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),ikc+g(0)t−k, c−g(0)ik
N(1−k)/2t−1,− c−f(0)
N(1−k)/2t1−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
−ikc+g(0)t−k− c−g(0)ik 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)ik N(1−k)/2t1−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=0bn(y)xn. From∆k(G) =0, we get the following recurrence relation
bn+2(y) =iky(n+1)bn+1(y)−kyb0n(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 bn(t) =0for all n≥0. This concludes that g= f|kωN.
which satisfiesL±(f,ψ,s) =L±(fψ,s). Next, we define severalL-functions as follows.
ΛN(f,ψ,s) = 2π
√ N
−s
(Γ(s)L+(f,ψ,s) +W1−k(s)L−(f,ψ,s)), ΞN(f,ψ,s) =
2π
√ N
−s
(Γ(s)L+(f,ψ,s) +W1−k(s)L−(f,ψ,s)), ΩN(f,ψ,s) =−2ΞN(f,ψ,s) +kΛN(f,ψ,s).
We can easily get the following relations.
ΛN(f,ψ,s) =ΛN(fψ,s), ΞN(f,ψ,s) =ΞN(fψ,s),ΩN(f,ψ,s) =ΩN(fψ,s).
Shankhadhar and Singh proved the converse theorem for harmonic Maass forms of polynomial growth. In this section, we will introduce their result.
Recall the set P given in Definition 11: P=PN is a set of odd prime number or 4 satifying the following two conditions:
• Any element ofPis prime toN.
• For any positive integersa,bwith(a,b) =1,P∩ {a+nb|n∈Z} 6=/0 .
Theorem 16 (Theorem 1.1 in [34]) Let k be a negetive integer, and N be a positive integer. Letχ be a Dirichlet character modulo N such that χ(−1) = (−1)k. Let f and g be two functions defined onH given by the formal Fourier series
f(z) =
∞
∑
n=0
c+f (n)qn+c−f(0)y1−k+
∑
n<0
c−f(n)Γ(1−k,−4πny)qn, g(z) =
∞
∑
n=0
c+g(n)qn+c−g(0)y1−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) The functions f and g belong to Hk#(N,χ)and Hk#(N,χ), respectively, and g= f|kωN. (2) (a) Statement (2) in Theorem 15 holds.
(b) For any primitive Dirichlet characterψwith conductor mψ∈P, each one of the L-functions ΛN(f,ψ,s),ΛN(g,ψ,s),ΩN(f,ψ,s),ΩN(g,ψ,s)can be analytically continued to the whole complex plane, bounded on any vertical strip, and satisfies the functional equation
ΛN(f,ψ,s) =ikCψΛN(g,ψ,k−s), ΩN(f,ψ,s) =−ikCψΩN(g,ψ,k−s), where
Cψ =CN,ψ =χ(m)ψ(−N)τ(ψ)/τ(ψ) =χ(m)ψ(−N)τ(ψ)2/m.
(Step 1) In this step, we prove the direct part of Theorem 16.
By putting f = fψ, g=Cψgψ, and N=Nm2ψ in Theorem 15, we have the following proposi- tion, which is an analogue of Proposition 1.
Proposition 10 Let k be a negative integer and N be a positive integer. Let f and g be two functions defined on H satisfy assumptions in Theorem 15. Let Cψ be a constant which may depend onψ. Then the following two statements are equivalent.
(1) fψ|kωNm2
ψ =Cψgψ.
(2) ΛN(f,ψ,s),ΛN(g,ψ,s),ΩN(f,ψ,s),ΩN(g,ψ,s)can be analytically continued to the whole complex plane, bounded on any vertical strip, and satisfies the functional equation
ΛN(f,ψ,s) =ikCψΛN(g,ψ,k−s), ΩN(f,ψ,s) =−ikCψΩN(g,ψ,k−s).
Next, for given harmonic Maass forms of polynomial growth f andg, we must show that their twist fψ andgψ are also harmonic Maass forms of polynomial growth. These claims can be justified by the following proposition, which is an analogue of Lemma 3 (1) and (2).
Proposition 11 (1) Let f be a smooth function defined onH satisfy assumptions in Theorem 15, and letψ be a primitive Dirichlet character of conductor m. Then for any integer k>0, we have
fψ=τ(ψ)−1
m
∑
u=1
ψ(u)(f|kTu/m) whereτ(ψ)is a Gauss sum ofψ.
(2) Letχ be a Dirichlet character modulo N with conductor mχ and f ∈Hk#(N,χ). Letψ be a primitive Dirichlet character of conductor mψ >1. Let M=lcm(N,m2ψ,mψmχ). Then
fψ ∈Hk#(M,χ ψ2).
Proof 31 (1) Using the Fourier expansion of harmonic Maass form of polynomial growth in- stead of modular form, this proof is the same as the proof of Lemma 3 (1).
(2) First, we need to prove the automorphy of fψ:, that is, fψ|kγ = χ ψ2(d)fψ for all γ =
a b
cN d
!
∈Γ0(N). This can be induced in exactly the same way as the proof of Lemma 3 using (1).
Next, by Lemma 11 (1), we have ∆k(f|kTu/mψ) = ∆k(f)|kTu/mψ =0. By (1), we have
∆k(fψ) =0.
Finally, applying Lemma 10 to (1), fψ also satisfies the polynomial growth.
Moreover, fromg= f|kωN, we need to find out the modularity relation between fψandgψ to use the above proposition. This can be
(Step 2) To prove the opposite direction, we need to show that f andgare annihilated by∆k, satisfy poly- nomial growth condition, and modularityg= f|kωN.
