# The Twelfth Moment of Hecke $L$-Functions in the Weight Aspect

Peter Humphries  Department of Mathematics, University of Virginia,
Charlottesville, VA 22904, USA<EMAIL_ADDRESS>https://sites.google.com/view/peterhumphries/ and  Rizwanur Khan  Department
of Mathematics, University of Mississippi, University, MS 38677, USA
<EMAIL_ADDRESS>
###### Abstract.

We prove an upper bound for the twelfth moment of Hecke $L$-functions
associated to holomorphic Hecke cusp forms of weight $k$ in a dyadic interval
$T\leq k\leq 2T$ as $T$ tends to infinity. This bound recovers the Weyl-
strength subconvex bound $L(1/2,f)\ll_{\varepsilon}k^{1/3+\varepsilon}$ and
shows that for any $\delta>0$, the sub-Weyl subconvex bound $L(1/2,f)\ll
k^{1/3-\delta}$ holds for all but $O_{\varepsilon}(T^{12\delta+\varepsilon})$
Hecke cusp forms $f$ of weight at most $T$. Our result parallels a related
result of Jutila for the twelfth moment of Hecke $L$-functions associated to
Hecke–Maaß cusp forms. The proof uses in a crucial way a spectral reciprocity
formula of Kuznetsov that relates the fourth moment of $L(1/2,f)$ weighted by
a test function to a dual fourth moment weighted by a different test function.

###### 2020 Mathematics Subject Classification:

11F66 (primary); 11F11 (secondary)

The first author was supported by the Simons Foundation (award 965056). The
second author was supported by the National Science Foundation Grants
DMS-2001183 and DMS-2140604 and the Simons Foundation (award 630985).

##  1\. Introduction

###  1.1. Main Result

Let $\mathcal{B}_{\mathrm{hol}}$ denote the set of holomorphic Hecke cusp
forms on the modular surface $\Gamma\backslash\mathbb{H}$, where
$\Gamma\coloneqq\operatorname{SL}_{2}(\mathbb{Z})$ denotes the modular group.
We denote by $k_{f}\in 2\mathbb{N}$ the weight of a holomorphic Hecke cusp
form $f\in\mathcal{B}_{\mathrm{hol}}$. We prove the following result
concerning the twelfth moment of the central value $L(1/2,f)$ of the Hecke
$L$-function of $f$ averaged over cusp forms of weight in a given dyadic
interval.

######  Theorem 1.1.

For $T\geq 1$, we have that

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{12}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{4+\varepsilon}.$

Note that the harmonic weight $1/L(1,\operatorname{ad}f)$ is positive and
satisfies the bounds

$\frac{1}{(\log k_{f})^{2}}\ll\frac{1}{L(1,\operatorname{ad}f)}\ll\log k_{f};$

in particular, the same bound $O_{\varepsilon}(T^{4+\varepsilon})$ holds for
the unweighted twelfth moment of $L(1/2,f)$. Upon dropping all but one term
and taking twelfth roots, we recover the well-known Weyl-strength subconvex
bound111Note, however, that we _input_ this Weyl-strength subconvex bound for
$L(1/2,f)$ during the course of the proof of Theorem 1.1; in particular,
Theorem 1.1 does not give a new proof of this bound.
$L(1/2,f)\ll_{\varepsilon}k_{f}^{1/3+\varepsilon}$; thus this should be
thought of as a _Weyl-strength_ bound for the twelfth moment of $L(1/2,f)$.
But in fact our result gives more information, showing for the first time that
the central values can only very rarely be as large as the Weyl bound. This is
a consequence of the density result given in Corollary 1.2.

###  1.2. Moments of Hecke $L$-Functions

Theorem 1.1 fits into an active field of the study of moments of central
values of $L$-functions in various families of $L$-functions (see in
particular [Sou21] for a recent survey on this and related topics). For the
family

$\mathcal{F}_{T}\coloneqq\\{f\in\mathcal{B}_{\mathrm{hol}}:T\leq k_{f}\leq
2T\\}$

of holomorphic cusp forms of weight $k_{f}$ lying in a dyadic interval
$[T,2T]$, asymptotic formulæ for the $k$-th moment

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{k}}{L(1,\operatorname{ad}f)}$

are known for $k\in\\{1,2\\}$ (see [BF21, Theorems 3.1, 6.4, 7.4, and 7.5] for
state of the art results); indeed, these are even known for the smaller family

$\mathcal{G}_{T}\coloneqq\\{f\in\mathcal{B}_{\mathrm{hol}}:k_{f}=T\\}$

with $T\in 4\mathbb{N}$. Essentially sharp upper bounds have been proven for
the third moment for $\mathcal{G}_{T}$ [Fro20, Theorem 1.3] (see also [Pen01,
Theorem 3.1.1]); current technology should also be able to obtain an
asymptotic for the third moment for $\mathcal{F}_{T}$ (cf. [Qi22, Theorem
1.4]). Essentially sharp upper bounds for the fourth moment for
$\mathcal{F}_{T}$ follow directly from the spectral large sieve, while
asymptotics for the fourth moment should be possible via current technology
(cf. [Ivi02]). Asymptotics for higher moments remain out of reach, though the
second author has proven an upper bound for the fifth moment for the family
$\mathcal{F}_{T}$ that is essentially sharp under the assumption of the
Selberg eigenvalue conjecture [Kha20, Theorem 1.1].

Tools from random matrix theory have been used to derive a conjectural recipe
for _arbitrary_ moments: [CFKRS05, Conjecture 1.5.5] posits an asymptotic
formula for

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\
k_{f}=T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{k}}{L(1,\operatorname{ad}f)}$

for each nonnegative integer $k$. In particular, it is conjectured that there
exists a positive arithmetic factor $a_{k}$ (given explicitly in terms of an
Euler product) and a positive geometric factor $g_{k}$ (given explicitly in
terms of an integral arising from random matrix theory) such that this $k$-th
moment is asymptotic to the product of these two factors times $T(\log
T)^{k(k-1)/2}$. Thus these random matrix theory heuristics suggest that
Theorem 1.1 is far from optimal: instead of an upper bound of size
$O_{\varepsilon}(T^{4+\varepsilon})$, we should expect an _asymptotic_
involving a main term of the form $c_{12}T^{2}(\log T)^{66}$ for some positive
constant $c_{12}$. On the other hand, unconditional improvements to Theorem
1.1 seemingly remain far out of reach, since any power-saving improvement
would break the Weyl barrier and yield a sub-Weyl subconvex bound of the form
$L(1/2,f)\ll k_{f}^{1/3-\delta}$ for some $\delta>0$, a well-known open
problem.

Theorem 1.1 parallels an analogous result of Jutila [Jut04, Theorem 2] on the
twelfth moment of Hecke $L$-functions associated to Hecke–Maaß cusp forms $f$
of spectral parameter $t_{f}$ in a dyadic interval $T\leq t_{f}\leq 2T$. We
also briefly note that there exist related results for twelfth moments of
other families of $L$-functions that similarly fall shy by a power from the
optimal asymptotic predicted by random matrix theory yet also yield Weyl-
strength subconvex bounds for individual $L$-functions. The quintessential
result in this regard is Heath-Brown’s bound for the twelfth moment of the
Riemann zeta function [H-B78, Theorem 1], which has subsequently been reproven
by different means by both Iwaniec [Iwa80, Theorem 4] and Jutila [Jut87,
Theorem 4.7] and extended more generally by Ivić [Ivi03, Chapter 8]. Finally,
bounds for the twelfth moment of Dirichlet $L$-functions have recently been
proven by Milićević and White [MW21, Theorem 1] and by Nunes [Nun21, Theorem
1.1]; the former is of Weyl-strength in the depth aspect, while the latter is
of Weyl-strength for smooth moduli. We remark that the twelfth moment bounds
for $\operatorname{GL}_{1}$ $L$-functions are somewhat easier than for
$\operatorname{GL}_{2}$ $L$-functions because they can be proven via short
interval second moment bounds, while the latter requires short fourth moment
bounds, as we discuss below.

###  1.3. Applications

Upper bounds or asymptotics for various moments of $L$-functions have many
applications, including information on the limiting distribution of
$L$-functions, upper bounds for the size of $L$-functions, and lower bounds
for the proportion of $L$-functions that are nonvanishing at the central
point; see [CS07] for further discussions of such applications. We highlight
two applications of Theorem 1.1 below.

The first application of Theorem 1.1 pertains to upper bounds for the
proportion of large values of $L(1/2,f)$, recalling the nonnegativity of
$L(1/2,f)$ [KZ81, Corollary 1]. Indeed, while Theorem 1.1 does not yield
improved _individual_ subconvex bounds for $L(1/2,f)$ beyond the Weyl bound
$L(1/2,f)\ll_{\varepsilon}k_{f}^{1/3+\varepsilon}$, it does show strong
_averaged_ subconvex bounds, in the form of the following density result (cf.
[H-B78, Theorem 2], [Ivi03, Corollary 8.1], [MW21, Theorem 2], and [Nun21,
Theorem 1.2]).

######  Corollary 1.2.

For $T\geq 1$ and $V>0$, we have that

(1.3) $\\#\left\\{f\in\mathcal{B}_{\mathrm{hol}}:T\leq k_{f}\leq 2T,\
L\left(\frac{1}{2},f\right)\geq V\right\\}\\\
\ll_{\varepsilon}\begin{dcases*}\frac{T^{2+\varepsilon}}{(1+V)^{4}}&for $0\leq
V\leq T^{2\vartheta}$,\\\ \frac{T^{2+2\vartheta+\varepsilon}}{V^{5}}&for
$T^{2\vartheta}\leq V\leq T^{\frac{2(1-\vartheta)}{7}}$,\\\
\frac{T^{4+\varepsilon}}{V^{12}}&for $T^{\frac{2(1-\vartheta)}{7}}\leq V\leq
T^{\frac{1}{3}}(\log T)^{\frac{13}{6}+\varepsilon}$,\\\ 0&for $V\geq
T^{\frac{1}{3}}(\log T)^{\frac{13}{6}+\varepsilon}$,\end{dcases*}$

Here $\vartheta=7/64$ denotes the best known bound towards the Selberg
eigenvalue conjecture [Kim03, Proposition 2 of Appendix 2]. The first range of
(1.3) follows from the standard fourth moment upper bound

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{2+\varepsilon},$

which is a straightforward consequence of the spectral large sieve, while the
second range is due to the second author’s fifth moment bound [Kha20, Theorem
1.1]

(1.4) $\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq
k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{5}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{2+2\vartheta+\varepsilon}.$

Theorem 1.1 yields the bounds (1.3) in the third range. The fourth range is a
consequence of Frolenkov’s Weyl-strength subconvex bound [Fro20, Corollary
1.4]

(1.5) $L\left(\frac{1}{2},f\right)\ll k_{f}^{\frac{1}{3}}(\log
k_{f})^{\frac{13}{6}}.$

In particular, (1.3) shows that for any $\delta>0$, the number of Hecke cusp
forms $f$ of weight $k_{f}\in[T,2T]$ that fail to satisfy the sub-Weyl
subconvex bound $L(1/2,f)\ll k^{1/3-\delta}$ is
$O_{\varepsilon}(T^{12\delta+\varepsilon})$.

Theorem 1.1 also has applications to improved $L^{4}$-norm bounds for
Hecke–Maaß cusp forms [HK22]. Here one must deal with mixed moments of the
form

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq t_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)L\left(\frac{1}{2},\operatorname{ad}g\otimes
f\right)}{L(1,\operatorname{ad}f)},$

where $g$ is a Hecke–Maaß cusp form (see in particular [HK22, Proposition
6.11]). While one could bound this mixed moment simply via the individual
Weyl-strength subconvex bound (1.5) for $L(1/2,f)$ together with pre-existing
bounds for the first moment of $L(1/2,\operatorname{ad}g\otimes f)$ (for which
see [HK22, Proposition 6.6]), stronger bounds can be attained by instead using
Hölder’s inequality and inputting Weyl-strength bounds for the twelfth moment
of $L(1/2,f)$, namely Theorem 1.1.

###  1.4. Sketch of Proof

The analogue of Theorem 1.1 for Hecke–Maaß cusp forms in place of holomorphic
Hecke cusp forms is known via work of Jutila [Jut04, Theorem 2]. The proof
that we give of Theorem 1.1 follows in broad strokes the same general strategy
of that of Jutila, but we need a crucial new idea to handle the holomorphic
case, as we now explain.

The approach to estimating the twelfth moment over the family
$\mathcal{F}_{T}$ involves understanding the fourth moment over smaller
families such as $\mathcal{G}_{T}$. To motivate the approach, define

(1.6)
$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\coloneqq\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\
k_{f}=k\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}$

for $k\in 2\mathbb{N}$ and consider the conjectural bound

(1.7)
$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\ll_{\varepsilon}k^{1+\varepsilon},$

which is an immediate consequence of the generalised Lindelöf hypothesis. This
conjectural bound, together with the very strong individual subconvex bound
$L(1/2,f)\ll_{\varepsilon}k_{f}^{1/4+\varepsilon}$ that it implies,
immediately yields Theorem 1.1. Although the proof of (1.7) is undoubtedly out
of reach of current technology, we are able to prove it in a mean square
sense:

(1.8) $\sum_{\begin{subarray}{c}T\leq k\leq 2T\\\ k\equiv
0\pmod{2}\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)^{2}\ll_{\varepsilon}T^{3+\varepsilon}.$

This is a special case of a more refined result, stated in Proposition 2.1,
where the mean square is restricted to
$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$ of a given size. It turns out
that such mean square estimates are enough to substitute for a result like
(1.7) as far as obtaining the twelfth moment bound is concerned. This is shown
in Section 2.

