TITLE: On extended Riemann Hypothesis and coefficients of Selberg Class L-functions
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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.