Taking complex conjugate on both sides, we obtain
χ(d)Lk(f)(z) =χ(d)Lk(f)(z) = (Lk(f)|k−2γ)(z) = (cz+d)−(k−2)Lk(f)(γz)
⇔Lk(f)(γz) =χ(d)(cz+d)k−2Lk(f)(z).
Thus we get
(ξkf|2−kγ)(z) = yk−2
|cz+d|2(k−2)
χ(d)(cz+d)k−2Lk(f)(z)
(cz+d)−(2−k)
=χ(d)yk−2Lk(f)(z)
=χ(d)(ξkf)(z).
This provesξk(f)∈M2−k! (N,χ).
To showξk(f(z))∈M2−k! (N,χ), we need to prove thatξk(f(z))is meromorphic at the cusps of Γ0(N). By the proof of (1), we know thatξk(f(z))is holomorphic onH and meromorphic at i∞.
For any cuspρ ofΓ0(N), the Fourier expansion of f at the cuspρis given in Lemma 9. Through a process similar to the proof of (1), we can calculate that the second series in(51)changes to the holomorphic part, and thus meromorphicity at other cusps follows directly.
Proving surjectivity uses the theory of algebraic geometry, especially the Serre duality. We omit this proof here, and we refer [36] for a detailed proof and [38] for more informations of Serre duality.
for anym≥0. Thus positive terms in the second series must be vanish. Consequently, if we assume that f has at most polynomial growth, we have the Fourier expansion of the form
f(z) =
∞
∑
n=0
c+f (n)qn+c−f(0)y1−k+
∑
n<0
c−f(n)Γ(1−k,−4πny)qn, (55) at the cuspi∞. Analogous expansions at all other cusps ofΓ0(N)follows from Lemma 9. We denote the subspace of all such forms byHk#(N,χ)and call themharmonic Maass forms of polynomial growth.
Proposition 9 Let k6=1and f ∈Hk#(N,χ). Then the followings are hold.
(1) f|kωN∈Hk#(N,χ).
(2) If k≤0, then D1−k(f)∈M2−k(N,χ).
(3) ξk(f)∈M2−k(N,χ).
(4) Suppose k≤0and f has Fourier expansion(55). Then c±f(n) =O(1)for all n∈Z. (5) For k>2, Hk#(N,χ) =Mk(N,χ).
Proof 26
(1) First, observe that for anyγ= a b c d
!
∈Γ0(N), we obtain
ωNγ= −cN d
aN bN
!
= d −c
−bN a
! ωN.
Then we have
(f|kωN)|kγ= f|kωNγ= f|k d −c
−bN a
!
ωN= f|k d −c
−bN a
!!
k
ωN
=χ(a)f|kωN
=χ(d)f|kωN. This proves modularity.
Next, by Lemma 11, we have
∆k(f|kωN) =∆k(f)|kωN=0.
Finally, by Lemma 10 we obtain that f|kωN has at most polynomial growth at every cusp.
(2), (3) By Proposition 7 and 8, we have D1−k(f)∈M2−k! (N,χ)andξk(f)∈M2−k! (N,χ). So we need to prove that D1−k(f)andξk(f)are both holomorphic at all cusps ofΓ0(N). By(52)and(54), we can derive the Fourier expansion of D1−k(f)andξk(f)for f ∈Hk#(N,χ)as follows.
D1−k(f) =−(4π)k−1(1−k)!c−f(0) +
∞
∑
n=0
c+f(n)n1−kqn,
and
ξk(f) = (1−k)c−f (0)−(4π)1−k
∞
∑
n=0
c−f(−n)n1−kqn.
These expansions show that D1−k(f)andξk(f)are holomorphic at the cusp i∞. Using Lemma 9, we can also check the holomorphy at the other cusps.
(4) By (2) and (3), D1−k(f)andξk(f)are weight2−k modular forms. Since the bound for coefficients of weight 2−k modular form is O(n1−k), we have c+f(n)n1−k =O(n1−k) and c−f(−n)n1−k = O(n1−k).
(5) Recall that there are no modular forms of negative weight (see (3) in Remark 1). So M2−k(N,χ) = {0}for k≥2. This implies that the coefficient c−f(−n)of the Fourier expansion ofξk(f)is zero for all n≥0given in the proof of (3). Therefore, for k>2the non-holomorphic part of f vanish and thus f is also an element of Hk(N,χ).
By (4) and (5) in the above Proposition, the only interesting case for our purpose isk≤0 andk=2.
Furthermore, by (2) and (3), the casek=2 is reduced since there are no modular forms of weight 0 by (3) in Remark 1. Therefore, we will only deal with the case ofk≤0 in the remaining sections.