To establish (1.8), we apply a spectral reciprocity formula due to Kuznetsov
and Motohashi (given in Theorem 3.1) to relate the fourth moment
$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$ to a fourth moment of Hecke–Maaß
cusp forms. We end up with a relation very roughly of the shape

$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\rightsquigarrow
k^{-1/2}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\ t_{f}\leq
k\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}k^{2it_{f}},$

where $\mathcal{B}_{0}$ denotes an orthonormal basis of Hecke–Maaß cusp forms
$f$ on $\Gamma\backslash\mathbb{H}$, while $t_{f}$ denotes the spectral
parameter of $f$. When the right hand side is squared and summed over $k$ in a
dyadic interval $[T,2T]$, the oscillatory weight $k^{2it_{f}}$ (which is an
imprecise version of the weight given in Lemma 4.21 with $K=k$ and $L=2$)
facilitates cancellation, as explicated in Lemma 5.6, and leads to the mean
square estimate.

In order to apply the spectral reciprocity formula, one must ensure that
(certain transforms of) the test functions involved (suppressed in our sketch
above) satisfy delicate decay properties. This requirement is well-known to
experts by now, but it was overlooked by Kuznetsov, invalidating parts of his
paper [Kuz89], as pointed out by Motohashi [Mot03]. It is no major challenge
to construct test functions satisfying these decay properties for the fourth
moment of Hecke–Maaß cusp forms provided that one uses the root number trick
(see [Mot97, Section 3.3]). For a purely holomorphic fourth moment such as
$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$, however, most natural choices of
test function are not valid, and so the spectral reciprocity formula of
Kuznetsov has never been successfully applied until now. Our key idea is to
introduce a carefully designed and rather non-obvious test function, given in
(4.1), that satisfies the required decay properties.

##  2\. Proof of Theorem 1.1

Theorem 1.1 is predicate on the following auxiliary result concerning the
second moment in dyadic intervals of large values of the fourth moment of
$L(1/2,f)$ in short intervals (cf. [Jut04, Theorem 1]).

######  Proposition 2.1.

For $k\in 2\mathbb{N}$, let $\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$ be as
in (1.6). Then for $T\geq 1$ and $V>0$, we have that

(2.2) $\undertilde{\mathcal{S}}(T,V)\coloneqq\sum_{\begin{subarray}{c}T\leq
k\leq 2T\\\ k\equiv 0\pmod{2}\\\
\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\geq
V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)^{2}\ll_{\varepsilon}\begin{dcases*}T^{3+\varepsilon}&if
$0<V<T$,\\\ \frac{T^{4+\varepsilon}}{V}&if $T\leq V\leq T^{\frac{4}{3}}(\log
T)^{7}$,\\\ 0&if $V>T^{\frac{4}{3}}(\log T)^{7}$.\end{dcases*}$

Note that the bound (2.2) is essentially sharp for $V<T^{1+\varepsilon}$,
whereas the generalised Lindelöf hypothesis shows that we should expect that
$\undertilde{\mathcal{S}}(T,V)=0$ for $V\geq T^{1+\varepsilon}$.

We postpone the proof of Proposition 2.1 for the time being and instead
proceed directly to the proof of Theorem 1.1.

###### Proof of Theorem 1.1.

We begin by breaking up the twelfth moment based on the size of $L(1/2,f)$. We
deal separately with those terms for which $L(1/2,f)<T^{(1-\vartheta)/7}$. By
the fifth moment bound (1.4), we have that

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\\\
L(\frac{1}{2},f)<T^{2(1-\vartheta)/7}\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{12}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{4+\varepsilon}.$

We then break up the remaining terms appearing in the twelfth moment via a
dyadic subdivision, namely into sums over cusp forms
$f\in\mathcal{B}_{\mathrm{hol}}$ for which $T\leq k_{f}\leq 2T$ and $V\leq
L(1/2,f)<2V$ with $V\geq T^{2(1-\vartheta)/7}$ a dyadic parameter. The Weyl-
strength subconvex bound (1.5) ensures that $V\leq k_{f}^{1/3}(\log
k_{f})^{7/3}$, say, so that

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{12}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{4+\varepsilon}+T^{\varepsilon}\max_{T^{2(1-\vartheta)/7}\leq
V\leq T^{1/3}(\log
T)^{7/3}}V^{8}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq
k_{f}\leq 2T\\\ V\leq
L(\frac{1}{2},f)<2V\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}.$

We introduce a further dyadic subdivision based on the number of positive even
integers $k\in[T,2T]$ for which the cardinality of the set

$\mathcal{B}(k;V)\coloneqq\left\\{f\in\mathcal{B}_{\mathrm{hol}}:k_{f}=k,\
V\leq L\left(\frac{1}{2},f\right)<2V\right\\}$

lies in a dyadic interval. Since $\\#\mathcal{B}(k;V)\leq\frac{k}{12}+1$, we
deduce that

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\\\ V\leq
L(\frac{1}{2},f)<2V\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{\varepsilon}\max_{1\leq
W\leq\frac{T}{6}+1}\sum_{\begin{subarray}{c}T\leq k\leq 2T\\\ k\equiv
0\pmod{2}\\\
W\leq\\#\mathcal{B}(k;V)<2W\end{subarray}}\sum_{f\in\mathcal{B}(k;V)}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}.$

If $W\leq\\#\mathcal{B}(k;V)<2W$ and $T\leq k\leq 2T$, then

$\frac{V^{4}W}{(\log
T)^{2}}\ll\sum_{f\in\mathcal{B}(k;V)}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$

as $L(1,\operatorname{ad}f)\ll(\log k_{f})^{2}$, where
$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$ is as in (1.6), and consequently

$\sum_{\begin{subarray}{c}T\leq k\leq 2T\\\ k\equiv 0\pmod{2}\\\
W\leq\\#\mathcal{B}(k;V)<2W\end{subarray}}\sum_{f\in\mathcal{B}(k;V)}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\ll\frac{(\log
T)^{3}}{V^{4}W}\undertilde{\mathcal{S}}\left(T,\frac{V^{4}W}{(\log
T)^{3}}\right),$

where $\undertilde{\mathcal{S}}(T,V)$ is as in (2.2). In this way, we see that

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{12}}{L(1,\operatorname{ad}f)}\ll_{\varepsilon}T^{4+\varepsilon}+T^{\varepsilon}\max_{T^{2(1-\vartheta)/7}\leq
V\leq T^{1/3}(\log T)^{7/3}}V^{4}\max_{1\leq
W\leq\frac{T}{6}+1}\frac{1}{W}\undertilde{\mathcal{S}}\left(T,\frac{V^{4}W}{(\log
T)^{3}}\right),$

at which point the result follows immediately from the bound (2.2), noting
that $\vartheta=7/64$ implies that $V^{4}W/(\log T)^{3}\gg T^{57/56}/(\log
T)^{3}>T$. ∎

We return to the proof of Proposition 2.1. The proof relies on the following
estimate, which is the technical heart of Theorem 1.1.

######  Lemma 2.3.

Fix $0<\delta<1/100$. For $T\geq 1$, $T^{1+\delta/2}\leq V\leq T^{4/3}(\log
T)^{7}$, and $L\coloneqq
2\left\lfloor\frac{1}{2}VT^{-1-\delta/4}\right\rfloor$, we have that

(2.4) $\sum_{\begin{subarray}{c}T\leq K\leq 2T\\\ K\equiv
0\pmod{2}\end{subarray}}\Bigg{(}\sum_{\begin{subarray}{c}K-L\leq k\leq K+L\\\
k\equiv 0\pmod{2}\\\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}\ll_{\varepsilon}T^{3+\varepsilon}.$

We defer the proof of Lemma 2.3, which takes up the bulk of this paper, to
Section 5. We use this to now prove Proposition 2.1.

###### Proof of Proposition 2.1.

Via a dyadic subdivision, in order to prove the bounds (2.2), it suffices to
prove the related bounds

(2.5)
$\undertilde{\mathcal{S}}^{\ast}(T,V)\coloneqq\sum_{\begin{subarray}{c}T\leq
k\leq 2T\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)^{2}\ll_{\varepsilon}\begin{dcases*}T^{3+\varepsilon}&if
$0<V<T$,\\\ \frac{T^{4+\varepsilon}}{V}&if $T\leq V\leq T^{\frac{4}{3}}(\log
T)^{7}$,\\\ 0&if $V>T^{\frac{4}{3}}(\log T)^{7}$.\end{dcases*}$

The bound (2.5) holds for $V>T^{4/3}(\log T)^{7}$ due to Frolenkov’s
refinement [Fro20, Theorem 1.3]

$\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\
k_{f}=k\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{3}}{L(1,\operatorname{ad}f)}\ll
k(\log k)^{\frac{9}{2}}$

of Peng’s third moment bound for $L(1/2,f)$ [Pen01, Theorem 3.1.1] together
with the individual Weyl-strength subconvex bound (1.5) for $L(1/2,f)$, since
these combine to show that

$\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\ll k^{\frac{4}{3}}(\log
k)^{\frac{20}{3}}.$

Moreover, the bound (2.5) trivially holds for $0<V<T^{1+\varepsilon/2}$. It
remains to deal with the range $T^{1+\varepsilon/2}\leq V\leq T^{4/3}(\log
T)^{7}$.

We begin by fixing $0<\delta<1/100$, taking $T^{1+\delta/2}\leq V\leq
T^{4/3}(\log T)^{7}$, and breaking up the sum over $T\leq k\leq 2T$ in the
definition of $\undertilde{\mathcal{S}}^{\ast}(T,V)$ to shorter intervals of
length $L\coloneqq 2\left\lfloor\frac{1}{2}VT^{-1-\delta/4}\right\rfloor$, so
that

$\undertilde{\mathcal{S}}^{\ast}(T,V)\leq\sum_{\ell=0}^{\left\lceil\frac{T}{L}\right\rceil-1}\sum_{\begin{subarray}{c}T+\ell
L\leq k\leq T+(\ell+1)L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)^{2}\leq
2V\sum_{\ell=0}^{\left\lceil\frac{T}{L}\right\rceil-1}\sum_{\begin{subarray}{c}T+\ell
L\leq k\leq T+(\ell+1)L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k).$

We introduce a dyadic subdivision based on the number of integers $\ell$ for
which the cardinality of

$\mathcal{P}(\ell;V)\coloneqq\left\\{T+\ell L\leq k\leq T+(\ell+1)L:k\equiv
0\pmod{2},\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq 2V\right\\}$

lies in a dyadic interval. If $\\#\mathcal{P}(\ell;V)\geq 2^{j}$, then

$\sum_{\begin{subarray}{c}T+\ell L\leq k\leq T+(\ell+1)L\\\ k\equiv
0\pmod{2}\\\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\geq 2^{j}V,$

and consequently

$\displaystyle\undertilde{\mathcal{S}}^{\ast}(T,V)$ $\displaystyle\leq
2V\sum_{j=0}^{\infty}\sum_{\begin{subarray}{c}\ell=0\\\
2^{j}\leq\\#\mathcal{P}(\ell;V)<2^{j+1}\end{subarray}}^{\left\lceil\frac{T}{L}\right\rceil-1}\sum_{\begin{subarray}{c}T+\ell
L\leq k\leq T+(\ell+1)L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)$
$\displaystyle\leq
2\sum_{j=0}^{\infty}2^{-j}\sum_{\begin{subarray}{c}\ell=0\\\
2^{j}\leq\\#\mathcal{P}(\ell;V)<2^{j+1}\end{subarray}}^{\left\lceil\frac{T}{L}\right\rceil-1}\Bigg{(}\sum_{\begin{subarray}{c}T+\ell
L\leq k\leq T+(\ell+1)L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}$
$\displaystyle\leq
4\sum_{\ell=0}^{\left\lceil\frac{T}{L}\right\rceil-1}\Bigg{(}\sum_{\begin{subarray}{c}T+\ell
L\leq k\leq T+(\ell+1)L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}.$

We next bound the sum over $\ell$. For each even integer $T+\ell L\leq K\leq
T+(\ell+1)L$, we have that

$\sum_{\begin{subarray}{c}T+\ell L\leq k\leq T+(\ell+1)L\\\ k\equiv
0\pmod{2}\\\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq\sum_{\begin{subarray}{c}K-L\leq
k\leq K+L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k).$

Upon squaring, summing over even integers $T+\ell L\leq K\leq T+(\ell+1)L$,
and dividing through by $L/2$, we obtain

$\displaystyle\undertilde{\mathcal{S}}^{\ast}(T,V)$
$\displaystyle\leq\frac{8}{L}\sum_{\ell=0}^{\left\lceil\frac{T}{L}\right\rceil-1}\sum_{\begin{subarray}{c}T+\ell
L\leq K\leq T+(\ell+1)L\\\ K\equiv
0\pmod{2}\end{subarray}}\Bigg{(}\sum_{\begin{subarray}{c}K-L\leq k\leq K+L\\\
k\equiv 0\pmod{2}\\\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}$
$\displaystyle\leq\frac{8}{L}\sum_{\begin{subarray}{c}T\leq K\leq 2T\\\
K\equiv 0\pmod{2}\end{subarray}}\Bigg{(}\sum_{\begin{subarray}{c}K-L\leq k\leq
K+L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}.$

The desired bound
$\undertilde{\mathcal{S}}^{\ast}(T,V)\ll_{\varepsilon}T^{4+\varepsilon}/V$ now
follows upon invoking (2.4). ∎

##  3\.
$\mathrm{GL}_{4}\times\mathrm{GL}_{2}\leftrightsquigarrow\mathrm{GL}_{4}\times\mathrm{GL}_{2}$
Spectral Reciprocity

We record in this section a particular form of spectral reciprocity: a
$\operatorname{GL}_{2}$ moment of
$\operatorname{GL}_{4}\times\operatorname{GL}_{2}$ Rankin–Selberg
$L$-functions is equal to a main term plus a dual moment, which is a
$\operatorname{GL}_{2}$ moment of
$\operatorname{GL}_{4}\times\operatorname{GL}_{2}$ Rankin–Selberg
$L$-functions. In the case that we are interested in, these
$\operatorname{GL}_{4}\times\operatorname{GL}_{2}$ Rankin–Selberg
$L$-functions factorise as the product of four $\operatorname{GL}_{2}$
standard $L$-functions, so that this is an explicit formula for the fourth
moment of $L(1/2,f)$ weighted by a test function $h^{\mathrm{hol}}(k_{f})$.
This formula is due to Kuznetsov [Kuz89, Kuz99], though the initial proof was
incomplete in parts and was subsequently completed by Motohashi [Mot03]. We
state the following version of Kuznetsov’s formula, which is implicit in the
work of Motohashi [Mot03]222Motohashi states a formulation of Theorem 3.1 for
which the left-hand side of (3.3) only involves a sum over Hecke–Maaß cusp
forms instead of including holomorphic Hecke cusp forms, as well as slightly
different assumptions on the pair of test functions $(h,h^{\mathrm{hol}})$;
nonetheless, an examination of the proof shows that the result remains valid
in the formulation (3.3) (cf. [Mot03, Remark, p. 4])..

