††thanks:  Keywords : Zeros of Dirichlet series, Hardy’s theorem, Half-
integral weight cusp forms 2020 Mathematics Subject Classification : Primary:
11F37, 11F66, 11M41. Department of Mathematics, Faculty of Sciences of
University of Porto, Rua do Campo Alegre, 687; 4169-007 Porto (Portugal).
E-mail<EMAIL_ADDRESS>
# On the number of zeros of $L-$functions attached to cusp forms of half-
integral weight

Pedro Ribeiro

###### Abstract

Meher et al. [Proc. Amer. Math. Soc. 147 (2019)] have recently established
that $L-$functions attached to certain cusp forms of half-integral weight have
infinitely many zeros on the critical line. Kim [J. Numb. Th. 253 (2023)]
obtained analogous results for $L-$functions attached to cusp forms twisted by
an additive character $e\left(\frac{p}{q}n\right)$,
$\frac{p}{q}\in\mathbb{Q}$. We extend the results of these authors by giving a
lower bound for the number of such zeros.

We start by developing a variant of a method of de la Valée Poussin which
seems to have interest as it avoids the evaluation of exponential sums. We
finish the paper with an improvement of our first estimate by using
Lekkerkerker’s variant of the Hardy-Littlewood method.

###### Contents

  1. 1 Introduction and Main Results
  2. 2 Preliminary results
  3. 3 Lemmas for the proof of Theorem 1.1
  4. 4 Proof of Theorem 1.1
  5. 5 Lemmas for the proof of Theorem 1.2
  6. 6 Proof of Theorem 1.2
    1. 6.1 Estimating $I(t)$
    2. 6.2 Estimating $J(t)$
      1. 6.2.1 Proof of Claim 6.1
  7. 7 Concluding Remarks

##  1 Introduction and Main Results

For $k,N\in\mathbb{N}$, let us denote by
$S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$ the space of cusp forms of
weight $k+\frac{1}{2}$ on the congruence subgroup $\Gamma_{0}(4N)$. This means
that any element $f(z)\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$
satisfies111Throughout this paper we define $\sqrt{z}=z^{1/2}$ so that
$-\frac{\pi}{2}<\arg(z^{1/2})\leq\frac{\pi}{2}$.

$f\left(\frac{az+b}{cz+d}\right)=\left(\frac{c}{d}\right)^{2k+1}\epsilon_{d}^{-2k-1}\left(cz+d\right)^{k+\frac{1}{2}}f(z),$
(1.1)

whenever $\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in\Gamma_{0}(4N)$. Here, $\left(\frac{c}{d}\right)$ is
Shimura’s extension of the Jacobi symbol [[25], p.442] and $\epsilon_{d}$ is
defined by

$\epsilon_{d}=\begin{cases}1&d\equiv 1\,\,\mod 4\\\ i&d\equiv 3\,\,\mod
4\end{cases}.$ (1.2)

The theory of modular forms of half-integral weight was extensively developed
by Shimura and we refer to the seminal paper [25] for some classical facts
about these modular forms. Since $\left(\begin{array}[]{cc}1&1\\\
0&1\end{array}\right)\in\Gamma_{0}(4N)$, $f(z)$ has a Fourier expansion of the
form

$f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi inz},$ (1.3)

where $z\in\mathbb{H}=\left\\{w\in\mathbb{C}\,:\,\text{Im}(w)>0\right\\}$. We
can attach a Dirichlet series $L(s,f)$ to $f(z)$,

$L(s,f)=\sum_{n=1}^{\infty}\frac{a_{f}(n)}{n^{s}},$ (1.4)

and show that, for sufficiently large $\text{Re}(s)$, this series converges
absolutely. The usual argument invoked to study the analytic continuation of
$L-$functions of cusp forms with integral weight can also be applied to the
half-integral case. Thus, $L(s,f)$ can be analytically continued to the whole
complex plane $\mathbb{C}$ as an entire function of $s$. Moreover, it will
satisfy Hecke’s functional equation.

In order to state its functional equation, let $W_{4N}$ denote the Fricke
involution acting on $S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$,

$\left(f|W_{4N}\right)(z)=i^{k+\frac{1}{2}}\left(4N\right)^{-\frac{k}{2}-\frac{1}{4}}z^{-k-\frac{1}{2}}f\left(-\frac{1}{4Nz}\right).$
(1.5)

Then the function $L(s,f)$ satisfies the functional equation

$\left(\frac{2\pi}{\sqrt{4N}}\right)^{-s}\Gamma\left(s\right)\,L\left(s,f\right)=\left(\frac{2\pi}{\sqrt{4N}}\right)^{-\left(k+\frac{1}{2}-s\right)}\Gamma\left(k+\frac{1}{2}-s\right)\,L\left(k+\frac{1}{2}-s,f|W_{4N}\right).$
(1.6)

Adopting the notation given in [21], let us note that (1.6) can be written in
the symmetric form

$\Lambda\left(s,f\right)=\Lambda\left(k+\frac{1}{2}-s,f|W_{4N}\right),$ (1.7)

where $\Lambda(s,f)$ represents the completed $L-$function

$\Lambda\left(s,f\right):=\left(\frac{2\pi}{\sqrt{4N}}\right)^{-s}\Gamma\left(s\right)\,L\left(s,f\right).$
(1.8)

Despite the formal similarities with the $L-$functions attached to a Hecke
eigenform with integral weight, $L-$functions associated with cusp forms
belonging to $S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$ do not possess an
Euler product. This imposes some restrictions and adds complications in
studying the location of zeros of such $L-$functions. We refer to the papers
[21] and [22] for nice overviews of the non-vanishing results for $L(s,f)$
available in the literature.

Our focus in this paper will be the study of the zeros of $L(s,f)$, $f\in
S_{k+\frac{1}{2}}(\Gamma_{0}(4N))$, on the critical line. According to [[21],
p.132], Yoshida [30] was the first mathematician to study the zeros of such
$L-$functions. To illustrate one of Yoshida’s examples, let

$\theta(z)=\sum_{n\in\mathbb{Z}}e^{2\pi
in^{2}z},\,\,\,\,\,\eta(z)=e^{\frac{\pi
iz}{12}}\prod_{n=1}^{\infty}\left(1-e^{2\pi
inz}\right),\,\,\,\,z\in\mathbb{H}.$

Then one can construct [[25], p.477] a cusp form $g(z)\in
S_{\frac{9}{2}}\left(\Gamma_{0}(4)\right)$ by taking

$g(z):=\theta(z)^{-3}\eta(2z)^{12}.$ (1.9)

Yoshida considered the $L-$function attached to $g$, $L(s,g)$, and from the
calculation of some of its zeros [[30], p.675], he showed that the analogue of
the Riemann hypothesis for $L(s,g)$ is false.

Nevertheless, since there are Dirichlet series (such as the Epstein zeta
function attached to a binary quadratic form) which do not satisfy the Riemann
hypothesis but still possess a positive proportion of zeros on the critical
line, one may study, in spite of Yoshida’s counterexample, analogues of
Hardy’s Theorem for this class of $L-$functions.

As far as we know, the work of J. Meher, S. Pujahari and K. Srinivas [22]
contains the first result of Hardy-type for $L-$functions associated with
half-integral weight cusp forms. They established the following theorem.

##### Theorem A [22]

Suppose that $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi inz}\in
S_{k+\frac{1}{2}}\left(\Gamma_{0}(4)\right)$ is an eigenform for all Hecke
operators $T_{n^{2}}$ and for the operator $W_{4}$ (1.5). Assume also that all
the Fourier coefficients of $f(z)$, $a_{f}(n)$, are either real or purely
imaginary numbers. Then the $L-$function (1.4) attached to $f$ has infinitely
many zeros on the critical line $\text{Re}(s)=\frac{k}{2}+\frac{1}{4}$.

We know that $\text{dim}S_{\frac{9}{2}}\left(\Gamma_{0}(4)\right)=1$ [[25],
p.477], so the cusp form $g(z)$ given by (1.9) is an eigenform for all the
Hecke operators and satisfies $\left(g|W_{4}\right)(z)=g(z)$. Thus, despite
violating the Riemann hypothesis, $L(s,g)$ has infinitely many zeros at the
critical line $\text{Re}(s)=\frac{9}{4}$.

Recently, Meher, Pujahari and Shankhadhar [21] improved Theorem A by extending
it to forms of higher level.

##### Theorem B [21]

Let $N$ be a perfect square and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi
inz}\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$. Assume also that all
the Fourier coefficients of $f(z)$, $a_{f}(n)$, are either real or purely
imaginary numbers. Then the function

$L\left(s,f\right)\pm L\left(s,f|W_{4N}\right)$

has infinitely many zeros of odd order on the critical line
$\text{Re}(s)=\frac{k}{2}+\frac{1}{4}$.

The proofs of Theorems A and B are different in nature. The proof of Theorem A
follows a classical argument first given by Landau [19] and its main idea is
to contrast the behavior of the integrals
$\left|\intop_{T}^{2T}\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\,dt\right|$
and
$\intop_{T}^{2T}\left|\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\right|\,dt$
as $T\rightarrow\infty$. Due to the oscillations coming from the
$\Gamma-$factor, the first integral
$\intop_{T}^{2T}\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\,dt$ has
substantial cancellation and an upper bound for it can be obtained by
estimating an exponential sum [[22], p.930, Lemma 3.2]. On the other hand, the
proof of Theorem B uses Wilton’s variant of Hardy’s theorem [28] and it is
much closer in spirit to Hardy’s original proof [12].

Kim studied the zeros of this class $L-$functions in Wilton’s setting. Using a
very elegant argument involving distributions, he directed his study towards
the zeros of the additively twisted $L-$function,

$L_{p/q}(s,f)=\sum_{n=1}^{\infty}\frac{a_{f}(n)\,e^{\frac{2\pi
ip}{q}n}}{n^{s}},\,\,\,\,\,\frac{p}{q}\in\mathbb{Q},\,\,\,\,f\in
S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right).$ (1.10)

Just like the Dirichlet series (1.4), $L_{p/q}(s,f)$ converges absolutely when
$\text{Re}(s)$ is sufficiently large. Moreover, in analogy to the integral
case [[29], Theorem 2.1 (a)], $L_{p/q}(s,f)$ satisfies Hecke’s functional
equation when $\frac{p}{q}$, $q>0$, is a rational number which is
$\Gamma_{0}(4N)-$equivalent to $i\infty$. This functional equation can be
explicitly written in the form [[18], p.174, Lemma 4.19]

$\left(\frac{2\pi}{q}\right)^{-s}\Gamma(s)\,L_{p/q}(s,f)=i^{k+\frac{1}{2}}\left(\frac{-q}{p}\right)^{-2k-1}\epsilon_{p}^{2k+1}\,\left(\frac{2\pi}{q}\right)^{-\left(k+\frac{1}{2}-s\right)}\Gamma\left(k+\frac{1}{2}-s\right)\,L_{-\overline{p}/q}\left(k+\frac{1}{2}-s,f\right),$
(1.11)

where $\overline{p}$ is such that $p\,\overline{p}\equiv 1\mod\,q$. Also,
$\left(\frac{-q}{p}\right)$ and $\epsilon_{p}$ are given in (1.1). Using
(1.11) and appealing to the notion of a distribution vanishing to infinite
order, Kim established the following theorem [[18], p.160, Theorem 1.17].

##### Theorem C [18]

Let $N$ be a positive integer and $\frac{p}{q}$, $q>0$, a rational number
which is $\Gamma_{0}(4N)-$equivalent to $i\infty$ and such that $p^{2}\equiv
1\mod q$. Moreover, let $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi inz}\in
S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$, with $a_{f}(n)$ being either
real or purely imaginary numbers.

Then the additively twisted $L-$function $L_{p/q}(s,f)$ (1.10) has infinitely
many zeros at the critical line $\text{Re}(s)=\frac{k}{2}+\frac{1}{4}$.

The main goal of our paper is to improve on Theorem B and Theorem C above by
giving explicit quantitative estimates for the number of zeros of
$L\left(\frac{k}{2}+\frac{1}{4}+it,f\right)$ and
$L_{p/q}\left(\frac{k}{2}+\frac{1}{4}+it,f\right)$. Like Theorem B above, our
first result uses the main setting of Wilton’s variant of Hardy’s theorem
[28].

######  Theorem 1.1.

Let $N$ be a perfect square and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi
inz}\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$. Assume also that the
Fourier coefficients of $f(z)$, $a_{f}(n)$, are either real or purely
imaginary numbers.

Let $N_{0}^{\pm}(T)$ represent the number of zeros of odd order of the
function $L(s,f)\pm L(s,f|W_{4N})$ written in the form
$s=\frac{k}{2}+\frac{1}{4}+it,$ $0\leq t\leq T$. Then we have that

$N_{0}^{\pm}(T)=\Omega\left(T^{\frac{1}{2}}\right),$ (1.12)

or, by other words, there exists some $d>0$ such that

$\limsup_{T\rightarrow\infty}\frac{N_{0}^{\pm}(T)}{T^{1/2}}>d.$ (1.13)

As far as we know, our Theorem 1.1 establishes for the first time a
quantitative lower bound on the number of critical zeros of $L-$functions
attached to half-integral weight cusp forms. As an immediate corollary, one
obtains.

######  Corollary 1.1.

Let $N$ be a perfect square and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi
inz}\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$ be such that
$f|W_{4N}=f$ or $f|W_{4N}=-f$. Assume also that the Fourier coefficients of
$f(z)$, $a_{f}(n)$, are either real or purely imaginary numbers.

Let $N_{0}(T)$ denote the number of zeros of odd order of $L(s,f)$ of the form
$s=\frac{k}{2}+\frac{1}{4}+it$, $0\leq t\leq T$. Then we have that

$N_{0}(T)=\Omega\left(T^{\frac{1}{2}}\right)\,\,\,\,\text{or, equivalently,
}\,\,\,\limsup_{T\rightarrow\infty}\frac{N_{0}(T)}{T^{1/2}}>d,$ (1.14)

for some $d>0$.

We remark that it is possible to get an explicit numerical value for $d$ in
(1.13) and (1.14). In fact, $d$ only depends on the Fourier expansion of a
function arising from the slash operator (see (4.16) and the proof of Lemma
3.2 for details).

Our proof of Theorem 1.1 is inspired by a beautiful argument due to de la
Vallée Poussin [5]. The belgian mathematician Charles-Jean de la Valeé Poussin
is mostly famous for his proof of the prime number theorem. However, his
contributions to the study of the zeros of $\zeta(s)$ on the line
$\text{Re}(s)=\frac{1}{2}$ are not so well-known. Shortly after the appearance
of the papers by Hardy and Landau [12, 19] on this topic, de la Vallée Poussin
published two short notes. In one of them, he proved that the number of zeros
of $\zeta(s)$ of the form $s=\frac{1}{2}+it,$ $0<t<T$, satisfies the estimate
$N_{0}(T)=\Omega\left(T^{1/2}\right)$ as $T\rightarrow\infty$. However, as he
remarks [[5], p.421] _“Cette méthode ne s’applique pas à toutes les fonctions
qui interviennent dans l’étude de la progression arithmétique”._ 222or by
other words, his method does not work for certain Dirichlet $L-$functions.__
In order to circumvent this difficulty, de la Vallée Poussin develops in a
later note an argument providing estimates for the difference
$N_{0}(T+H)-N_{0}(T)$, $\,\,T^{\frac{3}{4}+\epsilon}\leq H(T)\leq T$. His
second note was the first improvement of the results of Landau about the zeros
of Dirichlet $L-$functions.

Despite the beauty of the arguments presented in these short papers, the
results and methods of de la Vallée Poussin quickly faded into oblivion [3]
due to the much sharper estimates (proved around the same time) by Hardy and
Littlewood [13, 14]333The first improvements on the number of critical zeros
of $\zeta\left(s\right)$ by Hardy and Littlewood were in fact independent of
those of de la Vallée Poussin. An additional note to the paper [[13], p.196]
was written by Hardy and Littlewood just to acknowledge de la Vallée Poussin’s
contributions. and, decades later, by Selberg and Levinson. Even Titchmarsh’s
text on the zeta function, known for its very precise historical background,
omits de la Vallée Poussin’s contributions in this regard, notwithstanding the
fact that it exposes (just for the sake of historical interest) five different
proofs of Hardy’s theorem [[27], Chapter X]. 444It should be mentioned,
however, that de la Vallée Poussin’s papers [5, 6] appear as references in
Titchmarsh’s book but they are not mentioned in the main text.

With the proof of Theorem 1.1, we hope to bring de la Vallée’s Poussin elegant
arguments to a modern mathematical audience, as we believe that the method
followed in the proof of the result (1.12) has a theoretical interest in
itself. In a historical sense, it is closely connected to Hardy’s original
note [12], as well as to Wilton’s variant [28], which, as mentioned above, was
used in the proof of Theorem B. Thus, in a sense, our Theorem 1.1 can be seen
as a complement to Theorem B due to Meher, Pujahari and Shankhadhar. It has
also the advantage of giving a quantitative estimate for the number of zeros
of a Dirichlet series without depending on any evaluation of exponential sums,
which is rather unusual.

Similarly to de la Vallée Poussin’s case, the result (1.12) in itself can be
quickly surpassed by no more than the same ideas that Hardy and Littlewood
used to prove that the function $\zeta\left(\frac{1}{2}+it\right)$, $0\leq
t\leq T$, has $\gg T$ zeros [14]. Our next Theorem establishes an analogue of
Hardy-Littlewood’s famous result.

######  Theorem 1.2.

Let $N$ be a perfect square and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi
inz}\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$. Assume also that the
Fourier coefficients of $f(z)$, $a_{f}(n)$, are either real or purely
imaginary numbers.

Let $N_{0}^{\pm}(T)$ represent the number of zeros of odd order of the
function $L(s,f)\pm L(s,f|W_{4N})$ written in the form
$s=\frac{k}{2}+\frac{1}{4}+it,$ $0\leq t\leq T$.

Then we have that

$N_{0}^{\pm}(T)\gg T\,\,\,\,\,\text{or, equivalently,
}\,\,\,\liminf_{T\rightarrow\infty}\frac{N_{0}^{\pm}(T)}{T}>d,$ (1.15)

for some $d>0$.

By following the proof of Theorem 1.2, one sees that there are no extra
complications in adapting the method to the case of additively twisted
$L-$functions (1.10) (see Remark 6.1 below). Thus, the following extension of
Kim’s theorem holds.

######  Theorem 1.3.

Let $N$ be a positive integer and $\frac{p}{q}$, $q>0$, a rational number
which is $\Gamma_{0}(4N)-$equivalent to $i\infty$ and such that $p^{2}\equiv
1\mod q$. Moreover, let $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi inz}\in
S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$, with $a_{f}(n)$ being either
real or purely imaginary numbers.

If $N_{0,p/q}(T)$ denotes the number of zeros of $L_{p/q}(s,f)$ written in the
form $s=\frac{k}{2}+\frac{1}{4}+it,$ $0\leq t\leq T$, then we have that

$N_{0,p/q}(T)\gg T\,\,\,\,\,\text{or, equivalently,
}\,\,\,\liminf_{T\rightarrow\infty}\frac{N_{0,p/q}(T)}{T}>d,$ (1.16)

for some $d>0$.

Our proofs of Theorems 1.2 and 1.3 will rely on a variant of the Hardy-
Littlewood method developed by Lekkerkerker [20]. In his doctoral
dissertation, Lekkerkerker proved general theorems about the distribution of
zeros of Dirichlet series satisfying Hecke’s functional equation. The last
chapter of his thesis is devoted to prove a result of Hardy-Littlewood type
for entire Dirichlet series.

