FFOURIER INTERPOLATION ON THE REAL LINE
DANYLO RADCHENKO AND MARYNA VIAZOVSKA
Abstract.
We use weakly holomorphic modular forms for the Hecke theta group toconstruct an explicit interpolation formula for Schwartz functions on the real line. Theformula expresses the value of a function at any given point in terms of the values ofthe function and its Fourier transform on the set { , ±√ , ±√ , ±√ , . . . } . Introduction
Let f : R → R be an integrable function and let (cid:98) f be the Fourier transform of f : (cid:98) f ( ξ ) = (cid:90) ∞−∞ f ( x ) e − πiξx dx. The classical Whittaker-Shannon interpolation formula (see [16], [12]) states that if theFourier transform (cid:98) f is supported in [ − w/ , w/ f ( x ) = (cid:88) n ∈ Z f ( n/w ) sinc( wx − n ) , where sinc( x ) = sin( πx ) / ( πx ) is the cardinal sine function. In other words, the func-tions s n ( x ) = sinc( wx − n ) form an interpolation basis on the set w Z for the spaceof functions whose Fourier transform is supported in [ − w/ , w/
2] (the so-called Paley-Wiener space
P W w ). For a nice overview of history of the Whittaker-Shannon formula,its generalizations and other related results, see [10]. A generalization of the formula thatis useful for certain extremal problems in Fourier analysis is described in [13]Note that it is not possible to apply the Whittaker-Shannon formula directly to func-tions whose Fourier transform (cid:98) f has unbounded support, say, to f ( x ) = exp( − πx ).The main goal of this paper is to prove an interpolation formula that can be applied toarbitrary Schwartz functions on the real line. Theorem 1.
There exists a collection of even Schwartz functions a n : R → R with theproperty that for any even Schwartz function f : R → R and any x ∈ R we have (1) f ( x ) = ∞ (cid:88) n =0 a n ( x ) f ( √ n ) + ∞ (cid:88) n =0 (cid:98) a n ( x ) (cid:98) f ( √ n ) , where the right-hand side converges absolutely. As immediate corollary of Theorem 1, we get the following.
Corollary 1.
Let f : R → R be an even Schwartz function that satisfies f ( √ n ) = (cid:98) f ( √ n ) = 0 , n ∈ Z ≥ . Then f vanishes identically. Denote by s the vector space of all rapidly decaying sequences of real numbers, i.e.,sequences ( x n ) n ≥ such that for all k > n k x n → , n → ∞ . If we denote by S even a r X i v : . [ m a t h . N T ] J a n RADCHENKO AND VIAZOVSKA the space of even Schwartz functions on R (see Section 6), then there is a well-definedmap Ψ : S even → s ⊕ s given byΨ( f ) = ( f ( √ n )) n ≥ ⊕ ( (cid:98) f ( √ n )) n ≥ . Together with Theorem 1 the following result gives a complete description of what valuesan even Schwartz function and its Fourier transform can take at ±√ n for n ≥ Theorem 2.
The map Ψ is an isomorphism of the space of even Schwartz functions ontothe vector space ker L ⊂ s ⊕ s , where L : s ⊕ s → R is the linear functional L (( x n ) n ≥ , ( y n ) n ≥ ) = (cid:88) n ∈ Z x n − (cid:88) n ∈ Z y n . In the proof of Theorem 1 we will give an explicit construction of the interpolatingbasis { a n ( x ) } n ≥ . For instance, the Fourier invariant part of a n will be given by a n ( x ) + (cid:98) a n ( x ) = (cid:90) − g ( z ) e iπx z dz, where g is a certain weakly holomorphic modular form of weight 3 /
2, and the integral isover a semicircle in the upper half-plane. The anti-invariant part a n ( x ) − (cid:98) a n ( x ) will bedefined by a similar expression. For an explicit example, we define a ( x ) by a ( x ) = 14 (cid:90) − θ ( z ) e iπx z dz, where θ ( z ) is the classical theta series(2) θ ( z ) = (cid:88) n ∈ Z e iπn z . The modular transformation property of g is chosen in such a way that it complementsthe action of the Fourier transform on Gaussian functions: (cid:98) e z ( ξ ) = 1 √− iz e − /z ( ξ ) , where e z ( x ) = e iπzx , and the square root is chosen to be positive when z lies on theimaginary axis (this comment also applies whenever expression ( − iz ) α occurs throughoutthe paper). For instance, using the identity θ (cid:16) − z (cid:17) = √− iz θ ( z )and applying the change of variable z (cid:55)→ − /z in the integral that defines a ( x ) we seethat (cid:98) a = a . The general definition of a n needs some preparation, and will be given inSection 4. The plots of the first three functions are shown in Figure 1.An analogue of Theorem 1 holds also for odd Schwartz functions, but we postponeits formulation until Section 7. It is possible to combine the two results into a generalinterpolation theorem, but it is more convenient to work with the two cases separately. Remark.
Another way to interpret equation (1) is to think of it as a “deformation”of the classical Poisson summation formula(3) (cid:88) n ∈ Z f ( n ) = (cid:88) n ∈ Z (cid:98) f ( n ) , which will be a special case of (1) for x = 0 (more precisely, − a n (0) = (cid:99) a n (0) = 1for n ≥ a (0) = (cid:98) a (0) = 1 /
2, and all other values are zero).Our general approach fits into the framework of Eichler cohomology (see [5]; somerelevant results can also be found in [6] and [7]), but for the most part we avoid using its
OURIER INTERPOLATION ON THE REAL LINE 3 √ √ a a a √ √ (cid:98) a (cid:98) a (cid:98) a Figure 1.
Plots of a n and (cid:98) a n for n = 0 , , a n have recently beenused in [14] and [2] to solve the sphere packing problem in dimensions 8 and 24.The paper is organized as follows. In Section 2 we recall some known facts aboutmodular forms for the Hecke theta group Γ θ . In Section 3 we compute explicit basis ofa certain space of weakly holomorphic modular forms of weight 3 / θ .Then, in Section 4 we use these modular forms to construct an interpolation basis for theeven Schwartz functions and prove some of its properties. In the next section we prove anestimate on the growth of this sequence functions, this is by far the most technical partof the paper. In Section 6 we prove the main result for even functions, and in Section 7we define the interpolation basis and formulate corresponding statements for the oddfunctions. Acknowledgements.
The authors would like to thank Max Planck Institute for Math-ematics, Bonn for hospitality and support while this paper was being written. The firstauthor would also like to thank the Absus Salam International Center for TheoreticalPhysics, Trieste for the financial support.The authors are grateful to Don Zagier, Emanuel Carneiro, and Andrew Bakan formany helpful remarks and comments.2.
Hecke theta group
In this section we set up notation and collect facts about the Hecke theta group andrelated modular forms. Most of the material from this section can be found, in muchgreater detail, in [9]. For a motivated general introduction to the theory of modularforms, see [18].2.1.