######  Theorem 3.1 (Kuznetsov [Kuz89, Theorem 15], Motohashi [Mot03,
Theorem]).

Let $h(t)$ be an even function that is holomorphic in the horizontal strip
$|\Im(t)|\leq 1/2+\delta$ for some $\delta>0$ and satisfies
$h(t)\ll(1+|t|)^{-5}$, and let $h^{\mathrm{hol}}:2\mathbb{N}\to\mathbb{C}$ be
a sequence that satisfies $h^{\mathrm{hol}}(k)\ll k^{-5}$. Suppose
additionally that the function

(3.2) $H(x)\coloneqq\frac{i}{\pi}\int_{-\infty}^{\infty}\frac{r}{\cosh\pi
r}J_{2ir}(4\pi x)h(r)\,dr+\frac{1}{\pi}\sum_{\begin{subarray}{c}k=2\\\ k\equiv
0\pmod{2}\end{subarray}}^{\infty}(k-1)i^{-k}J_{k-1}(4\pi
x)h^{\mathrm{hol}}(k),$

where $J_{\alpha}(x)$ denotes the Bessel function of the first kind, is such
that its Mellin transform
$\widehat{H}(s)\coloneqq\int_{0}^{\infty}H(x)x^{s}\,\frac{dx}{x}$ is
holomorphic in the strip $-4<\Re(s)<1$, in which it satisfies the bounds
$\widehat{H}(s)\ll(1+|\Im(s)|)^{\Re(s)-4}$. Then

(3.3)
$\sum_{f\in\mathcal{B}_{0}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}h(t_{f})+\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}h(t)\,dt+\sum_{f\in\mathcal{B}_{\mathrm{hol}}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}h^{\mathrm{hol}}(k_{f})\\\
=\sum_{\pm}\sum_{f\in\mathcal{B}_{0}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\widetilde{h}^{\pm}(t_{f})+\sum_{\pm}\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}\widetilde{h}^{\pm}(t)\,dt+\sum_{f\in\mathcal{B}_{\mathrm{hol}}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\widetilde{h}^{\mathrm{hol}}(k_{f})\\\
+\lim_{(z_{1},z_{2},z_{3},z_{4})\to\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)}\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}.$

Here the transforms $\widetilde{h}^{\pm}$ and $\widetilde{h}^{\mathrm{hol}}$
of $h^{\mathrm{hol}}$ are defined by

(3.4) $\displaystyle\widetilde{h}^{\pm}(t)$ $\displaystyle=\frac{1}{2\pi
i}\int_{\sigma-i\infty}^{\sigma+i\infty}\widehat{H}(s)\mathscr{G}_{t}^{\pm}(s)\,ds,$
(3.5) $\displaystyle\widetilde{h}^{\mathrm{hol}}(k)$
$\displaystyle=\frac{1}{2\pi
i}\int_{\sigma-i\infty}^{\sigma+i\infty}\widehat{H}(s)\mathscr{G}_{k}^{\mathrm{hol}}(s)\,ds$

for $0<\sigma<1$, where

(3.6) $\displaystyle\mathscr{G}_{t}^{+}(s)$
$\displaystyle\coloneqq\frac{1}{\pi^{2}}(2\pi)^{s}\Gamma\left(\frac{s}{2}+it\right)\Gamma\left(\frac{s}{2}-it\right)\Gamma\left(\frac{1-s}{2}\right)^{4}\cos\frac{\pi
s}{2}\left(\sin^{2}\frac{\pi s}{2}+1\right),$ (3.7)
$\displaystyle\mathscr{G}_{t}^{-}(s)$
$\displaystyle\coloneqq\frac{2}{\pi^{2}}\cosh\pi
t(2\pi)^{s}\Gamma\left(\frac{s}{2}+it\right)\Gamma\left(\frac{s}{2}-it\right)\Gamma\left(\frac{1-s}{2}\right)^{4}\sin\frac{\pi
s}{2},$ (3.8) $\displaystyle\mathscr{G}_{k}^{\mathrm{hol}}(s)$
$\displaystyle\coloneqq\frac{i^{-k}}{\pi}(2\pi)^{s}\frac{\Gamma\left(\frac{s+k-1}{2}\right)}{\Gamma\left(\frac{1-s+k}{2}\right)}\Gamma\left(\frac{1-s}{2}\right)^{4}\left(\sin^{2}\frac{\pi
s}{2}+1\right).$

For $(z_{1},z_{2},z_{3},z_{4})$ in a sufficiently small neighbourhood of the
point $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$ excluding the lines
$z_{j}=1-z_{\ell}$ and $z_{j}=z_{\ell}$ with $1\leq j<\ell<4$, we have that

(3.9) $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}\coloneqq
2\sum_{j=1}^{4}\widehat{H}(2(1-z_{j}))\frac{\prod_{\begin{subarray}{c}\ell=1\\\
\ell\neq
j\end{subarray}}^{4}\zeta(1-z_{j}+z_{\ell})\prod_{\begin{subarray}{c}1\leq
m<n\leq 4\\\ m,n\neq
j\end{subarray}}\zeta(z_{m}+z_{n})}{\zeta\left(1-z_{j}+\sum_{\begin{subarray}{c}\ell=1\\\
\ell\neq j\end{subarray}}^{4}z_{\ell}\right)}\\\
-2\sum_{j=1}^{4}h(i(z_{j}-1))\frac{\prod_{\begin{subarray}{c}\ell=1\\\
\ell\neq
j\end{subarray}}^{4}\zeta(1-z_{j}+z_{\ell})\zeta(z_{j}+z_{\ell}-1)}{\zeta(3-2z_{j})}\\\
+\sum_{\pm}2\sum_{j=1}^{4}\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm}\frac{\prod_{\begin{subarray}{c}\ell=1\\\
\ell\neq
j\end{subarray}}^{4}\zeta(1+z_{j}-z_{\ell})\zeta\left(\sum_{\begin{subarray}{c}m=1\\\
m\neq
j,\ell\end{subarray}}^{4}z_{m}-1\right)}{\zeta\left(3+z_{j}-\sum_{\begin{subarray}{c}\ell=1\\\
\ell\neq j\end{subarray}}^{4}z_{\ell}\right)},$

where the transforms $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm}$ are
given by

$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{+}\coloneqq\frac{1}{2\pi^{3}i}\int_{\mathcal{C}_{j}}\widehat{H}(s)(2\pi)^{s}\Gamma\left(\frac{s}{2}+\sum_{\ell=1}^{4}z_{\ell}-z_{j}-2\right)\Gamma\left(\frac{s}{2}+z_{j}\right)\\\
\times\prod_{m=1}^{4}\Gamma\left(1-z_{m}-\frac{s}{2}\right)\sin\frac{\pi}{2}\left(s+\sum_{\ell=1}^{4}z_{\ell}-1\right)\\\
\times\left(\cos\frac{\pi}{2}(s+z_{1}+z_{2})\cos\frac{\pi}{2}(s+z_{3}+z_{4})+\cos\frac{\pi}{2}(z_{1}-z_{2})\cos\frac{\pi}{2}(z_{3}-z_{4})\right)\,ds$

and

$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{-}\coloneqq\frac{1}{2\pi^{3}i}\cos\frac{\pi}{2}\left(\sum_{\ell=1}^{4}z_{\ell}-2z_{j}\right)\int_{\mathcal{C}_{j}}\widehat{H}(s)(2\pi)^{s}\Gamma\left(\frac{s}{2}+\sum_{\ell=1}^{4}z_{\ell}-z_{j}-2\right)\Gamma\left(\frac{s}{2}+z_{j}\right)\\\
\times\prod_{m=1}^{4}\Gamma\left(1-z_{m}-\frac{s}{2}\right)\left(\cos\frac{\pi}{2}(z_{1}-z_{2})\cos\frac{\pi}{2}(s+z_{3}+z_{4})+\cos\frac{\pi}{2}(s+z_{1}+z_{2})\cos\frac{\pi}{2}(z_{3}-z_{4})\right)\,ds,$

where $\mathcal{C}_{j}$ is a contour from $-i\infty$ to $i\infty$ such that
$6+2z_{j}-2\sum_{\ell=1}^{4}z_{\ell}-2n$ and $2-2z_{j}-2n$ are to the left of
the contour and $-2z_{m}+2n$ is to the right of the contour for each $1\leq
m\leq 4$ and each $n\in\mathbb{N}$.

######  Remark 3.10.

There exist cuspidal analogues of Kuznetsov’s formula for the fourth moment of
$L(1/2,f)$, namely for the first moment of $L(1/2,F\otimes f)$ with $F$ an
automorphic form for $\operatorname{SL}_{4}(\mathbb{Z})$; Kuznetsov’s formula
corresponds to taking $F$ to be a minimal parabolic Eisenstein series. The
authors have proven an analogue of the identity (3.3) with $F$ an Eisenstein
series induced from a dihedral Hecke–Maaß cusp form, a quadratic Dirichlet
character, and the principal Dirichlet character [HK20, Proposition 7.1]; more
recently, the authors have proven an analogue of this identity with $F$ a
maximal parabolic Eisenstein series induced from a self-dual Hecke–Maaß cusp
form for $\operatorname{SL}_{3}(\mathbb{Z})$ [HK22, Theorem 4.1]. Blomer, Li,
and Miller have proven a _completely_ cuspidal version of this identity with
$F$ a Hecke–Maaß cusp form for $\operatorname{SL}_{4}(\mathbb{Z})$ [BLM19,
Theorem 1].

##  4\. Test Functions and Transforms

Our next step is to use Theorem 3.1 in order to prove Lemma 2.3. Naïvely, we
might hope to achieve this by using the identity (3.3) with the choice of pair
of test functions $(h,h^{\mathrm{hol}})$ given by

$\displaystyle h(t)$ $\displaystyle\coloneqq 0,$ $\displaystyle
h^{\mathrm{hol}}(k)$ $\displaystyle\coloneqq\begin{dcases*}1&if $K-L\leq k\leq
K+L$,\\\ 0&otherwise.\end{dcases*}$

This, however, does not meet the requirements of Theorem 3.1, since the Mellin
transform $\widehat{H}(s)$ of the transform $H(x)$ of the pair
$(h,h^{\mathrm{hol}})$ given by (3.2) does not decay sufficiently rapidly in
the vertical strip $-5<\Re(s)<1$. One might hope to get around this by instead
working with a smooth approximation to the indicator function of the short
interval $[K-L,K+L]$, namely $h^{\mathrm{hol}}(k)\coloneqq\Omega((k-K)/L)$
with $\Omega$ an arbitrary smooth function compactly supported on $[-3/2,3/2]$
that is nonnegative, equal to $1$ on $[-1,1]$, and has bounded derivatives,
but this nonetheless runs into the same issue.

We circumvent this obstacle via two tricks. The first is the root number
trick: since $L(1/2,f)=0$ whenever $k_{f}\equiv 2\pmod{4}$, as the root number
of $L(s,f)$ is $i^{k_{f}}$, we can instead choose our test function such that
$i^{k}h^{\mathrm{hol}}(k)$ approximates the indicator function of $[K-L,K+L]$.
The second trick is to carefully choose this approximation in such a way that
$\widehat{H}(s)$ is holomorphic on a wide vertical strip in which it decays
extremely rapidly.