Lekkerkerker’s results were mostly generalized by Berndt [1, 2], where the
assumption of Hecke’s functional equation was replaced by an equation with
more $\Gamma-$factors. Epstein, Hafner and Sarnak [8, 9], adapted
Lekkerkerker’s ideas to prove that the number of critical zeros of
$L-$functions attached to Maass cusp forms also satisfy (1.15). 555Hafner [11]
later generalized this result by proving the analogue of Selberg’s celebrated
theorem, showing that a positive proportion of the zeros of this $L-$functions
lie on the critical line $\mathbb{\text{Re}}(s)=\frac{1}{2}$.

It should be remarked that condition 3 on the statement of Theorem 13 of
Lekkerkerker’s dissertation is incomplete, because it is motivated by the
incorrect lemma 7 on Chapter III of [20]. In order to avoid some of the
incorrect steps motivated by this imprecision, we adapt to our case some of
Lekkerkerker’s arguments in a correct form and we present them in a way that
avoids the use of the incorrect lemma 7. An alternative approach of our Lemma
5.3, based on Fourier analysis, is also sketched at Remark 5.4 below and it
has the same spirit as the approach followed in [8].

Our paper is organized as follows. In the next section, we present some basic
facts that will be important throughout our exposition. Next, we present the
proof of Theorem 1.1 by modifying the argument of de la Vallée Poussin.
Section 5 is devoted to a careful adaptation of Lekkerkerker’s arguments to
our case. We finish the paper with an exposition of the classical Hardy-
Littlewood argument to prove Theorem 1.2 and with a remark on how the same
ideas can be used to prove Theorem 1.3.

##  2 Preliminary results

Like cusp forms with integral weight, cusp forms of half-integral weight also
satisfy uniform bounds on the upper half-plane [[15], Chapt.5, p.70]. We start
by quoting a lemma in Shimura’s textbook [[24], p.31, Lemma 6.2.].

If $f\in S_{k+\frac{1}{2}}(\Gamma_{0}(4N))$, with $N$ being a perfect square,
then $f\left(\frac{z}{\sqrt{4N}}\right)=\sum a_{f}(n)\,e^{2\pi inz/\ell}$ with
$\ell\in\mathbb{N}$. Then, by [[24], p.31, Lemma 6.2.], there exists a
positive constant $\mathcal{A}$ such that, for any $z\in\mathbb{H}$, the
uniform bound takes place666Shimura’s argument simply transforms the statement
on half-integral weight cusp forms into the same statement about cusp forms
with integral weight, which is well-known.

$\left|f\left(\frac{z}{\sqrt{4N}}\right)\right|\leq\frac{\mathcal{A}}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}.$
(2.1)

Connected with (2.1) is the mean value estimate for $a_{f}(m)$,

$\sum_{m=1}^{M}|a_{f}(m)|^{2}\ll M^{k+\frac{1}{2}},$ (2.2)

where, just like $\mathcal{A}$ in (2.1), the implied constant depends on $f$.
The usual Hecke argument can be developed by taking (2.1) as a starting point.
This classical procedure yields the bound

$a_{f}(n)\ll n^{\frac{k}{2}+\frac{1}{4}},$ (2.3)

where, again, the implied constant depends only on $f$. Of course, much better
bounds than (2.3) are currently available [4]. The uniform bound (2.1) will
play a crucial role in some estimates necessary for the proof of Theorem 1.2.

Like in the integral case, one may define slash operators acting on the space
$S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$. These are defined as follows
[[25], p.447]. Let $\alpha=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\in\text{GL}^{+}(2,\mathbb{R})$ and attach to it an
analytic function on $\mathbb{H}$, $\phi(z)$, such that

$\phi(z)^{2}=\tau\,\det(\alpha)^{-\frac{1}{2}}(cz+d),$ (2.4)

where $\tau$ is a complex number such that $|\tau|=1$. If we let $\mathcal{G}$
denote the set of all pairs $(\alpha,\phi)$, then
$\left(\mathcal{G},\star\right)$ forms a group, where the operation $\star$ is
defined by

$\left(\alpha,\phi\right)\star\left(\beta,\psi\right)=\left(\alpha\beta,\phi(\beta
z)\,\psi(z)\right).$

Under this setting, if $\xi=(\alpha,\phi(z))\in\mathcal{G}$ and $f\in
S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$, the slash operator is defined
as

$\left(f|\xi\right)(z)=f\left(\alpha z\right)\,\phi(z)^{-2k-1}.$ (2.5)

As the functional equations (1.6) and (1.11) already clarify, in several
occasions throughout this paper we shall need to estimate the asymptotic order
of certain integrals involving the Dirichlet series $L(s,f)$. To justify most
of the steps, we will often invoke the following version of Stirling’s formula

$\Gamma(\sigma+it)=(2\pi)^{\frac{1}{2}}\,t^{\sigma+it-\frac{1}{2}}\,e^{-\frac{\pi
t}{2}-it+\frac{i\pi}{2}(\sigma-\frac{1}{2})}\left(1+\frac{1}{12(\sigma+it)}+O\left(\frac{1}{t^{2}}\right)\right),$
(2.6)

as $t\rightarrow\infty$, uniformly for
$-\infty<\sigma_{1}\leq\sigma\leq\sigma_{2}<\infty$. A similar formula can be
written for $t<0$ as $t$ tends to $-\infty$ by using the fact that
$\Gamma(\overline{s})=\overline{\Gamma(s)}$. Of course, a direct consequence
of this exact version is

$|\Gamma(\sigma+it)|=(2\pi)^{\frac{1}{2}}\,|t|^{\sigma-\frac{1}{2}}\,e^{-\frac{\pi}{2}|t|}\left(1+O\left(\frac{1}{|t|}\right)\right),\,\,\,\,\,|t|\rightarrow\infty.$
(2.7)

Before proceeding further, let us briefly mention the usual reasoning to get
convex estimates for $L(\sigma+it,f)$. The estimate that we will be using in
this paper can be obtained through the familiar argument invoking the
classical Phragmén-Lindelöf theorem given in [[26], p.180, 5.65]. By Hecke’s
bound (2.3), we know that when $\sigma>\frac{k}{2}+\frac{5}{4}$,
$L\left(\sigma+it,f|W_{4N}\right)=O(1)$. Hence, from the functional equation
(1.6) and Stirling’s formula (2.6) we have, whenever
$\sigma<\frac{k}{2}-\frac{3}{4}$,

$|L\left(\sigma+it,f\right)|\ll\left|\frac{\Gamma\left(k+\frac{1}{2}-s\right)}{\Gamma(s)}L\left(k+\frac{1}{2}-s,f|W_{4N}\right)\right|\ll|t|^{k+\frac{1}{2}-2\sigma}.$

Thus, by the Phragmén-Lindelöf principle, we have the convex estimate for
$L(s,f)$,

$L\left(\sigma+it,f\right)\ll_{\epsilon}|t|^{\frac{k}{2}+\frac{5}{4}-\sigma+\epsilon},\,\,\,\,\,\,\,\,\,\,\,\frac{k}{2}-\frac{3}{4}-\epsilon<\sigma<\frac{k}{2}+\frac{5}{4}+\epsilon.$
(2.8)

##  3 Lemmas for the proof of Theorem 1.1

Since our first theorem concerns zeros of a combination of the form $L(s,f)\pm
L(s,f|W_{4N})$, we will define two important functions related to this
expression as

$R_{f}\left(t\right):=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)+\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f|W_{4N}\right)}{2},$
(3.1)

and

$I_{f}(t):=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)-\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f|W_{4N}\right)}{2i}.$
(3.2)

It follows from the functional equation for $\Lambda(f,s)$ (1.7) that
$R_{f}(t)$ can be written as

$R_{f}(t)=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)+\Lambda\left(\frac{k}{2}+\frac{1}{4}-it,f\right)}{2},$
(3.3)

which means that $R_{f}(-t)=R_{f}(t)$. In particular, if the coefficients of
$f(z)$, $a_{f}(n)$, are real numbers, then

$R_{f}(t)=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)+\overline{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)}}{2}=\text{Re}\left(\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\right).$
(3.4)

Hence, when we assume that $a_{f}(n)$ are real numbers, $R_{f}(t)$ represents
a real-valued and even function of $t$. Analogously, the use of the functional
equation for $L(s,f)$ allows to write the representation for $I_{f}(t)$

$I_{f}(t):=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)-\Lambda\left(\frac{k}{2}+\frac{1}{4}-it,f\right)}{2i},$
(3.5)

which means that $I_{f}(-t)=I_{f}(t)$. In particular, if the coefficients of
$f(z)$, $a_{f}(n)$, are real numbers, then

$I_{f}(t)=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)-\overline{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)}}{2i}=\text{Im}\left(\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\right).$
(3.6)

At last, we note that if the Fourier coefficients of $f(z)$, $a_{f}(n)$, are,
instead, purely imaginary numbers, then the roles of $R_{f}(t)$ and $I_{f}(t)$
are somewhat reversed, this is

$\displaystyle R_{f}(t)$
$\displaystyle=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)+\Lambda\left(\frac{k}{2}+\frac{1}{4}-it,f\right)}{2}=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)-\overline{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)}}{2}$
$\displaystyle=i\,\text{Im}\left(\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\right),$

while

$\displaystyle I_{f}(t)$
$\displaystyle=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)-\Lambda\left(\frac{k}{2}+\frac{1}{4}-it,f\right)}{2i}=\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)+\overline{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)}}{2i}$
$\displaystyle=-i\,\text{Re}\left(\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\right).$

Therefore, the difficulty in dealing with the case where $a_{f}(n)\in
i\mathbb{R}$ is exactly the same as dealing with the assumption
$a_{f}(n)\in\mathbb{R}$ and so, throughout this section and the next, we will
only give the details for the case $a_{f}(n)\in\mathbb{R}$.

######  Lemma 3.1.

Let $N$ be a perfect square and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi
inz}\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$. If the Fourier
coefficients of $f(z)$, $a_{f}(n)$, are real numbers, then the following
integral representations take place

$\intop_{0}^{\infty}R_{f}(t)\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt=\frac{\pi}{2}i^{\frac{k}{2}+\frac{1}{4}}e^{-i\left(\frac{k}{2}+\frac{1}{4}\right)u}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)+\frac{\pi}{2}i^{-\frac{k}{2}-\frac{1}{4}}e^{i\left(\frac{k}{2}+\frac{1}{4}\right)u}\,f\left(\frac{e^{iu}}{\sqrt{4N}}\right),$
(3.7)
$i\,\intop_{0}^{\infty}I_{f}(t)\,\sinh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt=\frac{\pi}{2}i^{\frac{k}{2}+\frac{1}{4}}\,e^{-i\left(\frac{k}{2}+\frac{1}{4}\right)u}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)-\frac{\pi}{2}i^{-\frac{k}{2}-\frac{1}{4}}e^{i\left(\frac{k}{2}+\frac{1}{4}\right)u}\,f\left(\frac{e^{iu}}{\sqrt{4N}}\right).$
(3.8)

###### Proof.

For any $-\frac{\pi}{2}<\omega<\frac{\pi}{2}$, we evaluate the integral

$\mathcal{J}(\omega):=\intop_{-\infty}^{\infty}R_{f}(t)\,e^{\omega t}dt,$
(3.9)

by using the Cahen-Mellin formula. Since $|\omega|<\frac{\pi}{2}$, it follows
from Stirling’s formula (2.7) and the Phragmén-Lindelöf principle (2.8) that
$|R_{f}(t)|e^{\omega t}\in L_{1}(\mathbb{R})$. By (3.3), we may write
$\mathcal{J}(\omega)$ as an integral over the vertical line
$\text{Re}(z)=\frac{k}{2}+\frac{1}{4}$, i.e.,

$\displaystyle\mathcal{J}(\omega)$
$\displaystyle=\intop_{-\infty}^{\infty}\frac{\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)+\Lambda\left(\frac{k}{2}+\frac{1}{4}-it,f\right)}{2}\,e^{\omega
t}dt$
$\displaystyle=\frac{1}{2}\intop_{-\infty}^{\infty}\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)e^{\omega
t}\,dt+\frac{1}{2}\intop_{-\infty}^{\infty}\Lambda\left(\frac{k}{2}+\frac{1}{4}+it,f\right)\,e^{-\omega
t}dt$
$\displaystyle=\frac{e^{i\omega\left(\frac{k}{2}+\frac{1}{4}\right)}}{2i}\,\intop_{\frac{k}{2}+\frac{1}{4}-i\infty}^{\frac{k}{2}+\frac{1}{4}+i\infty}\Lambda\left(z,f\right)\,e^{-i\omega
z}\,dz+\frac{e^{-i\omega\left(\frac{k}{2}+\frac{1}{4}\right)}}{2i}\,\intop_{\frac{k}{2}+\frac{1}{4}-i\infty}^{\frac{k}{2}+\frac{1}{4}+i\infty}\Lambda\left(z,f\right)\,e^{i\omega
z}\,dz.$ (3.10)

We will now evaluate the first integral, as the second can be analogously
computed by replacing $\omega$ by $-\omega$. We shift the line of integration
to $\text{Re}(z)=\frac{k+3}{2}\pm iT$: to do this we just need to integrate
integrating along a positively oriented rectangular contour $\mathcal{R}(T)$
with vertices $\frac{k}{2}+\frac{1}{4}\pm iT$ and $\frac{k+3}{2}\pm iT$,
$T>0$. Since $\Gamma\left(z\right)$ and $L(z,f)$ are analytic inside
$\mathcal{R}(T)$, an application of Cauchy’s theorem gives

$\left\\{\intop_{\frac{k}{2}+\frac{1}{4}-iT}^{\frac{k}{2}+\frac{1}{4}+iT}+\intop_{\frac{k}{2}+\frac{1}{4}+iT}^{\frac{k+3}{2}+iT}+\intop_{\frac{k+3}{2}-iT}^{\frac{k}{2}+\frac{1}{4}-iT}+\intop_{\frac{k+3}{2}+iT}^{\frac{k+3}{2}-iT}\right\\}\,\Lambda\left(z,f\right)\,e^{-i\omega
z}\,dz=0.$ (3.11)

Since $|\omega|<\frac{\pi}{2}$, using Stirling’s formula and the convex
estimates (2.8), we see that the integrals along the horizontal segments
$\left[\frac{k}{2}+\frac{1}{4}\pm iT,\frac{k+3}{2}\pm iT\right]$ can be
bounded as follows

$\left|\intop_{\frac{k}{2}+\frac{1}{4}+iT}^{\frac{k+3}{2}+iT}\Lambda\left(z,f\right)\,e^{-i\omega
z}\,dz\right|\ll_{\epsilon}T^{\frac{k}{2}+\frac{3}{4}+\epsilon}e^{-\left(\frac{\pi}{2}-\omega\right)T}$

and so they tend to zero as $T\rightarrow\infty$. Taking $T\rightarrow\infty$
in (3.11) and using the fact that the Dirichlet series (1.4) converges
absolutely when $\text{Re}(s)=\frac{k+3}{2}$, we deduce that

$\displaystyle\intop_{\frac{k}{2}+\frac{1}{4}-i\infty}^{\frac{k}{2}+\frac{1}{4}+i\infty}\Lambda\left(z,f\right)\,e^{-i\omega
z}\,dz$
$\displaystyle=\intop_{\frac{k+3}{2}-i\infty}^{\frac{k+3}{2}+i\infty}\left(\frac{2\pi}{\sqrt{4N}}\right)^{-z}\Gamma(z)\,L(z,f)\,e^{-i\omega
z}dz=2\pi i\sum_{n=1}^{\infty}\frac{a_{f}(n)}{2\pi
i}\intop_{\frac{k+3}{2}-i\infty}^{\frac{k+3}{2}+i\infty}\Gamma(z)\,\left(\frac{2\pi
ne^{i\omega}}{\sqrt{4N}}\right)^{-z}dz$ $\displaystyle=2\pi
i\,\sum_{n=1}^{\infty}a_{f}(n)\,\exp\left(-\frac{2\pi
ne^{i\omega}}{\sqrt{4N}}\right)=2\pi
i\,f\left(\frac{ie^{i\omega}}{\sqrt{4N}}\right).$

Note that, on the third equality, we have invoked the Cahen-Mellin integral,

$e^{-z}=\frac{1}{2\pi
i}\,\intop_{c-i\infty}^{c+i\infty}\Gamma(s)\,z^{-s}ds,\,\,\,\,\,c>0,\,\,\,\text{Re}(z)>0,$
(3.12)

which can be applied because $\text{Re}(e^{i\omega})=\cos(\omega)>0$ when
$|\omega|<\frac{\pi}{2}$. Thus, returning to (3.10) we find that

$\mathcal{J}(\omega)=\pi
e^{i\omega\left(\frac{k}{2}+\frac{1}{4}\right)}\,f\left(\frac{ie^{i\omega}}{\sqrt{4N}}\right)+\pi
e^{-i\omega\left(\frac{k}{2}+\frac{1}{4}\right)}\,f\left(\frac{ie^{-i\omega}}{\sqrt{4N}}\right).$
(3.13)

This implies immediately (3.7), after we substitute $\omega$ by
$\frac{\pi}{2}-u$ and use the fact that $R_{f}(t)=R_{f}(-t)$. The proof of the
second formula (3.8) uses the same computations but uses instead the fact that
$I_{f}(t)=-I_{f}(-t)$. ∎

######  Remark 3.1.

We can adapt the computations given in the evaluation of (3.9) to give an even
more general expression. In fact, for any $z\in\mathbb{H}$, the following
integral representation is valid

$\intop_{-\infty}^{\infty}R_{f}(t)\,(-iz)^{-it}dt=\pi
i^{-\frac{k}{2}-\frac{1}{4}}z^{\frac{k}{2}+\frac{1}{4}}\,\left\\{f\left(\frac{z}{\sqrt{4N}}\right)+\left(f|W_{4N}\right)\left(\frac{z}{\sqrt{4N}}\right)\right\\}.$
(3.14)

In the previous lemma, we have connected the functions $R_{f}(t)$ and
$I_{f}(t)$ (whose zeros we pretend to study) with an evaluation of the cusp
form $f\left(-e^{-iu}/\sqrt{4N}\right)$, for $0<u<\frac{\pi}{2}$. The next
lemma uses the slash operator (2.5) in order to establish a uniform bound for
any derivative of $f\left(-e^{-iu}/\sqrt{4N}\right)$ with respect to $u$.

