1 Introduction
Let
$f\in S_{k}(\unicode[STIX]{x1D6E4}_{1}(N))$
be a classical holomorphic modular form of weight
$k$
and level
$N$
. Assume that
$f$
is primitive, meaning that it is a normalized Hecke eigenform in the new subspace. Then it has a Fourier expansion of the shape
$$\begin{eqnarray}f(z)=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)n^{(k-1)/2}\text{e}^{2\unicode[STIX]{x1D70B}\text{i}nz},\end{eqnarray}$$
where the
$\unicode[STIX]{x1D706}_{f}(n)$
are multiplicative and satisfy the Ramanujan bound
$|\unicode[STIX]{x1D706}_{f}(n)|\leqslant d(n)$
. Let
$\unicode[STIX]{x1D6EC}_{f}(s)=\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s+(k-1)/2)L_{f}(s)$
denote the complete
$L$
-function of
$f$
, with analytic normalization, where
$$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)=2(2\unicode[STIX]{x1D70B})^{-s}\unicode[STIX]{x1D6E4}(s)\quad \text{and}\quad L_{f}(s)=\mathop{\sum }_{n=1}^{\infty }\frac{\unicode[STIX]{x1D706}_{f}(n)}{n^{s}},\end{eqnarray}$$
and let
$$\begin{eqnarray}N_{f}^{s}(T)=\#\{\unicode[STIX]{x1D70C}\in \mathbb{C}:\unicode[STIX]{x1D6EC}_{f}(\unicode[STIX]{x1D70C})=0,\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})\neq 0,|\Im (\unicode[STIX]{x1D70C})|\leqslant T\}\end{eqnarray}$$
be the number of simple zeros of
$\unicode[STIX]{x1D6EC}_{f}(s)$
with imaginary part in
$[-T,T]$
.
In [Reference Milinovich and Ng9], the second and third authors showed that if
$\unicode[STIX]{x1D6EC}_{f}(s)$
satisfies the generalized Riemann hypothesis, then
$$\begin{eqnarray}N_{f}^{s}(T)\geqslant T(\log T)^{-\unicode[STIX]{x1D700}}\end{eqnarray}$$
for any fixed
$\unicode[STIX]{x1D700}>0$
and all sufficiently large
$T>0$
. Unconditionally, when
$N=1$
and
$k=12$
, Conrey and Ghosh [Reference Conrey and Ghosh5] showed that
$$\begin{eqnarray}\text{for all}~\unicode[STIX]{x1D700}>0,\,\text{there exists}~T\geqslant \unicode[STIX]{x1D700}^{-1}~\text{such that}~N_{f}^{s}(T)\geqslant T^{1/6-\unicode[STIX]{x1D700}}.\end{eqnarray}$$
Moreover, their proof works more generally for
$N=1$
and arbitrary
$k$
, provided that
$N_{f}^{s}(T)$
is not identically 0. In light of the first author’s result [Reference Booker1] that
$N_{f}^{s}(T)\rightarrow \infty$
as
$T\rightarrow \infty$
, (1.1) holds for all primitive
$f$
of conductor 1.
In this paper, we aim to prove similar unconditional quantitative estimates of simple zeros for primitive forms of arbitrary conductor
$N$
. However, we encounter some obstacles that are reminiscent of the well-known difficulty of extending Hecke’s converse theorem to arbitrary conductor, and are not present for
$N=1$
. Taking inspiration from Weil’s generalization [Reference Weil11] of Hecke’s converse theorem, we consider character twists. For a Dirichlet character
$\unicode[STIX]{x1D712}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)$
, let
$f\otimes \unicode[STIX]{x1D712}$
denote the unique primitive form such that
$\unicode[STIX]{x1D706}_{f\otimes \unicode[STIX]{x1D712}}(n)=\unicode[STIX]{x1D706}_{f}(n)\unicode[STIX]{x1D712}(n)$
for all
$n$
coprime to
$q$
.
Theorem 1.1. Let
$f\in S_{k}(\unicode[STIX]{x1D6E4}_{1}(N))$
be a primitive form. Then there is a Dirichlet character
$\unicode[STIX]{x1D712}$
such that (1.1) holds with
$f\otimes \unicode[STIX]{x1D712}$
in place of
$f$
.
Next, for odd conductors we obtain a weaker but unconditional quantitative estimate for
$N_{f}^{s}(T)$
, without the twist. Moreover, we show that there is a sort of “Deuring–Heilbronn phenomenon” at play, so that if
$N_{f}^{s}(T)$
is unexpectedly small then we can substantially improve our result for
$N_{f\otimes \unicode[STIX]{x1D712}}^{s}(T)$
.
Theorem 1.2. Let
$f\in S_{k}(\unicode[STIX]{x1D6E4}_{1}(N))$
be a primitive form of odd conductor. Then for all
$\unicode[STIX]{x1D700}>0$
, there exists
$T\geqslant \unicode[STIX]{x1D700}^{-1}$
such that
$$\begin{eqnarray}\displaystyle N_{f}^{s}(T)\geqslant \left\{\begin{array}{@{}ll@{}}\exp ((\log T)^{1/3-\unicode[STIX]{x1D700}})\quad & \text{if}~k=1~\text{or}~f~\text{is a CM form},\\ \log \log \log T\quad & \text{otherwise}.\end{array}\right. & & \displaystyle \nonumber\end{eqnarray}$$
Further, if
$N_{f}^{s}(T)\ll 1+T^{\unicode[STIX]{x1D700}}$
for every
$\unicode[STIX]{x1D700}>0$
, then:
(i) there is a Dirichlet character
$\unicode[STIX]{x1D712}$
such that, for all
$\unicode[STIX]{x1D700}>0$
, there exists
$T\geqslant \unicode[STIX]{x1D700}^{-1}$
such that
$\unicode[STIX]{x1D6EC}_{f\otimes \unicode[STIX]{x1D712}}(s)$
has at least
$T^{1/2-\unicode[STIX]{x1D700}}$
simple zeros with real part
$\frac{1}{2}$
and imaginary part in
$[-T,T]$
;(ii)
$\unicode[STIX]{x1D6EC}_{f}(s)$
has simple zeros with real part arbitrarily close to
$1$
.
Remarks.
(1) The exponent
$\frac{1}{6}$
in (1.1) is related to the best known subconvexity estimate for modular form
$L$
-functions in the
$t$
aspect; it can be replaced by any
$\unicode[STIX]{x1D6FF}>0$
such that
$L_{f}(\frac{1}{2}+\text{i}t)\ll _{f,\unicode[STIX]{x1D700}}(1+|t|)^{1/2-\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D700}}$
holds for all primitive forms
$f$
and all
$\unicode[STIX]{x1D700}>0$
. In [Reference Booker, Milinovich and Ng3] we showed that
$\unicode[STIX]{x1D6FF}=\frac{1}{6}$
is admissible. Very recent work of Munshi [Reference Munshi10] improves this to
$\unicode[STIX]{x1D6FF}=\frac{1}{6}+\frac{1}{1200}$
for forms of level 1, with a corresponding improvement to (1.1) in that case.(2) In Theorem 1.1, one can take the conductor of
$\unicode[STIX]{x1D712}$
to be 1 or a prime number bounded by a polynomial function of
$N$
.(3) The proof of Theorem 1.2 makes use of the idea originating with Conrey and Ghosh [Reference Conrey and Ghosh5] of twisting the coefficients of
$L_{f}(s)$
by
$(-1)^{n}$
to prevent the main terms of our estimate from cancelling out. This relies implicitly on the fact that there is no primitive Dirichlet character of conductor 2, and is the ultimate reason for our restriction to odd
$N$
.(4) The improved estimate in Theorem 1.2 in the Galois and CM cases arises from Coleman’s Vinogradov-type zero-free region for Hecke
$L$
-functions [Reference Coleman4].
2 Dirichlet series
In order to establish the existence of simple zeros it is useful to study not only
$L_{f}(s)$
, but some related Dirichlet series and their additive twists. This is one of the central ideas in [Reference Conrey and Ghosh5]. A key role is played by the series
$$\begin{eqnarray}D_{f}(s)=L_{f}(s)\frac{d^{2}}{ds^{2}}\log L_{f}(s)=\mathop{\sum }_{n=1}^{\infty }c_{f}(n)n^{-s},\end{eqnarray}$$
which has a meromorphic continuation to
$\mathbb{C}$
with poles precisely at the simple zeros of
$L_{f}(s)$
(including the trivial zeros
$s=(1-k)/2-n$
for
$n=0,1,2,\ldots$
).
For
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
and
$\unicode[STIX]{x1D712}$
a Dirichlet character, let
$$\begin{eqnarray}L_{f}(s,\unicode[STIX]{x1D6FC})=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)e(\unicode[STIX]{x1D6FC}n)n^{-s}\quad \text{and}\quad L_{f}(s,\unicode[STIX]{x1D712})=\mathop{\sum }_{n=1}^{\infty }\unicode[STIX]{x1D706}_{f}(n)\unicode[STIX]{x1D712}(n)n^{-s}.\end{eqnarray}$$
Likewise, define
$$\begin{eqnarray}D_{f}(s,\unicode[STIX]{x1D6FC})=\mathop{\sum }_{n=1}^{\infty }c_{f}(n)e(\unicode[STIX]{x1D6FC}n)n^{-s}\quad \text{and}\quad D_{f}(s,\unicode[STIX]{x1D712})=\mathop{\sum }_{n=1}^{\infty }c_{f}(n)\unicode[STIX]{x1D712}(n)n^{-s}.\end{eqnarray}$$
Let
$\unicode[STIX]{x1D709}$
denote the nebentypus character of
$f$
. Set
$$\begin{eqnarray}Q(N)=\{1\}\cup \{q\text{ prime}:q\nmid N\},\end{eqnarray}$$
and for each
$q\in Q(N)$
, define the rational functions
$$\begin{eqnarray}P_{f,q}(x)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if}~q=1,\\ 1-\unicode[STIX]{x1D706}_{f}(q)x+\unicode[STIX]{x1D709}(q)x^{2}\quad & \text{otherwise}\end{array}\right.\end{eqnarray}$$
and
$$\begin{eqnarray}R_{f,q}(x)=\left\{\begin{array}{@{}ll@{}}0\quad & \text{if}~q=1,\\ {\displaystyle \frac{q\log ^{2}q}{q-1}}{\displaystyle \frac{x(\unicode[STIX]{x1D706}_{f}(q)-4\unicode[STIX]{x1D709}(q)x+\unicode[STIX]{x1D706}_{f}(q)\unicode[STIX]{x1D709}(q)x^{2})}{P_{f,q}(x)}}\quad & \text{if}~q\neq 1.\end{array}\right.\end{eqnarray}$$
These are such that, if
$$\begin{eqnarray}\unicode[STIX]{x1D712}_{0}(n)=\left\{\begin{array}{@{}ll@{}}1\quad & \text{if}~(n,q)=1,\\ 0\quad & \text{otherwise}\end{array}\right.\end{eqnarray}$$
denotes the trivial character mod
$q$
, then
$$\begin{eqnarray}L_{f}(s,\unicode[STIX]{x1D712}_{0})=P_{f,q}(q^{-s})L_{f}(s)\end{eqnarray}$$
and
$$\begin{eqnarray}D_{f}(s,\unicode[STIX]{x1D712}_{0})=P_{f,q}(q^{-s})D_{f}(s)-\frac{q-1}{q}R_{f,q}(q^{-s})L_{f}(s).\end{eqnarray}$$
For any
$a\in \mathbb{Z}$
coprime to
$q$
, we define
$$\begin{eqnarray}\displaystyle D_{f,a,q}(s) & = & \displaystyle D_{f}\bigg(s,\frac{a}{q}\bigg)-R_{f,q}(q^{-s})L_{f}(s)=\mathop{\sum }_{n=1}^{\infty }c_{f,a,q}(n)n^{-s},\nonumber\\ \displaystyle D_{f,a,q}^{\ast }(s) & = & \displaystyle D_{f,a,q}(s)+\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)L_{f}\bigg(s,\frac{a}{q}\bigg),\quad \text{where}~\unicode[STIX]{x1D713}(s)=\frac{\unicode[STIX]{x1D6E4}^{\prime }}{\unicode[STIX]{x1D6E4}}(s)\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}D_{f,a,q}(s,\unicode[STIX]{x1D6FC})=\mathop{\sum }_{n=1}^{\infty }c_{f,a,q}(n)e(\unicode[STIX]{x1D6FC}n)n^{-s}\quad \text{for}~\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }.\end{eqnarray}$$
To each of
$L_{f}$
,
$D_{f}$
,
$D_{f,a,q}$
,
$D_{f,a,q}^{\ast }$
and their twists, we define completed versions
$\unicode[STIX]{x1D6EC}_{f}$
,
$\unicode[STIX]{x1D6E5}_{f}$
,
$\unicode[STIX]{x1D6E5}_{f,a,q}$
,
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }$
obtained by multiplying by
$\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s+(k-1)/2)$
. By the Ramanujan bound
$|\unicode[STIX]{x1D706}_{f}(q)|\leqslant 2$
and [Reference Booker and Krishnamurthy2, Proposition 3.1],
$\unicode[STIX]{x1D6E5}_{f}(s,a/q)-\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
is holomorphic for
$\Re (s)>0$
. In turn, the analytic properties of
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
are described by the following proposition.
Proposition 2.1. Let
$f\in S_{k}(\unicode[STIX]{x1D6E4}_{0}(N),\unicode[STIX]{x1D709})$
be a primitive form,
$q\in Q(N)$
, and
$a\in \mathbb{Z}$
coprime to
$q$
. Then
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
is a ratio of entire functions of finite order, has at most simple poles, all of which are contained in the critical strip
$\{s\in \mathbb{C}:\Re (s)\in (0,1)\}$
, and satisfies the functional equation
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)=\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}^{\ast }(1-s),\end{eqnarray}$$
where
$\bar{f}\in S_{k}(\unicode[STIX]{x1D6E4}_{0}(N),\overline{\unicode[STIX]{x1D709}})$
is the dual of
$f$
,
$\unicode[STIX]{x1D716}\in \mathbb{C}^{\times }$
is the root number of
$f$
and
$\overline{Na}$
denotes a multiplicative inverse of
$Na\hspace{0.6em}({\rm mod}\hspace{0.2em}q)$
.
