# Selberg’s Central Limit Theorem for automorphic $L$-functions in the
$t$-aspect

Madhuparna Das  Department of Mathematics, University of Exeter, Stocker Rd,
Exeter EX4 4PY, UK<EMAIL_ADDRESS>
###### Abstract.

We present a proof of Selberg’s Central Limit Theorem for automorphic
$L$-functions of degree 2 using Radziwiłł and Soundararajan’s method.
Additionally, we prove the independence of the automorphic $L$-functions
associated with the sequence of primitive holomorphic cusp forms.

###### Key words and phrases:

Automorphic $L$-functions, value distribution, central limit theorem,
independence

###### 2010 Mathematics Subject Classification:

11M06

##  1\. Introduction

The value distribution (and moments) of the Riemann zeta function, as well as
$L$-functions has always been an interesting question for mathematicians (see
e.g. the introduction of Harper’s paper [9] for further references). In 1946,
Selberg [1, 2] observed that the logarithm of the Riemann zeta function
behaves like a random variable and it has normal distribution with mean 0 and
variance $\frac{1}{2}\log\log|t|$, where $t$ varies in $[T,2T]$ for
sufficiently large $T$. Further, he extended this result to all $L$-functions
that belong to the Selberg class [2].

In 2017, Radziwiłł and Sounadararajan [3] gave a new proof of Selberg’s
Central Limit Theorem. Rather than using complicated number theoretic
arguments, they have used basic facts of the Riemann zeta function which makes
their proof more elegant. Recently, P. Hsu and P. Wong [4] used the same
technique and proved this result for Dirichlet $L$-functions in $t$-aspect
(which means they have integrated over $[T,2T]$ instead of averaging over the
Dirichlet characters). In Radziwiłł Soundararajan’s paper [3], there was a
clear indication that this result could be extended for a higher degree
$L$-functions (of appropriate families).

In this paper, we give a simple proof of Selberg’s influential theorem
(formally stated below) for classical automorphic $L$-functions of degree 2,
by adapting Radziwiłł and Soundarajan’s technique.

######  Theorem 1.

Let $V$ be a fixed real number. Then for all large $T$,

$\frac{1}{T}meas\left\\{T\leq t\leq 2T:\log|L(f,\frac{1}{2}+it)|\geq
V\sqrt{\frac{1}{2}\log\log
T}\right\\}\sim\frac{1}{\sqrt{2\pi}}\int_{V}^{\infty}{e^{-u^{2}/2}du},$

where $f$ is a primitive holomorphic cusp form of weight $k\geq 1$, level $q$.

Well, some modifications are needed in the proof since we are working with
families of $L$-functions. In our case, we have a further twist
$\lambda_{f}(n)$ which is the Hecke-eigenvalue of the holomorphic cusp form
$f$. Except for the basic properties of the $L$-functions we also use a few
properties of $\lambda_{f}(n)$.

After the derivation of the central limit theorem Selberg [2] mentioned his
orthogonality conjecture following from which he further remarked that the
primitive $L$-functions belonging to the Selberg class are statistically
independent. Although, Selberg did not give any precise description of
independence. In the paper by P. H. Hsu and P. Wong [4], proved the same
result for the multiple Dirichlet $L$-functions associated with the primitive
Dirichlet character $\chi$.

######  Theorem 2.

Let $\lambda_{f_{1}}$ and $\lambda_{f_{2}}$ be two distinct holomorphic cusp
forms. For sufficiently large $T$ and $t\in[T,2T]$, the random vector

$\displaystyle\left(\log\left|L\left(f_{1},\frac{1}{2}+it\right)\right|,\log\left|L\left(f_{2},\frac{1}{2}+it\right)\right|\right)$

is approximately a bi-variate normal distribution with mean vector $0_{2}$ and
co-variance matrix $\frac{1}{2}(\log\log T)I_{2}$.
Consequently, $\log\left|L\left({f_{1}},\frac{1}{2}+it\right)\right|$ and
$\log\left|L\left({f_{2}},\frac{1}{2}+it\right)\right|$ are approximately
independent.

As a more generalized version of this theorem, we further prove that the
logarithm of the real part of the automorphic $L$-functions form a Gaussian
process. Since we intend to prove Theorem 2 by the method established in [3],
we need to prove our next theorem by studying the joint distribution of these
$L$-functions.

######  Theorem 3.

Let $(f_{j})_{j=1}^{n}$ be a sequence of distinct primitive holomorphic cusp
forms. Then for all large $T$ and $t\in[T,2T]$ the random vector

$\displaystyle\left(\log|L(f_{1},\frac{1}{2}+it)|,\ldots,\log|L(f_{j},\frac{1}{2}+it)|\right)$

is approximately $n$-variate normal distribution with mean vector $0_{n}$ and
co-variance matrix $\frac{1}{2}(\log\log T)I_{n}$, where $0_{n}$ is the zero
vector and $I_{n}$ is the $n\times n$ identity matrix.
Consequently, the random variables $\log|L(f_{j},\frac{1}{2}+it)|$’s (for
$j=1,\ldots,n$) are independent and each $\log|L(f_{j},\frac{1}{2}+it)|_{f\in
J}$ forms a Gaussian process [15] for any totally ordered set $J$ of distinct
primitive holomorphic cusp forms.

##  2\. Notation and Preliminaries

Throughout the paper we have use standard Landau-Vinogradov notation. We write
for the given functions $f(x)$ and $g(x)$, $f(x)=O(g(x))$ or, $f(x)\ll g(x)$
if there exists a constant $c>0$ such that $|f(x)|\leq cg(x)$ for all $x$,
where $c$ is the implied constant. We write $f(x)\sim g(x)$ (and
$f(x)=o(g(x))$) if the ration $f(x/g(x))$ tends to 1 (and 0 respectively) as
$x\to\infty$. The Möbius function is denoted as $\mu$ given by

$\displaystyle\mu(n)=\begin{cases}(-1)^{r}&\text{if $n$ is the product of $r$
distinct primes}\\\ 0&\text{otherwise.}\end{cases}$

The prime-omega function is denoted as $\Omega(n)$ which denote the number of
prime divisors of $n$ counted with multiplicity.

Let $f$ be primitive holomorphic cusp form of weight $k\geq 1$ and level $q$,
with nebentypus111Let $q\geq 1$ be an integer, and $\psi$ a Dirichlet
character modulo $q$ (not necessarily primitive). Clearly $\psi$ induces a
character of the modular group $\Gamma_{0}(q)$ by $\psi(g)=\psi(d)$ for
$g=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$. A modular form of weight $k$,
level $q$ and nebentypus (or character) $\psi$ is a holorphic function $f$ on
the upper-half plane $\mathbb{H}$ which satisfies
$f|_{k}\gamma=\psi(\gamma)f$, for all $\gamma\in\Gamma_{0}(q)$. $\psi$. Then
$f$ has the Fourier expansion

$\displaystyle f(z)=\sum_{n=1}^{\infty}n^{\frac{k-1}{2}}\lambda_{f}(n)e(nz)$

Then an $L$-series attached to $f$ is defined as

$\displaystyle
L(f,s)=\sum_{n}\frac{\lambda_{f}(n)}{n^{s}}=\prod_{p}\left(1-\frac{\lambda_{f}(p)}{p^{s}}-\frac{\psi(p)}{p^{2s}}\right)^{-1}$

where $\lambda_{f}(n)$ is the $n^{th}$ Hecke eigenvalue of $f$, with conductor
$q$ and gamma factor

$\displaystyle\gamma(f,s)=c_{k}(2\pi)^{-s}\Gamma\left(s+\frac{k-1}{2}\right)$

where $c_{k}=2^{(3-k)/2}\sqrt{\pi}$ by Legendre duplication formula.

Let

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

then $\Lambda(f,s)$ has analytic continuation to the entire complex plane and
satisfies the functional equation

$\displaystyle\Lambda(f,s)=q^{s/2}\gamma(f,s)L(f,s)=\varepsilon(f)\Lambda(\bar{f},1-s)$

where $\Lambda(f,s)$ denotes the complete $L$-function followed by the formula
given above and $\bar{f}$ is an object associated with $f$ (the dual of $f$)
for which $\lambda_{\bar{f}}(n)=\bar{\lambda}_{f}(n)$ and $\varepsilon(f)$ is
the complex number of absolute value 1 called the root number of $L(f,s)$.

For $s=\sigma+it$ with $t\neq 0$ we have the Stirling estimate (see 5.113 of
[6]) asserting that for any fixed $\sigma$,

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

Further from Stirling’s approximation we have for fixed $\delta>0$ (which is
sufficiently small)

$\displaystyle\frac{\Gamma(\mathfrak{z}+\alpha)}{\Gamma(\mathfrak{z}+\beta)}=\mathfrak{z}^{\alpha-\beta}\left(1+O\left(\frac{|(\alpha-\beta)(\alpha+\beta-1)|}{|\mathfrak{z}|}\right)\right),$

where $\alpha$ and $\beta$ are arbitrary constants and
$|\arg(\mathfrak{z})|\leq\pi-\delta$.

We require the following estimates [3]. For any $m,n\in\mathbb{N}$, one has

(1)
$\displaystyle\int_{T}^{2T}\left(\frac{m}{n}\right)^{it}dt=\begin{cases}T&\text{
if $m=n$};\\\ O\left(\min\\{T,\frac{1}{|\log(m/n)|}\\}\right)&\text{ if $m\neq
n$}.\end{cases}$

For $m\neq n$ one further has

(2) $\displaystyle\frac{1}{|\log(m/n)|}\ll\begin{cases}1&\text{ if $m\geq 2n$,
or $m\leq n/2$};\\\ \frac{m}{|m-n|}&\text{if $n/2<m<2n$};\\\ \sqrt{mn}&\text{
for all $m\neq n$}.\end{cases}$

For the proof of Theorem 2 and 3 we recall some standard results on normal
random variables. If $Z$ is a standard normal random variable and
$Z=\frac{X-\mu}{\sigma}$, then $X$ is a normal random variable with mean $\mu$
and variance $\sigma^{2}$ and we write $X\sim\mathcal{N}(\mu,\sigma^{2})$.

Let $X_{1}$ and $X_{2}$ be two random variables. The co-variance of $X_{1}$
and $X_{2}$ in terms of expectation is given by

$\displaystyle\text{Cov}(X_{1},X_{2})\coloneqq\mathbb{E}[X_{1}X_{1}]-\mathbb{E}[X_{1}]\mathbb{E}[X_{2}]$

Recall that $X_{1}$, $X_{2}$ is said to be uncorrelated if their co-variance
is 0. That means

$\displaystyle\text{Cov}(X_{1},X_{2})=0\Longleftrightarrow\mathbb{E}[X_{1}X_{2}]=\mathbb{E}[X_{1}]\mathbb{E}[X_{2}]$

The co-variance coefficient defined by

$\displaystyle\rho(X_{1},X_{2})\coloneqq\frac{\text{Cov}(X_{1},X_{2})}{\sqrt{\mathrm{Var}(X_{1})\mathrm{Var}(X_{2})}}.$

If $X_{1}$ and $X_{2}$ are uncorrelated then the co-variance coefficient
$\rho(X_{1},X_{2})$ is 0. Note that two independent random variables are
always uncorrelated, but the converse is not necessarily true. If $X_{1}$ and
$X_{2}$ are uncorrelated then

(3) $\displaystyle\textrm{Var}(X+Y)=\textrm{Var}(X)+\textrm{Var}(Y)$

######  Lemma 1.

Let $(X_{i})_{i=1}^{n}$ be a sequence of normal distributions. Then
$(X_{i})_{i=1}^{n}$ is an $n$-variate normal distribution if and only if any
linear combination of $X_{j}$’s is a normal distribution.

Suppose, further, that $(X_{i},X_{j})$ is a bivariate normal distribution.
Then $X_{k}$ and $X_{l}$ are independent if and only if they are uncorrelated.

In a later section, we prove that a finite sequence of $L$-functions
associated with distinct cusp forms, forms a Gaussian process [15].

A stochastic process $(X_{i})$ is called a Gaussian process if every finite
sub-sequence of $(X_{i})_{i\in J}$ has a multivariate normal distribution.

For some distributions, one can not trace the characteristic function but the
moments can be calculated for those distributions. In these cases, one can
prove the weak convergence of the distribution by establishing the moment
convergence under the condition in which the moments is uniquely determined.

######  Lemma 2.

Let $(\mathfrak{S},\mathfrak{F},\mathfrak{P})$ be a complete probability
space. We say that $(X_{n})$ converges to $X$ in distribution if
$\lim_{n\to\infty}\mathfrak{P}(s\in\mathfrak{S}:X_{n}(s)\leq
x)=\mathfrak{P}(s\in\mathfrak{S}:X(s)\leq x)$ for every $x$ such that
$\mathfrak{P}(s\in\mathfrak{S}:X(s)=x)=0$.

Suppose that the distribution of $X$ is determined by its moments of all
orders, and that
$\lim_{n\to\infty}\mathbb{E}[X_{n}^{r}]=\mathbb{E}[X^{r}]$ for $r=1,2,\ldots$.
Then $X_{n}\Rightarrow X$.

For the proof of this lemma see Theorem 30.2 of [16].