Upper half-plane and the action of SL ( R ) . Denote by H the complex upperhalf-plane { z ∈ C : Im( z ) > } . The group SL ( R ) of 2 × RADCHENKO AND VIAZOVSKA and determinant 1 acts on the upper half-plane on the left by Moebius transformations γz = az + bcz + d , γ = (cid:18) a bc d (cid:19) ∈ SL ( R ) . The kernel of this action coincides with the center {± I } of SL ( R ) and thus we can workwith the action of PSL ( R ) = SL ( R ) / {± I } instead.We will use special notation for the following elements of SL ( Z ) (or, by abuse ofnotation, of PSL ( Z )): I = (cid:18) (cid:19) , T = (cid:18) (cid:19) , S = (cid:18) −
11 0 (cid:19) . Recall that Γ(2) ⊂ SL ( Z ) is defined asΓ(2) = (cid:26) A ∈ SL ( Z ) (cid:12)(cid:12)(cid:12)(cid:12) A ≡ (cid:18) (cid:19) (mod 2) (cid:27) , and Γ θ is the subgroup of SL ( Z ) generated by S and T , or, equivalently,Γ θ = (cid:26) A ∈ SL ( Z ) (cid:12)(cid:12)(cid:12)(cid:12) A ≡ (cid:18) (cid:19) or (cid:18) (cid:19) (mod 2) (cid:27) . Note the obvious inclusions SL ( Z ) ⊃ Γ θ ⊃ Γ(2). The group Γ(2) has three cusps 0, 1,and ∞ , while the group Γ θ has only two cusps: 1 and ∞ . The standard fundamentaldomain for Γ θ is(4) D = { τ ∈ H : | τ | > , Re( τ ) ∈ ( − , } . Finally, we are going to use the “ θ -automorphy factor” on the group Γ θ , which wedefine for all z ∈ H and γ ∈ Γ θ by(5) j θ ( z, γ ) = θ ( z ) θ ( γz ) . From the definition it immediately follows that j θ ( z, γ γ ) = j θ ( z, γ ) j θ ( γ z, γ ), so j θ isindeed an automorphy factor on Γ θ . We have j θ ( z, T ) = 1 and j θ ( z, S ) = ( − iz ) − / , andin general we have j θ ( z, ( a bc d )) = ζ · ( cz + d ) − / for some suitable 8-th root of unity ζ (an explicit expression for ζ can be found in [9, Th. 7.1]). Using this automorphy factorwe define the following slash operator in weight k/ H )(6) (cid:0) f | k/ A (cid:1) ( z ) = j θ ( z, A ) k f (cid:16) az + bcz + d (cid:17) , where A = ( a bc d ) ∈ Γ θ . More generally, for ε ∈ {− , + } define a slash operator | εk/ by(7) f | εk/ A = χ ε ( A ) f | k/ A, where χ ε : Γ θ → {± } is the homomorphism defined by χ ε ( S ) = ε and χ ε ( T ) = 1. Theslash operator defines a group action, that is, f | AB = ( f | A ) | B . Another fact that wewill use is that for all ( a bc d ) ∈ SL ( R ) we have(8) Im (cid:16) aτ + bcτ + d (cid:17) = Im( τ ) | cτ + d | . For any real number a we will denote by q a the analytic function q a = q a ( z ) = exp(2 πiaz ) . Any T -periodic holomorphic function on H admits an expansion in powers of q /T (ingeneral as a Laurent series, but in our case such expansions will have only finitely many OURIER INTERPOLATION ON THE REAL LINE 5 negative powers). We will be using subscripts to indicate the main variable of q , i.e., q aτ is the same as q a ( τ ); by default the variable of q a is z .2.2. Modular forms for the group Γ θ . We begin by defining the classical Jacobi thetaseries (the so-called Thetanullwerte):Θ ( z ) = (cid:88) n ∈ Z + q n = 2 η (2 z ) η ( z ) , Θ ( z ) = (cid:88) n ∈ Z q n = η ( z ) η ( z/ η (2 z ) (= θ ( z )) , Θ ( z ) = (cid:88) n ∈ Z ( − n q n = η ( z/ η ( z ) , where η ( z ) = q / (cid:81) n ≥ (1 − q n ) is the Dedekind eta function. The functions Θ , Θ ,and Θ generate the ring of holomorphic modular forms on Γ(2) and satisfy the Jacobiidentity(9) Θ = Θ + Θ . The q -expansions of these forms at the cusp i ∞ are as follows:Θ ( z ) = 16 q / + 64 q / + 96 q / + O ( q ) , Θ ( z ) = 1 + 8 q / + 24 q + 32 q / + 24 q + 48 q / + O ( q ) , Θ ( z ) = 1 − q / + 24 q − q / + 24 q − q / + O ( q ) . Under the action of SL ( Z ) the theta functions transform as follows. Under the actionof S we have ( − iz ) − / Θ ( − /z ) = − Θ ( z ) , ( − iz ) − / Θ ( − /z ) = − Θ ( z ) , ( − iz ) − / Θ ( − /z ) = − Θ ( z ) , (10)and under the action of T we haveΘ ( z + 1) = e iπ/ Θ ( z ) , Θ ( z + 1) = Θ ( z ) , Θ ( z + 1) = Θ ( z )(11)Together with the q -series for Θ , Θ , and Θ , these transformations allow to computethe q -series expansion of any expression in theta functions at any of the three cuspsof Γ(2).Using these theta functions we can define the classical modular lambda invariant λ ( z ) = Θ ( z )Θ ( z ) = 16 q / − q + 704 q / + . . . , which is a Hauptmodul for Γ(2). In particular, we have λ (cid:16) az + bcz + d (cid:17) = λ ( z ) , (cid:18) a bc d (cid:19) ≡ (cid:18) (cid:19) (mod 2) , and any function with these transformation properties and with appropriate behavior atthe cusps can be expressed as a rational function of λ . From (9) – (11) we see that under RADCHENKO AND VIAZOVSKA the action of PSL ( Z ) the function λ ( z ) transforms as follows: λ (cid:16) − z (cid:17) = 1 − λ ( z ) ,λ ( z + 1) = λ ( z ) λ ( z ) − . (12)Since Θ , Θ , and Θ do not vanish in H (by the product expression using η ( z )), we getthe well-known fact that λ ( z ) omits the values 0 and 1.Using λ ( z ), define a Hauptmodul J for the group Γ θ (13) J ( z ) = 116 λ ( z )(1 − λ ( z )) = Θ ( z )Θ ( z )16Θ ( z ) = q / − q + 300 q / + . . . . Note that J ( z ) = η ( z/ η (2 z ) η ( z ) − , hence it does not have zeros in H . This functionsatisfies the transformation laws J (cid:16) − z (cid:17) = J ( z ) ,J ( z + 2) = J ( z ) , and it maps the fundamental domain D conformally onto the cut plane C (cid:114) [1 / , + ∞ ).Finally, note that 1 /J vanishes at the cusp 1, since(14) 1 J (1 − /z ) = − q − q + O ( q ) . Asymptotic notation.
We freely use the standard big O notation. In addition,we also use Vinogradov’s “ (cid:28) ” sign f (cid:28) ε,δ,... g ⇔ f = O ε,δ,... ( g ) . Notationally, we prefer to use “ O ” for sequences and additive remainders, while for mostinequalities with implied constants we use “ (cid:28) ”.3. Weakly holomorphic modular forms on Γ θ of weight / /
2. Namely, let { g + n ( z ) } n ≥ and { g − n ( z ) } n ≥ be two collections of holo-morphic functions on the upper half-plane H that satisfy the transformation properties g εn ( z + 2) = g εn ( z ) , ( − iz ) − / g εn ( − /z ) = εg εn ( z ) , (15)as well as the following behavior at the cusps g + n ( z ) = q − n/ + O ( q / ) , z → i ∞ ,g − n ( z ) = q − n/ + O (1) , z → i ∞ ,g εn (1 + i/t ) → , t → ∞ . (16)The reason behind these conditions will be made clear in the next section. We make thefollowing ansatz: g + n ( z ) = θ ( z ) P + n ( J − ( z )) ,g − n ( z ) = θ ( z )(1 − λ ( z )) P − n ( J − ( z )) , (17)where P ± n ∈ Q [ x ] are monic polynomials of degree n and P − n (0) = 0. The polynomials P ± n are uniquely determined by the first two conditions in (16), since J − has q -expansionstarting with q − / + 24 + O ( q / ), and thus the coefficients of P ± n can be found by OURIER INTERPOLATION ON THE REAL LINE 7 inverting an upper-triangular matrix. The transformation properties (15) follow from theproperties of J ( z ) and λ ( z ). The first few of these functions are g +0 = θ ,g +1 = θ · ( J − − ,g +2 = θ · ( J − − J − + 192) , g − = θ · (1 − λ ) · ( J − ) ,g − = θ · (1 − λ ) · ( J − − J − ) ,g − = θ · (1 − λ ) · ( J − − J − + 252 J − ) . The polynomials P ± n are analogues of Faber polynomials, see [3]. The main differencehere is that the group Γ θ has two cusps, so some additional normalization is needed. Inthe next theorem we give closed form expressions for generating functions of { g ± n } . Theorem 3.
