1 Introduction
Let
$H$
be the open upper half plane. The elements
$(\!\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\!)$
of the group
$\operatorname{PSL}(2,\mathbf{R})$
act on
$H$
by the rule
$z\rightarrow (az+b)/(cz+d)$
. Write
$$\begin{eqnarray}d\unicode[STIX]{x1D707}_{z}=\frac{dx\,dy}{y^{2}};\end{eqnarray}$$
this is the
$\operatorname{PSL}(2,\mathbf{R})$
-invariant measure on
$H$
.
Let
$\unicode[STIX]{x1D6E4}\subseteq \operatorname{PSL}(2,\mathbf{R})$
be a finite-volume Fuchsian group (see [Reference Iwaniec9, p. 40]), i.e.
$\unicode[STIX]{x1D6E4}$
acts discontinuously on
$H$
and it has a fundamental domain of finite volume (with respect to the measure
$d\unicode[STIX]{x1D707}_{z}$
). Let
$F$
be a fixed fundamental domain of
$\unicode[STIX]{x1D6E4}$
in
$H$
(it contains exactly one point of each
$\unicode[STIX]{x1D6E4}$
-equivalence class of
$H$
).
For
$z,w\in H$
, let
$$\begin{eqnarray}u(z,w)=\frac{|z-w|^{2}}{4\operatorname{Im}z\operatorname{Im}w};\end{eqnarray}$$
this is closely related to the hyperbolic distance
$\unicode[STIX]{x1D70C}(z,w)$
of
$z$
and
$w$
(see [Reference Iwaniec9, (1.3)]). For
$X>2$
, define
$$\begin{eqnarray}N(z,w,X):=|\{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}:4u(\unicode[STIX]{x1D6FE}z,w)+2\leqslant X\}|;\end{eqnarray}$$
the condition here is equivalent to
$\unicode[STIX]{x1D70C}(z,w)\leqslant \cosh ^{-1}(X/2)$
and hence
$N(z,w,X)$
is the number of points
$\unicode[STIX]{x1D6FE}z$
in the hyperbolic circle around
$w$
of radius
$\cosh ^{-1}(X/2)$
. Therefore, the estimation of
$N(z,w,X)$
is called the hyperbolic circle (or lattice point) problem. This is a classical problem; see the Introduction of [Reference Risager and Petridis11] for its history.
In order to give the main term in the asymptotic expansion of
$N(z,w,X)$
as
$X\rightarrow \infty$
, we have to introduce Maass forms.
The hyperbolic Laplace operator is given by
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}:=y^{2}\biggl(\frac{\unicode[STIX]{x2202}^{2}}{\unicode[STIX]{x2202}x^{2}}+\frac{\unicode[STIX]{x2202}^{2}}{\unicode[STIX]{x2202}y^{2}}\biggr).\end{eqnarray}$$
It is well known that
$\unicode[STIX]{x1D6E5}$
commutes with the action of
$\operatorname{PSL}(2,\mathbf{R})$
.
Let
$\{u_{j}(z):\;j\geqslant 0\}$
be a complete orthonormal system of Maass forms for
$\unicode[STIX]{x1D6E4}$
(the function
$u_{0}(z)$
is constant); let
$\unicode[STIX]{x1D6E5}u_{j}=\unicode[STIX]{x1D706}_{j}u_{j}$
, where
$\unicode[STIX]{x1D706}_{j}=s_{j}(s_{j}-1)$
,
$s_{j}={\textstyle \frac{1}{2}}+\text{i}t_{j}$
and
$\operatorname{Re}s_{j}={\textstyle \frac{1}{2}}$
or
${\textstyle \frac{1}{2}}<s_{j}\leqslant 1$
. Note that
$s_{j}=1$
if and only if
$j=0$
, and
${\textstyle \frac{1}{2}}<s_{j}<1$
holds only for finitely many
$j$
.
We can now define
$$\begin{eqnarray}M(z,w,X):=\sqrt{\unicode[STIX]{x1D70B}}\mathop{\sum }_{s_{j}\in (1/2,1 ]}\frac{\unicode[STIX]{x1D6E4}(s_{j}-\frac{1}{2})}{\unicode[STIX]{x1D6E4}(s_{j}+1)}u_{j}(z)\overline{u_{j}(w)}X^{s_{j}}.\end{eqnarray}$$
It is well known that
$$\begin{eqnarray}|N(z,w,X)-M(z,w,X)|=O_{z,w,\unicode[STIX]{x1D6E4}}(X^{2/3});\end{eqnarray}$$
see e.g. [Reference Iwaniec9, Theorem 12.1]. The error term here has never been improved for any group, but (as is noted in [Reference Iwaniec9]) it is conjectured that
$\frac{2}{3}$
might be lowered to any number greater than
$\frac{1}{2}$
.
It was proved recently in [Reference Risager and Petridis11] that in the case
$\unicode[STIX]{x1D6E4}=\operatorname{PSL}(2,\mathbf{Z})$
the error term
$X^{2/3}$
can be improved taking a certain local average. More precisely, it was proved that if
$f$
is a smooth non-negative function which is compactly supported on
$F$
, then
$$\begin{eqnarray}\int _{F}f(z)(N(z,z,X)-M(z,z,X))\,d\unicode[STIX]{x1D707}_{z}=O_{f,\unicode[STIX]{x1D716}}(X^{7/12+\unicode[STIX]{x1D716}})\end{eqnarray}$$
for any
$\unicode[STIX]{x1D716}>0$
.
In the present paper we show that for this local average the error term
$X^{2/3}$
can be improved in the case of any finite-volume Fuchsian group
$\unicode[STIX]{x1D6E4}$
. In this generality we get the exponent
$X^{5/8+\unicode[STIX]{x1D716}}$
, which is better than
$X^{2/3}$
but not as strong as the result of [Reference Risager and Petridis11] in the special case
$\unicode[STIX]{x1D6E4}=\operatorname{PSL}(2,\mathbf{Z})$
.
Theorem 1.1. Let
$f$
be a given smooth function on
$H$
such that it is compactly supported on
$F$
and, for
$X>2$
, let
$$\begin{eqnarray}N_{f}(X):=\int _{F}f(z)N(z,z,X)\,d\unicode[STIX]{x1D707}_{z}.\end{eqnarray}$$
Then
$$\begin{eqnarray}\displaystyle N_{f}(X) & = & \displaystyle \int _{F}f(z)\bigg(\sqrt{\unicode[STIX]{x1D70B}}\mathop{\sum }_{s_{j}\in (1/2,1]}\frac{\unicode[STIX]{x1D6E4}(s_{j}-\frac{1}{2})}{\unicode[STIX]{x1D6E4}(s_{j}+1)}X^{s_{j}}|u_{j}(z)|^{2}\bigg)\,d\unicode[STIX]{x1D707}_{z}\nonumber\\ \displaystyle & & \displaystyle +\,O_{f,\unicode[STIX]{x1D6E4},\unicode[STIX]{x1D716}}(X^{5/8+\unicode[STIX]{x1D716}})\nonumber\end{eqnarray}$$
for every given
$\unicode[STIX]{x1D716}>0$
.
Remark 1.1. As is noted in [Reference Risager and Petridis11, Remark 1.3], the proof there is valid only for groups similar to
$\operatorname{PSL}(2,\mathbf{Z})$
, as it requires strong arithmetic input not available for most groups. Our theorem is valid for any finite-volume Fuchsian group. In particular, it is valid for cocompact groups.
The main tool of our proof is our generalization of the Selberg trace formula [Reference Biró1], which is valid for every finite-volume Fuchsian group
$\unicode[STIX]{x1D6E4}$
. Note that the operator whose trace is studied in [Reference Biró1] has been used and analyzed also in a series of papers by Zelditch (see [Reference Zelditch12–Reference Zelditch14]).
Remark 1.2. As a very brief indication of the idea of our proof, we mention that
$$\begin{eqnarray}N(z,z,X)=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}k(u(z,\unicode[STIX]{x1D6FE}z)),\end{eqnarray}$$
where
$k$
is the characteristic function of the interval
$[0,x]$
with a large real
$x$
(in fact
$x=(X-2)/4$
). We use the decomposition
$$\begin{eqnarray}k(v)=k^{\ast }(v)+(k(v)-k^{\ast }(v)),\end{eqnarray}$$
where
$k^{\ast }$
is a certain smoothed version of
$k$
. We will estimate the contribution of
$k^{\ast }$
in (1.1) in the traditional way, using the spectral expansion of the automorphic kernel function given by
$k^{\ast }$
and estimating the Selberg–Harish-Chandra transform of
$k^{\ast }$
. However, the contribution of
$k(v)-k^{\ast }(v)$
to
$N_{f}(X)$
is estimated in a completely different way, using our generalization of the Selberg trace formula [Reference Biró1].
Remark 1.3. We have seen that
$N(z,z,X)$
is the number of points in the
$\unicode[STIX]{x1D6E4}$
-orbit of
$z$
in a hyperbolic circle around
$z$
of large radius. Note that the analogous quantity in the euclidean case (if we choose in place of
$\unicode[STIX]{x1D6E4}$
the group of translations on the euclidean plane with vectors having integer coordinates in place of
$\unicode[STIX]{x1D6E4}$
) is independent of
$z$
and hence averaging in
$z$
does not help in the euclidean case: the problem there remains the same.
We mention that different kinds of averages were considered by Chamizo in [Reference Chamizo3]. In particular, he proved a strong estimate for the integral with respect to
$z$
of the square of
$N(z,w,X)$
for a fixed
$w$
over the whole fundamental domain in the case of a cocompact group; see [Reference Chamizo3, Corollary 2.2.1].
Remark 1.4. The structure of the paper is the following. In §2 we introduce the necessary notation. In §3 we give the two main lemmas (Lemmas 3.3 and 3.4) needed for the proof of the theorem. Lemma 3.3 is our main new tool; this is a consequence of our generalization of the Selberg trace formula in [Reference Biró1]. Lemma 3.4 is the well-known spectral expansion of an automorphic kernel function. The proof of Theorem 1.1 is given in §4, using some results proved only later on special functions and automorphic functions in §§5 and 6, respectively.
2 Further notation
We fix a complete set
$A$
of inequivalent cusps of
$\unicode[STIX]{x1D6E4}$
and we will denote its elements by
$a$
,
$b$
or
$c$
, so e.g.
$\sum _{a}\sum _{c}$
or
$\bigcup _{a}$
will mean that
$a$
and
$c$
run over
$A$
. We say that
$\unicode[STIX]{x1D70E}_{a}$
is a scaling matrix of a cusp
$a$
if
$\unicode[STIX]{x1D70E}_{a}\infty =a$
,
$\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6E4}_{a}\unicode[STIX]{x1D70E}_{a}=B$
, where
$\unicode[STIX]{x1D6E4}_{a}$
is the stability group of
$a$
in
$\unicode[STIX]{x1D6E4}$
and
$B$
is the group of integer translations. The scaling matrix is determined up to composition with a translation from the right.