######  Lemma 3.2.

There exist four positive constants $A$, $B$, $C$ and $D$ (only depending on
the weight and level of $f$) such that, for any $0<u<\frac{\pi}{2}$ and
$p\in\mathbb{N}_{0}$,

$\left|\frac{d^{p}}{du^{p}}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right|<C\,\frac{2^{p}\,p!}{u^{p+k+\frac{1}{2}}}e^{-\frac{A}{u}}$
(3.15)

and

$\left|\frac{d^{p}}{du^{p}}f\left(\frac{e^{iu}}{\sqrt{4N}}\right)\right|<D\,\frac{2^{p}\,p!}{u^{p+k+\frac{1}{2}}}e^{-\frac{B}{u}}.$
(3.16)

###### Proof.

Let $0<u_{0}<\frac{\pi}{2}$. The purpose of our proof is to bound the absolute
value of
$\left[\frac{d^{p}}{du^{p}}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right]_{u=u_{0}}$.
The derivative of the function $f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)$ at
the point $u_{0}$ will be given by integrating along a positively oriented
circle with center $u_{0}$ and having radius $\lambda u_{0}$, $0<\lambda<1$.
Let us denote this circle by $C_{\lambda u_{0}}(u_{0})$: then for any $w\in
D_{\lambda u_{0}}(u_{0}):=\text{int}(C_{\lambda u_{0}}(u_{0}))$, one can
easily check that $\text{Im}\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)>0$. Hence,
$f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)$ must be analytic inside the circle
$C_{\lambda u_{0}}(u_{0})$ and so, by Cauchy’s formula,

$\left[\frac{d^{p}}{du^{p}}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right]_{u=u_{0}}=\frac{p!}{2\pi
i}\,\intop_{C_{\lambda
u_{0}}(u_{0})}\,\frac{f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)}{(w-u_{0})^{p+1}}\,dw.$
(3.17)

Our next task will be to bound the integral on the right-hand side of (3.17).
This will be done by considering the matrix

$\gamma=\left(\begin{array}[]{cc}-1&0\\\
\sqrt{4N}&-1\end{array}\right)\in\text{SL}(2,\mathbb{Z}),$ (3.18)

and using it to construct the slash operator (2.5)

$g(z):=\left(f|\gamma\right)(z)=\tau\,\left(\sqrt{4N}z-1\right)^{-k-\frac{1}{2}}f\left(\frac{z}{1-\sqrt{4N}z}\right),$
(3.19)

where $\tau$ is a complex number such that $|\tau|=1$. If we apply the
construction (3.19) with

$z:=-\frac{1}{\sqrt{4N}\left(e^{iw}-1\right)}\in\mathbb{H},$

then we see that

$f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)=\frac{1}{\tau}\left(-\frac{e^{iw}}{e^{iw}-1}\right)^{k+\frac{1}{2}}g\left(-\frac{1}{\sqrt{4N}\left(e^{iw}-1\right)}\right).$
(3.20)

Since $f(z)$ is a cusp form and, by hypothesis, $N$ is a perfect square,
$g(z)$ admits the Fourier expansion

$g(z):=\sum_{n=1}^{\infty}b(n)\,e^{\frac{2\pi in}{r}z},\,\,\,\,\text{for some
}r\in\mathbb{N},$ (3.21)

with $b(n)=O\left(n^{\alpha}\right)$ for some large $\alpha>0$. Using (3.20)
and (3.21) we obtain

$\displaystyle\left|f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)\right|$
$\displaystyle\leq\left(\frac{|e^{iw}|}{|e^{iw}-1|}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(\frac{\pi
n}{r\sqrt{N}}\,\text{Im}\left(\frac{1}{e^{iw}-1}\right)\right)$
$\displaystyle=\left(\frac{e^{-\text{Im}(w)/2}}{2\sqrt{\sin^{2}\left(\frac{\text{Re}(w)}{2}\right)+\sinh^{2}\left(\frac{\text{Im}(w)}{2}\right)}}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(-\frac{\pi
n}{2r\sqrt{N}}\,\text{Re}\left(\frac{1}{\tan\left(\frac{w}{2}\right)}\right)\right)$
$\displaystyle\leq\left(\frac{e^{\frac{\lambda
u_{0}}{2}}}{2\sin\left(\frac{\text{Re}(w)}{2}\right)}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(-\frac{\pi
n}{2r\sqrt{N}}\,\frac{\sin\left(\text{Re}(w)\right)}{\cosh\left(\text{Im}(w)\right)-\cos\left(\text{Re}(w)\right)}\right),$
(3.22)

where in the last step we just have used the fact that, for any $w\in
C_{\lambda u_{0}}(u_{0})$, $|\text{Im}(w)|\leq\lambda u_{0}$. Using now the
elementary Jordan inequality $\sin(x)>\frac{2x}{\pi}$, $0<x<\frac{\pi}{2}$,
and the fact that
$\frac{1}{2}(1-\lambda)u_{0}\leq\frac{1}{2}\text{Re}(w)\leq\frac{1}{2}(1+\lambda)u_{0}<\frac{\pi}{2}$,
we obtain

$\displaystyle\left|f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)\right|$
$\displaystyle<\left(\frac{\pi e^{\frac{\lambda
u_{0}}{2}}}{2\text{Re}(w)}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(-\frac{\pi
n}{2r\sqrt{N}}\,\frac{\sin\left(\text{Re}(w)\right)}{\cosh\left(\text{Im}(w)\right)-\cos\left(\text{Re}(w)\right)}\right)$
$\displaystyle\leq\left(\frac{\pi e^{\frac{\lambda
u_{0}}{2}}}{2\left(1-\lambda\right)u_{0}}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(-\frac{\pi
n}{2r\sqrt{N}}\,\frac{\sin\left(\text{Re}(w)\right)}{\cosh\left(\text{Im}(w)\right)-\cos\left(\text{Re}(w)\right)}\right).$
(3.23)

Our proof shall be concluded once we bound the terms of the exponential
function in (3.23). First, let us note that

$\displaystyle\cosh(\text{Im}(w))-\cos(\text{Re}(w))$
$\displaystyle=\intop_{0}^{\text{Im}(w)}\sinh(t)\,dt+\intop_{0}^{\text{Re}(w)}\sin(t)\,dt\leq\intop_{0}^{\text{Im}(w)}t\,e^{\frac{t^{2}}{6}}dt+\intop_{0}^{\text{Re}(w)}t\,dt$
$\displaystyle\leq\frac{\text{Im}(w)^{2}}{2}\,e^{\frac{\text{Im}(w)^{2}}{6}}+\frac{\text{Re}(w)^{2}}{2}<e^{\frac{\lambda^{2}}{6}u_{0}^{2}}\left(1+\lambda\right)^{2}u_{0}^{2},$
(3.24)

where we have used the fact that
$(1-\lambda)u_{0}\leq\text{Re}(w)\leq(1+\lambda)u_{0}$, $-\lambda
u_{0}\leq\text{Im}(w)\leq\lambda u_{0}$ and the well-known inequality
$\sinh(x)\leq x\,e^{x^{2}/6},\,\,x>0$. But since
$0<(1-\lambda)u_{0}\leq\text{Re}(w)\leq(1+\lambda)u_{0}<\pi$, another
application of Jordan’s inequality gives

$\sin(\text{Re}(w))>\begin{cases}\frac{2}{\pi}\text{Re}(w),&0<\text{Re}(w)<\frac{\pi}{2}\\\
2-\frac{2}{\pi}\text{Re}(w),&\frac{\pi}{2}\leq\text{Re}(w)<\pi\end{cases}\geq\frac{2}{\pi}\left(1-\lambda\right)u_{0}.$
(3.25)

Hence, the combination of (3.24) and (3.25) yields the lower bound

$\frac{\sin\left(\text{Re}(w)\right)}{\cosh\left(\text{Im}(w)\right)-\cos\left(\text{Re}(w)\right)}>\frac{2(1-\lambda)}{\pi
e^{\frac{\lambda^{2}}{6}u_{0}^{2}}\left(1+\lambda\right)^{2}u_{0}}>\frac{2(1-\lambda)e^{-\frac{\pi^{2}\lambda^{2}}{24}}}{\pi(1+\lambda)^{2}u_{0}}.$
(3.26)

Therefore, returning to (3.23) and denoting by $c$ the smallest integer such
that $b(c)\neq 0$,

$\displaystyle\left|f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)\right|$
$\displaystyle\leq\left(\frac{\pi e^{\frac{\lambda
u_{0}}{2}}}{2\left(1-\lambda\right)u_{0}}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(-\frac{\pi
n}{2r\sqrt{N}}\,\frac{\sin\left(\text{Re}(w)\right)}{\cosh\left(\text{Im}(w)\right)-\cos\left(\text{Re}(w)\right)}\right)$
$\displaystyle<\left(\frac{\pi e^{\frac{\lambda
u_{0}}{2}}}{2\left(1-\lambda\right)u_{0}}\right)^{k+\frac{1}{2}}\sum_{n=1}^{\infty}|b(n)|\,\exp\left(-\frac{n(1-\lambda)e^{-\frac{\pi^{2}\lambda^{2}}{24}}}{r\sqrt{N}\left(1+\lambda\right)^{2}u_{0}}\right)$
$\displaystyle=\left(\frac{\pi e^{\frac{\lambda
u_{0}}{2}}}{2\left(1-\lambda\right)u_{0}}\right)^{k+\frac{1}{2}}|b(c)|\,\exp\left(-\frac{(1-\lambda)\,c\,e^{-\frac{\pi^{2}\lambda^{2}}{24}}}{r\sqrt{N}\,\left(1+\lambda\right)^{2}u_{0}}\right)\,\sum_{n\geq
c}\left|\frac{b(n)}{b(c)}\right|\exp\left(-\frac{(1-\lambda)\,e^{-\frac{\pi^{2}\lambda^{2}}{24}}\,(n-c)}{r\sqrt{N}\left(1+\lambda\right)^{2}u_{0}}\right).$
(3.27)

Finally, since $0<u_{0}<\frac{\pi}{2}$ and $0<\lambda<1$ is arbitrary, we may
now take $\lambda=\frac{1}{2}$ and get the bound

$\displaystyle\left|f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)\right|$
$\displaystyle\leq|b(c)|\,\left(\frac{\pi
e^{\frac{\pi}{8}}}{2u_{0}}\right)^{k+\frac{1}{2}}\,\exp\left(-\frac{2ce^{-\frac{\pi^{2}}{96}}}{9r\sqrt{N}}\,\frac{1}{u_{0}}\right)\,\sum_{n\geq
c}\left|\frac{b(n)}{b(c)}\right|\exp\left(-\frac{4e^{-\frac{\pi^{2}}{96}}}{9\pi
r\sqrt{N}}(n-c)\right)$ (3.28)
$\displaystyle\leq\frac{C^{\prime}}{u_{0}^{k+\frac{1}{2}}}\exp\left(-\frac{A}{u_{0}}\right),$
(3.29)

where $C^{\prime}$ and $A$ only depend on the weight of the cusp form $f(z)$
and on the level $4N$. An explicit expression for $A$ is, therefore,

$A=\frac{2ce^{-\frac{\pi^{2}}{96}}}{9r\sqrt{N}},$ (3.30)

where, as already described, $r$ and $c$ only depend on the cusp form $g(z)$
given by (3.19).777Of course, there are other possible expressions for $A$
with a different choice of $\lambda$. Note that $C$ contains the numerical
value of the infinite series on (3.28), which is clearly convergent because
$b(n)=O(n^{\alpha})$. This proves (3.15) for $p=0$. For every $p\geq 1$, let
us return to Cauchy’s integral formula (3.17) with $\lambda=\frac{1}{2}$ and
use (3.29) to get

$\left|\left[\frac{d^{p}}{du^{p}}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right]_{u=u_{0}}\right|\leq\frac{2^{p}p!}{\pi
u_{0}^{p+1}}\,\intop_{C_{\frac{u_{0}}{2}}(u_{0})}\,\left|f\left(-\frac{e^{-iw}}{\sqrt{4N}}\right)\right|\,|dw|<C\,\frac{2^{p}p!}{u_{0}^{p+k+\frac{1}{2}}}e^{-\frac{A}{u_{0}}},$
(3.31)

which completes the proof of (3.15). The proof of the uniform bound (3.16) is
the same, the only difference being in taking the slash operator. Instead of
the matrix $\gamma$ given in (3.18), we consider

$\gamma^{\star}=\left(\begin{array}[]{cc}1&0\\\
\sqrt{4N}&1\end{array}\right)\in\text{SL}(2,\mathbb{Z}),\,\,\,\,\,\,\,\,\,h(z):=\left(f|\gamma^{\star}\right)(z)=\tau\,\left(\sqrt{4N}z+1\right)^{-k-\frac{1}{2}}f\left(\frac{z}{1+\sqrt{4N}z}\right),$
(3.32)

where, as in (3.19), $\tau$ is a complex number such that $|\tau|=1$. Applying
(3.32) with

$z:=\frac{1}{\sqrt{4N}\left(e^{-iw}-1\right)}\in\mathbb{H},$

we obtain, in analogy to (3.20),

$f\left(\frac{e^{iw}}{\sqrt{4N}}\right)=\frac{1}{\tau}\,\left(\frac{e^{-iw}}{e^{-iw}-1}\right)^{k+\frac{1}{2}}\,h\left(\frac{1}{\sqrt{4N}\left(e^{-iw}-1\right)}\right).$

Since $f$ is a cusp form and $N$ is a perfect square, we know that $h(z)$ also
admits the Fourier expansion

$h(z):=\sum_{n=1}^{\infty}b^{\star}(n)\,e^{\frac{2\pi
in}{r^{\prime}}z},\,\,\,\,\text{for some }r^{\prime}\in\mathbb{N}.$ (3.33)

From this point on, the proof of (3.16) is exactly the same as before and,
just like (3.30), we can find an explicit expression for the constant $B$ in
(3.16) in the form

$B=\frac{2c^{\prime}e^{-\frac{\pi^{2}}{96}}}{9r^{\prime}\sqrt{N}},$ (3.34)

where $c^{\prime}$ is the smallest integer such that
$b^{\star}(c^{\prime})\neq 0$ and $r^{\prime}$ is the integer appearing in the
Fourier expansion (3.33).

∎

##  4 Proof of Theorem 1.1

Throughout our proof we shall assume that $a_{f}(n)$ are real numbers. The
case where $a_{f}(n)$ are purely imaginary is analogous and the necessary
changes in the argument are already outlined in [21]. First we show that
$N_{0}^{+}(T)=\Omega\left(T^{1/2}\right)$, which is the first part of (1.12).
The proof that $N_{0}^{-}(T)=\Omega\left(T^{1/2}\right)$, as we shall see,
presents no extra difficulties.

Let $\left(\rho_{n}\right)_{n\in\mathbb{N}}$ be the sequence of zeros of odd
order of $L\left(s,f\right)+L\left(s,f|W_{4N}\right)$ such that
$\text{Re}(\rho_{n})=\frac{k}{2}+\frac{1}{4}$. Then we can write
$\rho_{n}:=\frac{k}{2}+\frac{1}{4}+i\tau_{n}$, with $\tau_{n}>0$ being an
increasing sequence888Note that if
$L\left(\frac{k}{2}+\frac{1}{4},f\right)+L\left(\frac{k}{2}+\frac{1}{4},f|W_{4N}\right)=0$,
we are excluding this real zero from the sequence
$(\tau_{n})_{n\in\mathbb{N}}$.. If we show that there is some $h>0$ such that,
for infinitely many values of $n$, $\tau_{n}<h\,n^{2}$, we are done. This is
the case because, if we choose the sequence $T_{n}:=hn^{2}$, then we find that
$N_{0}^{+}\left(T_{n}\right)\geq
N_{0}^{+}(\tau_{n})=n=\sqrt{\frac{T_{n}}{h}}$. Thus, we have built in this way
a sequence $\left(T_{n}\right)_{n\in\mathbb{N}}$ such that
$N_{0}^{+}\left(T_{n}\right)>\sqrt{\frac{T_{n}}{h}}$, which ultimately
establishes

$\limsup_{T\rightarrow\infty}\,\frac{N_{0}^{+}(T)}{\sqrt{T}}>\frac{1}{\sqrt{h}},\,\,\,\,\text{or,
equivalently, }\,\,\,N_{0}^{+}(T)=\Omega\left(T^{\frac{1}{2}}\right).$ (4.1)

Hence, for the sake of contradiction, let us assume that there is some $N_{0}$
such that, for every $n\geq N_{0}$ and any $h>0$, $\tau_{n}\geq h\,n^{2}$. We
will now show that there exists some (large enough) $h$ for which this
assumption is contradicted. Indeed, if we construct the entire function999The
infinite product can be written because, due to the result in [21] (Theorem B
above), we already know that $L\left(s,f\right)+L\left(s,f|W_{4N}\right)$ has
infinitely many zeros of the form $\frac{k}{2}+\frac{1}{4}+i\tau_{n}$.

$\varphi\left(y\right)=\prod_{j=1}^{\infty}\left(1-\frac{y^{2}}{\tau_{j}^{2}}\right)=\sum_{j=0}^{\infty}(-1)^{j}a_{2j}\,y^{2j},$
(4.2)

we see that $a_{0}=1$ and, for $j\geq 1$,

$a_{2j}=\sum_{r_{1}\geq
1}\sum_{r_{2}>r_{1}}...\sum_{r_{j}>r_{j-1}}\frac{1}{\tau_{r_{1}}^{2}\cdot...\cdot\tau_{r_{j}}^{2}}=\sum_{1\leq
r_{1}<r_{2}<...<r_{j}}\frac{1}{\tau_{r_{1}}^{2}\cdot...\cdot\tau_{r_{j}}^{2}},$
(4.3)

where we are summing over $(r_{1},...,r_{j})\in\mathbb{N}^{j}\text{ such that
}r_{1}<r_{2}<...<r_{j}$. Note that, in the $k^{\text{th}}$ nested series in
(4.3), the index $r_{k}$ always satisfies $r_{k}\geq k$, due to the condition
$r_{k}>r_{k-1}>...>r_{1}\geq 1$.