The generating functions for { g + n ( z ) } n ≥ and { g − n ( z ) } n ≥ are given by ∞ (cid:88) n =0 g + n ( z ) e iπnτ = θ ( τ )(1 − λ ( τ )) θ ( z ) J ( z ) J ( z ) − J ( τ ) =: K + ( τ, z ) , ∞ (cid:88) n =1 g − n ( z ) e iπnτ = θ ( τ ) J ( τ ) θ ( z )(1 − λ ( z )) J ( z ) − J ( τ ) =: K − ( τ, z ) . (18) Proof.
The proof follows the same lines as the proof of Theorem 2 from [4]. We onlyprove the statement for g + n , since the case of g − n is almost identical. From the q -expansionof J − and the fact that J ( z ) J ( z ) − J ( τ ) = (cid:88) n ≥ J n ( τ ) J − n ( z ) , it is clear that the g + n defined by (18) are also of the form θ ( z ) P n ( J − ( z )) for some monicpolynomial P n of degree n . The only thing that we need to check is that they satisfy g + n ( z ) = q − n/ + O ( q / ) , z → i ∞ , or, equivalently, that P n = P + n . By Cauchy’s theorem we know that g + n ( z ) = 12 (cid:90) τ +2 τ K + ( τ, z ) q − n/ τ dτ = 12 πi (cid:73) C K + ( τ, z ) q − ( n +1) / τ d ( q / τ ) , where τ ∈ H has sufficiently large imaginary part and C is a small enough loop around 0in the q / τ -plane. Using the identity(19) q / τ dJd ( q / τ ) ( τ ) = J (cid:48) ( τ ) πi = θ ( τ )(1 − λ ( τ )) J ( τ )we get that K + ( τ, z ) = q / τ dJd ( q / τ ) ( τ ) J ( z ) − J ( τ ) · θ ( z ) J ( z ) θ ( τ ) J ( τ ) , and thus changing the variable of integration we get g + n ( z ) = 12 πi (cid:73) ˜ C ( q / τ ( j )) − n J ( z ) − j · θ ( z ) J ( z ) θ ( τ ) j dj. Now recall that θ ( z ) P + n ( J − ( z )) = q − n/ + O ( q / ), so that ( θ ( τ ) P + n ( j − ) − q − n/ τ ( j )) /j is holomorphic in some small neighborhood of 0 in the j -plane. Therefore, for some small RADCHENKO AND VIAZOVSKA loop ˜ C around zero, we have g + n ( z ) = 12 πi (cid:73) ˜ C ( q / τ ( j )) − n J ( z ) − j · θ ( z ) J ( z ) θ ( τ ) j dj = θ ( z )2 πi (cid:73) ˜ C P + n ( j − ) j (cid:0) J ( z ) − j (cid:1) J ( z ) dj = − θ ( z )2 πi (cid:73) ˜ C P + n ( j − ) J ( z ) (cid:0) j − − J − ( z ) (cid:1) dj − = θ ( z ) P + n ( J − ( z )) . The last sign is changed since the contour for j − in the last application of Cauchy’sformula has the opposite orientation. (cid:3) Remark.
From (19) it also follows that K ε ( τ, z ) has a simple pole at z = τ withresidue iπ for all τ ∈ H . We also record here the following identities for K ε : K ε ( τ, − /z ) = ε ( − iz ) / K ε ( τ, z ) ,K ε ( − /τ, z ) = − ε ( − iτ ) / K ε ( τ, z ) . (20)Note that generating functions very similar to (18) have also been used in [17] in thecomputation of traces of singular moduli.4. Interpolation basis for even functions
Let us define a function b εm : R → R by the integral(21) b εm ( x ) = 12 (cid:90) − g εm ( z ) e iπx z dz, where the path of integration is chosen to lie in the upper half-plane and orthogonal tothe real line at the endpoints 1 and −
1. Since g εm has exponential decay at ±
1, the aboveintegral converges. Note that b εm is defined for m ≥ ε = +1 and for m ≥ ε = − b − ( x ) = 0. Proposition 1.
The function b εm : R → R is an even Schwartz function that satisfies (cid:99) b εm ( x ) = εb εm ( x ) and b εm ( √ n ) = δ n,m , n ≥ , m ≥ , where δ n,m is the Kronecker delta. In addition, we have b +0 (0) = 1 .Proof. Clearly, b εm is an even function, since e z ( x ) = e iπx z is even.Let us first prove that b εm belongs to the Schwartz class. We will only consider thecase “ ε = +”, but the same argument will work also in the case “ ε = − ”. Since g + n ( z ) = θ ( z ) P + n ( J − ( z )), it is enough to prove that for each n ∈ N the integral β n ( x ) = 12 (cid:90) − θ ( z ) J − n ( z ) e iπx z dz is a Schwartz function. On the circle arc from − J − ( z ) takes realvalues between 0 and 64, and moreover J − ( ± i/t ) ≤ C exp( − πt ) , t → ∞ , Re( t ) > . By taking the k -th derivative of β n ( x ) with respect to x under the integral we obtain β ( k ) n ( x ) = 12 (cid:90) − θ ( z ) J − n ( z ) Q k ( x, z ) e iπx z dz, OURIER INTERPOLATION ON THE REAL LINE 9 where Q k ( x, z ) are polynomials that satisfy Q n, = 1 and Q k +1 ( x, z ) = ∂∂x Q k ( x, z ) + (2 πizx ) Q k ( x, z ) . Clearly, there exists a constant C k such that | Q k ( x, z ) | ≤ C k (1 + | x | ) k (1 + | z | ) k , thus we get | β ( k ) n ( x ) | ≤ π k +3 C k (1 + | x | ) k (cid:90) / J − n ( e iπt ) e − πx sin( πt ) dt. Here we used a rather crude estimate | θ ( e iπt ) | < t ∈ (0 , / | x | is small, weestimate the above integral by 64 n , for all other values of x we estimate the integral bysplitting it into two parts (where we take δ = ( √ πx ) − ): (cid:90) / J − n ( e iπt ) e − πx sin( πt ) dt = (cid:90) δ J − n ( e iπt ) e − πx sin( πt ) dt + (cid:90) / δ J − n ( e iπt ) e − πx sin( πt ) dt ≤ Cδe − /δ + 64 n e − πδx = e − √ πx (64 n + C/ ( x √ π )) , from which it follows that β n is a Schwartz function.To check that (cid:99) b εm = εb εm we will use the fact that (cid:98) e z = ( − iz ) − / e − /z and the trans-formation property (15): (cid:99) b εm ( x ) = 12 (cid:90) − g εm ( z )( − iz ) − / e iπx ( − /z ) dz = 12 (cid:90) − − g εm ( z )( − iz ) / e iπx ( − /z ) d ( − /z )= 12 (cid:90) − εg εm ( − /z ) e iπx ( − /z ) d ( − /z ) = εb εm ( x ) . In the above computations we always choose the branch of ( − iz ) k/ that takes positivevalues for z on the imaginary semiaxis. Finally, note that b εm ( √ n ) = 12 (cid:90) − g εm ( z ) e iπnz dz is simply the coefficient of q − n/ in the q -expansion of g εm , so that (16) immediately implies b εm ( √ n ) = δ n,m and b +0 (0) = 1. (cid:3) Remark.