We also fix a complete set
$P$
of representatives of
$\unicode[STIX]{x1D6E4}$
-equivalence classes of the set
$$\begin{eqnarray}\{z\in H:\unicode[STIX]{x1D6FE}z=z\text{ for some }\text{id}\neq \unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}\}.\end{eqnarray}$$
For a
$p\in P$
, let
$m_{p}$
be the order of the stability group of
$p$
in
$\unicode[STIX]{x1D6E4}$
.
Let
$$\begin{eqnarray}P(Y)=\{z=x+iy:\;0<x\leqslant 1,y>Y\}\end{eqnarray}$$
and let
$Y_{\unicode[STIX]{x1D6E4}}$
be a constant (depending only on the group
$\unicode[STIX]{x1D6E4}$
) such that for any
$Y\geqslant Y_{\unicode[STIX]{x1D6E4}}$
the cuspidal zones
$F_{a}(Y)=\unicode[STIX]{x1D70E}_{a}P(Y)$
are disjoint and the fixed fundamental domain
$F$
of
$\unicode[STIX]{x1D6E4}$
is partitioned into
$$\begin{eqnarray}F=F(Y)\cup \mathop{\bigcup }_{a}F_{a}(Y),\end{eqnarray}$$
where
$F(Y)$
is the central part,
$$\begin{eqnarray}F(Y)=F\bigg\backslash\mathop{\bigcup }_{a}F_{a}(Y),\end{eqnarray}$$
and
$F(Y)$
has compact closure.
For
$j\geqslant 0$
and
$a\in A$
, we have the Fourier expansion
$$\begin{eqnarray}u_{j}(\unicode[STIX]{x1D70E}_{a}z)=\unicode[STIX]{x1D6FD}_{a,j}(0)y^{1-s_{j}}+\mathop{\sum }_{n\neq 0}\unicode[STIX]{x1D6FD}_{a,j}(n)W_{s_{j}}(nz),\end{eqnarray}$$
where
$W$
is the Whittaker function.
The Fourier expansion of the Eisenstein series (as in [Reference Iwaniec9, (3.20)]) is given by
$$\begin{eqnarray}E_{c}(\unicode[STIX]{x1D70E}_{a}z,s)=\unicode[STIX]{x1D6FF}_{ca}y^{s}+\unicode[STIX]{x1D719}_{c,a}(s)y^{1-s}+\mathop{\sum }_{n\neq 0}\unicode[STIX]{x1D719}_{a,c}(n,s)W_{s}(nz).\end{eqnarray}$$
For
$Y\geqslant Y_{\unicode[STIX]{x1D6E4}}$
, let us define the truncated Eisenstein series (as in [Reference Iwaniec9, pp. 95–96]) in the following way: for a given
$c\in A$
and every
$a\in A$
, let
$$\begin{eqnarray}E_{c}^{Y}(z,s)=E_{c}(z,s)-(\unicode[STIX]{x1D6FF}_{ca}(\operatorname{Im}\unicode[STIX]{x1D70E}_{a}^{-1}z)^{s}+\unicode[STIX]{x1D719}_{c,a}(s)(\operatorname{Im}\unicode[STIX]{x1D70E}_{a}^{-1}z)^{1-s})\quad \text{for }z\in F_{a}(Y),\end{eqnarray}$$
let
$$\begin{eqnarray}E_{c}^{Y}(z,s)=E_{c}(z,s)\quad \text{for }z\in F(Y)\end{eqnarray}$$
and finally let
$E_{c}^{Y}(\unicode[STIX]{x1D6FE}z,s)=E_{c}^{Y}(z,s)$
for
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
and
$z\in F$
.
For
$Y\geqslant Y_{\unicode[STIX]{x1D6E4}}$
and
$j\geqslant 0$
, let us also define the truncation of
$u_{j}$
in the following way: for every
$a\in A$
, let
$$\begin{eqnarray}u_{j}^{Y}(z)=u_{j}(z)-\unicode[STIX]{x1D6FD}_{a,j}(0)(\operatorname{Im}\unicode[STIX]{x1D70E}_{a}^{-1}z)^{1-s_{j}}\quad \text{for }z\in F_{a}(Y),\end{eqnarray}$$
let
$$\begin{eqnarray}u_{j}^{Y}(z)=u_{j}(z)\quad \text{for }z\in F(Y)\end{eqnarray}$$
and finally let
$u_{j}^{Y}(\unicode[STIX]{x1D6FE}z)=u_{j}^{Y}(z)$
for
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
and
$z\in F$
.
Let
$\{S_{l}:\;l\in L\}$
be the set of the poles in the half plane
$\operatorname{Re}s>{\textstyle \frac{1}{2}}$
of the Eisenstein series for
$\unicode[STIX]{x1D6E4}$
. Then
${\textstyle \frac{1}{2}}<S_{l}\leqslant 1$
for every
$l\in L$
, and
$L$
is a finite set. We have
$\unicode[STIX]{x1D6FD}_{a,j}(0)=0$
if
$u_{j}(z)$
is not a linear combination of the residues of Eisenstein series, so if
$j\geqslant 0$
is such that
$\unicode[STIX]{x1D6FD}_{a,j}(0)\neq 0$
for some
$a$
, then
$s_{j}=S_{l}$
for some
$l\in L$
. In particular,
$u_{j}^{Y}$
is the same as
$u_{j}$
for all but finitely many
$j.$
The constants in the symbols
$O$
will depend on the group
$\unicode[STIX]{x1D6E4}$
. For a function
$g$
, we will denote its
$j$
th derivative by
$g^{(j)}$
.
For
$\unicode[STIX]{x1D706}\leqslant 0$
, define the special function
$f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})$
in the following way:
$f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})$
is the unique even solution of the differential equation
$$\begin{eqnarray}f^{(2)}(\unicode[STIX]{x1D703})=\frac{\unicode[STIX]{x1D706}}{\cos ^{2}\unicode[STIX]{x1D703}}f(\unicode[STIX]{x1D703}),\quad \unicode[STIX]{x1D703}\in \bigg(-\frac{\unicode[STIX]{x1D70B}}{2},\frac{\unicode[STIX]{x1D70B}}{2}\bigg)\end{eqnarray}$$
with
$f_{\unicode[STIX]{x1D706}}(0)=1$
. Note that this differential equation (which appeared in [Reference Biró1] and also in [Reference Huber8], equations (10)–(11)) is the Laplacian on functions depending only on the hyperbolic distance from the imaginary real axis, i.e. if for
$z\in H$
we write
$z=r\text{e}^{\text{i}(\unicode[STIX]{x1D70B}/2+\unicode[STIX]{x1D703})}$
with
$r>0$
and
$\unicode[STIX]{x1D703}\in (-\unicode[STIX]{x1D70B}/2,\unicode[STIX]{x1D70B}/2)$
, then for
$$\begin{eqnarray}F(z):=f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})\end{eqnarray}$$
we have
$\unicode[STIX]{x1D6E5}F=\unicode[STIX]{x1D706}F$
.
For
$\unicode[STIX]{x1D706}\leqslant 0$
, define also the special function
$g_{\unicode[STIX]{x1D706}}(r)$
(
$r\in [0,\infty )$
) as the unique solution of
$$\begin{eqnarray}g^{(2)}(r)+\frac{\cosh r}{\sinh r}g^{(1)}(r)=\unicode[STIX]{x1D706}g(r)\end{eqnarray}$$
with
$g_{\unicode[STIX]{x1D706}}(0)=1$
. Note that it is well known (see e.g. [Reference Iwaniec9, (1.20)]) that this differential equation is the Laplacian on functions depending only on the hyperbolic distance
$\unicode[STIX]{x1D70C}(z,i)$
from the given point
$i$
, i.e. if for
$z\in H$
we write
$$\begin{eqnarray}G(z):=g_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D70C}(z,i)),\end{eqnarray}$$
then we have
$\unicode[STIX]{x1D6E5}G=\unicode[STIX]{x1D706}G$
.
Note that
$f_{0}(\unicode[STIX]{x1D703})$
and
$g_{0}(r)$
are the identically
$1$
functions.
If
$m$
is a compactly supported continuous function on
$[0,\infty )$
, let (see [Reference Iwaniec9, (1.62)])
$$\begin{eqnarray}g_{m}(a)=2q_{m}\bigg(\frac{\text{e}^{a}+\text{e}^{-a}-2}{4}\bigg)\quad \text{where }q_{m}(v)=\int _{0}^{\infty }\frac{m(v+\unicode[STIX]{x1D70F})}{\sqrt{\unicode[STIX]{x1D70F}}}\,d\unicode[STIX]{x1D70F},\end{eqnarray}$$
and let
$$\begin{eqnarray}h_{m}(r)=\int _{-\infty }^{\infty }g_{m}(a)\text{e}^{\text{i}ra}\,da.\end{eqnarray}$$
For
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
, denote by
$[\unicode[STIX]{x1D6FE}]$
the conjugacy class of
$\unicode[STIX]{x1D6FE}$
in
$\unicode[STIX]{x1D6E4}$
, i.e.
$$\begin{eqnarray}[\unicode[STIX]{x1D6FE}]=\{\unicode[STIX]{x1D70F}^{-1}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F}:\,\unicode[STIX]{x1D70F}\in \unicode[STIX]{x1D6E4}\}.\end{eqnarray}$$
We use the general notation
$$\begin{eqnarray}(F,G)=\int _{F}F(z)\overline{G(z)}\,d\unicode[STIX]{x1D707}_{z}.\end{eqnarray}$$
We write
$F(\!\begin{smallmatrix}\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FD}\\ \unicode[STIX]{x1D6FE}\end{smallmatrix};z)$
for the Gauss hypergeometric function. We use the notation
$\unicode[STIX]{x1D6E4}(X\pm Y)=\unicode[STIX]{x1D6E4}(X+Y)\unicode[STIX]{x1D6E4}(X-Y)$
.
3 Basic lemmas
Our two main results here are Lemmas 3.3 and 3.4, but first we have to prove two simple lemmas.
Lemma 3.1. Let
$a\in A$
. If
$Y$
is large enough (depending only on
$\unicode[STIX]{x1D6E4}$
), then for
$z\in F_{a}(Y)$
and
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
we have either
$$\begin{eqnarray}u(\unicode[STIX]{x1D6FE}z,z)\geqslant D_{\unicode[STIX]{x1D6E4}}Y^{2}\end{eqnarray}$$
with some constant
$D_{\unicode[STIX]{x1D6E4}}>0$
depending only on
$\unicode[STIX]{x1D6E4}$
or we have
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{a}$
.