##  3\. The Setup

In this section, we give a sketch of the proof of Theorem 1, then we will get
into the details. The proof contains four important steps. In the first step,
we take away the problem from the critical line by counting the distribution
of the zeros of the $L$-functions.

######  Proposition 1.

Let $T$ be large and suppose $T\leq t\leq 2T$. Then for any $\sigma>1/2$ we
have

$\int_{t-1}^{t+1}\left|{\log|L(f,\frac{1}{2}+iy)|}-{\log\left|L(f,\sigma+iy)\right|}\right|dy\ll(\sigma-\frac{1}{2})\log
T.$

Using the same notation as in [3] we fix the parameters,

$\displaystyle W=(\log\log\log T)^{4},\;X=T^{1/(\log\log\log
T)^{2}},\;Y=T^{(1/\log\log T)^{2}},\;\sigma_{0}=\frac{1}{2}+\frac{W}{\log T},$

where $T>0$ is sufficiently large so that $W\geq 3$.

As stated in [3] from Proposition 1, it follows that $\log|L(\frac{1}{2}+it)|$
and $\log|L(\sigma_{0}+it)|$ differ by at most $AW$ (where $\log
A=o(\sqrt{\log\log T})$) except on a set of measure $O(T/A)$. If $AW$ is small
compared to $\sqrt{\log\log T}$ then the difference between
$\log|L(\frac{1}{2}+it)|$ and $\log|L(\sigma_{0}+it)|$ is negligible and their
distributions is same. So from now on we can focus on studying the
distribution of $\log|L(\sigma_{0}+it)|$, which makes the problem more easier.

As the second step, consider the auxiliary series

$\displaystyle\mathcal{P}(f,s)=\mathcal{P}(f,s;X)=\sum_{2\leq n\leq
X}\frac{\Lambda_{f}(n)}{n^{s}\log n}$

By computing moments we determine the distribution of $\mathcal{P}(f,s)$.

######  Proposition 2.

As $t$ varies in $T\leq t\leq 2T$, the distribution of
$\Re(\mathcal{P}(f,\sigma_{0}+it))$ is approximately normal with mean 0 and
variance $\sim\frac{1}{2}\log\log T$.

Now it remains to connect the auxiliary series $\mathcal{P}(f,s)$ with the
$L$-functions to prove Theorem 1. For this, we use mollification technique.
Introducing the Dirichlet polynomial $M(f,s)$ given by

$\displaystyle M(f,s)=\sum_{n}\frac{\mu(n)a(n)\lambda_{f}(n)}{n^{s}}.$

where $a(n)$ is defined as

$\displaystyle a(n)=\begin{cases}1&\text{if $n$ is composed only primes below
$X$ and has at most $100\log\log T$ primes below $Y$,}\\\ &\text{and at most
$100\log\log\log T$ primes between $Y$ and $X$.}\\\ 0&\text{otherwise
}.\end{cases}$

By the definition of $a(n)$ it takes the value 0 except when $n\leq
Y^{100\log\log T}X^{100\log\log\log T}<T^{\epsilon}$. It is now evident that
$M(f,s)$ is short Dirichlet polynomial. Our target is to show that $M(f,s)$
can be approximated by $e^{-\mathcal{P}(f,s)}$.

######  Proposition 3.

For $T\leq t\leq 2T$

$\displaystyle M(f,\sigma_{0}+it)=(1+o(1))\exp(-\mathcal{P}(f,\sigma_{0}+it))$

except perhaps on a subset of measure $o(T)$.

Now it remains to connect the $L$-functions with the Dirichlet polynomial
$M(f,s)$ in order to prove the theorem. Roughly speaking from the definition
of $M(f,s)$ we can see that $L(f,s)$ and $M(f,s)$ are inverse to each other,
which we are going to prove as our final step.

######  Proposition 4.

For $T\leq t\leq 2T$,

$\displaystyle\int_{T}^{2T}\left|1-L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)\right|^{2}dt=o(1)$

So that for sufficiently large $t$ we have

$\displaystyle L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)=1+o(1),$

except perhaps on a set of measure $o(T)$.

Now we are ready to prove Theorem 1.

Proof of Theorem 1: Recalling Proposition 4, it says that for all $t\in[T,2T]$
(outside of a set of measure $o(T)$) we have

$\displaystyle L(f,\sigma_{0}+it)=(1+o(1))M(f,\sigma_{0}+it)^{-1}.$

By Proposition 3, for all $t\in[T,2T]$ (outside of a set of measure $o(T)$) we
know that

$\displaystyle
M(f,\sigma_{0}+it)^{-1}=(1+o(1))\exp(\mathcal{P}(f,\sigma_{0}+it))$

and by Proposition 2 we can conclude that $\log|L(f,\sigma_{0}+it)|$ is
normally distributed with mean 0 and variance $\frac{1}{2}\log\log T$. Finally
with the help of Proposition 1 we deduce that

$\displaystyle\int_{T}^{2T}\left|\log|L(f,\frac{1}{2}+it)-\log|L(f,\sigma_{0}+it)\right|dt\ll
T(\sigma_{0}-\frac{1}{2})\log T=WT,$

So outside of a set of measure $O(T/W)=o(T)$ we have

$\displaystyle\log|L(f,\frac{1}{2}+it)=\log|L(f,\sigma_{0}+it)+O(W^{2}).$

Since $W^{2}=o(\sqrt{\log\log T})$ it follows that similarly like
$\log|L(f,\sigma_{0}+it)$, $\log|L(f,\frac{1}{2}+it)$ has the normal
distribution with mean 0 and variance $\frac{1}{2}\log\log T$ as well, which
completes the proof of Theorem 1. ∎

##  4\. Proof of Proposition 1

Let $f$ be a primitive holomorphic cusp form of weight $k\geq 1$ and $q$ be
the conductor.

Set

(4)
$G(f,s)=q^{s/2}\gamma(f,s)=q^{s/2}c_{k}(2\pi)^{-s}\Gamma\left(s+\frac{k-1}{2}\right)$

where $\gamma(f,s)$ is the gamma factor of $L(f,s)$ with
$c_{k}=2^{(3-k)/2}\sqrt{\pi}$. If $t$ is sufficiently large and
$y\in[t-1,t+1]$, then by Stirling’s formula we will show that

$\left|\log\frac{G(f,\sigma+iy)}{G(f,1/2+iy)}\right|\ll\left(\sigma-\frac{1}{2}\right)\log
t$

Expanding the complex logarithm and Putting Stirling’s
approximation222Consider the Stirling formula given in notation. We can
separate the real and the argument part. For the real part we can write
$|\Gamma(\sigma+it)|\sim\sqrt{2\pi}|t|^{\sigma-\frac{1}{2}}e^{-\frac{\pi}{2}|t|}$.
The argument can be written as $\arg(\Gamma(\sigma+it))\sim t\log t-t$. for
Gamma function in (4) we have,

$\displaystyle\log\frac{G(f,\sigma+iy)}{G(f,1/2+iy)}$ $\displaystyle=$
$\displaystyle\log\left|\frac{G(f,\sigma+iy)}{G(f,1/2+iy)}\right|+i\arg\left(\frac{G(f,\sigma+iy)}{G(f,1/2+iy)}\right)$
$\displaystyle=$
$\displaystyle\log\frac{q^{\frac{\sigma}{2}}c_{k}{(2\pi)}^{-\sigma}|\Gamma{(\sigma+iy+\frac{k-1}{2}})|}{q^{\frac{(1/2)}{2}}c_{k}{(2\pi)}^{-1/2}|\Gamma{(1/2+iy+\frac{k-1}{2}})|}+i\arg\left(\frac{y/2\log
q\cdot y\log 2\pi\cdot(y-\log y)}{y/2\log q\cdot y\log 2\pi\cdot(y-\log
y)}\right).$

We know that $\arg(1)=0$. Since, $t$ is large enough and $y\in[t-1,t+1]$,
ignoring the $q$ term in the logarithm we write

$\displaystyle\left|\log\frac{G(f,\sigma+iy)}{G(f,1/2+iy)}\right|\ll\left|\log
t^{(\sigma-\frac{1}{2})}\right|\ll\left(\sigma-\frac{1}{2}\right)\log t.$

Recall the functional equation of the complete $L$-function

$\displaystyle\Lambda(f,s)=G(f,s)L(f,s).$

To prove Proposition 1 it is enough to prove that

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{\Lambda(f,1/2+iy)}{\Lambda(f,\sigma+iy)}\right|\right|dy\ll\left(\sigma-\frac{1}{2}\right)\log
T.$

Recalling Hadamard’s factorization formula (see e.g. Lemma 1), there exist
constants $a=a(f)$ and $b=b(f)$ (where $b(f)=-\sum_{\rho}\Re(1/{\rho})$) such
that

$\displaystyle(s(1-s))\Lambda(f,s)=e^{a+bs}\prod_{\rho\neq
0,1}\left(1-\frac{s}{\rho}\right)e^{s/\rho},$

where $\rho$ ranges over all zeros of $\Lambda(f,s)$ different from 0,1.

Assuming that $y$ is not the ordinate of a zero of $L(f,s)$ we can write

$\displaystyle\log\left|\frac{\Lambda(f,1/2+iy)}{\Lambda(f,\sigma+iy)}\right|$
$\displaystyle=$
$\displaystyle\sum_{\rho}\log\left|\frac{(1/2+iy)-\rho}{(\sigma+iy)-\rho}\right|.$

Suppose $\rho=\beta+i\gamma$ is a non-trivial zero of $L(f,s)$. Integrating
over
$y\in(t-1,t+1)$ we get

(5)
$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{\Lambda(f,\frac{1}{2}+iy)}{\Lambda(f,\sigma+iy)}\right|\right|dy\leq$
$\displaystyle\sum_{\rho}\int_{t-1}^{t+1}\left|\log\left|\frac{\frac{1}{2}+iy-\rho}{\sigma+iy-\rho}\right|\right|dy$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{\rho}\int_{t-1}^{t+1}\left|\log\frac{(\beta-\frac{1}{2})^{2}+(y-\gamma)^{2}}{(\beta-\sigma)^{2}+(y-\gamma)^{2}}\right|dy.$

If $|t-\gamma|\geq 2$ then for any $y\in(t-1,t+1)$ we have

$\displaystyle\left|\log\left|\frac{\frac{1}{2}+iy-\rho}{\sigma+iy-\rho}\right|\right|=\left|\Re\log\left(1-\frac{\sigma-\frac{1}{2}}{\sigma+iy-\rho}\right)\right|=$
$\displaystyle\left|\Re\frac{\sigma-\frac{1}{2}}{\sigma+iy-\rho}\right|+O\left(\frac{(\sigma-\frac{1}{2})^{2}}{(y-\gamma)^{2}}\right)$
$\displaystyle=$ $\displaystyle
O\left(\frac{(\sigma-\frac{1}{2})}{(y-\gamma)^{2}}\right)$

So we can write

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{(1/2+iy)-\rho}{(\sigma+iy)-\rho}\right|\right|dy\ll\frac{(\sigma-1/2)}{(t-\gamma)^{2}}.$

Then contribution of these zeros give

$\displaystyle\sum_{\begin{subarray}{c}\rho\\\ |t-\gamma|\geq
2\end{subarray}}\frac{(\sigma-\frac{1}{2})}{(t-\gamma)^{2}}\ll(\sigma-\frac{1}{2})\log
T.$

Now consider the range $|t-\gamma|\leq 2$ (which is basically the zeros near
$t$) we have

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{1/2+iy-\rho}{\sigma+iy-\rho}\right|\right|dy=$
$\displaystyle\frac{1}{2}\int_{t-1}^{t+1}\left|\log\frac{(\beta-\frac{1}{2})^{2}+(y-\gamma)^{2}}{(\beta-\sigma)^{2}+(y-\gamma)^{2}}\right|dy$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{-\infty}^{\infty}\left|\log\left|\frac{(\beta-\frac{1}{2})^{2}+x^{2}}{(\beta-\sigma)^{2}+x^{2}}\right|\right|dx$
$\displaystyle=$ $\displaystyle\pi\left(\sigma-\frac{1}{2}\right).$

So in this case the contribution of zeros is $\ll(\sigma-\frac{1}{2})\log T$.

Thus in either case

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{1/2+iy-\rho}{\sigma+iy-\rho}\right|\right|dy\ll\frac{(\sigma-\frac{1}{2})}{1+(t-\gamma)^{2}}.$

Inserting this in (5), from Theorem 5.38 of [6] we can conclude

$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{L(f,1/2+iy)}{L(f,\sigma+iy)}\right|\right|dy=$
$\displaystyle\int_{t-1}^{t+1}\left|\log\left|\frac{\Lambda(f,1/2+iy)}{\Lambda(f,\sigma+iy)}\right|\right|dy+O\left(\sigma-\frac{1}{2}\right)\log
T$ $\displaystyle\ll$ $\displaystyle\left(\sigma-\frac{1}{2}\right)\log T$

which completes the proof. ∎

##  5\. Proof of Proposition 2

To prove this proposition we restrict the sum $\mathcal{P}(f,s)$ to primes and
compute moments. We know that Gaussian distribution can uniquely be determined
by its moments.

For the terms involving higher power of primes i.e. $p^{k}$ (with $k\geq 3$)
we have333Note that
$\Lambda_{f}(p^{k})=(\alpha_{1}(p)^{k}+\alpha_{2}(p)^{k})\log p$ with
$|\alpha_{1}(p)|,|\alpha_{2}(p)|=1$, if $(p,q)=1$ (see Theorem 8.2 of [18]).