Note that (16) also implies that b + m (0) = δ m, , and using the explicit for-mula (18) for the kernel K − , we also get b − m (0) = (cid:40) − , m ≥ , , otherwise . Alternatively, this last equation follows from the Poisson summation formula (cid:88) n ∈ Z b − m ( n ) = (cid:88) n ∈ Z (cid:99) b − m ( n ) = − (cid:88) n ∈ Z b − m ( n ) . To establish other properties of the sequences { b εm ( x ) } m we will need to work withgenerating functions. Let D be the standard fundamental domain for the group Γ θ (asdefined in (4)). For a fixed x define a function F ε ( τ, x ) on the set { τ ∈ H : ∀ k ∈ Z , | τ − k | > } ⊃ D + 2 Z by(22) F ε ( τ, x ) = 12 (cid:90) − K ε ( τ, z ) e iπx z dz, where the contour is the semicircle in the upper half-plane that passes through − τ ) > F ε ( τ, x ) = ∞ (cid:88) n =0 b εn ( x ) e iπnτ , and the series converges absolutely. Our next task is to show that this identity holds forall τ ∈ H . Proposition 2.
For any ε ∈ { + , −} and x ∈ R the function F ε ( τ ; x ) admits an analyticcontinuation to H . Moreover, the analytic continuation satisfies the functional equations F ε ( τ ; x ) − F ε ( τ + 2; x ) = 0 ,F ε ( τ ; x ) + ε ( − iτ ) − / F ε (cid:16) − τ ; x (cid:17) = e iπτx + ε ( − iτ ) − / e iπ ( − /τ ) x . (24) Proof.
To prove the theorem, it is enough to show that there exists an analytic contin-uation to some open set Ω containing the boundary of D , on which the equations (24)hold. Indeed, we can then choose Ω such that D ⊂ Ω ⊂ D ∪ S D ∪ T D ∪ T − D Since ∪ g ∈ Γ θ g Ω = H , we can construct a continuation by repeatedly using (24). Since H is simply-connected, this gives a well-defined analytic function on H .The first functional equation in (24) is clearly satisfied, since the integral that defines F ε automatically defines a 2-periodic function on the open set { τ ∈ H : ∀ k ∈ Z , | τ − k | > } that contains the vertical lines Im( τ ) = ± F ε to some neighborhood of { z ∈ H : | z | = 1 , z (cid:54) = i } by changing thecontour of integration in (22). First, we rewrite the integral as2 F ε ( τ, x ) = (cid:90) i − K ε ( τ, z ) e iπx z dz + (cid:90) i K ε ( τ, z ) e iπx z dz = (cid:90) i − K ε ( τ, z ) e iπx z dz − (cid:90) i − K ε ( τ, − /z ) e iπx ( − /z ) z − dz = (cid:90) i − K ε ( τ, z )( e iπx z + ε ( − iz ) − / e iπx ( − /z ) ) dz, (25)where we have used the transformation property (20). Note, that if τ belongs to D ∪ S D ,then the only poles of K ε ( τ, z ) (as a function of z ) inside D ∪ S D are at z = τ and z = − /τ . Let γ be the circle arc from − i , and let γ be a simple smooth pathfrom − i that lies inside S D and strictly below γ . Denote by F the region enclosedbetween γ and γ . We will now build a continuation of F ε to F and show that it satisfiesthe functional equation. We define a continuation by the contour integral˜ F ε ( τ ; x ) = 12 (cid:90) γ K ε ( τ, z )( e iπx z + ε ( − iz ) − / e iπx ( − /z ) ) dz. OURIER INTERPOLATION ON THE REAL LINE 11 − iγ γ F D τ Sτ
Figure 2.
Fundamental domain for Γ θ and the contour of integration.Clearly, F ε = ˜ F ε for τ with big enough imaginary part, so ˜ F indeed defines an analyticcontinuation to F . For τ ∈ F we compute˜ F ε ( τ ; x ) + ε √− iτ F ε (cid:16) − τ ; x (cid:17) = ˜ F ε ( τ ; x ) − (cid:90) γ εK ε ( − /τ, z ) −√− iτ ( e iπx z + εe iπx ( − /z ) √− iz ) dz = ˜ F ε ( τ ; x ) − (cid:90) γ K ε ( τ, z )( e iπx z + ε ( − iz ) − / e iπx ( − /z ) ) dz = 12 (cid:90) ∂ F K ε ( τ, z )( e iπx z + ε ( − iz ) − / e iπx ( − /z ) ) dz = iπ (cid:88) x ∈F Res z = x (cid:0) K ε ( τ, z )( e iπx z + ε ( − iz ) − / e iπx ( − /z ) ) (cid:1) = e iπx τ + ε ( − iτ ) − / e iπx ( − /τ ) , which is precisely the functional equation that we needed. Similar computation works forthe arc from i to 1. The only thing that is left is to check that F ε has no pole at τ = i .For ε = 1 this follows from the second functional equation, while for ε = − λ ( z ) − e iπzr + ε ( − iz ) − / e iπ ( − /z ) r vanish at z = i , so that they cancel the double pole at i coming from J ( i ) − J ( z ), and hence the integral (25) converges at τ = i . (cid:3) As an immediate corollary, we obtain that formula (23) is valid for all τ ∈ H . Thisalready implies that for all δ > b εn ( x ) = O ((1 + δ ) n ). In the next section weprove a much stronger estimate.Note that the only properties of K ε that were used in the proof are the modularity in τ and in z , as well as the fact that the only poles are at z ∈ Γ θ τ , and that the residue at z = τ is equal to 1 / ( iπ ). 5. Growth estimate
The main result of this section is the following.
Theorem 4.
For any ε ∈ { + , −} the numbers b εn ( x ) satisfy | b εn ( x ) | = O ( n ) uniformly in x . To prove this we will use the following general result that goes back to Hecke (see, forexample, [1, Lemma 2.2, (ii)]).
Lemma 1.
If a -periodic analytic function f : H → C has a Fourier expansion f ( τ ) = (cid:80) n ≥ a n e iπnτ and for some α > it satisfies | f ( τ ) | ≤ C Im( τ ) − α for Im( τ ) < c, then for all sufficiently large n we have | a n | ≤ C (cid:0) eπα (cid:1) α n α . To prove Theorem 4 we will apply this lemma to the generating function F ε ( τ ; x ). Theestimate of | F ε ( τ ; x ) | naturally splits into two parts: combinatorial (estimate for valuesof a certain cocycle) and analytic (estimate for the value of a contour integral).To simplify notation, we will write F ε ( τ ) instead of F ε ( τ ; x ). Let us define a Γ θ -cocycle { φ A ( τ ) } A ∈ Γ θ by φ T ( τ ) = 0 ,φ S ( τ ) = e iπx τ + ε ( − iτ ) − / e iπx ( − /τ ) , (26)and extend it to Γ θ by the cocycle relation φ AB = φ B + φ A | B (we write | for | − ε / ). Sincethe only relation in the group Γ θ is S = 1, and φ S + φ S | S = 0, it indeed extends to aunique Γ θ -cocycle. The choice is motivated by the functional equations (24) which imply(27) F ε ( τ ) − ( F ε | − ε / A )( τ ) = φ A ( τ )for all A ∈ Γ θ .First, we need the following elementary lemma. Lemma 2.
