Hecke-type converse theorem

Applying this, we have

W_ν(s) = Z _∞

0

Γ(ν)e^−2x

ν−1

∑

l=0

(2x)^l l!

! e^xx^sdx

x

=Γ(ν)

ν−1

∑

l=0

2^l l!

Z _∞

0

e^−xx^s+ldx x

=Γ(ν)

ν−1

∑

l=0

2^l

l!Γ(s+l).

By Stirling’s approximation for gamma function, for anyµ>0we have Γ(s)∼√

2πt^σ−1/2e^−π|t|/2as|t| →∞, (s=σ+it).

W_ν(s)∼Γ(ν)

ν−1

∑

l=0

2^l l!(√

2πt^σ+l−1/2e^−π|t|/2) =O(t^σ+ν−3/2e^−π|t|/2) =O(t^−µ)as|t| →∞.

for fixedσ>0.

(3) By definition, we observe that W_ν(s)is the Mellin transform of Γ(ν,2x)e^x. So we only need to check the absolutely convergence of Γ(ν,2x)e^x for x>0. Clearly Γ(ν,2x)e^x is a continuous function. Since|Γ(ν,2x)e^x| ≤Γ(ν)e^xfor x≥0, we haveΓ(ν,2x)e^x=O(1)as x→0. In the proof of (1), we haveΓ(ν,2x)∼(2x)^ν−1e^−2xas|x| →∞, we get the required growth condition as x→∞.

Therefore the inversion integral ofΓ(ν,2x)e^xalong the vertical lineRe(s) =c>0exists.

Proof 28 As in the proof of Lemma 2, using the Euler-Gauss formula for gamma function, for n≥1we have

c⁺_f(n)≤C₁(−1)ⁿ

−v−1 n

,

for some constant C₁>0. Next, since1−k≥1we can use(56)as follows. For n≥1, we have Γ(1−k,4πny) =Γ(1−k)e^−4πny

−k

∑

l=0

(4πny)^l

l! =







O(e^−4πny), as y→0, O(y^−ke^−4πny), as y→∞.

Again by the Euler-Gauss formula, we have

|c⁻_f(−n)Γ(1−k,4πny)| ≤C₂(−1)ⁿe^−4πny

−k

∑

l=0

(4πny)^l l!

−v−l−1 n

for some constant C₂>0. Substituting these results to the Fourier expansion of f , we have

|f(z)| ≤C₁

∞

∑

n=0

(−1)ⁿ

−v−1 n

e^−2πny+c⁻_f(0)y^1−k+C2

−k

∑

l=0

(4πny)^l l!

∞

∑

n=1

(−1)ⁿ

−v−l−1 n

e^−2πny

≤C₁(1−e^−2πy)^−v−1+c⁻_f (0)y^1−k+C₂

−k

∑

l=0

(4πny)^l

l! ((1−e^−2πy)^−v−l−1−1). (57) This shows that the given series f(z)is convergent absolutely and uniformly on any compact subset of H.

Since c⁻_f(−n) =O(n^v), we obtain that the formal Fourier series (1−k)c⁻_f(0)−(4π)^1−k

∑

n≥1

c⁻_f(−n)n^1−kqⁿ

is uniformly convergent on any compact subset ofH and defines a holomorphic function. Moreover, by using Proposition 8 and(53), we have

∆k(f(z)) =−ξ_2−k(ξk(f(z))) =0.

Now Theorem 11 tells us that f(z)is a real analytic function onH.

Since1−k≥1and(1−e^−2πy)^−v−1=O(y^−v−1)as y→0, by using(57)we have f(z) =O(y^−v+k−1)as y→0.

Next, we have

f(z)−c⁺_f(0)−c⁻_f(0)y^1−k=e^2πizg(z), where

g(z) =

∞

∑

n=0

c⁺_f(n+1)qⁿ+

∑

n≤−2

c⁻_f(n+1)Γ(1−k,−4π(n+1)y)qⁿ.