###  4.1. Test Functions

We define the following pair of test functions $(h,h^{\mathrm{hol}})$:

(4.1) $\begin{split}h(t)&\coloneqq 0,\\\ h^{\mathrm{hol}}(k)&\coloneqq
i^{-k}\frac{\Gamma\left(\frac{L^{2}}{2}+1\right)^{2}\Gamma\left(K+\frac{L^{2}}{2}\right)\Gamma\left(\frac{K-L^{2}+k}{2}-1\right)}{\Gamma\left(K-\frac{L^{2}}{2}-1\right)\Gamma\left(\frac{L^{2}-K+k}{2}+1\right)\Gamma\left(\frac{K+L^{2}+k}{2}\right)\Gamma\left(\frac{K+L^{2}-k}{2}+1\right)}.\end{split}$

Here $K,L\in\mathbb{N}$ are auxiliary parameters such that $K\equiv L\equiv
0\pmod{2}$ and $L\leq K^{1/3}$. This particular function $h^{\mathrm{hol}}$ is
chosen to localise to short intervals: $i^{k}h^{\mathrm{hol}}(k)$ localises to
the interval $[K-L,K+L]$, as we now show.

######  Lemma 4.2.

Fix $0<\delta<1/100$. Let $(h,h^{\mathrm{hol}})$ be the pair of test functions
(4.1) with $K,L\in 2\mathbb{N}$ satisfying $K^{\delta/8}\leq L\leq
K^{1/3-\delta/8}$. Let $H$ be as in (3.2).

  1. (1)

The Mellin transform $\widehat{H}(s)$ of $H$ is such that for
$s=\sigma+i\tau$,

    1. (a)

$\widehat{H}(s)$ is holomorphic for $\sigma>L^{2}-K-1$,

    2. (b)

for $\sigma$ bounded, $\widehat{H}(s)$ satisfies the bounds

(4.3) $\widehat{H}(s)\ll\begin{dcases*}K^{\sigma}L&if
$|\tau|\leq\frac{K}{L}$,\\\
K^{\sigma}Le^{-\left(\frac{L|\tau|}{2K}\right)^{2}}&if
$\frac{K}{L}\leq|\tau|\leq K$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}e^{-\left(\frac{KL}{2|\tau|}\right)^{2}}&if
$K\leq|\tau|\leq KL$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}&if $|\tau|\geq
KL$.\end{dcases*}$

  2. (2)

The test function $h^{\mathrm{hol}}(k)$ is such that for $k\in 2\mathbb{N}$,

    1. (a)

$i^{k}h^{\mathrm{hol}}(k)>0$ for $K-L^{2}\leq k\leq K+L^{2}$ and
$h^{\mathrm{hol}}(k)=0$ otherwise,

    2. (b)

$i^{k}h^{\mathrm{hol}}(k)\asymp 1$ for $K-L\leq k\leq K+L$,

    3. (c)

$h^{\mathrm{hol}}(k)\ll e^{-\frac{(k-K)^{2}}{4L^{2}}}$ whenever
$L\leq|k-K|\leq L^{2}$.

###### Proof.

For $w_{1},w_{2}\in\mathbb{C}$ with $\Re(w_{2})>-1$ and
$\Re(w_{1})>-2\Re(w_{2})$, define

$H_{w_{1},w_{2}}^{\mathrm{hol}}(x)\coloneqq\frac{1}{\pi}\frac{\Gamma\left(\frac{w_{2}}{2}+1\right)^{2}\Gamma\left(w_{1}+\frac{w_{2}}{2}\right)}{\Gamma(w_{2}+1)\Gamma\left(w_{1}-\frac{w_{2}}{2}-1\right)}(2\pi
x)^{-w_{2}}J_{w_{1}-1}(4\pi x).$

By [GR15, 6.574.2], we have that

$\displaystyle(\mathscr{L}^{\mathrm{hol}}H_{w_{1},w_{2}}^{\mathrm{hol}})(k)$
$\displaystyle\coloneqq 2\pi
i^{-k}\int_{0}^{\infty}H_{w_{1},w_{2}}^{\mathrm{hol}}(x)J_{k-1}(4\pi
x)\,\frac{dx}{x}$
$\displaystyle=i^{-k}\frac{\Gamma\left(\frac{w_{2}}{2}+1\right)^{2}\Gamma\left(w_{1}+\frac{w_{2}}{2}\right)\Gamma\left(\frac{w_{1}-w_{2}+k}{2}-1\right)}{\Gamma\left(w_{1}-\frac{w_{2}}{2}-1\right)\Gamma\left(\frac{w_{2}-w_{1}+k}{2}+1\right)\Gamma\left(\frac{w_{1}+w_{2}+k}{2}\right)\Gamma\left(\frac{w_{1}+w_{2}-k}{2}+1\right)}.$

Choosing $w_{1}=K$ and $w_{2}=L^{2}$ with $K,L\in 2\mathbb{N}$, we obtain
$h^{\mathrm{hol}}(k)$. On the other hand, by [GR15, 6.574.2],

$\displaystyle(\mathscr{L}^{+}H_{w_{1},w_{2}}^{\mathrm{hol}})(t)$
$\displaystyle\coloneqq\frac{\pi i}{\sinh\pi
t}\int_{0}^{\infty}H_{w_{1},w_{2}}^{\mathrm{hol}}(x)\left(J_{2it}(4\pi
x)-J_{-2it}(4\pi x)\right)\,\frac{dx}{x}$
$\displaystyle=\frac{1}{\pi}\sin\frac{\pi(w_{1}-w_{2})}{2}\frac{\Gamma\left(\frac{w_{2}}{2}+1\right)^{2}\Gamma\left(w_{1}+\frac{w_{2}}{2}\right)}{\Gamma\left(w_{1}-\frac{w_{2}}{2}-1\right)}\prod_{\pm}\frac{\Gamma\left(\frac{w_{1}-w_{2}-1}{2}\pm
it\right)}{\Gamma\left(\frac{w_{1}+w_{2}+1}{2}\pm it\right)}.$

This vanishes if $w_{1}-w_{2}\in 2\mathbb{N}$, and in particular if $w_{1}=K$
and $w_{2}=L^{2}$ with $K,L\in 2\mathbb{N}$. It follows that
$(\mathscr{L}^{\mathrm{hol}}H_{K,L^{2}}^{\mathrm{hol}})(k)=h^{\mathrm{hol}}(k)$
and $(\mathscr{L}^{+}H_{K,L^{2}}^{\mathrm{hol}})(t)=h(t)$, so that
$H(x)=H_{K,L^{2}}^{\mathrm{hol}}(x)$ by the Sears–Titchmarsh inversion formula
[Iwa02, Appendix B.5].

  1. (1)

By [GR15, 6.561.14], we have that

(4.4)
$\widehat{H}(s)=(2\pi)^{-s-1}\frac{\Gamma\left(\frac{L^{2}}{2}+1\right)^{2}\Gamma\left(K+\frac{L^{2}}{2}\right)}{\Gamma(L^{2}+1)\Gamma\left(K-\frac{L^{2}}{2}-1\right)}\frac{\Gamma\left(\frac{K-L^{2}+s-1}{2}\right)}{\Gamma\left(\frac{K+L^{2}-s+1}{2}\right)}.$

This is holomorphic for $\sigma>L^{2}-K-1$. To prove the bounds (4.3), it
suffices to give an asymptotic expansion for $\Re(\log\widehat{H}(s))$. We
compute using Stirling’s formula the expansions

$\displaystyle\log\frac{\Gamma\left(\frac{L^{2}}{2}+1\right)^{2}}{\Gamma(L^{2}+1)}$
$\displaystyle=\log L-(L^{2}+1)\log 2+\cdots,$
$\displaystyle\log\frac{\Gamma\left(K+\frac{L^{2}}{2}\right)}{\Gamma\left(K-\frac{L^{2}}{2}-1\right)}$
$\displaystyle=(L^{2}+1)\log K+\cdots,$

where the ellipsis indicates lower order terms in $L$ or $K$. Next, we have
that

$\log\frac{\Gamma\left(\frac{K-L^{2}+s-1}{2}\right)}{\Gamma\left(\frac{K+L^{2}-s+1}{2}\right)}=\left(\frac{K-L^{2}+s-2}{2}\right)\log\left(\frac{K-L^{2}+s-1}{2e}\right)\\\
-\left(\frac{K+L^{2}-s}{2}\right)\log\left(\frac{K+L^{2}-s+1}{2e}\right)+\cdots.$

The asymptotic expansion of the first two terms depends on the size of
$|\tau|$.

For $|\tau|\leq K$, the real part of the first two terms in (4.15) may be
written in the form

$-\frac{|\tau|}{2}\arctan\left(\frac{2|\tau|(L^{2}-\sigma+1)}{K^{2}\left(1+\frac{\tau^{2}}{K^{2}}\right)\left(1-\frac{(L^{2}-\sigma+1)^{2}}{K^{2}+\tau^{2}}\right)}\right)+\frac{K-1}{4}\log\frac{1-\frac{2(L^{2}-\sigma+1)}{K\left(1+\frac{\tau^{2}}{K^{2}}\right)\left(1+\frac{(L^{2}-\sigma+1)^{2}}{K^{2}+\tau^{2}}\right)}}{1+\frac{2(L^{2}-\sigma+1)}{K\left(1+\frac{\tau^{2}}{K^{2}}\right)\left(1+\frac{(L^{2}-\sigma+1)^{2}}{K^{2}+\tau^{2}}\right)}}\\\
-\frac{L^{2}-\sigma+1}{2}\log\left(1+\frac{\tau^{2}}{K^{2}}\right)-\frac{L^{2}-\sigma+1}{2}\log\left(1+\frac{(L^{2}-\sigma+1)^{2}}{K^{2}+\tau^{2}}\right)\\\
-\frac{L^{2}-\sigma+1}{4}\log\left(1-\frac{4(L^{2}-\sigma+1)^{2}}{K^{2}\left(1+\frac{\tau^{2}}{K^{2}}\right)^{2}\left(1+\frac{(L^{2}-\sigma+1)^{2}}{K^{2}+\tau^{2}}\right)^{2}}\right)-(L^{2}-\sigma+1)\log\frac{K}{2e}.$

Via the Taylor series expansions of $\arctan$ and $\log$, we deduce that

$\Re\left(\log\widehat{H}(s)\right)=\log L+\sigma\log K+\begin{dcases*}O(1)&if
$|\tau|\leq\frac{K}{L}$,\\\
-\frac{\tau^{2}(L^{2}-\sigma+1)}{2K^{2}\left(1+\frac{\tau^{2}}{K^{2}}\right)}(1+o(1))&if
$\frac{K}{L}\leq|\tau|\leq K$.\end{dcases*}$

For $|\tau|\geq K$, we may instead write the real part of the first two terms
in (4.15) in the form

$-\frac{|\tau|}{2}\arctan\left(\frac{2(L^{2}-\sigma+1)}{|\tau|\left(1+\frac{K^{2}}{\tau^{2}}\right)\left(1-\frac{(L^{2}-\sigma+1)^{2}}{\tau^{2}+K^{2}}\right)}\right)+\frac{K-1}{4}\log\frac{1-\frac{2(L^{2}-\sigma+1)K}{\tau^{2}\left(1+\frac{K^{2}}{\tau^{2}}\right)\left(1+\frac{(L^{2}-\sigma+1)^{2}}{\tau^{2}+K^{2}}\right)}}{1+\frac{2(L^{2}-\sigma+1)K}{\tau^{2}\left(1+\frac{K^{2}}{\tau^{2}}\right)\left(1+\frac{(L^{2}-\sigma+1)^{2}}{\tau^{2}+K^{2}}\right)}}\\\
-\frac{L^{2}-\sigma+1}{2}\log\left(1+\frac{K^{2}}{\tau^{2}}\right)-\frac{L^{2}-\sigma+1}{2}\log\left(1+\frac{(L^{2}-\sigma+1)^{2}}{\tau^{2}+K^{2}}\right)\\\
-\frac{L^{2}-\sigma+1}{4}\log\left(1-\frac{4(L^{2}-\sigma+1)^{2}K^{2}}{\tau^{4}\left(1+\frac{K^{2}}{\tau^{2}}\right)\left(1+\frac{(L^{2}-\sigma+1)^{2}}{\tau^{2}+K^{2}}\right)^{2}}\right)-(L^{2}-\sigma+1)\log\frac{|\tau|}{2e}.$

Via the Taylor series expansions of $\arctan$ and $\log$, we deduce that

$\Re\left(\log\widehat{H}(s)\right)=\log
L+\sigma\log|\tau|-(L^{2}-\sigma+1)\log\frac{|\tau|}{K}\\\
+\begin{dcases*}-\frac{K^{2}(L^{2}-\sigma+1)}{2\tau^{2}\left(1+\frac{K^{2}}{\tau^{2}}\right)}(1+o(1))&if
$K\leq|\tau|\leq KL$,\\\ O(1)&if $|\tau|\geq KL$.\end{dcases*}$

  2. (2)

As $\Gamma(s)$ has simple poles whenever $s$ is a nonpositive integer,
$h^{\mathrm{hol}}(k)$ vanishes unless $K-L^{2}\leq k\leq K+L^{2}$, in which
case we have the simpler expression

$h^{\mathrm{hol}}(k)=i^{-k}\prod_{\ell=0}^{\frac{|K-k|}{2}-1}\frac{\frac{L^{2}-|K-k|}{2}+1+\ell}{\frac{L^{2}}{2}+1+\ell}\prod_{j=0}^{L^{2}}\frac{K-\frac{L^{2}}{2}-1+j}{\frac{K-L^{2}+k}{2}-1+j}$

by the fact that $\Gamma(n)=(n-1)!$ for $n\in\mathbb{N}$. In particular,
$i^{k}h^{\mathrm{hol}}(k)>0$ whenever $K-L^{2}\leq k\leq K+L^{2}$. To estimate
the size of $i^{k}h^{\mathrm{hol}}(k)$ in this range, we use Stirling’s
formula, which shows that

$\log(i^{k}h^{\mathrm{hol}}(k))=-\frac{(K-k)^{2}}{2L^{2}}(1+o(1))+\frac{(K-k)L^{2}}{2K}+O(1).$

for $K-L^{2}\leq k\leq K+L^{2}$. In particular, this is $O(1)$ for $K-L\leq
k\leq K+L$, so that $i^{k}h^{\mathrm{hol}}(k)\asymp 1$ in this range, while in
the range $L\leq|k-K|\leq L^{2}$, this asymptotic implies that
$h^{\mathrm{hol}}(k)\ll e^{-\frac{(k-K)^{2}}{4L^{2}}}$. ∎

###  4.2. Transforms

We determine the behaviour of the transforms
$(\widetilde{h}^{+},\widetilde{h}^{-},\widetilde{h}^{\mathrm{hol}})$ as in
(3.4) and (3.5) with $(h,h^{\mathrm{hol}})$ the pair of test functions (4.1).
The latter two transforms are readily estimated using the bounds (4.3).

