2. Auxiliary lemmas

In this section, we prepare some auxiliary results that are needed for the proofs of our theorems.

Proof. By symmetry, we may assume $\delta \ge 0$ . When $x+\delta \le 1$ , the assertion is trivial. When $x\ge 1$ , the assertion follows by

$$\begin{align*}|\log^{+}(x+\delta)-\log^{+}x| = \int_{x}^{x+\delta}\frac{du}{u} \le \delta. \end{align*}$$

For the remaining case $1-\delta <x<1$ , we have

$$\begin{align*}|\log^{+}(x+\delta)-\log^{+}x| = \log(x+\delta) \le \log(1+\delta) \le \delta. \end{align*}$$

This completes the proof.

Proof. Recalling that $\log ^+(xy)\le \log ^+x+\log ^+y$ and $|\log y|=\log ^+y+\log ^+(1/y)$ we have

$$ \begin{align*} \log^{+}(xy)-\log^{+}x\le \log^+x+\log^+y-\log^+x=\log^+y\le |\log y| \end{align*} $$

and

$$ \begin{align*} \log^+x-\log^{+}(xy)=\log^+(xy(1/y))-\log^+{(xy)}\le \log^+(1/y)\le |\log y|. \end{align*} $$

This completes the assertion.

Lemma 2.3. For a polynomial $p\in \mathbb {C}[X]$ and $r\ge 0$ , we have

$$\begin{align*}N(r,\zeta-p) = \log^{+}r. \end{align*}$$

Proof. Since $(\zeta -p)(s)$ has only one simple pole at $s=1$ , it is clear that

$$\begin{align*}N(r, \zeta) = \int_{0}^{r}\frac{n(t, \zeta-p)}{t}dt = \int_{1}^{r}\frac{dt}{t} = \log r \end{align*}$$

for $r\ge 1$ and $N(r,\zeta )=0$ for $0\le r<1$ .

Proof. By the Dirichlet series representation of $\zeta $ , we have

$$\begin{align*}\zeta(s) = 1+O\bigg(\sum_{n=2}^{\infty}\frac{1}{n^{\sigma}}\bigg) = 1+O\bigg(\frac{1}{2^{\sigma}}\sum_{n=2}^{\infty}\bigg(\frac{2}{n}\bigg)^{\sigma}\bigg) = 1+O\bigg(\frac{1}{2^{\sigma}}\bigg). \end{align*}$$

This gives the lemma.

Lemma 2.5. We have

$$\begin{align*}\zeta(s)=2^s\pi^{s-1}\sin\frac{s\pi}{2}\Gamma(1-s)\zeta(1-s). \end{align*}$$

Lemma 2.6. For $1<\sigma \le 2$ , we have

$$\begin{align*}\bigl|\log|\zeta(s)|\bigr| \ll \log\frac{1}{\sigma-1}+1. \end{align*}$$

Proof. By using the Dirichlet series representations

$$\begin{align*}\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}} \quad\text{and}\quad \frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}, \end{align*}$$

where $\mu (n)$ is the Möbius function, we get

$$\begin{align*}(\sigma-1) \ll \frac{1}{\zeta(\sigma)} \le |\zeta(s)| \le \zeta(\sigma) \ll \frac{1}{\sigma-1} \end{align*}$$

and so the assertion holds.

We write $\ll _{a_{1},\ldots ,a_{n}}$ to express that the implicit constant in $\ll $ depends on the parameters $a_{1},\ldots ,a_{n}$ .

Lemma 2.7. Let $C,\delta>0$ . For $|t|\ge 2$ , we have

(2.1) $$ \begin{align} \zeta(s) & \ll_{\delta}1\qquad\qquad\qquad\ \ \text{for } \sigma\in[1+\delta,\infty), \end{align} $$

(2.2) $$ \begin{align} \zeta(s) & \ll\log|t| & \text{for } \sigma\in[1,\infty), \end{align} $$

(2.3) $$ \begin{align} \zeta(s) & \ll|t|^{\frac{1}{2}}\log|t| & \text{for } \sigma\in[0,1], \end{align} $$

(2.4) $$ \begin{align} \zeta(s) & \ll_{C}|t|^{\frac{1}{2}-\sigma}\log|t| & \text{for } \sigma\in[-C,0], \end{align} $$

(2.5) $$ \begin{align} \zeta(s) & \ll_{C,\delta}|t|^{\frac{1}{2}-\sigma} & \text{for } \sigma\in[-C,-\delta]. \end{align} $$

Lemma 2.8. For $s=\sigma +it=re^{i\theta }$ with $\sigma \le \frac {1}{2}, r\ge 2$ and $\theta \in (0,\pi )$ , we have

$$ \begin{align*} \log |\zeta(s)| &= -(\log r-\log2\pi-1)r\cos\theta + \bigg(\theta-\frac{\pi}{2}\bigg)r\sin\theta + \frac{1}{2}\log r\\ & -\frac{1}{2}\log2\pi +\log|\zeta(1-s)| +\log|1-e^{is\pi}| +O\bigg(\frac{1}{r}\bigg). \end{align*} $$

Proof. By the functional equation, we deduce that

(2.6) $$ \begin{align} \begin{aligned} \log|\zeta(s)| &= r\cos\theta\log2\pi + \frac{\pi}{2}r\sin\theta + \log|\Gamma(1-s)|\\ & +\log|\zeta(1-s)|-\log2\pi+\log|e^{is\pi}-1| \end{aligned} \end{align} $$

for all s except negative even integers in view of the fact that

$$\begin{align*}\bigg|\sin\frac{s\pi}{2}\bigg| = \bigg|\frac{1}{2i}(e^{\frac{is\pi}{2}}-e^{-\frac{is\pi}{2}})\bigg| = \frac{1}{2}e^{\frac{r\pi \sin \theta}{2}}|e^{is\pi}-1|. \end{align*}$$