For any τ ∈ H with | τ | ≥ and any sequence of non-zero integers { n j } j ≥ define a sequence of numbers τ j ∈ H as follows: τ = τ,τ j = 2 n j − τ j − , j ≥ . Then the sequence { Im( τ j ) } j ≥ is strictly decreasing, and Im( τ j ) ≤ j − for all j ≥ .Proof. First, observe that | τ j | > j ≥ τ j ) ≥ Im( τ j +1 ) the follows from Im( τ j +1 ) = Im( τ j ) / | τ j | < Im( τ j ).For a, b ∈ R denote by D ( a, b ) the half-disk with center ( a + b ) / a and b . Let D be any such half-disk that does not intersect D ( − ,
1) and set D (cid:48) = SD . Then a simple calculation shows thatdiam( D (cid:48) ) ≤ diam( D )1 + diam( D ) . Note that τ ∈ (cid:83) n (cid:54) =0 D (2 n − , n + 1), so τ lies in some half-disk of diameter 2. Denotethis half-disk by D , and define D j +1 = 2 n j + SD j . Then τ j ∈ D j and all D j do notintersect D ( − , D j hasdiameter at most 2 / (2 j − τ j ) ≤ / (2 j − (cid:3) The following lemma allows us to estimate values of certain cocycles.
OURIER INTERPOLATION ON THE REAL LINE 13
Lemma 3.
Let { ψ A } A ∈ Γ θ be a cocycle (with respect to | = | − εk/ ) such that ψ T = 0 , | ψ S ( τ ) | ≤ | τ | α + Im( τ ) − β for some α, β ≥ . Let τ (cid:48) ∈ D , A ∈ Γ θ , and τ = Aτ (cid:48) ∈ H and suppose that Im( τ ) ≤ .Then | ψ A ( τ (cid:48) ) | ≤ | τ | α + Im( τ ) − α − + 2 Im( τ ) − β − . Proof.
Let us consider the case when A = ST n m ST n m − S . . . T n S. By applying the cocycle relation repeatedly, we get that ψ A = ψ S + ψ S | A + ψ S | A + · · · + ψ S | A m , where we write A j = T n j S . . . T n S . Hence | ψ A ( τ (cid:48) ) | ≤ m (cid:88) j =0 | ψ S ( τ j ) || c j τ + d j | k , where A j = ( a j b j c j d j ) and τ j are defined by (cid:40) τ = τ (cid:48) ,τ j = 2 n j − /τ j − . Under these definitions τ j = a j τ (cid:48) + b j c j τ (cid:48) + d j and τ = − /τ m . Multiplying both sides of the aboveinequality by Im( τ ) k/ we getIm( τ (cid:48) ) k/ | ψ ( τ (cid:48) ) | ≤ m (cid:88) j =0 Im( τ j ) k/ | ψ S ( τ j ) | Lemma 2 implies that Im( τ ) − ≥ m − τ j ) ≥ Im( τ ) for j = 0 , . . . , m . We alsohave | τ j | ≤ Im( τ ) − for j = 0 , . . . , m −
1, since Im( τ ) ≤ Im( τ j +1 ) = Im( τ j ) / | τ j | ≤ | τ j | − .ThereforeIm( τ (cid:48) ) k/ | ψ ( τ (cid:48) ) | ≤ m (cid:88) j =0 Im( τ j ) k/ ( | τ j | α + Im( τ j ) − β ) ≤ Im( τ (cid:48) ) k/ m (cid:88) j =0 ( | τ j | α + Im( τ j ) − β ) ≤ Im( τ (cid:48) ) k/ ( | τ | α + m Im( τ ) − α + ( m + 1)Im( τ ) − β ) ≤ Im( τ (cid:48) ) k/ ( | τ | α + Im( τ ) − α − + 2 Im( τ ) − β − ) , where in the last line we used m + 1 ≤ m − ≤ Im( τ ) − .The proof in the other cases, i.e., when A is not of the form ST n k ST n k − S . . . T n S can be completed using similar estimates. (cid:3) Next, we deal with the analytic part of the estimate. For Theorem 1 the case n = 0 ofthe lemma below will suffice, but we need the general form for the proof of Theorem 2. Lemma 4.
For each n, k ≥ there exists an absolute constant C n,k > such that theinequality (cid:12)(cid:12) x k d n dx n F ε ( τ ; x ) (cid:12)(cid:12) ≤ C n,k (1 + Im( τ ) − ( n + k +1) / ) holds for all τ ∈ D . Proof.
Let τ be any point in D . Since F ε ( it ) is real for all t >
0, from the Schwarzreflection principle we get that(28) F ε ( − τ ) = F ε ( τ ) . Using this symmetry we reduce the inequality to the case τ ∈ D , where D = { τ ∈D : Re( τ ) ∈ ( − , } . Observe that Im( J ( τ )) < τ ∈ D and Im( J ( τ )) ≥ τ ∈ D (cid:114) D . Indeed, since J is a Hauptmodul, the map J : D → C is injective. Theidentity (28) implies that for τ ∈ D the value F ε ( τ ) is real if and only if τ lies on theimaginary axis. It is easy to see from (13) that Im( J ( τ )) < τ ∈ D and Im( τ ) (cid:29) τ ∈ D .Define L = { w ∈ C | Re( w ) = J ( i ) = 1 / , Im( w ) > } , and let (cid:96) be the preimage of L under the map J : D → C (see Figure 5). Then (cid:96) is asmooth path contained in D (cid:114) D and goes from i to 1. We set γ to be the path S(cid:96) ∪ (cid:96) that goes from − | z | and | z | − are bounded on γ and that γ has finitelength (this fact will follow from the computations below).As in the proof of Proposition 1 let Q n ( x, z ) be a polynomial for which d n dx n e iπx z = Q n ( x, z ) e iπx z . We have x k d n dx n F ε ( τ ; x ) = 12 (cid:90) − K ε ( τ, z ) x k Q n ( x, z ) e iπx z dz. From (20) we find x k d n dx n F ε ( τ ; x ) = 12 (cid:90) i K ε ( τ, z ) x k (cid:0) Q n ( x, z ) e iπx z + ε ( − iz ) − / Q n ( x, − /z ) e iπx ( − /z ) (cid:1) dz. Without loss of generality, we may assume x ≥
0. Since | z | is bounded for z ∈ γ , anymonomial z α x β with 0 ≤ β ≤ n is majorized by 1 + x n , and thus for all such z wehave | x k Q n ( x, z ) | (cid:28) n,k,γ x n + k . Then (cid:12)(cid:12) x k d n dx n F ε ( τ ; x ) (cid:12)(cid:12) (cid:28) (cid:90) (cid:96) | K ε ( τ, z ) x k | (cid:12)(cid:12) Q n ( x, z ) e iπx z + ε ( − iz ) − / Q n ( x, − /z ) e iπx ( − /z ) (cid:12)(cid:12) | dz |(cid:28) (cid:90) (cid:96) | K ε ( τ, z ) | (1 + x k + n ) (cid:0) e − πx Im( z ) + | z | − / e − πx Im( − /z ) (cid:1) | dz | . (29)Next, we observe that (1 + x k + n ) e − πx Im( z ) (cid:28) k + n z ) − k − n . Note, that 1 ≤ | z | (cid:28) z ∈ (cid:96) . Hence, we get (cid:12)(cid:12) x k d n dx n F ε ( τ ; x ) (cid:12)(cid:12) (cid:28) (cid:90) (cid:96) | K ε ( τ, z ) | (cid:0) z ) − k − n + | z | − / Im( − z ) − k − n (cid:1) | dz | = (cid:90) (cid:96) | K ε ( τ, z ) | (cid:0) z ) − k − n + | z | k + n − / Im( z ) − k − n (cid:1) | dz |(cid:28) (cid:90) (cid:96) | K ε ( τ, z ) | (cid:0) z ) − k − n (cid:1) | dz | . (30)Without loss of generality, we may also assume that | τ − i | ≥ /
10, since we can recoverthe inequality in the region | τ − i | < /
10 by applying the maximum modulus principletogether with the functional equation for F ε . OURIER INTERPOLATION ON THE REAL LINE 15 − i D D (cid:114) D τ S(cid:96) (cid:96) Figure 3.