First, from the given formal Fourier series of f andg, we have∆k(f) =∆k(g) =0 immediately.
Next, notice that by Lemma 15, the given Fourier series of f and g is not holomorphic, but a real analytic. Furthermore, from Proposition 11 we derived twisted moduluarity between f and g, which is exactly the same as we used in the proof of Theorem 3. Conversely, we deduce that ordinary modularity of harmonic Maass forms can be induced in the same way as the proof of Theorem 7.
(Step 3) Finally, we will prove that f andgsatisfy the polynomial growth condition. This can be estab- lished by assuming modality.
Proposition 12 Let f :H →Cbe a smooth function. Assume that f satisfies f|kγ=χ(d)f for everyγ= a b
c d
!
, and∆k(f) =0. If f(z) =O(y−v)as y→0for some v≥0uniformly inRe(z), then f has at most polynomial growth at every cusp ofΓ0(N).
Proof 32 Recall that every cusp of Γ0(N) is in the set Q∪ {i∞}, and the cusp i∞ is Γ0(N)- equivalent to a rational cusp, it is enough to prove for rational cusps.
Letρ∈Qbe a cusp ofΓ0(N)with width t and parameterκ∈[0,1). Letγ= a b c d
!
∈SL(2,Z) such thatγ(i∞) =ρ. We have c6=0becauseγmaps i∞toρ∈Q.
Since c6=0, we have
Im(γz) = y
|cz+d|2 →0 as y→∞ uniformly on|Re(z)| ≤t/2. Then we have
(f|kγ)(z) = (cz+d)−kf(γz) =O(|cz+d|−kIm(γz)−v) =O(yv−k) as y→∞ (64) uniformly on|Re(z)| ≤t/2.
By Lemma 9, we have the (formal) Fourier series of f|kγ (f|kγ)(z) =
∞ n=−∞
∑
c+f(n)qn+κt +c−f(0)y1−kqκt +
∞ n=−∞
∑
n6=0
c−f (n)Γ(1−k,−4πny/t)qn+κt ,
which converges uniformly on any compact subset ofH. Consider the following integral, which computes the Fourier coefficients of f|kγwhich has period t.
1 t
Z z0+t z0
(f|kγ)(z)e−2πizn+κt dz, (z0=x0+iy0∈H). (65) Using(64), we have
1 t
Z z0+t z0
(f|kγ)(z)e−2πizn+κt dz=O 1
t Z z0+t
z0
yv−ke−2πi(x+iy)n+κt dz
=O(yv−ke2π(n+κ)y/t) as y→∞. (66) Now we compute the integral(65). If n6=0, we have
1 t
Z z0+t z0
(f|kγ)(z)e−2πizn+κt dz
=
∞ m=−∞
∑
c+f(m)e2π(n−m)y0/t 1
t Z x0+t
x0
e−2πi(n−m)x/t
dx
+c−f(0)y1−k0 e2πny0/t 1
t Z x0+t
x0
e−2πinx/tdx
+
∞ m=−∞
∑
n6=0
c−f (m)Γ(1−k,−4πmy0/t)e2π(n−m)y0/t 1
t Z x0+t
x0
e−2πi(n−m)x/tdx
=c+f(n) +c−f(n)Γ(1−k,−4πny0/t). (67)
If n=0, we have 1 t
Z z0+t z0
(f|kγ)(z)e−2πizκ/tdz
=
∞ m=−∞
∑
c+f(m) 1
t Z z0+t
z0
e2πinz/tdz
+c−f (0)y1−k 1
t Z z0+t
z0
dz
+
∞ m=−∞
∑
m6=0
c−f(m)Γ(1−k,−4πmy0/t)e−2πmy0/t 1
t Z x0+t
x0
e2πimx/tdx
=c+f(0) +c−f(0)y1−k0 .
In the above calculations, we used the fact that the contour of given complex integral is just a line from x0to x0+t with fixed imaginary part y0, and the following formula.
1 t
Z x0+t x0
e−2πi(n−m)x/t
dx=
0 if n6=m, 1 if n=m.
Combining(66)and(67), we have
c+f (n) +c−f(n)Γ(1−k,−4πny/t) =O(yv−ke2π(n+κ)y/t) as y→∞.
By definition of big-oh notation, we can rewrite this as
y→∞lim
c+f(n) +c−f (n)Γ(1−k,−4πny/t) yv−ke2π(n+κ)/t
<∞. (68)
By(50), we have
Γ(1−k,−4πny/t)∼(−4πny/t)−ke4πny/t as y→∞.
Then(68)can be write
y→∞lim
c+f(n) +c−f(n)(−4πny/t)−ke4πny/t yv−ke2π(n+κ)y/t
=lim
y→∞|c+f(n)y−(v−k)e−2π(n+κ)y/t
+c−f(n)(−4πn/t)−ky−ve(4πny−2π(n+κ)y)/t|<∞.
Now recall that k<0, v≥0, andκ∈[0,1). For n>0, the first term goes to0but the second term goes to∞as y→∞unless c−f(n) =0. Similarly, for n<0, the second term goes to0but the first term goes to∞as y→∞unless c+f(n) =0. Therefore we have
(f|kγ)(z) =
∞
∑
n=0
c+f(n)qn+κt +c−f(0)y1−kqκt +
∑
n<0
c−f(n)Γ(1−k,−4πny/t)qn+κt , and we conclude that f|kγhas a polynomial growth.