From this point on, we just need to find a suitable bound for $a_{2j}$. By
considering one of the nested infinite series above, we have two
possibilities: if, for some $1\leq k\leq j$, $r_{k}\geq N_{0}$, we know by the
contradiction hypothesis that $\tau_{r_{k}}^{-2}\leq\frac{r_{k}^{-4}}{h^{2}}$.
On the other hand, if $1\leq r_{k}\leq N_{0}-1$, then
$\tau_{r_{k}}^{-2}\leq\frac{r_{k}^{-4}}{h^{\star 2}}$, where
$h^{\star}:=\min_{1\leq n\leq
N_{0}-1}\left\\{\frac{\tau_{n}}{n^{2}}\right\\}$. Let us take a generic series
in (4.3): by the above reasoning, if $r_{k-1}\geq N_{0}-1$, then the
contradiction hypothesis gives

$\sum_{r_{k}>r_{k-1}}\frac{1}{\tau_{r_{k}}^{2}}=\sum_{r_{k}>r_{k-1}\geq
N_{0}-1}\frac{1}{\tau_{r_{k}}^{2}}\leq\frac{1}{h^{2}}\,\sum_{r_{k}>r_{k-1}}\frac{1}{r_{k}^{4}}.$
(4.4)

On the other hand, if $1\leq r_{k-1}<N_{0}-1$,

$\displaystyle\sum_{r_{k}>r_{k-1}}\frac{1}{\tau_{r_{k}}^{2}}$
$\displaystyle=\sum_{r_{k}=r_{k-1}+1}^{N_{0}-1}\frac{1}{\tau_{r_{k}}^{2}}+\sum_{r_{k}=N_{0}}^{\infty}\frac{1}{\tau_{r_{k}}^{2}}\leq\frac{1}{h^{\star
2}}\sum_{r_{k}=r_{k-1}+1}^{N_{0}-1}\frac{1}{r_{k}^{4}}+\frac{1}{h^{2}}\sum_{r_{k}=N_{0}}^{\infty}\frac{1}{r_{k}^{4}}$
$\displaystyle=\frac{1}{h^{2}}\sum_{r_{k}>r_{k-1}}\frac{1}{r_{k}^{4}}+\frac{1}{h^{2}}\left(\frac{h^{2}}{h^{\star
2}}-1\right)\sum_{r_{k}=r_{k-1}+1}^{N_{0}-1}\frac{1}{r_{k}^{4}}\leq\frac{1}{h^{2}}\sum_{r_{k}>r_{k-1}}\frac{1}{r_{k}^{4}}+\frac{1}{h^{2}}\left|\frac{h^{2}}{h^{\star
2}}-1\right|\sum_{j=1}^{N_{0}-1}\frac{1}{j^{4}}$
$\displaystyle\leq\frac{\mathscr{A}}{h^{2}}\,\sum_{r_{k}>r_{k-1}}\frac{1}{r_{k}^{4}},$
(4.5)

for some constant $\mathscr{A}$ depending on $h$ and $N_{0}$ but not on $k$.
We can explicitly choose this constant by taking the observation that there
exists some $A$ such that

$\left|\frac{h^{2}}{h^{\star
2}}-1\right|\sum_{j=1}^{N_{0}-1}\frac{1}{j^{4}}\leq A\,\sum_{j\geq
N_{0}}\frac{1}{j^{4}}\leq A\,\sum_{r_{k}>r_{k-1}}\frac{1}{r_{k}^{4}},$

where the last inequality is due to the fact that $1\leq r_{k-1}<N_{0}-1$.
Picking $\mathscr{A}:=1+A\geq 1$ gives (4.5). Since $r_{\ell}\geq\ell$ always,
a sufficient condition for $r_{k-1}\geq N_{0}-1$ is that $k\geq N_{0}$. Thus,
we have at most $N_{0}-1$ infinite series in (4.3) where we need to apply
(4.5). Hence, the sequence $a_{2j}$ can be bounded in a simple form

$\displaystyle a_{2j}=\sum_{r_{1}\geq
1}\sum_{r_{2}>r_{1}}...\sum_{r_{j}>r_{j-1}}\frac{1}{\tau_{r_{1}}^{2}\cdot...\cdot\tau_{r_{j}}^{2}}$
$\displaystyle=\sum_{r_{1}\geq
1}\sum_{r_{2}>r_{1}}...\sum_{r_{N_{0}}>r_{N_{0}-1}\geq
N_{0}-1}...\sum_{r_{j}>r_{j-1}}\frac{1}{\tau_{r_{1}}^{2}\cdot...\cdot\tau_{r_{j}}^{2}}$
$\displaystyle\leq\frac{\mathscr{A}^{N_{0}-1}}{h^{2j}}\sum_{r_{1}\geq
1}\sum_{r_{2}>r_{1}}...\sum_{r_{j}>r_{j-1}}\frac{1}{r_{1}^{4}\cdot...\cdot
r_{j}^{4}}$ $\displaystyle\leq\frac{\mathcal{K}}{h^{2j}}\,\sum_{r_{1}\geq
1}\sum_{r_{2}>r_{1}}...\sum_{r_{j}>r_{j-1}}\frac{1}{r_{1}^{4}\cdot...\cdot
r_{j}^{4}}=\frac{\mathcal{K}}{h^{2j}}\,\sum_{1\leq
r_{1}<r_{2}<...<r_{j}}\frac{1}{r_{1}^{4}\cdot...\cdot r_{j}^{4}},$ (4.6)

for some $\mathcal{K}$ only depending on $h$ and $N_{0}$.

We now see that a bound for $a_{2j}$ will depend on finding a bound for the
new coefficient

$b_{2j}:=\sum_{r_{1}\geq
1}\sum_{r_{2}>r_{1}}...\sum_{r_{j}>r_{j-1}}\frac{1}{r_{1}^{4}\cdot...\cdot
r_{j}^{4}}=\sum_{1\leq r_{1}<...<r_{j}}\,\frac{1}{r_{1}^{4}\cdot...\cdot
r_{j}^{4}}.$ (4.7)

Whoever is familiar with one of Euler’s proofs of the formula for $\zeta(2n)$
recognizes this sequence of numbers as coming from the Weierstrass
factorization of the sine function, which takes the form

$\frac{\sinh(\pi\sqrt{y})\sin(\pi\sqrt{y})}{\pi^{2}y}=\prod_{j=1}^{\infty}\left(1+\frac{y}{j^{2}}\right)\prod_{j=1}^{\infty}\left(1-\frac{y}{j^{2}}\right)=\prod_{j=1}^{\infty}\left(1-\frac{y^{2}}{j^{4}}\right):=1+\sum_{j=1}^{\infty}(-1)^{j}b_{2j}y^{2j}.$
(4.8)

Thus, we can get precise information about $b_{2j}$ (and, consequently, about
$a_{2j}$) by interpreting them as the Taylor coefficients of the function on
the left-hand side of (4.8).101010It is somewhat poetic that the same idea
that led Euler to have the first grasp on the nature of $\zeta(2n)$ can be
used to understand the zeros of the same function. Indeed, it is quite
simple111111Although these computations are standard, we did not find any
reference containing this Taylor expansion. In order to be self-contained, we
have decided to briefly present it. to see that, for any $y\in\mathbb{R}$,

$\displaystyle\sinh(\pi\sqrt{y})\,\sin(\pi\sqrt{y})$
$\displaystyle=-\frac{i}{2}\left[\cos\left((1-i)\pi\sqrt{y}\right)-\cos\left((1+i)\pi\sqrt{y}\right)\right]=-\frac{i}{2}\sum_{k=0}^{\infty}\frac{(-1)^{k}\pi^{2k}}{(2k)!}y^{k}\left\\{\left(1-i\right)^{2k}-(1+i)^{2k}\right\\}$
$\displaystyle=-\sum_{k=0}^{\infty}\frac{(-1)^{k}\pi^{2k}}{(2k)!}\left(2y\right)^{k}\sin\left(\frac{\pi}{2}k\right)=\sum_{j=0}^{\infty}\frac{(-1)^{j}2^{2j+1}\pi^{4j+2}}{(4j+2)!}y^{2j+1}.$
(4.9)

Thus, comparing (4.8) with (4.9) and returning to the inequality (4.6), we
find that

$b_{2j}:=\sum_{1\leq r_{1}<...<r_{j}}\,\frac{1}{r_{1}^{4}\cdot...\cdot
r_{j}^{4}}=\frac{2^{2j+1}\pi^{4j}}{(4j+2)!}\implies
a_{2j}\leq\mathcal{K}\,\,\frac{2^{2j+1}\pi^{4j}}{h^{2j}(4j+2)!}.$ (4.10)

Our proof will be concluded by seeing that (4.10) contradicts (3.7) and the
bounds for the derivatives of $f$ found in Lemma 3.2. By construction of
$\varphi(t)$, we know that $R_{f}(t)\varphi(t)$ must have constant sign for
any $t\in\mathbb{R}$. If $R_{f}(t)\varphi(t)\geq 0$ for any $t\in\mathbb{R}$,
the continuous function
$Q_{f}:\,\left(0,\,\frac{\pi}{2}\right)\longmapsto\mathbb{R}$ defined by the
integral

$Q_{f}(u):=\intop_{0}^{\infty}R_{f}(t)\varphi(t)\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt,\,\,\,\,0<u<\frac{\pi}{2},$

is positive decreasing. Analogously, if $R_{f}(t)\varphi(t)\leq 0$, $Q_{f}(u)$
will be negative increasing. In both cases, we have that the continuous
function

$|Q_{f}(u)|:=\left|\intop_{0}^{\infty}R_{f}(t)\varphi(t)\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt\right|,\,\,\,\,0<u<\frac{\pi}{2},$

will be positive decreasing. On the other hand, if we use the power series
(4.2) for $\varphi(t)$, we see that $Q_{f}(u)$ can be written as an infinite
series of the form

$Q_{f}(u)=\intop_{0}^{\infty}R_{f}(t)\,\varphi(t)\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt=\sum_{j=0}^{\infty}(-1)^{j}a_{2j}\,\intop_{0}^{\infty}R_{f}(t)\,t^{2j}\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt.$
(4.11)

Note that the interchange of the orders of summation and integration in (4.11)
is justified by Fubini’s theorem and the bounds for $a_{2j}$ given in (4.10):
indeed,

$\displaystyle\intop_{0}^{\infty}\sum_{j=0}^{\infty}|a_{2j}|t^{2j}\,|R_{f}(t)|\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt$
$\displaystyle\leq\mathcal{K}\intop_{0}^{\infty}\,\sum_{j=0}^{\infty}\frac{2^{2j+1}\pi^{4j}}{h^{2j}(4j+2)!}t^{2j}\,|R_{f}(t)|\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt$
$\displaystyle\leq
2\mathcal{K}\intop_{0}^{\infty}\,\sum_{j=0}^{\infty}\frac{1}{(4j)!}\left(\frac{2\pi^{2}t}{h}\right)^{2j}\,|R_{f}(t)|\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt$
$\displaystyle<2\mathcal{K}\intop_{0}^{\infty}\,\sum_{j=0}^{\infty}\frac{1}{j!}\left(\pi\sqrt{\frac{2t}{h}}\right)^{j}\,|R_{f}(t)|\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt$
$\displaystyle<2\mathcal{K}\intop_{0}^{\infty}\,|R_{f}(t)|\,\exp\left(\pi\sqrt{\frac{2t}{h}}+\left(\frac{\pi}{2}-u\right)t\right)\,dt$
$\displaystyle\leq
2\mathcal{K}\,C\,\intop_{0}^{\infty}\,t^{A}\,\exp\left(-ut+\pi\sqrt{\frac{2t}{h}}\right)\,dt<\infty,$
(4.12)

because $u>0$ by hypothesis. In the last step we have used Stirling’s formula
for $\Gamma(s)$ (2.7) as well as the convex estimate for $L(s,f)$ (2.8), which
show that $|R_{f}(t)|\leq C\,|t|^{A}e^{-\frac{\pi}{2}|t|}$ for sufficiently
large $C$ and $A$. Having assured that we can perform the operation (4.11), it
now follows from (4.12) that the Leibniz rule holds for the integral on the
right-hand side of (4.11) and, after using (3.7), we find the expression

$\displaystyle Q_{f}(u)$
$\displaystyle=\sum_{j=0}^{\infty}(-1)^{j}a_{2j}\,\frac{d^{2j}}{du^{2j}}\,\left\\{\intop_{0}^{\infty}R_{f}(t)\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt\right\\}$
$\displaystyle=\frac{\pi}{2}\,\sum_{j=0}^{\infty}(-1)^{j}\,a_{2j}\,\frac{d^{2j}}{du^{2j}}$
$\displaystyle\left[e^{\frac{i\pi}{2}\left(\frac{k}{2}+\frac{1}{4}\right)}\,e^{-i\left(\frac{k}{2}+\frac{1}{4}\right)u}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)+e^{-i\frac{\pi}{2}\left(\frac{k}{2}+\frac{1}{4}\right)}e^{i\left(\frac{k}{2}+\frac{1}{4}\right)u}\,f\left(\frac{e^{iu}}{\sqrt{4N}}\right)\right].$
(4.13)

Using Lemma 3.2 and our estimates for $a_{2j}$ (4.10), we can bound uniformly
the previous series with respect to $u$. Firstly, according to Lemma 3.2, we
have

$\displaystyle\left|\frac{d^{2j}}{du^{2j}}\,\left[e^{-i\left(\frac{k}{2}+\frac{1}{4}\right)u}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right]\right|$
$\displaystyle\leq\sum_{\ell=0}^{2j}\left(\begin{array}[]{c}2j\\\
\ell\end{array}\right)\,\left(\frac{k}{2}+\frac{1}{4}\right)^{2j-\ell}\left|\frac{d^{\ell}}{du^{\ell}}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right|$
$\displaystyle<C\frac{e^{-\frac{A}{u}}}{u^{k+\frac{1}{2}}}\sum_{\ell=0}^{2j}\frac{(2j)!}{(2j-\ell)!}\,\left(\frac{k}{2}+\frac{1}{4}\right)^{2j-\ell}\,\frac{2^{\ell}}{u^{\ell}}$
$\displaystyle\leq
C\frac{e^{-\frac{A}{u}}(2j)!}{u^{k+\frac{1}{2}}}\left(\frac{k}{2}+\frac{1}{4}\right)^{2j}\,\sum_{\ell=0}^{2j}\left(\frac{8}{(2k+1)u}\right)^{\ell}$
$\displaystyle=C\frac{e^{-\frac{A}{u}}(2j)!}{u^{k+2j+\frac{1}{2}}2^{4j}}\,\frac{8^{2j+1}-((2k+1)u)^{2j+1}}{8-(2k+1)u}.$

If we now pick $0<u<\frac{4}{2k+1}<\frac{\pi}{2}$, we see immediately that

$\left|\frac{d^{2j}}{du^{2j}}\,\left[e^{-i\left(\frac{k}{2}+\frac{1}{4}\right)u}f\left(-\frac{e^{-iu}}{\sqrt{4N}}\right)\right]\right|<C^{\prime}\,\frac{2^{2j}(2j)!e^{-A/u}}{u^{2j+k+\frac{1}{2}}},$

while

$\left|\frac{d^{2j}}{du^{2j}}\,\left[e^{i\left(\frac{k}{2}+\frac{1}{4}\right)u}\,f\left(\frac{e^{iu}}{\sqrt{4N}}\right)\right]\right|<D^{\prime}\,\frac{2^{2j}(2j)!e^{-B/u}}{u^{2j+k+\frac{1}{2}}}.$

Thus

$|Q_{f}(u)|\leq\frac{\pi}{2}\,\sum_{j=0}^{\infty}|a_{2j}|\frac{2^{2j}(2j)!}{u^{2j+k+\frac{1}{2}}}\,\left\\{C^{\prime}e^{-A/u}+D^{\prime}\,e^{-B/u}\right\\}\leq\frac{\pi\beta}{u^{k+\frac{1}{2}}}\,e^{-\alpha/u}\sum_{j=0}^{\infty}|a_{2j}|\frac{2^{2j}(2j)!}{u^{2j}},$
(4.14)

where $\beta=\max\left\\{C^{\prime},D^{\prime}\right\\}$ and
$\alpha=\min\left\\{A,B\right\\}$. To finish, we must estimate the infinite
series on the right-hand side of (4.14): we do this by employing (4.10), which
yields

$\displaystyle\sum_{j=0}^{\infty}|a_{2j}|\frac{2^{2j}(2j)!}{u^{2j}}$
$\displaystyle\leq\mathcal{K}\,\sum_{j=0}^{\infty}\frac{2^{4j+1}\pi^{4j}(2j)!}{h^{2j}(4j+2)!}\cdot\frac{1}{u^{2j}}=\frac{\sqrt{\pi}\,\mathcal{K}}{2}\,\sum_{j=0}^{\infty}\frac{\pi^{4j}}{\Gamma\left(2j+\frac{3}{2}\right)(2j+1)(hu)^{2j}}$
$\displaystyle<\sqrt{\pi}\,\mathcal{K}\,\sum_{j=0}^{\infty}\frac{\pi^{4j}}{(2j+1)!(hu)^{2j}}<\sqrt{\pi}\,\mathcal{K}\,\cosh\left(\frac{\pi^{2}}{h\,u}\right)\leq\sqrt{\pi}\,\mathcal{K}\exp\left(\frac{\pi^{2}}{h\,u}\right).$
(4.15)

Combining (4.14) and (4.15) gives

$|Q_{f}(u)|<\frac{\pi^{\frac{3}{2}}\beta\,\mathcal{K}}{u^{k+\frac{1}{2}}}\,\exp\left(-\left(\alpha-\frac{\pi^{2}}{h}\right)\frac{1}{u}\right).$

Hence, if we pick $h$ such that $h>\frac{\pi^{2}}{\alpha}$, then

$\lim_{u\rightarrow
0^{+}}\left|Q_{f}(u)\right|<\pi^{\frac{3}{2}}\beta\,\mathcal{K}\lim_{u\rightarrow
0^{+}}u^{-k-\frac{1}{2}}\exp\left(-\left(\alpha-\frac{\pi^{2}}{h}\right)\frac{1}{u}\right)=0,$

which contradicts the fact that $|Q_{f}(u)|$ is positive decreasing.
Consequently, we have found $h>\frac{\pi^{2}}{\alpha}$ such that
$\tau_{n}<h\,n^{2}$ for infinitely many values of $n$. This shows (1.13) and
so we complete the proof of our Theorem to $N_{0}^{+}(T)$.

Finally, we note that we can replace the value of $d$ in (1.13) by an explicit
constant depending on the properties of the cusp form $f$. Indeed, since
$\alpha=\min\left\\{A,B\right\\}$, we may assume without any loss of
generality that $\alpha=A$ and so $\alpha$ is explicitly given by (3.30).
Using the inequality in (4.1) and taking $h=\frac{4\pi^{2}}{\alpha}$, we find
that121212It is possible to improve the value of the explicit constant on
(4.16) by taking a more careful choice of $\lambda$ in (3.27). Recall that the
constant (3.30) was obtained by the choice $\lambda=\frac{1}{2}$.

$\limsup_{T\rightarrow\infty}\,\frac{N_{0}^{+}(T)}{\sqrt{T}}>\frac{e^{-\frac{\pi^{2}}{192}}}{6\pi}\sqrt{\frac{2c}{r\sqrt{N}}},$
(4.16)

where $c$ and $r$ are defined with respect to the Fourier expansion (3.21) of
$g(z):=\left(f|\gamma\right)(z)$, defined by the slash operator (3.19). This
completes the proof.

In order to study the number $N_{0}^{-}(T)$, the argument is the same with a
small modification. Replacing $R_{f}(t)$ by $I_{f}(t)$, instead of considering
$Q_{f}(u)$ we study the similar function

$P_{f}(u)=\intop_{0}^{\infty}tI_{f}(t)\,\varphi(t)\,\cosh\left(\left(\frac{\pi}{2}-u\right)t\right)\,dt,$

where $\varphi(t)$ is given by (4.2). Invoking (3.8) and using the bounds for
the coefficients $a_{2j}$ (4.10), we can see that $|P_{f}(u)|\rightarrow 0$ as
$u\rightarrow 0^{+}$, which again contradicts the first hypothesis over the
zeros of $N_{0}^{-}(T)$. $\blacksquare$

##  5 Lemmas for the proof of Theorem 1.2

Throughout this section, we will assume that $f(z)$ satisfies the conditions
of Theorem 1.2. Thus, $N$ will always be a perfect square and the Fourier
coefficients, $a_{f}(n)$, will always be real or purely imaginary numbers.

We start with a lemma which was proved by Wilton for Ramanujan’s Dirichlet
series (see Lemma 3 of [28]). Wilton used it to give a Voronoï-type formula
for the Riesz sum

$\sum_{n\leq x}{}^{{}^{\prime}}\tau(n)\,(x-n)^{\rho}\,e^{\frac{2\pi
ip}{q}n},\,\,\,\,\,0<p<q,\,\,\,\rho>0,\,\,\,x>0.$

To bring Wilton’s result into our study, we shall adapt a lemma in
Lekkerkerker’s thesis [[20], Chpt. 3, Lemma 3.8], which acts as a
generalization of Wilton’s lemma in the setting of Dirichlet series. Our proof
will take into consideration the uniform bound (2.1) for a generic cusp form
$f$.

######  Lemma 5.1.

Let $M$ be an integer $\geq 2$ and define $\delta:=\delta_{z,M}$ as follows

$\delta=\begin{cases}0&0<\text{Im}(z)\leq\frac{\sqrt{N}}{\pi M},\\\
1&\text{Im}(z)>\frac{\sqrt{N}}{\pi M}.\end{cases}$ (5.1)

Then we have that

$\sum_{m=1}^{M}a_{f}(m)\,e^{\frac{2\pi
im}{\sqrt{4N}}z}-\delta\,f\left(\frac{z}{\sqrt{4N}}\right)=O\left(e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(M+1)}\,M^{\frac{k}{2}+\frac{1}{4}}\log(M)\right)$
(5.2)

uniformly in $z$.

