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 103 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 101. 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 103 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 Lemma 100 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 Lemma 100 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 Lemmas 104 and 106 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 111 we know that

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

Thus we have that,