$\displaystyle\left|\sum_{\begin{subarray}{c}2\leq p^{k}\leq X\\\ k\geq
3\end{subarray}}\frac{\Lambda_{f}(p^{k})}{p^{ks}(k\log
p)}\right|\leq\sum_{\begin{subarray}{c}2\leq p^{k}\leq X\\\ k\geq
3\end{subarray}}\frac{1}{3p^{k\sigma_{0}}}=O(1).$

where $\Re(s)=\sigma_{0}>\frac{1}{2}$. As argued in Section 3 of [19]

$\displaystyle|\lambda_{f}(p^{2})-\psi(p)|^{2}=\left|\frac{\Lambda_{f}(p^{2})}{\log
p}\right|\leq\left|\frac{2\Lambda(p^{2})}{\log
p}\right|^{2}=\left|\frac{2\Lambda(p)}{\log p}\right|^{2}=4.$

By (1) and (2) contribution of the terms involving prime squares give

(6) $\displaystyle\int_{T}^{2T}\left|\sum_{2\leq p^{2}\leq
X}\frac{\lambda_{f}(p^{2})-\psi(p)}{p^{2(\sigma_{0}+it)}\cdot
2}\right|^{2}dt=\sum_{p_{1},p_{2}\leq\sqrt{X}}\int_{T}^{2T}\frac{1}{p_{1}^{2\sigma_{0}}p_{2}^{2\sigma_{0}}}dt$
$\displaystyle\ll
T\sum_{p\leq\sqrt{X}}\frac{1}{p^{4\sigma_{0}}}+\sum_{\begin{subarray}{c}p_{1},p_{2}\leq\sqrt{X}\\\
p_{1}\neq
p_{2}\end{subarray}}\frac{1}{p_{1}^{2\sigma_{0}}p_{2}^{2\sigma_{0}}}\sqrt{p_{1}p_{2}}\ll
T.$

Let $A(t;X)=A(t)\sum_{2\leq p\leq
X}\frac{\lambda_{f}(p^{2})-\psi(p)}{2p^{2\sigma_{0}}}$, from (6) and
Chebyshev’s inequality we have

$\displaystyle meas\\{T\leq t\leq
2T:|A(t)|>L\\}\leq\frac{1}{L^{2}}\int_{T}^{2T}|A(t)|^{2}dt\ll T/L^{2}.$

for any positive real number $L>1$. In other words we can say that the square
of primes in $\mathcal{P}(f,s)$ contribute a measure at most $O(T/L^{2})$.

Since we know that $\Lambda_{f}(p)=\lambda_{f}(p)\log p$ for all prime $p$ we
can write

$\displaystyle\mathcal{P}_{0}(f,\sigma_{0}+it)=\mathcal{P}_{0}(f,\sigma_{0}+it,X)=\sum_{p\leq
X}\frac{\lambda_{f}(p)}{p^{\sigma_{0}+it}}.$

Let us begin by studying the moments of $\mathcal{P}_{0}(f,\sigma_{0}+it)$.

######  Lemma 3.

Suppose that $k$ and $\ell$ are non-negative integers with $X^{k+\ell}\ll T$.
Then if $k\neq\ell$

$\displaystyle\int_{T}^{2T}\mathcal{P}_{0}(f,\sigma_{0}+it)^{k}\overline{\mathcal{P}_{0}(f,\sigma_{0}+it)}^{\ell}dt\ll
T.$

If $k=\ell$, we have

$\displaystyle\int_{T}^{2T}\left|\mathcal{P}_{0}(f,\sigma_{0}+it)\right|^{2k}dt=k!T(\log\log
T)^{k}+O_{k}(T(\log\log T)^{k-1+\epsilon})$

###### Proof.

Write $\mathcal{P}_{0}(f,s)^{k}=\sum_{n}\frac{a_{k}(n)\lambda_{f}(n)}{n^{s}}$
where

(7) $\displaystyle
a_{k}(n)=\begin{cases}\frac{k!}{\alpha_{1}!\cdots\alpha_{r}!}&{\text{ if
}}n=\prod_{j=1}^{r}{p_{j}}^{\alpha_{j}},p_{1}<\ldots<p_{r}<X,\sum_{j=1}^{r}\alpha_{j}=k.\\\
0&{\text{ otherwise}}.\end{cases}$

Therefore,

$\displaystyle\int_{T}^{2T}\mathcal{P}_{0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{0}(\sigma_{0}+it)}^{\ell}dt$
$\displaystyle=$
$\displaystyle\int_{T}^{2T}\sum_{n}\frac{a_{k}(n)\lambda_{f}(n)}{n^{s}}\sum_{n}\frac{a_{\ell}(n){\bar{\lambda}}_{f}(n)}{n^{\bar{s}}}dt$
$\displaystyle=$
$\displaystyle\sum_{n}a_{k}(n)a_{\ell}(n)\lambda_{f}(n)\bar{\lambda}_{f}(n)\int_{T}^{2T}\frac{1}{n^{2\sigma_{0}}}dt$
$\displaystyle=$ $\displaystyle
T\sum_{n}\frac{a_{k}(n)a_{\ell}(n)\lambda_{f}(n)\bar{\lambda}_{f}(n)}{n^{2\sigma_{0}}}+O\left(\sum_{m\neq
n}\frac{a_{k}(n)a_{\ell}(m)\lambda_{f}(n)\bar{\lambda}_{f}(m)}{(mn)^{\sigma_{0}}}\frac{1}{\log|m/n|}\right)$

Notice that if $k\neq{\ell}$ then $a_{k}(n)a_{\ell}(n)$ is 0 by definition. So
we don’t have to worry about the diagonal term contribution. For the off-
diagonal term from (2) we can see that the contribution of the denominator is
negligible since $\sigma_{0}$ is close to $\frac{1}{2}$. Applying Ramanujan-
Petersson conjecture, with $m\neq n$ we have the off-diagonal term
contribution given by

$\displaystyle\sum_{m\neq
n}a_{k}(n)a_{\ell}(m)\lambda_{f}(n)\bar{\lambda}_{f}(m)\ll
X^{k+{\ell}+\epsilon}\ll T.$

We conclude the first part of the Lemma.

For the second part of the lemma (which is the case for $k=l$) the diagonal
term contributes
$\sum_{n}\frac{a_{k}(n)^{2}\lambda_{f}(n)^{2}}{n^{2\sigma_{0}}}$. By the
definition for the given positive integers $\alpha_{1},\ldots,\alpha_{r}$ with
$\sum_{i=1}^{r}\alpha_{i}=k$, the contribution of the of $n$ of the form
$p_{1}^{\alpha_{1}},\dots,p_{r}^{\alpha_{r}}$ is given by

$\displaystyle\ll T\prod_{i=1}^{r}\left(\sum_{p\leq
X}\frac{\lambda_{f}(p)^{2}}{p^{2\sigma_{0}\alpha_{i}}}\right)\ll T(\log\log
T)^{r+\epsilon}.$

The terms with $n$ not being square free contributes (where $r\leq k-1$)
$O_{k}((\log\log T)^{k-1+\epsilon})$.

The square free terms $n$ give (see section 3 of [7]),

$\displaystyle k!\sum_{\begin{subarray}{c}p_{1},\ldots,p_{k}\leq X\\\
{\text{all $p_{j}$'s are
distinct}},(p_{j},q)=1\end{subarray}}\frac{\lambda_{f}(p_{1}\cdots
p_{k})^{2}}{(p_{1}\cdots p_{k})^{2\sigma_{0}}}=$ $\displaystyle
k!\left(\sum_{\begin{subarray}{c}p\leq X\\\
(p,q)=1\end{subarray}}\frac{\lambda_{f}(p)^{2}}{p^{2\sigma_{0}}}\right)^{k}$
$\displaystyle=$ $\displaystyle k!\left(\log\log
T\right)^{k}+O_{k}\left((\log\log T)^{k-1+\epsilon}\right).$

Recalling the definition of $X$, we conclude the proof. ∎

Proof of Proposition 2: Using Lemma 3 for any odd $k$,

$\displaystyle\int_{T}^{2T}\left(\Re\left(\mathcal{P}_{0}(f,\sigma_{0}+it)\right)\right)^{k}dt=$
$\displaystyle\int_{T}^{2T}\frac{1}{2^{k}}\left(\mathcal{P}_{0}(f,\sigma_{0}+it)+\overline{\mathcal{P}_{0}(f,\sigma_{0}+it)}\right)^{k}dt$
$\displaystyle=$
$\displaystyle\frac{1}{2^{k}}\sum_{{\ell}=0}^{k}\binom{k}{{\ell}}\int_{T}^{2T}\left(\mathcal{P}_{0}(f,\sigma_{0}+it)\right)^{\ell}\overline{\left(\mathcal{P}_{0}(f,\sigma_{0}+it)\right)^{k-{\ell}}}dt$
$\displaystyle\ll T.$

As it is impossible to have ${\ell}=k-{\ell}$ for any odd $k$. If $k$ is even,
then we apply Lemma 3 with ${\ell}=k-{\ell}=k/2$ to obtain,

$\displaystyle\frac{1}{T}\int_{T}^{2T}\left(\Re\left(\mathcal{P}_{0}(f,\sigma_{0}+it)\right)\right)^{k}dt=2^{-k}\binom{k}{k/2}\left(\frac{k}{2}\right)!(\log\log
T)^{\frac{k}{2}}+O_{k}\left((\log\log T)^{\frac{k}{2}-1+\epsilon}\right).$

The above equation matches with the distribution of the Gaussian random
variable with mean 0 and variance $\sim\frac{1}{2}\log\log T$, completing the
proof. ∎

##  6\. Proof of Proposition 3

To prove this proposition let us first decompose $\mathcal{P}(f,s)$ as
$\mathcal{P}_{1}(f,s)$ and $\mathcal{P}_{2}(f,s)$ where

$\displaystyle\mathcal{P}_{1}(f,s)=\sum_{2\leq n\leq
Y}\frac{\Lambda_{f}(n)}{n^{s}\log n};$
$\displaystyle\mathcal{P}_{2}(f,s)=\sum_{Y<n\leq
X}\frac{\Lambda_{f}(n)}{n^{s}\log n}.$

Set

$\displaystyle\mathcal{M}_{1}(f,s)=\sum_{0\leq k\leq 100\log\log
T}\frac{(-1)^{k}}{k!}\mathcal{P}_{1}(f,s)^{k};$
$\displaystyle\mathcal{M}_{2}(f,s)=\sum_{0\leq k\leq 100\log\log\log
T}\frac{(-1)^{k}}{k!}\mathcal{P}_{2}(f,s)^{k}.$

By recalling the definition of $X,Y$ it is obvious that $\mathcal{M}_{j}(f,s)$
(for $j=1,2.$) is a short Dirichlet polynomial of length $\ll T^{\epsilon}$.
Now we state the next lemma.

######  Lemma 4.

For $T\leq t\leq 2T$ we have

(8) $\displaystyle|\mathcal{P}_{1}(\sigma_{0}+it)|\leq\log\log T$
$\displaystyle|\mathcal{P}_{2}(\sigma_{0}+it)|\leq\log\log\log T.$

except perhaps for a set of measure $\frac{T}{\log\log\log T}$. We also have
that

(9)
$\displaystyle\mathcal{M}_{1}(\sigma_{0}+it)=\exp(-\mathcal{P}_{1}(\sigma_{0}+it))(1+O(\log
T)^{-99})$
$\displaystyle\mathcal{M}_{2}(\sigma_{0}+it)=\exp(-\mathcal{P}_{2}(\sigma_{0}+it))(1+O(\log\log
T)^{-99}).$

###### Proof.

Starting with the integration from (1) and (2), we have

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{1}(f,\sigma_{0}+it)|^{2}dt$
$\displaystyle\ll T\sum_{2\leq n_{1}=n_{2}\leq
Y}\frac{\Lambda_{f}{(n_{1})}\Lambda_{f}{(n_{2})}}{(n_{1}n_{2})^{\sigma_{0}}\log
n_{1}\log n_{2}}+\sum_{2\leq n_{1}\neq n_{2}\leq
Y}\frac{\Lambda_{f}{(n_{1})}\Lambda_{f}{(n_{2})}}{(n_{1}n_{2})^{\sigma_{0}}\log
n_{1}\log n_{2}}\sqrt{n_{1}n_{2}}$ $\displaystyle\ll T\log\log T$

Similarly, we have

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{2}(f,\sigma_{0}+it)|^{2}dt\ll
T\log\log\log T.$

and the first assertion follows.