Proof. Let
$z\in F_{a}(Y)=\unicode[STIX]{x1D70E}_{a}P(Y)$
; then
$z=\unicode[STIX]{x1D70E}_{a}w$
with some
$w\in P(Y)$
and, for
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
, we have
$$\begin{eqnarray}u(\unicode[STIX]{x1D6FE}z,z)=u(\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70E}_{a}w,w).\end{eqnarray}$$
Let
$\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70E}_{a}=(\!\begin{smallmatrix}\ast & \ast \\ C & D\end{smallmatrix}\!)$
. Assume that
$|C|>0$
; then
$\operatorname{Im}\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70E}_{a}w=\operatorname{Im}w/(|Cw+D|^{2})\leqslant 1/(C^{2}\operatorname{Im}w)$
. Since
$\operatorname{Im}w>Y$
, for large enough
$Y$
this implies that
$$\begin{eqnarray}u(\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70E}_{a}w,w)\geqslant \frac{|Y-1/C^{2}Y|^{2}}{4Y(1/C^{2}Y)}.\end{eqnarray}$$
Since
$$\begin{eqnarray}\min \bigg\{|C|>0:\bigg(\begin{array}{@{}cc@{}}\ast & \ast \\ C & \ast \end{array}\bigg)\in \unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6E4}\unicode[STIX]{x1D70E}_{a}\bigg\}\end{eqnarray}$$
exists (see [Reference Iwaniec9, p. 53]), (3.2) and (3.3) imply (3.1).
Assume that
$C=0$
. Then
$\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70E}_{a}\infty =\infty$
, so
$\unicode[STIX]{x1D6FE}a=a$
, because
$\unicode[STIX]{x1D70E}_{a}\infty =a$
; hence,
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}_{a}$
and the lemma is proved.◻
Lemma 3.2. Assume that
$m$
is a compactly supported function with bounded variation on
$[0,\infty )$
. Let
$a\in A$
and let
$Y$
be large enough depending on
$\unicode[STIX]{x1D6E4}$
and
$m$
. For
$z,w\in H$
, define
$$\begin{eqnarray}M(z,w):=\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}m\bigg(\frac{|z-\unicode[STIX]{x1D6FE}w|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}w}\bigg);\end{eqnarray}$$
then, for
$z\in F_{a}(Y)$
, we have
$$\begin{eqnarray}M(z,z)=4\operatorname{Im}w\int _{0}^{\infty }m(y^{2})\,dy+O_{\unicode[STIX]{x1D6E4},m}(1),\end{eqnarray}$$
where
$z=\unicode[STIX]{x1D70E}_{a}w$
.
Proof. It follows easily from Lemma 3.1 using (3.2) and
$\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6E4}_{a}\unicode[STIX]{x1D70E}_{a}=B$
that if
$Y$
is large enough, then
$$\begin{eqnarray}M(z,z)=\mathop{\sum }_{l=-\infty }^{\infty }m\bigg(\frac{l^{2}}{4\operatorname{Im}^{2}w}\bigg),\end{eqnarray}$$
where
$z=\unicode[STIX]{x1D70E}_{a}w$
. Then, using the inequality of Koksma [Reference Kuipers and Niederreiter10, Theorem 5.1], we get the lemma.◻
Lemma 3.3. Let
$u$
be a fixed
$\unicode[STIX]{x1D6E4}$
-automorphic eigenfunction of the Laplace operator with eigenvalue
$\unicode[STIX]{x1D706}=s(s-1)$
satisfying
$$\begin{eqnarray}\int _{F}|u(z)|\,d\unicode[STIX]{x1D707}_{z}<\infty\end{eqnarray}$$
and let
$\operatorname{Re}s={\textstyle \frac{1}{2}}$
or
${\textstyle \frac{1}{2}}<s\leqslant 1$
. Denote the Fourier expansion of
$u$
by
$$\begin{eqnarray}u(\unicode[STIX]{x1D70E}_{a}z)=\unicode[STIX]{x1D6FD}_{a}(0)y^{s}+\tilde{\unicode[STIX]{x1D6FD}}_{a}(0)y^{1-s}+\mathop{\sum }_{n\neq 0}\unicode[STIX]{x1D6FD}_{a}(n)W_{s}(nz).\end{eqnarray}$$
Introduce the notation
$$\begin{eqnarray}B_{u}=\mathop{\sum }_{a}\unicode[STIX]{x1D6FD}_{a}(0),\qquad \tilde{B}_{u}=\mathop{\sum }_{a}\tilde{\unicode[STIX]{x1D6FD}}_{a}(0).\end{eqnarray}$$
Assume that
$m$
is a compactly supported function with bounded variation on
$[0,\infty )$
and
$$\begin{eqnarray}\int _{0}^{\infty }\frac{m(v)}{\sqrt{v}}\,dv=0.\end{eqnarray}$$
Recalling the notation (3.4), we have that
$$\begin{eqnarray}\int _{F}M(z,z)u(z)\,d\unicode[STIX]{x1D707}_{z}=\unicode[STIX]{x1D6F4}_{\text{hyp}}+\unicode[STIX]{x1D6F4}_{\text{ell}}+\unicode[STIX]{x1D6F4}_{\text{par}}\end{eqnarray}$$
with the definition
$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{\text{hyp}}:=\mathop{\sum }_{\substack{ [\unicode[STIX]{x1D6FE}] \\ \unicode[STIX]{x1D6FE}~\text{hyperbolic}}}\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}u\,dS\bigg)\int _{-\unicode[STIX]{x1D70B}/2}^{\unicode[STIX]{x1D70B}/2}m\bigg(\frac{N(\unicode[STIX]{x1D6FE})+N(\unicode[STIX]{x1D6FE})^{-1}-2}{4\cos ^{2}\unicode[STIX]{x1D703}}\bigg)f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})\frac{d\unicode[STIX]{x1D703}}{\cos ^{2}\unicode[STIX]{x1D703}},\end{eqnarray}$$
where the summation is over the hyperbolic conjugacy classes of
$\unicode[STIX]{x1D6E4}$
,
$N(\unicode[STIX]{x1D6FE})$
is the norm of (the conjugacy class of)
$\unicode[STIX]{x1D6FE}$
,
$C_{\unicode[STIX]{x1D6FE}}$
is the closed geodesic obtained by factorizing the non-euclidean line connecting the fixed points of
$\unicode[STIX]{x1D6FE}$
by the action of the centralizer of
$\unicode[STIX]{x1D6FE}$
in
$\unicode[STIX]{x1D6E4}$
,
$dS=|dz|/y$
is the hyperbolic arc length,
$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{\text{ell}}:=\mathop{\sum }_{p\in P}\frac{2\unicode[STIX]{x1D70B}}{m_{p}}u(p)\mathop{\sum }_{l=1}^{m_{p}-1}\int _{0}^{\infty }m\bigg(\sin ^{2}\frac{l\unicode[STIX]{x1D70B}}{m_{p}}\sinh ^{2}r\bigg)g_{\unicode[STIX]{x1D706}}(r)\sinh r\,dr\end{eqnarray}$$
and, for
$s\neq 1$
, we write
$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{\text{par}}:=B_{u}2^{1-s}\unicode[STIX]{x1D701}(1-s)\int _{0}^{\infty }\frac{m(v)}{v^{(1+s)/2}}\,dv+\tilde{B}_{u}2^{s}\unicode[STIX]{x1D701}(s)\int _{0}^{\infty }\frac{m(v)}{v^{(2-s)/2}}\,dv;\end{eqnarray}$$
for
$s=1$
, we write
$$\begin{eqnarray}\unicode[STIX]{x1D6F4}_{\text{par}}:=\tilde{B}_{u}\int _{0}^{\infty }\frac{m(v)}{v^{1/2}}\log v\,dv,\end{eqnarray}$$
where
$\unicode[STIX]{x1D701}$
is the Riemann zeta function. The left-hand side of (3.7) is absolutely convergent and
$\unicode[STIX]{x1D6F4}_{\text{hyp}}$
is a finite sum.
Proof. This is essentially proved in [Reference Biró1] in the case
$s\neq 1$
and in [Reference Iwaniec9] in the case
$s=1$
(since for
$s=1$
this follows from the classical Selberg trace formula), but it is not stated there exactly in this form, so we explain how it follows from [Reference Biró1] and from [Reference Iwaniec9].
It follows from Lemma 3.2 and condition (3.6) that
$M(z,z)$
is bounded on
$H$
; hence, by (3.5), the left-hand side of (3.7) is absolutely convergent. Let us write
$$\begin{eqnarray}\displaystyle M_{\text{hyp}}(z) & := & \displaystyle \mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4} \\ \unicode[STIX]{x1D6FE}~\text{hyperbolic}}}m\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg),\nonumber\\ \displaystyle M_{\text{ell}}(z) & := & \displaystyle \mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4} \\ \unicode[STIX]{x1D6FE}~\text{elliptic}}}m\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg),\nonumber\\ \displaystyle M_{\text{par}}(z) & := & \displaystyle \mathop{\sum }_{\substack{ \unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4} \\ \unicode[STIX]{x1D6FE}~\text{parabolic}}}m\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg).\nonumber\end{eqnarray}$$
It is clear by Lemma 3.1 and the fact that
$\unicode[STIX]{x1D6E4}$
acts discontinuously on
$H$
(see [Reference Iwaniec9, p. 40]) that there are only finitely many
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
for which there is a
$z\in H$
such that the contribution of
$\unicode[STIX]{x1D6FE}$
to
$M_{\text{hyp}}(z)$
or
$M_{\text{ell}}(z)$
is non-zero. It follows then, on the one hand, that
$M_{\text{par}}(z)$
is also bounded on
$H$
and on the other hand that
$$\begin{eqnarray}\int _{F}M_{\text{hyp}}(z)u(z)\,d\unicode[STIX]{x1D707}_{z}=\mathop{\sum }_{\substack{ [\unicode[STIX]{x1D6FE}] \\ \unicode[STIX]{x1D6FE}~\text{hyperbolic}}}T_{\unicode[STIX]{x1D6FE}},\qquad \int _{F}M_{\text{ell}}(z)u(z)\,d\unicode[STIX]{x1D707}_{z}=\mathop{\sum }_{\substack{ [\unicode[STIX]{x1D6FE}] \\ \unicode[STIX]{x1D6FE}~\text{elliptic}}}T_{\unicode[STIX]{x1D6FE}},\end{eqnarray}$$
where the summation is over the hyperbolic and elliptic conjugacy classes of
$\unicode[STIX]{x1D6E4}$
, respectively, and
$$\begin{eqnarray}T_{\unicode[STIX]{x1D6FE}}:=\mathop{\sum }_{\unicode[STIX]{x1D6FF}\in [\unicode[STIX]{x1D6FE}]}\int _{F}m\bigg(\frac{|z-\unicode[STIX]{x1D6FF}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FF}z}\bigg)u(z)\,d\unicode[STIX]{x1D707}_{z}.\end{eqnarray}$$
Then it follows from [Reference Biró1, (3) and (4)] (and the reasoning there is valid also for the case
$s=1$
) that
$$\begin{eqnarray}\int _{F}M_{\text{hyp}}(z)u(z)\,d\unicode[STIX]{x1D707}_{z}=\unicode[STIX]{x1D6F4}_{\text{hyp}},\qquad \int _{F}M_{\text{ell}}(z)u(z)\,d\unicode[STIX]{x1D707}_{z}=\unicode[STIX]{x1D6F4}_{\text{ell}}.\end{eqnarray}$$
Since we have seen that
$M_{\text{par}}(z)$
is bounded,
$$\begin{eqnarray}\int _{F}M_{\text{par}}(z)u(z)\,d\unicode[STIX]{x1D707}_{z}=\unicode[STIX]{x1D6F4}_{\text{par}}\end{eqnarray}$$
follows from [Reference Biró1, Lemma 3] in the case
$s\neq 1$
and from [Reference Iwaniec9, (10.14) and (10.15)] in the case
$s=1$
(since in that case
$u$
is constant), taking into account that
$g_{m}(0)=0$
by our condition (3.6) (see (2.2)). The lemma is proved.◻
Lemma 3.4. Let
$m$
be a compactly supported continuous function on
$[0,\infty )$
. Assume that
$h_{m}(r)$
(defined in (2.2) and (2.3)) is even, it is holomorphic in the strip
$|\text{Im}\,r|\leqslant {\textstyle \frac{1}{2}}+\unicode[STIX]{x1D716}$
and
$h_{m}(r)=O((1+|r|)^{-2-\unicode[STIX]{x1D716}})$
in this strip for some
$\unicode[STIX]{x1D716}>0$
. Then for
$z\in H$
we have (using definition (3.4)) that
$$\begin{eqnarray}M(z,z)=\mathop{\sum }_{j=0}^{\infty }h_{m}(t_{j})|u_{j}(z)|^{2}+\mathop{\sum }_{a}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }h_{m}(r)\bigg|E_{a}\bigg(z,\frac{1}{2}+\text{i}r\bigg)\bigg|^{2}\,dr\end{eqnarray}$$
and this expression is absolutely and uniformly convergent on compact subsets of
$H$
.