Deforming the contour of integration.For τ with Im( τ ) ≥ / | τ − i | > /
10 we can estimate | K ε ( τ, z ) | (cid:28) | θ ( z ) | with aconstant independent of τ . Since | θ ( z ) | behaves like Im( z ) − e − π/ Im( z ) as z approaches 1,by splitting the integral into { z : Im( z ) ≥ /x } and { z : Im( z ) < /x } we obtain | F ε ( τ ; x ) | (cid:28) (1 + x ) e − cπx , which clearly implies the needed inequality.Now let Im( τ ) < /
2. To bound | K ε ( τ, z ) | we use the following estimates | θ ( z ) | (cid:28) | J ( z ) | − / Im( z ) − / , | − λ ( z ) | (cid:28) | J ( z ) | / , which hold for all z ∈ D near the cusp 1 (such z correspond to large values of | J ( z ) | ).The first inequality follows from the fact that θ ( z ) J ( z ) is a holomorphic modular form ofweight 4 for Γ θ (the term Im( z ) − / comes from the modular transformation). To provethe second inequality, simply note that (1 − λ ( z )) = 1 − J ( z ). Thus, we get | K + ( τ, z ) | (cid:28) Im( τ ) − / | J ( τ ) | / | J ( z ) | / Im( z ) − / | J ( z ) − J ( τ ) | , (31) | K − ( τ, z ) | (cid:28) Im( τ ) − / | J ( τ ) | / | J ( z ) | / Im( z ) − / | J ( z ) − J ( τ ) | . From now on, we make all estimates for z ∈ (cid:96) with Im( z ) < /
2, and we define t > J ( z ) = 1 /
64 + it . For such z we can use the following simple geometricestimate (recall that Im( J ( τ )) < | J ( τ ) − J ( z ) | (cid:29) (cid:112) | J ( τ ) | + | J ( z ) | . Let k : C (cid:114) [0 , ) → D be the inverse of J on D , so that z = k (1 /
64 + it ). We have J (cid:48) ( τ ) = iπf ( τ ) J ( τ ), where f ( τ ) = θ ( τ )(1 − λ ( τ )) is a holomorphic modular form ofweight 2. Since f does not vanish at the cusp 1, we have that | f ( z ) | (cid:29) Im( z ) − , and thus(33) | dz | = | k (cid:48) (1 /
64 + it ) | | dt | = | dt || J (cid:48) ( k ( + it )) | (cid:28) | dt || J ( z ) | · Im( z ) − . Note that this last estimate readily implies that (cid:96) has finite length.
Inequality (30) implies that it suffices to bound (cid:90) (cid:96) | K ε ( τ, z ) | Im( z ) − m | dz | for m ≥ . From inequalities (31), (32), (33) we deduce (cid:90) (cid:96) | K ε ( τ, z ) | Im( z ) − m | dz | (cid:28) (cid:90) ∞ | J ( τ ) | / t − / Im( z ) / − m Im( τ ) / (cid:112) t + | J ( τ ) | dt. We will also need the estimate | J ( z ) | (cid:29) e π/ Im( z ) for Im( z ) small enough. Indeed,this inequality follows from the q -expansion (14) of J ( z ) at the cusp 1. This implies thatIm( z ) − m (cid:28) m log m (1 + | J ( z ) | ). Thus, we have (cid:90) (cid:96) | K ε ( τ, z ) Im( z ) − m | dz | (cid:28) Im( τ ) − / (cid:90) ∞ | J ( τ ) | / t − / log m (1 + t ) dt (cid:112) | J ( τ ) | + t = Im( τ ) − / (cid:90) ∞ t − / log m (1 + t | J ( τ ) | ) dt √ t . By using an obvious inequality log(1 + ab ) ≤ log(1 + a ) + log(1 + b ), we estimate the lastintegral byIm( τ ) − / m (cid:88) j =0 (cid:18) mj (cid:19) log j (1 + | J ( τ ) | ) (cid:90) ∞ t − / log m − j (1 + t ) dt √ t (cid:28) m (cid:88) j =0 c j,m Im( τ ) − j − / , where c j,m = (cid:0) mj (cid:1) (cid:82) ∞ (1 + t ) − / t − / log m − j (1 + t ) dt are finite constants, and we haveused the reverse inequality log(1 + | J ( τ ) | ) (cid:28) Im( τ ) − , that also follows from (14).The estimates in the case “ ε = − ” are completely analogous, except that we need tochange the exponent 3 / / (cid:3) We are now ready to prove Theorem 4.
Proof of Theorem 4.
Let τ ∈ H be an arbitrary point in the upper half-plane withIm( τ ) ≤ D or anyof its translates by elements of Γ θ . Let τ = aτ (cid:48) + bcτ (cid:48) + d , where τ (cid:48) ∈ D and A = ( a bc d ) ∈ Γ θ .By (27) we have χ ε ( A ) j θ ( τ (cid:48) , A ) F ε (cid:16) aτ (cid:48) + bcτ (cid:48) + d (cid:17) = F ε ( τ (cid:48) ) − φ A ( τ (cid:48) ) . Combining the results of Lemma 4 and Lemma 3 (which we apply to ψ A = φ a with α = 0and β = 1 /
2) we obtain | F ε ( τ ) | ≤ Im( τ (cid:48) ) / Im( τ ) / | F ε ( τ (cid:48) ) | + Im( τ (cid:48) ) / Im( τ ) / | φ A ( τ (cid:48) ) |≤ C Im( τ (cid:48) ) / + Im( τ (cid:48) ) − / Im( τ ) / + Im( τ (cid:48) ) / (1 + Im( τ ) − / + 2 Im( τ ) − / ) . (Here C is the constant from Lemma 4.) If c = 0, then Im( τ (cid:48) ) = Im( τ ) and thus | F ε ( τ ) | ≤ C (1 + Im( τ ) − / ) + Im( τ ) / + Im( τ ) − + 2 Im( τ ) − / . If, on the other hand, c >
0, then we have Im( τ ) < Im( τ (cid:48) ) andIm( τ )Im( τ (cid:48) ) = Im( τ (cid:48) ) | cτ (cid:48) + d | ≤ , OURIER INTERPOLATION ON THE REAL LINE 17 and thus we again get the estimate | F ε ( τ ) | ≤ C Im( τ ) − / + Im( τ ) − / + Im( τ ) − / + 2 Im( τ ) − . Therefore, an application of Lemma 1 gives | b εn ( x ) | (cid:28) n . (cid:3) The exponent “2” in Theorem 4 is not optimal, but for the proof of Theorem 1 anypolynomial bound would suffice.6.
Proof of the main results
Now that we know that b εn ( x ) have polynomial growth in n , the proof of Theorem 1and Theorem 2 is not hard.Recall the definition of Schwartz functions: S = { f ∈ C ∞ ( R ) : (cid:107) f (cid:107) α,β < ∞ ∀ α, β ≥ } , where the seminorms (cid:107) · (cid:107) α,β are defined by (cid:107) f (cid:107) α,β = sup x ∈ R | x α f ( β ) ( x ) | . Proof of Theorem 1.
