My question: How does the bound \[ g_c(l)\ll 1/(xl)^2\hspace {10mm}\text {for }x^{1-2\epsilon }l>c^2\hspace {10mm}(1)\] towards the bottom of page 277 of http://matwbn.icm.edu.pl/ksiazki/aa/aa61/aa6134.pdf work?
How far I can get: Working upwards from the stated bound (you need pages 276 and 277), you can see that \[ g_c(l)=\sum _{l_1l_2=l}\int \int F(uc)F(vc)F(uvc^2)e\left (ul_1+vl_2\right )dudv=\frac {1}{c^2}\sum _{l_1l_2=l}\int \int F(u)F(v)F(uv)e\left (ul_1/c+vl_2/c\right )dudv.\] The function $F$ isn't given explicitly (you need page 276), and maybe that's where I get a bit confused. It seems maybe they're saying that $F'(t)=0$ outside of $(x-x^{1-\epsilon },x)$, and on that interval it is $\ll 1/x^{1-\epsilon }$? And I guess the higher derivatives then satisfy $F^{(n)}(t)\ll 1/x^{n(1-\epsilon )}$? So \[ \left (\frac {d}{dudv}\right )^n\Big \{ F(u)F(v)F(uv)\Big \} \ll \frac {x^n}{x^{2n(1-\epsilon )}}=\frac {1}{x^{n(1-2\epsilon )}}\] Integrating the above integral for $g_c(l)$, it seems to me that by each partial integration (of both variables) we get a factor $c^2/l$, so after $n$ integrations (of both variables) we are left with \[ g_c(l)=\frac {1}{c^2}\left (\frac {c^2}{l}\right )^n\int \int \left (\frac {d}{dudv}\right )^n\Big \{ F(u)F(v)F(uv)\Big \} e(...)\ll \frac {1}{c^2}\left (\frac {c^2}{lx^{1-2\epsilon }}\right )^n\int \int dudv\] but I don't know how to take this further - probably I've gone wrong already.
Can anyone help me fill out the details of (1) correctly?