There is the conjecture that Selberg Class L-functions satisfy RH.

So that an L-function needs to have its coefficient multiplicatives (plus other conditions: functional equation,...) in order to satisfy RH. But I would like to know if for a L-function of Selberg class it is possible to find another L-function having the same non trivial zeros (strictly in the critical strip) but having "the sum of its coefficient" bounded instead of "multiplicative coefficients".

Meaning that for a L-function of Selberg class $L(s)= \sum a_n n^s$ with multiplicative coefficient $a_n$ there is another L-function $L_1$ (of Selberg class or not) having same non trivial zeros as $L(s)$ such that $L_1(s)= \sum b_n n^s$ with $\exists K, \forall N, |\sum_{n=1}^{n=N} b_n| \le K$

For primitive Dirichlet L-functions the function is itself satisfying the condition. But for the Zeta function we need to multiply it by $(1-2^{1-s})$ in order to obtain the Dirichlet Eta function which has the same zeros (strictly in the critical strip) as Zeta with the sum of its coefficient bounded. (And no more multiplicative property for its coefficients)

Any existing results on this question ? May be it is obvious that such equivalence does not always exists?

  • $\begingroup$ I didn't see the strictly in the critical stripe part:( $\endgroup$
    – Marc Palm
    Oct 17, 2013 at 15:39
  • $\begingroup$ Marc, the "strictly" is there only because: $\eta (s) = (1-2^{1-s}) \zeta(s)$, so when $Re(s)=1$ the function $\eta$ has some zeros (for $1=2^{1-s}$) but they are not the one we are interested in... I would like to know if a solution of RH is theorically possible without using directly the multiplicative properties of coefficients but only for example the fact that their partial sum is bounded. (I mean: it is reasonable to think there is a demonstration of GRH not using multiplicative property of Dirichlet Characters? but only the fact that their partial sum is bounded?) $\endgroup$
    – Bertrand
    Oct 18, 2013 at 6:10
  • $\begingroup$ I'm not sure what you mean by the sum of the coefficients, but isn't this the same as saying that the sum converges at $s=1$? By multiplying any Selberg L-function by a power of $(1-2^{1-s})$, you can get it to be holomorphic at $s=1$. I'm not sure whether that always implies the sum will be convergent in the sense you desire. $\endgroup$
    – Will Sawin
    Oct 20, 2013 at 19:24
  • $\begingroup$ For $L(s)= \sum a_n n^{-s}$ you propose to multiply by $(1-2^{1-s})$ so to consider $\sum (-1)^n a_n n^{-s}$ but then do we have the partial sum bounded: $\forall N, |\sum_{n=1}^{N} (-1)^n a_n| \le K$ ? I am also not sure that this works for L function of Selberg class... I will clarify my question. $\endgroup$
    – Bertrand
    Oct 21, 2013 at 5:59
  • $\begingroup$ @WillSawin; any book reference for your comment so that I can investigate this? $\endgroup$
    – Bertrand
    Oct 21, 2013 at 6:13

2 Answers 2


I don't think this is possible with the strong notion of boundedness I believe you want. If you're willing to assume something weaker, it might be possible, but we'd be butting our heads against things that are quite hard. Also, a caveat: in what I discuss below, I want to assume that if we call something an $L$-function, then it's sufficiently nice (e.g., automorphic). This is different than your terminology.

First, what's the "right" notion of boundedness? For an $L$-function $L(s,f)=\sum a_f(n) n^{-s}$ of degree $d$, we know that the summatory function $S_f(X)$ of the coefficients satisfies $$ S_f(X):=\sum_{n \leq X} a_f(n) = M(X) + O(X^{\frac{d-1}{d+1}+\epsilon}),$$ where $M(X)$ is a term arising from the possible pole of $L(s,f)$ at $s=1$; if there is no pole, $M(X)=0$. We expect, though, to the best of my knowledge, we have no idea how to show, that the error can be improved to $O(X^{\frac{d-1}{2d}+\epsilon})$. Notice that, in either what we know or what we conjecture, if $d=1$, as is the case for the zeta function and Dirichlet $L$-functions, these bounds are $O(X^\epsilon)$, whence it's reasonable to ask whether $S_f(X)=M(X)+O(1)$. But if $d\geq 2$, then the error is bigger, and so it's probably not reasonable to ask for $S_f(X)=M(X)+O(1)$ (even if we tweak $L(s,f)$ suitably). However, we note that the conjectured error is always $O(X^{\frac{1}{2}-\delta})$ for some $\delta>0$; this is what I want to focus on. I suspect, though I'm by no means willing to conjecture it, that beating square root cancellation might imply automorphicity and hence RH (provided one adds some conditions so that it's not trivial -- e.g., require $L(s,|f|^2)$ to have a pole at $s=1$).

On to your specific question about "tweaking" an $L$-function. Given $L(s,f)$, if $M(X)=0$ (i.e., there is no pole at $s=1$), then, with the revised notion above, $L(s,f)$ itself works. If $L(s,f)$ does have a pole, it's easy to cook up a Dirichlet polynomial $g(s)$ that will kill the pole and not introduce zeros, and we can look at the series $g(s)L(s,f)$ and this will work with the revised notion. If we believe my suspicion, then this tweaking suffices to identify $L$-functions satisfying RH.