######  Lemma 4.5.

Fix $0<\delta<1/100$. Let $(h,h^{\mathrm{hol}})$ be the pair of test functions
(4.1) with $K,L\in 2\mathbb{N}$ satisfying $K^{\delta/8}\leq L\leq
K^{1/3-\delta/8}$. Then for $\widetilde{h}^{-}$ as in (3.4) and
$\widetilde{h}^{\mathrm{hol}}$ as in (3.5), we have that for $t\in\mathbb{R}$
and $k\in 2\mathbb{N}$,

(4.6) $\displaystyle\widetilde{h}^{-}(t)$ $\displaystyle\ll
K^{-100}L(1+|t|)^{-100},$ (4.7) $\displaystyle\widetilde{h}^{\mathrm{hol}}(k)$
$\displaystyle\ll\begin{dcases*}K^{1-k}L&if $k\leq 101$,\\\ K^{-100}L&if
$101\leq k\leq\frac{K}{L}$,\\\ k^{-100}&if $k\geq\frac{K}{L}$.\end{dcases*}$

###### Proof.

Let us first consider $\widetilde{h}^{-}(t)$. Provided that
$1+L^{2}-K<\sigma<1$ and that $s$ is a bounded distance away from the poles of
$\mathscr{G}_{t}^{-}(s)$ at $s=-2(\pm it+\ell)$ with $\ell$ a nonnegative
integer, we have by the bounds (4.3) for $\widehat{H}(s)$ and Stirling’s
formula applied to the definition (3.7) of $\mathscr{G}_{t}^{-}(s)$ that the
integrand in (3.4) with $\pm=-$ satisfies the bounds

$\widehat{H}(s)\mathscr{G}_{t}^{-}(s)\\\
\ll_{\sigma}((1+|\tau+2t|)(1+|\tau-2t|))^{\frac{\sigma-1}{2}}(1+|\tau|)^{-2\sigma}e^{-\frac{\pi}{2}|\tau|}\times\begin{dcases*}1&if
$|\tau|\leq 2|t|$,\\\ e^{-\frac{\pi}{2}(|\tau|-2|t|)}&if $|\tau|\geq
2|t|$,\end{dcases*}\\\ \times\begin{dcases*}K^{\sigma}L&if
$|\tau|\leq\frac{K}{L}$,\\\ K^{\sigma}Le^{-\frac{L^{2}\tau^{2}}{4K^{2}}}&if
$\frac{K}{L}\leq|\tau|\leq K$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}e^{-\frac{L^{2}K^{2}}{4\tau^{2}}}&if
$K\leq|\tau|\leq KL$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}&if $|\tau|\geq
KL$.\end{dcases*}$

In particular, the integral of this function along the line $\Re(s)=\sigma$
with $\sigma<0$ is

$\ll_{\sigma}K^{\sigma}L(1+|t|)^{\sigma-1}.$

Additionally, for $\ell$ a bounded nonnegative integer, we have by Stirling’s
formula that

$\operatorname*{Res}_{s=2(\pm
it-\ell)}\widehat{H}(s)\mathscr{G}_{t}^{-}(s)\ll_{\ell}(1+|t|)^{3\ell-\frac{1}{2}}e^{-\pi|t|}\times\begin{dcases*}K^{-2\ell}L&if
$|t|\leq\frac{K}{2L}$,\\\ K^{-2\ell}Le^{-\frac{L^{2}|t|^{2}}{K^{2}}}&if
$\frac{K}{2L}\leq|t|\leq\frac{K}{2}$,\\\
L|t|^{-2\ell}\left(\frac{2|t|}{K}\right)^{-L^{2}-1}e^{-\frac{L^{2}K^{2}}{16|t|^{2}}}&if
$\frac{K}{2}\leq|t|\leq\frac{KL}{2}$,\\\
L|t|^{-2\ell}\left(\frac{2|t|}{K}\right)^{-L^{2}-1}&if
$|t|\geq\frac{KL}{2}$.\end{dcases*}$

Thus by shifting the contour of integration in (3.4) to $\sigma=-100$, we
deduce the bounds (4.6).

Next, we consider $\widetilde{h}^{\mathrm{hol}}(k)$. Provided that
$1+L^{2}-K<\sigma<1$ and $s$ is a bounded distance away from the poles at
$s=1-k-2\ell$ with $\ell$ a nonnegative integer, we have by the bounds (4.3)
for $\widehat{H}(s)$ and Stirling’s formula applied to the definition (3.8) of
$\mathscr{G}_{k}^{\mathrm{hol}}(s)$ that the integrand in (3.5) satisfies the
bounds

$\widehat{H}(s)\mathscr{G}_{k}^{\mathrm{hol}}(s)\ll_{\sigma}(k+|\tau|)^{\sigma-1}(1+|\tau|)^{-2\sigma}\times\begin{dcases*}K^{\sigma}L&if
$|\tau|\leq\frac{K}{L}$,\\\ K^{\sigma}Le^{-\frac{L^{2}\tau^{2}}{4K^{2}}}&if
$\frac{K}{L}\leq|\tau|\leq K$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}e^{-\frac{L^{2}K^{2}}{4\tau^{2}}}&if
$K\leq|\tau|\leq KL$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}&if $|\tau|\geq
KL$.\end{dcases*}$

In particular, the integral of this function along the line $\Re(s)=\sigma$
with $\sigma<0$ is

$\ll_{\sigma}\begin{dcases*}L^{1+\sigma}&if $k\leq\frac{K}{L}$,\\\
K^{1-\sigma}L^{2\sigma}k^{\sigma-1}&if $k\geq\frac{K}{L}$.\end{dcases*}$

Additionally, for $\ell$ a bounded nonnegative integer, we have by Stirling’s
formula that

$\operatorname*{Res}_{s=1-k-2\ell}\widehat{H}(s)\mathscr{G}_{k}^{\mathrm{hol}}(s)\ll_{\ell}K^{1-k-2\ell}L.$

Thus by shifting the contour of integration in (3.5) to $\sigma=-\frac{100\log
K}{\log L}$, we deduce the bounds (4.7). ∎

Determining the behaviour of the transform $\widetilde{h}^{+}$ as in (3.4)
takes significantly more effort. Indeed, while $\widetilde{h}^{-}$ and
$\widetilde{h}^{\mathrm{hol}}$ are, in practice, negligibly small, this is
_not_ the case for $\widetilde{h}^{+}$ in certain ranges. Moreover, for the
purposes of proving Lemma 2.3, we require not only upper bounds for
$\widetilde{h}^{+}$ but additionally an _asymptotic_ in the range where it is
nonnegligible. As we shall presently show, in this nonnegligible range, the
function $\widetilde{h}^{+}$ is not so small but is highly oscillatory, a
property that we will later exploit in Lemma 5.6.

To determine the asymptotic behaviour of $\widetilde{h}^{+}$, we first
determine the behaviour of the integrand in the definition (3.4). From hereon,
we make use of the $\varepsilon$-convention: $\varepsilon$ denotes an
arbitrarily small positive constant whose value may change from occurrence to
occurrence; we will always assume that $\varepsilon$ is sufficiently small
with respect to $\delta$. In the proof of the following result, by a “lower
order term” in parameters $x_{1},\ldots x_{n}$, we will mean a function
$f(x_{1},x_{2},\ldots,x_{n})$ such that

$\frac{\partial^{j_{1}}}{\partial
x_{1}^{j_{1}}}\cdots\frac{\partial^{j_{n}}}{\partial
x_{n}^{j_{n}}}f(x_{1},\ldots,x_{n})\ll
K^{-\varepsilon}(1+|x_{1}|)^{-j_{1}}\cdots(1+|x_{n}|)^{-j_{n}}$

for any integers $j_{1},\ldots,j_{n}\geq 0$.

######  Lemma 4.8.

Fix $0<\delta<1/100$. Let $\widehat{H}(s)$ be given by (4.4) with $K,L\in
2\mathbb{N}$ satisfying $K^{\delta/8}\leq L\leq K^{1/3-\delta/8}$. Let
$K^{2/3+\varepsilon}<t<K(\log K)/L$. Let $s=\sigma+i(2t+y)$ with $\sigma>0$
fixed, where

(4.9) $\frac{t^{3}}{K^{2+\varepsilon}}<y<\frac{K\log K}{L}.$

For $\mathscr{G}_{t}^{+}(s)$ as in (3.6), we have that

(4.10)
$\widehat{H}(s)\mathscr{G}_{t}^{+}(s)=A(y,t,K)e^{iB(y,t,K)}+O(K^{-500}),$

where $A(y,t,K)$ and $B(y,t,K)$ are smooth functions (whose dependence on $L$
we suppress in the notation), complex-valued and real-valued respectively,
that satisfy

(4.11) $\frac{\partial^{j}}{\partial y^{j}}A(y,t,K)\ll\frac{K^{10}}{y^{j}},\ \
\frac{\partial}{\partial
y}B(y,t,K)=\log\frac{Ky^{\frac{1}{2}}(y+4t)^{\frac{1}{2}}}{(2t+y)^{2}}+O(1),\
\ \frac{\partial^{j+2}}{\partial y^{j+2}}B(y,t,K)(y)\ll\frac{1}{y^{j+1}}$

for any integer $j\geq 0$. If additionally

(4.12) $\frac{t^{3}}{K^{2+\varepsilon}}<y<\frac{t^{3}}{K^{2-\varepsilon}},$

then (4.10) holds with

(4.13) $\displaystyle A(y,t,K)$
$\displaystyle=\frac{L}{(ty)^{\frac{1}{2}}}\left(\frac{a_{1}K^{2}y}{t^{3}}\right)^{\frac{\sigma}{2}}\left(\frac{a_{2}K}{t}\right)^{2it}b_{1}(y,t,K),$
(4.14) $\displaystyle B(y,t,K)$
$\displaystyle=\frac{a_{3}t^{3}}{K^{2}}+\frac{y}{2}\left(\log\left(\frac{a_{4}K^{2}y}{t^{3}}\right)+b_{2}(y,t,K)\right),$

where $a_{i}$ are some real constants with $a_{1},a_{2},a_{4}>0$, and
$b_{1}(y,t,K)$, $b_{2}(y,t,K)$ are smooth functions (whose dependence on $L$
we suppress in the notation), complex-valued and real-valued respectively,
that satisfy

$\frac{\partial^{j_{1}}}{\partial y^{j_{1}}}\frac{\partial^{j_{2}}}{\partial
t^{j_{2}}}\frac{\partial^{j_{3}}}{\partial K^{j_{3}}}b_{i}(y,t,K)\ll
y^{-j_{1}}t^{-j_{2}}K^{-j_{3}}\times\begin{dcases*}1&if $i=1$,\\\
K^{-\varepsilon}&if $i=2$,\end{dcases*}$

for any $j_{1},j_{2},j_{3}\geq 0$.