###### Proof.

Note that, if we show (5.2) for $0\leq\text{Re}(z)<\sqrt{4N}$ we are done due
to the periodicity of the expansion on the left-hand side of (5.2). Therefore,
throughout our argument, we shall assume that $0\leq\text{Re}(z)<\sqrt{4N}$
and we will let $M$ be an integer greater than $2$. Consider the power series

$F(\zeta):=\sum_{n=1}^{\infty}a_{f}(n)\,\zeta^{n}.$

This infinite series converges in the circle $|\zeta|=1$ because
$f(z)=F\left(e^{2\pi iz}\right)$. Following Wilton and Lekkerkerker [20, 28],
we define a parameter $\beta$ in the following form

$\beta:=\begin{cases}\frac{1}{2}&\text{if }\text{Im}(z)>\frac{\sqrt{N}}{\pi
M}\\\ \frac{3}{2}&\text{if Im}(z)\leq\frac{\sqrt{N}}{\pi M}\end{cases},$ (5.3)

and, for each value of $\beta$, we consider a sequence of integrals,
$\left(I_{\beta,M}\right)_{M\in\mathbb{N}}$, defined as follows

$I_{\beta,M}=\frac{e^{\frac{2\pi iz}{\sqrt{4N}}(M+1)}}{2\pi
i}\,\intop_{|\zeta|=e^{-\beta/M}}\,\frac{F(\zeta)}{\zeta^{M+1}\left(\zeta-e^{2\pi
iz/\sqrt{4N}}\right)}\,d\zeta.$ (5.4)

The value of the integral will be dependent on the choice of the parameter
$\beta$ in (5.3). We are left with two possibilities:

  1. 1.

First, let $\beta=\frac{3}{2}$, so that $\text{Im}(z)\leq\frac{\sqrt{N}}{\pi
M}$. Then $\left|e^{2\pi
iz/\sqrt{4N}}\right|=e^{-\frac{\pi\text{Im}(z)}{\sqrt{N}}}\geq
e^{-\frac{1}{M}}>e^{-\frac{3}{2M}}$. Hence, the only pole that the above
integrand has inside the circle $|\zeta|=e^{-\frac{3}{2M}}$ is located at
$\zeta=0$ and its order is $M+1$. A standard computation of the residue at
this point gives

$I_{\frac{3}{2},M}=-\sum_{m=1}^{M}a_{f}(m)\,e^{\frac{2\pi iz}{\sqrt{4N}}m}.$
(5.5)

  2. 2.

Second, let $\beta=\frac{1}{2}$, so that $\text{Im}(z)>\frac{\sqrt{N}}{\pi
M}$. Therefore, $\left|e^{2\pi
iz/\sqrt{4N}}\right|=e^{-\frac{\pi\text{Im}(z)}{\sqrt{N}}}<e^{-\frac{1}{M}}<e^{-\frac{1}{2M}}$,
which shows that the point $e^{2\pi iz/\sqrt{4N}}$ is located inside the
circle $|\zeta|=e^{-\frac{1}{2M}}$. Thus, besides the pole at $\zeta=0$, the
integrand in (5.4) has an additional simple pole, $\zeta=e^{2\pi
iz/\sqrt{4N}}$. Once more, a standard application of the residue theorem
yields

$I_{\frac{1}{2},M}=F\left(e^{\frac{2\pi
iz}{\sqrt{4N}}}\right)-\sum_{m=1}^{M}a_{f}(m)\,e^{\frac{2\pi
iz}{\sqrt{4N}}m}=f\left(\frac{z}{\sqrt{4N}}\right)-\sum_{m=1}^{M}a_{f}(m)\,e^{\frac{2\pi
iz}{\sqrt{4N}}m}.$ (5.6)

Note that, in the cases (5.5) and (5.6), the symmetric value of the left-hand
side of (5.2) is displayed. Hence, if we show that $I_{\beta,M}$ satisfies the
bound at the right-hand side of (5.2), we finish our argument. Indeed,

$\displaystyle|I_{\beta,M}|$
$\displaystyle\leq\frac{e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(M+1)}e^{\frac{3}{2}(1+\frac{1}{M})}}{2\pi}\,\intop_{0}^{2\pi}\,\frac{\left|f\left(\frac{\theta}{2\pi}+\frac{i\beta}{2\pi
M}\right)\right|}{\left|1-e^{\frac{\beta}{M}+\frac{2\pi
iz}{\sqrt{4N}}-i\theta}\right|}\,d\theta$
$\displaystyle\leq\frac{K\,e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(M+1)}e^{\frac{3}{2}(1+\frac{1}{M})}}{2\pi}\left(\frac{\beta}{2\pi
M}\right)^{-\frac{k}{2}-\frac{1}{4}}\,\intop_{0}^{2\pi}\frac{d\theta}{\left|1-e^{\frac{\beta}{M}-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}+i\left(\frac{2\pi\text{Re}(z)}{\sqrt{4N}}-\theta\right)}\right|},$
(5.7)

where in the last step we have used the fact that
$|f(x+iy)|=O\left(y^{-\frac{k}{2}-\frac{1}{4}}\right)$ uniformly in $x$ (2.1).
Comparing (5.7) with our aim (5.2), in order to finish our proof we just need
to show that

$\intop_{0}^{2\pi}\frac{d\theta}{\left|1-e^{\frac{\beta}{M}-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}+i\left(\frac{2\pi\text{Re}(z)}{\sqrt{4N}}-\theta\right)}\right|}=O\left(\log(M)\right).$
(5.8)

To do so, let us take a new variable
$\eta:=\frac{\beta}{M}-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}$. Once more, we have
two possibilities on the range of $\eta$: if $\beta=\frac{1}{2}$, then
$|\eta|>\frac{1}{2M}$. On the other hand, if $\beta=\frac{3}{2}$, then
$|\eta|\geq\frac{1}{2M}$. Therefore, we shall estimate (5.8) only assuming
that $\beta=\frac{3}{2}$, as for the case $\beta=\frac{1}{2}$, the same
computations are allowed.

If we set $\varphi:=\frac{2\pi\text{Re}(z)}{\sqrt{4N}}$, then by the
hypothesis $0\leq\text{Re}(z)<\sqrt{4N}$, the range of $\varphi$ will be
$0\leq\varphi<2\pi$. Taking this change of variables, we just need to estimate
the integral

$J(\eta):=\intop_{\varphi-2\pi}^{\varphi}\frac{d\phi}{\left|1-e^{\eta+i\phi}\right|},$

subject to the condition that $|\eta|>\frac{1}{2M}$. First, if we assume that
$|\eta|\geq 1$, then $J(\eta)$ can be simply estimated as follows

$J(\eta)=\intop_{\varphi-2\pi}^{\varphi}\frac{d\phi}{\left|1-e^{\eta+i\phi}\right|}\leq\frac{2\pi}{\left|1-e^{\eta}\right|}\leq\frac{2\pi}{1-e^{-1}},$

which shows that our integral is uniformly bounded in this range. Thus, from
now on we may suppose that $\frac{1}{2M}\leq|\eta|<1$. Since

$\left|1-e^{\eta+i\phi}\right|=\sqrt{\left(1-e^{\eta}\right)^{2}+4e^{\eta}\sin^{2}\left(\frac{\phi}{2}\right)}\geq\sqrt{\left(1-e^{-\frac{1}{2M}}\right)^{2}+4e^{\eta}\sin^{2}\left(\frac{\phi}{2}\right)},$

we have that

$\displaystyle\intop_{\varphi-2\pi}^{\varphi}\frac{d\phi}{\left|1-e^{\eta+i\phi}\right|}$
$\displaystyle\leq
2\intop_{\frac{\varphi}{2}-\pi}^{\frac{\varphi}{2}}\frac{d\phi}{\sqrt{\left(1-e^{-\frac{1}{2M}}\right)^{2}+4e^{\eta}\sin^{2}\left(\psi\right)}}\leq
2\intop_{-\pi}^{\pi}\frac{d\psi}{\sqrt{\left(1-e^{-\frac{1}{2M}}\right)^{2}+4e^{\eta}\sin^{2}\left(\psi\right)}}$
$\displaystyle\leq
8\,\intop_{0}^{\frac{\pi}{2}}\,\frac{d\psi}{\sqrt{\left(\frac{1}{2M}-\frac{1}{8M^{2}}\right)^{2}+\frac{4}{e}\sin^{2}(\psi)}}<8\,\intop_{0}^{\frac{\pi}{2}}\,\frac{d\psi}{\sqrt{\frac{1}{16M^{2}}+\frac{4}{e}\sin^{2}(\psi)}},$
(5.9)

where we have used the fact that $M\geq 2$ and $1-e^{-x}\geq
x-\frac{x^{2}}{2}$, as well as the hypothesis $\frac{1}{2M}\leq|\eta|<1$. Now,
by Jordan’s inequality, $\sin\left(\psi\right)>\frac{2}{\pi}\psi$,
$0<\psi<\frac{\pi}{2}$, we finally reach the bound

$\displaystyle\intop_{0}^{\frac{\pi}{2}}\,\frac{d\psi}{\sqrt{\frac{1}{16M^{2}}+\frac{4}{e}\sin^{2}(\psi)}}$
$\displaystyle<\intop_{0}^{\frac{\pi}{2}}\,\frac{d\psi}{\sqrt{\frac{1}{16M^{2}}+\frac{16}{\pi^{2}e}\psi^{2}}}<4\intop_{0}^{\frac{\pi}{2}}\frac{d\psi}{\sqrt{\frac{1}{M^{2}}+\psi^{2}}}$
$\displaystyle=4\intop_{0}^{1/M}\frac{d\psi}{\sqrt{\frac{1}{M^{2}}+\psi^{2}}}+4\intop_{1/M}^{\frac{\pi}{2}}$
$\displaystyle\frac{d\psi}{\sqrt{\frac{1}{M^{2}}+\psi^{2}}}<4+4\intop_{1/M}^{\frac{\pi}{2}}\frac{d\psi}{\psi}=4\log\left(\frac{\pi
e}{2}M\right)=O\left(\log(M)\right),$ (5.10)

which finishes the proof of (5.8) and, consequently, the proof of the whole
lemma. ∎

######  Remark 5.1.

Consider the Fourier expansion of the Fricke involution of $f,$

$\left(f|W_{4N}\right)\left(\frac{z}{\sqrt{4N}}\right)=\sum_{n=1}^{\infty}a_{f|W_{4N}}(n)\,e^{\frac{2\pi
iz}{\sqrt{4N}}}.$

Then the previous lemma gives

$\sum_{m=1}^{M}a_{f|W_{4N}}(m)\,e^{\frac{2\pi
im}{\sqrt{4N}}z}-\delta\,\left(f|W_{4N}\right)\left(\frac{z}{\sqrt{4N}}\right)=O\left(e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(M+1)}\,M^{\frac{k}{2}+\frac{1}{4}}\log(M)\right)$

uniformly in $z$.

Since our proof of Theorem 1.2 will use Lekkerkerker’s variant of the Hardy-
Littlewood method, we shall reach a point in the argument below where we will
have to find a lower bound for the function

$J(t):=\intop_{t}^{t+H}\left|R_{f}(u)\right|e^{\left(\frac{\pi}{2}-\epsilon\right)u}du,$

where $R_{f}(u)$ is given by (3.3). The following simple lemma will be crucial
to achieve this lower bound. We omit its proof, as it is standard.

######  Lemma 5.2.

Let $c$ be the smallest positive integer for which $a_{f}(c)\neq 0$. Consider
the Dirichlet series

$L^{\star}(s,f):=c^{s}L(s,f)-a_{f}(c).$ (5.11)

Also, let $L(s,f_{1})$ be defined as

$L(s,f_{1}):=\sum_{n=c+1}^{\infty}\frac{a_{f}(n)}{\log\left(n/c\right)\,n^{s}},\,\,\,\,\,\text{Re}(s)>\frac{k}{2}+\frac{5}{4}.$
(5.12)

Then $L(s,f_{1})$ can be analytically continued as an entire function of $s$
and, for any $s\in\mathbb{C}$, the following relation holds

$c^{-s}\intop_{s}^{\infty}L^{\star}(z,f)\,dz=L(s,f_{1}).$ (5.13)

######  Remark 5.2.

The previous lemma also holds when we replace $f(z)$ by
$\left(f|W_{4N}\right)(z)$. Indeed, if $d$ is the smallest positive integer
such that $a_{f|W_{4N}}(d)\neq 0$, then, considering the Dirichlet series

$L^{\star}\left(s,f|W_{4N}\right):=d^{s}L\left(s,f|W_{4N}\right)-a_{f|W_{4N}}(d),$
(5.14)

as well as

$L\left(s,(f|W_{4N})_{1}\right):=\sum_{n=d+1}^{\infty}\frac{a_{f|W_{4N}}(n)}{\log\left(n/d\right)\,n^{s}},\,\,\,\,\,\text{Re}(s)>\frac{k}{2}+\frac{5}{4},$
(5.15)

one may see that $L\left(s,(f|W_{4N})_{1}\right)$ can be analytically
continued as an entire function of $s$. The relation analogous to (5.13) in
this case is

$d^{-s}\intop_{s}^{\infty}L^{\star}\left(z,f|W_{4N}\right)\,dz=L\left(s,(f|W_{4N})_{1}\right).$
(5.16)

The previous lemma defined a Dirichlet series, $L(s,f_{1})$, which possesses
certain characteristics that will be of interest in the forthcoming argument.
To get some of these properties, we need to attach it to a series resembling a
cusp form,

$f_{1}\left(\frac{z}{\sqrt{4N}}\right):=\sum_{n=c+1}^{\infty}\frac{a_{f}(n)}{\log(n/c)}\,e^{\frac{2\pi
in}{\sqrt{4N}}z}.$

Our next lemma will provide bounds for $f_{1}\left(z/\sqrt{4N}\right)$ in two
ranges of $\text{Im}(z)$. The estimates given below will be of crucial
importance for an application of Parseval’s formula in the proof of Claim 6.1
at the end of this section. The proof of the next lemma is essentially the
same as Lekkerkerker’s [[20], Chapt. 3, Lemma 10 and Theorem 12], but we
present it in a simpler way, avoiding the erroneous lemma 7 of his thesis. We
also remark that there is an alternative argument using Fourier analysis (see
Remark 5.4 below).

######  Lemma 5.3.

Let $z\in\mathbb{H}$ and consider

$f_{1}\left(\frac{z}{\sqrt{4N}}\right):=\sum_{n=c+1}^{\infty}\frac{a_{f}(n)}{\log(n/c)}\,e^{\frac{2\pi
in}{\sqrt{4N}}z},$ (5.17)

where $c$ is the smallest integer for which $a_{f}(c)\neq 0$. Then, for all
$\text{Im}(z)\geq\text{Im}(z_{0})>0$, $f_{1}\left(z/\sqrt{4N}\right)$
satisfies the estimate

$f_{1}\left(\frac{z}{\sqrt{4N}}\right)\ll
e^{-\frac{2\pi}{\sqrt{4N}}\text{Im}(z)}.$ (5.18)

Moreover, for small $\text{Im}(z)$, $f_{1}\left(z/\sqrt{4N}\right)$ satisfies

$f_{1}\left(\frac{z}{\sqrt{4N}}\right)\ll\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\,\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}.$
(5.19)

###### Proof.

Let us define the integer
$\ell:=1+\left[\frac{\sqrt{N}}{\pi\text{Im}(z)}\right]$. Moreover, for
$M\in\mathbb{N}_{0}$, consider $P_{M}\left(z/\sqrt{4N}\right)$ and
$Q_{M}\left(z/\sqrt{4N}\right)$ as the functions defined by

$P_{0}(0):=0,\,\,\,\,\,\,\,\,P_{M}\left(\frac{z}{\sqrt{4N}}\right)=\sum_{m=1}^{M}a_{f}(m)\,e^{\frac{2\pi
im}{\sqrt{4N}}z},$ (5.20)
$Q_{M}(z)=P_{M}(z)-f\left(\frac{z}{\sqrt{4N}}\right).$ (5.21)

Note that $P_{M}(z)=0$ if $M\leq c-1$. By estimate (5.2) given at lemma 5.1,
we have

$P_{M}\left(\frac{z}{\sqrt{4N}}\right)=O\left(e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(M+1)}\,M^{\frac{k}{2}+\frac{1}{4}}\log(M)\right),\,\,\,\text{for
}M=2,3,...,\ell-1,$ (5.22)
$Q_{M}\left(\frac{z}{\sqrt{4N}}\right)=O\left(e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(M+1)}\,M^{\frac{k}{2}+\frac{1}{4}}\log(M)\right),\,\,\,\text{for
}M=\ell,\ell+1,...,$ (5.23)

uniformly in $z$. Estimate (5.22) holds because if $2\leq
M\leq\ell-1\leq\frac{\sqrt{N}}{\pi\text{Im}(z)}$, then
$\text{Im}(z)\leq\frac{\sqrt{N}}{\pi M}$, and so we can apply (5.2) with
$\delta=0$. Analogously, the estimate (5.23) follows from (5.2) with
$\delta=1$.