If $|z|\leq k$ then using Stirling’s formula it is straight forward to check
that

$\displaystyle\left|e^{z}-\sum_{0\leq K\leq 100k}\frac{z^{k}}{k!}\right|\leq
e^{-99k}.$

Take $z=-\mathcal{P}_{1}(f,\sigma_{0}+it)$ and $k=\log\log T$ and (9) holds. ∎

As we have decomposed $\mathcal{P}(f,s)$ similarly, we decompose $M(f,s)$ as
$M_{1}(f,s)$ and $M_{2}(f,s)$. By the definition of $M(f,s)$ we need to
decompose $a(n)$ first. Set

$\displaystyle a_{1}(n)=\begin{cases}1&\text{if $n$ has at most $100\log\log
T$ prime factors with all $p\leq Y$, }\\\ 0&\text{otherwise }.\end{cases}$
$\displaystyle a_{2}(n)=\begin{cases}1&\text{if $n$ has at most
$100\log\log\log T$ prime factors with all $Y<p\leq X$, }\\\ 0&\text{otherwise
}.\end{cases}$

Therefore,

$\displaystyle M(f,s)=M_{1}(f,s)M_{2}(f,s),$ $\displaystyle
M_{1}(f,s)=\sum_{n}\frac{\mu(n)a_{1}(n)\lambda_{f}(n)}{n^{s}},$ $\displaystyle
M_{2}(f,s)=\sum_{n}\frac{\mu(n)a_{2}(n)\lambda_{f}(n)}{n^{s}}.$

######  Lemma 5.

We have

$\displaystyle\int_{T}^{2T}|\mathcal{M}_{1}(\sigma_{0}+it)-M_{1}(\sigma_{0}+it)|^{2}dt\ll
T(\log T)^{-60},$

and

$\displaystyle\int_{T}^{2T}|\mathcal{M}_{2}(\sigma_{0}+it)-M_{2}(\sigma_{0}+it)|^{2}dt\ll
T(\log\log T)^{-60}.$

###### Proof.

Write

$\displaystyle\mathcal{M}_{1}(f,s)=\sum_{n}\frac{b(n)\lambda_{f}(n)}{n^{s}}.$

where $b(n)$ satisfies the following properties.

  1. (1)

$|b(n)|\leq 1$ for all $n$.

  2. (2)

$b(n)=0$ unless $n\leq Y^{100\log\log T}$ has only prime factors below $Y$.

  3. (3)

$b(n)=\mu(n)a_{1}(n)$ unless $\Omega(n)>100\log\log T$ or, $p\leq Y$ s.t
$p^{k}|n$ with $p^{k}>Y$.

Set $c(n)=(b(n)-\mu(n)a_{1}(n))\lambda_{f}(n)$, from (1) and (2) we have

$\displaystyle\int_{T}^{2T}$
$\displaystyle|\mathcal{M}_{1}(f,\sigma_{0}+it)-M_{1}(f,\sigma_{0}+it)|^{2}dt$
$\displaystyle\ll
T\sum_{n_{1}=n_{2}}\frac{|c(n_{1})\overline{c(n_{2})}|}{(n_{1}n_{2})^{\sigma_{0}}}+\sum_{n_{1}\neq
n_{2}}\frac{|c(n_{1})\overline{c(n_{2})}|}{(n_{1}n_{2})^{\sigma_{0}}}\sqrt{n_{1}n_{2}}$
$\displaystyle\ll T(\log T)^{-60}.$

We note that our $a_{1}(n),a_{2}(n),b(n)$ are exactly the same as given in
Radziwiłł and Soundarajan’s paper [3]. The only difference is that now we have
a further twisting by $|\lambda_{f}(n)|\leq d(n)\ll n^{\epsilon}$, where
$d(n)$ is the divisor function. Since we know that $\sigma_{0}$ is close to
$\frac{1}{2}$, the off-diagonal terms with $n_{1}\neq n_{2}$ contribute

$\displaystyle\ll\sum_{n_{1}\neq n_{2}\leq Y^{100\log\log T}}1\ll
T^{\epsilon}.$

By recalling property (3) we can say that the diagonal term contributes

$\displaystyle\ll T\sum_{\begin{subarray}{c}p|n\implies p\leq Y\\\
\Omega(n)>100\log\log
T\end{subarray}}\frac{1}{n}+T\left(\sum_{\begin{subarray}{c}p\leq Y\\\
p>Y^{k}\end{subarray}}\frac{1}{p^{k}}\right)\left(\sum_{\begin{subarray}{c}p|n\implies
p\leq Y\end{subarray}}\frac{1}{n}\right).$

A small calculation shows that the second term above contribute $\ll T(\log
Y)/\sqrt{Y}\ll T(\log T)^{-60}$. For the first term above, note that for
$1<r<2$ the quantity $r^{\Omega(n)-100\log\log T}$ is always non-negative, in
fact it is $\geq 1$ for those $n$ with $\Omega(n)>100\log\log T$. Therefore,

$\displaystyle T\sum_{\begin{subarray}{c}p|n\implies p\leq Y\\\
\Omega(n)>100\log\log T\end{subarray}}\frac{1}{n}$ $\displaystyle\ll
Tr^{-100\log\log T}\prod_{p\leq
Y}\left(1+\frac{r}{p}+\frac{r^{2}}{p^{2}}+\cdots\right)$ $\displaystyle\ll
T(\log T)^{-100\log r}(\log T)^{r}.$

Choosing $r=e^{2/3}$, the above estimate becomes $\ll T(\log T)^{-60}$,
completing the proof of this lemma. ∎

Proof of Proposition 3: It follows from (9) that we have

$\displaystyle\mathcal{M}_{1}(f,\sigma_{0}+it)=\exp(-\mathcal{P}_{1}(f,\sigma_{0}+it))(1+O((\log
T)^{-99}))$

and by (8) we have

$\displaystyle(\log T)^{-1}\ll|\mathcal{M}_{1}(f,\sigma_{0}+it)|\ll\log T.$

So, we can conclude that, except on a set of measure $o(T)$,

$\displaystyle M_{1}(f,\sigma_{0}+it)=$
$\displaystyle\mathcal{M}_{1}(f,\sigma_{0}+it)+O((\log T)^{-25})$
$\displaystyle=$
$\displaystyle\exp(-\mathcal{P}_{1}(f,\sigma_{0}+it))(1+O((\log T)^{-20})).$

Similarly, except on a set of measure $o(T)$, we have

$\displaystyle
M_{2}(f,\sigma_{0}+it)=\exp(-\mathcal{P}_{2}(f,\sigma_{0}+it))(1+O((\log\log
T)^{-20})).$

Recalling the decomposition of $M(f,s)$ and $\mathcal{P}(f,s)$, by multiplying
these estimates we obtain

$\displaystyle
M(f,\sigma_{0}+it)=\exp(-\mathcal{P}f,(\sigma_{0}+it))(1+O((\log\log
T)^{-20})),$

completes the proof. ∎

##  7\. Proof of Proposition 4

In this section, we prove that the mollifier $M(f,s)$ and the $L$-function are
inverse to each other. As we have followed the method established in [3] to
prove the other propositions of this paper, this one is an exception. First,
we see that how does the mean square of $L$-function (associated with
holomorphic cusp forms) corresponds to the fourth moment of the Riemann zeta
function then we set our parameters in the proof given by Hughes and Young
[11]. We do not write the details of the proof as it is already given in [11]
but we have given all necessary information which to obtain the main term of
the twisted mean square of $L$-function.

By the approximate functional equation of $L$-function (see Theorem 5.13 of
[6]) we have,

$\displaystyle\int_{T}^{2T}L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)dt$
$\displaystyle=$ $\displaystyle\sum_{n\leq
T}\frac{\lambda_{f}(n)}{n^{\sigma}}\sum_{m}\frac{a(m)\mu(m)\lambda_{f}(m)}{m^{\sigma}}\int_{T}^{2T}(mn)^{-it}dt+O(T^{1/2+\epsilon})$
$\displaystyle=$ $\displaystyle T+O(T^{1/2+\epsilon}).$

Now expanding the integration in Proposition 4, we get

$\displaystyle\int_{T}^{2T}|1-L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)|^{2}dt$
$\displaystyle=$
$\displaystyle\int_{T}^{2T}|L(f,\sigma_{0}+it)M(f,\sigma_{0}+it)|^{2}dt-T+O(T^{1/2+\epsilon}).$

In order to prove Proposition 4, we need the mean value square estimate of the
$L$-function. Recall the functional equation of $L$-function,

$\displaystyle\Lambda(f,s)=\varepsilon(f)\Lambda(\bar{f},1-s),$

where $\bar{f}$ is an object associated with $f$ (the dual of $f$) for which
$\lambda_{\bar{f}}(n)=\bar{\lambda}_{f}(n)$ and $\varepsilon(f)$ is the
complex number of absolute value 1 called the root number of $L(f,s)$.

######  Lemma 6.

For $c>0$, define

$\displaystyle I(f,s)$ $\displaystyle\coloneqq\frac{1}{2\pi
i}\int_{(c)}\Lambda(f,z+s,)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z},$
$\displaystyle I(\bar{f},1-s)$ $\displaystyle\coloneqq\frac{1}{2\pi
i}\int_{(c)}\varepsilon(f)\Lambda(\bar{f},-z+(1-s))\Lambda(f,-z+(1-\bar{s}))e^{z^{2}}\frac{dz}{z}.$

where $(c)$ denotes the line from $c-i\infty$ to $c+i\infty$. Then we have

$\displaystyle|L(f,s)|^{2}=\frac{1}{|G(f,s)|^{2}}\left(I(f,s)+I(\bar{f},1-s)\right).$

###### Proof.

First we prove that $I(f,s)$ is independent of choice of $c$. Let
$\mathcal{R}$ be a rectangle on $\Re(s)>0$. We write

$\displaystyle\int_{\mathcal{R}}\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z}=0.$

Letting the length of the sizes that are parallel to imaginary axes to
infinity, we conclude that $I(f,s)$ is independent of choice of $c$. Using the
fact that $\varepsilon(\bar{f})=\bar{\varepsilon}(f)$, we deduce that

$\displaystyle\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})=\varepsilon(f)\Lambda(\bar{f},-z+(1-s))\varepsilon(\bar{f})\Lambda(f,-z+(1-\bar{s})).$

Let $\langle-c\rangle$ denotes the line from $(-c+i\infty,-c-i\infty)$, we
obtain

$\displaystyle\int_{\langle-c\rangle}\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z}=\int_{(c)}\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z},$

which is independent of the choice of $c$. Let $S$ be a square centered at the
origin equipped with positive orientation. A standard contour integral
together with the Cauchy integral formula implies

$\displaystyle\Lambda(f,s)\Lambda(\bar{f},\bar{s})=\frac{1}{2\pi
i}\int_{S}\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z}.$

By letting the length of the sides that are parallel to imaginary axis to
infinity, observe that the above equation equals

$\displaystyle\frac{1}{2\pi
i}\left(\int_{(c)}\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z}+\int_{\langle-c\rangle}\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})e^{z^{2}}\frac{dz}{z}\right),$

which is coming from the functional equation of $L(f,s)$. Notice that the
first term of the above equation is $I(f,s)$ and the second term is
$I(\bar{f},1-s)$, completes the proof. ∎

Note that Theorem 5.11 of [6] is a standard reference for the above lemma.

From Lemma 6 we can write

$\displaystyle\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}|L(f,\sigma+it)|^{2}dt=\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\frac{I(f,\sigma+it)}{|G(f,\sigma+it)|^{2}}dt+\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\frac{I(\bar{f},1-(\sigma+it))}{|G(f,\sigma+it)|^{2}}dt$

Denote $s=\sigma+it$ with $T\leq t\leq 2T$ and $\frac{1}{2}\leq\sigma\leq 1$.
If $z$ is a complex number with real part $c=1-\sigma+\frac{1}{\log T}$ then
by the Stirling’s formula,

$\displaystyle\frac{G(z+s,f)G(z+\bar{s},\bar{f})}{|G(f,s)|^{2}}=\left(\frac{\sqrt{q}t}{2\pi}\right)^{2z}\left(1+O\left(\frac{|z|^{2}}{T}\right)\right).$

Hence we have

$\displaystyle\frac{I(f,s)}{|G(f,s)|^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi
i}\int_{(c)}\frac{\Lambda(f,z+s)\Lambda(\bar{f},z+\bar{s})}{G(f,s)G(\bar{f},\bar{s})}\frac{e^{z^{2}}}{z}dz$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{(c)}\frac{G(f,z+s)}{G(f,s)}\cdot\frac{G(\bar{f},z+\bar{s})}{G(\bar{f},\bar{s})}\cdot
L(f,z+s)L(f,z+\bar{s})\frac{e^{z^{2}}}{z}dz$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi
i}\int_{(c)}\frac{e^{z^{2}}}{z}L(f,z+s)L(f,z+\bar{s})\left(\frac{\sqrt{q}t}{2\pi}\right)^{2z}dz+O(T^{-\sigma+\epsilon})$

where $\Lambda(f,s)=G(f,s)L(f,s)$ denotes the complete $L$-function.
Therefore,

$\displaystyle\int_{T}^{2T}\left(\frac{h}{k}\right)^{it}\frac{I(f,s)}{|G(f,s)|^{2}}dt$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\int_{(c)}\frac{e^{z^{2}}}{z}\sum_{m,n=1}^{\infty}\frac{\lambda_{f}(n)\bar{\lambda}_{f}(m)}{(mn)^{z+\sigma}}\left(\int_{T}^{2T}\left(\frac{hm}{kn}\right)^{it}\left(\frac{\sqrt{q}t}{2\pi}\right)^{2z}dt\right)dz+O(T^{1-\sigma+\epsilon}).$

Since we are in the region of absolute convergence of $L(f,s+z)$ and
$L(f,\bar{s}+z)$ we can interchange the summations and integrals.