By Stirling’s formula (see, e.g., [Reference Titchmarch14, pp. 150–151]), we have

$$\begin{align*}\log\Gamma(s) = \bigg(s-\frac{1}{2}\bigg)\log s-s+\frac{1}{2}\log2\pi+O_{\epsilon}\bigg(\frac{1}{|s|}\bigg) \end{align*}$$

uniformly for $-\pi +\epsilon \le \arg s\le +\pi -\epsilon $ and $|s|\ge \epsilon $ with $\epsilon>0$ and $\log s$ is the holomorphic branch of the logarithm with $-\pi <\arg s<+\pi $ . When $\sigma \le \frac {1}{2}$ and $r\ge 2$ , we have $1-\sigma \ge \frac {1}{2}$ and $|1-s|\asymp |s|$ . Thus, we can apply Stirling’s formula to obtain that

$$ \begin{align*} \log\Gamma(1-s) &= \bigg(\frac{1}{2}-s\bigg)\log (1-s) + s-1 + \frac{1}{2}\log2\pi + O\bigg(\frac{1}{|s|}\bigg) \\ &= \bigg(\frac{1}{2}-s\bigg)\log(-s) + s + \frac{1}{2}\log2\pi + O\bigg(\frac{1}{|s|}\bigg). \end{align*} $$

Since $\arg (-s)=\theta -\pi $ when $\theta \in (0,\pi )$ , this gives

$$ \begin{align*} \log|\Gamma(1-s)|&=\Re(\log\Gamma(1-s))\\ &= -r\cos\theta\log r +\frac{1}{2}\log r +r\sin\theta\arg(-s) +r\cos\theta +\frac{1}{2}\log2\pi +O\bigg(\frac{1}{r}\bigg)\\ &= -r\cos\theta\log r +(\theta-\pi)r\sin\theta +\frac{1}{2}\log r +r\cos\theta +\frac{1}{2}\log2\pi +O\bigg(\frac{1}{r}\bigg). \end{align*} $$

Combining this with (2.6) yields the assertion.

Lemma 2.9. For $s=\sigma +it=re^{i\theta }$ with $\theta \in (0,\pi )$ , we have

$$\begin{align*}\bigl|\log|1-e^{is\pi}|\bigr| \le \log(1-e^{-\pi t})^{-1}. \end{align*}$$

Proof. By using the Taylor expansion of $\log (1-z)$ , we have that for $|z|<1$ ,

$$\begin{align*}\bigl|\log|1-z|\bigr| \le |\log(1-z)| \le |z|+\frac{1}{2}|z|^{2}+\frac{1}{3}|z|^{3}+\cdots = \log(1-|z|)^{-1}. \end{align*}$$

Applying this to $z=e^{is\pi }$ with $|z|=e^{-\pi t}$ , we obtain the lemma.

For the proof of Theorem 1.2, we need the following lemmas.

Lemma 2.10 (Borel–Carathéodory theorem)

For a holomorphic function $f(z)$ on the disk $|z-z_{0}|\le R$ and $r\in (0,R)$ , we have

$$\begin{align*}\sup_{|z-z_{0}|\le r}|f(z)| \le \frac{2r}{R-r} \sup_{|z-z_{0}|=R}\Re(f(z)) + \frac{R+r}{R-r}|f(z_{0})|. \end{align*}$$

Lemma 2.11. For a holomorphic function $f(z)$ on $|z-z_{0}|\le R$ such that

(2.7) $$ \begin{align} \sup_{|z-z_{0}|=R}\bigg|\frac{f(z)}{f(z_{0})}\bigg|\le e^{M} \end{align} $$

and $r\in (0,R)$ , we have

$$\begin{align*}0\le \sum_{|\rho-z_{0}|\le r}\log\frac{R}{|\rho-z_{0}|} \le M \quad\text{and}\quad \sum_{|\rho-z_{0}|\le r}1 \le \frac{M}{\log\frac{R}{r}}, \end{align*}$$

where $\rho $ runs over zeros of $f(z)$ counted with multiplicity.

Proof. By Jensen’s formula (see [Reference Titchmarch14, Section 3.6, p. 125]), we have

$$\begin{align*}\sum_{|\rho-z_{0}|\le R}\log\frac{R}{|\rho-z_{0}|} = \frac{1}{2\pi}\int_{0}^{2\pi}\log\bigg|\frac{f(z_{0}+Re^{i\theta})}{f(z_{0})}\bigg|d\theta \end{align*}$$

and so the assertion follows by the assumption (2.7).

The next lemma is essentially the Poisson–Jensen formula combined with the Borel–Carathéodory theorem.

Lemma 2.12. Under the same assumption as Lemma 2.11, we have

$$\begin{align*}\log|f(z)| = - \sum_{|\rho-z_{0}|\le\frac{3}{4}R} \log\frac{R}{|z-\rho|} + \log|f(z_{0})| + O(M) \end{align*}$$

for $|z-z_{0}|\le \frac {1}{2}R$ , where $\rho $ runs over zeros of $f(z)$ counted with multiplicity.

Proof. We may assume $z_{0}=0$ without loss of generality. Let

$$\begin{align*}g(z):= f(z)\prod_{|\rho|\le\frac{3}{4}R}\frac{R^{2}-\overline{\rho}z}{R(z-\rho)}. \end{align*}$$

Then, $g(z)$ is holomorphic and non-zero in the disk $|z|\le \frac {3}{4}R$ . We can thus take a holomorphic logarithm $h(z)=\log (g(z)/g(0))$ in the disk $|z|\le \frac {3}{4}R$ with $h(0)=0$ . Also, on the circle $|z|=R$ , we have

$$\begin{align*}\bigg|\frac{g(z)}{g(0)}\bigg| = \bigg|\frac{f(z)}{f(0)}\bigg| \prod_{|\rho|\le\frac{3}{4}R}\frac{|\rho|}{R} \le e^{M} \end{align*}$$

and so, by the maximum modulus principle, on the circle $|z|=\frac {3}{4}R$ , we have

$$\begin{align*}\Re h(z) = \log\bigg|\frac{g(z)}{g(0)}\bigg| \le M. \end{align*}$$

By applying the Borel–Carathéodory theorem (Lemma 2.10) to $h(z)$ with replacing $R,r$ by $\frac {3}{4}R,\frac {1}{2}R$ , respectively, we get

(2.8) $$ \begin{align} \Biggl|\log\bigg|\frac{g(z)}{g(0)}\bigg|\Biggr| = |\Re(h(z))| \le |h(z)| \le 4\sup_{|z-z_{0}|=\frac{3}{4}R}\Re(h(z)) \le 4M \end{align} $$

in the disk $|z|\le \frac {1}{2}R$ . On the other hand, by the definition of g, we have

$$\begin{align*}\log\bigg|\frac{g(z)}{g(0)}\bigg| = \log\bigg|\frac{f(z)}{f(0)}\bigg| + \sum_{|\rho|\le\frac{3}{4}R} \log\bigg|\frac{R^{2}-\overline{\rho}z}{R(z-\rho)}\bigg| - \sum_{|\rho|\le\frac{3}{4}R} \log\frac{R}{|\rho|}. \end{align*}$$

