# Large values of $\zeta(1/2+it)$ from sums of short moments

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)?

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).