho ')^2{\lt}16/9+25/36=89/36. \]

Which implies that

\begin{equation} \label{pickupPoint5} 0\leq \frac{12}{89} {\lt}\frac{1+\delta -\Re \rho '}{(1+\delta -\Re \rho ')^2+(t-\mathfrak {I}\rho ')^2} =\Re \left(\frac{1}{1+\delta +it-\rho '}\right). \end{equation}

Note that, from Lemma 5.3.14, we have

\[ \sum _{\rho \in \mathcal{Z}_t}m_\zeta (\rho )\, \Re \left(\frac{1}{1+\delta +it-\rho }\right) -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +it)\right) \leq \left|\frac{\zeta '}{\zeta }(1+\delta +it) -\sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{1+\delta +it-\rho }\right| \ll \log |t|. \]

Since \(m_\zeta (\rho )\geq 0\) for all \(\rho '\in \mathcal{Z}_t\), 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

\[ \frac{m_\zeta (\rho )}{\Re (1+\delta +it-\rho )} -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +it)\right)\ll \log |t|. \]

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

\[ -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +it)\right) \leq -\frac{m_\zeta (\rho )}{1+\delta -\sigma }+C\log |t|. \]

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 \(1\leq m_\zeta (\rho )\) completes the proof.

Lemma 5.3.17 ShiftZeroBound

For all \(\delta \in (0,1)\) we have

\[ -\Re \left(\frac{\zeta '}{\zeta }(1+\delta )\right)\leq \frac{1}{\delta }+O(1). \]

Proof ▶

From Theorem 4.4.13 we know that

\[ -\frac{\zeta '}{\zeta }(s)=\frac{1}{s-1}+O(1). \]

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

\[ -\Re \left(\frac{\zeta '}{\zeta }(1+\delta )\right)\leq \left| -\frac{\zeta '}{\zeta }(1+\delta )\right|\leq \frac{1}{\delta }+O(1). \]

Lemma 5.3.18 ThreeFourOneTrigIdentity

We have that

\[ 0\leq 3+4\cos \theta +\cos 2\theta \]

for all \(\theta \in \mathbb {R}\).

Proof ▶

We know that \(\cos (2\theta )=2\cos ^2\theta -1\), thus

\[ 3+4\cos \theta +\cos 2\theta =2+4\cos \theta +2\cos ^2\theta =2\, (1+\cos \theta )^2. \]

Noting that \(0\leq 1+\cos \theta \) completes the proof.

Theorem 5.3.11 ZeroInequality

There exists a constant \(0 {\lt} E{\lt}1\) such that for all \(\rho =\sigma +it\) with \(\zeta (\rho )=0\) and \(|t|\geq 2\), one has

\[ \sigma \leq 1-\frac{E}{\log |t|}. \]

Proof ▶

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

\[ -\frac{\zeta '}{\zeta }(s)=\sum _{1\leq n}\frac{\Lambda (n)}{n^s}. \]

Thus,

\[ -3\, \frac{\zeta '}{\zeta }(1+\delta ) -4\, \frac{\zeta '}{\zeta }(1+\delta +it) -\frac{\zeta '}{\zeta }(1+\delta +2it) =\sum _{1\leq n}\Lambda (n)\, n^{-(1+\delta )}\left(3+4n^{-it}+n^{-2it}\right). \]

Now applying Euler’s identity

\begin{align*} -3\, \Re \left(\frac{\zeta '}{\zeta }(1+\delta )\right)& -4\, \Re \left(\frac{\zeta '}{\zeta }(1+\delta +it)\right) -\Re \left(\frac{\zeta '}{\zeta }(1+\delta +2it)\right) \\ & \qquad \qquad \qquad =\sum _{1\leq n}\Lambda (n)\, n^{-(1+\delta )} \left(3+4\cos (-it\log n)+\cos (-2it\log n)\right) \end{align*}