For $|z|\le \frac {1}{2}R$ , we have that $\frac {5}{8}\le |\frac {R^{2}-\overline {\rho }z}{R^{2}}|\le \frac {11}{8}$ and thus

$$\begin{align*}\log\bigg|\frac{g(z)}{g(0)}\bigg| = \log\bigg|\frac{f(z)}{f(0)}\bigg| + \sum_{|\rho|\le\frac{3}{4}R} \log\frac{R}{|z-\rho|} + O\bigg( \sum_{|\rho|\le\frac{3}{4}R} \log\frac{R}{|\rho|} \bigg). \end{align*}$$

By combining this formula with (2.8), we obtain

$$\begin{align*}\log|f(z)| = - \sum_{|\rho|\le\frac{3}{4}R} \log\frac{R}{|z-\rho|} + \log|f(0)| + O\bigg( M + \sum_{|\rho|\le\frac{3}{4}R} \log\frac{R}{|\rho|} \bigg). \end{align*}$$

By using Lemma 2.11, we obtain the assertion.

Lemma 2.13. Let $a>0$ and $\varphi $ be an even integrable real-valued function on $(-a,+a)$ which is non-increasing on $(0,a)$ . Then, for any Lebesgue measurable set $E\subset (-a,+a)$ with the Lebesgue measure $2b$ , we have

$$\begin{align*}\int_{E}\varphi(x)dx \le \int_{-b}^{+b}\varphi(x)dx. \end{align*}$$

3. Proof of Theorem 1.1

We may assume r is large since otherwise the assertion is trivial. By the definition of $m(r, \zeta )$ and noting that $\overline {\zeta (re^{i\theta })}=\zeta (re^{-i\theta })$ , we can write

(3.1) $$ \begin{align} m(r, \zeta) &= \frac{1}{\pi}\int_{0}^{\pi}\log^{+}|\zeta(re^{i\theta})|d\theta\nonumber\\ &= \frac{1}{\pi} \Biggl( \int_{2<\sigma\leq r} + \int_{1<\sigma\le 2} + \int_{0\le\sigma\le 1} + \int_{-r\le\sigma<0} \Biggr) \nonumber\\ &:= I_1+I_2+I_3+I_4, \end{align} $$

where $s=\sigma +it=re^{i\theta }$ and we used an abuse of the notation for the integrals over the obvious intervals of corresponding $\theta $ with $0\le \theta \le \pi $ .

For the integral $I_{1}$ , by Lemmas 2.1 and 2.4, we have

(3.2) $$ \begin{align} I_{1} \ll \int_{0}^{\arccos\frac{2}{r}}\frac{d\theta}{2^{r\cos\theta}} = \int_{\arcsin\frac{2}{r}}^{\frac{\pi}{2}}\frac{d\theta}{2^{r\sin\theta}} \le \int_{\arcsin\frac{2}{r}}^{\frac{\pi}{2}}\frac{d\theta}{2^{\frac{2}{\pi}r\theta}} \ll \frac{1}{r} \end{align} $$

since $\sin \theta \ge \frac {2}{\pi }\theta $ for $\theta \in [0,\frac {\pi }{2}]$ .

For the integral $I_{2}$ , we change variables via $\sigma =r\cos \theta $ so that $d\theta =-\frac {d\sigma }{r\sin \theta }$ . By Lemma 2.6 and $\sin \theta \gg 1$ in $I_{2}$ , we obtain that

(3.3) $$ \begin{align} I_{2} = \frac{1}{\pi r}\int_{1}^{2}\log^{+}|\zeta(re^{i\theta})|\frac{d\sigma}{\sin\theta} \ll \frac{1}{\pi r} \int_{1}^{2} \bigg(\log\frac{1}{\sigma-1}+1\bigg)d\sigma \ll \frac{1}{r}. \end{align} $$

In the integral $I_{3}$ , we again change variables via $\sigma =r\cos \theta $ . By using $\sin \theta =1+O(\frac {1}{r^{2}})$ for $0\le \sigma \le 1$ and Lemma 2.7, we have

(3.4) $$ \begin{align} I_{3} &= \frac{1}{\pi r} \int_{0}^{1} \log^{+}|\zeta(re^{i\theta})|\frac{d\sigma}{\sin\theta}\nonumber\\ &= \frac{1}{\pi r} \int_{0}^{1} \log^{+}|\zeta(re^{i\theta})|d\sigma + O\bigg(\frac{\log r}{r^{3}}\bigg). \end{align} $$

In this integral, we have

(3.5) $$ \begin{align} re^{i\theta} = \sigma+ir\sin\theta = \sigma+ir+O\bigg(\frac{1}{r}\bigg). \end{align} $$

Let $\delta :=\frac {1}{\log r}$ . By Lemma 2.7, we have $\zeta (s)\ll |t|^{\frac {1}{2}+\delta }\log |t|\ll r^{\frac {1}{2}}\log r$ for $\sigma \in [-\delta , 1+\delta ]$ and $|t-r|\le 1$ . Set

$$\begin{align*}g(s)=\frac{\zeta(\sigma+ir+s)-\zeta(\sigma+ir)}{2M}, \end{align*}$$

where M is the maximum of $\zeta (\sigma +ir+s)$ in $|s|\le \delta $ , which satisfies $M=O(r^{\frac {1}{2}}\log r)$ . Noting that $g(0)=0$ and $|g(s)|\le 1$ in $|s|\le \delta $ , we obtain by the Schwarz lemma (see, e.g., [Reference Titchmarch13, p. 168]) that $|g(s)|\le \frac {|s|}{\delta }=|s|\log r$ for $|s|\le \delta $ , which clearly implies, in view of (3.5), that

$$\begin{align*}\zeta(re^{i\theta}) = \zeta(\sigma+ir) + O\bigg(\frac{(\log r)^{2}}{r^{\frac{1}{2}}}\bigg) \end{align*}$$

in $I_{3}$ . On inserting this formula into (3.4) and using Lemma 2.1, we obtain

(3.6) $$ \begin{align} I_{3} = \frac{1}{\pi r}\int_{0}^{1}\log^{+}|\zeta(\sigma+ir)|d\sigma + O\bigg(\frac{(\log r)^{2}}{r^{\frac{3}{2}}}\bigg). \end{align} $$

