Prime Number Theorem And ...

4 Third Approach

4.1 Strong PNT

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

Theorem 24 BorelCaratheodory
#

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

\[ \sup _{|z|\leq r}|f(z)|\leq \frac{2Mr}{R-r}. \]
Proof

Let

\[ f_M(z)=\frac{f(z)/z}{2M-f(z)}. \]

Note that \(2M-f(z)\neq 0\) because \(\mathfrak {R}(2M-f(z))=2M-\mathfrak {R}f(z)\geq M{\gt}0\). Additionally, since \(f(z)\) has a zero at \(0\), we know that \(f(z)/z\) is analytic on \(|z|\leq R\). Likewise, \(f_M(z)\) is analytic on \(|z|\leq R\).

Now note that \(|f(z)|\leq |2M-f(z)|\) since \(\mathfrak {R}f(z)\leq M\). Thus we have that

\[ |f_M(z)|=\frac{|f(z)|/|z|}{|2M-f(z)|}\leq \frac{1}{|z|}. \]

Now by the maximum modulus principle, we know the maximum of \(|f_M|\) must occur on the boundary where \(|z|=R\). Thus, \(|f_M(z)|\leq 1/R\) for all \(|z|\leq R\). So for \(|z|=r\) we have

\[ |f_M(z)|=\frac{|f(z)|/r}{|2M-f(z)|}\leq \frac{1}{R}\implies R\, |f(z)|\leq r\, |2M-f(z)|\leq 2Mr+r\, |f(z)|. \]

Which by algebraic manipulation gives

\[ |f(z)|\leq \frac{2Mr}{R-r}. \]

Once more, by the maximum modulus principle, we know the maximum of \(|f|\) must occur on the boundary where \(|z|=r\). Thus, the desired result immediately follows

Lemma 99 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 \(\mathfrak {R}f(z)\leq M\) 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 Cauchy’s integral formula 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}
1

Now applying Theorem 24, 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 25 BorelCaratheodoryDeriv
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Let \(R,\, M{\gt}0\). Let \(f\) be analytic on \(|z|\leq R\) such that \(f(0)=0\) and suppose \(\mathfrak {R}f(z)\leq M\) 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 99 with \(r'=(R+r)/2\), and noting that \(r {\lt} R\), we have that

\[ |f'(z)|\leq \frac{4M(R+r)^2}{(R-r)^3}\leq \frac{16MR^2}{(R-r)^3}. \]
Theorem 26 LogOfAnalyticFunction
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Let \(0 {\lt} r {\lt} R{\lt}1\). 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 \(J_B:\overline{\mathbb {D}_r}\to \mathbb {C}\) that is analytic on neighborhoods of points in \(\overline{\mathbb {D}_r}\) such that

  • \(J_B(0)=0\)

  • \(J_B'(z)=B'(z)/B(z)\)

  • \(\log |B(z)|-\log |B(0)|=\mathfrak {R}J_B(z)\)

for all \(z\in \overline{\mathbb {D}_r}\).

Proof

We let \(J_B(z)=\mathrm{Log}\, B(z)-\mathrm{Log}\, B(0)\). Then clearly, \(J_B(0)=0\) and \(J_B'(z)=B'(z)/B(z)\). Showing the third property is a little more difficult, but by no standards terrible. Exponentiating \(J_B(z)\) we have that

\[ \exp (J_B(z))=\exp (\mathrm{Log}\, B(z)-\mathrm{Log}\, B(0))=\frac{B(z)}{B(0)}\implies B(z)=B(0)\exp (J_B(z)). \]

Now taking the modulus

\[ |B(z)|=|B(0)|\cdot |\exp (J_B(z))|=|B(0)|\cdot \exp (\mathfrak {R}J_B(z)). \]

Taking the real logarithm of both sides and rearranging gives the third point.

Definition 20 SetOfZeros
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Let \(R{\gt}0\) and \(f:\overline{\mathbb {D}_R}\to \mathbb {C}\). Define the set of zeros \(\mathcal{K}_f(R)=\{ \rho \in \mathbb {C}:|\rho |\leq R,\, f(\rho )=0\} \).

Definition 21 ZeroOrder
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Let \(0 {\lt} R{\lt}1\) and \(f:\mathbb {C}\to \mathbb {C}\) be analtyic on neighborhoods of points in \(\overline{\mathbb {D}_1}\). For any zero \(\rho \in \mathcal{K}_f(R)\), we define \(m_f(\rho )\) as the order of the zero \(\rho \) w.r.t \(f\).