By Lemma 5.3.18 we know that the series on the right hand side is bounded below by \(0\), and by Lemmas 5.3.15, 5.3.16, and 5.3.17 we have an upper bound on the left hand side. So,

\[ 0\leq \frac{3}{\delta }+3A-\frac{4}{1+\delta -\sigma }+4B\log |t|+C\log |t| \]

where \(A\), \(B\), and \(C\) are the implied constants coming from Lemmas 5.3.17, 5.3.16, and 5.3.15 respectively. By choosing \(D\geq 3A/\log 2+4B+C\) we have

\[ \frac{4}{1+\delta -\sigma }\leq \frac{3}{\delta }+D\log |t| \]

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

\[ \frac{4}{1-\sigma +1/(2D\log |t|)}\leq 7D\log |t|. \]

So with some manipulation we have that

\[ \sigma \leq 1-\frac{1}{14D\log |t|}. \]

This is exactly the desired result with the constant \(E=(14D)^{-1}\)

Definition 5.3.9 DeltaT

✓

Let \(\delta _t=E/\log |t|\) where \(E\) is the constant coming from Theorem 5.3.11.

Lemma 5.3.19 DeltaRange

For all \(t\in \mathbb {R}\) with \(|t|\geq 2\) we have that

\[ \delta _t{\lt}1/14. \]

Proof ▶

Note that \(\delta _t=E/\log |t|\) where \(E\) is the implied constant from Lemma 5.3.11. But we know that \(E=(14D)^{-1}\) where \(D\geq 3A/\log 2+4B+C\) where \(A\), \(B\), and \(C\) are the constants coming from Lemmas 5.3.17, 5.3.16, and 5.3.15 respectively. Thus,

\[ E\leq \frac{1}{14\, (3A/\log 2+4B+C)}. \]

But note that \(A\geq 0\) and \(B\geq 0\) by Lemmas 5.3.17 and 5.3.16 respectively. However, we have that

\[ C\geq 2+\frac{2\log ((13\, \zeta (3/2))/(3\, \zeta (3)))}{\log 2} \]

by Theorem 5.3.10 with Lemmas 5.3.14 and 5.3.15. So, by a very lazy estimate we have \(C\geq 2\) and \(E\leq 1/28\). Thus,

\[ \delta _t=\frac{E}{\log |t|}\leq \frac{1}{28\, \log 2}{\lt}\frac{1}{14}. \]

Lemma 5.3.20 SumBoundII

For all \(t\in \mathbb {R}\) with \(|t|\geq 2\) and \(z=\sigma +it\) where \(1-\delta _t/3\leq \sigma \leq 3/2\), we have that

\[ \left|\frac{\zeta '}{\zeta }(z) -\sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{z-\rho }\right|\ll \log |t|. \]

Proof ▶

By Lemma 5.3.19 we have that

\[ -11/21{\lt}-1/2-\delta _t/3\leq \sigma -3/2\leq 0. \]

We apply Theorem 5.3.10 where \(r'=2/3\), \(r=3/4\), \(R'=5/6\), and \(R=8/9\). Thus for all \(z\in \overline{\mathbb {D}_{2/3}}\setminus \mathcal{K}_f(5/6)\) we have that

\[ \left|\frac{\zeta '}{\zeta }(z+3/2+it) -\sum _{\rho \in \mathcal{K}_f(5/6)}\frac{m_f(\rho )}{z-\rho }\right|\ll \log |t| \]

where \(f(z)=\zeta (z+3/2+it)\) for \(t\in \mathbb {R}\) with \(|t|\geq 3\). Now if we let \(z=\sigma -3/2\), then \(z\in (-11/21,0)\subseteq \overline{\mathbb {D}_{2/3}}\). Additionally, \(f(z)=\zeta (\sigma +it)\), where \(\sigma +it\) lies in the zero free region given by Lemma 5.3.11 since \(\sigma \geq 1-\delta _t/3\geq 1-\delta _t\). Thus, \(z\not\in \mathcal{K}_f(5/6)\). So,

\[ \left|\frac{\zeta '}{\zeta }(\sigma +it) -\sum _{\rho \in \mathcal{K}_f(5/6)}\frac{m_f(\rho )}{\sigma -3/2-\rho }\right|\ll \log |t|. \]