By(50), for any n≤ −2we have

|Γ(1−k,−4π(n+1)y)qⁿ| ≤((−4π(n+1)y)^−ke^4π(n+1)y)e^−2πny

= (−4π(n+1)y)^−ke^2π(n+2)y as y→∞. Then we have

|g(z)| ≤C₁

∞ n=0

∑

(−1)ⁿ⁺¹

−v−1 n+1

e^−2πny+

∑

n≤−2

(−(n+1))^v(−4π(n+1)y)^−ke^2π(n+2)y

≤C₁((1−e^−2πy)^−v−1−1) +

∑

n≤−2

(−(n+1))^v(−4π(n+1)y)^−ke^2π(n+2)y. This shows that g(z)is bounded in a neighborhood of i∞. Therefore we have

f(z)−c⁺_f(n)−c⁻_f (n)y^1−k=O(e^−2πy)as y→∞.

The following lemma and the two specific functions serve to combine the two L-functionsΛN and ΩNwell in the proof of the converse theorem.

Lemma 16 Let k∈Z. Let f and g be two real analytic function onH such that g=f|_kωN. Define H(z) =2iy∂f

∂x(z) +k f(z), I(z) =2iy∂g

∂x(z) +kg(z).

Then for any t>0, we have

H|_kωN(it) =N^−k/2(it)^−kH

− 1 Nit

=−I(it).

Proof 29 Since g(z) = f|_kωN(z) =N^k/2(Nz)^−kf(−1/Nz), we have

∂g

∂x(z) =N^−k/2∂f

∂x

z^−kf

− 1 Nz

=N^−k/2

−kz^−k−1f

− 1 Nz

+z^−k∂f

∂x

− 1 Nz

1 Nz²

=−kN^−k/2z^−k−1f

− 1 Nz

+N^−k/2z^−k∂f

∂x

− 1 Nz

1 Nz². Multiplying both side by2iy and evaluating at z=it (x=0,y=t) for t>0. Then we get

2iy∂g

∂x(z) =−2kN^−k/2iyz^−k−1f

− 1 Nz

+2N^−k/2iyz^−k∂f

∂x

− 1 Nz

1 Nz²

⇒2iy∂g

∂x(z) +kN^−k/2iyz^−k−1f

− 1 Nz

=−kN^−k/2iyz^−k−1f

− 1 Nz

+2N^−k/2iyz^−k∂f

∂x

− 1 Nz

1 Nz²

⇒2it∂g

∂x(it) +kN^−k/2(it)^−kf

− 1 Nit

=−kN^−k/2(it)^−kf

− 1 Nit

−2iN^−k/2(it)^−k∂f

∂x

− 1 Nit

1 Nt.

Now the left-hand side is 2it∂g

∂x(it) +k

N^−k/2(it)^−kf

− 1 Nit

=2it∂g

∂x(it) +kg(it) =I(it), and the right-hand side is

−N^−k/2(it)^−k

k f

− 1 Nit

+2i 1

Nt

∂f

∂x

− 1 Nit

=−N^−k/2(it)^−k=−H|_kωN(it).

This proves the lemma.

Using the above lemmas, we will prove the following Hecke-type converse theorem for the harmonic Maass forms of polynomial growth.

Theorem 15 (Theorem 4.5 in [34]) Let k be a negative integer and N be a positive integer. Let f and g be two functions defined onH by the formal Fourier series

f(z) =

∞ n=0

∑

c⁺_f (n)qⁿ+c⁻_f(0)y^1−k+

∑

n<0

c⁻_f(n)Γ(1−k,−4πny)qⁿ, g(z) =

∞

∑

n=0

c⁺_g(n)qⁿ+c⁻_g(0)y^1−k+

∑

n<0

c⁻_g(n)Γ(1−k,−4πny)qⁿ, with c^±_f (n),c^±_g(n) =O(|n|^v), n∈Zfor some v≥0. Then the followings are equivalent.

(1) g= f|_kωN

(2) The L-functionsΛN(f,s),ΛN(g,s),ΩN(f,s), andΩN(g,s)satisfy the following conditions (a) ΛN(f,s), ΛN(g,s), ΩN(f,s), and ΩN(g,s) admits meromorphic continuation to the whole

complex plane,

(b) the following functions are entire and bounded in vertical strips Λ^∗_N(f,s) =ΛN(f,s) +c⁺_f(0)

s +c⁺_g(0)i^k

k−s + c⁻_f (0) N^(1−k)/2

1

s−k+1+ c⁻_g(0)i^k N^(1−k)/2

1 1−s, Λ^∗_N(g,s) =ΛN(g,s) +c⁺_g(0)

s +c⁺_f(0)i^−k

k−s + c⁻_g(0) N^(1−k)/2

1

s−k+1+c⁻_f(0)i^−k N^(1−k)/2

1 1−s, Ω^∗_N(f,s) =ΩN(f,s) +k c⁺_f(0)

s −c⁺_g(0)i^k

k−s + c⁻_f (0) N^(1−k)/2

1

s−k+1− c⁻_g(0)i^k N^(1−k)/2

1 1−s

! ,

Ω^∗_N(g,s) =ΩN(g,s) +k c⁺_g(0)

s −c⁺_f(0)i^−k

k−s + c⁻_g(0) N^(1−k)/2

1

s−k+1−c⁻_f(0)i^−k N^(1−k)/2

1 1−s

! .