Proof. For
$q=1$
, the result follows immediately from [Reference Booker1, (3.1)], so we may assume that
$q$
is prime. Let
$\unicode[STIX]{x1D712}$
be a Dirichlet character of conductor
$q$
. Then the complete twisted
$L$
-function
$\unicode[STIX]{x1D6EC}_{f}(s,\unicode[STIX]{x1D712})$
satisfies the functional equation
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}(s,\unicode[STIX]{x1D712})=\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)\unicode[STIX]{x1D712}(N)\frac{\unicode[STIX]{x1D70F}(\unicode[STIX]{x1D712})^{2}}{q}(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6EC}_{\bar{f}}(1-s,\overline{\unicode[STIX]{x1D712}}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D716}\in \mathbb{C}^{\times }$
is the root number of
$f$
. Applying [Reference Booker1, (3.1)] to
$f\otimes \unicode[STIX]{x1D712}$
, we thus have
$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D712})-\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)\unicode[STIX]{x1D712}(N)\frac{\unicode[STIX]{x1D70F}(\unicode[STIX]{x1D712})^{2}}{q}(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s,\overline{\unicode[STIX]{x1D712}})\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D6EC}_{f}(s,\unicode[STIX]{x1D712})\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg).\end{eqnarray}$$
Next, we have
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}\bigg(s,\frac{a}{q}\bigg)=\unicode[STIX]{x1D6E5}_{f}(s)-\frac{q}{q-1}\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D712}_{0})+\frac{1}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}\;(\text{mod }q) \\ \unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}}\unicode[STIX]{x1D70F}(\overline{\unicode[STIX]{x1D712}})\unicode[STIX]{x1D712}(a)\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D712}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D712}_{0}$
is the trivial character mod
$q$
. Combining this with (2.1) we get
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f,a,q}(s)=\bigg(1-\frac{q}{q-1}P_{f,q}(q^{-s})\bigg)\unicode[STIX]{x1D6E5}_{f}(s)+\frac{1}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}\;(\text{mod }q) \\ \unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}}\unicode[STIX]{x1D70F}(\overline{\unicode[STIX]{x1D712}})\unicode[STIX]{x1D712}(a)\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D712}).\end{eqnarray}$$
Note in particular that
$\unicode[STIX]{x1D6E5}_{f,a,q}(s)$
is a ratio of entire functions of finite order, and all of its poles in
$\{s\in \mathbb{C}:\Re (s)>0\}$
are simple and located at simple zeros of either
$\unicode[STIX]{x1D6EC}_{f}(s)$
or
$\unicode[STIX]{x1D6EC}_{f}(s,\unicode[STIX]{x1D712})$
for some
$\unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}$
.
Note that
$P_{f,q}$
satisfies the functional equation
$$\begin{eqnarray}1-\frac{q}{q-1}P_{f,q}(q^{-s})=\unicode[STIX]{x1D709}(q)q^{1-2s}\bigg(1-\frac{q}{q-1}P_{\bar{f},q}(q^{s-1})\bigg),\end{eqnarray}$$
and thus, by [Reference Booker1, (3.1)],
$$\begin{eqnarray}\displaystyle & & \displaystyle \!\bigg(1-\frac{q}{q-1}P_{f,q}(q^{-s})\bigg)\unicode[STIX]{x1D6E5}_{f}(s)\nonumber\\ \displaystyle & & \displaystyle \!\!\!\qquad -\,\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{1/2-s}\bigg(1-\frac{q}{q-1}P_{\bar{f},q}(q^{s-1})\bigg)\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s)\nonumber\\ \displaystyle & & \displaystyle \!\!\!\quad =\bigg(1-\frac{q}{q-1}P_{f,q}(q^{-s})\bigg)\unicode[STIX]{x1D6EC}_{f}(s)\bigg(\!\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg).\;\qquad\end{eqnarray}$$
Thus, replacing
$f$
by
$\bar{f}$
,
$s$
by
$1-s$
,
$a$
by
$-\overline{Na}$
and
$\unicode[STIX]{x1D712}$
by
$\overline{\unicode[STIX]{x1D712}}$
in (2.4), we get
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(1-s) & = & \displaystyle \bigg(1-\frac{q}{q-1}P_{\bar{f},q}(q^{s-1})\bigg)\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}\;(\text{mod}~q) \\ \unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}}\unicode[STIX]{x1D70F}(\unicode[STIX]{x1D712})\unicode[STIX]{x1D712}(-Na)\unicode[STIX]{x1D6E5}_{\bar{f}}(1-s,\overline{\unicode[STIX]{x1D712}}).\nonumber\end{eqnarray}$$
Applying the functional equations (2.3) and (2.5), together with the relation
$\unicode[STIX]{x1D70F}(\unicode[STIX]{x1D712})\unicode[STIX]{x1D70F}(\overline{\unicode[STIX]{x1D712}})=\unicode[STIX]{x1D712}(-1)q$
, we thus have
$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E5}_{f,a,q}(s)-\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(1-s)\nonumber\\ \displaystyle & & \displaystyle \quad =\left[\bigg(1-\frac{q}{q-1}P_{f,q}(q^{-s})\bigg)\unicode[STIX]{x1D6EC}_{f}(s)+\frac{1}{q-1}\mathop{\sum }_{\substack{ \unicode[STIX]{x1D712}\;(\text{mod}~q) \\ \unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}}}\unicode[STIX]{x1D70F}(\overline{\unicode[STIX]{x1D712}})\unicode[STIX]{x1D712}(a)\unicode[STIX]{x1D6EC}_{f}(s,\unicode[STIX]{x1D712})\right]\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg).\nonumber\end{eqnarray}$$
Applying the classical Voronoi formula [Reference Kowalski, Michel and VanderKam8, p. 179, (A.10)]
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)=\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6EC}_{\bar{f}}\bigg(1-s,-\frac{\overline{Na}}{q}\bigg),\end{eqnarray}$$
we arrive at (2.2).
Finally, by (2.4) and the non-vanishing of automorphic
$L$
-functions for
$\Re (s)\geqslant 1$
[Reference Jacquet and Shalika7],
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
is holomorphic for
$\Re (s)\geqslant 1$
. This conclusion applies to
$\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}^{\ast }(s)$
as well, so by (2.2), all poles of
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
have real part in
$(0,1)$
.◻
Fix, for the remainder of this section, a choice of
$f,a,q$
as in Proposition 2.1, and
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
. We define
$$\begin{eqnarray}N_{f,a,q}^{s}(T)=\#\Big\{\unicode[STIX]{x1D70C}\in \mathbb{C}:|\Im (\unicode[STIX]{x1D70C})|\leqslant T,\underset{s=\unicode[STIX]{x1D70C}}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)\neq 0\Big\}\end{eqnarray}$$
and
$$\begin{eqnarray}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})=\mathop{\sum }_{\unicode[STIX]{x1D70C}\in \mathbb{C}}\underset{s=\unicode[STIX]{x1D70C}}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)(y-\text{i}\unicode[STIX]{x1D6FC})^{-\unicode[STIX]{x1D70C}-(k-1)/2}\quad \text{for}~y\in \mathbb{R}_{{>}0},\end{eqnarray}$$
where
$(y-\text{i}\unicode[STIX]{x1D6FC})^{-\unicode[STIX]{x1D70C}-(k-1)/2}$
is defined in terms of the principal branch of
$\log (y-\text{i}\unicode[STIX]{x1D6FC})$
. Our goal is to derive the following expression for the Mellin transform of
$S_{f,a,q}(y,\unicode[STIX]{x1D6FC})$
, up to a holomorphic function on
$\{s\in \mathbb{C}:\Re (s)>0\}$
.
Proposition 2.2. Define
$$\begin{eqnarray}\displaystyle H_{f,a,q,\unicode[STIX]{x1D6FC}}(s) & = & \displaystyle \unicode[STIX]{x1D6E5}_{f,a,q}(s,\unicode[STIX]{x1D6FC})\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(\text{i}\operatorname{sgn}\unicode[STIX]{x1D6FC})^{k}(Nq^{2}\unicode[STIX]{x1D6FC}^{2})^{s-1/2}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}\bigg(s,-\frac{1}{Nq^{2}\unicode[STIX]{x1D6FC}}\bigg)\qquad\end{eqnarray}$$
and
$$\begin{eqnarray}I_{f,a,q,\unicode[STIX]{x1D6FC}}(s)=\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})y^{s+(k-1)/2}\frac{dy}{y}.\end{eqnarray}$$
Then,
$I_{f,a,q,\unicode[STIX]{x1D6FC}}(s)-H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
has analytic continuation to
$\Re (s)>0$
. Moreover, if
$$\begin{eqnarray}\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S_{f,a,q}(y,\unicode[STIX]{x1D6FC})|y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}<\infty\end{eqnarray}$$
for some
$\unicode[STIX]{x1D70E}\geqslant 0$
, then
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
is holomorphic for
$\Re (s)>\unicode[STIX]{x1D70E}$
.
The proof will be carried out in several lemmas, and involves the following auxiliary functions defined on
$\mathbb{H}=\{z\in \mathbb{C}:\Im (z)>0\}$
:
$$\begin{eqnarray}\displaystyle & \displaystyle F(z)=2\mathop{\sum }_{n=1}^{\infty }c_{f,a,q}(n)n^{(k-1)/2}e(nz), & \displaystyle \nonumber\\ \displaystyle & \displaystyle \overline{F}(z)=2\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)n^{(k-1)/2}e(nz), & \displaystyle \nonumber\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \displaystyle A(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=k/2}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)+\unicode[STIX]{x1D713}^{\prime }\bigg(s-\frac{k-1}{2}\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,(-\text{i}z)^{-s-(k-1)/2}\,ds,\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}B(z)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=k/2}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\frac{\unicode[STIX]{x1D70B}^{2}}{\sin ^{2}(\unicode[STIX]{x1D70B}(s+(k-1)/2))}(-\text{i}z)^{-s-(k-1)/2}\,ds.\end{eqnarray}$$
We first derive the following expression for
$S_{f,a,q}$
.
Lemma 2.3. For
$z=\unicode[STIX]{x1D6FC}+\text{i}y\in \mathbb{H}$
, we have
$$\begin{eqnarray}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})=F(z)-\frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)}{(-\text{i}\sqrt{N}qz)^{k}}\overline{F}\bigg(-\frac{1}{Nq^{2}z}\bigg)+A(z)-B(z).\end{eqnarray}$$
Proof. Let
$0<\unicode[STIX]{x1D700}<\frac{1}{2}$
. For
$z\in \mathbb{H}$
we define
$$\begin{eqnarray}\displaystyle & \displaystyle I_{R}(z)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=1+\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6E5}_{f,a,q}(s)(-\text{i}z)^{-s-(k-1)/2}\,ds, & \displaystyle \nonumber\\ \displaystyle & \displaystyle I_{L}(z)=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=-\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6E5}_{f,a,q}(s)(-\text{i}z)^{-s-(k-1)/2}\,ds. & \displaystyle \nonumber\end{eqnarray}$$
For the remainder of the proof we let
$z=\unicode[STIX]{x1D6FC}+\text{i}y$
.