Consider finally the integral $I_{4}$ in (3.1) over $-r\le \sigma <0$ , i.e., $\frac {\pi }{2}<\theta \le \pi $ . In order to estimate $I_{4}$ , we use Lemma 2.8. Let

$$\begin{align*}Z(s) := \frac{\zeta(s)}{1-e^{is\pi}} \quad\text{so}\quad \zeta(s)=Z(s)(1-e^{is\pi}). \end{align*}$$

By Lemma 2.2, we then have

$$\begin{align*}I_{4} = \frac{1}{\pi}\int_{\frac{\pi}{2}}^{\pi}\log^{+}|Z(re^{i\theta})|d\theta + O\bigg( \int_{\frac{\pi}{2}}^{\pi}\bigl|\log|1-e^{is\pi}|\bigr|d\theta \bigg). \end{align*}$$

For the error term, by using Lemma 2.9, we deduce that

(3.7) $$ \begin{align} \int_{\frac{\pi}{2}}^{\pi}\bigl|\log|1-e^{is\pi}|\bigr|d\theta &\le \int_{\frac{\pi}{2}}^{\pi} \log(1-e^{-\pi r\sin\theta})^{-1}d\theta\nonumber\\ &= \int_{0}^{\frac{\pi}{2}} \log(1-e^{-\pi r\sin\theta})^{-1}d\theta\nonumber\\ &\le \int_{0}^{\frac{\pi}{2}} \log(1-e^{-2r\theta})^{-1}d\theta\nonumber\\ &\le \frac{1}{2r} \int_{0}^{\infty} \log(1-e^{-\theta})^{-1}d\theta \ll \frac{1}{r} \end{align} $$

since $\sin \theta \ge \frac {2}{\pi }\theta $ for $\theta \in [0,\frac {\pi }{2}]$ . We thus have

(3.8) $$ \begin{align} I_{4} = \frac{1}{\pi}\int_{\frac{\pi}{2}}^{\pi}\log^{+}|Z(re^{i\theta})|d\theta + O\bigg(\frac{1}{r}\bigg). \end{align} $$

For the function $Z(s)$ , Lemma 2.8 gives

(3.9) $$ \begin{align} \begin{aligned} \log |Z(s)| &= -(\log r-\log2\pi-1)r\cos\theta + \bigg(\theta-\frac{\pi}{2}\bigg)r\sin\theta\\ & + \frac{1}{2}\log r -\frac{1}{2}\log2\pi +\log|\zeta(1-s)| +O\bigg(\frac{1}{r}\bigg). \end{aligned} \end{align} $$

When $\theta \in [\frac {\pi }{2},\pi ]$ , we have

$$\begin{align*}-(\log r-\log2\pi-1)r\cos\theta\ge 0 \quad\text{and}\quad \bigg(\theta-\frac{\pi}{2}\bigg)r\sin\theta\ge0 \end{align*}$$

and so, by Lemma 2.7, we have

$$\begin{align*}\log|Z(s)|\ge\frac{1}{2}\log r+O(\log\log r)\ge 0. \end{align*}$$

Thus, (3.8) becomes

(3.10) $$ \begin{align} I_{4} = \frac{1}{\pi}\int_{\frac{\pi}{2}}^{\pi}\log|Z(re^{i\theta})|d\theta + O\bigg(\frac{1}{r}\bigg). \end{align} $$

We then integrate the right-hand side of (3.9) termwise to obtain that

(3.11) $$ \begin{align} \begin{aligned} \frac{1}{\pi} \int_{\frac{\pi}{2}}^{\pi} (-(\log r-\log2\pi-1)r\cos\theta)d\theta &= \frac{1}{\pi}r\log r - \frac{1}{\pi}(1+\log 2\pi)r,\\ \frac{1}{\pi} \int_{\frac{\pi}{2}}^{\pi} \bigg(\theta-\frac{\pi}{2}\bigg)r\sin\theta d\theta &=\frac{1}{\pi}\bigg(\frac{\pi}{2}-1\bigg)r,\\ \frac{1}{\pi} \int_{\frac{\pi}{2}}^{\pi} \bigg(\frac{1}{2}\log r-\frac{1}{2}\log2\pi\bigg)d\theta &= \frac{1}{4}\log r-\frac{1}{4}\log2\pi. \end{aligned} \end{align} $$

By arguing similarly to $I_{1},I_{2}$ , we also deduce that

(3.12) $$ \begin{align} \int_{\frac{\pi}{2}}^{\pi}\log|\zeta(1-s)|d\theta \ll \int_{-r\le\sigma<-2} + \int_{-2\le\sigma<0} \ll \frac{1}{r}. \end{align} $$

Combining (3.9), (3.10), (3.11), and (3.12), we thus obtain that

(3.13) $$ \begin{align} \begin{aligned} I_{4} &= \frac{1}{\pi}r\log r -\bigg(2+\log2\pi-\frac{\pi}{2}\bigg)\frac{r}{\pi}\\ & + \frac{1}{4}\log r - \frac{1}{4}\log2\pi + O\bigg(\frac{1}{r}\bigg). \end{aligned} \end{align} $$

By substituting (3.2), (3.3), (3.6), and (3.13) into (3.1), we obtain

(3.14) $$ \begin{align} \begin{aligned} m(r, \zeta) &= \frac{1}{\pi}r\log r - \bigg(2+\log2\pi-\frac{\pi}{2}\bigg)\frac{r}{\pi} + \frac{1}{4}\log r - \frac{1}{4}\log2\pi\\ & + \frac{1}{\pi r}\int_{0}^{1}\log^{+}|\zeta(\sigma+ir)|d\sigma + O\bigg(\frac{1}{r}\bigg). \end{aligned} \end{align} $$

Recall that $T(r,\zeta )=m(r,\zeta )+N(r,\zeta )$ . By Lemma 2.3 and using Lemma 2.7 to bound the integral in (3.14), we obtain (1.4). This completes the proof.

