Degree 1 Elements of the Selberg Class K

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Expo. Math. 23 (2005) 65–70 www.elsevier.de/exmath

Degree 1 elements of the Selberg class K. Soundararajan Department of Mathematics, University of Michigan, Ann Arbor, MI 48109, USA Received 16 June 2003; received in revised form 26 January 2004

Abstract We provide a short and simple proof of a beautiful result of Kaczorowski and Perelli classifying the elements of degree one in the Selberg class.

MSC 2000: primary 11M06; secondary 11M41

Keywords: L-functions; Selberg class

In [5], Selberg axiomatized properties expected of L-functions and introduced the “Selberg class.” We recall that anelement F of the Selberg class S satisﬁes the following axioms.

Axiom 1. Inthe half-plane > 1 the function F(s) is givenby anabsolutely convergent ∞ a(n)n−s a( ) = a(n)>n > Dirichlet series n=1 with 1 1and for every 0.

Axiom 2. There is a natural number m such that (s − 1)mF(s) extends to an analytic function of ﬁnite order in the entire complex plane.

E-mail address: [email protected] (K. Soundararajan).

Axiom 3. There is a function (s) = QsG(s)F (s) where Q>0and r G(s) = (j s + j ) with j > 0andRej 0 j=1 such that

(s) = (1 − s), ||= f(s)=f(s) d := r where 1 and for any function f we denote .Welet 2 j=1 j denote the “degree” of F.

Axiom 4. We may express log F(s)by a Dirichlet series

∞ b(n) (n) log F(s)= , ns log n n=2

b(n)>nϑ ϑ < 1 b(n) = where for some 2 . Set 0ifn is not a prime power.

A fundamental conjecture asserts that the degree of an element in the Selberg class is an integer. From the work of Richert [4] it follows that there are no elements in the Selberg class with degree 0 Riemann zeta function and shifts of Dirichlet L-functions. Subsequently, in [3] they showed that there are no elements of the Selberg class with degree

Theorem. Suppose F satisﬁes Axioms 1–3 and that the degree of F is 1. Then there exists a positive integer q and a real number A such that a(n)n−iA is periodic (mod q). If in addition F satisﬁes Axiom 4 then there is a primitive Dirichlet character (mod q) such that F(s)= L(s + iA, ).

We remark that Kaczorowski and Perelli obtain their results without assuming the hypoth- esis a(n)>n. We could restructure our proof to avoid this assumption, but have preferred not to do so in the interest of keeping the exposition transparent. Our method may also be modified and combined with the ideas in [3] to give a simplificationof their result for 0. Let be positive and T 1. Define

T 2 t 1 it log −i F( ,T)= √ F(1/2 + it)e 2 e 4 dt (2a) T and set (it will follow from our proof that the limit below is well deﬁned)

1 F( ) = lim F( ,T). (2b) T →∞ T 1+iA

Lemma. For any real number t and all X 1 we have ∞ a(n) 1 −n/X 1+ −1+ 1 + −|t| F + it = e + O((1 +|t|) X + X 2 e ). 2 n 1 +it n=1 2

c> 1 Proof. Consider for 2 c+i∞ 1 1 w F + it + w X (w) dw. 2 i c−i∞ 2

Expanding F(1 + it + w) into its Dirichlet series and integrating term by term we see that 2 − 1 −it −n/X this equals n a(n)n 2 e . Next we move the line of integration to Re(w) =−1 + . w = F(1 + t) w = 1 − t The pole at 0 leaves the residue 2 i . The possible pole at 2 i leaves a 1 + 1 + −|t| >X 2 ( +|t|) |( 1 + t)|>X 2 ( 1 + t) residue 1 2 i e due to the rapid decay of 2 i . Note that by the functional equation and Stirling’s formula 1 F + it + w 2 |G( 1 − t − w)| 1−2 Re w 2 i 1 1+ = Q F − it − w >(1 +|t|+|w|) |G( 1 + t + w)| 2 2 i for any w onthe lineRe w =−1 + . Hence the integral on the line Re (w) =−1 + is >X−1+(1 +|t|)1+ .