Lemma 100 ZeroFactorization
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Let \(f:\overline{\mathbb {D}_1}\to \mathbb {C}\) be analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(f(0)\neq 0\). For all \(\rho \in \mathcal{K}_f(1)\) there exists \(h_\rho (z)\) that 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 22 CFunction
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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 \(f(0)\neq 0\). We define a function \(C_f:\overline{\mathbb {D}_R}\to \mathbb {C}\) 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 100.

Definition 23 BlaschkeB
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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 \(f(0)\neq 0\). We define a function \(B_f:\overline{\mathbb {D}_R}\to \mathbb {C}\) 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 101 BlaschkeOfZero
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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

\[ |B_f(0)|=|f(0)|\prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{|\rho |}\right)^{m_f(\rho )}. \]
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 23,

\[ |B_f(0)|=|C_f(0)|\prod _{\rho \in \mathcal{K}_f(r)}R^{m_f(\rho )}=|f(0)|\prod _{\rho \in \mathcal{K}_f(r)}\left(\frac{R}{|\rho |}\right)^{m_f(\rho )}. \]
Lemma 102 DiskBound
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Let \(B{\gt}1\) and \(0 {\lt} R{\lt}1\). If \(f:\mathbb {C}\to \mathbb {C}\) is a function analytic on neighborhoods of points in \(\overline{\mathbb {D}_1}\) with \(|f(z)|\leq B\) 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 23,

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

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 \(|z|\leq R\).

Theorem 27 ZerosBound
#

Let \(B{\gt}1\) and \(0{\lt} r {\lt} R{\lt}1\). 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 \(f(0)=1\), 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 23,

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

whereby Lemma 102 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 24 JBlaschke
#

Let \(B{\gt}1\) and \(0 {\lt} R{\lt}1\). 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 26 and \(B_f\) is from Definition 23.

Lemma 103 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 \(\overline{\mathbb {D}_1}\). 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 100. Thus, substituting this into Definition 23,

\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}
4

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 23,

\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}
5

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 28 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 102 we immediately know that \(|B_f(z)|\leq B\) for all \(|z|\leq R\). Now since \(L_f=J_{B_f}\) by Definition 24, by Theorem 26 we know that

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

for all \(|z|\leq r\). So by Theorem 25, it follows that

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

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

Theorem 29 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 22 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 23 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

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

By Definition 24 and Theorem 26 we recall that

\[ L_f(z)=J_{B_f}(z)=\mathrm{Log}\, B_f(z)-\mathrm{Log}\, B_f(0). \]

Taking the derivative of both sides we have that \(L_f'(z)=(B_f'/B_f)(z)\). Thus,

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

Now since \(z\in \overline{\mathbb {D}_{R'}}\) and \(\rho \in \mathcal{K}_f(R')\), we know that \(R^2/R'-R'\leq |z-R^2/\rho |\). 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 27 and 28 we get our desired result with a little algebraic manipulation.

Theorem 30 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 \(\mathfrak {R}s{\gt}1\)

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

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 104 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 105 ZetaAltFormulaAnalytic
#

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

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 106 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 31 GlobalBound
#

For all \(s\in \mathbb {C}\) with \(|s|\leq 1\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\), 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 \(|t|\geq 3\), we know \(z\in S\). Thus, from Lemma 106 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,

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

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 32 LogDerivZetaFinalBound
#

Let \(t\in \mathbb {R}\) with \(|t|\geq 3\) 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 30 and 31. Thus by Theorem 29 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

\[ C\geq 1+\frac{\log ((13\, \zeta (3/2))/(3\, \zeta (3)))}{\log 3}. \]
Definition 25 ZeroWindows
#

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

Lemma 107 SumBound
#

For all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\) we have

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

We apply Theorem 32 where \(r'=1/2\), \(r=2/3\), \(R'=5/6\), and \(R=8/9\). Thus, for all \(z\in \overline{\mathbb {D}_{5/6}}\setminus \mathcal{K}_f(5/6)\) 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 \(z\in (-1/2,1/2)\subseteq \overline{\mathbb {D}_{5/6}}\). 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

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

Since \(\log |t|\leq \log (2+|t|)\) for all \(t\in \mathbb {R}\), the result follows.