4. Proof of Theorem 1.2

By Jensen’s formula (see, e.g., [Reference Goldberg and Ostrovskii2, p. 13]) and Lemma 2.3, we have

$$\begin{align*}N\bigg(r,\frac{1}{\zeta-p}\bigg) = \frac{1}{2\pi} \int_{0}^{2\pi} \log|\zeta-p|(re^{i\theta})d\theta + \log r + \log\frac{1}{|c|}, \end{align*}$$

where c is the first nonzero coefficient in the Taylor expansion of $\zeta -p$ at $s=0$ . We thus only need to estimate the integral

$$\begin{align*}I = I(p):= \frac{1}{2\pi}\int_{0}^{\pi} \log|\zeta-p|(re^{i\theta}) d\theta \end{align*}$$

since the integral over $\pi $ to $2\pi $ is reduced to the integral over $0$ to $\pi $ in view of the fact that $(\zeta -p)(re^{-i\theta })=\overline {\zeta (re^{i\theta })-\overline {p}(re^{i\theta })}$ (cf. (3.1)), that is,

(4.1) $$ \begin{align} N\bigg(r,\frac{1}{\zeta-p}\bigg) = I(p)+I(\overline{p})+\log r+\log\frac{1}{|c|}. \end{align} $$

We shall calculate $I=I(p)$ . Let

$$\begin{align*}\sigma_{\pm}:= -\bigg(d-\frac{1}{2}\bigg)\pm\frac{B}{\log r}, \end{align*}$$

where $B\ge 0$ will be chosen later. We then decompose the integral I as

(4.2) $$ \begin{align} I &= \frac{1}{2\pi} \Biggl( \int_{2<\sigma\leq r} + \int_{\sigma_{+}<\sigma\le2} + \int_{\sigma_{-}<\sigma\le\sigma_{+}} + \int_{-r\le\sigma\le\sigma_{-}} \Biggr)\nonumber\\ &:= I_1+I_2+I_3+I_{4}. \end{align} $$

We assume that r is large in terms of B and so our choice of B will be independent of r.

We first estimate $I_{1}$ . For $\sigma \ge 2$ , we have $|\zeta (s)|\le \zeta (2)$ . We also have

(4.3) $$ \begin{align} |p(re^{i\theta})|=|a_d|r^{d}\bigg(1+O\bigg(\frac{1}{r}\bigg)\bigg), \end{align} $$

where $a_d$ is the leading coefficient of the polynomial p. In this proof, all implicit constants may depend on p. Hence, we have that

$$\begin{align*}\log|\zeta-p|(re^{i\theta}) = \log|p(re^{i\theta})| + \log\bigg|1-\frac{\zeta}{p}\bigg|(re^{i\theta}) = \log|p(re^{i\theta})|+O\bigg(\frac{1}{r}\bigg) \end{align*}$$

for $2<\sigma \le r$ and thus that

(4.4) $$ \begin{align} I_{1} = \frac{1}{2\pi}\int_{2<\sigma\le r}\log|p(re^{i\theta})|d\theta + O\bigg(\frac{1}{r}\bigg). \end{align} $$

We then estimate $I_{2}$ . By Lemma 2.7, we have

$$\begin{align*}\zeta(s) \ll |t|^{\max(\frac{1}{2}-\sigma,\frac{1}{2})}\log|t| \ll r^{\frac{3}{4}}\log r \ll r^{d}e^{-B} \quad\text{for } -\tfrac{1}{4}<\sigma\le2 \end{align*}$$

and

$$\begin{align*}\zeta(s) \ll |t|^{\frac{1}{2}-\sigma} \ll r^{d}e^{-B} \quad\text{for }\sigma_{+}<\sigma\le-\tfrac{1}{4} \end{align*}$$

if r is large in terms of B (but the implicit constants here are independent of B). Thus, by taking B large enough, we get

$$\begin{align*}|\zeta(s)|\le\frac{1}{2}|p(s)| \quad\text{and so}\quad \log|\zeta-p|(re^{i\theta}) = \log|p(re^{i\theta})| + O(1) \end{align*}$$

for large r. Note also that the argument $\theta $ of s on the arc for the integral over $\sigma _{+}<\sigma \le 2$ varies by a quantity $O(\frac {1}{r})$ . Thus, we deduce that

(4.5) $$ \begin{align} I_{2} = \frac{1}{2\pi} \int_{\sigma_{+}<\sigma\le 2} \log|p(re^{i\theta})|d\theta + O\bigg(\frac{1}{r}\bigg) \end{align} $$

for large r.

We next estimate $I_{3}$ . By changing the variable via $\sigma =r\cos \theta $ , we have

(4.6) $$ \begin{align} I_{3} = \frac{1}{2\pi r} \int_{\sigma_{-}}^{\sigma_{+}} \log|\zeta-p|(re^{i\theta})\frac{d\sigma}{\sin\theta}. \end{align} $$

Let $\phi _{-}$ be the acute angle satisfying

$$\begin{align*}r\sin\phi_{-} = d-\frac{1}{2}+\frac{B}{\log r} = -\sigma_{-}. \end{align*}$$

Then, $\sin \theta \ge \sin (\frac {\pi }{2}+\phi _{-})=\cos \phi _{-}\gg 1$ in $I_{3}$ . We use Lemma 2.12 with

$$\begin{align*}f(z):=(\zeta-p)(z),\quad z_{0}:= 2+ir,\quad R:= 2(d+3). \end{align*}$$

For the bound (2.7), by Lemma 2.7, we can obtain

(4.7) $$ \begin{align} \begin{aligned} \frac{f(z)}{f(z_{0})} &= \frac{(\zeta-p)(z)}{(\zeta-p)(2+ir)}\\ &\ll \frac{r^{2d+\frac{9}{2}}}{|\zeta(2+ir)-p(2+ir)|} \ll \frac{r^{2d+\frac{9}{2}}}{|p(2+ir)|-\zeta(2)} \ll r^{d+\frac{9}{2}} \end{aligned} \end{align} $$

for $|z-z_{0}|\le R$ and large r. We can thus take $M:=(d+\frac {9}{2})\log r$ . Since now $\log R=O(1)$ , Lemmas 2.11 and 2.12 then gives

(4.8) $$ \begin{align} \log|\zeta-p|(z) = \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} \log|z-\rho| + O(\log r) \end{align} $$

for $|z-(2+ir)|\le d+3$ . For large r, we have $\sigma \in [-d,-(d-1)]$ and $t\in [r-1,r]$ in the integral $I_{3}$ and so the range of integration is covered by the disk $|s-(2+ir)|\le d+3$ . We can thus use (4.8), $\sigma _{+}-\sigma _{-}\ll _{B}\frac {1}{\log r}$ and $\sin \theta \gg 1$ in (4.6) to get

(4.9) $$ \begin{align} I_{3} = \frac{1}{2\pi r} \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} \int_{\sigma_{-}}^{\sigma_{+}} \log|re^{i\theta}-\rho| \frac{d\sigma}{\sin\theta} + O_{B}\bigg(\frac{1}{r}\bigg), \end{align} $$

where the subscript B of the Landau symbol expresses that the implicit constant here depends on B. Since $|re^{i\theta }-\rho |\ll 1$ in (4.9), by using Lemma 2.11 with (4.7), we have

