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I came across the following modification of the Selberg class in some of my work (see below), and while I've moved on in some sense -- I submitted the paper in question -- I can't get it off of my mind.

The question: Consider the following modification of the axioms of the Selberg class. Let $L(s)$ be a function such that:

1) $L(s)$ possesses a Dirichlet series and Euler product as in the standard axioms for the Selberg class (thus, the Dirichlet coefficients are bounded by $n^\epsilon$, e.g.).

2) There are rational $a_j$ such that the completion $\Lambda(s):=L(s) Q^s \prod_{j=1}^k \Gamma\left(\lambda_j s + \mu_j\right)^{a_j}$ satisfies the standard sort of functional equation, namely $\Lambda(1-s) = \epsilon \cdot \overline{\Lambda}(s)$ valid when $s,1-s$ are both in the region of analyticity (see 3). This is the first place where the definition differs from the Selberg class. There, the $a_j$'s are taken to be positive integers, whereas here they could be both negative and non-integral, provided that they are still rational.

3) Via the functional equation in 2, suppose that $L(s)$ can be analytically continued to all of $\mathbb{C}$, with the possible exception of a simple pole at $s=1$, and allowing for the (finitely many) branch cuts $(-\infty+iy_j,1/2+iy_j)$ where each $y_j$ is the imaginary part of $\mu_j$.

An example of something which is almost in this class, but fails 3, is $L(s,f)^{1/2}$ where $L(s,f)$ is a primitive element of the Selberg class. Conjecturally, $L(s,f)$ will have infinitely many simple zeros, so there would need to be infinitely many branches to continue $L(s,f)^{1/2}$ beyond the half-line.

My question is what you'd probably expect: if $L(s)$ satisfies the above properties, must it actually be in the Selberg class? Roughly speaking, you might expect this for a couple of reasons. First, the analytic properties of the class of functions satisfying 1-3 are similar enough to those of the Selberg class that, believing that elements of the Selberg class come from arithmetic, probably the same is true for this class. Second, believing in a proof of the classification of the Selberg class -- nevermind that such a thing is a long way off -- the proof must only rely on the analytic properties of the functions (obviously! what else could it depend on?), and probably any particular proof can be modified to work for this extension. In particular, I did this for the $d=1$ case for my work, and I'd imagine that $0\leq d < 2$ should also work, but I haven't looked at this.

I expect that this question is hard (unless it's false), so I'd be happy with an answer that assumes big conjectures.

Also, maybe this is more of a meta-question, but is this interesting for a reason beyond my application? That is, are there interesting consequences if this turns out to be true? One can deduce some simple consequences about divisibility, e.g., that if the zero-set of one element (with multiplicity) is contained in the zero-set of another, then the first function must divide the second. But I'm not sure that's so interesting.

My application: If we look at multiplicative functions $|f(n)|\leq 1$, complex-valued, what must be true if $\sum_{n\leq x} f(n) \ll x^{1/2-\delta}$ for some fixed $\delta>0$? Certainly we want to assume that $f(n)$ is not too small (i.e., we want to rule out things like $f(n)=1/n$), so let's impose the condition that $\sum_{n\leq x} |f(n)| \gg x$. If such a function exhibits more than square root cancellation, must it come from a Dirichlet character? I don't know of a counterexample, and I suspect that this is true (but I will also readily admit that I'm not sure I'm thinking about the right pathologies). Believing that it's true, it's probably very hard to prove.

Based on this, we want to look at a class of multiplicative functions for which we can see Dirichlet characters naturally appearing, and for which we can provide an answer. The class of functions I looked at depended on the arithmetic of a finite, Galois extension $K/\mathbb{Q}$. In particular, look at functions satisfying the following properties:

1) $f(n)$ is completely multiplicative, 2) $|f(p)|=1$ for all primes $p$ that split completely in $K$, and 3) if $\mathrm{Frob}_p = \mathrm{Frob}_q$ (up to conjugacy -- crucially, we'll want $K$ to be non-abelian), then $f(p)=f(q)$.

We can see Dirichlet characters arising by taking $K$ to be cyclotomic. It turns out that for functions defined as above, more than square root cancellation implies that $f(n)$ coincides with a Dirichlet character. The argument essentially shows that some product $$\prod_{\rho} L(s,\rho)^{a_\rho}$$ of Artin $L$-functions associated to the irreducible representations of $\mathrm{Gal}(K/\mathbb{Q})$, each $a_\rho\in\mathbb{Q}$, is in the above defined class. The restriction that $|f(p)|=1$ for a prime that splits completely actually forces the degree to be 1, and following the proof of the classification of degree 1 elements of the Selberg class, one obtains the result.

However, there is a natural objection to the above setup. We need to talk about non-abelian number fields for things that are legitimately different from Dirichlet characters to arise, but once we're in this setting, the condition that $|f(p)|=1$ for primes that split completely is totally unnatural. If the above class is actually equal to the Selberg class, then one could remove this condition and conclude that the Dirichlet series $L(s,f)$ is in the Selberg class. Maybe that's not so interesting, but I think it would be more satisfying than what I have now.

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If $f$ is totally multiplicative, $|f(p)| = 1$ at all but finitely many primes, and $\sum_{n \leq x} f(n)| \gg x$, then a theorem of Halasz says that $f$ "pretends" to be something like a Dirichlet character, or rather a Hecke character on the rationals. Google Halasz's theorem or "pretentious character"; there is work on this theme by Granville and Soundaraajan that you'll find. –  KConrad Nov 30 '12 at 22:51
    
I find the use of rational powers of an $L$-function to be strange, particularly if you want to talk about analytic continuation. What is your definition of $L(s,f)^{1/2}$? When Artin proved that "rational powers" of Artin $L$-functions have a meromorphic continuation, what he really proved is that some positive integral power of any Artin $L$-function has a meromorphic continuation. Brauer later improved this to meromorphic continuation of the Artin $L$-function itself. Rational powers directly seem pretty awkward. –  KConrad Nov 30 '12 at 22:53
    
Right, this absolutely falls under the purview of pretentiousness; in fact, my question about more than square root cancellation can be viewed as a counterpart to Halasz's theorem which says that anything with large sums must come from (pretend to be) one of a natural set of examples. There, though, the natural examples are not Dirichlet characters. Instead, they are the additive characters, $n^{it}$. Indeed, non-principal Dirichlet characters don't have large partial sums - the partial sums are bounded! Thus, they are the natural examples for exceptional (more than square root) cancellation. –  rlo Nov 30 '12 at 23:29
    
You're absolutely right that there are issues with $L(s,f)^{1/2}$, which is why it's not actually something I want to consider. I brought it up mostly to clarify points 1 and 2. Maybe a prototype question would be this: Can $L(s,f)^{1/2}L(s,g)^{1/2}$ ever be sensibly continued to an entire function, where $L(s,f)$ and $L(s,g)$ are primitive elements of the Selberg class? It's known that each has zeros disjoint from the other, but maybe all the simple zeros coincide, or are there are no simple zeros, or... Conjecturally, this can't happen, but that's the sort of thing I'm imagining. –  rlo Nov 30 '12 at 23:37
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