We shall use these bounds to estimate the modified “polynomial”

$P_{1,M}\left(\frac{z}{\sqrt{4N}}\right):=\sum_{m=c+1}^{M}\frac{a_{f}(m)}{\log(m/c)}e^{\frac{2\pi
im}{\sqrt{4N}}z}.$ (5.24)

Note that, as $M\rightarrow\infty$,
$P_{1,M}\left(z/\sqrt{4N}\right)\rightarrow f_{1}\left(z/\sqrt{4N}\right)$.
Due to the choice of $\ell$, we have two possibilities: $\ell\geq c+1$ or
$1\leq\ell\leq c$. By definition of
$\ell:=1+\left[\frac{\sqrt{N}}{\pi\text{Im}(z)}\right]$,
$\text{Im}(z)\leq\frac{\sqrt{N}}{\pi c}$ in the first case and, in the second,
$\text{Im}(z)\geq\frac{\sqrt{N}}{\pi c}$. Using partial summation and assuming
first that $\ell\geq c+1$, we find that

$\displaystyle
P_{1,M}\left(\frac{z}{\sqrt{4N}}\right)=\sum_{m=c+1}^{M}\frac{a_{f}(m)}{\log(m/c)}e^{\frac{2\pi
im}{\sqrt{4N}}z}$
$\displaystyle=\sum_{m=c+1}^{\ell}\frac{a_{f}(m)}{\log(m/c)}e^{\frac{2\pi
im}{\sqrt{4N}}z}+\sum_{m=\ell+1}^{M}\frac{a_{f}(m)}{\log(m/c)}e^{\frac{2\pi
im}{\sqrt{4N}}z}$
$\displaystyle=\sum_{m=c+1}^{\ell}\left\\{P_{m}\left(\frac{z}{\sqrt{4N}}\right)-P_{m-1}\left(\frac{z}{\sqrt{4N}}\right)\right\\}\frac{1}{\log(m/c)}$
$\displaystyle+\sum_{m=\ell+1}^{M}\left\\{Q_{m}\left(\frac{z}{\sqrt{4N}}\right)-Q_{m-1}\left(\frac{z}{\sqrt{4N}}\right)\right\\}\,\frac{1}{\log(m/c)}$
$\displaystyle=\sum_{m=c+1}^{\ell-1}P_{m}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}$
$\displaystyle+P_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\,\frac{1}{\log(\ell/c)}-P_{c}\left(\frac{z}{\sqrt{4N}}\right)\frac{1}{\log\left((c+1)/c\right)}$
$\displaystyle+\sum_{m=\ell}^{M-1}Q_{m}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}$
$\displaystyle+Q_{M}\left(\frac{z}{\sqrt{4N}}\right)\,\frac{1}{\log\left(M/c\right)}-Q_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\,\frac{1}{\log\left((\ell+1)/c\right)}.$
(5.25)

Recalling that
$P_{\ell}(z/\sqrt{4N})-Q_{\ell}(z/\sqrt{4N})=f\left(z/\sqrt{4N}\right)$, we
deduce

$\displaystyle P_{1,M}\left(\frac{z}{\sqrt{4N}}\right)$
$\displaystyle=\sum_{m=c+1}^{\ell-1}P_{m}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}+\frac{f\left(z/\sqrt{4N}\right)}{\log(\ell/c)}+\frac{Q_{M}\left(z/\sqrt{4N}\right)}{\log(M/c)}$
$\displaystyle+Q_{\ell}\left(\frac{z}{\sqrt{4N}}\right)$
$\displaystyle\left\\{\frac{1}{\log(\ell/c)}-\frac{1}{\log\left((\ell+1)/c\right)}\right\\}+\sum_{m=\ell}^{M-1}Q_{m}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}-\frac{P_{c}\left(z/\sqrt{4N}\right)}{\log\left((c+1)/c\right)}.$
(5.26)

Analogously, if $1\leq\ell\leq c$, we have

$\displaystyle P_{1,M}\left(\frac{z}{\sqrt{4N}}\right)$
$\displaystyle=\sum_{m=c+1}^{M}\frac{a_{f}(m)}{\log(m/c)}e^{\frac{2\pi
im}{\sqrt{4N}}z}=\sum_{m=\ell+1}^{M}\left\\{Q_{m}\left(\frac{z}{\sqrt{4N}}\right)-Q_{m-1}\left(\frac{z}{\sqrt{4N}}\right)\right\\}\,\frac{1}{\log(m/c)}$
$\displaystyle=\sum_{m=c+1}^{M-1}Q_{m}\left(\frac{z}{\sqrt{4N}}\right)$
$\displaystyle\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}+Q_{M}\left(\frac{z}{\sqrt{4N}}\right)\,\frac{1}{\log\left(M/c\right)}-Q_{c}\left(\frac{z}{\sqrt{4N}}\right)\,\frac{1}{\log\left((c+1)/c\right)}.$
(5.27)

Taking $M\rightarrow\infty$ in both expressions (5.26) and (5.27) and also
invoking (5.23), we see that

$\displaystyle f_{1}\left(\frac{z}{\sqrt{4N}}\right)$
$\displaystyle=\sum_{m=c+1}^{\ell-1}P_{m}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}+\frac{f\left(z/\sqrt{4N}\right)}{\log(\ell/c)}-P_{c}\left(\frac{z}{\sqrt{4N}}\right)\frac{1}{\log\left((c+1)/c\right)}$
$\displaystyle+\sum_{m=\ell}^{\infty}\,Q_{m}\left(\frac{z}{\sqrt{4N}}\right)\,\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}+Q_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log(\ell/c)}-\frac{1}{\log\left((\ell+1)/c\right)}\right\\},$
(5.28)

whenever $\ell\geq c+1$, while

$f_{1}\left(\frac{z}{\sqrt{4N}}\right)=\sum_{m=c+1}^{\infty}Q_{m}\left(\frac{z}{\sqrt{4N}}\right)\left\\{\frac{1}{\log\left(m/c\right)}-\frac{1}{\log\left((m+1)/c\right)}\right\\}-Q_{c}\left(\frac{z}{\sqrt{4N}}\right)\,\frac{1}{\log\left((c+1)/c\right)},$
(5.29)

if $1\leq\ell\leq c$.

The rest of the proof will consist in estimating the terms on the right-hand
sides of (5.28) and (5.29). We start by dealing with (5.28) and so we are
under the hypothesis that $\ell\geq c+1$. Using the mean value theorem for the
function $F(x)=1/\log(x/c)$, as well as the inequalities (5.22) and (5.23), we
see that there exists some constant $C_{1}>0$ such that

$\left|f_{1}\left(\frac{z}{\sqrt{4N}}\right)\right|<C_{1}\left\\{\sum_{m=2}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(m+1)}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+\frac{\left|f\left(z/\sqrt{4N}\right)\right|}{\log(\ell/c)}+\frac{\left|Q_{\ell}\left(z/\sqrt{4N}\right)\right|}{\ell\log^{2}\left(\ell/c\right)}+\frac{\left|P_{c}\left(z/\sqrt{4N}\right)\right|}{\log\left((c+1)/c\right)}\right\\}.$
(5.30)

If $\frac{\sqrt{N}}{\pi c}\geq\text{Im}(z)\geq\text{Im}(z_{0})>0$, then
$P_{c}(z/\sqrt{4N})\ll e^{-2\pi\text{Im}(z)/\sqrt{4N}}$. Thus, (5.30) gives

$\displaystyle\left|f_{1}\left(\frac{z}{\sqrt{4N}}\right)\right|$
$\displaystyle<C\,\left\\{\sum_{m=2}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(m+1)}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+\frac{\left|f\left(z/\sqrt{4N}\right)\right|}{\log(\ell/c)}+\frac{\left|Q_{\ell}\left(z/\sqrt{4N}\right)\right|}{\ell\log^{2}\left(\ell/c\right)}\right\\}$
$\displaystyle\leq
C\,\left\\{e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}}\sum_{m=2}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}m^{\frac{k}{2}-\frac{3}{4}}+\frac{\left|f\left(z/\sqrt{4N}\right)\right|}{\log(\ell/c)}+\frac{\left|Q_{\ell}\left(z/\sqrt{4N}\right)\right|}{\ell\log^{2}\left(\ell/c\right)}\right\\}$
$\displaystyle<C^{\prime}\,\left\\{\frac{e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}}}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}+\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-1}\left|f\left(\frac{z}{\sqrt{4N}}\right)\right|+\frac{\text{Im}(z)\left|Q_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\right|}{\sqrt{N}}\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-2}\right\\}$
$\displaystyle<D\,e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}},$ (5.31)

for some positive $D$. At the final step we have used the fact that
$\ell>\frac{\sqrt{N}}{\pi\text{Im}(z)}$, as well as the elementary estimate

$\sum_{m=2}^{\infty}m^{h-1}e^{-m\alpha}=O\left(\alpha^{-h}\right),\,\,\,\,h,\alpha>0.$
(5.32)

Moreover, the bounds for $\left|f\left(z/\sqrt{4N}\right)\right|$ and
$\left|Q_{\ell}(z/\sqrt{4N})\right|$ are obtained using the elementary
observation that, for $\frac{\sqrt{N}}{\pi
c}\geq\text{Im}(z)\geq\text{Im}(z_{0})>0$,

$\displaystyle\left|f\left(\frac{z}{\sqrt{4N}}\right)\right|$
$\displaystyle\leq
e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}}\,\sum_{n=1}^{\infty}|a_{f}(n)|\,e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(n-1)}$
$\displaystyle\leq
e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}}\,\sum_{n=1}^{\infty}|a_{f}(n)|\,e^{-\frac{2\pi\text{Im}(z_{0})}{\sqrt{4N}}(n-1)}=O\left(e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}}\right).$
(5.33)

Analogously, for the case where $1\leq\ell\leq c$ we already know that
$\text{Im}(z)\geq\frac{\sqrt{N}}{\pi c}$, so we can bound (5.29) using (5.33)
with $\text{Im}(z_{0})$ replaced by $\frac{\sqrt{N}}{\pi c}$. Therefore,
(5.31) holds also in the second case $1\leq\ell\leq c$. This gives our first
bound (5.18).

We will now get (5.19). From now on we shall assume that
$0<\text{Im}(z)<\min\left\\{\frac{\sqrt{N}}{3\pi},\frac{\sqrt{N}}{\pi
c}\right\\}$ and our goal will be to prove that the estimate

$f_{1}\left(\frac{z}{\sqrt{4N}}\right)\ll\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\,\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}$
(5.34)

holds in this range. By the hypothesis
$0<\text{Im}(z)<\min\left\\{\frac{\sqrt{N}}{3\pi},\frac{\sqrt{N}}{\pi
c}\right\\}$, we know that $\ell\geq\max\\{4,c+1\\}$ by definition of $\ell$.
Thus, in order to prove (5.34), we will use the expression for
$f_{1}(z/\sqrt{4N})$ holding for $\ell\geq c+1$, which is (5.28). Recalling
the starting point (5.30) and some of the steps given in (5.31), we know that
$f_{1}(z/\sqrt{4N})$ can be bounded in the form

$\displaystyle\left|f_{1}\left(\frac{z}{\sqrt{4N}}\right)\right|\ll
e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}}\sum_{m=2}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}m^{\frac{k}{2}-\frac{3}{4}}+\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-1}\left|f\left(\frac{z}{\sqrt{4N}}\right)\right|$
$\displaystyle+\frac{\text{Im}(z)}{\sqrt{N}}\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-2}\left|Q_{\ell}(z/\sqrt{4N})\right|+\frac{\left|P_{c}\left(z/\sqrt{4N}\right)\right|}{\log\left((c+1)/c\right)}.$
(5.35)

From now on, we shall proceed by estimating each term individually. First, by
(5.20), $P_{c}\left(z/\sqrt{4N}\right)=O(1)$ in the range
$0<\text{Im}(z)<\min\left\\{\frac{\sqrt{N}}{3\pi},\frac{\sqrt{N}}{\pi
c}\right\\}$. Furthermore, using the uniform bound (2.1) for
$f\left(z/\sqrt{4N}\right)$, we know that

$\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-1}\left|f\left(\frac{z}{\sqrt{4N}}\right)\right|\ll\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}.$
(5.36)

In the meantime, in order to treat $|Q_{\ell}\left(z/\sqrt{4N}\right)|$, we
use (5.21) and (2.1) to obtain

$\displaystyle\left|Q_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\right|$
$\displaystyle\leq\left|f\left(\frac{z}{\sqrt{4N}}\right)\right|+\left|P_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\right|\leq\frac{\mathcal{A}}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}+\mathcal{B}_{1}\ell^{\frac{k}{2}+\frac{1}{4}}\log(\ell)\,e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}(\ell+1)}$
$\displaystyle\leq\frac{\mathcal{A}}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}+\frac{\mathcal{B}_{2}}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\log\left(\frac{1}{\text{Im}(z)}\right),$

which, in its turn, provides the estimate

$\displaystyle\frac{\text{Im}(z)}{\sqrt{N}}\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-2}\left|Q_{\ell}\left(\frac{z}{\sqrt{4N}}\right)\right|$
$\displaystyle\ll\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-2}\left\\{\frac{1}{\text{Im}(z)^{\frac{k}{2}-\frac{3}{4}}}+\frac{1}{\text{Im}(z)^{\frac{k}{2}-\frac{3}{4}}}\log\left(\frac{1}{\text{Im}(z)}\right)\right\\}$
$\displaystyle\ll\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}.$
(5.37)

If we now return to (5.35) and combine (5.36) with (5.37), we deduce that

$f_{1}\left(\frac{z}{\sqrt{4N}}\right)=O\left(\sum_{m=2}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}\right)+O\left(\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}\right)+O(1).$
(5.38)

The second $O$ term already contains the desired bound (5.34). We shall now
estimate the first $O$ term containing the infinite series by using (5.32),
from which we deduce

$\displaystyle\sum_{m=2}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}$
$\displaystyle=\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+\sum_{m=\ell+1}^{\infty}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}$
$\displaystyle<\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+\frac{1}{\log(\ell)}\sum_{m=\ell+1}^{\infty}m^{\frac{k}{2}-\frac{3}{4}}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}$
$\displaystyle<\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+O\left\\{\left(\log\left(\frac{\sqrt{N}}{\pi\text{Im}(z)}\right)\right)^{-1}\text{Im}(z)^{-\frac{k}{2}-\frac{1}{4}}\right\\}$
$\displaystyle=\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+O\left\\{\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}\text{Im}(z)^{-\frac{k}{2}-\frac{1}{4}}\right\\}.$
(5.39)

The second term of (5.39) already contains what we want, this is, (5.34).
Therefore, if we show that a similar estimate takes place for the finite sum
in (5.39), we are done. This final estimate will depend on the range of $k$,
for which we have two possibilities: $k=1$ or $k\geq 2$.

  1. 1.

If $k=1$, we need to sum over the expression
$\frac{1}{m^{\frac{1}{4}}\log(m)}$. Since the real function
$f(x)=\frac{1}{x^{1/4}\log(x)}$ is steadily decreasing for $x\geq 2$, we have

$\displaystyle\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{-\frac{1}{4}}}{\log(m)}$
$\displaystyle<\sum_{m=2}^{\ell}\frac{m^{-\frac{1}{4}}}{\log(m)}<\frac{2^{-\frac{1}{4}}}{\log(2)}+\intop_{2}^{\ell}\frac{x^{-1/4}}{\log(x)}\,dx$
$\displaystyle=\frac{2^{-\frac{1}{4}}}{\log(2)}$
$\displaystyle+\intop_{2}^{\sqrt{\ell}}\frac{x^{-1/4}}{\log(x)}\,dx+\intop_{\sqrt{\ell}}^{\ell}\frac{x^{-1/4}}{\log(x)}\,dx$
$\displaystyle<\frac{2^{-\frac{1}{4}}}{\log(2)}$
$\displaystyle+\frac{4\ell^{\frac{3}{8}}}{3\log(2)}+\frac{8\ell^{\frac{3}{4}}}{3\log(\ell)}<\frac{4\ell^{\frac{3}{8}}}{3\log(2)}+\frac{16\ell^{\frac{3}{4}}}{3\log(\ell)}.$
(5.40)

The real function $h(x):=x^{-\frac{3}{8}}\log(x)$ has a maximum equal to
$\frac{8}{3e}$ which is attained at the point $x=e^{\frac{8}{3}}$. Thus, for
every $\ell\geq 4$,

$\ell^{-\frac{3}{8}}<\frac{16}{3e\,\log(\ell)}.$ (5.41)

Using (5.41) in (5.40), one finds that

$\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{-\frac{1}{4}}}{\log(m)}<\frac{4\ell^{\frac{3}{8}}}{3\log(2)}+\frac{16\ell^{\frac{3}{4}}}{3\log(\ell)}<\left(1+\frac{4}{3\log(2)}\right)\frac{16\,\ell^{\frac{3}{4}}}{3\log(\ell)}:=\frac{A_{1}}{\log(\ell)}\ell^{\frac{3}{4}}.$
(5.42)

Recalling once more that
$\ell:=1+\left[\frac{\sqrt{N}}{\pi\text{Im}(z)}\right]$, (5.42) yields

$\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{-\frac{1}{4}}}{\log(m)}<\frac{A_{1}}{\log(\ell)}\ell^{\frac{3}{4}}=O\left(\frac{1}{\text{Im}(z)^{\frac{3}{4}}}\left\\{\log\left(\frac{1}{\text{Im}(z)}\right)\right\\}^{-1}\right),$
(5.43)

which proves (5.34) for $k=1$.

  2. 2.

For $k\geq 2$, it is even easier to get an estimate of the form (5.43).
Indeed,

$\displaystyle\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}$
$\displaystyle<\sum_{m=2}^{\ell}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}=\sum_{m=2}^{\sqrt{\ell}}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}+\sum_{m=\sqrt{\ell}+1}^{\ell}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}$
$\displaystyle<\frac{\ell^{\frac{k}{4}+\frac{1}{8}}}{\log(2)}+2\frac{\ell^{\frac{k}{2}+\frac{1}{4}}}{\log(\ell)}.$

As in (5.41), we know that $h(x):=x^{-\frac{k}{4}-\frac{1}{8}}\log(x)$ has a
maximum equal to $\frac{8}{(2k+1)e}$, which is attained at the point
$x=e^{\frac{8}{2k+1}}$. Therefore, for every $\ell\geq 4$, the inequality
takes place

$\ell^{-\frac{k}{4}-\frac{1}{8}}<\frac{16}{(2k+1)e\,\log(\ell)},$

and so, since $\ell=1+\left[\frac{\sqrt{N}}{\pi\text{Im}(z)}\right]$,

$\displaystyle\sum_{m=2}^{\ell}e^{-\frac{2\pi\text{Im}(z)}{\sqrt{4N}}m}\frac{m^{\frac{k}{2}-\frac{3}{4}}}{\log(m)}$
$\displaystyle<\frac{\ell^{\frac{k}{4}+\frac{1}{8}}}{\log(2)}+2\frac{\ell^{\frac{k}{2}+\frac{1}{4}}}{\log(\ell)}<\left(1+\frac{8}{(2k+1)e\log(2)}\right)\,\frac{2\ell^{\frac{k}{2}+\frac{1}{4}}}{\log(\ell)}$
$\displaystyle=O\left(\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\left\\{\log\left(\frac{1}{\text{Im}(z)}\right)\right\\}^{-1}\right),$

which finally proves (5.34).

∎

######  Remark 5.3.

In the same lines as those of Remark 5.2, it is clear from the properties of
the Fricke involution that, if we consider

$\left(f|W_{4N}\right)_{1}\left(\frac{z}{\sqrt{4N}}\right):=\sum_{n=d+1}^{\infty}\frac{a_{f|W_{4N}}(n)}{\log(n/d)}\,e^{\frac{2\pi
in}{\sqrt{4N}}z},$

then, for small $\text{Im}(z)$, $\left(f|W_{4N}\right)_{1}\left(z\right)$
obeys to the estimate

$\left(f|W_{4N}\right)_{1}\left(\frac{z}{\sqrt{4N}}\right)\ll\frac{1}{\text{Im}(z)^{\frac{k}{2}+\frac{1}{4}}}\,\left(\log\left(\frac{1}{\text{Im}(z)}\right)\right)^{-1}.$

######  Remark 5.4.

An alternative proof of the above lemma, which is short but invokes Fourier
analysis, is given for $L-$functions associated with Maass cusp forms in [[8],
pp.126-127]. Finding a bound for (5.17) is equivalent to finding a bound for

$f_{1}\left(z\right):=\sum_{n=c+1}^{\infty}\frac{a_{f}(n)}{\log(n/c)}\,e^{2\pi
inz}.$

It is clear that $f_{1}(x+iy)$ can be written in terms of the convolution

$f_{1}(x+iy)=\intop_{0}^{1}f\left(t+iy\right)\mathcal{H}(x-t)\,dt,$

where

$\mathcal{H}(x):=\sum_{n=c+1}^{\infty}\frac{\cos(2\pi
nx)}{\log\left(n/c\right)}.$ (5.44)

The convexity of the sequence $(b_{n})_{n\geq c+1}:=\frac{1}{\log(n/c)}$
assures that $\mathcal{H}\in L^{1}\left([0,2\pi)\right)$ [[17], p.23]. Hence,
the proof of lemma 5.3 follows from an estimate of the $L^{1}$ modulus of
continuity of $\mathcal{H}(x)$ combined with an estimate of the form (5.2).
See [[8], p.127] for details.

