In a now [classical paper](https://zbmath.org/?q=an:0438.10030), 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)?**



Large values of $\zeta(1/2+it)$ from sums of short moments
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](https://arxiv.org/abs/math/0512016) 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).