We know that the Hecke-eigenvalues of $GL(2)$ $L$-function (associated with a
cusp form) can be approximated by the divisor function which is
$\lambda_{f}(n)\ll d(n)$. So, we can write

$\displaystyle\sum_{m,n=1}^{\infty}\frac{\lambda_{f}(n)\bar{\lambda}_{f}(m)}{(mn)^{z+\sigma}}\ll\sum_{m,n=1}^{\infty}\frac{d(n)d(m)}{(mn)^{z+\sigma}}$

Observing the above equation we can write in terms of shifted parameters by
taking $m=n+r$. Then the above equation become

$\displaystyle\sum_{n\leq X}\frac{d(n)d(n+r)}{(n(n+r))^{z+\sigma}}$

as $X\to\infty$. Observing the above equation it is now clear that the mean
square of the $L$-function (associated with a holomorphic cusp form) roughly
corresponds to the fourth moment of the Riemann zeta function. Heath-Brown’s
result [12] says that

$\displaystyle\int_{0}^{T}|\zeta(1/2+it)|^{4}dt=\frac{T}{2\pi^{2}}(\log
T)^{4}+O(T(\log T)^{3}).$

We can directly use Heath-Brown’s argument to evaluate the mean square
estimate for the $L$-function by shifting the parameters. But we have an
additive twist in our problem which is the mollifier. So we use the argument
given by C. P. Hughes and M. P. Young [11] where they have calculated the
twisted fourth moment of the Riemann zeta function with the additive twist of
the mollifier which is a small Dirichlet polynomial of size $T^{\epsilon}$.
They have estimated the asymptotic formula for

$\displaystyle\int_{0}^{T}|M(\frac{1}{2}+it)|^{2}|\zeta(\frac{1}{2}+it)|^{4}dt.$

We give the same argument as given in [11] since we are dealing with the
$L$-function we set the parameters according to our case.

By expanding the mollifier $M(f,s)$ the twisted mean square of $L$-function
reduces to

$\displaystyle
I(h,k)=\int_{\infty}^{\infty}\left(\frac{h}{k}\right)^{-it}L(\frac{1}{2}+\alpha+it)L(\frac{1}{2}+\beta-
it)w(t)dt$

where $\alpha,\beta$ are very small complex numbers $\ll(\log T)^{-1}$ and $w$
is a smooth nice function. The main term is written in the shifted products of
the $L$-function as well as some finite Euler products.

Set

$\displaystyle
A_{\alpha,\beta}(s)=\frac{L(f,1+s+\alpha)L(f,1+s+\beta)}{L(f,2+2s+\alpha+\beta)}$

Further we set

$\displaystyle\sigma_{\alpha,\beta}(f,n)=\sum_{n=n_{1}n_{2}}\frac{\lambda_{f}(n_{1})\lambda_{f}(n_{2})}{{n_{1}}^{\alpha}{n_{2}}^{\beta}}.$

Suppose $(h,k)=1$, $p^{h_{p}}||h$ and $p^{k_{p}}||k$, then define

$\displaystyle
B_{\alpha,\beta,h,k}(s)=\prod_{p|h}\left(\frac{\sum_{j=0}^{\infty}\sigma_{\alpha}(p^{j})\sigma_{\beta}(p^{j+h_{p}})p^{-j(s+1)}}{\sum_{j=0}^{\infty}\sigma_{\alpha}(p^{j})\sigma_{\beta}(p^{j})p^{-j(s+1)}}\right)\times\prod_{p|k}\left(\frac{\sum_{j=0}^{\infty}\sigma_{\alpha}(p^{j}+k_{p})\sigma_{\beta}(p^{j})p^{-j(s+1)}}{\sum_{j=0}^{\infty}\sigma_{\alpha}(p^{j})\sigma_{\beta}(p^{j})p^{-j(s+1)}}\right)$

Let

$\displaystyle
Z_{\alpha,\beta,h,k}(s)=A_{\alpha,\beta}(s)B_{\alpha,\beta,h,k}(s)$

######  Lemma 7.

Let

$\displaystyle
I(h,k)=\int_{\infty}^{\infty}\left(\frac{h}{k}\right)^{-it}L(f,\frac{1}{2}+\alpha+it)L(f,\frac{1}{2}+\beta-
it)w(t)dt$

where $w(t)$ is a smooth, non-negative function supported in
$[\frac{T}{2},4T]$, satisfying $w^{(j)}(t)\ll_{j}{T_{0}}^{-j}$ for all
$j=0,1,2,\dots$ where $T^{\frac{1}{2}+\epsilon}\ll T_{0}\ll T$. Suppose
$(h,k)=1$, $hk\leq T^{\epsilon}$ and $\alpha,\beta\ll(\log T)^{-1}$. Then

$\displaystyle
I(h,k)=\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}w(t)\left(Z_{\alpha,\beta,h,k}(f,0)+\left(\frac{\sqrt{q}t}{2\pi}\right)^{-\alpha-\beta}Z_{-\alpha,-\beta,h,k}(f,0)\right)dt$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+O\left(T^{1/2+\epsilon}(hk)^{7/8}(T/T_{0})^{9/4}\right).$

###### Proof.

By a similar proof of Theorem 1.1 of [11] we can prove this lemma with some
different parameters.

From the functional equation of the $L$-function we can write

$\displaystyle\Lambda_{\alpha,\beta,t}(f,s)=\Lambda_{-\alpha,-\beta,t}(f,s)$

where
$\Lambda_{\alpha,\beta,t}(f,s)=\Lambda(f,1/2+s+\alpha+it)\Lambda(f,1/2+s+\beta-
it)$.

In our case the formula for the twisted mean value square of $L$-function is

$\displaystyle I(h,k)=$
$\displaystyle\sum_{m,n}\frac{\sigma_{\alpha}(f,m)\sigma_{\beta}(f,n)}{(mn)^{\frac{1}{2}}}\int_{-\infty}^{\infty}\left(\frac{hm}{kn}\right)^{-it}V_{\alpha,\beta,t}(\pi^{2}mn)w(t)dt$
$\displaystyle+$
$\displaystyle\sum_{m,n}\frac{\sigma_{-\alpha}(f,m)\sigma_{-\beta}(f,n)}{(mn)^{\frac{1}{2}}}\int_{-\infty}^{\infty}\left(\frac{hm}{kn}\right)^{-it}X_{\alpha,\beta,t}V_{-\alpha,-\beta,t}(\pi^{2}mn)w(t)dt$
$\displaystyle=$
$\displaystyle\sum_{m,n}\frac{\lambda_{f}(m)\lambda_{f}(n)}{m^{1/2+\alpha}n^{1/2+\beta}}\int_{-\infty}^{\infty}\left(\frac{hm}{kn}\right)^{-it}V_{\alpha,\beta,t}(\pi^{2}mn)w(t)dt$
$\displaystyle+\sum_{m,n}\frac{\lambda_{f}(m)\lambda_{f}(n)}{m^{1/2-\alpha}n^{1/2-\beta}}\int_{-\infty}^{\infty}\left(\frac{hm}{kn}\right)^{-it}X_{\alpha,\beta,t}V_{-\alpha,-\beta,t}(\pi^{2}mn)w(t)dt$

where

$\displaystyle V_{\alpha,\beta,t}(x)=\frac{1}{2\pi
i}\int_{(1)}\frac{G(s)}{s}g_{\alpha,\beta,t}(s,t)x^{-s}ds.$

where $G(s)$ be an even entire function with rapid decay as $|s|\to\infty$ in
any fixed strip $|\Re(s)|\leq C$. Let

$\displaystyle
I^{(1)}(h,k)=\sum_{m,n}\frac{\lambda_{f}(m)\lambda_{f}(n)}{m^{1/2+\alpha}n^{1/2+\beta}}\int_{-\infty}^{\infty}\left(\frac{hm}{kn}\right)^{-it}V_{\alpha,\beta,t}(\pi^{2}mn)w(t)dt$

and

$\displaystyle
I^{(2)}(h,k)=\sum_{m,n}\frac{\lambda_{f}(m)\lambda_{f}(n)}{m^{1/2-\alpha}n^{1/2-\beta}}\int_{-\infty}^{\infty}\left(\frac{hm}{kn}\right)^{-it}X_{\alpha,\beta,t}V_{-\alpha,-\beta,t}(\pi^{2}mn)w(t)dt$

The diagonal term contribution for $I^{(1)}(h,k)$ with $hm=kn$ is given by

$\displaystyle{I_{D}}^{(1)}(h,k)=\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}w(t)\frac{1}{2\pi
i}\int_{(1)}\frac{G(s)}{s}(\pi^{2}hk)^{-s}g_{\alpha,\beta}(s,t)\sum_{n=1}^{\infty}\frac{\sigma_{\alpha}(f,kn)\sigma_{\beta}(f,hn)}{n^{1+2s}}dsdt$

where

$\displaystyle
g_{\alpha,\beta}(s,t)=t^{2s}\left(1+O\left(\frac{1}{t}\right)\right)$

We compute the Dirichlet series

$\displaystyle\sum_{n=1}^{\infty}\frac{\sigma_{\alpha}(f,kn)\sigma_{\beta}(f,hn)}{n^{1+s}}$

using that $\sigma_{\alpha}(f,kn)\sigma_{\beta}(f,hn)$ is multiplicative in
$n$. In terms of the Euler product it is

$\displaystyle\left(\prod_{p,hk}\sum_{j=0}^{\infty}\frac{\sigma_{\alpha}(f,p^{j})\sigma_{\beta}(f,p^{j})}{p^{j(1+s)}}\right)$
$\displaystyle\times\left(\prod_{p|h}\sum_{j=0}^{\infty}\frac{\sigma_{\alpha}(f,p^{j})\sigma_{\beta}(f,p^{h_{p}+j})}{p^{j(1+s)}}\right)$
$\displaystyle\times\left(\prod_{p|k}\sum_{j=0}^{\infty}\frac{\sigma_{\alpha}(f,p^{k_{p}+j})\sigma_{\beta}(f,p^{j})}{p^{j(1+s)}}\right)$

which equals $Z_{\alpha,\beta,h,k}(f,s)$ since

$\displaystyle\sum_{n=1}^{\infty}\frac{\sigma_{\alpha}(f,n)\sigma_{\beta}(f,n)}{n^{1+s}}=A_{\alpha,\beta}(f,s).$

Hence

$\displaystyle{I_{D}}^{(1)}(h,k)=\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}w(t)\frac{1}{2\pi
i}\int_{\epsilon}\frac{G(s)}{s}(\pi^{2}hk)^{-s}g_{\alpha,\beta}(s,t)Z_{\alpha,\beta,h,k}(f,2s)dsdt$

Applying Stirling’s approximation on $g_{\alpha,\beta}(s,t)$ we have

$\displaystyle{I_{D}}^{(1)}(h,k)=\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}w(t)\frac{1}{2\pi
i}\int_{\epsilon}\frac{G(s)}{s}\left(\frac{t^{2}}{\pi^{2}hk}\right)^{s}(s,t)Z_{\alpha,\beta,h,k}(f,2s)dsdt+O((hk)^{-\frac{1}{2}}T^{\epsilon}).$

Note that ${I_{D}}^{(1)}(h,k)$ is holomorphic in $\alpha,\beta$ sufficiently
small.

Now we move $\Re(s)$ to $-\frac{1}{4}+\epsilon$, crossing a pole at $s=0$ as
well as a pole at $2s=-\alpha-\beta$. The integral on the new line is

$\displaystyle\ll(hk)^{-\frac{1}{4}+\epsilon}T^{\epsilon}\int_{T/2}^{4T}t^{-\frac{1}{2}+\epsilon}|w(t)|dt\ll(hk)^{-\frac{1}{4}+\epsilon}T^{\frac{1}{2}+\epsilon},$

uniformly in terms of the shift parameters. The pole at $s=0$ gives

$\displaystyle\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}Z_{\alpha,\beta,h,k}(f,0)w(t)dt$

The pole at $s=-\alpha-\beta$ gives

$\displaystyle
J_{\alpha,\beta}^{(1)}=\frac{\mathrm{Res}_{s=\frac{-\alpha-\beta}{2}}(Z_{\alpha,\beta,h,k}(f,2s))}{(hk)^{\frac{1}{2}-\frac{\alpha+\beta}{2}}}\frac{G(\frac{-\alpha-\beta}{2})}{\frac{-\alpha-\beta}{2}}\int_{-\infty}^{\infty}\left(\frac{\sqrt{qt}}{2\pi}\right)^{-\alpha-\beta}w(t)dt$

Since we have

$\displaystyle
X_{\alpha,\beta,t}=\left(\frac{\sqrt{q}t}{2\pi}\right)^{-\alpha-\beta}\left(1+O\left(\frac{1}{t}\right)\right)$

The computation of the diagonal terms of ${I_{D}}^{(2)}(h,k)$ is similar. We
get

$\displaystyle{I_{D}}^{(2)}(h,k)=$
$\displaystyle\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}\left(\frac{\sqrt{q}t}{2\pi}\right)^{-\alpha-\beta}w(t)\frac{1}{2\pi
i}\int_{\epsilon}\frac{G(s)}{s}\left(\frac{t^{2}}{\pi^{2}hk}\right)^{s}(s,t)Z_{-\alpha,-\beta,h,k}(f,2s)dsdt$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+O((hk)^{-\frac{1}{2}}T^{\epsilon})$

