In a now classical paper, Iwaniec proved the following theorem.

Theorem. Let $T \geq 2$, $T^{1/2} < T_0 \leq T$ and $T \leq t_1 < t_2 < \cdots < t_R \leq 2T$, $t_{r+1} - t_r \geq T_0$. Then $$ \tag{1} \sum_{r=1}^R \int_{t_r}^{t_r+T_0} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt \ll \left( R T_0 + R^{1/2}T_0^{-1/2} T\right) T^\varepsilon. $$ Immediately after the statement, Iwaniec says that this leads to the estimate $$ \tag{2} \int_{T}^{2T} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{12} dt \ll T^{2+\varepsilon}, $$ but gives no details of this calculation. Thus my question: how does (2) follow from (1)?


1 Answer 1


I wrestled with this question for an annoyingly long time. Now understanding things much better, I'm a bit embarrassed that this took me so long to figure out (such is the process of learning though). I was quite frustrated that I couldn't find the details of this anywhere, aside from some helpful comments in this paper of Ivić. To spare others my frustration, I'm posting the answer in a Q&A style, as well as for my own reference.

It suffices to show that if $V > 0$ and $t_1,\ldots, t_L \in [T,2T]$ with $t_{l+1}-t_l \geq 2$ and $|\zeta(\frac{1}{2}+it_l)| \geq V$, then $$ L \ll T^{2+\varepsilon}V^{-12}, $$ for then the estimate for the twelfth moment follows by the argument originally given by Heath-Brown in his paper first proving the estimate (2). Note that we may also assume that $V > T^{1/8+\varepsilon/2}$, since the contribution to the integral (2) of the $t$ with $|\zeta(\frac{1}{2}+it)| \leq T^{1/8+\varepsilon/2}$ is $$ \leq T^{1+4\varepsilon} \int_{T}^{2T} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{4} dt \ll T^{2+5\varepsilon} $$ by the standard estimate for the fourth moment of $\zeta$. In particular, we may assume $V^4 > T^{1/2+2\varepsilon}$.

Cover the interval $[T,2T]$ by $R$ disjoint intervals $I_1,\ldots,I_{R}$ of length $V^4T^{-2\varepsilon}$ ($> T^{1/2}$) such that each point $t_l$ lies in some interval $I_r$. Then $$ R \leq L \leq \sum_{r=1}^R \sum_{t_l \in I_r} \frac{\left|\zeta\left(\tfrac{1}{2}+it_l\right) \right|^4}{V^4}. $$ At this point, we need that for $t \in [T,2T]$, $$ \tag{3} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{2k} \ll \log T \int_{t-1/3}^{t+1/3} \left|\zeta\left(\tfrac{1}{2}+iu\right) \right|^{2k} du, $$ which can be found in the paper of Ivić linked above. Since the points $t_l$ are spaced by at least 2, $$ \sum_{t_l \in I_r} \left|\zeta\left(\tfrac{1}{2}+it_l\right) \right|^4 \ll \log T \int_{I_r'} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt, $$ where $I_r'$ is the interval $I_r$ enlarged by $1/3$ on either end. Applying the estimate (1), we find that \begin{align*} \sum_{r=1}^R \log T \int_{I_r'} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt &\ll \left(RV^4 T^{-2\varepsilon} + R^{1/2}V^{-2}T^{1+\varepsilon} \right)T^{\varepsilon} \\ &\ll R T^{-\varepsilon} V^4 + R^{1/2}T^{1+2\varepsilon}V^{-2}. \end{align*} and so in particular $$ \tag{4} R \leq L \ll RT^{-\varepsilon} + R^{1/2}T^{1+2\varepsilon}V^{-6} $$ Since $RT^{-\varepsilon} < cR$ for any constant $c > 0$ and $T$ sufficiently large, we must have $$ R\ll R^{1/2}T^{1+2\varepsilon}V^{-6}. $$ Thus $$ R \ll T^{2+4\varepsilon} V^{-12}, $$ and the required bound for $L$ now follows from (4).

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    $\begingroup$ Good question, and good answer! $\endgroup$
    – GH from MO
    Commented Mar 20, 2023 at 22:40
  • 2
    $\begingroup$ Yes, there is some serious irony in the point that one usually feels "slow"... In my observation, the only people who feel otherwise are delusional in one way or another. :) $\endgroup$ Commented May 30, 2023 at 17:48
  • $\begingroup$ @SamHopkins Thanks for pointing that out. Twas an instance of several papers running together in my mind. The error has been fixed. $\endgroup$ Commented Jun 2, 2023 at 4:11
  • $\begingroup$ A sketch of this result is given in Section 6 of the following paper of Jutila: doi.org/10.4064/aa-57-2-93-114 $\endgroup$ Commented Sep 15, 2023 at 16:51

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