###### Proof.

By Stirling’s formula, we have that

(4.15)
$\log\frac{\Gamma\left(\frac{K-L^{2}+s-1}{2}\right)}{\Gamma\left(\frac{K+L^{2}-s+1}{2}\right)}=\left(\frac{K-L^{2}+s-2}{2}\right)\log\left(\frac{K-L^{2}+s-1}{2e}\right)\\\
-\left(\frac{K+L^{2}-s}{2}\right)\log\left(\frac{K+L^{2}-s+1}{2e}\right)+\cdots+O(K^{-1000}),$

where the ellipsis indicates lower order terms in $s,K,L$. We follow the same
convention below, with parameters that will be obvious from the context. Next,
we insert the power series expansion

$\log\left(\frac{K\pm(s-L^{2}-1)}{2e}\right)=\log\left(\frac{K}{2}\right)-1+\sum_{j=1}^{\infty}\frac{(-1)^{j+1}}{j}\left(\pm\frac{s-L^{2}-1}{K}\right)^{j}.$

The contribution to (4.15) of $-1$ and the term $j=1$ is

$\left(\frac{K-L^{2}+s-2}{2}\right)\left(-1+\frac{s-L^{2}-1}{K}\right)-\left(\frac{K+L^{2}-s}{2}\right)\left(-1-\frac{s-L^{2}-1}{K}\right)=-\frac{s-L^{2}-1}{K}.$

Similarly computing the contributions of the terms with $j$ even and odd
separately, we get that

$\log\frac{\Gamma\left(\frac{K-L^{2}+s-1}{2}\right)}{\Gamma\left(\frac{K+L^{2}-s+1}{2}\right)}=(s-L^{2}-1)\log\left(\frac{K}{2}\right)-\frac{s-L^{2}-1}{K}\\\
-\sum_{j\in
2\mathbb{N}}\frac{K}{j(j+1)}\left(\frac{s-L^{2}-1}{K}\right)^{j+1}+\cdots+O(K^{-1000}).$

We also compute using Stirling’s formula the expansions

$\log\frac{\Gamma\left(\frac{L^{2}}{2}+1\right)^{2}}{\Gamma(L^{2}+1)}=\log
L-(L^{2}+1)\log 2+\cdots+O(K^{-1000})$

and

$\log\frac{\Gamma\left(K+\frac{L^{2}}{2}\right)}{\Gamma\left(K-\frac{L^{2}}{2}-1\right)}=(L^{2}+1)\log
K+\cdots+O(K^{-1000}).$

Recalling the definition (4.4) of $\widehat{H}(s)$ and the assumption
$K^{\frac{2}{3}+\varepsilon}<t<\frac{K^{1+\varepsilon}}{L}$, we deduce that

$\widehat{H}(s)=(2\pi)^{-s-1}L\exp\left(s\log\left(\frac{K}{2}\right)-\sum_{j\in
2\mathbb{N}}\frac{K}{j(j+1)}\left(\frac{s-K^{2}-1}{K}\right)^{j+1}+\cdots\right)+O(K^{-750}).$

Consider first contribution of $j=2$, which is
$-\frac{1}{6}\frac{(\sigma+i(2t+y)-L^{2}-1)^{3}}{K^{2}}$. We expand out the
cube and put aside the terms $\frac{i4t^{3}}{3K^{2}}$ and
$\frac{-(2t+y)^{2}L^{2}}{2K^{2}}$. The rest of the terms are either real and
of lower order or imaginary. The imaginary terms divided by $iy$ are real and
of lower order. Expanding out the contribution of $j\geq 4$, we see that the
real terms are of lower order and the imaginary terms divided by $iy$ are real
and of lower order, using (4.9). In this way, we get that $\widehat{H}(s)$ is
equal to

(4.16)
$L\left(\frac{K}{4\pi}\right)^{\sigma+2it}\exp\left(\frac{-(2t+y)^{2}L^{2}}{2K^{2}}\right)\exp\left(\frac{i4t^{3}}{3K^{2}}\right)\exp\left(iy\left(\log\left(\frac{K}{4\pi}\right)+\cdots\right)+\cdots\right)+O(K^{-750}).$

Next, we turn to the function $\mathscr{G}_{t}^{+}(s)$. We write
$\mathscr{G}_{t}^{+}(s)=\mathscr{H}(s)\mathscr{I}_{t}(s)$, where

$\displaystyle\mathscr{H}(s)$
$\displaystyle\coloneqq\frac{1}{\pi^{2}}(2\pi)^{2s}\Gamma\left(\frac{1-s}{2}\right)^{4}\left(\sin^{2}\frac{\pi
s}{2}+1\right),$ $\displaystyle\mathscr{I}_{t}(s)$
$\displaystyle\coloneqq(2\pi)^{-s}\Gamma\left(\frac{s}{2}+it\right)\Gamma\left(\frac{s}{2}-it\right)\cos\frac{\pi
s}{2}.$

By Stirling’s formula, we get that

(4.17)
$\mathscr{H}(s)=-\left(\frac{2t+y}{4\pi}\right)^{-2\sigma}\exp\left((-2iy-4it)\log\left(\frac{2t+y}{4\pi
e}\right)+\cdots\right)+O(K^{-1000}).$

In the range (4.12), we can further simplify this to

(4.18) $\mathscr{H}(s)=-e^{-2\sigma}\left(\frac{t}{2\pi
e}\right)^{-2\sigma-4it}\exp\left(-2iy\left(\log\left(\frac{t}{2\pi}\right)+\cdots\right)+\cdots\right)+O(K^{-1000}).$

Finally, we also compute using Stirling’s formula the expansion

(4.19)
$\mathscr{I}_{t}(s)=-\frac{i}{2}\left(\frac{y}{4\pi}\right)^{\frac{\sigma-1}{2}}\left(\frac{y+4t}{4\pi}\right)^{\frac{\sigma-1}{2}}\exp\left(\frac{iy}{2}\log\left(\frac{y}{4\pi
e}\right)+\frac{i(y+4t)}{2}\log\left(\frac{y+4t}{4\pi
e}\right)+\cdots\right)\\\ +O(K^{-1000}).$

In the range (4.12), we have that

(4.20)
$\mathscr{I}_{t}(s)=-\frac{i}{2}e^{\frac{\sigma-1}{2}}\left(\frac{y}{4\pi}\right)^{\frac{\sigma-1}{2}}\left(\frac{t}{\pi
e}\right)^{\frac{\sigma-1}{2}+2it}\exp\left(\frac{iy}{2}\left(\log\left(\frac{yt}{4\pi^{2}e}\right)+\cdots\right)+\cdots\right)+O(K^{-1000}).$

Taking (4.16), (4.17), (4.19) together gives (4.10). Taking (4.16), (4.18),
(4.20) together gives (4.13), after collecting the lower order terms into the
functions $b_{1}(y,t,K)$ and $b_{2}(y,t,K)$. The factor
$\exp\left(\frac{-(2t+y)^{2}L^{2}}{2K^{2}}\right)$, which decays for
$t\gg(2t+y)\gg K/L$, is also absorbed into $b_{1}(y,t,K)$. ∎

Now we are ready to analyse $\widetilde{h}^{+}(t)$. The main tool is
stationary phase analysis.

######  Lemma 4.21.

Fix $0<\delta<1/100$. Let $(h,h^{\mathrm{hol}})$ be the pair of test functions
(4.1) with $K,L\in 2\mathbb{N}$ satisfying $K^{\delta/8}\leq L\leq
K^{1/3-\delta/8}$. Then for $\widetilde{h}^{+}$ as in (3.4), we have that for
$t\in\mathbb{R}$,

(4.22)
$\widetilde{h}^{+}(t)=\begin{dcases*}O\left(L^{1+\varepsilon}(1+|t|)^{-\frac{1}{2}}\right)&for
$|t|\leq K^{\frac{2}{3}+\varepsilon}$,\\\ g(t,K,L)+O(K^{-100})&for
$K^{\frac{2}{3}+\varepsilon}<|t|<\frac{K\log K}{L}$,\\\ O(|t|^{-1000})&for
$|t|\geq\frac{K\log K}{L}$,\end{dcases*}$

where

(4.23)
$g(t,K,L)\coloneqq\sum_{\pm}\frac{L}{|t|^{\frac{1}{2}}}\left(\frac{c_{1}K}{|t|}\right)^{\pm
2it}\exp\left(\pm i\frac{c_{2}t^{3}}{K^{2}}(1+d_{1}(\pm t,K))\right)d_{2}(\pm
t,K)$

with $c_{i}$ some real constants for which $c_{1}>0$ and $d_{1}(t,K)$ and
$d_{2}(t,K)$ smooth functions (whose dependence on $L$ we suppress in the
notation), real-valued and complex-valued respectively, that satisfy

(4.24) $\frac{\partial^{j_{1}}}{\partial
t^{j_{1}}}\frac{\partial^{j_{2}}}{\partial K^{j_{2}}}d_{i}(t,K)\ll
t^{-j_{1}}K^{-j_{2}}\times\begin{dcases*}K^{-\varepsilon}&if $i=1$,\\\ 1&if
$i=2$,\end{dcases*}$

for any $j_{1},j_{2}\geq 0$.

###### Proof.

The integrand in (3.4) with $\pm=+$ is meromorphic in the open half-plane
strip $\Re(s)>1+L^{2}-K$ with simple poles at $s=2(\pm it-\ell)$ and poles of
order $4$ at $s=2\ell+1$ for each $\ell\in\mathbb{N}_{0}$. For
$s=\sigma+i\tau$ a bounded distance away from such a pole with $\sigma$
bounded,

(4.25) $\widehat{H}(s)\mathscr{G}_{t}^{+}(s)\\\
\ll_{\sigma}((1+|\tau+2t|)(1+|\tau-2t|))^{\frac{\sigma-1}{2}}(1+|\tau|)^{-2\sigma}\times\begin{dcases*}e^{-\frac{\pi}{2}(2|t|-|\tau|)}&if
$|\tau|\leq 2|t|$,\\\ 1&if $|\tau|\geq 2|t|$,\end{dcases*}\\\
\times\begin{dcases*}K^{\sigma}L&if $|\tau|\leq\frac{K}{L}$,\\\
K^{\sigma}Le^{-\frac{L^{2}\tau^{2}}{4K^{2}}}&if $\frac{K}{L}\leq|\tau|\leq
K$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}e^{-\frac{L^{2}K^{2}}{4\tau^{2}}}&if
$K\leq|\tau|\leq KL$,\\\
L|\tau|^{\sigma}\left(\frac{|\tau|}{K}\right)^{-L^{2}-1}&if $|\tau|\geq
KL$.\end{dcases*}$

In particular, the integral of this function along the line $\Re(s)=\sigma$
with $\sigma<-1$ is

$\ll_{\sigma}L^{1+\sigma}+K^{\sigma}L(1+|t|)^{-\frac{3\sigma}{2}-\frac{1}{2}}.$

Moreover,

$\operatorname*{Res}_{s=2(\pm
it-\ell)}\widehat{H}(s)\mathscr{G}_{t}^{+}(s)\ll_{\ell}(1+|t|)^{3\ell-\frac{1}{2}}\times\begin{dcases*}K^{-2\ell}L&if
$|t|\leq\frac{K}{2L}$,\\\ K^{-2\ell}Le^{-\frac{L^{2}|t|^{2}}{K^{2}}}&if
$\frac{K}{2L}\leq|t|\leq\frac{K}{2}$,\\\
L|t|^{-2\ell}\left(\frac{|t|}{K}\right)^{-L^{2}-1}e^{-\frac{L^{2}K^{2}}{16|t|^{2}}}&if
$\frac{K}{2}\leq|t|\leq\frac{KL}{2}$,\\\
L|t|^{-2\ell}\left(\frac{|t|}{K}\right)^{-L^{2}-1}&if
$|t|\geq\frac{KL}{2}$.\end{dcases*}$

Thus for $|t|\leq K^{2/3-\varepsilon}$, we shift the contour of integration to
the line $\Re(s)=-\frac{1}{2}\frac{\log K}{\log L}$. The dominant contribution
comes from the poles at $s=\pm 2it$, so that $\widetilde{h}^{+}(t)\ll
L(1+|t|)^{-1/2}$.

For $|t|\geq K(\log K)/L$, we keep the contour of integration at
$\sigma\in(0,1)$ and simply estimate the integral via (4.25). This shows that
$\widetilde{h}^{+}(t)\ll|t|^{-1000}$ in this range.

Now suppose that $K^{2/3-\varepsilon}<|t|<K(\log K)/L$. We move the line of
integration in (3.4) far to the right. In doing so, we cross poles of order
$4$ of $\mathscr{G}_{t}^{+}(s)$ at $s=2\ell-1$ for each $\ell\in\mathbb{N}$.
The total contribution of the residues of these poles is $O(K^{-1000})$ by the
bounds (4.3) for $\widehat{H}(s)$ together with Stirling’s formula.

On the new line of integration, we write $s=\sigma+i\tau$ for $\sigma$ a large
fixed even integer, say. By the bounds (4.25), the integrand is negligibly
small outside the range

(4.26) $2|t|-(\log K)^{2}<|\tau|<\frac{K\log K}{L}.$

In this range, we have that

(4.27)
$\widehat{H}(s)\mathscr{G}_{t}^{+}(s)\ll_{\sigma}\frac{L}{(1+|\tau+2t|)^{\frac{1}{2}}(1+|\tau-2t|)^{\frac{1}{2}}}\left(\frac{K^{2}(1+|\tau+2t|)(1+|\tau-2t|)}{|t|^{4}}\right)^{\frac{\sigma}{2}}.$

Thus for $\sigma$ sufficiently large with respect to $\delta$ and
$\varepsilon$, we see that the contribution of the ranges $|\tau\pm
2t|\leq|t|^{3}/K^{2+\varepsilon}$ is $O(K^{-1000})$. Together with (4.26),
this implies that if we define $y\coloneqq\tau-2t$, then we may restrict to
the range

(4.28) $\frac{t^{3}}{K^{2+\varepsilon}}<y<\frac{K\log K}{L}$

in the case that $t$ is positive, and the analogous range with $y$ negative in
the case that $t$ is negative. Since both cases are entirely similar (indeed,
they correspond to the $\pm$ cases in (4.22)), we henceforth assume that $t$
is positive and that (4.28) holds.