Proof. This follows from [Reference Iwaniec9, Theorem 7.4]. ◻
4 Proof of the theorem
Let
$x$
be a large positive number and let
$1<d<x/\log x$
(say). We will choose later the parameter
$d$
optimally.
Let
$$\begin{eqnarray}k(y)=1\quad \text{for }0\leqslant y\leqslant x,\qquad k(y)=0\quad \text{for }y>x\end{eqnarray}$$
and let
$k^{\ast }$
be a smoothed version of
$k$
, more precisely,
$$\begin{eqnarray}k^{\ast }(y):=\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})k(y\text{e}^{\unicode[STIX]{x1D70F}})\,d\unicode[STIX]{x1D70F},\end{eqnarray}$$
where the function
$\unicode[STIX]{x1D702}$
will be a smooth even function satisfying
$\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})=0$
for
$|\unicode[STIX]{x1D70F}|>d/x$
. We define
$\unicode[STIX]{x1D702}$
precisely below.
But before defining
$\unicode[STIX]{x1D702}$
, let us remark that our goal is to achieve
$$\begin{eqnarray}\int _{0}^{\infty }(k^{\ast }(u)-k(u))u^{-1/2}\,du=0,\end{eqnarray}$$
since we would like to apply Lemma 3.3 for this difference. By (4.2), we have
$$\begin{eqnarray}\int _{0}^{\infty }k^{\ast }(u)u^{-1/2}\,du=\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}/2}\,d\unicode[STIX]{x1D70F}\int _{0}^{\infty }k(v)v^{-1/2}\,dv\end{eqnarray}$$
and so we want to have
$$\begin{eqnarray}\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}/2}\,d\unicode[STIX]{x1D70F}=1.\end{eqnarray}$$
We will take
$$\begin{eqnarray}\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F}):=\frac{x}{d}\unicode[STIX]{x1D702}_{0}\bigg(\frac{x}{d}\unicode[STIX]{x1D70F}\bigg),\end{eqnarray}$$
where the function
$\unicode[STIX]{x1D702}_{0}$
will be a smooth even function satisfying
$\unicode[STIX]{x1D702}_{0}(\unicode[STIX]{x1D70F})=0$
for
$|\unicode[STIX]{x1D70F}|>1$
. Then
$$\begin{eqnarray}\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}/2}\,d\unicode[STIX]{x1D70F}=\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}_{0}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}(d/2x)}\,d\unicode[STIX]{x1D70F},\end{eqnarray}$$
so we need
$$\begin{eqnarray}\int _{-\infty }^{\infty }{\unicode[STIX]{x1D702}_{0}}_{}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}(d/2x)}\,d\unicode[STIX]{x1D70F}=1.\end{eqnarray}$$
We now define
$\unicode[STIX]{x1D702}_{0}$
. First let
$\unicode[STIX]{x1D713}_{0}$
be a given smooth even non-negative function on the real line such that
$$\begin{eqnarray}\unicode[STIX]{x1D713}_{0}(\unicode[STIX]{x1D70F})=0\quad \text{for }|\unicode[STIX]{x1D70F}|>1\end{eqnarray}$$
and
$$\begin{eqnarray}\int _{-\infty }^{\infty }{\unicode[STIX]{x1D713}_{0}}_{}(\unicode[STIX]{x1D70F})\,d\unicode[STIX]{x1D70F}=1.\end{eqnarray}$$
Then
$$\begin{eqnarray}I_{d,x}:=\int _{-\infty }^{\infty }{\unicode[STIX]{x1D713}_{0}}_{}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}(d/2x)}\,d\unicode[STIX]{x1D70F}=1+O\bigg(\bigg(\frac{d}{x}\bigg)^{2}\bigg)\end{eqnarray}$$
with implied absolute constant. If
$x$
is large enough, then clearly
$1/2<I_{d,x}<2$
. Let
$$\begin{eqnarray}\unicode[STIX]{x1D702}_{0}(\unicode[STIX]{x1D70F})=\frac{{\unicode[STIX]{x1D713}_{0}}_{}(\unicode[STIX]{x1D70F})}{I_{d,x}}\end{eqnarray}$$
for real
$\unicode[STIX]{x1D70F}$
; then we have (4.7). Note that
$\unicode[STIX]{x1D702}_{0}$
slightly depends on
$d$
and
$x$
, but we do not denote it. Formulas (4.9), (4.10) and (4.5) define
$\unicode[STIX]{x1D702}$
and, by (4.7), (4.4) and (4.6), we get (4.3).
Note that, by the definitions for any integer
$j\geqslant 0$
, we have that
$$\begin{eqnarray}\int _{-\infty }^{\infty }|\unicode[STIX]{x1D702}^{(j)}(\unicode[STIX]{x1D70F})|\,d\unicode[STIX]{x1D70F}\ll _{j}\bigg(\frac{x}{d}\bigg)^{j}\end{eqnarray}$$
and we also have (by (4.8), (4.9), (4.10) and (4.5)) that
$$\begin{eqnarray}\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\,d\unicode[STIX]{x1D70F}=1+O\bigg(\bigg(\frac{d}{x}\bigg)^{2}\bigg).\end{eqnarray}$$
So, the smoothed version
$k^{\ast }$
of
$k$
is now defined. As is mentioned in Remark 1.2, we will use the decomposition (1.2); we will apply Lemma 3.4 for the first term there and we will apply Lemma 3.3 for the second term. To apply these lemmas we need estimates for the function transforms occurring in those lemmas. We give such estimates in the next three lemmas.
For simplicity, introduce the abbreviations
$q^{\ast }=q_{k^{\ast }}$
,
$g^{\ast }=g_{k^{\ast }}$
and
$h^{\ast }=h_{k^{\ast }}$
(see (2.2) and (2.3)).
Lemma 4.1. For every integer
$j\geqslant 2$
, we have for
$r\geqslant 1$
that
$$\begin{eqnarray}|h^{\ast }(r)|\ll _{j}\frac{d^{3/2}}{x}\bigg(\frac{x}{dr}\bigg)^{j}.\end{eqnarray}$$
We also have for every complex
$r$
that
$$\begin{eqnarray}|h^{\ast }(r)|\ll x^{1/2+|\text{Im}\,r|}\log x.\end{eqnarray}$$
Proof. By (2.2) and (4.1), we have
$$\begin{eqnarray}q_{k}(y)=2\sqrt{x-y}\quad \text{for }0\leqslant y\leqslant x,\qquad q_{k}(y)=0\quad \text{for }y>x.\end{eqnarray}$$
It is easy to see by (2.2) and (4.2) that we have
$$\begin{eqnarray}q^{\ast }(v)=\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\text{e}^{-\unicode[STIX]{x1D70F}/2}q_{k}(v\text{e}^{\unicode[STIX]{x1D70F}})\,d\unicode[STIX]{x1D70F}\end{eqnarray}$$
and, by the substitution
$\text{e}^{\unicode[STIX]{x1D707}}=v\text{e}^{\unicode[STIX]{x1D70F}}$
, we can also write
$$\begin{eqnarray}q^{\ast }(v)=\sqrt{v}\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D707}-\log v)\text{e}^{-\unicode[STIX]{x1D707}/2}q_{k}(\text{e}^{\unicode[STIX]{x1D707}})\,d\unicode[STIX]{x1D707}.\end{eqnarray}$$
Then, for any integer
$j\geqslant 0$
, we have on the one hand that
$$\begin{eqnarray}(q^{\ast }(v))^{(j)}=\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\text{e}^{(j-1/2)\unicode[STIX]{x1D70F}}q_{k}^{(j)}(v\text{e}^{\unicode[STIX]{x1D70F}})\,d\unicode[STIX]{x1D70F}\end{eqnarray}$$
and we have on the other hand that
$$\begin{eqnarray}(q^{\ast }(v))^{(j)}=\sqrt{v}\int _{-\infty }^{\infty }\frac{\mathop{\sum }_{l=0}^{j}c_{l,j}\unicode[STIX]{x1D702}^{(l)}(\unicode[STIX]{x1D707}-\log v)}{v^{j}}\text{e}^{-\unicode[STIX]{x1D707}/2}q_{k}(\text{e}^{\unicode[STIX]{x1D707}})\,d\unicode[STIX]{x1D707}\end{eqnarray}$$
with some constants
$c_{l,j}$
.
It is clear from (4.15) and (4.16) that for
$v\geqslant x\text{e}^{d/x}$
, we have
$$\begin{eqnarray}q^{\ast }(v)=0.\end{eqnarray}$$
It is also clear by the same formulas that for
$0\leqslant v\leqslant x\text{e}^{d/x}$
, we have
$$\begin{eqnarray}0\leqslant q^{\ast }(v)\ll \sqrt{x}.\end{eqnarray}$$
The estimate (4.14) follows at once from (4.19), (4.20), (2.2) and (2.3).