(4.10) $$ \begin{align} I_{3} \le O_{B}\bigg( \frac{1}{r\log r\cos\phi_{-}} \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} 1 + \frac{1}{r} \bigg) = O_{B}\bigg(\frac{1}{r}\bigg). \end{align} $$

For the lower bound, note that in (4.9), we have

$$\begin{align*}\log|re^{i\theta}-\rho| = -\log\frac{1}{|re^{i\theta}-\rho|} \ge -\log\frac{1}{|\sigma-\beta|} \ge -\log^{+}\frac{1}{|\sigma-\beta|}, \end{align*}$$

where we write $\beta :=\Re \rho $ . Thus, by changing the variable, we obtain

(4.11) $$ \begin{align} I_{3} &\ge - \frac{1}{2\pi r\cos\phi_{-}} \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} \int_{\sigma_{-}}^{\sigma_{+}} \log^{+}\frac{1}{|\sigma-\beta|}d\sigma + O_{B}\bigg(\frac{1}{r}\bigg)\nonumber\\ &= - \frac{1}{2\pi r\cos\phi_{-}} \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} \int_{\sigma_{-}-\beta}^{\sigma_{+}-\beta} \log^{+}\frac{1}{|\sigma|}d\sigma + O_{B}\bigg(\frac{1}{r}\bigg). \end{align} $$

Note that the function $\log ^{+}\frac {1}{|\sigma |}$ is integrable over $(-a,+a)$ for any $a>0$ and non-increasing for $(0,a)$ . Also, the Lebesgue measure of the interval $[\sigma _{-}-\beta ,\sigma _{+}-\beta ]$ is $2B/\log r$ . Thus, by using Lemma 2.11 with (4.7) and Lemma 2.13 in (4.11), we get

(4.12) $$ \begin{align} I_{3} &\ge - \frac{1}{2\pi r\cos\phi_{-}} \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} \int_{-\frac{B}{\log r}}^{+\frac{B}{\log r}} \log^{+}\frac{1}{|\sigma|}d\sigma + O_{B}\bigg(\frac{1}{r}\bigg)\nonumber\\ &= O_{B}\bigg(\frac{\log\log r}{r\log r} \sum_{|\rho-(2+ir)|\le\frac{3}{4}R} 1 + \frac{1}{r} \bigg) = O_{B}\bigg(\frac{\log\log r}{r}\bigg). \end{align} $$

By (4.10) and (4.12), we obtain

(4.13) $$ \begin{align} I_{3} \ll_{B} \frac{\log\log r}{r}. \end{align} $$

We finally consider $I_{4}$ . We take acute angles $\psi ,\eta $ satisfying

$$ \begin{gather*} r\sin\psi = d+\frac{5}{2},\quad r\sin\eta = \frac{1}{r(\log r)^{2}}, \end{gather*} $$

respectively. Then, $\frac {\pi }{2}+\phi _{-}<\frac {\pi }{2}+\psi <\pi -\eta $ for large r and the range of the angle $\theta $ in the integral $I_{4}$ is $[\frac {\pi }{2}+\phi _{-},\pi ]$ . We then decompose $I_{4}$ as

(4.14) $$ \begin{align} I_{4} = I_{41}+I_{42}+I_{43}+I_{44}, \end{align} $$

where

$$ \begin{align*} I_{41} &:= \frac{1}{2\pi} \int_{\frac{\pi}{2}+\phi_{-}}^{\pi-\eta} \log|\zeta(re^{i\theta})|d\theta,\\ I_{42} &:= \frac{1}{2\pi} \int_{\frac{\pi}{2}+\phi_{-}}^{\frac{\pi}{2}+\psi} \log\bigg|1-\frac{p}{\zeta}\bigg|(re^{i\theta})d\theta,\\ I_{43} &:= \frac{1}{2\pi} \int_{\frac{\pi}{2}+\psi}^{\pi-\eta} \log\bigg|1-\frac{p}{\zeta}\bigg|(re^{i\theta})d\theta,\\ I_{44} &:= \frac{1}{2\pi} \int_{\pi-\eta}^{\pi} \log|\zeta-p|(re^{i\theta})d\theta. \end{align*} $$

For $I_{41}$ , we integrate the right-hand side of Lemma 2.8 termwise. We can use (3.7), (3.11), and (3.12) obtained in the proof of Theorem 1.1 together with

$$ \begin{align*} &\frac{1}{2\pi} \int_{\frac{\pi}{2}}^{\frac{\pi}{2}+\phi_{-}} (-(\log r-\log2\pi-1)r\cos\theta)d\theta\\ &= \frac{1}{4\pi}\phi_{-}^{2}(\log r-\log2\pi-1)r+O_{B}(\phi_{-}^{4}r\log r)\\ &= \frac{1}{4\pi}\bigg(d-\frac{1}{2}\bigg)^{2}\frac{\log r}{r} +O_{B}\bigg(\frac{1}{r}\bigg) \end{align*} $$

and

$$ \begin{gather*} \frac{1}{2\pi} \int_{\frac{\pi}{2}}^{\frac{\pi}{2}+\phi_{-}} \bigg(\theta-\frac{\pi}{2}\bigg)r\sin\theta d\theta \ll_{B} \frac{1}{r},\\ \frac{1}{2\pi} \int_{\frac{\pi}{2}}^{\frac{\pi}{2}+\phi_{-}} \bigg(\frac{1}{2}\log r-\frac{1}{2}\log2\pi\bigg)d\theta = \frac{1}{4\pi}\bigg(d-\frac{1}{2}\bigg) \frac{\log r}{r}+O_{B}\bigg(\frac{1}{r}\bigg),\\ r\log r\int_{\pi-\eta}^{\pi}d\theta \ll \frac{1}{r} \end{gather*} $$

to obtain

(4.15) $$ \begin{align} \begin{aligned} I_{41} &= \frac{1}{2\pi}r\log r - \bigg(2+\log2\pi-\frac{\pi}{2}\bigg)\frac{r}{2\pi} \\ & + \frac{1}{8}\log r - \frac{1}{8}\log2\pi - \frac{1}{4\pi}\bigg(d^{2}-\frac{1}{4}\bigg)\frac{\log r}{r} +O_{B}\bigg(\frac{1}{r}\bigg). \end{aligned} \end{align} $$