We may write

$\widetilde{h}^{+}(t)=\frac{1}{2\pi}\int_{\frac{t^{3}}{K^{2+\varepsilon}}}^{\frac{K\log
K}{L}}A(y,t,K)e^{iB(y,t,K)}\,dy+O(K^{-100}),$

where $A(y,t,K)$ and $B(y,t,K)$ are as in Lemma 4.8. We then perform a dyadic
subdivision on this integral by inserting a collection of weight functions of
the form $W(\frac{y}{R})$ for

(4.29) $\frac{t^{3}}{K^{2+\varepsilon}}<R<\frac{K\log K}{L},$

where $W(y)$ is a smooth function compactly supported on the interval $(1,2)$
with bounded derivatives. After making the substitution $y\mapsto Ry$, we are
left with considering integrals of the form

(4.30) $\frac{R}{2\pi}\int_{0}^{\infty}W(y)A(yR,t,K)e^{iB(yR,t,K)}\,dy.$

By (4.11), we have that

$\frac{\partial}{\partial
y}B(yR,t,K)=R\log\frac{K(yR)^{\frac{1}{2}}(yR+4t)^{\frac{1}{2}}}{(2t+yR)^{2}}+O(R).$

The key point is that this first derivative is sizeable when $R$ is
sufficiently large. Specifically, if $R>t^{3}/K^{2-\varepsilon}$, then by
(4.11), we have that

$\left|\frac{\partial}{\partial y}B(yR,t,K)\right|\gg
R,\qquad\frac{\partial^{j+2}}{\partial y^{j+2}}B(yR,t,K)\ll
R,\qquad\frac{\partial^{j}}{\partial y^{j}}W(y)A(yR,t,K)\ll K^{10}$

for any $j\geq 0$. It follows by integration by parts (for example, by [BKY13,
Lemma 8.1] with parameters $R$ as in (4.29), $Y=R$, and $U=Q=1$) that the
integral (4.30) is $O(K^{-1000})$ when $R>t^{3}/K^{2-\varepsilon}$. We have
therefore shown that

(4.31)
$\widetilde{h}^{+}(t)=\frac{1}{2\pi}\int_{0}^{\infty}\Omega\left(y\frac{K^{2}}{t^{3}}\right)A(y,t,K)e^{iB(y,t,K)}\,dy+O(K^{-100}),$

where $A(y,t,K)$ and $B(y,t,K)$ are as in Lemma 4.8 and $\Omega$ is a smooth
function compactly supported on the interval

(4.32) $\left(K^{-\varepsilon},K^{\varepsilon}\right)$

with derivatives satisfying $\Omega^{(j)}(y)\ll_{j}(K^{\varepsilon})^{j}$.

We now address the range $K^{2/3-\varepsilon}<t<K^{2/3+\varepsilon}$. In this
case, the integral in (4.31) is restricted to the range
$K^{-\varepsilon}<y<K^{\varepsilon}$, and simply bounding trivially using
(4.27) gives the required bound.

We are left to consider the range $K^{2/3+\varepsilon}<t<K(\log K)/L$. Making
the substitution $y\mapsto y\frac{t^{3}}{K^{2}}$ in (4.31) and inserting the
expansions (4.13) and (4.14) for $A(y,t,K)$ and $B(y,t,K)$, we see that up to
a negligible error term, $\widetilde{h}^{+}(t)$ is equal to

(4.33)
$\frac{Lt}{K}\left(\frac{a_{2}K}{t}\right)^{2it}\exp\left(i\frac{a_{3}t^{3}}{K^{2}}\right)\\\
\times\int_{0}^{\infty}\Omega(y)(a_{1}y)^{\frac{\sigma-1}{2}}b_{1}\left(y\frac{t^{3}}{K^{2}},t,K\right)\exp\left(\frac{iyt^{3}}{2K^{2}}\left(\log(a_{4}y)+b_{2}\left(y\frac{t^{3}}{K^{2}},t,K\right)\right)\right)\,dy.$

Letting

$\phi_{t,K}(y)\coloneqq\frac{yt^{3}}{2K^{2}}\left(\log(a_{4}y)+b_{2}\left(y\frac{t^{3}}{K^{2}},t,K\right)\right)$

denote the phase of the integrand, we have that

$\displaystyle\phi_{t,K}^{\prime}(y)$
$\displaystyle=\frac{t^{3}}{2K^{2}}\left(\log(ea_{4}y)+b_{2}\left(y\frac{t^{3}}{K^{2}},t,K\right)+y\frac{\partial}{\partial
y}b_{2}\left(y\frac{t^{3}}{K^{2}},t,K\right)\right)$
$\displaystyle=\frac{t^{3}}{2K^{2}}\left(\log(ea_{4}y)+O(K^{-\varepsilon})\right),$
$\displaystyle\phi_{t,K}^{(j)}(y)$
$\displaystyle=\frac{1}{y^{j-1}}\frac{t^{3}}{2K^{2}}\left(1+O(K^{-\varepsilon})\right)$

for $j\geq 2$. In particular, for $K$ sufficiently large,
$\phi_{t,K}^{\prime}(y)$ strictly increases from a negative to positive value
through the interval (4.32). Thus there is a unique stationary point

$y_{0}=y_{0}(t,K)\coloneqq\frac{1}{ea_{4}}+r(t,K)=\frac{1}{ea_{4}}+O(K^{-\varepsilon}),$

say. Moreover, since $r(t,K)$ is defined implicitly by $F(t,K,r(t,K))=0$ for

$F(t,K,x)\coloneqq\log(1+ea_{4}x)+b_{2}\left(\left(\frac{1}{ea_{4}}+x\right)\frac{t^{3}}{K^{2}},t,K\right)+\left(\frac{1}{ea_{4}}+x\right)\frac{\partial}{\partial
x}b_{2}\left(\left(\frac{1}{ea_{4}}+x\right)\frac{t^{3}}{K^{2}},t,K\right),$

we can use the implicit function theorem to get that $r(t,K)$ is smooth, and
to compute

$\frac{\partial}{\partial t}r(t,K)=-\frac{\left.\frac{\partial}{\partial
t}\right|_{x=r(t,K)}F(t,K,x)}{\left.\frac{\partial}{\partial
x}\right|_{x=r(t,K)}F(t,K,x)}$

and the partial derivative with respect to $K$ and higher derivatives
similarly. From this, we get that

$\frac{\partial^{j_{1}}}{\partial t^{j_{1}}}\frac{\partial^{j_{2}}}{\partial
K^{j_{2}}}r(t,K)\ll K^{-\varepsilon}t^{-j_{1}}K^{-j_{2}}$

for $j_{1},j_{2}\geq 0$. The final step is to derive the stationary phase
expansion of the integral in (4.33), with leading term of size
$\phi_{t,K}^{{}^{\prime\prime}}(y_{0})^{-\frac{1}{2}}\asymp(\frac{t^{3}}{K^{2}})^{-\frac{1}{2}}$
and phase

$\displaystyle\phi_{t,K}(y_{0})$
$\displaystyle=\left(\frac{1}{ea_{4}}+r(t,K)\right)\frac{t^{3}}{2K^{2}}\left(-1-\left(\frac{1}{ea_{4}}+r(t,K)\right)\left.\frac{\partial}{\partial
y}\right|_{y=\frac{1}{ea_{4}}+r(t,K)}b_{2}\left(y\frac{t^{3}}{K^{2}},t,K\right)\right)$
$\displaystyle\asymp\frac{t^{3}}{K^{2}}.$

The full form of the stationary phase expansion can be obtained using [BKY13,
Proposition 8.2] with the main parameter being $Y=\frac{t^{3}}{K^{2}}$ and the
other parameters $V,X,Q$ lying in the interval
$(K^{-\varepsilon},K^{\varepsilon})$. This yields the desired result, where
all lower order terms have been collected into the functions $d_{1}(t,K)$ and
$d_{2}(t,K)$. ∎

Our final task is to prove upper bounds for the last term on the right-hand
side of (3.3).

######  Lemma 4.34.

Fix $0<\delta<1/100$. Let $(h,h^{\mathrm{hol}})$ be the pair of test functions
(4.1) with $K,L\in 2\mathbb{N}$ satisfying $K^{\delta/8}\leq L\leq
K^{1/3-\delta/8}$. Then for $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$ as in
(3.9), we have that

(4.35)
$\lim_{(z_{1},z_{2},z_{3},z_{4})\to\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)}\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}\ll_{\varepsilon}K^{1+\varepsilon}L.$

###### Proof.

Let $0<|z|\leq\varepsilon$ and $z_{j}=\frac{1}{2}+jz$ for $1\leq j\leq 4$, a
choice that ensures that the point $(z_{1},z_{2},z_{3},z_{4})$ does not lie on
the polar lines $z_{n}=z_{m}$ and $z_{n}+z_{m}=1$ for $1\leq n<m\leq 4$. We
need to find an upper bound for the limit of
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$ as $z$ tends to $0$ in terms of $K$
and $L$.

In the definition (3.9) of $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$, we see
that the second sum vanishes because $h(t)$ is identically zero by
construction; see (4.1). The third sum involves the integrals
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm}$. We deform, in a rightward
direction, the contours $\mathcal{C}_{j}$ of these integrals to the vertical
line $\Re(s)=1+100\varepsilon$. We denote the shifted integrals by
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm,\mathrm{shifted}}$ and note
that they are holomorphic for $|z|\leq\varepsilon$. In deforming the contour
of integration, we cross simple poles of
$\Gamma\left(1-z_{m}-\frac{s}{2}\right)$ at $s=2-2z_{m}$ for $1\leq m\leq 4$,
and this gives rise to residues, which we denote
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm,\mathrm{residue}}$. These
residues are equal to the product of $\widehat{H}(2(1-z_{m}))$ and other
meromorphic functions of $z$ that do not depend on $K$ and $L$.

We now have an expression for $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$ as the
first sum in (3.9) plus sums like the third one in (3.9) but with
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm}$ replaced by the sum of
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm,\mathrm{shifted}}$ and
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm,\mathrm{residue}}$. Although
various individual terms in these sums may have poles at $z=0$, arising from
the presence of $\zeta(1+nz)$ and $\Gamma(nz)$ values for nonzero integers
$n$, the total expression $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$ cannot
have a pole at $z=0$ because (3.3) shows that the limit $\lim\limits_{z\to
0}\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$ is finite. By Cauchy’s integral
formula, we can compute this limit as

$\frac{1}{2\pi
i}\oint_{|z|=\varepsilon}\frac{\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}}{z}\,dz,$

and so it suffices to bound $\widetilde{h}_{z_{1},z_{2},z_{3},z_{4}}$ for
$|z|=\varepsilon$. To do this, we may bound all functions that do not depend
on $K$ and $L$ by a constant. The only functions that depend on $K$ and $L$
and arise from the contribution of the first sum in (3.9) and
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm,\mathrm{residue}}$ are
$\widehat{H}(2(1-z_{m}))$ for $1\leq m\leq 4$. These are values of
$\widehat{H}$ evaluated close to $1$, and by Lemma 4.2 we may bound them by
$O_{\varepsilon}(K^{1+\varepsilon}L)$. The only function with $K$ and $L$
dependence that arises from the contribution of
$\widetilde{h}_{z_{1},z_{2},z_{3},z_{4};j}^{\pm,\mathrm{shifted}}$ is
$\widehat{H}(s)$, appearing in the integral over the line
$\Re(s)=1+100\varepsilon$. By the rapid decay of the gamma function in the
integrand, this integral can be truncated to $|\Im(s)|\leq K^{\varepsilon}$ up
to a negligible error. Thus $\widehat{H}(s)$ is also a value of $\widehat{H}$
evaluated close to $1$, and we may bound it by
$O_{\varepsilon}(K^{1+100\varepsilon}L)$. Since $\varepsilon>0$ was arbitrary,
we deduce the requisite bound (4.35). ∎

##  5\. Proof of Lemma 2.3

With the estimates and asymptotics for $\widetilde{h}^{\pm}(t)$ and
$\widetilde{h}^{\mathrm{hol}}(k)$ in hand via Lemmata 4.5 and 4.21, we may now
proceed towards the proof of Lemma 2.3. Our strategy is to apply Kuznetsov’s
formula (3.3) with the choice of pair of test functions $(h,h^{\mathrm{hol}})$
given by (4.1). To control the size of the dual fourth moments appearing on
the right-hand side of (3.3), we require the following well-known moment
bounds.

######  Lemma 5.1.

For $T\geq 1$, we have that

(5.2) $\begin{drcases*}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\ T\leq
t_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}&\\\
\frac{1}{2\pi}\int\limits_{T\leq|t|\leq
2T}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}\,dt&\\\
\sum_{\begin{subarray}{c}f\in\mathcal{B}_{\mathrm{hol}}\\\ T\leq k_{f}\leq
2T\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}&\end{drcases*}\ll_{\varepsilon}T^{2+\varepsilon}.$

###### Proof.

Each of these estimates is a standard application of the spectral large sieve.
∎

We now use Kuznetsov’s formula (3.3) to give an upper bound for large values
of the fourth moment of $L(1/2,f)$ in short intervals.

######  Lemma 5.3.

Fix $0<\delta<1/100$. Let $T\geq 1$, $K\in 2\mathbb{N}$ with $T\leq K\leq 2T$,
$T^{1+\delta/2}\leq V\leq T^{4/3}(\log T)^{7}$, and define $L\coloneqq
2\left\lfloor\frac{1}{2}VT^{-1-\delta/4}\right\rfloor$. Then

(5.4) $\sum_{\begin{subarray}{c}K-L\leq k\leq K+L\\\ k\equiv 0\pmod{2}\\\
V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\\\
\ll\Bigg{|}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\
T^{\frac{2}{3}+\varepsilon}<t_{f}<\frac{T\log
T}{L}\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}g(t_{f},K,L)\Bigg{|}+\Bigg{|}\int\limits_{T^{\frac{2}{3}+\varepsilon}<|t|<\frac{T\log
T}{L}}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}g(t,K,L)\,dt\Bigg{|}$

with $g(t,K,L)$ as in (4.23).