Assume that
$0\leqslant v\leqslant x\text{e}^{-2d/x}$
. Then, for
$\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\neq 0$
, we have
$$\begin{eqnarray}v\text{e}^{\unicode[STIX]{x1D70F}}\leqslant v\text{e}^{d/x}\leqslant v+\frac{x-v}{2},\end{eqnarray}$$
since this latter inequality is easily seen to be equivalent to
$$\begin{eqnarray}2\text{e}^{d/x}-1\leqslant \frac{x}{v},\end{eqnarray}$$
which is true, since
$x/v\geqslant \text{e}^{2d/x}$
. So, for any integer
$j\geqslant 1$
, we have by (4.15) that
$$\begin{eqnarray}|q_{k}^{(j)}(v\text{e}^{\unicode[STIX]{x1D70F}})|\,\ll _{j}\,(x-v\text{e}^{\unicode[STIX]{x1D70F}})^{1/2-j}\,\ll _{j}\,(x-v)^{1/2-j};\end{eqnarray}$$
hence, by (4.17), we get that
$$\begin{eqnarray}|(q^{\ast }(v))^{(j)}|\ll _{j}(x-v)^{1/2-j}\end{eqnarray}$$
for
$0\leqslant v\leqslant x\text{e}^{-2d/x}$
and
$j\geqslant 1$
.
Now let
$x\text{e}^{-2d/x}\leqslant v\leqslant x\text{e}^{d/x}$
. Then we use (4.18). If the integrand here is non-zero, then we must have
$|\unicode[STIX]{x1D707}-\log v|\leqslant d/x$
, so
$$\begin{eqnarray}x\text{e}^{-3d/x}\leqslant \text{e}^{\unicode[STIX]{x1D707}}\leqslant x\text{e}^{2d/x};\end{eqnarray}$$
hence, by (4.15), one has
$$\begin{eqnarray}q_{k}(\text{e}^{\unicode[STIX]{x1D707}})\ll \sqrt{d}\end{eqnarray}$$
and by the upper and lower bounds for
$\text{e}^{\unicode[STIX]{x1D707}}$
and
$v$
, one also has
$$\begin{eqnarray}x\ll \text{e}^{\unicode[STIX]{x1D707}}\ll x,\qquad x\ll v\ll x.\end{eqnarray}$$
Using these estimates, by (4.18) and (4.11), we get for any integer
$j\geqslant 1$
that
$$\begin{eqnarray}|(q^{\ast }(v))^{(j)}|\ll _{j}d^{1/2-j}\end{eqnarray}$$
for
$x\text{e}^{-2d/x}\leqslant v\leqslant x\text{e}^{d/x}$
.
We see by (2.2) for every
$j\geqslant 1$
and real
$a$
that
$$\begin{eqnarray}|(g^{\ast }(a))^{(j)}|\ll _{j}\mathop{\sum }_{l=1}^{j}\bigg|\bigg(q^{\ast }\bigg(\sinh ^{2}\frac{a}{2}\bigg)\bigg)^{(l)}\bigg|\text{e}^{l|a|}.\end{eqnarray}$$
By (2.3), we have by repeated partial integration for every
$j\geqslant 1$
and
$r\geqslant 1$
that
$$\begin{eqnarray}|h^{\ast }(r)|\ll _{j}\frac{1}{r^{j}}\int _{-\infty }^{\infty }|(g^{\ast }(a))^{(j)}|\,da.\end{eqnarray}$$
By (4.19), (4.21), (4.22) and (4.23), we obtain (4.13). The lemma is proved. ◻
Lemma 4.2. For
${\textstyle \frac{1}{100}}\leqslant \text{i}t\leqslant {\textstyle \frac{1}{2}}$
(say), we have that
$$\begin{eqnarray}h^{\ast }(t)=\sqrt{\unicode[STIX]{x1D70B}}\frac{\unicode[STIX]{x1D6E4}(\text{i}t)2^{2\text{i}t+1}}{\unicode[STIX]{x1D6E4}(\frac{3}{2}+\text{i}t)}x^{1/2+\text{i}t}+O\bigg(x\bigg(\frac{d}{x}\bigg)^{2}\bigg)+O(x^{1/2}).\end{eqnarray}$$
Proof. Assume first that
$\text{i}t<{\textstyle \frac{1}{2}}$
. By [Reference Iwaniec9, (1.62
$^{\prime }$
)] (the function
$F_{s}(u)$
is defined by the formulas of [Reference Iwaniec9, p. 26, line 7 and (B.23)]) and by [Reference Gradshteyn and Ryzhik5, p. 995, 9.113], we have that
$$\begin{eqnarray}h^{\ast }(t)=\frac{2}{\text{i}\unicode[STIX]{x1D6E4}(\frac{1}{2}\pm \text{i}t)}\int _{(\unicode[STIX]{x1D70E})}\frac{\unicode[STIX]{x1D6E4}(\frac{1}{2}\pm \text{i}t+S)\unicode[STIX]{x1D6E4}(-S)}{\unicode[STIX]{x1D6E4}(1+S)}\bigg(\int _{0}^{\infty }k^{\ast }(u)u^{S}\,du\bigg)\,dS,\end{eqnarray}$$
where
$\text{i}t-{\textstyle \frac{1}{2}}<\unicode[STIX]{x1D70E}<0$
; so, by (4.1) and (4.2), we get
$$\begin{eqnarray}h^{\ast }(t)=\frac{2}{\text{i}\unicode[STIX]{x1D6E4}(\frac{1}{2}\pm \text{i}t)}\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})\!\int _{(\unicode[STIX]{x1D70E})}\frac{\unicode[STIX]{x1D6E4}(\frac{1}{2}\pm \text{i}t+S)\unicode[STIX]{x1D6E4}(-S)}{\unicode[STIX]{x1D6E4}(1+S)}\frac{(x/\text{e}^{\unicode[STIX]{x1D70F}})^{S+1}}{S+1}\,dS\,d\unicode[STIX]{x1D70F}.\end{eqnarray}$$
Shifting the integration to the left, we get (and observe that it is also true for
$\text{i}t={\textstyle \frac{1}{2}}$
) that
$$\begin{eqnarray}h^{\ast }(t)=\frac{4\unicode[STIX]{x1D70B}}{\unicode[STIX]{x1D6E4}(\frac{1}{2}+\text{i}t)}\frac{\unicode[STIX]{x1D6E4}(2\text{i}t)}{\unicode[STIX]{x1D6E4}(\frac{3}{2}+\text{i}t)}\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})(x/\text{e}^{\unicode[STIX]{x1D70F}})^{1/2+\text{i}t}\,d\unicode[STIX]{x1D70F}+O(x^{1/2})\end{eqnarray}$$
and taking into account the properties of
$\unicode[STIX]{x1D702}$
and the duplication formula for the
$\unicode[STIX]{x1D6E4}$
-function [Reference Iwaniec9, (B.5)], we obtain the lemma.
If
$m$
is a compactly supported continuous function on
$[0,\infty )$
,
$\unicode[STIX]{x1D706}\leqslant 0$
and
$T>0$
, we introduce the notation
$$\begin{eqnarray}M_{m,\unicode[STIX]{x1D706}}(T):=\int _{-\unicode[STIX]{x1D70B}/2}^{\unicode[STIX]{x1D70B}/2}m\bigg(\frac{T}{\cos ^{2}\unicode[STIX]{x1D703}}\bigg)f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})\frac{d\unicode[STIX]{x1D703}}{\cos ^{2}\unicode[STIX]{x1D703}}.\end{eqnarray}$$
Note that this kind of transform appeared in [Reference Biró1, equation (7)], and also in [Reference Huber8, equation (31)].
We obviously have
$$\begin{eqnarray}M_{k^{\ast },\unicode[STIX]{x1D706}}(T)=\int _{-\infty }^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{\unicode[STIX]{x1D70F}})\,d\unicode[STIX]{x1D70F}.\end{eqnarray}$$
◻
Lemma 4.3. (i) For every
$T>0$
, the functions
$$\begin{eqnarray}M_{k,-1/4-t^{2}}(T),\qquad M_{k^{\ast },-1/4-t^{2}}(T)\end{eqnarray}$$
are entire functions of
$t$
.
(ii) Let
$0<\unicode[STIX]{x1D6FF}<{\textstyle \frac{1}{2}}$
be given. There is a constant
$A_{\unicode[STIX]{x1D6FF}}>0$
depending only on
$\unicode[STIX]{x1D6FF}$
such that for every
$t$
satisfying
${\textstyle \frac{1}{4}}+t^{2}\geqslant 0$
or
$|\text{Re}(\text{i}t)|\leqslant {\textstyle \frac{1}{2}}-\unicode[STIX]{x1D6FF}$
one has the following statements with the notation
$\unicode[STIX]{x1D706}=-{\textstyle \frac{1}{4}}-t^{2}$
.
For
$T\geqslant x\text{e}^{d/x}$
, we have that
$$\begin{eqnarray}M_{k^{\ast },\unicode[STIX]{x1D706}}(T)=M_{k,\unicode[STIX]{x1D706}}(T)=0,\end{eqnarray}$$
for
$x\text{e}^{-2d/x}\leqslant T\leqslant x\text{e}^{d/x}$
, we have that
$$\begin{eqnarray}|M_{k^{\ast },\unicode[STIX]{x1D706}}(T)|+|M_{k,\unicode[STIX]{x1D706}}(T)|\ll _{\unicode[STIX]{x1D6FF}}(1+|t|)^{A_{\unicode[STIX]{x1D6FF}}}\bigg(\frac{d}{x}\bigg)^{1/2}\end{eqnarray}$$
and for
$0<T\leqslant x\text{e}^{-2d/x}$
, we have that
$$\begin{eqnarray}M_{k^{\ast },\unicode[STIX]{x1D706}}(T)-M_{k,\unicode[STIX]{x1D706}}(T)\ll _{\unicode[STIX]{x1D6FF}}\frac{(1+|t|)^{A_{\unicode[STIX]{x1D6FF}}}\,d^{2}}{T^{1/2}(x-T)^{3/2}}.\end{eqnarray}$$
Proof. Note first that we can give an explicit formula for
$f_{\unicode[STIX]{x1D706}}$
, namely,
$$\begin{eqnarray}f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})=F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-\frac{\sin ^{2}\unicode[STIX]{x1D703}}{\cos ^{2}\unicode[STIX]{x1D703}}\bigg)\end{eqnarray}$$
for
$\unicode[STIX]{x1D703}\in (-\unicode[STIX]{x1D70B}/2,\unicode[STIX]{x1D70B}/2)$
, where
$\unicode[STIX]{x1D706}=-{\textstyle \frac{1}{4}}-t^{2}$
. This can be proved in the following way. One has
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{1}{\unicode[STIX]{x1D70B}^{1/2}}\int _{-\unicode[STIX]{x1D70B}/2}^{\unicode[STIX]{x1D70B}/2}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-\frac{\sin ^{2}\unicode[STIX]{x1D703}}{\cos ^{2}\unicode[STIX]{x1D703}}\bigg)\cos ^{2s}\unicode[STIX]{x1D703}\frac{d\unicode[STIX]{x1D703}}{\cos ^{2}\unicode[STIX]{x1D703}}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\unicode[STIX]{x1D6E4}(s-1/4+\text{i}t/2)\unicode[STIX]{x1D6E4}(s-1/4-\text{i}t/2)}{\unicode[STIX]{x1D6E4}^{2}(s)}\nonumber\end{eqnarray}$$
for
$\operatorname{Re}s>{\textstyle \frac{1}{2}}$
, as one can see by the substitution
$y=\sin ^{2}\unicode[STIX]{x1D703}/\cos ^{2}\unicode[STIX]{x1D703}$
and by [Reference Gradshteyn and Ryzhik5, p. 807, 7.512.10]. By [Reference Biró1, Lemma 11], it follows that
$$\begin{eqnarray}\int _{0}^{\unicode[STIX]{x1D70B}/2}\bigg(F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-\frac{\sin ^{2}\unicode[STIX]{x1D703}}{\cos ^{2}\unicode[STIX]{x1D703}}\bigg)-f_{\unicode[STIX]{x1D706}}(\unicode[STIX]{x1D703})\bigg)\cos ^{n}\unicode[STIX]{x1D703}\,d\unicode[STIX]{x1D703}=0\end{eqnarray}$$
for every non-negative integer
$n$
, which easily implies (4.28).
