I'm reading the proof of Hardy and Littlewood's theorem in the book Analytic Number Theory, written by Henryk Iwaniec and Emmanuel Kowalski (p. 547):
Theorem (Hardy and Littlewood): Let $N_0(T)$ be the number of non-trivial zeros of the Riemann zeta function $s=\sigma+i t$, with $0<t\leq T$. Then $$N_0(T)\gg T. $$
The idea of the proof is clear and is based in two lemmas. Define $$f(u)=\frac{g(1/2+iu)}{|g(1/2+iu)|}\zeta(1/2+iu),$$ and $$I(t)=\int_t^{t+\Delta}f(u)\,du,$$ where $\Delta>0$ a fixed number and $$g(s)=\pi^{-s/2}\Gamma(s/2).$$ The book asserts:
Lemma: Let $\Delta\geq 1$. Then, if $T\geq \Delta^6$ $$\int_T^{2T}|I(t)|^2\, dt\ll \Delta T,$$ where the implied constant is absolute.
In certain point (the beginning) the authors say $$\begin{align*} I&=\int_0^\Delta\int_0^\Delta\int_T^{2T}f(t+u_1)\overline{f}(t+u_2)\,dt\,du_1\,du_2\\ &=\int_0^\Delta\int_0^\Delta\int_T^{2T}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2+O(\Delta^3 T^{1/2}).\\ \end{align*}$$ This step is not clear for me. The book say this follows by use of the convexity bound $|\zeta(1/2+it)|\ll |t|^{1/4}$, and the expressions suggest the change $t+u_1\to t$ which make appear the integrand but change the limits of integration. What I got $$\begin{align*} I&=\int_0^\Delta\int_0^\Delta\int_T^{2T}f(t+u_1)\overline{f}(t+u_2)\,dt\,du_1\,du_2\\ &=\int_0^\Delta\int_0^\Delta\int_{T+u_1}^{2T+u_1}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2.\\ &=\int_0^\Delta\int_0^\Delta\int_{T}^{2T}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\\ &\quad\quad+\left(\int_0^\Delta\int_0^\Delta\int_{T+u_1}^{2T+u_1}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\right.\\ &\quad\quad\left.-\int_0^\Delta\int_0^\Delta\int_{T}^{2T}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\right)\\ &=:M+E. \end{align*}$$ Hence I need to prove $$\begin{align*} E&=\int_0^\Delta\int_0^\Delta\int_{T+u_1}^{2T+u_1}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\\ &\quad\quad-\int_0^\Delta\int_0^\Delta\int_{T}^{2T}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\ll \Delta^3 T^{1/2}. \end{align*}$$
Question: How to prove this estimate using the Convexity bound?
It can be possible that this is not the approach, the thing is that I have time trying but didn't get the desired result. Thanks in advance.
Finally I got a solution. It's a very simple calculation. Write $E$ as \begin{align*} E&=\int_0^\Delta\int_0^\Delta\int_{T}^{T+u_1}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\\ &\quad\quad+\int_0^\Delta\int_0^\Delta\int_{2T}^{2T+u_1}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2=:E_1+E_2. \end{align*} We will bound each one by separate. \begin{align*} E_1&=\int_0^\Delta\int_0^\Delta\int_{T}^{T+u_1}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2\\ &\leq \int_0^\Delta\int_0^\Delta\int_{T}^{T+\Delta}f(t)\overline{f}(t+u_2-u_1)\,dt\,du_1\,du_2,\\ \end{align*} by convexity \begin{align*} E_1&\ll \int_0^\Delta\int_0^\Delta\int_{T}^{T+\Delta}|t|^{1/4}|t+u_2-u_1|^{1/4}\,dt\,du_1\,du_2,\\ &\ll \int_0^\Delta\int_0^\Delta\int_{T}^{T+\Delta}|T+\Delta|^{1/4}|T+2\Delta|^{1/4}\,dt\,du_1\,du_2,\\ &\ll \Delta^3|T+\Delta|^{1/4}|T+2\Delta|^{1/4}. \end{align*} Since $T\geq \Delta^6$ then $T+\Delta$ and $T+2\Delta$ are $\ll T$. Hence the error term. In the same fashion we can prove $E_2\ll \Delta^3 T^{1/2}$.