This follows directly from the definition of \(g\).

We need to show that \(g\) is complex differentiable at every point in \(s\). For \(z\neq 0\), this follows directly from the definition of \(g\) and the fact that \(f\) is analytic on \(s\). For \(z=0\), we use the definition of the derivative and the fact that \(f(0)=0\):

where the last equality follows from the definition of the derivative of \(f\) at \(0\). Thus, \(g\) is complex differentiable at \(0\) with derivative \(0\), completing the proof.

The proof is similar to that of Lemma 100, but we need to consider two cases: when \(x\) is on the boundary of the closed ball and when it is in the interior. In the first case, we take a small open ball around \(x\) that lies entirely within the closed ball, and apply Lemma 100 on this smaller ball. In the second case, we can take the entire open ball centered at \(0\) with radius \(R\), and again apply Lemma 100. In both cases, we use the fact that \(f(0)=0\) to ensure that the removable singularity at \(0\) is handled correctly.

This follows directly from Lemma 101 and the fact that the difference of two analytic functions is analytic.

We square both sides and simplify to obtain the equivalent inequality

which follows directly from the assumption \(\Re x\leq M\) and the positivity of \(M\).

Let

Note that \(2M-f(z)\neq 0\) because \(\Re (2M-f(z))=2M-\Re f(z)\geq M{\gt}0\). 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 \(\Re f(z)\leq M\). 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 25, and noting that \(r'-r\leq |r'e^{it}-z|\), we have that

Substituting this into Equation (3) and evaluating the integral completes the proof.

Using Lemma 104 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 25,

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

Thus, substituting this into Definition 25,

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 25,

whereby Lemma 107 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 105. Thus, substituting this into Definition 25,

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 (6) 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 25,

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 (7) 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 107 we immediately know that \(|B_f(z)|\leq B\) for all \(|z|\leq R\). Now since \(L_f=J_{B_f}\) by Definition 26, by Theorem 27 we know that

for all \(|z|\leq r\). So by Theorem 26, 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 24 we know that

Substituting this into Definition 25 we have that

Taking the complex logarithm of both sides we have that

Taking the derivative of both sides we have that

By Definition 26 and Theorem 27 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 28 and 29 we get our desired result with a little algebraic manipulation.

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

Thus, we have that

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

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

Note that 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 \(1/2\leq \Re z\); additionally, as \(|t|\geq 3\), we know \(z\in S\). Thus, from Lemma 111 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 31 and 32. Thus by Theorem 30 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 33 where \(r'=2/3\), \(r=3/4\), \(R'=5/6\), 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

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 \(\Re \rho \in (2/3,7/3)\) and \(\mathfrak {I}\rho \in (2t-5/6,2t+5/6)\). Thus, we have that

Which implies that

Note that, from Lemma 112, we have

Since \(m_\zeta (\rho )\geq 0\) for all \(\rho \in \mathcal{Z}_{2t}\), the inequality from Equation (8) tells us that by subtracting the sum from both sides we have

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

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

Now since \(\rho '\in \mathcal{Z}_t\), we have that \(|\rho -(3/2+it)|\leq 5/6\). 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

Which implies that

Note that, from Lemma 112, we have

Since \(m_\zeta (\rho )\geq 0\) for all \(\rho '\in \mathcal{Z}_t\), the inequality from Equation (9) 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 \(\Re (1+\delta +it-\rho )=1+\delta -\sigma \). 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 \(\Re s{\gt}1\) we have

Thus,

Now applying Euler’s identity

By Lemma 116 we know that the series on the right hand side is bounded below by \(0\), and by Lemmas 113, 114, and 115 we have an upper bound on the left hand side. So,

where \(A\), \(B\), and \(C\) are the implied constants coming from Lemmas 115, 114, and 113 respectively. By choosing \(D\geq 3A/\log 3+4B+C\) we have

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

So with some manipulation we have that

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

Note that \(\delta _t=E/\log |t|\) where \(E\) is the implied constant from Lemma 34. But we know that \(E=(14D)^{-1}\) where \(D\geq 3A/\log 3+4B+C\) where \(A\), \(B\), and \(C\) are the constants coming from Lemmas 115, 114, and 113 respectively. Thus,

But note that \(A\geq 0\) and \(B\geq 0\) by Lemmas 115 and 114 respectively. However, we have that

by Theorem 33 with Lemmas 112 and 113. So, by a very lazy estimate we have \(C\geq 2\) and \(E\leq 1/28\). Thus,

By Lemma 117 we have that

We apply Theorem 33 where \(r'=2/3\), \(r=3/4\), \(R'=5/6\), 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 \(z=\sigma -3/2\), then \(z\in (-43/84,0)\subseteq \overline{\mathbb {D}_{5/6}}\). Additionally, \(f(z)=\zeta (\sigma +it)\), where \(\sigma +it\) lies in the zero free region given by Lemma 34 since \(\sigma \geq 1-\delta _t/3\geq 1-\delta _t\). 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 \(|\rho |\leq 5/6\). Additionally, note that \(m_f(\rho )=m_\zeta (\rho +3/2+it)\). So changing variables using these facts gives us that

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

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

So by taking reciprocals and multiplying through by a constant we have that \(\delta _t\leq 2\delta _{t'}\). Now note that since \(\rho \in \mathcal{Z}_t\) we know that \(\sigma '\leq 1-\delta _{t'}\) by Theorem 34 (here we use the fact that \(|t|\geq 4\) to give us that \(|t'|\geq 3\)). Thus,

Take \(F=E/3\) where \(E\) comes from Theorem 34. Then we have that \(\sigma \geq 1-\delta _t/3\). So, we apply Lemma 118, which gives us that

Using the reverse triangle inequality and rearranging, we have that

where \(C\) is the implied constant in Lemma 118. Now applying Lemma 119 we have that

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

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

Thus we change variables to have that

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

by Theorems 31 and 32. Thus by Theorem 28 we have that

where \(D\) is taken to be sufficiently large. Recall, by definition that, \(\delta _t=E/\log |t|\) with \(E\) coming from Theorem 34. By using this fact and the above, we have that

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.