##  6 Proof of Theorem 1.2

As is customary, any variant of the Hardy-Littlewood method starts by taking a
positive number $H$ (to be chosen at a later stage of the proof), some
$0<\epsilon<\frac{\pi}{2}$ (to be very small later), and by considering the
integrals (with $t\in\mathbb{R}$)

$I(t):=\intop_{t}^{t+H}R_{f}(u)e^{\left(\frac{\pi}{2}-\epsilon\right)u}\,du$
(6.1)

and

$J(t):=\intop_{t}^{t+H}\left|R_{f}(u)\right|e^{\left(\frac{\pi}{2}-\epsilon\right)u}du.$
(6.2)

The idea of the proof is to contrast the behavior of the integrals $I(t)$ and
$J(t)$ as $t\rightarrow\infty$.

We start the argument by estimating $I(t)$.

###  6.1 Estimating $I(t)$

The estimate of $I(t)$ is done via the representation (3.14), in which we
consider $\xi$ real, pick $0<\epsilon<\frac{\pi}{2}$ and take
$z=-\exp\left\\{\xi-i\epsilon\right\\}$ (which clearly satisfies the condition
$\text{Im}(z)>0$). We can therefore rewrite (3.14) as the Fourier transform

$\displaystyle\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}R_{f}(t)e^{\left(\frac{\pi}{2}-\epsilon\right)t}\,e^{-i\xi
t}\,dt=\sqrt{\frac{\pi}{2}}\,\exp\left\\{\left(\frac{k}{2}+\frac{1}{4}\right)\xi+i\left(\frac{k}{2}+\frac{1}{4}\right)\left(\frac{\pi}{2}-\epsilon\right)\right\\}$
$\displaystyle\times\left[f\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)+\left(f|W_{4N}\right)\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)\right].$
(6.3)

From the Phragmén-Lindelöf principle (2.8) and Stirling’s formula (2.7), we
know that, as $|t|\rightarrow\infty$,

$R_{f}(t)e^{\left(\frac{\pi}{2}-\epsilon\right)t}=O\left(|t|^{A}\,\exp\left\\{-\frac{\pi}{2}\left(|t|-t\right)-\epsilon
t\right\\}\right),$ (6.4)

for some $A>0$. This estimate proves that
$|R_{f}(t)|\,e^{\left(\frac{\pi}{2}-\epsilon\right)t}$ and
$|R_{f}(t)|^{2}\,e^{\left(\pi-2\epsilon\right)t}$ are integrable over
$\mathbb{R}$ and the same must be true for the function $I(t)$ defined by
(6.1), as well as to $|I(t)|$ and $|I(t)|^{2}$. An integration by parts gives
the Fourier transform

$\displaystyle\hat{I}(\xi)$
$\displaystyle:=\frac{1}{\sqrt{2\pi}}\,\intop_{-\infty}^{\infty}I(t)\,e^{-i\xi
t}dt=\frac{1}{\sqrt{2\pi}}\,\intop_{-\infty}^{\infty}e^{-i\xi
t}\,\intop_{t}^{t+H}R_{f}(u)\,e^{\left(\frac{\pi}{2}-\epsilon\right)u}\,du\,dt$
$\displaystyle=\frac{1}{i\xi\,\sqrt{2\pi}}\,\intop_{-\infty}^{\infty}e^{-i\xi
t}\,\left\\{R_{f}(t+H)\,e^{\left(\frac{\pi}{2}-\epsilon\right)(t+H)}-R_{f}(t)\,e^{\left(\frac{\pi}{2}-\epsilon\right)t}\right\\}\,dt$
$\displaystyle=\sqrt{\frac{\pi}{2}}\,\frac{e^{i\xi
H}-1}{i\xi}\,\exp\left\\{\left(\frac{k}{2}+\frac{1}{4}\right)\xi+i\left(\frac{k}{2}+\frac{1}{4}\right)\left(\frac{\pi}{2}-\epsilon\right)\right\\}\left[f\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)+\left(f|W_{4N}\right)\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)\right],$
(6.5)

and so, by Parseval’s formula,

$\intop_{-\infty}^{\infty}|I(t)|^{2}dt=2\pi\intop_{-\infty}^{\infty}\frac{\sin^{2}\left(\frac{\xi
H}{2}\right)}{\xi^{2}}e^{\left(k+\frac{1}{2}\right)\xi}\,\left|f\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)+\left(f|W_{4N}\right)\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)\right|^{2}d\xi.$
(6.6)

We now give an estimate for the latter integral: invoking the uniform bound
for cusp forms (2.1), we know that there exist two positive constants $A_{1}$
and $A_{2}$ such that

$\left|f\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)\right|\leq\frac{A_{1}}{\sin^{\frac{k}{2}+\frac{1}{4}}(\epsilon)}e^{-\left(\frac{k}{2}+\frac{1}{4}\right)\xi}$
(6.7)

and

$\left|\left(f|W_{4N}\right)\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)\right|\leq\frac{A_{2}}{\sin^{\frac{k}{2}+\frac{1}{4}}(\epsilon)}e^{-\left(\frac{k}{2}+\frac{1}{4}\right)\xi}.$
(6.8)

Supposing, without any loss of generality, that $A_{1}\geq A_{2}$ and
returning to (6.7), we find the mean inequality for $I(t)$,

$\intop_{-\infty}^{\infty}|I(t)|^{2}dt<\frac{8\pi
A_{1}^{2}}{\sin^{k+\frac{1}{2}}(\epsilon)}\,\intop_{-\infty}^{\infty}\frac{\sin^{2}\left(\frac{\xi
H}{2}\right)}{\xi^{2}}d\xi<K_{1}\epsilon^{-k-\frac{1}{2}}H,$ (6.9)

where we have used the fact that $0<\epsilon<\frac{\pi}{2}$ and invoked
Jordan’s inequality, $\sin(\epsilon)>\frac{2}{\pi}\epsilon$. The constant
$K_{1}$ is, of course, absolute, not depending on $H$ or $\epsilon$.

###  6.2 Estimating $J(t)$

Now, let us define the Dirichlet series associated with $R_{f}(t)$ (3.1),

$\varphi_{f}(s):=\frac{1}{2}\left(L(s,f)+L(s,f|W_{4N})\right)=\frac{1}{2}\sum_{n=1}^{\infty}\frac{a_{f}(n)+a_{f|W_{4N}}(n)}{n^{s}}:=\sum_{n=1}^{\infty}\frac{\alpha_{f}(n)}{n^{s}},$

and assume that $r$ is the smallest positive integer for which
$\alpha_{f}(r)\neq 0$. We will apply the reasoning of Lemma 5.2 to the
Dirichlet series

$\varphi_{f}^{\star}(s):=r^{s}\varphi_{f}(s)-\alpha_{f}(r).$

From Stirling’s formula, we know that there exist $T_{1},K_{2}>0$ such that

$|R_{f}(t)|>K_{2}\,|t|^{\frac{k}{2}-\frac{1}{4}}\,e^{-\frac{\pi}{2}|t|}\,\left|\varphi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)\right|,\,\,\,\,\,|t|\geq
T_{1}.$

With the construction of this number $T_{1}$, let

$T>T_{0}:=\max\left\\{1,6H,T_{1}\right\\},$ (6.10)

and assume that $t\in[T,2T]$ (recall that this is the argument of $I(t)$ and
$J(t)$). If we now take the choice $\epsilon=T^{-1}$ in our integrals (6.1)
and (6.2), the condition $t\leq u\leq t+H$ implies that $\epsilon
u<\epsilon(2T+H)<3$ and
$u^{\frac{k}{2}-\frac{1}{4}}>T^{\frac{k}{2}-\frac{1}{4}}$. Consequently, we
find the inequality

$\displaystyle J(t)$
$\displaystyle=\intop_{t}^{t+H}\left|R_{f}(u)\right|e^{\left(\frac{\pi}{2}-\epsilon\right)u}du>K^{\prime}\,T^{\frac{k}{2}-\frac{1}{4}}\,\intop_{t}^{t+H}\left|\varphi_{f}\left(\frac{k}{2}+\frac{1}{4}+iu\right)\right|\,du$
$\displaystyle=K^{\prime}\,T^{\frac{k}{2}-\frac{1}{4}}\,\intop_{t}^{t+H}\left|\alpha_{f}(r)\,r^{-\frac{k}{2}-\frac{1}{4}-iu}+r^{-\frac{k}{2}-\frac{1}{4}-iu}\varphi_{f}^{\star}\left(\frac{k}{2}+\frac{1}{4}+iu\right)\right|\,du$
$\displaystyle=K^{\prime}|\alpha_{f}(r)|\,r^{-\frac{k}{2}-\frac{1}{4}}\,T^{\frac{k}{2}-\frac{1}{4}}\,\intop_{t}^{t+H}\left|1+\frac{1}{\alpha_{f}(r)}\varphi_{f}^{\star}\left(\frac{k}{2}+\frac{1}{4}+iu\right)\right|\,du$
$\displaystyle\geq
K^{\prime}|\alpha_{f}(r)|\,r^{-\frac{k}{2}-\frac{1}{4}}\,T^{\frac{k}{2}-\frac{1}{4}}\,\left\\{H+\text{Re}\intop_{t}^{t+H}\frac{1}{\alpha_{f}(r)}\varphi_{f}^{\star}\left(\frac{k}{2}+\frac{1}{4}+iu\right)\,du\right\\}$
$\displaystyle=K_{3}\,T^{\frac{k}{2}-\frac{1}{4}}\left\\{H+\text{Re}\intop_{t}^{t+H}\frac{1}{\alpha_{f}(r)}\varphi_{f}^{\star}\left(\frac{k}{2}+\frac{1}{4}+iu\right)\,du\right\\}.$
(6.11)

Analogously to $L(s,f_{1})$ in (5.12), we can construct the Dirichlet series

$\psi_{f}(s):=\sum_{n=r+1}^{\infty}\frac{\alpha_{f}(n)}{\log(n/r)\,n^{s}},\,\,\,\,\,\text{Re}(s)>\frac{k}{2}+\frac{5}{4},$
(6.12)

and, due to Lemma 5.2, say that $\psi_{f}(s)$ can be continued to the whole
complex plane as an entire function which satisfies

$r^{-s}\intop_{s}^{\infty}\varphi_{f}^{\star}(z)\,dz=\psi_{f}(s).$ (6.13)

Employing (6.13), we find the representation

$\frac{1}{\alpha_{f}(r)}\intop_{t}^{t+H}\varphi_{f}^{\star}\left(\frac{k}{2}+\frac{1}{4}+iu\right)\,du=-\frac{ir^{\frac{k}{2}+\frac{1}{4}}}{\alpha_{f}(r)}\left\\{r^{it}\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)-r^{i(t+H)}\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+i(t+H)\right)\right\\}.$
(6.14)

Let $\Psi(t)$ denote the right-hand side of (6.14): then (6.11) says that

$J(t)\geq
K_{3}\,T^{\frac{k}{2}-\frac{1}{4}}\left\\{H+\text{Re}\,\Psi(t)\right\\},$
(6.15)

with $K_{3}$ not depending on any of the parameters $H,$ $t$ or $T$. If, at
last, we deduce a suitable estimate for $\intop_{T}^{2T}|\Psi(t)|^{2}dt$, we
can finish the application of the Hardy-Littlewood method. The next claim
contains the necessary fact to conclude the proof of our result.

######  Claim 6.1.

For large $T$, there exists some absolute constant $K_{4}>0$ such that

$\intop_{T}^{2T}|\Psi(t)|^{2}dt\leq K_{4}\,T.$ (6.16)

Before giving a proof of this claim, we show how this assertion can be used to
complete the proof of Theorem 1.2. From now on, the reasoning is standard and
the corresponding known argument for the zeta case can be found in
Titchmarsh’s text [[27], pp.267-268]. However, for completeness, we shall also
present it. Let $S$ be the subset of $(T,2T)$ where $|I(t)|=J(t)$. Note that
this happens if and only if $R_{f}(u)$ does not have a zero of odd order on
the interval $(t,t+H)$. We shall estimate the measure of $S$, $m(S)$, by
comparing (6.11) and (6.15). Indeed, using (6.15) and claim 6.1, we get

$\displaystyle\intop_{S}J(t)\,dt$ $\displaystyle\geq
K_{3}T^{\frac{k}{2}-\frac{1}{4}}\left\\{Hm(S)+\intop_{S}\text{Re}\,\Psi(t)\,dt\right\\}\geq
K_{3}HT^{\frac{k}{2}-\frac{1}{4}}\,m(S)-K_{3}T^{\frac{k}{2}+\frac{1}{4}}\left\\{\intop_{T}^{2T}|\Psi(t)|^{2}dt\right\\}^{1/2}$
$\displaystyle\geq
K_{3}HT^{\frac{k}{2}-\frac{1}{4}}\,m(S)-K_{3}K_{4}^{\frac{1}{2}}\,T^{\frac{k}{2}+\frac{3}{4}}.$
(6.17)

On the other hand, we can bound the previous integral by above if we use (6.9)
with $\epsilon=T^{-1}$, which gives

$\intop_{S}J(t)\,dt=\intop_{S}|I(t)|\,dt\leq\intop_{T}^{2T}|I(t)|\,dt\leq
T^{1/2}\left\\{\intop_{T}^{2T}|I(t)|^{2}dt\right\\}^{1/2}\leq
K_{1}^{\frac{1}{2}}H^{\frac{1}{2}}T^{\frac{k}{2}+\frac{3}{4}}.$ (6.18)

Combining (6.17) and (6.18), we find an upper bound for $m(S)$ of the form

$K_{3}H\,m(S)<K_{1}^{\frac{1}{2}}H^{\frac{1}{2}}T+K_{3}K_{4}^{\frac{1}{2}}\,T.$

Now, we can choose $H$ so large that

$K_{4}^{\frac{1}{2}}H^{-1}+K_{1}^{\frac{1}{2}}K_{3}^{-1}H^{-\frac{1}{2}}<\frac{1}{12},$

and, with this choice of $H$, conclude that

$m(S)<\frac{T}{12}.$ (6.19)

The argument is completed if we subdivide $(T,2T)$ into $\left[T/2H\right]$
pairs of abutting subintervals in the form

$(T,2T)=\bigcup_{j=1}^{[T/2H]}\left(I_{1}^{(j)}\cup I_{2}^{(j)}\right),$

with $|I_{1}^{(j)}|=H$, $j=1,...,\left[\frac{T}{2H}\right]$ and
$|I_{2}^{(j)}|=H$, $j=1,...,\left[\frac{T}{2H}\right]-1$. Suppose that, for
some $\nu\in\mathbb{N}$, the intervals $I_{1}^{(j_{1})},$…,
$I_{1}^{(j_{\nu})}$ consist entirely of points of $S$. Since each of these
intervals has length $H$ and there are $\nu$ such intervals, then $\nu\leq
m(S)/H$. Of the remaining $\left[\frac{T}{2H}\right]-\nu$ pairs of
subintervals of $(T,2T)$, either $I_{1}^{(j)}$ or the corresponding
$I_{2}^{(j)}$ contains a zero of odd order of $R_{f}(t)$. By our choice of $T$
(6.10) and the measure of $S$ (6.19), we find that

$\left[\frac{T}{2H}\right]-\nu>\frac{T}{3H}-\frac{m(S)}{H}>\frac{T}{3H}-\frac{T}{12H}=\frac{T}{4H}.$

Hence, the interval $(T,2T)$ has at least $T/4H$ zeros of odd order of
$R_{f}(t)$. The same is evidently true for
$L\left(\frac{k}{4}+\frac{1}{2}+it,f\right)+L\left(\frac{k}{4}+\frac{1}{2}+it,f|W_{4N}\right)$
and, with the necessary changes, the same can be concluded for
$L\left(\frac{k}{4}+\frac{1}{2}+it,f\right)-L\left(\frac{k}{4}+\frac{1}{2}+it,f|W_{4N}\right)$.
This completes the proof. $\blacksquare$

####  6.2.1 Proof of Claim 6.1

Since, by definition,

$\Psi(t):=-\frac{ir^{\frac{k}{2}+\frac{1}{4}}}{\alpha_{f}(r)}\left\\{r^{it}\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)-r^{i(t+H)}\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+i(t+H)\right)\right\\},$

in order to prove (6.16) it is enough to show that

$\intop_{T}^{2T}\left|\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)\right|^{2}dt\ll
T,$ (6.20)

where

$\psi_{f}(s):=\sum_{n=r+1}^{\infty}\frac{\alpha_{f}(n)}{\log(n/r)\,n^{s}},\,\,\,\,\,\text{Re}(s)>\frac{k}{2}+\frac{5}{4}.$

Just like in the case of (5.17), we can construct a series resembling a cusp
form attached to $\psi_{f}(s)$ of the form

$\sum_{n=r+1}^{\infty}\frac{\alpha_{f}(n)}{\log(n/r)}e^{\frac{2\pi
inz}{\sqrt{4N}}}:=g_{1}\left(\frac{z}{\sqrt{4N}}\right).$ (6.21)

By adapting the proof of lemma 5.3, one can see that
$g_{1}\left(z/\sqrt{4N}\right)$ must satisfy the bounds given in (5.18) and
(5.19), which are dependent on the range of $\text{Im}(z)$. We will find the
estimate (6.20) as a consequence of a $L^{2}$ estimate of the function

$\Phi_{f}\left(t\right):=\left(\frac{2\pi}{\sqrt{4N}}\right)^{-\frac{k}{2}-\frac{1}{4}-it}\Gamma\left(\frac{k}{2}+\frac{1}{4}+it\right)\,\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right),$
(6.22)

which, in its turn, will be found by using the estimates for
$g_{1}(z/\sqrt{4N})$.

Just like (3.14), the inversion formula for the Mellin transform gives

$\frac{1}{2\pi}\,\intop_{-\infty}^{\infty}\Phi_{f}\left(t\right)(-iz)^{-\frac{k}{2}-\frac{1}{4}-it}dt=g_{1}\left(\frac{z}{\sqrt{4N}}\right),\,\,\,\,\forall
z\in\mathbb{H}.$ (6.23)

Let us now take (as above) $\xi\in\mathbb{R}$, $0<\epsilon<\frac{\pi}{2}$ and
write $z=-\exp\left\\{\xi-i\epsilon\right\\}$: then (6.23) can be viewed in
terms of the Fourier transform

$\frac{1}{\sqrt{2\pi}}\,\intop_{-\infty}^{\infty}\Phi_{f}(t)e^{\left(\frac{\pi}{2}-\epsilon\right)t}\,e^{-i\xi
t}\,dt=\sqrt{2\pi}\,\exp\left\\{\left(\frac{k}{2}+\frac{1}{4}\right)\xi+i\left(\frac{k}{2}+\frac{1}{4}\right)\left(\frac{\pi}{2}-\epsilon\right)\right\\}\,g_{1}\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)$

and so, by Parseval’s formula,

$\intop_{-\infty}^{\infty}|\Phi_{f}(t)|^{2}e^{\left(\pi-2\epsilon\right)t}\,dt=2\pi\,\intop_{-\infty}^{\infty}\exp\left\\{\left(k+\frac{1}{2}\right)\xi\right\\}\,\left|g_{1}\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{\sqrt{4N}}\right)\right|^{2}d\xi=2\pi\,\intop_{0}^{\infty}y^{k-\frac{1}{2}}\left|g_{1}\left(-\frac{ye^{-i\epsilon}}{\sqrt{4N}}\right)\right|^{2}\,dy$

By lemma 5.3 (adapted to $g_{1}(z/\sqrt{4N})$), we know that there exists some
$\eta\in(0,1)$ such that

$g_{1}\left(-\frac{ye^{-i\epsilon}}{\sqrt{4N}}\right)\ll\frac{\left\\{\log\left(\frac{1}{y\sin(\epsilon)}\right)\right\\}^{-1}}{(y\sin(\epsilon))^{\frac{k}{2}+\frac{1}{4}}},\,\,\,\,0<y\sin(\epsilon)<\eta<1,$
(6.24)

while

$g_{1}\left(-\frac{ye^{-i\epsilon}}{\sqrt{4N}}\right)\ll
e^{-\frac{2\pi\sin(\epsilon)}{\sqrt{4N}}y},\,\,\,\,y\sin(\epsilon)\geq\eta.$
(6.25)

Thus, invoking (6.24) and (6.25) in the respective ranges of $y$, we deduce
that

$\displaystyle\intop_{-\infty}^{\infty}|\Phi_{f}(t)|^{2}e^{\left(\pi-2\epsilon\right)t}\,dt=2\pi\,\intop_{0}^{\infty}y^{k-\frac{1}{2}}\left|g_{1}\left(-\frac{ye^{-i\epsilon}}{\sqrt{4N}}\right)\right|^{2}\,dy$
$\displaystyle<C\left\\{\intop_{0}^{\eta/\sin(\epsilon)}\frac{\left\\{\log\left(\frac{1}{y\sin(\epsilon)}\right)\right\\}^{-2}}{\sin^{k+\frac{1}{2}}(\epsilon)}\,\frac{dy}{y}+\intop_{\eta/\sin(\epsilon)}^{\infty}y^{k-\frac{1}{2}}e^{-\frac{4\pi\sin(\epsilon)}{\sqrt{4N}}y}\,dy\right\\}$
$\displaystyle<\frac{D}{\sin^{k+\frac{1}{2}}(\epsilon)}\left\\{\intop_{0}^{\eta}\frac{dv}{v\log^{2}(v)}+\intop_{\eta}^{\infty}v^{k-\frac{1}{2}}e^{-\frac{4\pi
v}{\sqrt{4N}}}\,dv\right\\}<D^{\prime}\epsilon^{-k-\frac{1}{2}},$ (6.26)

for some absolute constant $D^{\prime}>0$.