Using the functional equation in (2a) we see that

T 2 G( / − t) t −2it 1 2 i it log −i F( ,T)= √ F(1/2 − it)Q e 2 e 4 dt T G(1/2 + it) and using (1) this is iB 2 T e − t A √ F(1/2 − it)( CQ2 ) i ti (1 + O(1/T))dt. T 68 K. Soundararajan / Expo. Math. 23 (2005) 65–70

We now input our lemma above with X = T 4/3 to deduce that iB 2 T it e a(m) −m/X m A F( ,T)= √ √ e ti T m CQ2 m 1 2 + × + O t + O(T 3 ) 1 T d iB a(m) 2 T m it e −m/X iA 2 + = √ √ e t dt + O(T 3 ). (3) m 2 m T CQ

If x = 1 thenintegrationby parts gives that 2 T iA 2iT iA iT t A (2 T) x − ( T) x xi ti dt = T i log x 2 T it x A− 1 − iAti 1 dt> , (4a) T i log x | log x| while if x = 1 we have that 2 T 1+iA 1+iA A (2 T) − ( T) ti dt = . (4b) T 1 + iA

Using (4a) and (4b) in (3) we obtain that

− − A F( ) = lim T 1 i F( ,T) T →∞ 2 iA 1+iA B a( CQ ) 2 − 1 = ei ( CQ2 ∈ N) √ CQ + A 1 i − + |a(m)| −m/X + O lim T 1 √ e T →∞ m m 2 iA 1+iA B a( CQ ) 2 − 1 = ei ( CQ2 ∈ N) √ , (5) CQ 1 + iA where ( CQ2 ∈ N) = 1if CQ2 ∈ N and is 0 otherwise. We now present a different way of evaluating F( ,T) which will show that F( ) is periodic in with period 1. Using our lemma with X = T 4/3 again, we see that

T a(n) 2 t 1 −n/X it log −i 2 + F( ,T)= √ √ e e 2 en 4 dt + O(T 3 ). (6) n n T

The oscillatory integral in (6) above is estimated by familiar techniques, see Lemmas 4.2 and 4.6 of Titchmarsh [6]. For 2 n>3T we use Lemma 4.2 of Titchmarsh which shows 2 + that the integral is >1. Thus the contribution of such n to (6) is >T 3 . For smaller n we K. Soundararajan / Expo. Math. 23 (2005) 65–70 69 use Lemma 4.6 of Titchmarsh. Inthe range T 2 n2T we obtainthat the integral is √ √ 2 1 n e(−n ) + O T 5 + T, 2 min | (T / n)| log 2 √ 1 + min T, . | log(T / n)| For 2 n below T or between2 T and 3T the integral is bounded by the error terms above. Piecing this together we conclude that / + F( ,T)= 2 a(n)e(−n ) + O(T 9 10 ). T 2 n 2T Hence F( ) = F( + 1) which leads by (5) to

( CQ2 ∈ N)a( CQ2 ) iA A = ( CQ2( + 1) ∈ N)a( CQ2( + 1))( + 1)i . (7) From (7) we deduce immediately that CQ2 = q must be a positive integer and further that a(n)niA must be periodic (mod q). This proves the first part of our theorem. Suppose now that F also satisfies Axiom 4 so that the coefficients a(n) are multiplicative. Periodicity and multiplicativity together imply that for all n coprime to q, a(n)n−iA must equal (n) for a Dirichlet character (mod q). Let (mod q) denote the primitive character inducing . Thenthe ratio F (s)/L(s + iA, ) is anEuler product over the finitelymany primes dividing q, and by Axiom (4) the logarithm of this Euler product converges absolutely > ϑ ϑ < 1 a = ( − (− ))/ inthe half-plane (recall that 2 ). Thus, with 1 1 2, QsG(s)F (s) H(s) := s s+ A+a (q/ ) 2 ( i )L(s + A, ) 2 i is anentirefunctioninRe (s) > ϑ and H(s) is also entire in this region. Further both functions are free of zeros in this region. The functional equations for F and L now show H(s) H(s) (s) < − ϑ ϑ < 1 that and are entire and free of zeros in the region Re 1 . Since 2 we deduce that H(s) and H(s) are entire functions in all of C and that they never vanish. Since H is the ratio of two entire functions1 of order 1 it follows by Hadamard’s theorem that H(s) = aebs for some constants a and b. The functional equation connecting H(s) and H(1 − s) now mandates that b = 0andsoH is a constant. Examining the behaviour of H(1/2 + it) for large t it follows easily that F(s)= L(s + iA, ), proving our theorem.

Acknowledgements

The author is partially supported by the American Institute of Mathematics and by the National Science Foundation.

1 If is the trivial character then multiply the numerator and denominator in the deﬁnition of H by s(s − 1) to make them regular at 1. 70 K. Soundararajan / Expo. Math. 23 (2005) 65–70

References

[1] J.B. Conrey, A. Ghosh, On the Selberg class of Dirichlet series: small degrees, Duke Math. J. 72 (1993) 673–693. [2] J. Kaczorowski, A. Perelli, Onthe structure of the Selberg class, I : 0 d 1, Acta. Math. 182 (1999) 207–241. [3] J. Kaczorowski, A. Perelli, Onthe structure of the Selberg class, V : 1