Since
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
is a ratio of entire functions of finite order with at most simple poles, by the calculus of residues we have
$$\begin{eqnarray}\underset{s=0}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}(s)(-\text{i}z)^{-s-(k-1)/2}+S_{f,a,q}(y,\unicode[STIX]{x1D6FC})=I_{R}(z)-I_{L}(z).\end{eqnarray}$$
Note that the residue term at
$s=0$
vanishes unless
$k=1$
. We have
$$\begin{eqnarray}\displaystyle I_{R}(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=1+\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6E4}_{\mathbb{C}}\bigg(s+\frac{k-1}{2}\bigg)D_{f,a,q}(s)(-\text{i}z)^{-s-(k-1)/2}\,ds\nonumber\\ \displaystyle & = & \displaystyle 2(-2\unicode[STIX]{x1D70B}\text{i}z)^{-(k-1)/2}\mathop{\sum }_{n=1}^{\infty }c_{f,a,q}(n)\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=1+\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6E4}\bigg(s+\frac{k-1}{2}\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,(-2\unicode[STIX]{x1D70B}\text{i}nz)^{-s}\,ds.\nonumber\end{eqnarray}$$
Using the identity
$$\begin{eqnarray}\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=1+\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6E4}\bigg(s+\frac{k-1}{2}\bigg)z^{-s}\,ds=z^{(k-1)/2}\text{e}^{-z}\quad \text{for}~\Re (z)>0,\end{eqnarray}$$
it follows that
$$\begin{eqnarray}I_{R}(z)=2\mathop{\sum }_{n=1}^{\infty }c_{f,a,q}(n)n^{(k-1)/2}e(nz)=F(z).\end{eqnarray}$$
By the functional equation, we have
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6E5}_{f,a,q}(s) & = & \displaystyle \unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(1-s)\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg),\nonumber\end{eqnarray}$$
so
$I_{L}(z)=I_{L1}(z)+I_{L2}(z)$
, where
$$\begin{eqnarray}\displaystyle I_{L1}(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=-\unicode[STIX]{x1D700}}\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{1/2-s}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(1-s)(-\text{i}z)^{-s-(k-1)/2}\,ds,\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle I_{L2}(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=-\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,(-\text{i}z)^{-s-(k-1)/2}\,ds.\end{eqnarray}$$
Making the substitution
$s\mapsto 1-s$
in (2.10), we get
$$\begin{eqnarray}\displaystyle I_{L1}(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=1+\unicode[STIX]{x1D700}}\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{s-1/2}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(s)(-\text{i}z)^{s-(k+1)/2}\,ds\nonumber\\ \displaystyle & = & \displaystyle 2\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{-1/2}(-\text{i}z)^{-(k+1)/2}(2\unicode[STIX]{x1D70B})^{-(k-1)/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=1+\unicode[STIX]{x1D700}}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(s)\bigg(\frac{2\unicode[STIX]{x1D70B}}{-\text{i}Nq^{2}z}\bigg)^{-s}\,ds\nonumber\\ \displaystyle & = & \displaystyle 2\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(Nq^{2})^{-1/2}(-\text{i}z)^{-(k+1)/2}(2\unicode[STIX]{x1D70B})^{-(k-1)/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)\bigg(\frac{2\unicode[STIX]{x1D70B}n}{-\text{i}Nq^{2}z}\bigg)^{(k-1)/2}e\bigg(-\frac{n}{Nq^{2}z}\bigg)\nonumber\\ \displaystyle & = & \displaystyle \frac{2\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)}{(-\text{i}\sqrt{N}qz)^{k}}\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)n^{(k-1)/2}e\bigg(-\frac{n}{Nq^{2}z}\bigg)\nonumber\\ \displaystyle & = & \displaystyle \frac{\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)}{(-\text{i}\sqrt{N}qz)^{k}}\overline{F}\bigg(-\frac{1}{Nq^{2}z}\bigg).\end{eqnarray}$$
Next, note that the integrand in (2.11) is holomorphic for
$-(k-1)/2<\Re (s)<(k+1)/2$
. Moving the contour to
$\Re (s)=k/2$
, we get a contribution from the pole at
$s=0$
(present only when
$k=1$
) of
$$\begin{eqnarray}\displaystyle & & \displaystyle \underset{s=0}{\text{Res}}\,\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)(-\text{i}z)^{-s-(k-1)/2}\nonumber\\ \displaystyle & & \displaystyle \quad =-\underset{s=0}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}\bigg(s,\frac{a}{q}\bigg)(-\text{i}z)^{-s-(k-1)/2}.\nonumber\end{eqnarray}$$
Thus
$$\begin{eqnarray}\displaystyle & & \displaystyle I_{L2}(z)+\underset{s=0}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}\bigg(s,\frac{a}{q}\bigg)(-\text{i}z)^{-s-(k-1)/2}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=k/2}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,(-\text{i}z)^{-s-(k-1)/2}\,ds.\nonumber\end{eqnarray}$$
The reflection formula for
$\unicode[STIX]{x1D6E4}$
implies that
$\unicode[STIX]{x1D713}^{\prime }(1-s)+\unicode[STIX]{x1D713}^{\prime }(s)=\unicode[STIX]{x1D70B}^{2}/(\sin ^{2}(\unicode[STIX]{x1D70B}s))$
, so
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D713}^{\prime }\bigg(\frac{k+1}{2}-s\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg) & = & \displaystyle \frac{\unicode[STIX]{x1D70B}^{2}}{\sin ^{2}(\unicode[STIX]{x1D70B}(s+(k-1)/2))}\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)-\unicode[STIX]{x1D713}^{\prime }\bigg(s-\frac{k-1}{2}\bigg).\nonumber\end{eqnarray}$$
Therefore,
$I_{L2}(z)+\underset{s=0}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}(s,a/q)(-\text{i}z)^{-s-(k-1)/2}=I_{L2B}(z)-I_{L2A}(z)$
, where
$$\begin{eqnarray}\displaystyle I_{L2A}(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=k/2}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\bigg(\unicode[STIX]{x1D713}^{\prime }\bigg(s+\frac{k-1}{2}\bigg)+\unicode[STIX]{x1D713}^{\prime }\bigg(s-\frac{k-1}{2}\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,(-\text{i}z)^{-s-(k-1)/2}\,ds,\nonumber\\ \displaystyle I_{L2B}(z) & = & \displaystyle \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=k/2}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\frac{\unicode[STIX]{x1D70B}^{2}}{\sin ^{2}(\unicode[STIX]{x1D70B}(s+(k-1)/2))}\nonumber\\ \displaystyle & & \displaystyle \times \,(-\text{i}z)^{-s-(k-1)/2}\,ds.\nonumber\end{eqnarray}$$
Hence,
$$\begin{eqnarray}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})=I_{R}(z)-I_{L1}(z)+I_{L2A}(z)-I_{L2B}(z).\end{eqnarray}$$
By applying (2.9), (2.12) and by setting
$A(z)=I_{L2A}(z)$
and
$B(z)=I_{L2B}(z)$
, we establish Lemma 2.3.◻
Next we evaluate
$\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})y^{s+(k-1)/2}\,dy/y$
, considering each term on the right-hand side of (2.8) in turn.
Lemma 2.4.
$\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}F(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s+(k-1)/2}\,dy/y-\unicode[STIX]{x1D6E5}_{f,a,q}(s,\unicode[STIX]{x1D6FC})$
continues to an entire function of
$s$
.
Proof. From the definition of
$F$
we compute that
$$\begin{eqnarray}\int _{0}^{\infty }F(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s+(k-1)/2}\frac{dy}{y}=\unicode[STIX]{x1D6E5}_{f,a,q}(s,\unicode[STIX]{x1D6FC}).\end{eqnarray}$$
Moreover,
$F(\unicode[STIX]{x1D6FC}+\text{i}y)$
decays exponentially as
$y\rightarrow \infty$
, so the contribution to the integral from
$y>|\unicode[STIX]{x1D6FC}|/4$
is entire.◻
Lemma 2.5. For any
$M\in \mathbb{Z}_{{\geqslant}0}$
,
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}(-\text{i}\sqrt{N}q(\unicode[STIX]{x1D6FC}+\text{i}y))^{-k}\overline{F}\bigg(-\frac{1}{Nq^{2}(\unicode[STIX]{x1D6FC}+\text{i}y)}\bigg)y^{s+(k-1)/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad -\,(\text{i}\operatorname{sgn}\unicode[STIX]{x1D6FC})^{k}\mathop{\sum }_{m=0}^{M-1}(-\text{i}\unicode[STIX]{x1D6FC})^{-m}\binom{s+m-(k+1)/2}{m}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,(Nq^{2}\unicode[STIX]{x1D6FC}^{2})^{s-1/2+m}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}\bigg(s+m,-\frac{1}{Nq^{2}\unicode[STIX]{x1D6FC}}\bigg)\end{eqnarray}$$
continues to a holomorphic function on
$\{s\in \mathbb{C}:\Re (s)>1-M\}$
.
Proof. As the proof of this lemma is very similar to that of [Reference Booker1, Lemma 3.3], we just provide a sketch and refer to the appropriate parts of loc. cit. for the relevant details. Fix
$y\in (0,|\unicode[STIX]{x1D6FC}|/4]$
, and set
$z=\unicode[STIX]{x1D6FC}+\text{i}y$
,
$\unicode[STIX]{x1D6FD}=-1/Nq^{2}\unicode[STIX]{x1D6FC}$
, and
$u=y/\unicode[STIX]{x1D6FC}$
. It may be checked that
$$\begin{eqnarray}-\frac{1}{Nq^{2}z}=\unicode[STIX]{x1D6FD}+\text{i}|\unicode[STIX]{x1D6FD}u|-\frac{\unicode[STIX]{x1D6FD}u^{2}}{1+\text{i}u}.\end{eqnarray}$$
Therefore,
$$\begin{eqnarray}\displaystyle (-\text{i}\sqrt{N}qz)^{-k}\overline{F}\bigg(-\frac{1}{Nq^{2}z}\bigg) & = & \displaystyle 2(-\text{i}\sqrt{N}q\unicode[STIX]{x1D6FC})^{-k}\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)n^{(k-1)/2}e(\unicode[STIX]{x1D6FD}n)\nonumber\\ \displaystyle & & \displaystyle \times \,\text{e}^{-2\unicode[STIX]{x1D70B}n|\unicode[STIX]{x1D6FD}u|}(1+\text{i}u)^{-k}e\bigg(-\frac{n\unicode[STIX]{x1D6FD}u^{2}}{1+\text{i}u}\bigg).\nonumber\end{eqnarray}$$
It was shown in [Reference Booker1, p. 820] that
$$\begin{eqnarray}(1+\text{i}u)^{-k}e\bigg(-\frac{n\unicode[STIX]{x1D6FD}u^{2}}{1+\text{i}u}\bigg)=\mathop{\sum }_{m=0}^{\infty }(-\text{i}u)^{m}\mathop{\sum }_{j=0}^{m}\binom{m+k-1}{m-j}\frac{(-2\unicode[STIX]{x1D70B}n|\unicode[STIX]{x1D6FD}u|)^{j}}{j!},\end{eqnarray}$$
and for
$M,K\in \mathbb{Z}_{{\geqslant}0}$
and
$|u|\leqslant \frac{1}{4}$
,
$$\begin{eqnarray}\mathop{\sum }_{m=M}^{\infty }(-\text{i}u)^{m}\mathop{\sum }_{j=0}^{m}\binom{m+k-1}{m-j}\frac{(-2\unicode[STIX]{x1D70B}n|\unicode[STIX]{x1D6FD}u|)^{j}}{j!}\ll _{\unicode[STIX]{x1D6FC},M,K}|u|^{M-K}n^{-K}\text{e}^{2\unicode[STIX]{x1D70B}n|\unicode[STIX]{x1D6FD}u|}.\end{eqnarray}$$
Thus, we obtain
$$\begin{eqnarray}\displaystyle & & \displaystyle (-\text{i}\sqrt{N}qz)^{-k}\overline{F}\bigg(-\frac{1}{Nq^{2}z}\bigg)=O_{M,K}\bigg(y^{M-K}\mathop{\sum }_{n=1}^{\infty }|c_{\bar{f},-\overline{Na},q}(n)|n^{(k-1)/2-K}\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad +\,2(-\text{i}\sqrt{N}q\unicode[STIX]{x1D6FC})^{-k}\mathop{\sum }_{m=0}^{M-1}\bigg(-\frac{\text{i}y}{\unicode[STIX]{x1D6FC}}\bigg)^{m}\mathop{\sum }_{j=0}^{m}\binom{m+k-1}{m-j}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)n^{(k-1)/2}e(\unicode[STIX]{x1D6FD}n)\frac{1}{j!}\bigg(-\frac{2\unicode[STIX]{x1D70B}ny}{Nq^{2}\unicode[STIX]{x1D6FC}^{2}}\bigg)^{j}\text{e}^{-2\unicode[STIX]{x1D70B}ny/Nq^{2}\unicode[STIX]{x1D6FC}^{2}}.\end{eqnarray}$$
By the choice
$K=\lfloor (k-1)/2\rfloor +2$
, the error term converges and is
$O_{M}(y^{M-K})$
. For the other term note that
$$\begin{eqnarray}\displaystyle & & \displaystyle 2y^{m}\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)n^{(k-1)/2}e(\unicode[STIX]{x1D6FD}n)\frac{1}{j!}\bigg(-\frac{2\unicode[STIX]{x1D70B}ny}{Nq^{2}\unicode[STIX]{x1D6FC}^{2}}\bigg)^{j}\text{e}^{-2\unicode[STIX]{x1D70B}ny/Nq^{2}\unicode[STIX]{x1D6FC}^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =2\frac{y^{j+m}}{j!}\frac{d^{j}}{dy^{j}}\mathop{\sum }_{n=1}^{\infty }c_{\bar{f},-\overline{Na},q}(n)n^{(k-1)/2}e(\unicode[STIX]{x1D6FD}n)\text{e}^{-2\unicode[STIX]{x1D70B}ny/Nq^{2}\unicode[STIX]{x1D6FC}^{2}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{y^{j+m}}{j!}\frac{d^{j}}{dy^{j}}\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=m+2}(Nq^{2}\unicode[STIX]{x1D6FC}^{2})^{s+(k-1)/2}\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(s,\unicode[STIX]{x1D6FD})y^{-s-(k-1)/2}\,ds\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=2}\binom{-s-\binom{k-1}{2}-m}{j}(Nq^{2}\unicode[STIX]{x1D6FC}^{2})^{s+(k-1)/2+m}\nonumber\\ \displaystyle & & \displaystyle \qquad \times \,\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}(s+m,\unicode[STIX]{x1D6FD})y^{-s-(k-1)/2}\,ds.\nonumber\end{eqnarray}$$
Inserting this in the last term of (2.14) and using the Chu–Vandermonde identity
$$\begin{eqnarray}\mathop{\sum }_{j=0}^{m}\binom{m+k-1}{m-j}\binom{-s-\binom{k-1}{2}-m}{j}=\binom{-s+\binom{k-1}{2}}{m}=(-1)^{m}\binom{s+m-\binom{k+1}{2}}{m},\end{eqnarray}$$
we arrive at
$$\begin{eqnarray}\displaystyle (-\text{i}\sqrt{N}qz)^{-k}\overline{F}\bigg(-\frac{1}{Nq^{2}z}\bigg) & = & \displaystyle O_{M}(y^{M-\lfloor (k+3)/2\rfloor })+(\text{i}\operatorname{sgn}\unicode[STIX]{x1D6FC})^{k}\mathop{\sum }_{m=0}^{M-1}\frac{(-\text{i}\unicode[STIX]{x1D6FC})^{-m}}{2\unicode[STIX]{x1D70B}\text{i}}\nonumber\\ \displaystyle & & \displaystyle \times \,\int _{\Re (s)=2}\binom{s+m-\frac{k+1}{2}}{m}(Nq^{2}\unicode[STIX]{x1D6FC}^{2})^{s-1/2+m}\nonumber\\ \displaystyle & & \displaystyle \times \,\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q}\bigg(s+m,-\frac{1}{Nq^{2}\unicode[STIX]{x1D6FC}}\bigg)y^{-s-(k-1)/2}\,ds.\nonumber\end{eqnarray}$$
We multiply both sides by
$y^{s+(k-1)/2-1}$
and integrate over
$y\in (0,|\unicode[STIX]{x1D6FC}|/4]$
. The error term yields a holomorphic function for
$\Re (s)>2-M$
. As for the sum over
$m$
, by shifting the contour to the right, we see that each term decays rapidly as
$y\rightarrow \infty$
, so the integral over
$(0,|\unicode[STIX]{x1D6FC}|/4]$
differs from the full Mellin transform by an entire function. By Mellin inversion, it follows that (2.13) is holomorphic for
$\Re (s)>2-M$
. Finally, replacing
$M$
by
$M+1$
and discarding the final term of the sum concludes the proof of the lemma.◻
Lemma 2.6.