Using (6.26) and the relation between $\Phi_{f}(t)$ and
$\psi_{f}\left(\frac{k}{4}+\frac{1}{2}+it\right)$, we are ready to establish
(6.20). By Stirling’s formula, there exists some $T_{0}>0$ for which the
following inequality holds

$\left|\Gamma\left(\frac{k}{2}+\frac{1}{4}+it\right)\right|\geq\sqrt{\frac{\pi}{2}}\,\,|t|^{\frac{k}{2}-\frac{1}{4}}e^{-\frac{\pi}{2}|t|},\,\,\,\,\,\,|t|\geq
T_{0}.$ (6.27)

Thus, if we pick $T>\max\left\\{T_{0},\frac{2}{\pi}\right\\}$ and put
$\epsilon=T^{-1}$, we have from (6.22) and (6.22),

$\intop_{T}^{2T}|\Phi_{f}(t)|^{2}e^{\left(\pi-\frac{2}{T}\right)t}\,dt>A\,\intop_{T}^{2T}t^{k-\frac{1}{2}}\left|\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)\right|^{2}dt>A\,T^{k-\frac{1}{2}}\intop_{T}^{2T}\left|\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)\right|^{2}dt,$

for some positive constant $A$. Combining this inequality and (6.26) with
$\epsilon=T^{-1}$, we finally deduce

$\intop_{T}^{2T}\left|\psi_{f}\left(\frac{k}{2}+\frac{1}{4}+it\right)\right|^{2}dt<\frac{1}{AT^{k-\frac{1}{2}}}\,\intop_{T}^{2T}|\Phi_{f}(t)|^{2}e^{\left(\pi-\frac{2}{T}\right)t}\,dt<\frac{1}{AT^{k-\frac{1}{2}}}\intop_{-\infty}^{\infty}|\Phi_{f}(t)|^{2}e^{\left(\pi-\frac{2}{T}\right)t}\,dt<\frac{D^{\prime}}{A}\,T,$

which proves (6.20) and establishes our claim.

######  Remark 6.1.

The proof of the result stated as Theorem 1.3 follows the same ideas. Let

$\eta_{p/q}\left(s,f\right):=\left(\frac{2\pi}{q}\right)^{-s}\Gamma(s)\,L_{p/q}(s,f).$

We start by recalling the functional equation for $L_{p/q}(s,f)$ (1.11),

$\left(\frac{2\pi}{q}\right)^{-s}\Gamma(s)\,L_{p/q}(s,f)=i^{k+\frac{1}{2}}\left(\frac{-q}{p}\right)^{-2k-1}\epsilon_{p}^{2k+1}\,\left(\frac{2\pi}{q}\right)^{-\left(k+\frac{1}{2}-s\right)}\Gamma\left(k+\frac{1}{2}-s\right)\,L_{-\overline{p}/q}\left(k+\frac{1}{2}-s,f\right),$

where $p\,\overline{p}\equiv 1\mod\,q$ and the symbols
$\left(\frac{-q}{p}\right)$ and $\epsilon_{p}$ are defined by (1.1). One can
see that, if $p^{2}\equiv 1\mod q$ and the the Fourier coefficients of $f(z)$,
$a_{f}(n)$, are real numbers (resp. purely imaginary numbers) then the
function

$Z_{f,p/q}\left(t\right):=i^{-\frac{k}{2}-\frac{1}{4}}\left(\frac{-q}{p}\right)^{k+\frac{1}{2}}\epsilon_{p}^{k+\frac{1}{2}}\,\eta_{p/q}\left(\frac{k}{2}+\frac{1}{4}+it,f\right)$

is always real for any $t\in\mathbb{R}$ (resp. purely imaginary for any
$t\in\mathbb{R}$). Analogously to Lemma 3.1, it is then straightforward to
deduce the integral representation

$\intop_{-\infty}^{\infty}Z_{f,p/q}\left(t\right)\,(-iz)^{-it}dt=2\pi\,i^{k+\frac{1}{2}}\,\left(\frac{-q}{p}\right)^{k+\frac{1}{2}}\epsilon_{p}^{k+\frac{1}{2}}z^{\frac{k}{2}+\frac{1}{4}}f\left(\frac{z+p}{q}\right),\,\,\,\,\,z\in\mathbb{H}.$
(6.28)

If we consider $\xi\in\mathbb{R}$ and let $0<\epsilon<\frac{\pi}{2}$, then
substituting $z=-\exp\left\\{\xi-i\epsilon\right\\}$ in (6.28) yields the
Fourier transform

$\frac{1}{\sqrt{2\pi}}\intop_{-\infty}^{\infty}Z_{f,p/q}\left(t\right)e^{\left(\frac{\pi}{2}-\epsilon\right)t}e^{-i\xi
t}dt=\sqrt{2\pi}\,\left(\frac{-q}{p}\right)^{k+\frac{1}{2}}\epsilon_{p}^{k+\frac{1}{2}}\,\exp\left\\{\left(\frac{k}{2}+\frac{1}{4}\right)\xi-i\left(\frac{k}{2}+\frac{1}{4}\right)\epsilon\right\\}f\left(-\frac{\exp\left\\{\xi-i\epsilon\right\\}}{q}+\frac{p}{q}\right),$

which is analogous to (6.3). Since Lemmas 5.1-5.3 also apply to
$L_{p/q}(s,f)$, the bounds (6.9) and (6.16) are also valid in this case and
the conclusion of Theorem 1.3 must follow from them.

##  7 Concluding Remarks

Similar results to the ones described here can be established for
$L-$functions attached to cusp forms with integral weight and level $N$. In
this case, let us recall the definition of the Fricke involution as

$\left(f|W_{N}\right)(z)=i^{k}N^{-k/2}z^{-k}f\left(-\frac{1}{Nz}\right).$

If we adapt the proof of our Theorem 1.2, we can deduce the following integral
analogue.

######  Theorem 7.1.

Let $N$ be a perfect square and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi
inz}\in S_{k}\left(\Gamma_{0}(N)\right)$, with $a_{f}(n)$ being either real or
purely imaginary numbers.

If $\mathcal{N}_{0}^{\pm}(T)$ represents the number of zeros of odd order of
$L(s,f)\pm L(s,f|W_{N})$ written in the form $s=\frac{k}{2}+it,$ $0\leq t\leq
T$, then there exists some $d>0$ such that

$\liminf_{T\rightarrow\infty}\,\frac{\mathcal{N}_{0}^{\pm}(T)}{T}>d.$ (7.1)

Taking into consideration the simple modifications outlined in remark 6.1, a
similar extension holds for the twisted $L-$functions studied by Kim.

######  Theorem 7.2.

Let $N$ be a positive integer and $\frac{p}{q}$ a rational number which is
$\Gamma_{0}(N)-$equivalent to $i\infty$ and such that $p^{2}\equiv 1\mod q$.
Moreover, let $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi inz}\in
S_{k}\left(\Gamma_{0}(N)\right)$, with $a_{f}(n)$ being either real or purely
imaginary numbers.

If $\mathcal{N}_{0,p/q}(T)$ denotes the number of zeros of $L_{p/q}(s,f)$
written in the form $s=\frac{k}{2}+it,$ $0\leq t\leq T$, then there exists
some $d>0$ such that

$\liminf_{T\rightarrow\infty}\,\frac{\mathcal{N}_{0,p/q}(T)}{T}>d.$ (7.2)

Of course, when $f$ is a cusp form of weight $k$ with respect to the full
modular group, then (7.1) and (7.2) are given in Lekkerkerker’s thesis [[20],
Chapter IV, Theorems 15 and 16]. It is also well-known that Lekkerkerker’s
result was superseded by Hafner [10], who proved that $L-$functions attached
to holomorphic cusp forms of integral weight have a positive proportion of
their zeros at the critical line $\text{Re}(s)=\frac{k}{2}$.

When $f\in S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$, one might study the
distribution of the zeros of $L(s,f)$ at the line
$\text{Re}(s)=\frac{k}{2}+\frac{1}{4}$ by taking another point of view. For
instance, one may consider the problem of finding suitable bounds for the gaps
between consecutive zeros of $L\left(\frac{k}{2}+\frac{1}{4}+it,f\right)$.
When $f(z)$ is a cusp form of integral weight for the full modular group with
real or purely imaginary coefficients, it was found by Jutila [[16], p. 140,
Theorem 5] that, for any fixed $\epsilon>0$ and $T\geq T_{0}(\epsilon)$, there
is always a zero $\frac{k}{2}+i\tau$ of $L\left(s,f\right)$ such that
$\tau\in[T,T+T^{\frac{1}{3}+\epsilon}]$. The proof of Jutila is particularly
beautiful because it involves the transformation of some exponential sums via
Voronoï’s summation formula. Yogananda [[29], p. 21, Theorem 4.1] extended
Jutila’s result to certain $L-$functions attached to cusp forms belonging to
$S_{k}\left(\Gamma_{0}(N)\right)$.

An examination of Yogananda’s argument (which is motivated by a remark at the
end of Jutila’s paper [[16], p. 156]) shows that only Rankin’s mean value
estimate for the coefficients of $f$ (and not something stronger like
Deligne’s bound) is needed to establish this gap between zeros of $L(s,f)$.
Motivated by this, it might be plausible (provided that all transformation
formulas of Voronoï type are valid in this case) to conjecture an analogous
result for $L-$functions in the half-integral case.

######  Conjecture 7.1.

Let $N\in\mathbb{N}$ and $f(z)=\sum_{n=1}^{\infty}a_{f}(n)\,e^{2\pi inz}\in
S_{k+\frac{1}{2}}\left(\Gamma_{0}(4N)\right)$ be such that $f|W_{4N}=f$ or
$f|W_{4N}=-f$. Assume also that $a_{f}(n)$ are either real or purely imaginary
numbers. Then, for any $\epsilon>0$, there exists $T_{0}(\epsilon)$ such that,
for all $T\geq T_{0}(\epsilon)$, $L(s,f)$ has a zero of the form
$s=\frac{k}{2}+\frac{1}{4}+i\tau$ with
$\tau\in[T,T+T^{\frac{1}{3}+\epsilon}]$.

There are yet other directions of research that can be taken. Together with
Yakubovich [23], the author of this paper derived the following result.

##### Theorem D [23]:

Let $f(z)$ be a cusp form of weight $k$ for the full modular group with real
Fourier coefficients $a_{f}(n)$. Also, let $L(s,f)$ represent the
corresponding Dirichlet series.

If $(c_{j})_{j\in\mathbb{N}}$ is a sequence of non-zero real numbers such that
$\sum_{j=1}^{\infty}\,|c_{j}|<\infty$,
$\left(\lambda_{j}\right)_{j\in\mathbb{N}}$ is a bounded sequence of distinct
real numbers131313In the statement of Theorem 1.4. of [23] we require that the
sequence $\left(\lambda_{j}\right)_{j\in\mathbb{N}}$ attains its bounds.
However, in view of Remark 5.4 of [23], this condition is not necessary when
we are working with combinations involving entire Dirichlet series. and $z$ is
a complex number satisfying the condition

$z\in\mathscr{D}:=\left\\{z\in\mathbb{C}\,:\,|\text{Re}(z)|<2\sqrt{\pi},\,\,|\text{Im}(z)|<2\sqrt{\pi}\right\\},$
(7.3)

then the function

$\sum_{j=1}^{\infty}c_{j}\,\left(2\pi\right)^{-s-i\lambda_{j}}\Gamma(s+i\lambda_{j})\,L\left(s+i\lambda_{j},f\right)\,\left\\{{}_{1}F_{1}\left(k-s-i\lambda_{j};\,k;\,\frac{z^{2}}{4}\right)+\,_{1}F_{1}\left(k-\overline{s}+i\lambda_{j};\,k;\,\frac{\overline{z}^{2}}{4}\right)\right\\}$
(7.4)

has infinitely many zeros on the critical line $\text{Re}(s)=\frac{k}{2}$.

In the statement above, ${}_{1}F_{1}(a;c;w)$ represents the confluent
hypergeometric function. The previous result is, in fact, a generalization of
a Theorem of Dixit, Kumar, Maji and Zaharescu [7]. Like the proof presented in
[7], deriving it requires the use of a generalization of the theta
transformation formula. We conclude this paper by mentioning that, using our
variant of de la Vallée Poussin’s method given in the proof of Theorem 1.1, it
is possible to obtain an estimate for the number of critical zeros of the
function (7.4). The details of this study and related topics will be presented
elsewhere.

Acknowledgements: This work was partially supported by CMUP, member of LASI,
which is financed by national funds through FCT - Fundação para a Ciência e a
Tecnologia, I.P., under the projects with reference UIDB/00144/2020 and
UIDP/00144/2020. We also acknowledge the support from FCT (Portugal) through
the PhD scholarship 2020.07359.BD. The author would like to thank to Semyon
Yakubovich for unwavering support and guidance throughout the writing of this
paper.

## References

  * [1] B. C. Berndt, On the Zeros of a Class of Dirichlet Series I, Illinois J. Mathematics, 14 (1970), 244–258.
  * [2] ———, On the Zeros of a Class of Dirichlet Series II, Illinois J. Mathematics, 14 (1970), 678–691.
  * [3] J. C. Burkill, Charles-Joseph de la Vallée Poussin, Journal of the London Math. Soc. 39 (1964), 165–175.
  * [4] J. B. Conrey and H. Iwaniec, The cubic moment of central values of automorphic $L-$functions, Ann. of Math. 151 (2000), 1175-–1216.
  * [5] De la Vallée Poussin, Sur les zeros de $\zeta(s)$ de Riemann, Comptes rendus, 163 (1916), 418–421.
  * [6] De la Vallée Poussin, Sur les zeros de $\zeta(s)$ de Riemann, Comptes rendus, 163 (1916), 471–473.
  * [7] A. Dixit, R. Kumar, B. Maji and A. Zaharescu, Zeros of combinations of the Riemann $\Xi$-function and the confluent hypergeometric function on bounded vertical shifts, J. Math. Anal. Appl. 466 (2018), 307–323.
  * [8] C. Epstein, J. Hafner, P. Sarnak, Zeros of $L-$functions attached to Maass forms, Math. Zeit., 190 (1985) 113–128.
  * [9] James L. Hafner, Some remarks on odd Maass wave forms (and a correction to: “Zeros of $L-$functions attached to Maass forms”), Math. Zeit., 196 (1987), 129-–132.
  * [10] James L. Hafner, Zeros on the critical line for Dirichlet series attached to certain cusp forms, Math. Ann. 264 (1983), 21–37.
  * [11] James L. Hafner, Zeros on the critical line for Maass wave form $L-$functions, J. Reine Angew. Math., 377 (1987), 127–158.
  * [12] G. H. Hardy, Sur les zeros de la fonction $\zeta(s)$ de Riemann, Comptes rendus, 158 (1914), 1012–1014.
  * [13] G. H. Hardy and J. E. Littlewood, Contributions to the theory of the Riemann zeta-function and the distribution of primes, Acta Math., 41 (1916), 119-–196.
  * [14] G. H. Hardy and J. E. Littlewood, The zeros of Riemann’s zeta-function on the critical line, Math. Zeit. 10 (1921), 283-–317.
  * [15] H. Iwaniec, Topics in classical automorphic forms, Graduate Studies in Mathematics, vol. 17, American Mathematical Society, Providence, RI, 1997.
  * [16] M. Jutila, Transformation Formulae for Dirichlet Polynomials, Journal of Number Theory, 18 (1984), 135–156.
  * [17] Y. Katznelson. An introduction to harmonic analysis, Dover Publications Inc., New York, corrected edition, 1976.
  * [18] D. Kim, Infinitely many zeros of additively twisted $L-$functions on the critical line, J. Numb. Theory, 253 (2023), 157–187.
  * [19] E. Landau, Über die Hardysche Entdeckung unendlich vieler Nullstellen der Zetafunktion mit reellem Teil $\frac{1}{2}$, Math. Annalen, 76 (1915), 212–243.
  * [20] C. G. Lekkerkerker, On the zeros of a class of Dirichlet series, Dissertation, Utrecht, 1955.
  * [21] J. Meher, S. Pujahari and K. Shankhadhar, Zeros of $L-$functions attached to cusp forms of half-integral weight, Proceedings of the American Mathematical Society,  147 (2019), 131–143.
  * [22] J. Meher, S. Pujahari, S. Kotyada, Zeros of $L-$functions attached to modular forms of half-integral weight, Bull. Lond. Math. Soc. 49 (2017), 926-–936.
  * [23] P. Ribeiro and S. Yakubovich, Certain extensions of results of Siegel, Wilton and Hardy, preprint, arXiv:2304.09545.
  * [24] Goro Shimura, Modular forms: basics and beyond, Springer Monographs in Mathematics, Springer, New York, 2012.
  * [25] Goro Shimura, On modular forms of half integral weight, Ann. Math. 97 (1973), 440–-481.
  * [26] E. C. Titchmarsh, The Theory of Functions, 2nd ed., Oxford University Press, London, 1939.
  * [27] E. C. Titchmarsh, The Theory of The Riemann Zeta-Function, 2nd ed., The Clarendon Press, Oxford, 1986.
  * [28] J. R. Wilton, A note on Ramanujan’s arithmetical function $\tau$(n), Math. Proc. Cambridge Philos. Soc. 25 (1929), 121–129.
  * [29] C. S. Yogananda, Transformation formula for exponential sums involving Fourier coefficients of modular forms, Proc. Indian Acad. Sci. (Math. Sci.), 103 (04.1993), 1–25.
  * [30] H. Yoshida, On calculations of zeros of various $L-$functions, J. Math. Kyoto Univ., 35 (1995), 663–696.