We then estimate $I_{42}$ . In the integral $I_{42}$ , since $r\sin \theta \ge r\sin (\frac {\pi }{2}+\psi )\ge 1$ for large r, Lemma 2.9 gives $\bigl |\log |1-e^{is\pi }|\bigr |\le \log (1-e^{-\pi })^{-1}$ . Also, we have $r\cos \theta \le \sigma _{-}$ and $(\theta -\frac {\pi }{2})\sin \theta \ge 0$ . Thus, by (2.1) and Lemma 2.8, we have

(4.16) $$ \begin{align} \log|\zeta(s)| &\ge -(\log r-\log2\pi-1)\sigma_{-}+\frac{1}{2}\log r+O(1)\nonumber\\ &\ge d\log r+B+O\bigg(1+\frac{B}{\log r}\bigg) \ge \log|p(s)|+1 \end{align} $$

by taking B sufficiently large if r is large in terms of B. We thus obtain

(4.17) $$ \begin{align} I_{42} = \frac{1}{2\pi} \int_{\frac{\pi}{2}+\phi_{-}}^{\frac{\pi}{2}+\psi} \log\bigg|1-\frac{p}{\zeta}\bigg|(re^{i\theta})d\theta \ll \psi-\phi_{-} \ll \frac{1}{r}. \end{align} $$

In the integral $I_{43}$ , Lemma 2.9 and $r\sin \theta \ge r\sin (\pi -\eta )=\frac {1}{r(\log r)^{2}}$ give

$$\begin{align*}\bigl|\log|1-e^{is\pi}|\bigr|\le\log(1-e^{-\frac{\pi}{r(\log r)^{2}}})^{-1}=\log r+O(\log\log r). \end{align*}$$

Thus, similarly to (4.16), we get

$$ \begin{align*} \log|\zeta(s)| &\ge (\log r-\log2\pi-1)\bigg(d+\frac{5}{2}\bigg)-\frac{1}{2}\log r+O(\log\log r)\\ &\ge (d+2)\log r+O(\log\log r) \ge \log|p(s)|+\log r \end{align*} $$

and so

(4.18) $$ \begin{align} I_{43} = \frac{1}{2\pi} \int_{\frac{\pi}{2}+\psi}^{\pi-\eta} \log\bigg|1-\frac{p}{\zeta}\bigg|(re^{i\theta})d\theta \ll \frac{1}{2\pi} \int_{\frac{\pi}{2}+\psi}^{\pi-\eta} \bigg|\frac{p}{\zeta}\bigg|(re^{i\theta})d\theta \ll \frac{1}{r}. \end{align} $$

For $I_{44}$ , by changing the variable via $r\sin \theta =t$ so $d\theta =\frac {dt}{r\cos \theta }$ , we have

(4.19) $$ \begin{align} I_{44} = \frac{1}{2\pi r} \int_{0}^{\frac{1}{r(\log r)^{2}}} \log|\zeta-p|(re^{i\theta})\frac{dt}{|\cos\theta|}. \end{align} $$

Take $m\in \mathbb {N}$ by $2m-1\le r\le 2m+1$ . We then use Lemma 2.12 with

$$\begin{align*}f(z):=(\zeta-p)(z),\quad z_{0}:= -2m,\quad R:= 4. \end{align*}$$

For the bound (2.7), since Lemma 2.8 give

$$\begin{align*}\log|\zeta(z)| \le r\log r+\log(1+e^{-\pi t}) + O(r) = O(r\log r) \end{align*}$$

for $|z-z_{0}|\le R$ , by using $\zeta (-2m)=0$ , we can obtain

(4.20) $$ \begin{align} \bigg|\frac{f(z)}{f(z_{0})}\bigg| = \bigg|\frac{(\zeta-p)(z)}{p(-2m)}\bigg| \le e^{O(r\log r)} \end{align} $$

for $|z-z_{0}|\le R$ and large r. We can thus take $M=O(r\log r)$ . Since now $\log R=O(1)$ , Lemmas 2.11 and 2.12 then gives

(4.21) $$ \begin{align} \log|\zeta-p|(z) = \sum_{|\rho+2m|\le3} \log|z-\rho| + O(r\log r) \end{align} $$

for $|z+2m|\le 2$ . For large r, we have $\sigma \in [-2m-\frac {3}{2},-2m+\frac {3}{2}]$ and $t\in [0,\frac {1}{2}]$ in the integral $I_{44}$ and so the range of integration is covered by the disk $|s+2m|\le 2$ . We can thus insert (4.21) into (4.19) to get

(4.22) $$ \begin{align} I_{44} = \frac{1}{2\pi r} \sum_{|\rho+2m|\le 3} \int_{0}^{\frac{1}{r(\log r)^{2}}} \log|re^{i\theta}-\rho|\frac{dt}{|\cos\theta|} + O\bigg(\frac{1}{r}\bigg). \end{align} $$

Since $|re^{i\theta }-\rho |\ll 1$ in (4.22), by using Lemma 2.11 with (4.20) and the bound $|\cos \theta |\ge \cos \eta \gg 1$ , we have

(4.23) $$ \begin{align} I_{44} \le O\bigg( \frac{1}{(r\log r)^{2}} \sum_{|\rho+2m|\le3} 1 + \frac{1}{r} \bigg) = O\bigg(\frac{1}{r}\bigg). \end{align} $$

For the lower bound, note that in (4.22), we have

$$\begin{align*}\log|re^{i\theta}-\rho| = -\log\frac{1}{|re^{i\theta}-\rho|} \ge -\log\frac{1}{|t-\gamma|} \ge -\log^{+}\frac{1}{|t-\gamma|}, \end{align*}$$

where we write $\gamma :=\Im \rho $ , and so, by changing the variable, we obtain

(4.24) $$ \begin{align} I_{44} &\ge - \frac{1}{2\pi r\cos\eta} \sum_{|\rho+2m|\le3} \int_{0}^{\frac{1}{r(\log r)^{2}}} \log^{+}\frac{1}{|t-\gamma|}dt + O\bigg(\frac{1}{r}\bigg)\nonumber\\ &= - \frac{1}{2\pi r\cos\eta} \sum_{|\rho+2m|\le3} \int_{-\gamma}^{\frac{1}{r(\log r)^{2}}-\gamma} \log^{+}\frac{1}{|t|}dt + O\bigg(\frac{1}{r}\bigg). \end{align} $$