$\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)^{-1}\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}A(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}(dy/y)$
continues to an entire function of
$s$
.
Proof. Let
$\unicode[STIX]{x1D6F7}(s)=\unicode[STIX]{x1D713}^{\prime }(s+(k-1)/2)+\unicode[STIX]{x1D713}^{\prime }(s-(k-1)/2)$
. By the identity
$\unicode[STIX]{x1D713}^{\prime }(s)=\int _{1}^{\infty }(\log x/(x-1))x^{-s}\,dx$
, we have
$\unicode[STIX]{x1D6F7}(s)=\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)x^{-s-(k-1)/2}\,dx$
for
$\Re (s)>(k-1)/2$
, where
$\unicode[STIX]{x1D719}(x)=(x^{k-1}+1)/(x-1)\log x$
. It follows that
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}(s)\unicode[STIX]{x1D6E4}\bigg(s+\frac{k-1}{2}\bigg) & = & \displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}(x)x^{-s-(k-1)/2}\,dx\int _{0}^{\infty }\text{e}^{-y}y^{s+(k-1)/2}\,dy\nonumber\\ \displaystyle & = & \displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\int _{0}^{\infty }\text{e}^{-y}\bigg(\frac{y}{x}\bigg)^{s+(k-1)/2}\frac{dy}{y}.\nonumber\end{eqnarray}$$
By the variable change
$y\mapsto xy$
we obtain
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F7}(s)\unicode[STIX]{x1D6E4}\bigg(s+\frac{k-1}{2}\bigg) & = & \displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\int _{0}^{\infty }\text{e}^{-xy}y^{s+(k-1)/2}\frac{dy}{y}\nonumber\\ \displaystyle & = & \displaystyle \int _{0}^{\infty }\bigg(\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\text{e}^{-xy}\,dx\bigg)y^{s+(k-1)/2}\frac{dy}{y}.\nonumber\end{eqnarray}$$
By Mellin inversion,
$$\begin{eqnarray}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\text{e}^{-xy}\,dx=\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=2}\unicode[STIX]{x1D6F7}(s)\unicode[STIX]{x1D6E4}\bigg(s+\frac{k-1}{2}\bigg)y^{-s-(k-1)/2}\,ds.\end{eqnarray}$$
Observe that
$L_{\bar{f}}(s,-\overline{Na}/q)=\sum _{n=1}^{\infty }b_{n}n^{-s}$
, where
$b_{n}=\unicode[STIX]{x1D706}_{\bar{f}}(n)e(-\overline{Na}/q)$
. Thus for
$z\in \mathbb{H}$
,
$$\begin{eqnarray}\displaystyle A(z) & = & \displaystyle 2\mathop{\sum }_{n=1}^{\infty }b_{n}\cdot \frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=2}\unicode[STIX]{x1D6F7}(s)\unicode[STIX]{x1D6E4}\bigg(s+\frac{k-1}{2}\bigg)(-2\unicode[STIX]{x1D70B}\text{i}nz)^{-s-(k-1)/2}\,ds\nonumber\\ \displaystyle & = & \displaystyle 2\mathop{\sum }_{n=1}^{\infty }b_{n}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)\text{e}^{2\unicode[STIX]{x1D70B}\text{i}nxz}\,dx,\nonumber\end{eqnarray}$$
where the last step follows from (2.15). For
$z=\unicode[STIX]{x1D6FC}+\text{i}y$
this simplifies to
$$\begin{eqnarray}A(\unicode[STIX]{x1D6FC}+\text{i}y)=2\mathop{\sum }_{n=1}^{\infty }b_{n}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)e(\unicode[STIX]{x1D6FC}nx)\text{e}^{-2\unicode[STIX]{x1D70B}nxy}\,dx.\end{eqnarray}$$
Using this expression, it follows that
$$\begin{eqnarray}\displaystyle \int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}\frac{dy}{y} & = & \displaystyle 2\mathop{\sum }_{n=1}^{\infty }b_{n}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)e(\unicode[STIX]{x1D6FC}nx)\int _{0}^{\infty }\text{e}^{-2\unicode[STIX]{x1D70B}nxy}y^{s}\frac{dy}{y}\,dx\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)\mathop{\sum }_{n=1}^{\infty }b_{n}n^{-s}\int _{1}^{\infty }\unicode[STIX]{x1D719}(x)e(\unicode[STIX]{x1D6FC}nx)x^{-s}\,dx.\end{eqnarray}$$
For
$j=0,1,2,\ldots ,$
define the sequence of functions
$\unicode[STIX]{x1D719}_{j}(x,s)$
by
$$\begin{eqnarray}\unicode[STIX]{x1D719}_{0}(x,s)=\unicode[STIX]{x1D719}(x),\qquad \unicode[STIX]{x1D719}_{j+1}(x,s)=x\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}_{j}}{\unicode[STIX]{x2202}x}(x,s)-(s+j)\unicode[STIX]{x1D719}_{j}(x,s).\end{eqnarray}$$
Integrating by parts,
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}_{j}(x,s)e(\unicode[STIX]{x1D6FC}nx)x^{-s-j}\,dx\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{e(\unicode[STIX]{x1D6FC}n)\unicode[STIX]{x1D719}_{j}(1,s)}{2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D6FC}n}-\frac{1}{2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D6FC}n}\int _{1}^{\infty }\unicode[STIX]{x1D719}_{j+1}(x,s)e(\unicode[STIX]{x1D6FC}nx)x^{-s-j-1}\,dx.\nonumber\end{eqnarray}$$
Repeated application of this yields
$$\begin{eqnarray}\displaystyle \int _{1}^{\infty }\unicode[STIX]{x1D719}(x)e(\unicode[STIX]{x1D6FC}nx)x^{-s}\,dx & = & \displaystyle e(\unicode[STIX]{x1D6FC}n)\mathop{\sum }_{j=0}^{m-1}\frac{\unicode[STIX]{x1D719}_{j}(1,s)}{(-2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D6FC}n)^{j+1}}\nonumber\\ \displaystyle & & \displaystyle +\,(-2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D6FC}n)^{-m}\int _{1}^{\infty }\unicode[STIX]{x1D719}_{m}(x,s)e(\unicode[STIX]{x1D6FC}nx)x^{-s-m}\,dx\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
for
$m\in \mathbb{Z}_{{\geqslant}0}$
. By (2.17) and (2.18) it follows that
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)}\int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{j=0}^{m-1}\frac{\unicode[STIX]{x1D719}_{j}(1,s)}{(-2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D6FC})^{j+1}}L_{\bar{f}}\bigg(s+j+1,-\frac{\overline{Na}}{q}+\unicode[STIX]{x1D6FC}\bigg)\nonumber\\ \displaystyle & & \displaystyle \qquad +\,(-2\unicode[STIX]{x1D70B}\text{i}\unicode[STIX]{x1D6FC})^{-m}\mathop{\sum }_{n=1}^{\infty }\frac{b_{n}}{n^{s+m}}\int _{1}^{\infty }\unicode[STIX]{x1D719}_{m}(x,s)e(\unicode[STIX]{x1D6FC}nx)x^{-s-m}\,dx.\nonumber\end{eqnarray}$$
Each term in the sum extends to an entire function of
$s$
, by [Reference Booker and Krishnamurthy2, Proposition 3.1]. Furthermore, it may be checked that
$\unicode[STIX]{x1D719}_{m}(x,s)\ll _{m,k}(1+|s|)^{m}x^{k-1}$
. Therefore, the last integral is holomorphic for
$\Re (s)>k-m$
. Letting
$m\rightarrow \infty$
shows that
$\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)^{-1}\int _{0}^{\infty }A(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}(dy/y)$
continues to an entire function.
Finally, from (2.16) we see that
$A(\unicode[STIX]{x1D6FC}+\text{i}y)$
decays exponentially as
$y\rightarrow \infty$
, and hence
$\int _{|\unicode[STIX]{x1D6FC}|/4}^{\infty }A(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}(dy/y)$
is entire. This completes the proof.◻
Lemma 2.7.
$\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)^{-1}\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}B(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}(dy/y)$
continues to an entire function of
$s$
.
Proof. Following the proof of [Reference Booker1, Lemma 3.4], we obtain
$$\begin{eqnarray}B(\unicode[STIX]{x1D6FC}+\text{i}y)=\mathop{\sum }_{j=0}^{M-1}P_{j}(\unicode[STIX]{x1D6FC})y^{j}+O_{M}(y^{M})\quad \text{for all}~M\in \mathbb{Z}_{{\geqslant}0},y\in \bigg(0,\frac{|\unicode[STIX]{x1D6FC}|}{4}\bigg],\end{eqnarray}$$
where
$$\begin{eqnarray}\displaystyle P_{j}(\unicode[STIX]{x1D6FC}) & = & \displaystyle \frac{(-\text{i}\unicode[STIX]{x1D6FC})^{-j}}{2\unicode[STIX]{x1D70B}\text{i}}\int _{\Re (s)=k/2}\text{e}^{\text{i}(\unicode[STIX]{x1D70B}/2)\operatorname{sgn}(\unicode[STIX]{x1D6FC})(s+(k-1)/2)}|\unicode[STIX]{x1D6FC}|^{-s-(k-1)/2}\nonumber\\ \displaystyle & & \displaystyle \times \,\binom{-s-\binom{k-1}{2}}{j}\unicode[STIX]{x1D6EC}_{f}\bigg(s,\frac{a}{q}\bigg)\frac{\unicode[STIX]{x1D70B}^{2}}{\sin ^{2}(\unicode[STIX]{x1D70B}(s+(k-1)/2))}\,ds.\nonumber\end{eqnarray}$$
Hence,
$$\begin{eqnarray}\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}B(\unicode[STIX]{x1D6FC}+\text{i}y)y^{s}\frac{dy}{y}-\mathop{\sum }_{j=0}^{M-1}P_{j}(\unicode[STIX]{x1D6FC})\frac{|\unicode[STIX]{x1D6FC}/4|^{s+j}}{s+j}\end{eqnarray}$$
is holomorphic for
$\Re (s)>-M$
. Note that the sum over
$j$
is entire apart from at most simple poles at the poles of
$\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)$
. Dividing by
$\unicode[STIX]{x1D6E4}_{\mathbb{C}}(s)$
and taking
$M\rightarrow \infty$
concludes the proof.◻
Proof of Proposition 2.2.
Combining Lemmas 2.3–2.7 and taking
$M=1$
, we see that
$I_{f,a,q,\unicode[STIX]{x1D6FC}}(s)-H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
has analytic continuation to
$\Re (s)>0$
. If
$\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S_{f,a,q}(y,\unicode[STIX]{x1D6FC})|y^{\unicode[STIX]{x1D70E}+(k-1)/2}(dy/y)<\infty$
for some
$\unicode[STIX]{x1D70E}\geqslant 0$
, then the integral defining
$I_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
converges absolutely for
$\Re (s)>\unicode[STIX]{x1D70E}$
, and hence
$I_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
is holomorphic in that region.◻
3 Estimates for
$N_{f,a,q}^{s}(T)$
Fix
$f$
,
$a$
,
$q$
as in Proposition 2.1, and let
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
. In this section, we derive estimates for
$N_{f,a,q}^{s}(T)$
based on Proposition 2.2.
Lemma 3.1. Let
$f\in S_{k}(\unicode[STIX]{x1D6E4}_{1}(N))$
be a primitive form. For
$\unicode[STIX]{x1D70C}=\unicode[STIX]{x1D6FD}+\text{i}\unicode[STIX]{x1D6FE}$
a zero of
$\unicode[STIX]{x1D6EC}_{f}(s)$
, we have
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})\ll _{f}(2+|\unicode[STIX]{x1D6FE}|)^{k/2+|\unicode[STIX]{x1D6FD}-1/2|/3-1/6}\log ^{2}(2+|\unicode[STIX]{x1D6FE}|)\text{e}^{-(\unicode[STIX]{x1D70B}/2)|\unicode[STIX]{x1D6FE}|}.\end{eqnarray}$$
Proof. We begin by establishing, for
$s=\unicode[STIX]{x1D70E}+\text{i}t$
and
$\unicode[STIX]{x1D70E}\in [\frac{1}{2},1]$
,
$$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{\mathbb{C}}\bigg(s+\frac{k-1}{2}\bigg)L_{f}^{\prime }(s)\ll \unicode[STIX]{x1D70F}^{k/2-(1-\unicode[STIX]{x1D70E})/3}\text{e}^{-(\unicode[STIX]{x1D70B}/2)|t|}\log ^{2}\unicode[STIX]{x1D70F},\end{eqnarray}$$
where
$\unicode[STIX]{x1D70F}=|t|+2$
. By [Reference Booker, Milinovich and Ng3, Theorem 1.1], we have
$$\begin{eqnarray}L_{f}\big({\textstyle \frac{1}{2}}+\text{i}t\big)\ll \unicode[STIX]{x1D70F}^{1/3}\log \unicode[STIX]{x1D70F}.\end{eqnarray}$$
By the Phragmén–Lindelöf principle, using
$$\begin{eqnarray}L_{f}\bigg(-\frac{1}{\log \unicode[STIX]{x1D70F}}+\text{i}t\bigg)\ll \unicode[STIX]{x1D70F}\log \unicode[STIX]{x1D70F}\quad \text{and}\quad L_{f}\bigg(1+\frac{1}{\log \unicode[STIX]{x1D70F}}+\text{i}t\bigg)\ll \log \unicode[STIX]{x1D70F},\end{eqnarray}$$
it follows that
$L_{f}(\unicode[STIX]{x1D70E}+\text{i}t)\ll \unicode[STIX]{x1D70F}^{1/3}\log \unicode[STIX]{x1D70F}$
when
$|\unicode[STIX]{x1D70E}-{\textstyle \frac{1}{2}}|\leqslant 1/\log \unicode[STIX]{x1D70F}$
. An application of the Cauchy integral formula then yields
$L_{f}^{\prime }({\textstyle \frac{1}{2}}+\text{i}t)\ll \unicode[STIX]{x1D70F}^{1/3}\log ^{2}\unicode[STIX]{x1D70F}$
. By Cauchy’s inequality and Rankin’s estimate
$\sum _{n\leqslant x}|\unicode[STIX]{x1D706}_{f}(n)|^{2}\ll x$
, we get
$$\begin{eqnarray}\displaystyle |L_{f}^{\prime }(1+\unicode[STIX]{x1D700}+\text{i}t)| & {\leqslant} & \displaystyle \mathop{\sum }_{n=1}^{\infty }\frac{|\unicode[STIX]{x1D706}_{f}(n)|\log n}{n^{1+\unicode[STIX]{x1D700}}}\leqslant \bigg(\mathop{\sum }_{n=1}^{\infty }\frac{|\unicode[STIX]{x1D706}_{f}(n)|^{2}}{n^{1+\unicode[STIX]{x1D700}}}\bigg)^{1/2}\unicode[STIX]{x1D701}^{\prime \prime }(1+\unicode[STIX]{x1D700})^{1/2}\nonumber\\ \displaystyle & \ll & \displaystyle \unicode[STIX]{x1D700}^{-1/2}\unicode[STIX]{x1D700}^{-3/2}=\unicode[STIX]{x1D700}^{-2}\nonumber\end{eqnarray}$$
for
$\unicode[STIX]{x1D700}>0$
. Another application of the Phragmén–Lindelöf principle yields
$$\begin{eqnarray}L_{f}^{\prime }(\unicode[STIX]{x1D70E}+\text{i}t)\ll \unicode[STIX]{x1D70F}^{(2/3)(1-\unicode[STIX]{x1D70E})}\log ^{2}\unicode[STIX]{x1D70F}\end{eqnarray}$$
for
$\unicode[STIX]{x1D70E}\in [\frac{1}{2},1]$
. This, together with the Stirling formula estimate
$$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{\mathbb{C}}\bigg(s+\frac{k-1}{2}\bigg)\ll \unicode[STIX]{x1D70F}^{\unicode[STIX]{x1D70E}+k/2-1}\text{e}^{-(\unicode[STIX]{x1D70B}/2)|t|},\end{eqnarray}$$
yields (3.2).