We can see part (i) at once.
To show part (ii), note that using (4.24), (4.28) and the substitution
$$\begin{eqnarray}Y=\log \frac{1}{\cos ^{2}\unicode[STIX]{x1D703}},\end{eqnarray}$$
one has for
$T\leqslant x$
that
$$\begin{eqnarray}M_{k,\unicode[STIX]{x1D706}}(T)=\int _{0}^{\log x-\log T}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};1-\text{e}^{Y}\bigg)\frac{\text{e}^{Y}\,dY}{\sqrt{\text{e}^{Y}-1}}\end{eqnarray}$$
and, for
$T>x$
, we have
$M_{k,\unicode[STIX]{x1D706}}(T)=0$
. Then (4.26) is obvious by (4.25). We also have by (4.29) and Lemma 5.1 for
$T\leqslant x$
that
$$\begin{eqnarray}M_{k,\unicode[STIX]{x1D706}}(T)\ll _{\unicode[STIX]{x1D6FF}}(1+|t|)^{A_{\unicode[STIX]{x1D6FF}}}\bigg(\frac{x}{T}-1\bigg)^{1/2}\end{eqnarray}$$
and then (using also (4.25)) (4.27) follows.
By (4.25), (4.12) and since
$\unicode[STIX]{x1D702}$
is even, we have that
$$\begin{eqnarray}M_{k^{\ast },\unicode[STIX]{x1D706}}(T)-M_{k,\unicode[STIX]{x1D706}}(T)\end{eqnarray}$$
equals
$$\begin{eqnarray}\int _{0}^{\infty }\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})(M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{\unicode[STIX]{x1D70F}})+M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{-\unicode[STIX]{x1D70F}})-2M_{k,\unicode[STIX]{x1D706}}(T))\,d\unicode[STIX]{x1D70F}+O\bigg(\bigg(\frac{d}{x}\bigg)^{2}\bigg)M_{k,\unicode[STIX]{x1D706}}(T).\end{eqnarray}$$
Assuming that
$T\text{e}^{\unicode[STIX]{x1D70F}}\leqslant x$
, by (4.29) we have that
$$\begin{eqnarray}\displaystyle & & \displaystyle M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{\unicode[STIX]{x1D70F}})+M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{-\unicode[STIX]{x1D70F}})-2M_{k,\unicode[STIX]{x1D706}}(T)\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{\log (x/T)}^{\log (x/T)+\unicode[STIX]{x1D70F}}\left(\frac{F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};1-\text{e}^{Y}\bigg)\text{e}^{Y}}{\sqrt{\text{e}^{Y}-1}}\right.\nonumber\\ \displaystyle & & \displaystyle \left.\qquad -\,\frac{F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};1-\text{e}^{Y-\unicode[STIX]{x1D70F}}\bigg)\text{e}^{Y-\unicode[STIX]{x1D70F}}}{\sqrt{\text{e}^{Y-\unicode[STIX]{x1D70F}}-1}}\right)\,dY.\nonumber\end{eqnarray}$$
For
$0<T\leqslant x\text{e}^{-2d/x}$
and
$|\unicode[STIX]{x1D70F}|\leqslant d/x$
, we then have by the mean-value theorem and by Lemma 5.1 that
$$\begin{eqnarray}\displaystyle & & \displaystyle M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{\unicode[STIX]{x1D70F}})+M_{k,\unicode[STIX]{x1D706}}(T\text{e}^{-\unicode[STIX]{x1D70F}})-2M_{k,\unicode[STIX]{x1D706}}(T)\nonumber\\ \displaystyle & & \displaystyle \quad \ll _{\unicode[STIX]{x1D6FF}}(1+|t|)^{A_{\unicode[STIX]{x1D6FF}}}\unicode[STIX]{x1D70F}^{2}\bigg(\frac{x}{T}\bigg)^{1/2}\bigg(1+\frac{1}{\log (x/T)}\bigg)^{3/2}.\nonumber\end{eqnarray}$$
So, by (4.30), (4.11) with
$j=0$
, (4.31) and (4.32), using that
$\unicode[STIX]{x1D702}(\unicode[STIX]{x1D70F})=0$
for
$|\unicode[STIX]{x1D70F}|>d/x$
, we get for
$0<T\leqslant x\text{e}^{-2d/x}$
that
$$\begin{eqnarray}M_{k^{\ast },\unicode[STIX]{x1D706}}(T)-M_{k,\unicode[STIX]{x1D706}}(T)\ll _{\unicode[STIX]{x1D6FF}}(1+|t|)^{A_{\unicode[STIX]{x1D6FF}}}\bigg(\frac{d}{x}\bigg)^{2}\bigg(\frac{x}{T}\bigg)^{1/2}\bigg(\frac{x}{x-T}\bigg)^{3/2}.\end{eqnarray}$$
The lemma is proved. ◻
Remark 4.4. We now compare (4.28) to the formulas (86) and (87) in [Reference Fay4]. Indeed, by the notation used there, we see that
$$\begin{eqnarray}N_{1/2+\text{i}t,0}^{0}\bigg(\unicode[STIX]{x1D703}+\frac{\unicode[STIX]{x1D70B}}{2}\bigg)\quad \text{and}\quad N_{1/2-\text{i}t,0}^{0}\bigg(\unicode[STIX]{x1D703}+\frac{\unicode[STIX]{x1D70B}}{2}\bigg)\end{eqnarray}$$
are solutions of our differential equation (2.1) for
$-\unicode[STIX]{x1D70B}/2<\unicode[STIX]{x1D703}<0$
. Using [Reference Fay4, (87)] and the quadratic transformation [Reference Gradshteyn and Ryzhik5, p. 999, 9.134.1], one sees that
$$\begin{eqnarray}N_{1/2+\text{i}t,0}^{0}\bigg(\unicode[STIX]{x1D703}+\frac{\unicode[STIX]{x1D70B}}{2}\bigg)=\bigg(\frac{\cos ^{2}\unicode[STIX]{x1D703}}{\sin ^{2}\unicode[STIX]{x1D703}}\bigg)^{1/4+\text{i}t/2}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{3}{4}+\frac{\text{i}t}{2}\\ 1+\text{i}t\end{array};-\frac{\cos ^{2}\unicode[STIX]{x1D703}}{\sin ^{2}\unicode[STIX]{x1D703}}\bigg)\end{eqnarray}$$
for
$-\unicode[STIX]{x1D70B}/2<\unicode[STIX]{x1D703}<0$
. Using the same relation with
$-t$
in place of
$t$
, we see by [Reference Gradshteyn and Ryzhik5, p. 999, 9.132.2], that the right-hand side of (4.28) is a linear combination of
$N_{1/2+\text{i}t,0}^{0}(\unicode[STIX]{x1D703}+\unicode[STIX]{x1D70B}/2)$
and
$N_{1/2-\text{i}t,0}^{0}(\unicode[STIX]{x1D703}+\unicode[STIX]{x1D70B}/2)$
on the interval
$-\unicode[STIX]{x1D70B}/2<\unicode[STIX]{x1D703}<0$
; hence, it is also a solution of (2.1). Since it is an even function, it gives a solution of (2.1) on the whole interval
$-\unicode[STIX]{x1D70B}/2<\unicode[STIX]{x1D703}<\unicode[STIX]{x1D70B}/2$
.
Continuing the proof of Theorem 1.1, let
$$\begin{eqnarray}m_{1}(v)=k^{\ast }(v),\qquad m_{2}(v)=k(v)-k^{\ast }(v).\end{eqnarray}$$
For
$m_{1}(v)$
, we will apply Lemma 3.4 and, for
$m_{2}(v)$
, we will apply Lemma 3.3.
We can see e.g. by [Reference Iwaniec9, (1.62
$^{\prime }$
)] (the function
$F_{s}(u)$
is defined by the formulas on p. 26, line 7, and [Reference Iwaniec9, (B.23)]) and by [Reference Biró2, Lemma 6.2] that the conditions of Lemma 3.4 are satisfied by writing
$m_{1}$
in place of
$m$
; hence, for
$z\in H$
, we have that
$$\begin{eqnarray}\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}m_{1}\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg)\end{eqnarray}$$
equals
$$\begin{eqnarray}\mathop{\sum }_{j=0}^{\infty }h_{m_{1}}(t_{j})|u_{j}(z)|^{2}+\mathop{\sum }_{a}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }h_{m_{1}}(r)\bigg|E_{a}\bigg(z,\frac{1}{2}+\text{i}r\bigg)\bigg|^{2}\,dr.\end{eqnarray}$$
Then, applying Lemma 4.1 (we apply (4.13) with
$j=2$
for
$1\leqslant r<x/d$
, we apply (4.13) with
$j=3$
for
$r\geqslant x/d$
and finally we apply (4.14) for
$|\text{Re}\,r|<1,|\text{Im}\,r|<1/100$
), Lemma 4.2 and [Reference Iwaniec9, Proposition 7.2], for every
$z\in H$
satisfying
$f(z)\neq 0$
and for every
$\unicode[STIX]{x1D716}>0$
, we get that (4.34) equals
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j,\text{i}t_{j}>0}\sqrt{\unicode[STIX]{x1D70B}}\frac{\unicode[STIX]{x1D6E4}(\text{i}t_{j})2^{2\text{i}t_{j}+1}}{\unicode[STIX]{x1D6E4}(\frac{3}{2}+\text{i}t_{j})}x^{1/2+\text{i}t_{j}}|u_{j}(z)|^{2}\nonumber\\ \displaystyle & & \displaystyle \quad +\,O_{f,\unicode[STIX]{x1D716}}\bigg(x^{\unicode[STIX]{x1D716}}\bigg(\frac{x}{\sqrt{d}}+x^{1/2+1/100}+\frac{d^{2}}{x}\bigg)\bigg),\end{eqnarray}$$
where we took into account that
$f$
is compactly supported on
$F$
.
