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The coefficient of Selberg Class L-function satisfy:

$a_n <M_{\epsilon} n^{\epsilon}$ (for any $\epsilon >0$) and the $a_n$ are multiplicative.

So I would like to know if it can be shown that we also have the following partial sum bounded by a constant (maybe using also other properties of $a_n$ coefficients):

$\sum_{n=1}^{2N} a_n - a_2 a_n < M$

(As $a_n$ are multiplicative, we see that some terms disappear from the sum as $a_{2k}=a_2 a_k$ for k odd, but is there a chance to have this sum bounded by a constant?)

More generally what can you advise to read on the characterisation of coefficent of Selberg Class L-function?

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    $\begingroup$ Bertrand, if you like my answer (after studying Landau's oscillation theorem), then please accept it officially. Thanks! $\endgroup$
    – GH from MO
    Commented Sep 10, 2013 at 12:45

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I think this fails for the Dirichlet coefficients of $L(s):=\zeta(s+i\gamma)\zeta(s-i\gamma)$ whenever $\gamma>0$ satisfies $2^{i\gamma}+2^{-i\gamma}\neq 1$. Indeed, for such $\gamma$, the inequality would imply that $S(x):=\sum_{n\leq x}a_n$ is bounded by a constant either from below or from above. However, $L(s)$ has simple poles at $1\pm i\gamma$ and it is holomorphic elsewhere, hence a quantitative form of Landau's oscillation theorem yields $S(x)=\Omega_{\pm}(x)$. See Theorem 11.8 in Bateman-Diamond: Analytic number theory - An introductory course.

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  • $\begingroup$ Thanks, it helps (i will need to read carrefully your reference) $\endgroup$
    – Bertrand
    Commented Sep 10, 2013 at 12:29
  • $\begingroup$ In fact a characterisation like, they are constants A and M such that $|\sum_{n=1}^{2N} a_n - A \sum_{n=1}^{N} a_{2n}|<M$ seems more reasonable. $\endgroup$
    – Bertrand
    Commented Sep 10, 2013 at 12:50
  • $\begingroup$ @Bertrand: I think your second proposed inequality fails already for $\zeta(s)^2=\sum a_nn^{-s}$. In this case the sum of $a_n$ over the odd (resp. even) integers up to $x$ is of the form $(1/4)x\log x+c_1 x$ (resp. $(3/4)x\log x+c_2 x$) plus lower order terms, where $c_1$ and $c_2$ are linearly independent over $\mathbb{Q}$ (I haven't checked this). So no $A$ is capable of neutralizing the $x\log x$ and $x$ terms simultaneously. $\endgroup$
    – GH from MO
    Commented Sep 10, 2013 at 14:23
  • $\begingroup$ Yes your example is very interesting, thanks for your help, showing there is no such dependance between odd and even coefficients. $\endgroup$
    – Bertrand
    Commented Sep 11, 2013 at 12:55

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