Setting
$s=\unicode[STIX]{x1D70C}=\unicode[STIX]{x1D6FD}+\text{i}\unicode[STIX]{x1D6FE}$
with
$\unicode[STIX]{x1D6FD}\geqslant \frac{1}{2}$
in (3.2) gives
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})=\unicode[STIX]{x1D6E4}_{\mathbb{ C}}\bigg(\unicode[STIX]{x1D70C}+\frac{k-1}{2}\bigg)L_{f}^{\prime }(\unicode[STIX]{x1D70C})\ll (2+|\unicode[STIX]{x1D6FE}|)^{k/2-(1-\unicode[STIX]{x1D6FD})/3}\log ^{2}(2+|\unicode[STIX]{x1D6FE}|)\text{e}^{-(\unicode[STIX]{x1D70B}/2)|\unicode[STIX]{x1D6FE}|}.\end{eqnarray}$$
Now suppose
$\unicode[STIX]{x1D6FD}<\frac{1}{2}$
. Differentiating the functional equation we obtain
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})=-\unicode[STIX]{x1D716}N^{1/2-\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D6EC}_{\bar{f}}^{\prime }(1-\unicode[STIX]{x1D70C}),\end{eqnarray}$$
where
$|\unicode[STIX]{x1D716}|=1$
. Applying (3.3) to
$\unicode[STIX]{x1D6EC}_{\bar{f}}^{\prime }(1-\unicode[STIX]{x1D70C})$
it follows that
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})\ll (2+|\unicode[STIX]{x1D6FE}|)^{k/2-\unicode[STIX]{x1D6FD}/3}\log ^{2}(2+|\unicode[STIX]{x1D6FE}|)\text{e}^{-(\unicode[STIX]{x1D70B}/2)|\unicode[STIX]{x1D6FE}|}.\end{eqnarray}$$
Lemma 3.2. For any fixed
$\unicode[STIX]{x1D700}>0$
and all
$\unicode[STIX]{x1D70E}\in [\unicode[STIX]{x1D700},2]$
,
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S_{f,a,q}(y,\unicode[STIX]{x1D6FC})|y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \mathop{\sum }_{\substack{ \unicode[STIX]{x1D70C}=\unicode[STIX]{x1D6FD}+\text{i}\unicode[STIX]{x1D6FE} \\ \text{a pole of }\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)}}(2+|\unicode[STIX]{x1D6FE}|)^{(1+|\unicode[STIX]{x1D6FD}-(1/2)|)/3-\unicode[STIX]{x1D70E}}\log ^{2}(2+|\unicode[STIX]{x1D6FE}|).\nonumber\end{eqnarray}$$
Proof. Throughout this proof we let
$\unicode[STIX]{x1D70C}=\unicode[STIX]{x1D6FD}+\text{i}\unicode[STIX]{x1D6FE}$
denote a pole of
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
, and we set
$\unicode[STIX]{x1D70F}=2+|\unicode[STIX]{x1D6FE}|$
. Recalling (2.6), observe that
$(y-\text{i}\unicode[STIX]{x1D6FC})^{-\unicode[STIX]{x1D70C}-(k-1)/2}=\text{e}^{\text{i}(\unicode[STIX]{x1D70B}/2)\operatorname{sgn}(\unicode[STIX]{x1D6FC})(\unicode[STIX]{x1D70C}+(k-1)/2)}|\unicode[STIX]{x1D6FC}|^{-\unicode[STIX]{x1D70C}-(k-1)/2}(1+\text{i}y/\unicode[STIX]{x1D6FC})^{-\unicode[STIX]{x1D70C}-(k-1)/2}$
and
$$\begin{eqnarray}\displaystyle \bigg|\bigg(1+\text{i}\frac{y}{\unicode[STIX]{x1D6FC}}\bigg)^{-(\unicode[STIX]{x1D6FD}+\text{i}\unicode[STIX]{x1D6FE}+(k-1)/2)}\bigg| & = & \displaystyle \big|\text{e}^{-((1/2)\log (1+(y/\unicode[STIX]{x1D6FC})^{2})+\text{i}\arctan (y/\unicode[STIX]{x1D6FC}))(\unicode[STIX]{x1D6FD}+\text{i}\unicode[STIX]{x1D6FE}+(k-1)/2)}\big|\nonumber\\ \displaystyle & = & \displaystyle \bigg(1+\bigg(\frac{y}{\unicode[STIX]{x1D6FC}}\bigg)^{2}\bigg)^{-\unicode[STIX]{x1D6FD}/2-(k-1)/4}\text{e}^{\unicode[STIX]{x1D6FE}\arctan (y/\unicode[STIX]{x1D6FC})}.\end{eqnarray}$$
Therefore,
$$\begin{eqnarray}(y-\text{i}\unicode[STIX]{x1D6FC})^{-\unicode[STIX]{x1D70C}-(k-1)/2}\ll \text{e}^{\unicode[STIX]{x1D6FE}\operatorname{sgn}(\unicode[STIX]{x1D6FC})(\arctan (y/|\unicode[STIX]{x1D6FC}|)-\unicode[STIX]{x1D70B}/2)}.\end{eqnarray}$$
Next, we treat the residue in (2.6). By (2.4), the poles of
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
arise from poles of
$\unicode[STIX]{x1D6E5}_{f}(s)$
and
$\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D712})$
with
$\unicode[STIX]{x1D712}\neq \unicode[STIX]{x1D712}_{0}$
. The contributrion of an individual term of (2.4) to
$\text{Res}_{s=\unicode[STIX]{x1D70C}}\,\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
, if non-zero, is of the form
$$\begin{eqnarray}-\bigg(1-\frac{q}{q-1}P_{f,q}(q^{-\unicode[STIX]{x1D70C}})\bigg)\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C})\quad \text{or}\quad -\frac{\unicode[STIX]{x1D70F}(\overline{\unicode[STIX]{x1D712}})\unicode[STIX]{x1D712}(a)}{q-1}\unicode[STIX]{x1D6EC}_{f}^{\prime }(\unicode[STIX]{x1D70C},\unicode[STIX]{x1D712}).\end{eqnarray}$$
Applying Lemma 3.1 (possibly replacing
$f$
by
$f\otimes \unicode[STIX]{x1D712}$
) to each of these expressions, it follows that
$$\begin{eqnarray}\underset{s=\unicode[STIX]{x1D70C}}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)\ll \unicode[STIX]{x1D70F}^{k/2+|\unicode[STIX]{x1D6FD}-1/2|/3-1/6}(\log ^{2}\unicode[STIX]{x1D70F})\text{e}^{-(\unicode[STIX]{x1D70B}/2)|\unicode[STIX]{x1D6FE}|}.\end{eqnarray}$$
It follows from (2.6), (3.6) and (3.7) that
$$\begin{eqnarray}\displaystyle S_{f,a,q}(y,\unicode[STIX]{x1D6FC}) & \ll & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D70F}^{k/2+|\unicode[STIX]{x1D6FD}-1/2|/3-1/6}(\log ^{2}\unicode[STIX]{x1D70F})\nonumber\\ \displaystyle & & \displaystyle \times \,\text{e}^{|\unicode[STIX]{x1D6FE}|[\operatorname{sgn}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FE})\arctan (y/|\unicode[STIX]{x1D6FC}|)-(\unicode[STIX]{x1D70B}/2)(1+\operatorname{sgn}(\unicode[STIX]{x1D6FC}\unicode[STIX]{x1D6FE}))]}.\nonumber\end{eqnarray}$$
By considering cases and using the bound
$\arctan u\geqslant u/2$
for
$0\leqslant u\leqslant \frac{1}{4}$
, we have
$$\begin{eqnarray}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})\ll \mathop{\sum }_{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D70F}^{k/2+|\unicode[STIX]{x1D6FD}-1/2|/3-1/6}(\log ^{2}\unicode[STIX]{x1D70F})\text{e}^{-c|\unicode[STIX]{x1D6FE}|y}\quad \text{for}~y\in \bigg(0,\frac{|\unicode[STIX]{x1D6FC}|}{4}\bigg],\end{eqnarray}$$
where
$c=1/2|\unicode[STIX]{x1D6FC}|>0$
. We deduce from this
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S_{f,a,q}(y,\unicode[STIX]{x1D6FC})|y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}\nonumber\\ \displaystyle & & \displaystyle \quad \ll \mathop{\sum }_{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D70F}^{k/2+|\unicode[STIX]{x1D6FD}-1/2|/3-1/6}\log ^{2}\unicode[STIX]{x1D70F}\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}\text{e}^{-c|\unicode[STIX]{x1D6FE}|y}y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}.\nonumber\end{eqnarray}$$
Now
$$\begin{eqnarray}\displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}\text{e}^{-c|\unicode[STIX]{x1D6FE}|y}y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y} & \ll & \displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}\text{e}^{-c\unicode[STIX]{x1D70F}y}y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \int _{0}^{\infty }\text{e}^{-c\unicode[STIX]{x1D70F}y}y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}.\nonumber\end{eqnarray}$$
By the variable change
$u=c\unicode[STIX]{x1D70F}y$
, the last integral equals
$$\begin{eqnarray}\frac{1}{(c\unicode[STIX]{x1D70F})^{\unicode[STIX]{x1D70E}+(k-1)/2}}\int _{0}^{\infty }\text{e}^{-u}u^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{du}{u}=\frac{\unicode[STIX]{x1D6E4}(\unicode[STIX]{x1D70E}+(k-1)/2)}{(c\unicode[STIX]{x1D70F})^{\unicode[STIX]{x1D70E}+(k-1)/2}}\ll \unicode[STIX]{x1D70F}^{-\unicode[STIX]{x1D70E}-(k-1)/2},\end{eqnarray}$$
and thus
$$\begin{eqnarray}\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}S_{f,a,q}(y,\unicode[STIX]{x1D6FC})y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y}\ll \mathop{\sum }_{\unicode[STIX]{x1D70C}}\unicode[STIX]{x1D70F}^{(1+|\unicode[STIX]{x1D6FD}-1/2|)/3-\unicode[STIX]{x1D70E}}\log ^{2}\unicode[STIX]{x1D70F}.\end{eqnarray}$$
For a meromorphic function
$h$
on
$\{s\in \mathbb{C}:\Re (s)>1\}$
, define
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(h)=\inf \{\unicode[STIX]{x1D703}\geqslant 0:h~\text{continues analytically to}~\{s\in \mathbb{C}:\Re (s)>\unicode[STIX]{x1D703}\}\}.\end{eqnarray}$$
We also set
$$\begin{eqnarray}\unicode[STIX]{x1D703}_{f,a,q}(T)=\sup \Big(\!\{0\}\cup \Big\{\Re (\unicode[STIX]{x1D70C}),1-\Re (\unicode[STIX]{x1D70C}):\unicode[STIX]{x1D70C}\in \mathbb{C},|\Im (\unicode[STIX]{x1D70C})|\leqslant T,\underset{s=\unicode[STIX]{x1D70C}}{\text{Res}}\,\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)\neq 0\Big\}\Big)\end{eqnarray}$$
and
$$\begin{eqnarray}\unicode[STIX]{x1D703}_{f,a,q}=\lim _{T\rightarrow \infty }\unicode[STIX]{x1D703}_{f,a,q}(T).\end{eqnarray}$$
By Proposition 2.1, we have
$$\begin{eqnarray}\unicode[STIX]{x1D703}_{f,a,q}=\max (\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,q}),\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},-\overline{Na},q})).\end{eqnarray}$$
Proposition 3.3. If
$\unicode[STIX]{x1D6E9}(H_{f,a,q,\unicode[STIX]{x1D6FC}})>0$
then
$\unicode[STIX]{x1D703}_{f,a,q}\geqslant \frac{1}{2}$
and
$$\begin{eqnarray}N_{f,a,q}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{(1/3)(1-\unicode[STIX]{x1D703}_{f,a,q})+\unicode[STIX]{x1D6E9}(H_{f,a,q,\unicode[STIX]{x1D6FC}})-1/2-\unicode[STIX]{x1D700}})\quad \text{for all}~\unicode[STIX]{x1D700}>0.\end{eqnarray}$$
Further, if
$\unicode[STIX]{x1D6E9}(H_{f,a,q,\unicode[STIX]{x1D6FC}})=\frac{1}{2}$
and
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
has a pole with real part
$\frac{1}{2}$
, then
$$\begin{eqnarray}N_{f,a,q}^{s}(T)=\unicode[STIX]{x1D6FA}\!\bigg(\frac{T^{(1/3)(1-\unicode[STIX]{x1D703}_{f,a,q}(T))}}{(1-\unicode[STIX]{x1D703}_{f,a,q}(T))\log ^{2}T}\bigg),\end{eqnarray}$$
and there are arbitrarily large
$T>0$
such that
$$\begin{eqnarray}N_{f,a,q}^{s}(T)\geqslant \log \log \log T.\end{eqnarray}$$
Proof. Let
$\unicode[STIX]{x1D6FD}_{n}+\text{i}\unicode[STIX]{x1D6FE}_{n}$
run through the poles of
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
, in increasing order of
$|\unicode[STIX]{x1D6FE}_{n}|$
. For brevity, we write
$I(s)$
,
$H(s)$
,
$\unicode[STIX]{x1D6E9}$
,
$S(y)$
,
$N(t)$
,
$\unicode[STIX]{x1D703}(t)$
and
$\unicode[STIX]{x1D703}$
for
$I_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
,
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
,
$\unicode[STIX]{x1D6E9}(H_{f,a,q,\unicode[STIX]{x1D6FC}})$
,
$S_{f,a,q}(y,\unicode[STIX]{x1D6FC})$
,
$N_{f,a,q}^{s}(t)$
,
$\unicode[STIX]{x1D703}_{f,a,q}(t)$
and
$\unicode[STIX]{x1D703}_{f,a,q}$
, respectively. By Lemma 3.2, we have
$$\begin{eqnarray}\displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S(y)|y^{\unicode[STIX]{x1D70E}+(k-1)/2}\frac{dy}{y} & \ll & \displaystyle \mathop{\sum }_{n\geqslant 1}(2+|\unicode[STIX]{x1D6FE}_{n}|)^{(1+|\unicode[STIX]{x1D6FD}_{n}-1/2|)/3-\unicode[STIX]{x1D70E}}\log ^{2}(2+|\unicode[STIX]{x1D6FE}_{n}|)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \mathop{\sum }_{n\geqslant 1}(2+|\unicode[STIX]{x1D6FE}_{n}|)^{\unicode[STIX]{x1D703}(|\unicode[STIX]{x1D6FE}_{n}|)/3+1/6-\unicode[STIX]{x1D70E}}\log ^{2}(2+|\unicode[STIX]{x1D6FE}_{n}|).\nonumber\end{eqnarray}$$
If
$\unicode[STIX]{x1D6E9}>0$
then by Proposition 2.2, the integral must diverge for sufficiently small
$\unicode[STIX]{x1D70E}>0$
, and thus the right-hand side has infinitely many terms. Thus
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
has poles, so
$\unicode[STIX]{x1D703}\geqslant \frac{1}{2}$
.
