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I have a question which may look naive for many experts here:

For any primitive holomorphic form $f$ of level $M$ ($M\in \mathbb{N}$), whether or not one has the lower bound that:

$$\sum_{X<n\le 2X} \lambda^2_f(n)\gg X^{1-\epsilon} \text{ for any } X\ge 2?$$ Here, $\lambda_f(n)$ denotes the $n$-th normalized Hecke eigenvalue attached to $f$.

Many thanks in advance.

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Something much stronger is true. There exists an effective constant $c_f>0$ such that

$$\sum_{X<n\leq 2X}\lambda_f(n)^2 \sim c_f L(1,\mathrm{Ad}\,f) X.$$

If the level $M_f$ of $f$ is squarefree, then

$$c_f = \prod_{\gcd(p,M_f)=1}\Big(1-\frac{1}{p^{2}}\Big).$$

This holds even if $f$ is a real-analytic Hecke-Maass newform. See Section 5.11 of Iwanienc and Kowalski (one of several possible references).

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  • $\begingroup$ Many thanks for kindly reply! $\endgroup$
    – hofnumber
    Commented Nov 3, 2023 at 11:08
  • $\begingroup$ 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? $\endgroup$
    – hofnumber
    Commented Nov 3, 2023 at 11:42
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    $\begingroup$ If $X$ is small compared to the conductor of $f$ then we do not have such a strong lower bound. This is a well-known problem for constructing efficient amplifiers. There is a weaker lower bound of size around $X^{1/2}$ due to Iwaniec, using the Hecke relation. $\endgroup$
    – Matt Young
    Commented Nov 3, 2023 at 12:43
  • $\begingroup$ @MattYoung Thanks for kindly comments. $\endgroup$
    – hofnumber
    Commented Nov 3, 2023 at 14:10

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