Let
$f(z)$
be as in the theorem and consider the integral
$$\begin{eqnarray}\int _{F}f(z)\bigg(\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}m_{2}\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg)\bigg)\,d\unicode[STIX]{x1D707}_{z}.\end{eqnarray}$$
By (4.3) and Lemma 3.2, we see that the function in the bracket here is bounded. Then it follows from the spectral theorem [Reference Iwaniec9, Theorems 4.7 and 7.3] that (4.36) equals
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=0}^{\infty }(f,u_{j})\int _{F}u_{j}(z)\bigg(\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}m_{2}\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg)\bigg)\,d\unicode[STIX]{x1D707}_{z}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\mathop{\sum }_{a}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)\int _{F}E_{a}\bigg(z,\frac{1}{2}+\text{i}r\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\bigg(\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}m_{2}\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg)\bigg)\,d\unicode[STIX]{x1D707}_{z}\,dr.\end{eqnarray}$$
By (4.3), we see that the conditions of Lemma 3.3 are satisfied writing
$m_{2}$
in place of
$m$
and writing
$u=u_{j}$
or
$u=E_{a}(\ast ,{\textstyle \frac{1}{2}}+\text{i}r)$
. By Lemmas 6.3 and 5.2, (4.2) and (4.11) with
$j=0$
, using also [Reference Iwaniec9, (6.28)] and a convexity bound for the Riemann zeta function, we get that after applying Lemma 3.3, the contribution of
$\unicode[STIX]{x1D6F4}_{\text{ell}}$
and
$\unicode[STIX]{x1D6F4}_{\text{par}}$
to (4.37) is
$O_{f,\unicode[STIX]{x1D716}}(x^{1/2+\unicode[STIX]{x1D716}})$
for every
$\unicode[STIX]{x1D716}>0$
.
Therefore, for a hyperbolic
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
, introducing the notation
$$\begin{eqnarray}T(\unicode[STIX]{x1D6FE})=\frac{N(\unicode[STIX]{x1D6FE})+N(\unicode[STIX]{x1D6FE})^{-1}-2}{4}\end{eqnarray}$$
and recalling (4.24), we get that (4.36) equals
$$\begin{eqnarray}O_{f,\unicode[STIX]{x1D716}}(x^{1/2+\unicode[STIX]{x1D716}})+\mathop{\sum }_{\substack{ [\unicode[STIX]{x1D6FE}] \\ \unicode[STIX]{x1D6FE}~\text{hyperbolic}}}(\unicode[STIX]{x1D6F4}_{1}(\unicode[STIX]{x1D6FE})+\unicode[STIX]{x1D6F4}_{2}(\unicode[STIX]{x1D6FE}))\end{eqnarray}$$
with the notation
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{1}(\unicode[STIX]{x1D6FE}) & := & \displaystyle \mathop{\sum }_{j=0}^{\infty }(f,u_{j})\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}u_{j}^{Y_{\unicode[STIX]{x1D6E4}}}\,dS\bigg)M_{m_{2},-1/4-t_{j}^{2}}(T(\unicode[STIX]{x1D6FE}))\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{a}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}E_{a}^{Y_{\unicode[STIX]{x1D6E4}}}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\,dS\bigg)M_{m_{2},-1/4-r^{2}}(T(\unicode[STIX]{x1D6FE}))\,dr,\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{2}(\unicode[STIX]{x1D6FE}) & := & \displaystyle \mathop{\sum }_{j=0}^{\infty }(f,u_{j})\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}(u_{j}-u_{j}^{Y_{\unicode[STIX]{x1D6E4}}})\,dS\bigg)M_{m_{2},-1/4-t_{j}^{2}}(T(\unicode[STIX]{x1D6FE}))\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{a}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}D_{a}^{Y_{\unicode[STIX]{x1D6E4}}}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\,dS\bigg)M_{m_{2},-1/4-r^{2}}(T(\unicode[STIX]{x1D6FE}))\,dr,\nonumber\end{eqnarray}$$
where for simplicity we wrote
$$\begin{eqnarray}D_{a}^{Y_{\unicode[STIX]{x1D6E4}}}(\ast ,{\textstyle \frac{1}{2}}+\text{i}r):=E_{a}(\ast ,{\textstyle \frac{1}{2}}+\text{i}r)-E_{a}^{Y_{\unicode[STIX]{x1D6E4}}}(\ast ,{\textstyle \frac{1}{2}}+\text{i}r).\end{eqnarray}$$
Observe that for any
$a\in A$
and any
$y>0$
, we have by [Reference Iwaniec9, (6.22) and (6.27)] that
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{c}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{c}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\bigg(\unicode[STIX]{x1D6FF}_{ca}y^{1/2+\text{i}r}+\unicode[STIX]{x1D719}_{c,a}\bigg(\frac{1}{2}+\text{i}r\bigg)y^{1/2-\text{i}r}\bigg)M_{m_{2},-1/4-r^{2}}(T(\unicode[STIX]{x1D6FE}))\,dr\nonumber\end{eqnarray}$$
equals
$$\begin{eqnarray}\frac{1}{2\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}-\text{i}r\bigg)\bigg)y^{1/2-\text{i}r}M_{m_{2},-1/4-r^{2}}(T(\unicode[STIX]{x1D6FE}))\,dr.\end{eqnarray}$$
We then see using the notation of Lemma 6.2 (taking into account also that
$E_{a}(\ast ,{\textstyle \frac{1}{2}}-\text{i}r)$
and
$E_{a}(\ast ,{\textstyle \frac{1}{2}}+\text{i}r)$
are conjugates of each other for real
$r$
) that
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{2}(\unicode[STIX]{x1D6FE}) & = & \displaystyle \mathop{\sum }_{j=0}^{\infty }(f,u_{j})\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}(u_{j}-u_{j}^{Y_{\unicode[STIX]{x1D6E4}}})\,dS\bigg)M_{m_{2},-1/4-t_{j}^{2}}(T(\unicode[STIX]{x1D6FE}))\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{a}\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{1/2}\bigg(\int _{F}f(z)E_{a}(z,s)\,d\unicode[STIX]{x1D707}_{z}\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}A_{a}(\ast ,s)\,dS\bigg)M_{m_{2},s(s-1)}(T(\unicode[STIX]{x1D6FE}))\,ds.\nonumber\end{eqnarray}$$
Since
$M_{m_{2},s(s-1)}(T(\unicode[STIX]{x1D6FE}))$
is analytic in
$s$
by Lemma 4.3, so shifting the line of integration to the right, using Lemma 6.4 to see that the residues cancel out and using also Lemma 6.3(iii) and Lemma 4.3, we get that
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6F4}_{2}(\unicode[STIX]{x1D6FE}) & = & \displaystyle (f,u_{0})\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}(u_{0}-u_{0}^{Y_{\unicode[STIX]{x1D6E4}}})\,dS\bigg)M_{m_{2},0}(T(\unicode[STIX]{x1D6FE}))\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{a}\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{(1-\unicode[STIX]{x1D6FF})}\bigg(\int _{F}f(z)E_{a}(z,s)\,d\unicode[STIX]{x1D707}_{z}\bigg)\nonumber\\ \displaystyle & & \displaystyle \times \,\bigg(\int _{C_{\unicode[STIX]{x1D6FE}}}A_{a}(\ast ,s)\,dS\bigg)M_{m_{2},s(s-1)}(T(\unicode[STIX]{x1D6FE}))\,ds,\end{eqnarray}$$
where
$\unicode[STIX]{x1D6FF}>0$
is a small number chosen in such a way that
$1-\unicode[STIX]{x1D6FF}>S_{l}$
for every
$l\in L$
satisfying
$S_{l}<1$
.
Choosing
$\unicode[STIX]{x1D6FF}$
small enough in terms of
$\unicode[STIX]{x1D716}$
, applying Lemmas 4.3, 6.2 and 6.3 and using (4.38), (4.39) and (4.40), we get that (4.36) equals
$$\begin{eqnarray}\displaystyle \begin{array}{@{}l@{}}\displaystyle O_{f,\unicode[STIX]{x1D716}}(x^{1/2+\unicode[STIX]{x1D716}})\\ \displaystyle \quad +\,O_{f,\unicode[STIX]{x1D716}}\biggl(x^{\unicode[STIX]{x1D716}}\mathop{\sum }_{\substack{ [\unicode[STIX]{x1D6FE}] \\ \unicode[STIX]{x1D6FE}~\text{hyperbolic},\,T(\unicode[STIX]{x1D6FE})\leqslant x\text{e}^{-2d/x}}}\frac{d^{2}\log N(\unicode[STIX]{x1D6FE})}{T(\unicode[STIX]{x1D6FE})^{1/2}(x-T(\unicode[STIX]{x1D6FE}))^{3/2}}\biggr)\\ \displaystyle \quad +\,O_{f,\unicode[STIX]{x1D716}}\biggl(x^{\unicode[STIX]{x1D716}}\bigg(\frac{d}{x}\bigg)^{1/2}\mathop{\sum }_{\substack{ [\unicode[STIX]{x1D6FE}] \\ \unicode[STIX]{x1D6FE}~\text{hyperbolic},\,x\text{e}^{-2d/x}\leqslant T(\unicode[STIX]{x1D6FE})\leqslant x\text{e}^{d/x}}}\log N(\unicode[STIX]{x1D6FE})\biggr).\end{array} & & \displaystyle \nonumber\end{eqnarray}$$
From the prime geodesic theorem [Reference Iwaniec9, Theorem 10.5], we then get, assuming that
$$\begin{eqnarray}d\geqslant x^{3/4},\end{eqnarray}$$
that (4.36) equals
$$\begin{eqnarray}O_{f,\unicode[STIX]{x1D716}}\bigg((x^{\unicode[STIX]{x1D716}})\bigg(x^{1/2}+\frac{d^{2}}{x}+\frac{d^{3/2}}{\sqrt{x}}\bigg)\bigg).\end{eqnarray}$$
Then, by (4.33), (4.34), (4.36), (4.41) and (4.35), we get for
$x^{3/4}\leqslant d\leqslant x/\log x$
that
$$\begin{eqnarray}\int _{F}f(z)\bigg(\mathop{\sum }_{\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}}k\bigg(\frac{|z-\unicode[STIX]{x1D6FE}z|^{2}}{4\operatorname{Im}z\operatorname{Im}\unicode[STIX]{x1D6FE}z}\bigg)\bigg)\,d\unicode[STIX]{x1D707}_{z}\end{eqnarray}$$
equals
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{F}f(z)\bigg(\mathop{\sum }_{j,\text{i}t_{j}>0}\sqrt{\unicode[STIX]{x1D70B}}\frac{\unicode[STIX]{x1D6E4}(\text{i}t_{j})2^{2\text{i}t_{j}+1}}{\unicode[STIX]{x1D6E4}(\frac{3}{2}+\text{i}t_{j})}x^{1/2+\text{i}t_{j}}|u_{j}(z)|^{2}\bigg)\,d\unicode[STIX]{x1D707}_{z}\nonumber\\ \displaystyle & & \displaystyle \quad +\,O_{f,\unicode[STIX]{x1D716}}\bigg((x^{\unicode[STIX]{x1D716}})\bigg(\frac{x}{\sqrt{d}}+\frac{d^{3/2}}{\sqrt{x}}\bigg)\bigg).\nonumber\end{eqnarray}$$
Choosing
$d=x^{3/4}$
and
$x=(X-2)/4$
, we get the theorem.
