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

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.

Using Lemma 99 with \(r'=(R+r)/2\), and noting that \(r {\lt} R\), we have that

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.

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

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

Thus, substituting this into Definition 23,

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 \(|z|\leq R\).

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.

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.

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

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.

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.

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.

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.

This is an immediate consequence of the identity theorem.

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 \(g(z)=\zeta (z+3/2+it)/\zeta (3/2+it)\). 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

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.

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.

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.

From Theorem 16 we know that

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

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

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

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 \((14D)^{-1}\).