But now note that if \(\rho \in \mathcal{K}_f(5/6)\), then \(\zeta (\rho +3/2+it)=0\) and \(|\rho |\leq 5/6\). Additionally, note that \(m_f(\rho )=m_\zeta (\rho +3/2+it)\). So changing variables using these facts gives us that

\[ \left|\frac{\zeta '}{\zeta }(\sigma +it) -\sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{\sigma +it-\rho }\right|\ll \log |t|. \]

Lemma 5.3.21 GapSize

Let \(t\in \mathbb {R}\) with \(|t|\geq 3\) and \(z=\sigma +it\) where \(1-\delta _t/3\leq \sigma \leq 3/2\). Additionally, let \(\rho \in \mathcal{Z}_t\). Then we have that

\[ |z-\rho |\geq \delta _t/6. \]

Proof ▶

Let \(\rho =\sigma '+it'\) and note that since \(\rho \in \mathcal{Z}_t\), we have \(t'\in (t-5/6,t+5/6)\). Thus, if \(t{\gt}1\) we have

\[ \log |t'|\leq \log |t+5/6|\leq \log |2t|=\log 2+\log |t|\leq 2\log |t|. \]

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

\[ \log |t'|\leq \log |t-5/6|\leq \log |2t|=\log 2+\log |t|\leq 2\log |t|. \]

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 5.3.11 (here we use the fact that \(|t|\geq 3\) to give us that \(|t'|\geq 2\)). Thus,

\[ \delta _t/6\leq \delta _{t'}-\delta _t/3 =1-\delta _t/3-(1-\delta _{t'})\leq \sigma -\sigma '\leq |z-\rho |. \]

Lemma 5.3.22 LogDerivZetaUniformLogSquaredBoundStrip

There exists a constant \(F\in (0,1/2)\) such that for all \(t\in \mathbb {R}\) with \(|t|\geq 3\) one has

\[ 1-\frac{F}{\log |t|}\leq \sigma \leq 3/2 \implies \left|\frac{\zeta '}{\zeta }(\sigma +it)\right|\ll \log ^2|t| \]

where the implied constant is uniform in \(\sigma \).

Proof ▶

Take \(F=E/3\) where \(E\) comes from Theorem 5.3.11. Then we have that \(\sigma \geq 1-\delta _t/3\). So, we apply Lemma 5.3.20, which gives us that

\[ \left|\frac{\zeta '}{\zeta }(z) -\sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{z-\rho }\right|\ll \log |t|. \]

Using the reverse triangle inequality and rearranging, we have that

\[ \left|\frac{\zeta '}{\zeta }(z)\right| \leq \sum _{\rho \in \mathcal{Z}_t}\frac{m_\zeta (\rho )}{|z-\rho |}+C\, \log |t| \]

where \(C\) is the implied constant in Lemma 5.3.20. Now applying Lemma 5.3.21 we have that

\[ \left|\frac{\zeta '}{\zeta }(z)\right| \leq \frac{6}{\delta _t}\sum _{\rho \in \mathcal{Z}_t}m_\zeta (\rho )+C\, \log |t|. \]

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

\[ 0=\zeta (\rho )=\zeta (\rho '+3/2+it)=f(\rho ') \]

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

\[ 5/6\geq |\rho -(3/2+it)|=|\rho '|. \]

Thus we change variables to have that

\[ \left|\frac{\zeta '}{\zeta }(z)\right| \leq \frac{6}{\delta _t}\sum _{\rho '\in \mathcal{K}_f(5/6)}m_f(\rho ')+C\, \log |t|. \]