The pole at $s=0$ gives

$\displaystyle\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}\left(\frac{\sqrt{q}t}{2\pi}\right)^{-\alpha-\beta}Z_{-\alpha,-\beta,h,k}(f,0)w(t)dt$

The pole at $2s=\alpha+\beta$ gives

$\displaystyle
J_{\alpha,\beta}^{(2)}=\frac{\mathrm{Res}_{s=\frac{\alpha+\beta}{2}}(Z_{-\alpha,-\beta,h,k}(f,2s))}{(hk)^{\frac{1}{2}+\frac{\alpha+\beta}{2}}}\frac{G(\frac{\alpha+\beta}{2})}{\frac{\alpha+\beta}{2}}\int_{-\infty}^{\infty}\left(\frac{\sqrt{qt}}{2\pi}\right)^{-\alpha-\beta}w(t)dt$

We have the following results:

$\displaystyle{I_{D}}^{(1)}(h,k)=\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}Z_{\alpha,\beta,h,k}(f,0)w(t)dt+J_{\alpha,\beta}^{(1)}+O\left(\frac{T^{\frac{1}{2}\epsilon}}{(hk)^{\frac{1}{4}}}\right),$

and

$\displaystyle{I_{D}}^{(2)}(h,k)=\frac{1}{\sqrt{hk}}\int_{-\infty}^{\infty}\left(\frac{\sqrt{q}t}{2\pi}\right)^{-\alpha-\beta}Z_{-\alpha,-\beta,h,k}(f,0)w(t)dt+J_{\alpha,\beta}^{(2)}+O\left(\frac{T^{\frac{1}{2}\epsilon}}{(hk)^{\frac{1}{4}}}\right).$

As argued in [11], $G$ can be chosen from a wide class of functions, in the
terms of the form $J^{(1)}$ and $J^{(2)}$ should not contribute to $I(h,k)$.
Further note that $J^{(1)}$ and $J^{(2)}$ are not holomorphic in terms of the
shift parameters because there are various poles but of course the poles must
cancel when $J$’s and the term involving $Z$ are summed.

Now it concerns us to calculate the off-diagonal term. Since our formula for
$\sigma_{\alpha,\beta}(f,s)$ involve the Hecke-eigenvalue of $f$, we can not
follow the calculation of [11]. In fact, our error term in this lemma is
different than the error term given in Theorem 1.1 of [11]. Obviously, we need
the help of the shifted convolution problem to calculate the error term for
our case. The error term calculation follows from [14]. Since they have
calculated the error term for $L$-functions attached to holomorphic cusp form
we are not repeating the same calculation. ∎

Remarks.

  * •

Note that $Z_{\alpha,\beta,h,k}(0)$ has a pole at $\beta=-\alpha$ but the
symmetries of the expression imply that the poles cancel to form a holomorphic
function. Lemma 2.5.1 of [13] exhibits an integral representation for the
permutation sum that proves the holomorphy.

  * •

We would also suggest the reading of result given by P. Kühn, N. Robles and D.
Zeindler [14] where they have calculated the mean value of the twisted
automorphic $L$-function with the twist of the mollifier. To be precise
Theorem 2.1 of [14] is the type result we have obtained. The parameters we
have set here matches with their method except for the mollifier. This type of
result can be obtained by their method as well.

  * •

The error term of $I(h,k)$ can be improved but that is not the main focus of
our work here. We have considered $T_{0}=T^{1-\epsilon}$ to obtain the main
and the error term. The choice of $h,k$ is important. We consider the size of
$h,k$ as considered in [3]. It’s not a problem since $T$ is sufficiently large
$h,k$ is very small in comparison.

In this paragraph we give an overview of the off-diagonal term calculations,
since it is already given in [20, 14].

######  Lemma 8.

Let $\epsilon>0$, $0<\gamma<1$, $\alpha,\beta\ll(\log T)^{-1}$ be complex
numbers and $h,k\leq T^{\nu}$ be positive integers. Then for all real number
$A>0$ we have

$\displaystyle I_{O}^{(1)}(h,k)=\sum_{\begin{subarray}{c}hm\neq kn\\\ mn\ll
T^{2+\epsilon}\\\ \left|\frac{hm}{kn}-1\right|\ll
T^{\gamma}\end{subarray}}\frac{\lambda_{f}(m)\lambda_{f}(n)}{m^{\frac{1}{2}+\alpha}n^{\frac{1}{2}+\beta}}\int_{-\infty}^{\infty}w(t)\left(\frac{hm}{kn}\right)^{-it}V_{\alpha,\beta}(mn,t)dt+O(T^{-A}).$

For the proof of this lemma see Lemma 3 of [20].

Let us fix an arbitrary smooth function $\rho:]0,\infty[\to\mathbb{R}$ which
is compactly supported in $[1,2]$ and satisfies

$\displaystyle\sum_{\ell=-\infty}^{\infty}\rho\left(2^{-\ell/2}x\right)=1$

To build such a function see section 5 of [21]. For each integer $\ell$ we
define

$\displaystyle\rho_{\ell}(x)=\rho\left(\frac{x}{A_{\ell}}\right)\text{ with
$A_{\ell}2^{\ell/}T^{\gamma}$}.$

Then by Lemma 4 of [20], one has

$\displaystyle
I_{O}^{(1)}(h,k)=\sum_{\begin{subarray}{c}A_{\ell_{1}}A_{\ell_{1}}\ll
hkT^{2+\epsilon}\\\ A_{\ell_{1}}\asymp A_{\ell_{2}}\\\
A_{\ell_{1}},A_{\ell_{2}}\gg T^{\gamma}\end{subarray}}\sum_{0<|h|\ll
T^{-\gamma}\sqrt{A_{\ell_{1}}A_{\ell_{2}}}}\sum_{hm-
kn=r}\lambda_{f}(m)\lambda_{f}(n)F_{r;{\ell_{1}},{\ell_{2}}}(hm,kn)+O(T^{-A})$

where

$\displaystyle
F_{r;{\ell_{1}},{\ell_{2}}}(x,y)=\frac{h^{\frac{1}{2}+\alpha}k^{\frac{1}{2}+\beta}}{x^{\frac{1}{2}+\alpha}y^{\frac{1}{2}+\beta}}\int_{-\infty}^{\infty}w(t)\left(1+\frac{r}{y}\right)^{-it}V_{\alpha,\beta}\left(\frac{xy}{hk},t\right)dt\times\rho_{\ell_{1}}(x)\rho_{\ell_{2}}(y)$

with positive integers $h,k\leq T^{\nu}$ and $0<\gamma<1$. The key ingredient
of the proof of this lemma depends on a strong result of shifted convolution
sums on average due to Bloomer (see Theorem 2 of [22]). A straightforward
adaption of Bloomer’s result and the argument given by Bernard prove the
following lemma.

######  Lemma 9 (Bernard).

Let ${\ell}_{1},{\ell}_{2},H$ and $h_{1}$ be positive integers. Let
$M_{1},M_{2},P_{1},P_{2}$ be real numbers greater than 1. Let $\\{g_{r}\\}$ be
a family of smooth functions, supported on
$[M_{1},2M_{1}]\times[M_{2},2M_{2}]$ such that
$|g_{h}^{(ij)}|_{\infty}\ll_{i,j}(p_{1}/M_{1})^{i}(P_{2}/M_{2})^{j}$ for all
$i,j\geq 0$. Let $(a(r))$ be sequence of complex numbers such that

$\displaystyle a(r)\neq 0\implies r\leq
H,h_{1}|r\mathrm{and}\left(h_{1},\frac{r}{h_{1}}\right)=1.$

If ${\ell}_{1}M_{1}\asymp{\ell}_{2}M_{2}\asymp A$ and if there exits
$\epsilon>0$ such that

$\displaystyle
H\ll\frac{A}{\mathrm{max}\\{P_{1},P_{2}\\}}\frac{1}{({\ell}_{1}{\ell}_{2}M_{1}M_{2}P_{1}P_{2})^{\epsilon}},$

then, for all real numbers $\epsilon>0$, we have

$\displaystyle\sum_{r=1}^{H}a(r)\sum_{\begin{subarray}{c}m_{1}m_{2}\geq 1\\\
{\ell}_{1}m_{1}-{\ell}_{2}m_{2}=r\end{subarray}}\lambda_{f}(m_{1})\bar{\lambda}_{f}(m_{2})g_{1}(m_{1},m_{2})\ll
A^{\frac{1}{2}}h_{1}^{\theta}||a||_{2}(P_{1}+P_{2})^{\frac{3}{2}}$
$\displaystyle\times\left[\sqrt{P_{1}+P_{2}}+\left(\frac{A}{\mathrm{max}\\{P_{1},P_{2}\\}}\right)^{\theta}\left(1+\sqrt{\frac{(h_{1},{\ell_{1}}{\ell_{2}})H}{h_{1},{\ell_{1}}{\ell_{2}}}}\right)\right]({\ell}_{1}{\ell}_{2}M_{1}M_{2}P_{1}P_{2})^{\epsilon}.$

where $\theta$ is the exponent in the Ramanujan-Petersson conjecture.

The next step is to determined the required bound for the test function. Let
$\alpha,\beta\ll(\log T)^{-1}$ be complex numbers and let $\sigma$ be any
positive real number. For all non-negative integers $i,j$ one has

(10) $\displaystyle
x^{i}y^{j}\frac{\partial^{i+j}F_{r;{\ell_{1}},{\ell_{2}}}}{\partial
x^{i}\partial
y^{j}}(x,y)\ll_{i,j}\left(\frac{a}{A_{\ell_{1}}}\right)^{\frac{1}{2}+\Re\alpha+\sigma}\left(\frac{b}{A_{\ell_{2}}}\right)^{\frac{1}{2}+\Re\beta+\sigma}T^{1+2\sigma}(\ln{T})^{j}$

where the implicit constant does not depend on $r$. The trivial bound for
shifted convolution sums can yield

$\displaystyle\sum_{{\ell_{1}}-m_{1}{\ell_{2}}m_{2}=r}\lambda_{f}(m_{1})\lambda_{f}(m_{2})\ll_{\epsilon}\min\\{M_{1},M_{2}\\}(M_{1}M_{2})^{\epsilon}.$

This trivial bound together with (10) proves the following:

$\displaystyle I_{O}^{(1)}(h,k)\ll_{\epsilon}\min\\{h,k\\}T^{1+\epsilon}$

which is not very useful for us. As argued in [20] using Theorem 6.3 of [23]
we get

$\displaystyle
I_{O}^{(1)}(h,k)\ll_{\epsilon}\min(hk)^{3/4+\theta/2}T^{3/2+2\epsilon}.$

Instead of using the trivial bound if we use Theorem 1.3 of [24] along with
(10) then we have

$\displaystyle
I_{O}^{(1)}(h,k)\ll_{\epsilon}\min(hk)^{3/4+\theta/2}T^{1/2+2\epsilon}.$

By Lemma 9 with $H=T^{-\gamma}\sqrt{A_{\ell_{1}}A_{\ell_{2}}}$, $h_{1}=1$ and
$a(r)=\begin{cases}1&\text{ if }r\leq H\\\ 0&\text{otherwise}.\end{cases}$
Therefore, from Lemma 8 we get

$\displaystyle
I_{O}^{(1)}(h,k)\ll_{\epsilon}(hk)^{(1+\theta)/2}T^{1/2+\theta+\epsilon}.$

With a similar argument and from Corollary 5 of [20] one has

$\displaystyle
I_{O}^{(2)}(h,k)\ll_{\epsilon}(hk)^{(1+\theta)/2}T^{1/2+\theta+\epsilon}.$

Then we can trivially bound the off-diagonal term by

$\displaystyle\sum_{h,k\leq T^{\epsilon}}I_{O}^{(1)}(h,k)+I_{O}^{(2)}(h,k)\ll
T^{1/2+\theta+\epsilon}\sum_{h,k\leq T^{\epsilon}}(hk)^{(1+\theta)/2}\ll
T^{1-\epsilon}=o(T)$

We can use Delinge’s [18] bound, where $\theta=\frac{7}{64}$.

The calculation of the diagonal and the off-diagonal term completes the proof
of this proposition.

Proof of Proposition 4: From Lemma 7 and the argument given in [11], we can
obtain the main term of the twisted mean square of $L$-function is
$\sim(hk)^{-\frac{1}{2}}T$. Since $h$ and $k$ is very small in our case
(because $T$ is sufficiently large) we can say that the main term is $\sim T$,
which completes the proof of the proposition. ∎

##  8\. Independence of the automorphic $L$-functions

In this section, we prove Theorem 2 and 3. For the sake of the calculations we
start with the proof of Theorem 2, the computation for the next theorem is
rather complicated.