If you really want to get bounded summatory functions, I think you're out of luck (though I can't prove it, and from here on out, this answer will get progressively more and more wishy-washy). Let's say $G(s)$ is a Dirichlet series that has the same non-trivial zeros in the critical strip as $L(s,f)$. If $G(s)$ is reasonably nice (e.g., is in the extended Selberg class), then $G(s)/L(s,f)$ will behave a lot like a degree 0 element of the Selberg class. Such elements are known to be Dirichlet polynomials, and it would likely be possible, assuming $G(s)$ is sufficiently nice, to show that this is the case for $G(s)/L(s,f)$. A Dirichlet polynomial can kill a pole at $s=1$, but it's extremely unlikely that it'll cancel out the error terms (in essence, they are "legitimately" error, so they shouldn't satisfy any fixed relation as would arise from a Dirichlet polynomial). If $G(s)$ is not sufficiently nice, I don't know how to argue, except to say that there really shouldn't be any coherent way to force the error terms to cancel more.

  • $\begingroup$ Thanks, your comment shows that the question is not obvious. Your approach is analytic and I think you may be right, multiplying by a Dirichlet polynomial will not always work, but it works for Zeta which transform in to Eta function (with a finite Dirichlet polynomial) because Zeta is fully multiplicative and all $a_p=1$ for p prime, so a finite polynomial is sufficient to transform it into another L-function having same zeros and bounded partial sum of coeffcients. In your approach you consider only finite polynomials and you do not use the fact that coefficient are multiplicative. $\endgroup$
    – Bertrand
    Oct 22, 2013 at 7:46
  • $\begingroup$ In general case, if such a correspondance exists then i think maybe an infinite sum is required to take care of all the $a_p$ coefficients (p prime) that determine the rest of the coefficients (as coefficient are multiplicative). $\endgroup$
    – Bertrand
    Oct 22, 2013 at 7:50
  • $\begingroup$ Even with an infinite sum, I think it's impossible if $d>1$. Let $G(s)=L(s,f)H(s)$. We need to balance three things: 1) The sum of $f*h$ being bounded, 2) $H(s)$ analytic in $\Re(s)>0$, maybe by asking for $S_h(X) \ll X^\epsilon$, and 3) The non-existence of zeros of $H(s)$ in $\Re(s)>0$. It's not even obvious to me that 1 and 2 are compatible once $d>1$: each is equivalent to a system of inequalities on $h(n)$. 1 has coefficients $S_f(X/n)$, whereas 2 has coefficients all one. Once $S_f(X)$ has size (as in $d>1$), these systems are on different scales, so a common solution is not obvious. $\endgroup$
    – rlo
    Oct 23, 2013 at 1:03
  • $\begingroup$ Thanks for your answer, you should be right: in general case it seems difficult to have equivalence I suggested. $\endgroup$
    – Bertrand
    Oct 24, 2013 at 7:58

Here's an argument that I think demonstrates why no such tweak is possible if the degree is strictly bigger than $1$. For simplicity let us just consider the case when $L(s,f)$ has no pole.
Suppose there is some tweaked Dirichlet series $H(s,f)=\sum_{n=1}^{\infty} b(n)n^{-s}$ with coefficients that have bounded partial sum and such that $H(s,f)$ and $L(s,f)$ have the same zeros in $\sigma >0$.
Then $L(s,f)/H(s,f)$ is an analytic function in $\sigma >0$, and in some the half-plane $\sigma >2$ say this function is bounded. Now we can use a complex function theory argument (as in Riemann hypothesis implies Lindel{\" o}f) to show that $L(s,f)/H(s,f)$ is bounded by $(1+|s|)^{\epsilon}$ for all $s$ with Re$(s)\ge \delta >0$ for any fixed positive $\delta$.

Next, we note that the Plancherel formula can be used to show that $$ \int_0^{\infty} \Big| \sum_{n\le e^{x} } b(n) \Big|^2 e^{-2\sigma x} dx = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{|H(\sigma+it)|^2}{|\sigma+it|^2}dt, $$ and the formula holds for all $\sigma >0$ since we are assuming that the partial sums of $b(n)$ are bounded. However the right hand side is, in view of our previous observation, $$ \gg \int_{-\infty}^{\infty} \frac{|L(\sigma+it,f)|^2}{|\sigma+it|^{2+\epsilon}} dt, $$ if $\sigma \ge \delta >0$. Now the idea is to show that the functional equation for $L(s,f)$ will prevent this from happening. Basically the functional equation shows that for $0<\sigma <1/2$, $L(s,f)$ roughly is of size $|s|^{d(1/2-\sigma)}$ times $L(1-s,f)$ (which in mean square can be estimated). In other words, $|L(\sigma+it,f)|^2$ will roughly grow like $|s|^{d(1-2\sigma)}$ and if $d$ is $>1$ for $\sigma$ suitably close to zero it will not be possible for $$ \int_{-\infty}^{\infty} \frac{|L(\sigma+it,f)|^2}{|\sigma+it|^{2+\epsilon}} dt $$ to converge.

This last part was sketched quickly, but it is a standard way of producing $\Omega$ results for arithmetic functions. What it shows is that the partial sums of the coefficients of the actual $L$-function must be $\Omega(x^{(d-1)/(2d)-\epsilon})$ which is the kind of result mentioned in rlo's answer.


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