Suppose that (3.9) does not hold. Then there exists
$\unicode[STIX]{x1D700}\in (0,\unicode[STIX]{x1D6E9})$
such that
$N(t)=o(t^{(1/3)(1-\unicode[STIX]{x1D703})+\unicode[STIX]{x1D6E9}-1/2-\unicode[STIX]{x1D700}})$
. Choosing
$\unicode[STIX]{x1D70E}=\unicode[STIX]{x1D6E9}-\unicode[STIX]{x1D700}/3$
and using the estimate
$\log ^{2}(2+|\unicode[STIX]{x1D6FE}_{n}|)\ll (2+|\unicode[STIX]{x1D6FE}_{n}|)^{\unicode[STIX]{x1D700}/3}$
, we have
$$\begin{eqnarray}\displaystyle \int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S(y)|y^{\unicode[STIX]{x1D6E9}-\unicode[STIX]{x1D700}/3+(k-1)/2}\frac{dy}{y} & \ll & \displaystyle \mathop{\sum }_{n\geqslant 1}(2+|\unicode[STIX]{x1D6FE}_{n}|)^{\unicode[STIX]{x1D703}/3+1/6-\unicode[STIX]{x1D6E9}+(2/3)\unicode[STIX]{x1D700}}\nonumber\\ \displaystyle & \ll & \displaystyle 1+\int _{1}^{\infty }t^{\unicode[STIX]{x1D703}/3+1/6-\unicode[STIX]{x1D6E9}+(2/3)\unicode[STIX]{x1D700}}\,dN(t)\nonumber\\ \displaystyle & \ll & \displaystyle 1+\int _{1}^{\infty }t^{\unicode[STIX]{x1D703}/3+1/6-\unicode[STIX]{x1D6E9}+(2/3)\unicode[STIX]{x1D700}-1}N(t)\,dt\nonumber\\ \displaystyle & \ll & \displaystyle 1+\int _{1}^{\infty }t^{-1-\unicode[STIX]{x1D700}/3}\,dt\ll 1.\nonumber\end{eqnarray}$$
By Proposition 2.2, it follows that
$H(s)$
is holomorphic for
$\Re (s)>\unicode[STIX]{x1D6E9}-\unicode[STIX]{x1D700}/3$
. This is a contradiction, so (3.9) must hold.
Next suppose that
$\unicode[STIX]{x1D6E9}=\frac{1}{2}$
, and let
$\unicode[STIX]{x1D70C}$
be a pole of
$H(s)$
with
$\Re (\unicode[STIX]{x1D70C})=\frac{1}{2}$
. Then for sufficiently small
$\unicode[STIX]{x1D6FF}>0$
, by Proposition 2.2, we have
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}^{-1}\ll |H(\unicode[STIX]{x1D70C}+\unicode[STIX]{x1D6FF})|\ll 1+|I(\unicode[STIX]{x1D70C}+\unicode[STIX]{x1D6FF})|\leqslant 1+\int _{0}^{|\unicode[STIX]{x1D6FC}|/4}|S(y)|y^{\unicode[STIX]{x1D6FF}+k/2}\frac{dy}{y},\end{eqnarray}$$
where we understand the right-hand side to be
$\infty$
if the integral diverges. Applying Lemma 3.2, we thus have
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}^{-1}\ll 1+\mathop{\sum }_{n\geqslant 1}(2+|\unicode[STIX]{x1D6FE}_{n}|)^{(\unicode[STIX]{x1D703}(|\unicode[STIX]{x1D6FE}_{n}|)-1)/3-\unicode[STIX]{x1D6FF}}\log ^{2}(2+|\unicode[STIX]{x1D6FE}_{n}|).\end{eqnarray}$$
In particular, the right-hand side must have infinitely many terms. Applying integration by parts, we get
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}^{-1} & \ll & \displaystyle 1+\int _{1}^{\infty }t^{(\unicode[STIX]{x1D703}(t)-1)/3-\unicode[STIX]{x1D6FF}}\log ^{2}t\,dN(t)\nonumber\\ \displaystyle & = & \displaystyle 1-\int _{1}^{\infty }N(t)d(t^{(\unicode[STIX]{x1D703}(t)-1)/3-\unicode[STIX]{x1D6FF}}\log ^{2}t)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 1+\int _{1}^{\infty }N(t)\bigg(\frac{1-\unicode[STIX]{x1D703}(t)}{3}+\unicode[STIX]{x1D6FF}\bigg)t^{(\unicode[STIX]{x1D703}(t)-1)/3-\unicode[STIX]{x1D6FF}-1}\log ^{2}t\,dt,\nonumber\end{eqnarray}$$
where for the last inequality we have used the fact that
$\unicode[STIX]{x1D703}(t)$
is non-decreasing and
$$\begin{eqnarray}\displaystyle d(t^{(\unicode[STIX]{x1D703}(t)-1)/3-\unicode[STIX]{x1D6FF}}\log ^{2}t) & = & \displaystyle t^{(\unicode[STIX]{x1D703}(t)-1)/3-\unicode[STIX]{x1D6FF}-1}(\log t)\bigg[2-\bigg(\frac{1-\unicode[STIX]{x1D703}(t)}{3}+\unicode[STIX]{x1D6FF}\bigg)\log t\bigg]\,dt\nonumber\\ \displaystyle & & \displaystyle +\,\frac{1}{3}\log ^{3}t\,d\unicode[STIX]{x1D703}(t).\nonumber\end{eqnarray}$$
Suppose that (3.10) is false, so that the function
$\unicode[STIX]{x1D700}(t)=N(t)t^{(1/3)(\unicode[STIX]{x1D703}(t)-1)}(1-\unicode[STIX]{x1D703}(t))\log ^{2}t$
satisfies
$\lim _{t\rightarrow \infty }\unicode[STIX]{x1D700}(t)=0$
. Then we have
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}^{-1}\ll 1+\int _{1}^{\infty }\bigg(\frac{1}{3}+\frac{\unicode[STIX]{x1D6FF}}{1-\unicode[STIX]{x1D703}(t)}\bigg)\unicode[STIX]{x1D700}(t)t^{-1-\unicode[STIX]{x1D6FF}}\,dt.\end{eqnarray}$$
By the standard zero-free region [Reference Iwaniec and Kowalski6, Theorem 5.10], we have
$$\begin{eqnarray}\frac{1}{1-\unicode[STIX]{x1D703}(t)}\ll \log \max (t,2),\end{eqnarray}$$
so that
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6FF}^{-1} & \ll & \displaystyle 1+\int _{1}^{\infty }(1+\unicode[STIX]{x1D6FF}\log t)\unicode[STIX]{x1D700}(t)t^{-1-\unicode[STIX]{x1D6FF}}\,dt\nonumber\\ \displaystyle & = & \displaystyle 1+\unicode[STIX]{x1D6FF}^{-1}\int _{0}^{\infty }\unicode[STIX]{x1D700}(\text{e}^{u/\unicode[STIX]{x1D6FF}})(1+u)\text{e}^{-u}\,du=o(\unicode[STIX]{x1D6FF}^{-1}).\nonumber\end{eqnarray}$$
This is a contradiction, so (3.10) holds.
Finally, suppose (3.11) is false, so that
$N(T)<\log \log \log T$
for all sufficiently large
$T$
. Then there exists
$n_{0}\geqslant \mathbb{Z}_{{>}0}$
such that
$|\unicode[STIX]{x1D6FE}_{n}|>\exp \exp \exp n$
for all
$n\geqslant n_{0}$
. Since the terms from
$n<n_{0}$
contribute a bounded amount to (3.12), we have
$$\begin{eqnarray}1+\mathop{\sum }_{n=n_{0}}^{\infty }|\unicode[STIX]{x1D6FE}_{n}|^{-\unicode[STIX]{x1D6FF}}\log ^{2}|\unicode[STIX]{x1D6FE}_{n}|\gg \unicode[STIX]{x1D6FF}^{-1}\end{eqnarray}$$
for all sufficiently small
$\unicode[STIX]{x1D6FF}>0$
.
Next we claim that there are infinitely many
$m\geqslant n_{0}$
such that
$$\begin{eqnarray}\log \log |\unicode[STIX]{x1D6FE}_{m+1}|\geqslant {\textstyle \frac{13}{5}}\log \log |\unicode[STIX]{x1D6FE}_{m}|.\end{eqnarray}$$
If not then there exists
$n_{1}\geqslant n_{0}$
such that (3.13) fails for all
$m\geqslant n_{1}$
, and by induction it follows that
$$\begin{eqnarray}\log \log |\unicode[STIX]{x1D6FE}_{n}|\leqslant \big({\textstyle \frac{13}{5}}\big)^{n-n_{1}}\log \log |\unicode[STIX]{x1D6FE}_{n_{1}}|=c\big({\textstyle \frac{13}{5}}\big)^{n}\quad \text{for}~n\geqslant n_{1},\end{eqnarray}$$
where
$c=(13/5)^{-n_{1}}\log \log |\unicode[STIX]{x1D6FE}_{n_{1}}|>0$
. Hence,
$$\begin{eqnarray}n<\log \log \log |\unicode[STIX]{x1D6FE}_{n}|\leqslant \log c+n\log {\textstyle \frac{13}{5}}.\end{eqnarray}$$
Since
$\log \frac{13}{5}<1$
, this is false for sufficiently large
$n$
, proving the claim.
Choose a large
$m\geqslant n_{0}$
satisfying (3.13), and set
$\unicode[STIX]{x1D6FF}_{m}=(\log |\unicode[STIX]{x1D6FE}_{m}|)^{-12/5}$
. Then using the trivial bound
$\text{e}^{e^{e}}\leqslant |\unicode[STIX]{x1D6FE}_{n}|\leqslant |\unicode[STIX]{x1D6FE}_{m}|$
for
$n_{0}\leqslant n\leqslant m$
, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{n=n_{0}}^{m}|\unicode[STIX]{x1D6FE}_{n}|^{-\unicode[STIX]{x1D6FF}_{m}}\log ^{2}|\unicode[STIX]{x1D6FE}_{n}| & {\leqslant} & \displaystyle m\log ^{2}|\unicode[STIX]{x1D6FE}_{m}|<(\log \log \log |\unicode[STIX]{x1D6FE}_{m}|)\log ^{2}|\unicode[STIX]{x1D6FE}_{m}|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (\log |\unicode[STIX]{x1D6FE}_{m}|)^{11/5}=\unicode[STIX]{x1D6FF}_{m}^{-11/12},\nonumber\end{eqnarray}$$
since
$\log \log x\leqslant x^{1/5}$
for all
$x>1$
.