(c) satisfy the functional equations

ΛN(f,s) =i^kΛN(g,k−s), ΩN(f,s) =−i^kΩN(g,k−s).

Proof 30

(1)⇒(2) For n≥1, we have c⁻_f(−n)

Z _∞

0

Γ(1−k,4πnt/√

N)e^−2πnt/

√ Nt^sdt

t =c⁻_f(−n) 2πn

√N

−sZ _∞

0

Γ(1−k,2t)e^−tt^sdt t

=c⁻_f(−n) 2πn

√ N

−s

W_1−k(s).

Then forRe(s)>v+1, we obtain Z _∞

0

f it

√N

−c⁺_f(0)−c⁻_f (0) t

√N 1−k!

t^sdt t

= Z _∞

0

∞

∑

n=1

c⁺_f (n)e^−2πnt/

√ N

! t^sdt

t + Z _∞

0

∞

∑

n=1

c⁻_f (−n)Γ(1−k,4πnt/√

N)e^−2πnt/

√ N

! t^sdt

t

= 2π

√ N

−s

(Γ(s)L⁺(f,s) +W_1−k(s)L⁻(f,s))

=ΛN(f,s). (58)

Now forRe(s)>v+1, we have ΛN(f,s) =

Z _∞

0

f it

√ N

−c⁺_f(0)−c⁻_f(0) t

√ N

1−k! t^sdt

t

= _{Z ∞}

1

+ Z ₁

0

f

it

√ N

−c⁺_f(0)−c⁻_f (0) t

√ N

1−k! t^sdt

t

= Z _∞

1

f it

√ N

−c⁺_f(0)−c⁻_f(0) t

√ N

1−k! t^sdt

t +

( Z ₁

0

f it

√ N

t^sdt

t −c⁺_f(0)

s − c⁻_f(0)

√N^1−k 1 s−k+1

)

= Z _∞

1

f it

√ N

−c⁺_f(0)−c⁻_f(0) t

√ N

1−k! t^sdt

t +i^k

Z _∞

1

f i

√ Nt

t^k−sdt

t −c⁺_f(0)

s − c⁻_f(0) N^(1−k)/2

1 s−k+1. Using (1), forRe(s)>max{k,v+1}we have

ΛN(f,s) = Z _∞

1

f it

√ N

−c⁺_f(0)−c⁻_f (0) t

√ N

1−k! t^sdt

t +i^k

Z _∞

1

g it

√ N

t^k−sdt

t −c⁺_f(0)

s − c⁻_f(0) N^(1−k)/2

1 s−k+1

= Z _∞

1

f it

√ N

−c⁺_f(0)−c⁻_f (0) t

√ N

1−k! t^sdt

t +i^k

Z _∞

1

g it

√N

−c⁺_g(0)−c⁻_g(0) t

√N 1−k!

t^k−sdt t

−c⁺_f(0)

s − c⁻_f(0) N^(1−k)/2

1

s−k+1−c⁺_g(0)i^k

k−s − c⁻_g(0)i^k N^(1−k)/2

1

1−s. (59)

By Lemma 15,Λ^∗_N(f,s)is entire and bounded in vertical strips. By the same computation, we see that Λ^∗_N(g,s) can be expressed by a similar integral representation as(59) and so is entire and bounded in vertical strips. Comparing two integral representations forΛN(f,s)andΛN(g,s), we obtainΛN(f,s) =i^kΛN(g,k−s).

Next, we have

∂f

∂x(z) =2πi

∞

∑

n=0

nc⁺_f (n)qⁿ+

∑

n<0

nc⁻_f(n)Γ(1−k,−4πny)qⁿ

! ,

∂g

∂x(z) =2πi

∞

∑

n=0

nc⁺_g(n)qⁿ+

∑

n<0

nc⁻_g(n)Γ(1−k,−4πny)qⁿ

! .