Now note that \(f(0)=1\) and for \(|z|\leq 8/9\) we have

\[ |f(z)|=\frac{|\zeta (z+3/2+it)|}{|\zeta (3/2+it)|} \leq \frac{\zeta (3/2)}{\zeta (3)}\cdot (7+2\, |t|)\leq \frac{13\, \zeta (3/2)}{3\, \zeta (3)}\, |t| \]

by Theorems 5.3.8 and 5.3.9. Thus by Theorem 5.3.5 we have that

\[ \sum _{\rho '\in \mathcal{K}_f(5/6)}m_f(\rho ') \leq \frac{\log |t|+\log (13\, \zeta (3/2)/(3\, \zeta (3)))}{\log ((8/9)/(5/6))}\leq D\log |t| \]

where \(D\) is taken to be sufficiently large. Recall, by definition that, \(\delta _t=E/\log |t|\) with \(E\) coming from Theorem 5.3.11. By using this fact and the above, we have that

\[ \left|\frac{\zeta '}{\zeta }(z)\right|\ll \log ^2|t|+\log |t| \]

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

Theorem 5.3.12 LogDerivZetaUniformLogSquaredBound

✓

There exists a constant \(F\in (0,1/2)\) such that for all \(t\in \mathbb {R}\) with \(|t|\geq 3\) one has

\[ 1-\frac{F}{\log |t|}\leq \sigma \implies \left|\frac{\zeta '}{\zeta }(\sigma +it)\right|\ll \log ^2|t| \]

where the implied constant is uniform in \(\sigma \).

Proof ▶

Note that

\[ \left|\frac{\zeta '}{\zeta }(\sigma +it)\right| =\sum _{1\leq n}\frac{\Lambda (n)}{|n^{\sigma +it}|}=\sum _{1\leq n}\frac{\Lambda (n)}{n^\sigma } =-\frac{\zeta '}{\zeta }(\sigma )\leq \left|\frac{\zeta '}{\zeta }(\sigma )\right|. \]

From Theorem 4.4.13, and applying the triangle inequality we know that

\[ \left|\frac{\zeta '}{\zeta }(s)\right|\leq \frac{1}{|s-1|}+C. \]

where \(C{\gt}0\) is some constant. Thus, for \(\sigma \geq 3/2\) we have that

\[ \left|\frac{\zeta '}{\zeta }(\sigma +it)\right| \leq \left|\frac{\zeta '}{\zeta }(\sigma )\right| \leq \frac{1}{\sigma -1}+C\leq 2+C\ll 1\ll \log ^2|t|. \]

Putting this together with Lemma 5.3.22 completes the proof.

Theorem 5.3.13 LogDerivZetaLogSquaredBoundSmallt

For \(T{\gt}0\) and \(\sigma '=1-\delta _T/3=1-F/\log T\), if \(|t|\leq T\) then we have that

\[ \left|\frac{\zeta '}{\zeta }(\sigma '+it)\right|\ll \log ^2(2+T). \]

Proof ▶

Note that if \(|t|\geq 3\) then from Theorem 5.3.12 we have that

\[ \left|\frac{\zeta '}{\zeta }(\sigma '+it)\right|\ll \log ^2|t|\leq \log ^2T\leq \log ^2(2+T). \]

Otherwise, if \(|t|\leq 3\), then from Theorem 4.4.13 and applying the triangle inequality we know

\[ \left|\frac{\zeta '}{\zeta }(\sigma '+it)\right| \leq \frac{1}{|(\sigma '-1)+it|}+C\leq \frac{\log T}{F}+C \]

where \(C\geq 0\). Thus, we have that

\[ \left|\frac{\zeta '}{\zeta }(\sigma '+it)\right| \leq \left(\frac{\log T}{F\, \log 2}+\frac{C}{\log 2}\right)\, \log (2+|t|) \leq \left(\frac{\log (2+T)}{F\, \log 2}+\frac{C}{\log 2}\right)\log (2+T) \ll \log ^2(2+T). \]