To estimate the contribution from
$n>m$
we apply integration by parts. Set
$g(t)=t^{-\unicode[STIX]{x1D6FF}_{m}}\log ^{2}t$
. Then
$g^{\prime }(t)<0$
for
$t>\text{e}^{2/\unicode[STIX]{x1D6FF}_{m}}=\exp (2(\log |\unicode[STIX]{x1D6FE}_{m}|)^{12/5})$
; in particular, if
$m$
is sufficiently large then, by (3.13),
$g^{\prime }(t)<0$
for
$t\geqslant |\unicode[STIX]{x1D6FE}_{m+1}|$
. Hence, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{n=m+1}^{\infty }g(|\unicode[STIX]{x1D6FE}_{n}|) & = & \displaystyle \lim _{\unicode[STIX]{x1D700}\rightarrow 0^{+}}\int _{|\unicode[STIX]{x1D6FE}_{m+1}|-\unicode[STIX]{x1D700}}^{\infty }g(t)\,dN(t)=\int _{|\unicode[STIX]{x1D6FE}_{m+1}|}^{\infty }(-g^{\prime }(t))(N(t)-m)\,dt\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \int _{|\unicode[STIX]{x1D6FE}_{m+1}|}^{\infty }(-g^{\prime }(t))(\log \log \log t)\,dt\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \unicode[STIX]{x1D6FF}_{m}\int _{|\unicode[STIX]{x1D6FE}_{m+1}|}^{\infty }t^{-\unicode[STIX]{x1D6FF}_{m}-1}(\log t)^{11/5}\,dt\nonumber\\ \displaystyle & = & \displaystyle \unicode[STIX]{x1D6FF}_{m}\int _{\log |\unicode[STIX]{x1D6FE}_{m+1}|}^{\infty }\text{e}^{-\unicode[STIX]{x1D6FF}_{m}u}u^{11/5}\,du.\nonumber\end{eqnarray}$$
Applying integration by parts three times and using that
$\unicode[STIX]{x1D6FF}_{m}\log |\unicode[STIX]{x1D6FE}_{m+1}|\gg 1$
, we get
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{m}\int _{\log |\unicode[STIX]{x1D6FE}_{m+1}|}^{\infty }\text{e}^{-\unicode[STIX]{x1D6FF}_{m}u}u^{11/5}\,du\ll |\unicode[STIX]{x1D6FE}_{m+1}|^{-\unicode[STIX]{x1D6FF}_{m}}(\log |\unicode[STIX]{x1D6FE}_{m+1}|)^{11/5}.\end{eqnarray}$$
Note that
$\unicode[STIX]{x1D6FF}_{m}=(\log |\unicode[STIX]{x1D6FE}_{m}|)^{-12/5}\geqslant (\log |\unicode[STIX]{x1D6FE}_{m+1}|)^{-12/13}$
, so
$|\unicode[STIX]{x1D6FE}_{m+1}|^{-\unicode[STIX]{x1D6FF}_{m}}\leqslant \exp (-(\log |\unicode[STIX]{x1D6FE}_{m+1}|)^{1/13})$
. Hence, we conclude that
$$\begin{eqnarray}\mathop{\sum }_{n=m+1}^{\infty }g(|\unicode[STIX]{x1D6FE}_{n}|)\ll \exp (-(\log |\unicode[STIX]{x1D6FE}_{m+1}|)^{1/13})(\log |\unicode[STIX]{x1D6FE}_{m+1}|)^{11/5}\ll 1.\end{eqnarray}$$
Thus, altogether we have
$$\begin{eqnarray}\unicode[STIX]{x1D6FF}_{m}^{-1}\ll 1+\mathop{\sum }_{n=n_{0}}^{\infty }|\unicode[STIX]{x1D6FE}_{n}|^{-\unicode[STIX]{x1D6FF}_{m}}\log ^{2}|\unicode[STIX]{x1D6FE}_{n}|\ll 1+\unicode[STIX]{x1D6FF}_{m}^{-11/12}.\end{eqnarray}$$
This is false for sufficiently large
$m$
, so (3.11) must hold for some arbitrarily large
$T$
.◻
4 Proofs of Theorems 1.1 and 1.2
We begin with an overview of the argument. By Proposition 3.3,
$N_{f,a,q}^{s}(T)$
is sometimes large if there exists
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
for which
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
has a pole with large real part. The main obstacle to showing this is that
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
is defined as the difference of two functions (cf. (2.7)), whose poles could in principle cancel out. However, as we show, there are some dependencies between
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
for various choices of
$(a,q,\unicode[STIX]{x1D6FC})$
, from which it follows that there is a suitable pole for at least one choice of inputs. More specifically, in Lemma 4.1 we exhibit a relationship between
$H_{f,1,1,a/p}(s)$
and
$H_{f,a,q,-a/q}(s)$
, where
$p$
and
$q$
are primes satisfying
$pq\equiv -1\hspace{0.6em}({\rm mod}\hspace{0.2em}Na)$
. For any prime
$p\nmid N$
, we show that there is some choice of
$a\in \mathbb{Z}$
for which this leads to poles at the simple zeros of
$\unicode[STIX]{x1D6EC}_{f}(s)$
, and thanks to [Reference Booker1, Theorem 1.1], those exist in abundance. Ultimately this implies that at least one of
$N_{f}^{s}(T)$
,
$N_{f,a,p}^{s}(T)$
,
$N_{f,a,q}^{s}(T)$
is large, which yields Theorem 1.1. Choosing
$p=2$
and appealing to the second and third conclusions of Proposition 3.3 yields Theorem 1.2.
Proceeding, given a prime
$p$
and
$a\in \mathbb{Z}$
coprime to
$p$
, define
$$\begin{eqnarray}C_{f,a,p}(s)=\unicode[STIX]{x1D6E5}_{f,a,p}(s)-\unicode[STIX]{x1D709}(p)p^{1-2s}\unicode[STIX]{x1D6E5}_{f}(s).\end{eqnarray}$$
Lemma 4.1. Let
$a\in \mathbb{Z}$
, and let
$p$
and
$q$
be prime numbers such that
$pq\equiv -1\hspace{0.6em}({\rm mod}\hspace{0.2em}Na)$
. Then:
(i)
$C_{f,a,p}(s)-(H_{f,1,1,a/p}(s)-\unicode[STIX]{x1D709}(p)p^{1-2s}H_{f,a,q,-a/q}(s))$
is holomorphic for
$\Re (s)>0$
;(ii)
$\sum _{b=1}^{p-1}C_{f,b,p}(s)=-P_{f,p}(p^{1-s})\unicode[STIX]{x1D6E5}_{f}(s)$
.
Proof. We first consider
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
, where
$\unicode[STIX]{x1D6FC}=-a/q$
. We have
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f,a,q}(s,\unicode[STIX]{x1D6FC})-\unicode[STIX]{x1D6E5}_{f}(s)=-R_{f,q}(q^{-s})\unicode[STIX]{x1D6EC}_{f}(s),\end{eqnarray}$$
which is holomorphic for
$\Re (s)>0$
. Set
$a^{\prime }=-(1+pq)/Na$
, so that
$a^{\prime }/q-1/Nq^{2}\unicode[STIX]{x1D6FC}=-p/Na$
. Let
$r_{\bar{f},q}(j)$
be the numbers such that
$$\begin{eqnarray}R_{\bar{f},q}(x)=\mathop{\sum }_{j=1}^{\infty }r_{\bar{f},q}(j)x^{j}.\end{eqnarray}$$
By Fourier inversion, we have
$$\begin{eqnarray}\mathop{\sum }_{\substack{ j\geqslant 1 \\ j\equiv t\;(\text{mod}~\unicode[STIX]{x1D711}(Na))}}r_{\bar{f},q}(j)x^{j}=\frac{1}{\unicode[STIX]{x1D711}(Na)}\mathop{\sum }_{\ell =1}^{\unicode[STIX]{x1D711}(Na)}e\bigg(-\frac{\ell t}{\unicode[STIX]{x1D711}(Na)}\bigg)R_{\bar{f},q}\bigg(e\bigg(\frac{\ell }{\unicode[STIX]{x1D711}(Na)}\bigg)x\bigg).\end{eqnarray}$$
Thus,
$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },q}\bigg(s,-\frac{1}{Nq^{2}\unicode[STIX]{x1D6FC}}\bigg)-\unicode[STIX]{x1D6E5}_{\bar{f}}\bigg(s,-\frac{p}{Na}\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad =-\mathop{\sum }_{j=1}^{\infty }r_{\bar{f},q}(j)q^{-js}\unicode[STIX]{x1D6EC}_{\bar{f}}\bigg(s,\frac{q^{j-1}}{Na}\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad =-\mathop{\sum }_{t=1}^{\unicode[STIX]{x1D711}(Na)}\unicode[STIX]{x1D6EC}_{\bar{f}}\bigg(s,\frac{q^{t-1}}{Na}\bigg)\mathop{\sum }_{\substack{ j\geqslant 1 \\ j\equiv t\;(\text{mod}~\unicode[STIX]{x1D711}(Na))}}r_{\bar{f},q}(j)q^{-js}\nonumber\\ \displaystyle & & \displaystyle \quad =-\frac{1}{\unicode[STIX]{x1D711}(Na)}\mathop{\sum }_{t=1}^{\unicode[STIX]{x1D711}(Na)}\unicode[STIX]{x1D6EC}_{\bar{f}}\bigg(s,\frac{q^{t-1}}{Na}\bigg)\mathop{\sum }_{\ell =1}^{\unicode[STIX]{x1D711}(Na)}e\bigg(-\frac{\ell t}{\unicode[STIX]{x1D711}(Na)}\bigg)R_{\bar{f},q}\bigg(e\bigg(\frac{\ell }{\unicode[STIX]{x1D711}(Na)}\bigg)q^{-s}\bigg),\nonumber\end{eqnarray}$$
which is again holomorphic for
$\Re (s)>0$
. Hence, up to a holomorphic function,
$H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)$
is
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}(s)-\unicode[STIX]{x1D716}\unicode[STIX]{x1D709}(q)(-\text{i}\operatorname{sgn}a)^{k}(Na^{2})^{s-1/2}\unicode[STIX]{x1D6E5}_{\bar{f}}\bigg(s,-\frac{p}{Na}\bigg).\end{eqnarray}$$
Next note that
$$\begin{eqnarray}\displaystyle C_{f,a,p}(s)-H_{f,1,1,a/p}(s) & = & \displaystyle \unicode[STIX]{x1D716}(\text{i}\operatorname{sgn}a)^{k}\bigg(\frac{Na^{2}}{p^{2}}\bigg)^{s-1/2}\unicode[STIX]{x1D6E5}_{\bar{f}}\bigg(s,-\frac{p}{Na}\bigg)\nonumber\\ \displaystyle & & \displaystyle -\,\unicode[STIX]{x1D709}(p)p^{1-2s}\unicode[STIX]{x1D6E5}_{f}(s)-R_{f,q}(q^{-s})\unicode[STIX]{x1D6EC}_{f}(s).\nonumber\end{eqnarray}$$
Therefore, since
$\unicode[STIX]{x1D709}(p)\unicode[STIX]{x1D709}(q)=\unicode[STIX]{x1D709}(-1)=(-1)^{k}$
, we see that
$$\begin{eqnarray}C_{f,a,p}(s)-H_{f,1,1,a/p}(s)+\unicode[STIX]{x1D709}(p)p^{1-2s}H_{f,a,q,\unicode[STIX]{x1D6FC}}(s)\end{eqnarray}$$
is holomorphic for
$\Re (s)>0$
.
Finally, by (2.4) we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{b=1}^{p-1}C_{f,b,p}(s) & = & \displaystyle (p-1)\bigg[1-\frac{p}{p-1}P_{f,p}(p^{-s})-\unicode[STIX]{x1D709}(p)p^{1-2s}\bigg]\unicode[STIX]{x1D6E5}_{f}(s)\nonumber\\ \displaystyle & = & \displaystyle -P_{f,p}(p^{1-s})\unicode[STIX]{x1D6E5}_{f}(s).\nonumber\end{eqnarray}$$
In the following, we shall make frequent use of the observation that for any pair
$h_{1},h_{2}$
of meromorphic functions,
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(h_{1}+h_{2})\leqslant \max (\unicode[STIX]{x1D6E9}(h_{1}),\unicode[STIX]{x1D6E9}(h_{2}))\quad \text{with equality when}~\unicode[STIX]{x1D6E9}(h_{1})\neq \unicode[STIX]{x1D6E9}(h_{2}).\end{eqnarray}$$
Fix a prime
$p\nmid N$
. By [Reference Booker1, Theorem 1.1] and the functional equation,
$\unicode[STIX]{x1D6E5}_{f}(s)$
has a pole with real part
${\geqslant}\frac{1}{2}$
, and thus
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f})=\unicode[STIX]{x1D703}_{f,1,1}\geqslant {\textstyle \frac{1}{2}}.\end{eqnarray}$$
Since all zeros of
$P_{f,p}(p^{1-s})$
have real part 1, this is also true of
$P_{f,p}(p^{1-s})\unicode[STIX]{x1D6E5}_{f}(s)$
. Hence, by Lemma 4.1(ii), there exists
$a\in \{1,\ldots ,p-1\}$
such that
$C_{f,a,p}(s)$
has a pole with real part
${\geqslant}\frac{1}{2}$
and satisfies
$\unicode[STIX]{x1D6E9}(C_{f,a,p})\geqslant \unicode[STIX]{x1D703}_{f,1,1}$
. By (4.1) and (4.2), it follows that
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(C_{f,a,p})=\max (\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,p}),\unicode[STIX]{x1D703}_{f,1,1}).\end{eqnarray}$$
Let
$q$
be a prime satisfying
$pq\equiv -1\hspace{0.6em}({\rm mod}\hspace{0.2em}Na)$
, and set
$a^{\prime }=-(1+pq)/(Na)$
.