######  Theorem 4.

Let $f_{1}$ and $f_{2}$ be distinct primitive holomorphic cusp forms. Let $V$
be a fixed real number. As $T\to\infty$, for $a_{1},a_{2}\in\mathbb{R}$, for
all $v\in[-V,V]$,

$\displaystyle\frac{1}{T}meas$
$\displaystyle\left\\{t\in[T,2T]:\log|\mathcal{L}_{a_{1},a_{2}}\left(\frac{1}{2}+it\right)|\geq
v\sqrt{\frac{{a_{1}}^{2}+{a_{2}}^{2}}{2}\log\log
T}\right\\}\sim\frac{1}{\sqrt{2\pi}}\int_{v}^{\infty}e^{-\frac{x^{2}}{2}}dx,$

where

$\displaystyle\mathcal{L}_{a_{1},a_{2}}(s)=\mathcal{L}(f_{1},f_{2},s;a_{1},a_{2})\coloneqq|L(f_{1},s)|^{a_{1}}|L(f_{2},s)|^{a_{2}}.$

The proof of this theorem is similar to the proof of Theorem 1. So we define
the auxiliary series444For the convenience of the reader we write
$\mathcal{P}_{a_{1},a_{2},0}(s)$ instead of
$\mathcal{P}_{a_{1},a_{2},0}(f_{1},f_{2},s)$.,

$\displaystyle\mathcal{P}_{a_{1},a_{2},0}(s)=\mathcal{P}_{0}(f_{1},f_{2},s;X)\coloneqq\sum_{p\leq
X}\frac{a_{1}\lambda_{f_{1}}(p)+a_{2}\lambda_{f_{2}}(p)}{p^{s}},$

where $\lambda_{f_{1}},\lambda_{f_{2}}$ are the Fourier coefficients of the
primitive holomorphic cusp forms $f_{1},f_{2}$ respectively. Similarly to the
proof of Theorem 1, we need moment calculation for
$\mathcal{P}_{a_{1},a_{2},0}(s)$.

######  Lemma 10.

Let $f_{1},f_{2}$ be distinct primitive holomorphic cusp forms. Suppose,
$k,\ell$ are non-negative integers with $X^{k+\ell}\ll T$. Then for any real
numbers $a_{1},a_{2}$, we have,

$\displaystyle\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{\ell}}dt\ll
T,$

for $k\neq\ell$, and

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)|^{2k}dt=k!T((a_{1}^{2}+a_{2}^{2})\log\log
T)^{k}+O_{k}(T(\log\log T)^{k-1+\epsilon}).$

###### Proof.

Set

$\displaystyle\psi(p)=\psi_{a_{1},a_{2}}(p)\coloneqq
a_{1}\lambda_{f_{1}}(p)+a_{2}\lambda_{f_{2}}(p)$

and

$\displaystyle\Psi_{k}(n)\coloneqq\prod_{j=1}^{r}\psi(p_{j})^{\alpha_{j}}$

where $n=p_{1}^{\alpha_{1}}\cdots p_{r}^{\alpha_{r}}$ with
$\alpha_{1}+\cdots+\alpha_{r}=k$, we can write

$\displaystyle\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k}=\sum_{n}\frac{a_{k}(n)\Psi_{k}(n)}{n^{\sigma_{0}+it}},$

where $a_{k}(n)$ is same as defined in (7). Therefore,

(11)
$\displaystyle\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{\ell}}dt$
$\displaystyle=$ $\displaystyle
T\sum_{n}\frac{a_{k}(n)a_{\ell}(n)\Psi_{k}(n)\overline{\Psi_{\ell}(n)}}{n^{2\sigma_{0}}}+O\left(\sum_{n\neq
m}\frac{a_{k}(n)a_{\ell}(m)|\Psi_{k}(n)\overline{\Psi_{\ell}(m)}|}{(nm)^{\sigma_{0}-\frac{1}{2}}}\right).$

From the definition of $\Psi_{k}(n)$, for
$n=\prod_{j=1}^{r}p_{j}^{\alpha_{j}}$ we have

$\displaystyle|\Psi_{k}(n)|=\prod_{j=1}^{r}|\psi(p_{j})|^{\alpha_{j}}\leq\prod_{j=1}^{r}(|a_{1}\lambda_{f_{1}}(p_{j})|+|a_{2}\lambda_{f_{2}}(p_{j})|)=X(|a_{1}|+|a_{2}|)^{k}.$

Thus, the big-$O$ term is at most

$\displaystyle\sum_{n\neq
m}a_{k}(n)a_{\ell}(m)X^{2}(|a_{1}|+|a_{2}|)^{k+\ell}\ll X^{(k+1)+(\ell+1)}\ll
T.$

By the definition $a_{k}(n)a_{\ell}(n)=0$ if $k\neq\ell$, we conclude the
first assertion of Lemma 10. It remains to prove the lemma for $k=\ell$.
For $n=\prod_{j=1}^{r}p_{j}^{\alpha_{j}}$, we write

$\displaystyle\Psi_{k}(n)\overline{\Psi_{k}(n)}=\prod_{j=1}^{r}\left(a_{1}^{2}+a_{2}^{2}+a_{1}a_{2}\lambda_{f}(p_{j})+a_{1}a_{2}\bar{\lambda}_{f}(p_{j})\right)^{\alpha_{j}},$

where $\lambda_{f}\coloneqq\lambda_{f_{1}}\bar{\lambda}_{f_{2}}$. From the
above equation and the fact that $\sum_{j}\alpha_{j}=k$, for the non-square
free $n$ we have

$\displaystyle\left|\sum_{n\text{ non square
free}}\frac{a_{k}(n)a_{k}(n)\Psi_{k}(n)\overline{\Psi_{k}(n)}}{n^{2\sigma_{0}}}\right|\leq(|a_{1}|+|a_{2}|)^{2k}\sum_{n\text{
non square free}}\frac{a_{k}(n)^{2}\lambda_{f}(n)^{2}}{n^{2\sigma_{0}}}.$

Consider the each term $\prod_{j=1}^{r}a_{1}a_{2}\lambda_{f}(p_{j})$. From the
fundamental theorem of arithmetic each $p_{j}$’s are distinct primes. Since,
$\lambda_{f}(n)$ is a real valued function and gcd of each pair
$(p_{j},p_{i})$ equals to 1 for $i\neq j$ we can write
$\prod_{j=1}^{r}|(a_{1}a_{2}\lambda_{g}(p_{j}))|^{\alpha_{j}}=|a_{1}a_{2}|^{k}\prod_{j=1}^{r}|\lambda_{g}(p_{j})|^{\alpha_{j}}=|a_{1}a_{2}|^{k}(p_{1}^{\alpha_{1}}\cdots
p_{j}^{\alpha_{j}})^{\delta}=|a_{1}a_{2}|^{k}n^{\delta}$, where $\delta$ is
the appropriate constant satisfying $|\lambda_{g}(n)|\leq n^{\delta}$.

Therefore as argued in [4] the non-square free term contributes a quantity of
order $O((\log\log T)^{k-1+\epsilon})$. For $n$ square free we can express
$\Psi(n)\bar{\Psi}(n)$ as

(12)
$\displaystyle\sum(a_{1}^{2}+a_{2}^{2})^{\beta}(a_{1}a_{2})^{k-\beta}\lambda_{f}(m)\bar{\lambda}_{f}(m^{\prime})$

where the sum is over $\beta+\Omega(m)+\Omega(m^{\prime})=k$ such that
$0\leq\beta\leq k$and $n=mm^{\prime}$ where $n$ has the prime composition of
$p_{1},\ldots,p_{k}$. From (12) we can express that

$\displaystyle\sum_{0\leq\beta\leq
k}(a_{1}^{2}+a_{2}^{2})^{\beta}(a_{1}a_{2})^{k-\beta}\sum_{n}\frac{a_{k}(n)^{2}\lambda_{f}(m)\bar{\lambda}_{f}(m^{\prime})}{n^{2\sigma_{0}}}$

where the inner sum runs over $n=mm^{\prime}$ with $(m,m^{\prime})=1$ and
$\beta+\Omega(m)+\Omega(m^{\prime})=k$. Since $\lambda_{f}(n)$ is real
multiplicative function we have

$\displaystyle k!\sum_{0\leq\beta\leq
k}\frac{k!}{(k-\beta)!\beta!}\cdot\sum_{\begin{subarray}{c}p_{1},\ldots,p_{k}\leq
X\\\ \text{all $p_{j}$'s are
distinct}\end{subarray}}\frac{\lambda_{f}(p_{1}\cdots
p_{\beta})\bar{\lambda}_{f}(p_{\beta+1}\cdots p_{k})}{(p_{1}\cdots
p_{k})^{2\sigma_{0}}}$

which is

$\displaystyle k!\sum_{0\leq\beta\leq
k}\frac{k!}{(k-\beta)!\beta!}\left(\sum_{p\leq
X}\frac{\lambda_{f}(p)}{p^{2\sigma_{0}}}\right)^{\beta}\left(\sum_{p\leq
X}\frac{\bar{\lambda}_{f}(p)}{p^{2\sigma_{0}}}\right)^{k-\beta}.$

From [7] we conclude that the sum in (11) has its main contribution from
$\beta=k$, which is

$\displaystyle k!((a_{1}^{2}+a_{2}^{2})\log\log T)^{k}+O_{k}((\log\log
T)^{k-1+\epsilon}),$

completes the proof. ∎

Let $X_{j}=\log|L(f_{j},\sigma_{0}+it)|$ for $j=1,2$. We are now ready to
prove Theorem 2.

Proof of Theorem 4: Observe that
$\log|\mathcal{L}_{a_{1},a_{2}(s)}|=a_{1}X_{1}+a_{2}X_{2}$. Like an similar
argument of Proposition 1 we have

(13)
$\displaystyle\int_{t-1}^{t+1}\left|\log|\mathcal{L}_{a_{1},a_{2}(\frac{1}{2}+iy)}|-|\mathcal{L}_{a_{1},a_{2}(\sigma_{0}+iy)}|\right|dy\ll\left(\sigma_{0}-\frac{1}{2}\right)\log
T.$

By a similar argument of the proof of Proposition 1 the above equation is
true. However, we can study the function
$\log|\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)|$ away from the critical line
$\Re(s)=\frac{1}{2}$.

Define $\mathcal{P}_{a_{1},a_{2}}(s)\coloneqq
a_{1}\mathcal{P}(f_{1},s)+a_{2}\mathcal{P}(f_{2},s)$ where
$\mathcal{P}(f_{j},s)$’s are defined as the auxiliary series
$\mathcal{P}(f,s)$. Similarly, the contribution of the higher order terms of
the primes is at most $O(1)$. For the terms involving $p^{2}$ we can consider
$\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)$, as defined. Now by Lemma 10 for
$X^{k}\ll T$ and odd $k$,

$\displaystyle\int_{T}^{2T}(\Re(\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)))^{k}dt$
$\displaystyle=$
$\displaystyle\frac{1}{2^{k}}\sum_{\ell=0}^{k}\binom{k}{\ell}\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{\ell}\overline{\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)^{k-\ell}}dt\ll
T.$

Also, by Lemma 10, for even $k$,

$\displaystyle\int_{T}^{2T}(\Re(\mathcal{P}_{a_{1},a_{2},0}(\sigma_{0}+it)))^{k}dt$
$\displaystyle=$ $\displaystyle
2^{-k}\binom{k}{k/2}\left(\frac{k}{2}\right)!\left(\frac{a_{1}^{2}+a_{2}^{2}}{2}\log\log
T\right)^{\frac{k}{2}}+O_{k}((\log\log T)^{\frac{k}{2}-1+\epsilon}).$

It proves that the real part of $\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it)$ has
an approximately normal distribution with mean 0 and variance
$\frac{a_{1}^{2}+a_{2}^{2}}{2}\log\log T$.

Now it remains to connect $\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)$ with
$\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it)$. From the similar argument of
Proposition 3 and Proposition 4,
$|L(g_{j},\sigma_{0}+it)|=(1+o(1))\exp(\Re(\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it)))$,
except for a possible set of measure $o(T)$, for each $j$. Since,
$\mathcal{L}_{a_{1},a_{2}}(s)=|L(f_{1},s)|^{a_{1}}|L(f_{2},s)|^{a_{2}}$ and
$\mathcal{P}_{a_{1},a_{2}}(s)=a_{1}\mathcal{P}(f_{1},s)+a_{2}\mathcal{P}(f_{2},s)$,
we have

$\displaystyle\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)=(1+o(1))\exp(\Re(\mathcal{P}_{a_{1},a_{2}}(\sigma_{0}+it))).$

It shows $\log|\mathcal{L}_{a_{1},a_{2}}(\sigma_{0}+it)|$ has Gaussian
distribution with mean 0 and variance $\frac{a_{1}^{2}+a_{2}^{2}}{2}\log\log
T$, combining with (13) we conclude the proof. ∎

Finally, we are ready to prove Theorem 2.

Proof of Theorem 2: Let
$\mathbf{X}\coloneqq(X_{j})_{j=1}^{2}=\log|L(f_{j},\frac{1}{2}+it)|_{j=1}^{2}$.
Theorem 4 and Lemma 1 (see [16]) imply that $\mathbf{X}$ is an approximate bi-
variate normal distribution. Now we apply Theorem 4 with $a_{1}=a_{2}=1$, from
(3) we deduce that $X_{1}$ and $X_{2}$ are uncorrelated. Hence $X_{1},X_{2}$
are independent. ∎

##  9\. Gaussian Process for automorphic $L$-functions

In this section, we prove Theorem 3. The proof follows similarly to the proof
of Theorem 2. Since we are proving a more generalized version of it we need to
face more complicated computations.