From here out we closely follow our previous proof of the Medium PNT and we modify it using our new estimate in Theorem 5.3.12. Recall Definition 4.5.2; for fixed \(\varepsilon {\gt}0\) and a bump function \(\nu \) supported on \([1/2,2]\) we have

\[ \psi _\varepsilon (X) =\frac{1}{2\pi i}\int _{(\sigma )}\left(-\frac{\zeta '}{\zeta }(s)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(s)\, X^s\, ds \]

where \(\sigma =1+1/\log X\). Let \(T{\gt}3\) be a large constant to be chosen later, and we take \(\sigma '=1-\delta _T/3=1-F/\log T\) with \(F\) coming from Theorem 5.3.12. We integrate along the \(\sigma \) vertical line, and we pull contours accumulating the pole at \(s=1\) when we integrate along the curves

\(I_1\): \(\sigma -i\infty \) to \(\sigma -iT\)
\(I_2\): \(\sigma '-iT\) to \(\sigma -iT\)
\(I_3\): \(\sigma '-iT\) to \(\sigma '+iT\)
\(I_4\): \(\sigma '+iT\) to \(\sigma +iT\)
\(I_5\): \(\sigma +iT\) to \(\sigma +i\infty \).

Definition 5.3.10 I1New

✓

Let

\[ I_1(\nu ,\varepsilon ,X,T)= \frac{1}{2\pi i}\int _{-\infty }^{-T}\left(-\frac{\zeta '}{\zeta }(\sigma +it)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma +it)\, X^{\sigma +it}\, dt. \]

Definition 5.3.11 I5New

✓

Let

\[ I_5(\nu ,\varepsilon ,X,T)= \frac{1}{2\pi i}\int _T^\infty \left(-\frac{\zeta '}{\zeta }(\sigma +it)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma +it)\, X^{\sigma +it}\, dt. \]

Lemma 5.3.23 I1NewBound

✓

We have that

\[ |I_1(\nu ,\varepsilon ,X,T)|\ll \frac{X}{\varepsilon \sqrt{T}}. \]

Proof ▶

Note that \(|I_1(\nu ,\varepsilon ,X,T)|=\)

\[ \left|\frac{1}{2\pi i}\int _{-\infty }^{-T}\left(-\frac{\zeta '}{\zeta }(\sigma +it)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma +it)\, X^{\sigma +it}\, dt\right| \ll \int _{-\infty }^{-T}\left|\frac{\zeta '}{\zeta }(\sigma +it)\right| \cdot |\mathcal{M}(\tilde{1}_\varepsilon )(\sigma +it)|\cdot X^\sigma \, dt. \]

Applying Theorem 5.3.12 and Lemma 4.3.10, we have that

\[ |I_1(\nu ,\varepsilon ,X,T)| \ll \int _{-\infty }^{-T}\log ^2|t|\cdot \frac{X^\sigma }{\varepsilon \, |\sigma +it|^2}\, dt \ll \frac{X}{\varepsilon }\int _T^\infty \frac{\sqrt{t}\, dt}{t^2} \ll \frac{X}{\varepsilon \sqrt{T}}. \]

Here we are using the fact that \(\log ^2 t\) grows slower than \(\sqrt{t}\), \(|\sigma +it|^2\geq t^2\), and \(X^\sigma =X\cdot X^{1/\log X}=eX\).

Lemma 5.3.24 I5NewBound

✓

We have that

\[ |I_5(\nu ,\varepsilon ,X,T)|\ll \frac{X}{\varepsilon \sqrt{T}}. \]

Proof ▶

By symmetry, note that

\[ |I_1(\nu ,\varepsilon ,X,T)|=|\overline{I_5(\nu ,\varepsilon ,X,T)}|=|I_5(\nu ,\varepsilon ,X,T)|. \]

Applying Lemma 5.3.23 completes the proof.