We aim to prove that
$$\begin{eqnarray}\max (N_{f}^{s}(T),N_{f,a,p}^{s}(T),N_{f,a,q}^{s}(T))=\unicode[STIX]{x1D6FA}(T^{1/6-\unicode[STIX]{x1D700}})\quad \text{for all}~\unicode[STIX]{x1D700}>0.\end{eqnarray}$$
To that end, we will show that at least one of the following inequalities holds for some
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
:
(i)
$\max (\unicode[STIX]{x1D6E9}(H_{f,1,1,\unicode[STIX]{x1D6FC}}),\unicode[STIX]{x1D6E9}(H_{\bar{f},1,1,\unicode[STIX]{x1D6FC}}))\geqslant \unicode[STIX]{x1D703}_{f,1,1}\geqslant \frac{1}{2}$
;(ii)
$\max (\unicode[STIX]{x1D6E9}(H_{f,a,p,\unicode[STIX]{x1D6FC}}),\unicode[STIX]{x1D6E9}(H_{\bar{f},a^{\prime },p,\unicode[STIX]{x1D6FC}}))\geqslant \unicode[STIX]{x1D703}_{f,a,p}\geqslant \frac{1}{2}$
;(iii)
$\max (\unicode[STIX]{x1D6E9}(H_{f,a,q,\unicode[STIX]{x1D6FC}}),\unicode[STIX]{x1D6E9}(H_{\bar{f},a^{\prime },q,\unicode[STIX]{x1D6FC}}))\geqslant \unicode[STIX]{x1D703}_{f,a,q}\geqslant \frac{1}{2}$
.
To see that this suffices, suppose for instance that (iii) holds. By Proposition 2.1, we have
$N_{f,a,q}^{s}(T)=N_{\bar{f},a^{\prime },q}^{s}(T)$
and
$\unicode[STIX]{x1D703}_{f,a,q}=\unicode[STIX]{x1D703}_{\bar{f},a^{\prime },q}$
. Thus, applying Proposition 3.3 to either
$(f,a,q)$
or
$(\bar{f},a^{\prime },q)$
, we conclude that
$$\begin{eqnarray}\displaystyle N_{f,a,q}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D700}}),\quad \text{where}~\unicode[STIX]{x1D6FD} & {\geqslant} & \displaystyle \frac{1}{3}(1-\unicode[STIX]{x1D703}_{f,a,q})+\unicode[STIX]{x1D703}_{f,a,q}-\frac{1}{2}\nonumber\\ \displaystyle & = & \displaystyle \frac{2\unicode[STIX]{x1D703}_{f,a,q}}{3}-\frac{1}{6}\geqslant \frac{1}{6}.\nonumber\end{eqnarray}$$
If, instead, (i) or (ii) holds, then by a similar argument we find that
$N_{f,1,1}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D700}})$
or
$N_{f,a,p}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D700}})$
for some
$\unicode[STIX]{x1D6FD}\geqslant \frac{1}{6}$
. Hence, (4.5) follows in any case.
Let us suppose that conditions (i) and (iii) are false for all
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
and show that this leads to (ii). Since (i) is false, in view of (4.3) we must have
$\unicode[STIX]{x1D703}_{f,1,1}>\unicode[STIX]{x1D6E9}(H_{f,1,1,a/p})$
. In turn, by (4.4) this implies that
$\unicode[STIX]{x1D6E9}(C_{f,a,p})>\unicode[STIX]{x1D6E9}(H_{f,1,1,a/p})$
. Hence, by Lemma 4.1(i) and (4.2), we have
$\unicode[STIX]{x1D6E9}(H_{f,a,q,-a/q})=\unicode[STIX]{x1D6E9}(C_{f,a,p})$
. By (4.4), this implies
$\unicode[STIX]{x1D6E9}(H_{f,a,q,-a/q})\geqslant \unicode[STIX]{x1D703}_{f,1,1}>0$
, and thus
$\unicode[STIX]{x1D703}_{f,a,q}\geqslant \frac{1}{2}$
, by Proposition 3.3.
Next, by (3.8) we have
$\unicode[STIX]{x1D703}_{f,a,p}=\max (\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,p}),\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },p}))$
. If
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },p})\leqslant \max (\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,p}),\unicode[STIX]{x1D703}_{f,1,1})=\unicode[STIX]{x1D6E9}(H_{f,a,q,-a/q})\end{eqnarray}$$
then it follows that
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(H_{f,a,q,-a/q})=\max (\unicode[STIX]{x1D703}_{f,a,p},\unicode[STIX]{x1D703}_{f,1,1}).\end{eqnarray}$$
Suppose now that (4.6) is false. Then
$\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },p})>\max (\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,p}),\unicode[STIX]{x1D703}_{f,1,1})$
, so that
$$\begin{eqnarray}\unicode[STIX]{x1D703}_{f,a,p}=\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },p})>\unicode[STIX]{x1D703}_{f,1,1}.\end{eqnarray}$$
Since (i) is false, this implies that
$\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },p})>\max (\unicode[STIX]{x1D6E9}(H_{\bar{f},1,1,a^{\prime }/p}),\unicode[STIX]{x1D703}_{f,1,1})$
. By (4.1) and Lemma 4.1(i) with
$(\bar{f},a^{\prime })$
in place of
$(f,a)$
, it follows from (4.2) that
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(H_{\bar{f},a^{\prime },q,-a^{\prime }/q})=\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },p})=\max (\unicode[STIX]{x1D703}_{f,a,p},\unicode[STIX]{x1D703}_{f,1,1}).\end{eqnarray}$$
Therefore, since at least one of (4.7) and (4.8) must hold, we have
$$\begin{eqnarray}\max (\unicode[STIX]{x1D6E9}(H_{f,a,q,-a/q}),\unicode[STIX]{x1D6E9}(H_{\bar{f},a^{\prime },q,-a^{\prime }/q}))\geqslant \max (\unicode[STIX]{x1D703}_{f,a,p},\unicode[STIX]{x1D703}_{f,1,1}).\end{eqnarray}$$
Since (iii) is false, this implies that
$\unicode[STIX]{x1D703}_{f,a,q}>\max (\unicode[STIX]{x1D703}_{f,a,p},\unicode[STIX]{x1D703}_{f,1,1})$
. Hence, by (3.8), either
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,q})>\max (\unicode[STIX]{x1D703}_{f,a,p},\unicode[STIX]{x1D703}_{f,1,1})\quad \text{or}\quad \unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },q})>\max (\unicode[STIX]{x1D703}_{f,a,p},\unicode[STIX]{x1D703}_{f,1,1}).\end{eqnarray}$$
Suppose that the first inequality in (4.9) holds. Then by (4.1) (with
$q$
in place of
$p$
) and (4.2), we have
$\unicode[STIX]{x1D6E9}(C_{f,a,q})=\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,q})>\unicode[STIX]{x1D703}_{f,1,1}$
. Since (i) is false, this implies
$\unicode[STIX]{x1D6E9}(C_{f,a,q})>\unicode[STIX]{x1D6E9}(H_{f,1,1,a/q})$
. On the other hand, by Lemma 4.1(i) (with the roles of
$p$
and
$q$
reversed) and (4.2), we have
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(H_{f,a,p,-a/p})=\unicode[STIX]{x1D6E9}(C_{f,a,q})=\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{f,a,q})>\unicode[STIX]{x1D703}_{f,a,p}.\end{eqnarray}$$
This also implies that
$\unicode[STIX]{x1D6E9}(H_{f,a,p,-a/p})\,>\,0$
, whence
$\unicode[STIX]{x1D703}_{f,a,p}\,\geqslant \,\frac{1}{2}$
, by Proposition 3.3.
If, instead, the second inequality holds in (4.9), then running through the same argument with
$(\bar{f},a^{\prime })$
in place of
$(f,a)$
, we find that
$$\begin{eqnarray}\unicode[STIX]{x1D6E9}(H_{\bar{f},a^{\prime },p,-a^{\prime }/p})=\unicode[STIX]{x1D6E9}(C_{\bar{f},a^{\prime },q})=\unicode[STIX]{x1D6E9}(\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },q})>\unicode[STIX]{x1D703}_{\bar{f},a^{\prime },p}\geqslant {\textstyle \frac{1}{2}}.\end{eqnarray}$$
Hence, in either case we see that (ii) holds, and this concludes the proof of (4.5).
Now, by (4.5) and (2.4), it follows that there is a character
$\unicode[STIX]{x1D712}$
of conductor 1,
$p$
or
$q$
such that
$N_{f\otimes \unicode[STIX]{x1D712}}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{1/6-\unicode[STIX]{x1D700}})$
for all
$\unicode[STIX]{x1D700}>0$
. This implies Theorem 1.1.
For the proof of Theorem 1.2, we may assume that
$N_{f}^{s}(T)\ll 1+T^{\unicode[STIX]{x1D700}}$
for all
$\unicode[STIX]{x1D700}>0$
, since the result is trivial otherwise. To avoid contradicting Proposition 3.3, it must therefore be the case that
$\max (\unicode[STIX]{x1D6E9}(H_{f,1,1,\unicode[STIX]{x1D6FC}}),\unicode[STIX]{x1D6E9}(H_{\bar{f},1,1,\unicode[STIX]{x1D6FC}}))\leqslant \frac{1}{2}$
for all
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
.
Since
$N$
is odd, we can take
$p=2$
and
$a=1$
in the above, and choose any suitable prime
$q$
. Then by Lemma 4.1(ii), we have
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f,a,p}(s)=(\unicode[STIX]{x1D709}(p)p^{1-2s}-P_{f,p}(p^{1-s}))\unicode[STIX]{x1D6E5}_{f}(s),\end{eqnarray}$$
and it follows that
$N_{f,a,p}^{s}(T)\leqslant N_{f}^{s}(T)$
and
$\max (\unicode[STIX]{x1D6E9}(H_{f,a,p,\unicode[STIX]{x1D6FC}}),\unicode[STIX]{x1D6E9}(H_{\bar{f},a^{\prime },p,\unicode[STIX]{x1D6FC}}))\leqslant \frac{1}{2}$
for all
$\unicode[STIX]{x1D6FC}\in \mathbb{Q}^{\times }$
. Thus, by (4.5),
$N_{f,a,q}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{1/6-\unicode[STIX]{x1D700}})$
for all
$\unicode[STIX]{x1D700}>0$
. Therefore, by Proposition 2.1, at least one of
$\unicode[STIX]{x1D6E5}_{f,a,q}(s),\unicode[STIX]{x1D6E5}_{\bar{f},a^{\prime },q}(s)$
has a pole in the region
$\{s\in \mathbb{C}:\Re (s)\geqslant \frac{1}{2}\}$
that is not a pole of
$\unicode[STIX]{x1D6E5}_{f}(s)$
. By (4.1) and Lemma 4.1(i), the same applies to one of
$H_{f,1,1,a/q}(s)$
,
$H_{f,a,p,-a/p}(s)$
,
$H_{\bar{f},1,1,a^{\prime }/q}(s)$
, or
$H_{\bar{f},a^{\prime },p,-a^{\prime }/p}(s)$
.
Since
$$\begin{eqnarray}N_{f,a,p}^{s}(T)=N_{\bar{f},a^{\prime },p}^{s}(T)\leqslant N_{f,1,1}^{s}(T)=N_{\bar{f},1,1}^{s}(T),\end{eqnarray}$$
whichever function has the pole, we can apply Proposition 3.3 to see that
$N_{f}^{s}(T)$
satisfies the second and third conclusions. In particular,
$N_{f}^{s}(T)\geqslant \log \log \log T$
for some arbitrarily large
$T$
, and if
$k=1$
or
$f$
is a CM form then Coleman’s theorem [Reference Coleman4] implies that
$$\begin{eqnarray}1-\unicode[STIX]{x1D703}_{f,a,p}(T)\geqslant 1-\unicode[STIX]{x1D703}_{f,1,1}(T)\gg (\log T)^{-2/3}(\log \log T)^{-1/3}\quad \text{for all}~T\geqslant 3,\end{eqnarray}$$
whence
$N_{f}^{s}(T)=\unicode[STIX]{x1D6FA}(\exp ((\log T)^{1/3-\unicode[STIX]{x1D700}}))$
for all
$\unicode[STIX]{x1D700}>0$
. Moreover, since
$N_{f}^{s}(T)\ll 1+T^{\unicode[STIX]{x1D700}}$
, we must have
$\unicode[STIX]{x1D703}_{f,1,1}=1$
, so
$\unicode[STIX]{x1D6EC}_{f}(s)$
has simple zeros with real part arbitrarily close to 1.
Finally, by Lemma 4.1(ii) we have
$\unicode[STIX]{x1D6E9}(C_{f,a,p})=1$
. Since
$\unicode[STIX]{x1D6E9}(H_{f,1,1,a/p})\leqslant \frac{1}{2}$
, Lemma 4.1(i) and (4.2) imply that
$\unicode[STIX]{x1D6E9}(H_{f,a,q,-a/q})=1$
. Applying Proposition 3.3, it follows that
$N_{f,a,q}^{s}(T)=\unicode[STIX]{x1D6FA}(T^{1/2-\unicode[STIX]{x1D700}})$
for all
$\unicode[STIX]{x1D700}>0$
. By Lemma 4.1(i),
$C_{f,a,q}(s)$
and
$C_{\bar{f},a^{\prime },q}(s)$
are holomorphic for
$\Re (s)>\frac{1}{2}$
. Hence, by (4.1) and Proposition 2.1, all poles of
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
that are not poles of
$\unicode[STIX]{x1D6E5}_{f}^{\ast }(s)$
lie on the line
$\{s\in \mathbb{C}:\Re (s)=\frac{1}{2}\}$
. Since
$N_{f}^{s}(T)\ll 1+T^{\unicode[STIX]{x1D700}}$
,
$\unicode[STIX]{x1D6E5}_{f,a,q}^{\ast }(s)$
must have
$\unicode[STIX]{x1D6FA}(T^{1/2-\unicode[STIX]{x1D700}})$
poles with real part
$\frac{1}{2}$
and imaginary part in
$[-T,T]$
. By (2.4), the same applies to
$\unicode[STIX]{x1D6E5}_{f}(s,\unicode[STIX]{x1D712})$
for some
$\unicode[STIX]{x1D712}\hspace{0.6em}({\rm mod}\hspace{0.2em}q)$
.
Acknowledgements
We thank the anonymous referee for helpful suggestions and corrections, and the Banff International Research Station for hosting us for a Research in Teams Meeting (15rit201). A significant portion of this project was completed during that week and we appreciated the excellent working conditions.