For $(a_{j})_{j=1}^{n}\subset\mathbb{R}^{N}$ we consider,

$\displaystyle\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(s)=\sum_{p\leq
X}\frac{a_{1}\lambda_{f_{1}}(p)+\cdots+a_{N}\lambda_{f_{N}}(p)}{p^{s}}.$

Following the tradition, to prove Theorem 3 we need the next lemma.

######  Lemma 11.

Let $(\lambda_{f_{J}})_{j=1}^{N}$ be a sequence of distinct primitive
holomorphic cusp forms. Assume that $k,\ell$ are non-negative integers with
$X^{k+\ell}\leq T$. Then for any real numbers $(a_{j})_{j=1}^{n}$, we have for
$k\neq\ell$

$\displaystyle\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(\sigma_{0}+it)^{\ell}}\ll
T.$

For $k=\ell$

$\displaystyle\int_{T}^{2T}|\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(\sigma_{0}+it)|^{2k}dt=k!T\left(\left(\sum_{j=1}^{N}a_{j}^{2}\right)\log\log
T\right)^{k}+O_{k}(T(\log\log T)^{k-1+\epsilon}).$

###### Proof.

Similarly like the proof of Lemma 10 we start the proof by setting

$\displaystyle\psi(p)\coloneqq\sum_{j=1}^{N}a_{j}\lambda_{f_{j}}(p)$

and

$\displaystyle\Psi_{k}(n)\coloneqq\prod_{j=1}^{r}\psi(p_{j})^{\alpha_{j}}$

where $n=p_{1}^{\alpha_{1}}\cdots p_{r}^{\alpha_{r}}$ and
$\alpha_{1}+\cdots+\alpha_{r}=k$. Then we define

$\displaystyle\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(\sigma_{0}+it)^{k}=\sum_{n}\frac{a_{k}(n)\Psi_{k}(n)}{n^{\sigma_{0}+it}}$

where $a_{k}(n)$ is defined in (7). Therefore,

(14)
$\displaystyle\int_{T}^{2T}\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(\sigma_{0}+it)^{k}\overline{\mathcal{P}_{a_{1},a_{2},\ldots,a_{N},0}(\sigma_{0}+it)^{\ell}}dt$
$\displaystyle=$ $\displaystyle
T\sum_{n}\frac{a_{k}(n)a_{\ell}(n)\Psi_{k}(n)\overline{\Psi_{\ell}(n)}}{n^{2\sigma_{0}}}+O\left(\sum_{n\neq
m}\frac{a_{k}(n)a_{\ell}(m)|\Psi_{k}(n)\overline{\Psi_{\ell}(m)}|}{(nm)^{\sigma_{0}-\frac{1}{2}}}\right).$

From the definition of $\Psi_{k}(n)$ for $n=\prod_{j=1}^{r}$ we have

$\displaystyle|\Psi_{k}(n)|=\prod_{j=1}^{r}|\psi(p_{j})|^{\alpha_{j}}\leq\prod_{j=1}^{r}(|a_{1}|+\cdots+|a_{N}|)^{\alpha_{j}}=(|a_{1}|+\cdots+|a_{N}|)^{k}.$

Thus the big-$O$ term in (14) contributes

$\displaystyle\sum_{n\neq
m}a_{k}(n)a_{\ell}(m)(|a_{1}|+\cdots+|a_{N}|)^{k+\ell}\ll
X^{(k+1)+(\ell+1)}\ll T.$

As we know that by the definition $a_{k}(n)a_{\ell}(n)$ is 0 if $k\neq\ell$,
we conclude the first assertion of the lemma.

Now we need to prove the next part of the lemma which is the case for
$k=\ell$. For $n=\prod_{j=1}^{r}p_{j}^{\alpha_{j}}$ we write

(15)
$\displaystyle\Psi_{k}(n)\overline{{\Psi}_{k}(n)}=\prod_{j=1}^{r}\left(\left(\sum_{j=1}^{N}a_{j}^{2}\right)\sum_{i=1}^{N}\sum_{i\neq
i^{\prime}}a_{i}a_{i^{\prime}}(\lambda_{f_{i}}\bar{\lambda}_{f_{i^{\prime}}}(p_{j})+\lambda_{f_{i^{\prime}}}\bar{\lambda}_{f_{i}}(p_{j}))\right)^{\alpha_{j}}.$

We set $\lambda_{f}(n)=\sum_{i=1}^{N}\lambda_{f_{i}}\bar{\lambda}_{f_{i}}$. By
(15) and the fact that $\sum_{j}\alpha_{j}=k$ we have

$\displaystyle\left|\sum_{\begin{subarray}{c}n\text{ non square
free}\end{subarray}}\frac{a_{k}(n)a_{k}(n)\Psi_{k}(n)\overline{\Psi_{k}(n)}}{n^{2\sigma_{0}}}\right|\leq\left(\sum_{j=1}^{N}|a_{j}|\right)^{2k}\sum_{n\text{
non square free}}\frac{a_{k}(n)^{2}\lambda_{f}(n)^{2}}{n^{2\sigma_{0}}}.$

As argued in the proof of Lemma 10 the non square free $n$ terms in (15) give
$O_{k}((\log\log T)^{k-1+\epsilon})$. For square free $n$ we write

$\displaystyle\Psi_{k}(n)\overline{\Psi_{k}(n)}=$
$\displaystyle\sum_{0\leq\beta\leq
k}{A_{N}^{\beta}(a_{1}a_{2})^{\Omega(m_{1})+\Omega(m_{1}^{\prime})}\cdots(a_{N-1}a_{N})^{\Omega(m_{(N-1)N/2})+\Omega(m^{\prime}_{(N-1)N/2})}}$
$\displaystyle\cdot\sum_{n}\frac{a_{k}(n)^{2}\lambda_{f_{1}}\bar{\lambda}_{f_{2}}(m_{1})\lambda_{f_{2}}\bar{\lambda}_{f_{1}}(m^{\prime}_{1})\cdots\lambda_{f_{N}}\bar{\lambda}_{f_{N-1}(m^{\prime}_{(N-1)N/2})}}{n^{2\sigma_{0}}}$

where $A_{N}=\sum_{j=1}^{N}a_{j}^{2}$ and the second sum is over
$n=\prod_{j=1}^{(N-1)N/2}m_{j}m^{\prime}_{j}$ where $(m_{j},m^{\prime}_{j})=1$
for all $j=1,\ldots,N$. Also,
$\beta+\sum_{j=1}^{(N-1)N/2}(\Omega(m_{j})+\Omega(m^{\prime}_{j}))=k$. Since
each $\lambda_{f_{j}}$ is a real multiplicative function and by the change of
variables $\Omega(m_{j})\mapsto\gamma_{2j-1}$,
$\Omega(m^{\prime}_{j})\mapsto\gamma_{2j}$ we can write the second sum of the
above equation as

(16) $\displaystyle k!$
$\displaystyle\sum_{0\leq\gamma_{1},\ldots,\gamma_{(N-1)N}\leq
k}\frac{k!}{\gamma_{1}!\cdots\gamma_{(N-1)N}!(k-\gamma_{1}!-\cdots-\gamma_{(N-1)N})!}$
$\displaystyle\cdot\sum_{\begin{subarray}{c}p_{1},\ldots,p_{k}\leq X\\\ \text{
$p_{j}$'s are
distinct}\end{subarray}}\frac{\lambda_{f_{1}}\bar{\lambda}_{f_{2}}(p_{1}\cdots
p_{\gamma_{1}})\cdots\lambda_{f_{N}}\bar{\lambda}_{N-1}(p_{\gamma_{(N-1)N}}\cdots
p_{k})}{(p_{1}\cdots p_{k})^{2\sigma_{0}}}$ $\displaystyle=k!$
$\displaystyle\sum_{0\leq\gamma_{1},\ldots,\gamma_{(N-1)N}\leq
k}\frac{k!}{\gamma_{1}!\cdots\gamma_{(N-1)N}!(k-\gamma_{1}!-\cdots-\gamma_{(N-1)N})!}$
$\displaystyle\cdot\left(\sum_{p\leq
X}\frac{\lambda_{f_{1}}\bar{\lambda}_{f_{2}}(p)}{p^{2\sigma_{0}}}\right)^{\gamma_{1}}\cdots\left(\sum_{p\leq
X}\frac{\lambda_{f_{N}}\bar{\lambda}_{f_{N-1}}(p)}{p^{2\sigma_{0}}}\right)^{\gamma_{(N-1)N}}.$

By (16),(15) and the similar argument of the proof of Lemma 10, the first sum
of in (14) is mainly contributed by $k=\gamma_{1}+\cdots+\gamma_{(N-1)N}$
which is

$\displaystyle k!\left(\left(\sum_{j=1}^{N}a_{j}^{2}\right)\log\log
T\right)^{k}+O_{k}((\log\log T)^{k-1+\epsilon}),$

completes the proof of the lemma. ∎

In Lemma 11, we have computed the moments of the auxiliary series
$\mathcal{P}_{a_{1},\ldots,a_{N},0}(s)$, with this calculation in hand in the
next theorem we study the joint distribution of the automorphic $L$-functions.

######  Theorem 5.

Let $(f_{j})_{j=1}^{N}$ be a sequence of distinct primitive holomorphic cusp
forms. Let $V$ be a fixed positive real number. For any real numbers
$(a_{j})_{j=1}^{N}$, $\log|\mathcal{L}_{a_{1},\ldots,a_{N}}(\frac{1}{2}+it)|$
has normal distribution with mean 0 and variance
$\frac{a_{1}^{2}+\cdots+a_{N}^{2}}{2}\log\log T$ as $T\to\infty$ and

$\displaystyle\mathcal{L}_{a_{1},\ldots,a_{N}}(s)=\mathcal{L}(f_{1},\ldots,f_{N},a_{1},\ldots,a_{N},s)\coloneqq|L(f_{1},s)|^{a_{1}}\cdots|L(f_{N},s)|^{a_{N}}.$

###### Proof.

The proof of this theorem is the same as the proof of Theorem 4, instead of
Lemma 10 we have to consider Lemma 11, that’s why we can omit it without
repeating the same steps. ∎

Now we are ready to prove Theorem 3.

Proof of Theorem 3: Let $J$ denotes a totally ordered set of distinct
primitive holomorphic cusp forms. For any finite ordered subset
$\\{f_{1},\ldots,f_{N}\\}$ of $J$ we consider
${\bf{X}}\coloneqq(\log|L(f_{j},\frac{1}{2}+it)|)_{j=1}^{N}$. From Theorem 5
and Lemma 1 (see [16]), it follows that $\bf{X}$ is an approximate $N$-variate
normal distribution. Thus, we see that any finite linear combination of
elements in $(\log|L(f,\frac{1}{2}+it)|)_{f\in J}$ is a multivariate normal
distribution. Hence, $(\log|L(f,\frac{1}{2}+it)|)_{f\in J}$ forms a Gaussian
process.
Moreover, the components in $\bf{X}$ are mutually independent since are pair-
wise independent by Theorem 2. ∎

##  10\. Conclusion

We have followed the method of Radziwiłł and Soudararajan [3] to prove Theorem
1 but the proof for the $L$-functions need some modifications. Like the method
established in [3], we have used very basic properties of $L$-functions. But
the proof of Proposition 4 does not follow the method established in [3].
Since we need to calculate the mean square of twisted automorphic $L$-function
we can not apply the method established in [3], directly. So we have taken the
help of Hughes and Young’s [11] work instead. The most important fact we have
relied on is the Ramanujan-Peterson conjecture. Also, we have taken the help
of P. Kühn et al result [14]. It is possible to obtain a better error term
that we have obtained. But the main focus of this paper concerns the proof of
Selberg’s central limit theorem for $L$-functions. Any improvement on the
error term for the mean square estimate of $L$-function will not effect our
result.

Selberg [2] mentioned the concept of statistical independence of $L$-functions
in his paper. As Hsu-Wong [4] proved this result for Dirichlet $L$-functions,
we have proved this result for the families of automorphic $L$-functions
associated with the sequence of primitive holomorphic cusp forms. Such a
result can be proved for $q$-aspect as well by fixing the range of $t$.

As mentioned in [3] Proposition 1-3 can be established for the higher degree
$L$-functions as well. For Proposition 4, we rely on the information
contributed by shifted convolution problem (to calculate the error term of the
mean square of $L$-functions) which is not known for degree 3 or higher.
However, some hybrid results for $GL(3)$ $L$-functions are possible. Moreover,
one can extend Selberg’s Central limit theorem for the imaginary part of the
Riemann zeta function as well as $L$-functions. As mentioned in [3] the
mollification technique is not helpful but one can use the method developed by
J. Najnudel [10] instead.

Keating and Snaith[8] have conjectured that central values of $L$-functions in
families have the log-normal distribution with an appropriate mean and
variance depending on their families. We have proved the main result for
$t$-aspect. The hybrid aspect for this type of result is not possible because
we are proving Theorem 1 on the critical line. In the hybrid aspect if we
average over the Dirichlet characters without fixing the range of $t$ then by
the assumption of Grand Riemann Hypothesis the statement of these theorem and
propositions do not hold, since $\log 0$ is undefined. But the proof on weight
aspect for such result is possible. For such calculations Lau’s [17] work
might be helpful.

## Acknowledgments

I would like to express my gratitude to my primary supervisors, Dr Gihan
Marasingha and Prof. Julio Andrade, who guided me throughout this project.

I am also grateful to Dr Dirk Zeindler for his encouragement and help.

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