Definition 5.3.12 I2New

✓

Let

\[ I_2(\nu ,\varepsilon ,X,T) =\frac{1}{2\pi i}\int _{\sigma '}^\sigma \left(-\frac{\zeta '}{\zeta }(\sigma _0-iT)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma _0-iT)\, X^{\sigma _0-iT}\, d\sigma _0. \]

Definition 5.3.13 I4New

✓

Let

\[ I_4(\nu ,\varepsilon ,X,T) =\frac{1}{2\pi i}\int _{\sigma '}^\sigma \left(-\frac{\zeta '}{\zeta }(\sigma _0+iT)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma _0+iT)\, X^{\sigma _0+iT}\, d\sigma _0. \]

Lemma 5.3.25 I2NewBound

We have that

\[ |I_2(\nu ,\varepsilon ,X,T)|\ll \frac{X}{\varepsilon \sqrt{T}}. \]

Proof ▶

Note that \(|I_2(\nu ,\varepsilon ,X,T)|=\)

\[ \left|\frac{1}{2\pi i}\int _{\sigma '}^\sigma \left(-\frac{\zeta '}{\zeta }(\sigma _0-iT)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma _0-iT)\, X^{\sigma _0-iT}\, d\sigma _0\right| \ll \int _{\sigma '}^\sigma \left|\frac{\zeta '}{\zeta }(\sigma _0-iT)\right| \cdot |\mathcal{M}(\tilde{1}_\varepsilon )(\sigma _0-iT)|\cdot X^{\sigma _0}\, d\sigma _0. \]

Applying Theorem 5.3.12 and Lemma 4.3.10, we have that

\[ |I_2(\nu ,\varepsilon ,X,T)|\ll \int _{\sigma '}^\sigma \log ^2 T \cdot \frac{X^{\sigma _0}}{\varepsilon \, |\sigma _0-iT|^2}\, d\sigma _0 \ll \frac{X\, \log ^2T}{\varepsilon \, T^2}\int _{\sigma '}^\sigma d\, \sigma _0 =\frac{X\, \log ^2T}{\varepsilon \, T^2}\, (\sigma -\sigma '). \]

Here we are using the fact that \(X^{\sigma _0}\leq X^\sigma =X\cdot X^{1/\log X}=eX\) and \(|\sigma _0-iT|^2\geq T^2\). Now note that

\[ |I_2(\nu ,\varepsilon ,X,T)|\ll \frac{X\, \log ^2T}{\varepsilon \, T^2}\, (\sigma -\sigma ') =\frac{X\, \log ^2T}{\varepsilon \, T^2\, \log X}+\frac{FX\, \log T}{\varepsilon \, T^2} \ll \frac{X}{\varepsilon \sqrt{T}}. \]

Here we are using the fact that \(\log T\ll T^{3/2}\), \(\log ^2T\ll T^{3/2}\), and \(X/\log X\ll X\).

Lemma 5.3.26 I4NewBound

✓

We have that

\[ |I_4(\nu ,\varepsilon ,X,T)|\ll \frac{X}{\varepsilon \sqrt{T}}. \]

Proof ▶

By symmetry, note that

\[ |I_2(\nu ,\varepsilon ,X,T)|=|\overline{I_4(\nu ,\varepsilon ,X,T)}|=|I_4(\nu ,\varepsilon ,X,T)|. \]

Applying Lemma 5.3.25 completes the proof.

Definition 5.3.14 I3New

✓

Let

\[ I_3(\nu ,\varepsilon ,X,T) =\frac{1}{2\pi i}\int _{-T}^T\left(-\frac{\zeta '}{\zeta }(\sigma '+it)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma '+it)\, X^{\sigma '+it}\, dt. \]

Lemma 5.3.27 I3NewBound

We have that

\[ |I_3(\nu ,\varepsilon ,X,T)|\ll \frac{X^{1-F/\log T}\sqrt{T}}{\varepsilon }. \]

Proof ▶

Note that \(|I_3(\nu ,\varepsilon ,X,T)|=\)

\[ \left|\frac{1}{2\pi i}\int _{-T}^T\left(-\frac{\zeta '}{\zeta }(\sigma '+it)\right) \, \mathcal{M}(\tilde{1}_\varepsilon )(\sigma '+it)\, X^{\sigma '+it}\, dt\right| \ll \int _{-T}^T\left|\frac{\zeta '}{\zeta }(\sigma '+it)\right| \cdot |\mathcal{M}(\tilde{1}_\varepsilon )(\sigma '+it)|\cdot X^{\sigma '}\, dt. \]

