@MattYoung Thanks for kindly comments.

Dear professor, I am still curious, as you pointed out in the post (URL: mathoverflow.net/questions/445006/…) that, if the primitive cusp form $f$ is of prime level $p$, then $$\sum_{n\leq x}|\lambda_{f}(n)|^2 = \frac{L(1,\mathrm{Ad}~f)}{\zeta^{(p)}(2)}x+O((p x)^{1/2+\epsilon}) = \frac{6L(1,\mathrm{Ad}~f)}{\pi^2}\Big(1-\frac{1}{p^2}\Big)^{-1}x+O((p x)^{1/2+\epsilon}).$$ It appears that one would get the lower bund $\gg x$ only if $p<x$. If $p$ is very large, can we still have the lower bund?

Many thanks for kindly reply!

@WillSawin Dear professor, many thanks for so kindly help without any rewards! I need a quantitative bound for this sum now. The obstacle is that there exist some factor of $q$ in the numerator of the exponential sum above which may bring some troubles, I think. I don't know if one may study the Newton polyhedron nondegeneracy of the corresponding Newton polyhedron, or the works of Fouvry-Kowalski-Michel can be put into use.

@WillSawin Dear professor, I also wanna ask, for fixed $h$, if we have the bound $\ll h (qh)^{1+\varepsilon} $ for this triple sum, by plugging the sum over $h$ outside, and then using the estimate $\sideset{_{}^{}}{^{\ast}_{}}\sum _{x,y \bmod {c}} e\left (({x+\overline{xy} +\overline{y}+y })/{c}\right)\ll c^{1+\varepsilon}$ for inner hyper-Kloosterman sum over $x,y$? Notice that there is still a factor of $q$ in the numerator. I am not sure if this would work.

@WillSawin Dear Sawin, thanks for timely reply. Maybe I didn't state my purpose clearly. My point of the departure is to give a bound for this sum. But, I find the known estimate for double hyper-Kloosterman sum in the post above can not be put into use. So I speculate if we have a further decomposition by changing variables.

@PeterHumphries Great thanks for comments.

@WindomEarle Dear professor, I have anther request. Could you tell me you real name so that I can mention you in the Acknowledgments of my work which I contained the ideas you told me before ? Many many thanks. You may also can told me in the E-mail: [email protected]. Thanks.

@WindomEarle Dear professor, sorry to disturb again. How did it go? Last year, you spent you valuable time share with me how to derive the Voronoi formula for symmetric lift by appealing to Corbett's paper. Much obliged for all you have done. By the way, I wanna ask an extra question: dear professor, the argument you told me seems to work with arbitrary level $N$, not necessary the square-free level. Whether or not we can get the Voronoi formula for the level $N$ being a positive integer which just satisfies that $(N,c)=1$ in {\bf{Theorem}} above?