Let us define a n ( x ) := b + n ( x ) + b − n ( x )2 . Lemma 1 implies that (cid:98) a n ( x ) = b + n ( x ) − b − n ( x )2 . Our aim is to show that (1) holds for all f ∈ S even . Theorem 4 implies that the series onthe right-hand side of ∞ (cid:88) n =0 a n ( x ) f ( √ n ) + ∞ (cid:88) n =0 (cid:98) a n ( x ) (cid:98) f ( √ n )converges absolutely. Moreover, it follows from the definition of b εn and the functionalequations (24) that for any τ ∈ H we have(34) e τ ( x ) = ∞ (cid:88) n =0 a n ( x ) e τ ( √ n ) + ∞ (cid:88) n =0 (cid:98) a n ( x ) (cid:98) e τ ( √ n ) , where e τ ( x ) = e iπτx .Let S even be the space of even Schwartz functions. For x ≥ φ x on S even given by φ x ( f ) := f ( x ) − ∞ (cid:88) n =0 a n ( x ) f ( √ n ) − ∞ (cid:88) n =0 (cid:98) a n ( x ) (cid:98) f ( √ n ) . It follows from Theorem 4 that φ x is a tempered distribution, i.e., it is continuous withrespect to convergence in S even . From equation (34) we see that φ x vanishes on thesubspace spanned by { e τ } τ ∈ H . Our goal is to show that φ x vanishes on the whole S even .Let C be the space of compactly supported even C ∞ functions on R . Recall, that C dense in S even (see [15, pp. 74-75]). Therefore, it suffices to show (1) for f ∈ C . Let f bea function in C . We may assume that f ( x ) = F ( x ) e − πx where F is a C ∞ function with compact support on R . Consider the one-dimensionalFourier transform of F (cid:98) F ( s ) := ∞ (cid:90) −∞ F ( t ) e − πist dt. Note, that (cid:98) F is a Schwartz function. By the Fourier transform inversion formula we have f ( x ) = F ( x ) e − πx = ∞ (cid:90) −∞ (cid:98) F ( s ) e πisx − πx d s = ∞ (cid:90) −∞ (cid:98) F ( s ) e i +2 s ( x ) d s. Define h T := T (cid:90) − T (cid:98) F ( s ) e i +2 s ( x ) d s. It is easy to see that for all seminorms (cid:107) · (cid:107) α,β (cid:107) f − h T (cid:107) α,β → T → ∞ . Therefore, for all x ≥ φ x ( f − h T ) → T → ∞ . On the other hand, we have φ x ( h T ) = T (cid:90) − T (cid:98) F ( s ) φ x ( e i +2 s ) d s = 0 . This finishes the proof of Theorem 1. (cid:3)
We are also ready to prove Theorem 2.
Proof of Theorem 2.
First, we observe that the image of Ψ is contained in the kernel of L .Indeed, the Poisson summation formula implies (cid:88) n ∈ Z f ( n ) = (cid:88) n ∈ Z (cid:98) f ( n )for all f ∈ S as well as f ∈ S even . This identity is equivalent to L ◦ Ψ( f ) = 0.Next, we construct the function Φ : ker L → S even such that Ψ ◦ Φ = I ker L . To this endwe consider the mapΦ : ker L → S even , (( x n ) , ( y n )) (cid:55)→ (cid:88) n x n a n ( x ) + y n (cid:98) a n ( x ) . We need to show that Φ is well-defined. Since S is complete with respect to the family ofnorms (cid:107) · (cid:107) α,β it is enough to prove that for any fixed α, β ≥ (cid:107) a n (cid:107) α,β ) n and ( (cid:107) (cid:98) a n (cid:107) α,β ) n have at most polynomial growth in n . Equivalently, it is enough to provethat the sequences ( (cid:107) b εn (cid:107) α,β ) n have polynomial growth.As before, let Q k ( x, z ) be the polynomial defined by d k dx k e iπx z = Q k ( x, z ) e iπx z . Let U ( τ ; x ) be the generating function U ( τ ; x ) = x α d β dx β F ε ( τ ; x ) = x α ∞ (cid:88) n =0 d β dx β b εn ( x ) e iπnτ . OURIER INTERPOLATION ON THE REAL LINE 19
Then, following the proof of Proposition 2, we see that the generating function U satisfiesthe functional equation U ( τ ) − ( U | − ε / A )( τ ) = φ A ( τ ) , where φ A is the cocycle defined by φ T ( τ ) = 0 ,φ S ( τ ) = x α Q β ( x, τ ) e iπx τ + ε ( − iτ ) − / x α Q β ( x, − /τ ) e iπx ( − /τ ) . Using the estimates | x k τ l e iπx τ | (cid:28) | τ | l Im( τ ) − k/ < Im( τ ) − k + | τ | l and | x k τ − l e iπx ( − /τ ) | (cid:28) | τ | k − l Im( τ ) − k/ < Im( τ ) − k + | τ | k − l , and in case k < l the replacing | τ | k − l by Im( τ ) k − l , we see that Lemma 3 can be appliedto { φ A } A ∈ Γ θ (for some choice of α and β in Lemma 3). Lemma 4 implies that for τ ∈ D we have U ( τ ; x ) (cid:28) τ ) − ( α + β +1) / . Arguing the same way as in the proof of Theorem 4 we obtain that for some
C > τ ∈ H with Im( τ ) < | U ( τ ; x ) | (cid:28) Im( τ ) − C , which implies that (cid:107) b εn (cid:107) α,β (cid:28) n C .Therefore, the map Φ is well-defined.Now Theorem 1 implies that Φ ◦ Ψ = I S even and Proposition 1 implies that Ψ ◦ Φ = I ker L .This finishes the proof. (cid:3) Interpolation basis for odd functions
The case of odd Schwartz functions is very similar to the even case. The proofs areeasy enough to adapt to this case, so we will just give the general outline. The role ofthe Gaussian e τ ( x ) = e iπτx is played by the Schwartz function o τ ( x ) = xe iπτx , that satisfies (cid:98) o τ ( ξ ) = − i ( − iτ ) − / o − /τ ( ξ ) . To construct the interpolation basis for odd Schwartz functions we use the same idea asbefore: to get an eigenfunction we integrate o τ over τ with some “modular weight”. Moreprecisely, let h εn : H → C be holomorphic functions with the following properties: h εn ( z + 2) = h εn ( z ) , ( − iz ) − / h εn ( − /z ) = εh εn ( z ) ,h + n ( z ) = q − n/ + O ( q / ) , z → i ∞ ,h − n ( z ) = q − n/ + O (1) , z → i ∞ ,h εn (1 + i/t ) → , t → ∞ . Once again, we may assume that they have the form h + n ( z ) = θ ( z ) Q + n ( J − ( z )) ,h − n ( z ) = θ ( z )(1 − λ ( z )) Q − n ( J − ( z )) , (35) where Q ± n ∈ Q [ x ] are monic of degree n and Q − n has no constant term. The first few ofthese functions are h +0 = θ,h +1 = θ · ( J − − ,h +2 = θ · ( J − − J − + 76) , h − = θ · (1 − λ ) · ( J − ) ,h − = θ · (1 − λ ) · ( J − − J − ) ,h − = θ · (1 − λ ) · ( J − − J − + 168 J − ) . By the same arguments as in the even case, we establish generating functions for h εn ,which turn out to be the same, except that τ and z are now switched. Theorem 5.
The generating functions for { h + n ( z ) } n ≥ and { h − n ( z ) } n ≥ are given by ∞ (cid:88) n =0 h + n ( z ) e iπnτ = θ ( τ )(1 − λ ( τ )) θ ( z ) J ( z ) J ( z ) − J ( τ ) = − K − ( z, τ ) , ∞ (cid:88) n =1 h − n ( z ) e iπnτ = θ ( τ ) J ( τ ) θ ( z )(1 − λ ( z )) J ( z ) − J ( τ ) = − K + ( z, τ ) . (36)Similarly, define c εm : R → R by d εm ( x ) = 12 (cid:90) − h εm ( z ) xe iπx z dz. Proposition 3.
The function d εm : R → R is odd, belongs to the Schwartz class, andsatisfies (cid:99) d εm ( x ) = ( − iε ) d εm ( x ) and d εm ( √ n ) = δ n,m √ n, n ≥ , where δ n,m is the Kronecker delta. Moreover, lim x → d + m ( x ) x = δ m, . Furthermore, we have the following estimate on the growth of d ± n ( x ) as a function of n . Theorem 6.