Applying Theorem 5.3.13 and Lemma 4.3.10, we have that

\[ |I_3(\nu ,\varepsilon ,X,T)|\ll \int _{-T}^T\log ^2(2+T) \cdot \frac{X^{\sigma '}}{\varepsilon \, |\sigma '+it|^2}\, dt \ll \frac{X^{1-F/\log T}\, \sqrt{T}}{\varepsilon }\int _0^T\frac{dt}{|\sigma '+it|^2}. \]

Here we are using the fact that this integrand is symmetric in \(t\) about \(0\) and that \(\log ^2(2+T)\ll \sqrt{T}\) for sufficiently large \(T\). Now note that, by Lemma 5.3.19, we have

\[ \frac{1}{|\sigma '+it|^2}=\frac{1}{(1-\delta _T/3)^2+t^2}{\lt}\frac{1}{(41/42)^2+t^2}. \]

Thus,

\[ |I_3(\nu ,\varepsilon ,X,T)| \ll \frac{X^{1-F/\log T}\sqrt{T}}{\varepsilon }\int _0^T\frac{dt}{|\sigma '+it|^2} \leq \frac{X^{1-F/\log T}\sqrt{T}}{\varepsilon }\int _0^\infty \frac{dt}{(41/42)^2+t^2}. \]

The integral on the right hand side evaluates to \(21\pi /41\), which is just a constant, so the desired result follows.

Theorem 5.3.14 SmoothedChebyshevPull3

✓

We have that

\[ \psi _\varepsilon (X)=\mathcal{M}(\tilde{1}_\varepsilon )(1)\, X^1+I_1-I_2+I_3+I_4+I_5. \]

Proof ▶

Pull contours and accumulate the pole of \(\zeta '/\zeta \) at \(s=1\).

Theorem 5.3.15 StrongPNT

We have

\[ \sum _{n\leq x}\Lambda (n)=x+O\left(x\exp (-c\sqrt{\log x})\right). \]

Proof ▶

By Theorem 4.5.3 and 5.3.14 we have that

\[ \mathcal{M}(\tilde{1}_\varepsilon )(1)\, x^1+I_1-I_2+I_3+I_4+I_5 =\psi (x)+O(\varepsilon x\log x). \]

Applying Theorem 4.3.11 and Lemmas 5.3.23, 5.3.25, 5.3.27, 5.3.26, and 5.3.24 we have that

\[ \psi (x)=x+O(\varepsilon x)+O(\varepsilon x\log x) +O\left(\frac{x}{\varepsilon \sqrt{T}}\right) +O\left(\frac{x^{1-F/\log T}\sqrt{T}}{\varepsilon }\right). \]

We absorb the \(O(\varepsilon x)\) term into the \(O(\varepsilon x\log x)\) term and balance the last two terms in \(T\).

\[ \frac{x}{\varepsilon \sqrt{T}} =\frac{x^{1-F/\log T}\sqrt{T}}{\varepsilon }\implies T =\exp (\sqrt{F\log x}). \]

Thus,

\[ \psi (x)=x+O(\varepsilon x\log x) +O\left(\frac{x}{\displaystyle \varepsilon \exp ((1/2)\cdot \sqrt{F\log x})}\right). \]

Now we balance the last two terms in \(\varepsilon \).

\[ \varepsilon x\log x =\frac{x}{\displaystyle \varepsilon \exp ((1/2)\cdot \sqrt{F\log x})} \implies \varepsilon \log x =\frac{\displaystyle \sqrt{\log x}}{\displaystyle \exp ((1/4)\cdot \sqrt{F\log x})}. \]

Thus,

\[ \psi (x)=x+O\left(x\exp (-(\sqrt{F}/4)\cdot \sqrt{\log x})\sqrt{\log x}\right). \]

Absorbing the \(\displaystyle \sqrt{\log x}\) into the \(\displaystyle \exp (-(\sqrt{F}/4)\cdot \sqrt{\log x})\) completes the proof.