Lemma 108 ShiftTwoBound
#

For all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\) we have

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

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

\begin{align*} \mathfrak {R}\left(\frac{1}{1+\delta +2it-\rho }\right)& =\mathfrak {R}\left(\frac{1+\delta -2it-\overline{\rho }}{(1+\delta +2it-\rho )(1+\delta -2it-\overline{\rho })}\right) \\ & =\frac{\mathfrak {R}(1+\delta -2it-\overline{\rho })}{|1+\delta +2it-\rho |^2}=\frac{1+\delta -\mathfrak {R}\rho }{(1+\delta -\mathfrak {R}\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 \(\mathfrak {R}\rho \in (2/3,7/3)\) and \(\mathfrak {I}\rho \in (2t-5/6,2t+5/6)\). Thus, we have that

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

Which implies that

\begin{equation} \label{pickupPoint4} 0\leq \frac{12}{89}{\lt}\frac{1+\delta -\mathfrak {R}\rho }{(1+\delta -\mathfrak {R}\rho )^2+(2t-\mathfrak {I}\rho )^2}=\mathfrak {R}\left(\frac{1}{1+\delta +2it-\rho }\right). \end{equation}
6

Note that, from Lemma 107, we have

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

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

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

Noting that \(\log (2+|2t|)\leq \log (4+|2t|)=\log (2)+\log (2+|t|)\leq 2\log (2+|t|)\) completes the proof.

Lemma 109 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

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

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

\begin{align*} \mathfrak {R}\left(\frac{1}{1+\delta +it-\rho '}\right)& =\mathfrak {R}\left(\frac{1+\delta -it-\overline{\rho '}}{(1+\delta +it-\rho ')(1+\delta -it-\overline{\rho '})}\right) \\ & =\frac{\mathfrak {R}(1+\delta -it-\overline{\rho '})}{|1+\delta +it-\rho '|^2}=\frac{1+\delta -\mathfrak {R}\rho '}{(1+\delta -\mathfrak {R}\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 \(\mathfrak {R}\rho '\in (2/3,7/3)\) and \(\mathfrak {I}\rho '\in (t-5/6,t+5/6)\). Thus we have that

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

Which implies that

\begin{equation} \label{pickupPoint5} 0\leq \frac{12}{89}{\lt}\frac{1+\delta -\mathfrak {R}\rho '}{(1+\delta -\mathfrak {R}\rho ')^2+(t-\mathfrak {I}\rho ')^2}=\mathfrak {R}\left(\frac{1}{1+\delta +it-\rho '}\right). \end{equation}
7

Note that, from Lemma 107, we have

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

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

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

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

\[ -\mathfrak {R}\left(\frac{\zeta '}{\zeta }(1+\delta +it)\right)\leq -\frac{m_\zeta (\rho )}{1+\delta -\sigma }+C\log (2+|t|). \]

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

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

From Theorem 16 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

\[ -\mathfrak {R}\left(\frac{\zeta '}{\zeta }(1+\delta )\right)\leq \left|-\frac{\zeta '}{\zeta }(1+\delta )\right|\leq \frac{1}{\delta }+O(1). \]
Lemma 111 ThreeFourOneTrigIdentity
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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 33 ZeroInequality
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There exists a constant \(0 {\lt} C{\lt}1\) such that for all \(\rho =\sigma +it\) with \(\zeta (\rho )=0\) and \(|t|\geq 3\), one has

\[ \frac{1}{\log (2+|t|)}\ll 1-\sigma . \]
Proof

From Theorem 18 when \(\mathfrak {R}s{\gt}1\) 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\, \mathfrak {R}\left(\frac{\zeta '}{\zeta }(1+\delta )\right)& -4\, \mathfrak {R}\left(\frac{\zeta '}{\zeta }(1+\delta +it)\right)-\mathfrak {R}\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 111 we know that the series on the right hand side is bounded below by \(0\), and by Lemmas 108, 109, and 110 we have an upper bound on the left hand side. So,

\[ 0\leq \frac{3}{\delta }+3A-\frac{4}{1+\delta -\sigma }+4B\log (2+|t|)+C\log (2+|t|) \]

where \(A\), \(B\), and \(C\) are the implied constants coming from Lemmas 110, 109, and 108 respectively. By choosing \(D\geq 3A/\log 5+4B+C\) we have

\[ \frac{4}{1+\delta -\sigma }\leq \frac{3}{\delta }+D\log (2+|t|) \]

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

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

So with some manipulation we have that

\[ \frac{1}{14D\log (2+|t|)}\leq 1-\sigma . \]

This is exactly the desired result with an implied constant of \((14D)^{-1}\).