For any ε ∈ { + , −} the numbers d εn ( x ) satisfy d εn ( x ) = O ( n / ) uniformly in x . The proof of this estimate is also based on analyzing the growth as Im( τ ) → G ε ( τ, x ) = (cid:88) n ≥ d εn ( x ) e iπnτ . In this case the functional equations for G ε are G ε ( τ ; x ) − G ε ( τ + 2; x ) = 0 ,G ε ( τ ; x ) + ε ( − iτ ) − / G ε (cid:16) − τ ; x (cid:17) = xe iπτx + ε ( − iτ ) − / xe iπ ( − /τ ) x . (37)The difference in exponents of ( − iτ ) come from the fact that the weight of K ε ( z, τ ) in τ is now 3 / /
2, but with appropriate changes the proof still goes through.Finally, we get the following interpolation theorem for odd Schwartz functions.
OURIER INTERPOLATION ON THE REAL LINE 21
Theorem 7.
For any odd Schwartz function f : R → R and any x ∈ R we have (38) f ( x ) = d +0 ( x ) f (cid:48) (0) + i (cid:98) f (cid:48) (0)2 + ∞ (cid:88) n =1 c n ( x ) f ( √ n ) √ n − ∞ (cid:88) n =1 (cid:98) c n ( x ) (cid:98) f ( √ n ) √ n , where c n ( x ) = ( d + n ( x ) + d − n ( x )) / . As in the even case, the functional equations for G ε show that (38) holds for o τ ( x ), soone only needs to show that o τ are dense in the space of odd Schwartz functions, whichcan be done by an approximation argument, similarly to the proof of Theorem 1. Remark.
As in the even case, using the explicit formula for the kernels, we get d + m (cid:48) (0) = δ m, ,d − m (cid:48) (0) = − r ( m ) , m ≥ , where r ( m ) is the number of representations of m as the sum of squares of 3 integers.Taking x = 0 in (38) we get the following identity f (cid:48) (0) + ∞ (cid:88) n =1 r ( n ) f ( √ n ) √ n = i (cid:98) f (cid:48) (0) + ∞ (cid:88) n =1 r ( n ) i (cid:98) f ( √ n ) √ n , valid for arbitrary odd Schwartz functions.8. Open questions and concluding remarks
Let us indicate some further directions and observations related to Theorem 1.
Function space.
While in this paper we have only worked with the space of Schwartzfunctions, it is interesting to ask in what generality the interpolation formula (1) holds.The best possible scenario would be a positive answer to the following question.
Question 1.
Do the results of Theorems 1 and 7 hold whenever the right-hand sideconverges absolutely? Here we implicitly assume that f and (cid:98) f are continuous, so thatpoint evaluation is well-defined. However, even to find explicit conditions for when the convergence is absolute, onewould need to obtain exact bounds on the growth of b εn ( x ), which appears to be difficult.Let us outline a simple approximation argument that shows that the interpolation formulais true whenever both f and (cid:98) f decay sufficiently fast: Proposition 4.
Let f be an even integrable function. If f ( x ) and (cid:98) f ( x ) are both boundedby (1 + | x | ) − , then the summation formula (1) holds.Proof sketch. Indeed, for every
T > R T thattakes values in S : R T ( f )( x ) = T / e i/T · ( e iT ∗ f )( x ) = T / e − πx /T (cid:90) R f ( x − y ) e − πT y dy. The Fourier transform is then given by (cid:92) R T ( f )( x ) = T / e iT ∗ ( e i/T · (cid:98) f )( x ) = T / (cid:90) R f ( x − y ) e − πT y − π ( x − y ) /T dy. Then a routine calculation shows that |R T ( f )( x ) − f ( x ) | (cid:28) (1 − e − πx /T ) | f ( x ) | + T − / max y ∈ [ x − ,x +1] | f (cid:48) ( y ) | , and similarly | (cid:92) R T ( f )( x ) − (cid:98) f ( x ) | (cid:28) (cid:16) − e − πx T/ (1+ T ) √ T − (cid:17) | (cid:98) f ( x ) | + T − / max y ∈ [ x − ,x +1] | (cid:98) f (cid:48) ( y ) | . By summing up these estimates for x = √ n over n ≥ T → ∞ wesee that the proof will be complete if we can show that f (cid:48) ( x ) and (cid:98) f (cid:48) ( x ) decay as (1 + | x | ) − l for some l > a n ( x ) = O ( n )). We consider only bounding f (cid:48) ( x ), since one canobtain the other estimate by interchanging f and (cid:98) f . It was pointed out to the authorsby Emanuel Carneiro that this can be done by using the following simple observation:if g is a C -smooth function on [1 , ∞ ) that satisfies | g ( x ) | (cid:28) x − k and | g (cid:48)(cid:48) ( x ) | (cid:28) | g (cid:48) ( x ) | (cid:28) x − k/ . Indeed, then by the Fourier inversion formula we have | f (cid:48)(cid:48) ( x ) | (cid:28) | f (cid:48) ( x ) | (cid:28) (1 + | x | ) − / , and this proves theproposition.To prove the above observation: let | g (cid:48)(cid:48) ( x ) | ≤ | g ( x ) | ≤ Cx − k . Then Taylor’stheorem with remainder in the Lagrange form implies that for any ∆ ≥ | g ( x + ∆) − g ( x ) − g (cid:48) ( x )∆ | ≤ ∆ , from which we get, taking ∆ = 2 √ Cx − k , that | g (cid:48) ( x ) | ≤ ∆2 + 2 Cx − k ∆ = 2 √ C x − k/ , as required. (cid:3) Of course, the number “13” in the above proposition can be improved by using morecareful estimates.
Relation to the Laplace transform.
The basis functions that we have constructedare all of the form f ( x ) = 12 (cid:90) − g ( z ) e iπx z dz for some weakly holomorphic modular form g (in the odd case, f is multiplied by x ). Toget an alternative expression for f we can shift the contour of integration to the rectanglepassing through − − iT , 1 + iT , and 1. A simple computation then shows that f ( x ) = sin( πx ) (cid:90) T g (1 + it ) e − πx t dt + e − πx T (cid:90) − g ( s + iT ) e iπsx ds. If we take T to infinity, then we see that for all x greater than the order of the pole of g at i ∞ we have f ( x ) = sin( πx ) (cid:90) ∞ g (1 + it ) e − πx t dt. The integral on the right is simply the Laplace transform of g (1+ it ) evaluated at πx . Thiscan be used to show that all but finitely many real zeros of b ± m ( x ) are of the form ±√ n .Combined with the q -expansion of g at infinity, this also implies that b ± m extends analyti-cally to an entire function. Alternatively, this also follows directly from the definition (21). OURIER INTERPOLATION ON THE REAL LINE 23
Sine-sinh ratio.
The function d +0 ( x ) is quite special. Recall that it is defined by d +0 ( x ) = 12 (cid:90) − θ ( z ) xe iπx z dz. Changing the contour of integration as before, we get d +0 ( x ) = x sin( πx ) (cid:90) ∞ θ (1 + it ) e − πx t dt. Next, integrating the q -expansion of θ term by term and using the identity (cid:88) n ∈ Z ( − n π ( x + n ) = 1 x sinh( πx )we find that d +0 ( x ) is, in fact, an elementary function: d +0 ( x ) = sin( πx )sinh( πx ) . Note that d +0 ( x ) and its Fourier transform (cid:99) d +0 ( x ) = ( − i ) d +0 ( x ) both vanish at x = ±√ n for all n ≥
0. It follows from Theorems 1 and 7 that any Schwartz function with thisproperty must be a multiple of d +0 .It appears that this function was first considered by Ramanujan in [11], where hestudies a number of integrals involving similar expressions, and, in particular, shows theFourier invariance of d +0 (see [11, eq. 34]). It is also directly related to the so-calledMordell integral [8], which played an important role in Zwegers’s seminal work on mocktheta functions [19]. References [1] B. C. Berndt, M. I. Knopp,
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