From the Fourier expansion of f∈Hk#(N,χ)given by (55), we define two Dirichlet series corresponding to holomorphic and non-holomorphic parts as follows.
Definition 21 Let k≤0and let f ∈Hk#(N,χ) has a Fourier expansion (55). Define Dirichlet series associated to holomorphic part f+ and non-holomorphic part f−as
L+(f,s) =
∞
∑
n=1
c+f (n)n−s, L−(f,s) =
∞
∑
n=1
c−f(−n)n−s.
Suppose k≤0. If f ∈Hk#(N,χ)with Fourier series expansion (55), then by Proposition 9 (4), the Dirichlet seriesL±(f,s)converge absolutely in the half-plane Re(s)>1.
Now we shall define L-function associated to f ∈Hk#(N,χ). We can expect that the L-function of L+(f,s) will be similar in shape to the L-function of modular forms. We can also predict that the L-function ofL−(f,s)will be similar in shape to theL-function of Maass forms, but with some appro- priate modification of the gamma factor. Then theL-function associated to f ∈Hk#(N,χ)will be define as a linear combination of them.
Furthermore, our goal is to prove the converse theorem of the harmonic Maass form. In the process of proving the converse theorem of the Maass form (Theorem 13), the proof was completed by applying two equations derived from the even and odd Maass forms.
Definition 22 Let k6=0and let f ∈Hk#(N,χ). Define L-function associated to f as ΛN(f,s) =
2π
√ N
−s
(Γ(s)L+(f,s) +W1−k(s)L−(f,s)), where
Wν(s) = Z ∞
0
Γ(ν,2x)exxsdx
x , (Re(s)>0) forν∈R. In addition, we define the following two auxiliary L-functions.
ΞN(f,s) = 2π
√ N
−s
(Γ(s)L+(f,s) +W1−k(s)L−(f,s)), ΩN(f,s) =−2ΞN(f,s) +kΛN(f,s).
Note that in the above definition, the functionWν is derived from the incomplete gamma function in the non-holomorphic parts of f (we will compute this in the next section), and acts like the gamma factor of L−(f,s). We will end this section with some properties of the functionWν.
Lemma 14 (1) For anyν>0, the function Wν(s)is analytic in the half-planeRe(s)>0.
(2) Let ν be a positive integer and let µ >0. Then on any vertical line Re(s) =σ >0, we have Wν(s) =O(Im(s)−µ)as|Im(s)| →∞.
(3) For any x>0and c>0, we have the following inverse Mellin transform Γ(ν,2x)e−x= 1
2πi Z c+i∞
c−i∞
Wν(s)x−sds.
Proof 27
(1) For x≥0, we have
|Γ(ν,2x)|=
Z ∞
2x
e−ttνdt t
≤ Z ∞
0
e−ttνdt
t =Γ(ν).
In the above inequality, we used the fact that e−ttν−1>0for t>0. By(50), we haveΓ(ν,2x)∼ (2x)ν−1e−2xas|x| →∞.
Wν(s) = Z ∞
0
Γ(ν,2x)exxsdx
x ∼2ν−1 Z ∞
0
e−xxs+ν−1dx
x =2ν−1Γ(s+ν−1)as|x| →∞.
Thus Wν(s)is an analytic function in the half-planeRe(s)>0.
(2) First, note that for any positive integer n and x∈R, the incomplete gamma function has the following property.
Γ(n,x) =Γ(n)e−x
n−1
∑
l=0
xl
l!. (56)
Applying this, we have
Wν(s) = Z ∞
0
Γ(ν)e−2x
ν−1
∑
l=0
(2x)l l!
! exxsdx
x
=Γ(ν)
ν−1
∑
l=0
2l l!
Z ∞
0
e−xxs+ldx x
=Γ(ν)
ν−1
∑
l=0
2l
l!Γ(s+l).
By Stirling’s approximation for gamma function, for anyµ>0we have Γ(s)∼√
2πtσ−1/2e−π|t|/2as|t| →∞, (s=σ+it).
Wν(s)∼Γ(ν)
ν−1
∑
l=0
2l l!(√
2πtσ+l−1/2e−π|t|/2) =O(tσ+ν−3/2e−π|t|/2) =O(t−µ)as|t| →∞.
for fixedσ>0.
(3) By definition, we observe that Wν(s)is the Mellin transform of Γ(ν,2x)ex. So we only need to check the absolutely convergence of Γ(ν,2x)ex for x>0. Clearly Γ(ν,2x)ex is a continuous function. Since|Γ(ν,2x)ex| ≤Γ(ν)exfor x≥0, we haveΓ(ν,2x)ex=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)exalong the vertical lineRe(s) =c>0exists.