4 Third Approach
4.1 Strong PNT
This upstreamed from https://github.com/math-inc/strongpnt/tree/main
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 \(0 {\lt} r {\lt} R\),
Let
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
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
Which by algebraic manipulation gives
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
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
for all \(|z|\leq r\).
By Cauchy’s integral formula we know that
Thus,
Now applying Theorem 24, 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.
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 \(0 {\lt} r {\lt} R\),
for all \(|z|\leq r\).
Using Lemma 99 with \(r'=(R+r)/2\), and noting that \(r {\lt} R\), we have that
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 \(z\in \overline{\mathbb {D}_r}\).
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
Now taking the modulus
Taking the real logarithm of both sides and rearranging gives the third point.
Let \(R{\gt}0\) and \(f:\overline{\mathbb {D}_R}\to \mathbb {C}\). Define the set of zeros \(\mathcal{K}_f(R)=\{ \rho \in \mathbb {C}:|\rho |\leq R,\, f(\rho )=0\} \).
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 \(f\).
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 \(f(z)=(z-\rho )^{m_f(\rho )}\, h_\rho (z)\).
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
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 \(\overline{\mathbb {D}_r}\).
where \(h_z(z)\) comes from Lemma 100.
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.
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
Since \(f(0)\neq 0\), we know that \(0\not\in \mathcal{K}_f(r)\). Thus,
Thus, substituting this into Definition 23,
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.
For \(|z|=R\), we know that \(z\not\in \mathcal{K}_f(r)\). Thus,
Thus, substituting this into Definition 23,
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 \(|z|\leq R\).
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
Since \(f(0)=1\), we know that \(0\not\in \mathcal{K}_f(r)\). Thus,
Thus, substituting this into Definition 23,
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.
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 \(z\in \overline{\mathbb {D}_r}\).
Suppose that \(z\in \mathcal{K}_f(r)\). Then we have that
where \(h_z(z)\neq 0\) according to Lemma 100. Thus, substituting this into Definition 23,
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 23,
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.
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 \(|z|\leq r'\)
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
for all \(|z|\leq r\). So by Theorem 25, it follows that
for all \(|z|\leq r'\).
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
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
Substituting this into Definition 23 we have that
Taking the complex logarithm of both sides we have that
Taking the derivative of both sides we have that
By Definition 24 and Theorem 26 we recall that
Taking the derivative of both sides we have that \(L_f'(z)=(B_f'/B_f)(z)\). Thus,
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
Now by Theorem 27 and 28 we get our desired result with a little algebraic manipulation.
For all \(t\in \mathbb {R}\) one has
From the Euler product expansion of \(\zeta \), we have that for \(\mathfrak {R}s{\gt}1\)
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.
Let
We have that \(\zeta (s)=\zeta _0(s)\) for \(\sigma {\gt}1\).
Note that for \(\sigma {\gt}1\) we have
Thus
Now we note that
So, substituting this we have
But noting that \(\lfloor x\rfloor =x-\{ x\} \) we have that
Evaluating the first integral completes the result.
We have that \(\zeta _0(s)\) is analytic for all \(s\in S\) where \(S=\{ s\in \mathbb {C}:\mathfrak {R}s{\gt}0,\, s\neq 1\} \).
Note that we have
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.
We have that
for all \(s\in S\).
This is an immediate consequence of the identity theorem.
For all \(s\in \mathbb {C}\) with \(|s|\leq 1\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\), we have that
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
by applying the triangle inequality. Now note that \(|z-1|\geq 1\). Likewise,
Thus we have that,
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
Let \(g(z)=\zeta (z+3/2+it)/\zeta (3/2+it)\). Note that \(g(0)=1\) and for \(|z|\leq R\)
by Theorems 30 and 31. Thus by Theorem 29 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
Let \(\mathcal{Z}_t=\{ \rho \in \mathbb {C}:\zeta (\rho )=0,\, |\rho -(3/2+it)|\leq 5/6\} \).
For all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\) we have
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
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,
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
Since \(\log |t|\leq \log (2+|t|)\) for all \(t\in \mathbb {R}\), the result follows.
For all \(\delta \in (0,1)\) and \(t\in \mathbb {R}\) with \(|t|\geq 3\) we have
Note that, for \(\rho \in \mathcal{Z}_{2t}\)
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
Which implies that
Note that, from Lemma 107, 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 \(\log (2+|2t|)\leq \log (4+|2t|)=\log (2)+\log (2+|t|)\leq 2\log (2+|t|)\) completes the proof.
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
Note that for \(\rho '\in \mathcal{Z}_t\)
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
Which implies that
Note that, from Lemma 107, we have
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
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 \(1\leq m_\zeta (\rho )\) completes the proof.
For all \(\delta \in (0,1)\) we have
From Theorem 16 we know that
Changing variables \(s\mapsto 1+\delta \) and applying the triangle inequality we have that
We have that
for all \(\theta \in \mathbb {R}\).
We know that \(\cos (2\theta )=2\cos ^2\theta -1\), thus
Noting that \(0\leq 1+\cos \theta \) completes the proof.
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
From Theorem 18 when \(\mathfrak {R}s{\gt}1\) we have
Thus,
Now applying Euler’s identity
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,
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
by some manipulation. Now if we choose \(\delta =(2D\log (2+|t|))^{-1}\) then we have
So with some manipulation we have that
This is exactly the desired result with an implied constant of \((14D)^{-1}\).