For n≥1, we have

2πinc⁻_f(−n) Z _∞

0

Γ(1−k,4πnt/√

N)e^2πnt/

√N

t^s+1dt t

=2πinc⁻_f(−n) 2πn

√N

−s−1Z _∞

0

Γ(1−k,2t)e^tt^s+1dt t

=i

√

Nc⁻_f(−n) 2πn

√ N

−s

W_1−k(s+1).

So we have

Z _∞

0

H it

√ N

−kc⁺_f(0)−kc⁻_f(0) t

√ N

1−k! t^sdt

t

= 2i

√N Z _∞

0

∂f

∂x it

√N

t^s+1dt t +k

Z _∞

0

f it

√ N

−c⁺_f(0)−c⁻_f (0) t

√ N

1−k! t^sdt

t

=−2Ξ_N(f,s) +kΛN(f,s)

=ΩN(f,s).

In a similar way to the calculation ofΛN(f,s)with Lemma 16, we can obtain an integral repre- sentation ofΩN(f,s), and by Lemma 15, we can see thatΩ^∗_N(f,s)andΩ^∗_N(g,s) are entire and bounded in any vertical strips. Comparing the integral representations ofΩN(f,s)andΩN(g,s), we obtainΩN(f,s) =−i^kΩN(g,k−s).

(2)⇒(1) From(58), for any t>0andα >v+1we have an inverse Mellin transform 1

2πi Z _α+i∞

α−i∞

t^−sΛN(f,s)ds= f it

√ N

−c⁺_f(0)−c⁻_f(0) t

√ N

1−k

.

Combining Stirling’s approximation, Lemma 14 (2), and Phragmén-Lindelöf principle with the proof of Theorem 5, we obtain the following boundary condition forΛN(f,s)in the specific verti- cal strip as follows. Chooseβ such that k−β>v+1. Then for anyµ>0,

ΛN(f,s) =O(|Im(s)|^−µ) as|Im(s)| →∞onβ ≤Re(s)≤α. (60)

Now assume thatα >1(and soβ <k−1). The function t^−sΛN(f,s)has possibly simple poles at s=0,k,1,k−1with the residues−c⁺_f(0),i^kc⁺_g(0)t^−k, c⁻_g(0)i^k

N^(1−k)/2t⁻¹,− c⁻_f(0)

N^(1−k)/2t^1−krespectively.

So we can shift the integral path fromRe(s) =α toRe(s) =β using(60), we obtain 1

2πi Z _β+i∞

β−i∞

ΛN(f,s)t^−sds= f it

√ N

−i^kc⁺_g(0)t^−k− c⁻_g(0)i^k N^(1−k)/2t⁻¹. By the functional equation forΛN, we have

f|_kωN(it) =g(it). (61)

Similar calculations gives us that, for any t>0andα >v+1we have 1

2πi Z _β+i∞

β−i∞ ΩN(f,s)t^−sds=− 2t i√ N

∂f

∂x it

√ N

+k f

it

√ N

−kc⁺_f (0)−kc⁻_f(0)i^k N^(1−k)/2t^1−k, and by the functional equation forΩN, we have

H|_kωN(it) =−I(it). (62)

Finally, we will finish the proof to apply the method of proof of Theorem 12.

In the proof of Lemma 15, we observed that the given Fourier expansion of f and g satisfies

∆k(f) =∆k(g) =0. Letting G= f|_kωN−g, we have∆k(G) =0. Since∆k is elliptic, by Theo- rem 11 the function G(z)is real analytic and so G has a power series expansion in x of the form G(z) =∑^∞_n=0b_n(y)xⁿ. From∆k(G) =0, we get the following recurrence relation

b_n+2(y) =iky(n+1)b_n+1(y)−kyb⁰_n(y)−y²bn(y)

(n+1)(n+2)y² . (63)

To prove G(z) =0, it is sufficient to show that b₀(y) =b₁(y) =0. But we have b₀(t) =G(it) = f|_kωN(it)−g(it),

b₁(t) =∂G

∂x(it) = ∂

∂x

N^−k/2z^−kf

− 1 Nz

(it)−∂g

∂x(it).

By(61)and(62), we have b0(t) =b₁(t) =0. Thus by(63)we have b_n(t) =0for all n≥0. This concludes that g= f|_kωN.