Note that the function $\log ^{+}\frac {1}{|t|}$ is integrable over $(-a,+a)$ for any $a>0$ and non-increasing for $(0,a)$ . Thus, by using Lemma 2.11 with (4.20), Lemma 2.13, and $\cos \eta \gg 1$ in (4.24), we get

(4.25) $$ \begin{align} I_{44} &\ge - \frac{1}{2\pi r\cos\eta} \sum_{|\rho+2m|\le3} \int_{-\frac{1}{2r(\log r)^{2}}}^{+\frac{1}{2r(\log r)^{2}}} \log^{+}\frac{1}{|t|}dt + O\bigg(\frac{1}{r}\bigg)\nonumber\\ &= O\bigg( \frac{1}{r^{2}\log r} \sum_{|\rho+2m|\le3} 1 + \frac{1}{r} \bigg) = O\bigg(\frac{1}{r}\bigg). \end{align} $$

By (4.23) and (4.25), we obtain

(4.26) $$ \begin{align} I_{44} \ll \frac{1}{r}. \end{align} $$

On inserting (4.15), (4.17), (4.18), and (4.26) into (4.14), we get

(4.27) $$ \begin{align} \begin{aligned} I_{4} &= \frac{1}{2\pi}r\log r - \bigg(2+\log2\pi-\frac{\pi}{2}\bigg)\frac{r}{2\pi}\\ & + \frac{1}{8}\log r - \frac{1}{8}\log2\pi - \frac{1}{4\pi}\bigg(d^{2}-\frac{1}{4}\bigg)\frac{\log r}{r} +O_{B}\bigg(\frac{1}{r}\bigg). \end{aligned} \end{align} $$

Finally, on inserting (4.4), (4.5), (4.13), and (4.27) into (4.2) and fixing a large B, we obtain

$$ \begin{align*} I(p) &= \frac{1}{2\pi}r\log r - \bigg(2+\log2\pi-\frac{\pi}{2}\bigg)\frac{r}{2\pi} \\ & + \frac{1}{8}\log r - \frac{1}{8}\log2\pi - \frac{1}{4\pi}\bigg(d^{2}-\frac{1}{4}\bigg)\frac{\log r}{r}\\ & + \frac{1}{2\pi} \int_{\sigma_{+}\leq\sigma\le r} \log|p(re^{i\theta})|d\theta + O\bigg(\frac{\log\log r}{r}\bigg). \end{align*} $$

For the last integral, by using (4.3), we have

$$ \begin{align*} &\frac{1}{2\pi} \int_{\sigma_{+}\leq\sigma\le r} \log|p(re^{i\theta})|d\theta\\ &= \frac{1}{2\pi} \int_{\sigma_{+}\leq\sigma\le r} (d\log r+\log|a_{d}|)d\theta + O\bigg(\frac{1}{r}\bigg)\\ &= \frac{1}{2\pi} \bigg(\frac{\pi}{2} + \arcsin\bigg(\frac{d-\frac{1}{2}}{r}-\frac{B}{r\log r}\bigg) \bigg) d\log r + \frac{1}{4}\log|a_{d}| + O\bigg(\frac{1}{r}\bigg)\\ &= \frac{d}{4}\log r + \frac{1}{4}\log|a_{d}| + \frac{1}{2\pi}d\bigg(d-\frac{1}{2}\bigg)\frac{\log r}{r} + O\bigg(\frac{1}{r}\bigg). \end{align*} $$

Thus, we obtain

$$ \begin{align*} I(p) &= \frac{1}{2\pi}r\log r - \bigg(2+\log2\pi-\frac{\pi}{2}\bigg)\frac{r}{2\pi} + \bigg(\frac{d}{4}+\frac{1}{8}\bigg)\log r\\ & - \frac{1}{8}\log2\pi + \frac{1}{4}\log|a_{d}| + \frac{1}{4\pi}\bigg(d-\frac{1}{2}\bigg)^{2}\frac{\log r}{r} + O\bigg(\frac{\log\log r}{r}\bigg). \end{align*} $$

Finally, combining this with (4.1), we arrive at the conclusion.

5. Remarks on the distribution of fixed points

As mentioned in §1, the distribution of fixed points of $\zeta $ is different from that of zeros of $\zeta $ , for which all the nontrivial zeros are located only inside the critical strip and conjectured to lie on the critical line $\sigma =\frac {1}{2}$ by the Riemann hypothesis. We include here a further remark on the distribution of fixed points of $\zeta $ or, more generally, zeros of $\zeta -p$ by comparing the size of $\zeta $ and the polynomial p. From the estimates on $\zeta -p$ used in the proof of Theorem 1.2, (see the arguments for $I_{3}$ and $I_{44}$ in the proof of Theorem 1.2), one can verify that: All zeros of $\zeta -p$ , except those bounded by some constant depending on p, lie only in a shrinking vertical strip

$$\begin{align*}-\bigg(d-\frac{1}{2}\bigg)-\frac{B}{\log(|t|+4)} \le \sigma \le -\bigg(d-\frac{1}{2}\bigg)+\frac{B}{\log(|t|+4)} \end{align*}$$

for some constant $B>0$ depending on p, which has the vertical axis

$$\begin{align*}\sigma=-\bigg(d-\frac{1}{2}\bigg) \end{align*}$$

as its asymptote and in a shrinking horizontal strip

$$\begin{align*}|t|\le\frac{1}{(|\sigma|+1)(\log(|\sigma|+1))^{2}} \quad\text{with}\quad \sigma\le 0 \end{align*}$$

or, more strongly,

(5.1) $$ \begin{align} |t|\le\exp(-(1+o(1))|\sigma|\log|\sigma|)\quad(\sigma\to-\infty) \quad\text{with}\quad \sigma\le 0 \end{align} $$

which has the real axis as its asymptote. This phenomenon for fixed points of $\zeta $ contrasts with the case for zeros of $\zeta $ , since it is unknown if the nontrivial zeros of $\zeta $ are even asymptotic to the critical line $\sigma =\frac {1}{2}$ . Note also that with a standard application of Rouché’s theorem, one can show an even stronger claim on the zeros of $\zeta -p$ in the above horizontal strip: All zeros of $\zeta -p$ in the range (5.1), except those bounded by some constant depending on p, are simple and associated with each of the trivial zeros $s=-2m$ of $\zeta $ . Clearly, if p is real (e.g., in the case $p(s)=s$ of fixed points), then all zeros of $\zeta -p$ , with the exceptions mentioned above, are real.