###### Proof.

From Lemma 4.2 (2) (2)(b) and the fact that $L(1/2,f)=0$ whenever $k_{f}\equiv
2\pmod{4}$, we have that

$\sum_{\begin{subarray}{c}K-L\leq k\leq K+L\\\ k\equiv
0\pmod{2}\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\\\
\ll\sum_{f\in\mathcal{B}_{0}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}h(t_{f})+\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}h(t)\,dt+\sum_{f\in\mathcal{B}_{\mathrm{hol}}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}h^{\mathrm{hol}}(k_{f})$

with $(h,h^{\mathrm{hol}})$ the pair of test functions as in (4.1). We apply
Theorem 3.1 with this choice of test functions and bound each term on the
right-hand side of (3.3) individually. Via the bounds (4.6) and (4.7) for
$\widetilde{h}^{-}(t)$ and $\widetilde{h}^{\mathrm{hol}}(k)$ and the spectral
large sieve bounds (5.2), we have that

$\begin{drcases*}\sum_{f\in\mathcal{B}_{0}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\widetilde{h}^{-}(t_{f})&\\\
\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}\widetilde{h}^{-}(t)\,dt&\\\
\sum_{f\in\mathcal{B}_{\mathrm{hol}}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\widetilde{h}^{\mathrm{hol}}(k_{f})&\end{drcases*}\ll
T^{-10}.$

Next, the bounds and asymptotics (4.22) for $\widetilde{h}^{+}(t)$ together
with the spectral large sieve bounds (5.2) yield

(5.5)
$\sum_{f\in\mathcal{B}_{0}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\widetilde{h}^{+}(t_{f})+\frac{1}{2\pi}\int_{-\infty}^{\infty}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}\widetilde{h}^{+}(t)\,dt\\\
=\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\
T^{\frac{2}{3}+\varepsilon}<t_{f}<\frac{T\log
T}{L}\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}g(t_{f},K,L)+\frac{1}{2\pi}\int\limits_{T^{\frac{2}{3}+\varepsilon}<|t|<\frac{T\log
T}{L}}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}g(t,K,L)\,dt\\\
+O_{\varepsilon}(T^{1+\varepsilon}L).$

Finally, the last term on the right-hand side of (3.3) is
$O_{\varepsilon}(T^{1+\varepsilon}L)$ by (4.35). We end by noting that the
left-hand side of (5.4) is either $0$ or at least as large as $V$. Since
$T^{1+\delta/2}\leq V\leq T^{4/3}(\log T)^{7}$ and
$L=2\left\lfloor\frac{1}{2}VT^{-1-\delta/4}\right\rfloor$, so that
$T^{1+\varepsilon}L=o(V)$, it follows that the only terms that may contribute
towards an upper bound for (5.4) are those appearing in (5.5). ∎

We are interested in the second moment of the left-hand side of (5.4) averaged
over even integers $K$ in the interval $[T,2T]$. Bearing in mind the upper
bound given by the two terms on the right-hand side of (5.4), we are led to
studying the average behaviour of $g(t_{1},K,L)\overline{g(t_{2},K,L)}$. It is
here that we take advantage of the oscillatory behaviour of $g(t,K,L)$.

######  Lemma 5.6.

Fix $0<\delta<1/100$. Let $T\geq 1$, $T^{1+\delta/2}\leq V\leq T^{4/3}(\log
T)^{7}$, and define $L\coloneqq
2\left\lfloor\frac{1}{2}VT^{-1-\delta/4}\right\rfloor$. Let $\Omega$ be a
smooth function compactly supported on $(1/2,5/2)$ that is nonnegative, equal
to $1$ on $[1,2]$, and has bounded derivatives. Then for
$T^{2/3+\varepsilon}<|t_{1}|,|t_{2}|<T(\log T)/L$, we have that

(5.7) $\sum_{\begin{subarray}{c}K=2\\\ K\equiv
0\pmod{2}\end{subarray}}^{\infty}\Omega\left(\frac{K}{T}\right)g(t_{1},K,L)\overline{g(t_{2},K,L)}\ll_{A}\frac{TL^{2}}{\sqrt{|t_{1}t_{2}|}}\left(1+\sum_{\pm}|t_{1}\pm
t_{2}|\right)^{-A}$

for any $A\geq 0$, where $g(t,K,L)$ is as in (4.23).

###### Proof.

By the Poisson summation formula, the left-hand side of (5.7) is equal to

$\sum_{\ell=-\infty}^{\infty}\frac{T}{2}\int_{-\infty}^{\infty}\Omega(x)g(t_{1},Tx,L)\overline{g(t_{2},Tx,L)}\left(1+e\left(\frac{Tx}{2}\right)\right)e(-\ell
Tx)\,dx$

upon making the change of variables $x\mapsto Tx$.

For $\ell\neq 0$, we repeatedly integrate by parts, antidifferentiating
$e(-\ell Tx)$ and differentiating the rest. Recalling the definition (4.23) of
$g(t,K,L)$ and the derivative bounds (4.24), we see that each summand with
$\ell\neq 0$ is

$\ll_{A}\frac{TL^{2}}{\sqrt{|t_{1}t_{2}|}}\left(\frac{|\ell|T}{|t_{1}|}+\frac{|\ell|T}{|t_{2}|}\right)^{-A}$

for any $A\geq 0$. Since $|t_{1}|,|t_{2}|<T(\log T)/L$ and $L\geq
T^{\delta/8}$, we deduce that

$\sum_{\begin{subarray}{c}\ell=-\infty\\\ \ell\neq
0\end{subarray}}^{\infty}\frac{T}{2}\int_{-\infty}^{\infty}\Omega(x)g(t_{1},Tx,L)\overline{g(t_{2},Tx,L)}\left(1+e\left(\frac{Tx}{2}\right)\right)e(-\ell
Tx)\,dx\\\
\ll_{A}\frac{TL^{2}}{\sqrt{|t_{1}t_{2}|}}\left(1+\sum_{\pm}|t_{1}\pm
t_{2}|\right)^{-A}.$

For $\ell=0$ and $|t_{1}-t_{2}|\ll 1$, we simply bound the integral trivially,
so that

$\frac{T}{2}\int_{-\infty}^{\infty}\Omega(x)g(t_{1},Tx,L)\overline{g(t_{2},Tx,L)}\left(1+e\left(\frac{Tx}{2}\right)\right)\,dx\ll\frac{TL^{2}}{\sqrt{|t_{1}t_{2}|}}.$

Finally, for $\ell=0$ and $|t_{1}-t_{2}|\gg 1$, we instead repeatedly
integrate by parts, antidifferentiating $x^{\pm_{1}2it_{1}\mp_{2}2it_{2}}$ and
differentiating the rest. We deduce that

$\displaystyle\frac{T}{2}\int_{-\infty}^{\infty}\Omega(x)g(t_{1},Tx,L)\overline{g(t_{2},Tx,L)}$
$\displaystyle\left(1+e\left(\frac{Tx}{2}\right)\right)\,dx$
$\displaystyle\ll_{A}\frac{TL^{2}}{\sqrt{|t_{1}t_{2}|}}\left(\sum_{\pm}|t_{1}\pm
t_{2}|+\frac{T^{2}}{|t_{1}|^{2}}+\frac{T^{2}}{|t_{2}|^{2}}\right)^{-A}$
$\displaystyle\ll_{A}\frac{TL^{2}}{\sqrt{|t_{1}t_{2}|}}\sum_{\pm}|t_{1}\pm
t_{2}|^{-A}$

for any $A\geq 0$, having again recalled that $|t_{1}|,|t_{2}|<T(\log T)/L$
and $L\geq T^{\delta/8}$. ∎

We are now in a position to complete the proof of Lemma 2.3.

###### Proof of Lemma 2.3.

From Lemmata 5.3 and 5.6 and Young’s inequality, we have that

$\displaystyle\sum_{\begin{subarray}{c}T\leq K\leq 2T\\\ K\equiv
0\pmod{2}\end{subarray}}$ $\displaystyle\hskip
28.45274pt\Bigg{(}\sum_{\begin{subarray}{c}K-L\leq k\leq K+L\\\ k\equiv
0\pmod{2}\\\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}$
$\displaystyle\ll\sum_{\begin{subarray}{c}K=2\\\ K\equiv
0\pmod{2}\end{subarray}}^{\infty}\Omega\left(\frac{K}{T}\right)\Bigg{|}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\
T^{\frac{2}{3}+\varepsilon}<t_{f}<\frac{T\log
T}{L}\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}g(t_{f},K,L)\Bigg{|}^{2}$
$\displaystyle\qquad+\sum_{\begin{subarray}{c}K=2\\\ K\equiv
0\pmod{2}\end{subarray}}^{\infty}\Omega\left(\frac{K}{T}\right)\Bigg{|}\int\limits_{T^{\frac{2}{3}+\varepsilon}<|t|<\frac{T\log
T}{L}}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{2}g(t,K,L)\,dt\Bigg{|}^{2}$
$\displaystyle\ll_{A}TL^{2}\sum_{\begin{subarray}{c}f_{1},f_{2}\in\mathcal{B}_{0}\\\
T^{\frac{2}{3}+\varepsilon}<t_{f_{1}},t_{f_{2}}<\frac{T\log
T}{L}\end{subarray}}\frac{L\left(\frac{1}{2},f_{1}\right)^{4}L\left(\frac{1}{2},f_{2}\right)^{4}}{L(1,\operatorname{ad}f_{1})L(1,\operatorname{ad}f_{2})}\frac{1}{\sqrt{t_{f_{1}}t_{f_{2}}}}(1+|t_{f_{1}}-t_{f_{2}}|)^{-A}$
$\displaystyle\qquad+TL^{2}\iint\limits_{T^{\frac{2}{3}+\varepsilon}<t_{1},t_{2}<\frac{T\log
T}{L}}\left|\frac{\zeta\left(\frac{1}{2}+it_{1}\right)^{4}\zeta\left(\frac{1}{2}+it_{2}\right)^{4}}{\zeta(1+2it_{1})\zeta(1+2it_{2})}\right|^{2}\frac{1}{\sqrt{t_{1}t_{2}}}(1+|t_{1}-t_{2}|)^{-A}\,dt_{1}\,dt_{2}$
$\displaystyle\ll_{A}TL^{2}\iint\limits_{T^{\frac{2}{3}+\varepsilon}<x_{1},x_{2}<\frac{T\log
T}{L}}\Bigg{(}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\ x_{1}\leq
t_{f}\leq
x_{1}+1\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\Bigg{)}^{2}\frac{1}{\sqrt{x_{1}x_{2}}}(1+|x_{1}-x_{2}|)^{-A}\,dx_{1}\,dx_{2}$
$\displaystyle\qquad+TL^{2}\iint\limits_{T^{\frac{2}{3}+\varepsilon}<t_{1},t_{2}<\frac{T\log
T}{L}}\left|\frac{\zeta\left(\frac{1}{2}+it_{1}\right)^{4}}{\zeta(1+2it_{1})}\right|^{4}\frac{1}{\sqrt{t_{1}t_{2}}}(1+|t_{1}-t_{2}|)^{-A}\,dt_{1}\,dt_{2}.$

The integral over $x_{2}$ is $O(x_{1}^{-1/2})$ and the integral over $t_{2}$
is $O(t_{1}^{-1/2})$, so that

$\sum_{\begin{subarray}{c}T\leq K\leq 2T\\\ K\equiv
0\pmod{2}\end{subarray}}\Bigg{(}\sum_{\begin{subarray}{c}K-L\leq k\leq K+L\\\
k\equiv 0\pmod{2}\\\ V\leq\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\leq
2V\end{subarray}}\undertilde{\mathcal{M}}^{\mathrm{hol}}(k)\Bigg{)}^{2}\\\ \ll
TL^{2}\int_{T^{\frac{2}{3}+\varepsilon}}^{\frac{T\log
T}{L}}\Bigg{(}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\ x\leq t_{f}\leq
x+1\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\Bigg{)}^{2}\,\frac{dx}{x}+TL^{2}\int_{T^{\frac{2}{3}+\varepsilon}}^{\frac{T\log
T}{L}}\left|\frac{\zeta\left(\frac{1}{2}+it\right)^{4}}{\zeta(1+2it)}\right|^{4}\,\frac{dt}{t}.$

To bound these two terms, we dyadically subdivide both integrals, additionally
using the classical lower bound $|\zeta(1+it)|\gg 1/\log(3+|t|)$ for
$t\in\mathbb{R}$ for the latter. We then appeal to the bounds

$\displaystyle\int_{U}^{2U}\Bigg{(}\sum_{\begin{subarray}{c}f\in\mathcal{B}_{0}\\\
x\leq t_{f}\leq
x+1\end{subarray}}\frac{L\left(\frac{1}{2},f\right)^{4}}{L(1,\operatorname{ad}f)}\Bigg{)}^{2}\,dx\ll_{\varepsilon}U^{3+\varepsilon},$
$\displaystyle\int_{U}^{2U}\left|\zeta\left(\frac{1}{2}+it\right)\right|^{16}\,dt\ll_{\varepsilon}U^{\frac{283}{108}+\varepsilon}$

for any $U\geq 1$, with the former due to Jutila [Jut04, Theorem 1] and the
latter due to Ivić [Ivi03, Theorem 8.3]. In this way, we deduce the desired
upper bound $O_{\varepsilon}(T^{3+\varepsilon})$. ∎

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