5 Lemmas on special functions
Lemma 5.1. Let
$0<\unicode[STIX]{x1D6FF}<{\textstyle \frac{1}{2}}$
be given. There is a constant
$A_{\unicode[STIX]{x1D6FF}}>0$
depending only on
$\unicode[STIX]{x1D6FF}$
such that for every
$t$
satisfying
${\textstyle \frac{1}{4}}+t^{2}\geqslant 0$
or
$|\text{Re}(\text{i}t)|\leqslant {\textstyle \frac{1}{2}}-\unicode[STIX]{x1D6FF}$
and for every
$X\geqslant 0$
, one has that
$$\begin{eqnarray}\displaystyle & & \displaystyle \bigg|F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-X\bigg)\bigg|+\bigg|(1+X)\frac{d}{dX}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-X\bigg)\bigg|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant A_{\unicode[STIX]{x1D6FF}}(1+|t|)^{A_{\unicode[STIX]{x1D6FF}}}.\nonumber\end{eqnarray}$$
Proof. Note that
$$\begin{eqnarray}\frac{d}{dX}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-X\bigg)=-2\bigg(\frac{1}{16}+\frac{t^{2}}{4}\bigg)F\bigg(\begin{array}{@{}c@{}}\frac{5}{4}+\frac{\text{i}t}{2},\frac{5}{4}-\frac{\text{i}t}{2}\\ \frac{3}{2}\end{array};-X\bigg)\end{eqnarray}$$
and so
$$\begin{eqnarray}(1+X)\frac{d}{dX}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{1}{2}\end{array};-X\bigg)=-2\bigg(\frac{1}{16}+\frac{t^{2}}{4}\bigg)F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ \frac{3}{2}\end{array};-X\bigg),\end{eqnarray}$$
where we used the third line of [Reference Gradshteyn and Ryzhik5, p. 998, 9.131.1]. Then it is trivial by [Reference Gradshteyn and Ryzhik5, p. 995, 9.111], that the statement of the lemma is true for
$|t|<1/10$
(say). The statement of the lemma is also trivial for every
$t$
and for
$|X|<1/10(1+|t|)^{2}$
(say) by estimating trivially the series definition of the hypergeometric function.
For
$j=0,1$
and
$X>0$
, one has that
$$\begin{eqnarray}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+j+\frac{\text{i}t}{2},\frac{1}{4}+j-\frac{\text{i}t}{2}\\ \frac{1}{2}+j\end{array};-X\bigg)\end{eqnarray}$$
equals the sum of
$$\begin{eqnarray}\displaystyle & & \displaystyle \frac{\unicode[STIX]{x1D6E4}(\frac{1}{2}+j)\unicode[STIX]{x1D6E4}(\text{i}t)\unicode[STIX]{x1D6E4}(1-\text{i}t)}{\unicode[STIX]{x1D6E4}(\frac{1}{4}\pm \text{i}t/2)\unicode[STIX]{x1D6E4}(\frac{3}{4}-\text{i}t/2)\unicode[STIX]{x1D6E4}(\frac{1}{4}+j+\text{i}t/2)}\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\int _{0}^{1}y^{-1/4-\text{i}t/2}(1-y)^{-3/4-\text{i}t/2}(X+y)^{-1/4-j+\text{i}t/2}\,dy\nonumber\end{eqnarray}$$
and the same expression writing
$-t$
in place of
$t$
; this follows from [Reference Gradshteyn and Ryzhik5, p. 999, 9.132.2] and [Reference Gradshteyn and Ryzhik5, p. 995, 9.111]. Then the statement follows for the case
$|t|\geqslant 1/10$
,
$|X|\geqslant 1/10(1+|t|)^{2}$
. The lemma is proved.◻
Lemma 5.2. Let
$x>0$
and let the function
$k$
be defined by (4.1). Let
$0<a<1$
. There is an absolute constant
$A_{0}>0$
such that for every
$t$
satisfying
${\textstyle \frac{1}{4}}+t^{2}\geqslant 0$
, one has, writing
$\unicode[STIX]{x1D706}=-{\textstyle \frac{1}{4}}-t^{2}$
, that
$$\begin{eqnarray}\bigg|\int _{0}^{\infty }k(a\sinh ^{2}r)g_{\unicode[STIX]{x1D706}}(r)\sinh r\,dr\bigg|\leqslant A_{0}(1+|t|)^{A_{0}}\bigg(1+\frac{x}{a}\bigg)^{1/2}.\end{eqnarray}$$
Proof. Note first that we can give an explicit formula for
$g_{\unicode[STIX]{x1D706}}$
, namely,
$$\begin{eqnarray}g_{\unicode[STIX]{x1D706}}(r)=F\bigg(\begin{array}{@{}c@{}}\frac{3}{4}+\frac{\text{i}t}{2},\frac{3}{4}-\frac{\text{i}t}{2}\\ 1\end{array};-\text{sinh}^{2}r\bigg)\cosh r\end{eqnarray}$$
for
$r\geqslant 0$
, where
$\unicode[STIX]{x1D706}=-{\textstyle \frac{1}{4}}-t^{2}$
. This can be proved in the following way. One has
$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{0}^{\infty }F\bigg(\begin{array}{@{}c@{}}\frac{3}{4}+\frac{\text{i}t}{2},\frac{3}{4}-\frac{\text{i}t}{2}\\ 1\end{array};-\text{sinh}^{2}r\bigg)\cosh r\sinh ^{1-2s}r\,dr\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{\unicode[STIX]{x1D6E4}(s-\frac{1}{4}\pm \text{i}t/2)\unicode[STIX]{x1D6E4}(1-s)}{2\unicode[STIX]{x1D6E4}(\frac{3}{4}\pm \text{i}t/2)\unicode[STIX]{x1D6E4}(s)}\nonumber\end{eqnarray}$$
for
${\textstyle \frac{1}{2}}<\operatorname{Re}s<1$
, as one can see by the substitution
$y=\sinh ^{2}r$
and by [Reference Gradshteyn and Ryzhik5, p. 806, 7.511]. By [Reference Biró1, Lemma 11], it follows that
$$\begin{eqnarray}\int _{0}^{\infty }\bigg(g_{\unicode[STIX]{x1D706}}(r)-F\bigg(\begin{array}{@{}c@{}}\frac{3}{4}+\frac{\text{i}t}{2},\frac{3}{4}-\frac{\text{i}t}{2}\\ 1\end{array};-\text{sinh}^{2}r\bigg)\cosh r\bigg)\sinh ^{1-2s}r\,dr=0\end{eqnarray}$$
for every
${\textstyle \frac{1}{2}}<\operatorname{Re}s<1$
, which easily implies (5.1).
Then, by the substitution
$y=\sinh ^{2}r$
and by the particular shape of
$k$
, we see that
$$\begin{eqnarray}\int _{0}^{\infty }k(a\sinh ^{2}r)g_{\unicode[STIX]{x1D706}}(r)\sinh r\,dr\end{eqnarray}$$
equals
$$\begin{eqnarray}\frac{1}{2}\int _{0}^{x/a}F\bigg(\begin{array}{@{}c@{}}\frac{3}{4}+\frac{\text{i}t}{2},\frac{3}{4}-\frac{\text{i}t}{2}\\ 1\end{array};-y\bigg)\,dy.\end{eqnarray}$$
Since the integrand here equals
$$\begin{eqnarray}(1+y)^{-1/2}\frac{1}{\unicode[STIX]{x1D70B}}\int _{0}^{1}q^{-1/2}(1-q)^{-1/2}F\bigg(\begin{array}{@{}c@{}}\frac{1}{4}+\frac{\text{i}t}{2},\frac{1}{4}-\frac{\text{i}t}{2}\\ 1/2\end{array};-qy\bigg)\,dq\end{eqnarray}$$
by [Reference Gradshteyn and Ryzhik5, p. 807, 7.512.11], and the third line of [Reference Gradshteyn and Ryzhik5, p. 998, 9.131.1], so using Lemma 5.1 the present lemma is proved. ◻
6 Lemmas on automorphic functions
Lemma 6.1. There is a constant
$A_{\unicode[STIX]{x1D6E4}}>0$
depending only on
$\unicode[STIX]{x1D6E4}$
such that if
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
is hyperbolic and
$z\in H$
is a point on the non-euclidean line connecting the fixed points of
$\unicode[STIX]{x1D6FE}$
, then, for every
$a\in A$
, one has
$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FF}z\notin P(A_{\unicode[STIX]{x1D6E4}}\sqrt{N(\unicode[STIX]{x1D6FE})})\end{eqnarray}$$
for every
$\unicode[STIX]{x1D6FF}\in \unicode[STIX]{x1D6E4}$
.
Proof. By the conditions on
$\unicode[STIX]{x1D6FE}$
and
$z$
, we have (see [Reference Iwaniec9, p. 19]) that
$$\begin{eqnarray}\unicode[STIX]{x1D70C}(z,\unicode[STIX]{x1D6FE}z)=\log N(\unicode[STIX]{x1D6FE}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D70C}$
is the distance function on
$H$
defined in [Reference Iwaniec9, (1.2)]. Then, by [Reference Iwaniec9, (1.3)], we have that
$$\begin{eqnarray}u(z,\unicode[STIX]{x1D6FE}z)=\frac{N(\unicode[STIX]{x1D6FE})+N(\unicode[STIX]{x1D6FE})^{-1}-2}{4}\end{eqnarray}$$
and so for every
$\unicode[STIX]{x1D6FF}\in \unicode[STIX]{x1D6E4}$
one has
$$\begin{eqnarray}u(\unicode[STIX]{x1D6FF}z,(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FF}^{-1})\unicode[STIX]{x1D6FF}z)=\frac{N(\unicode[STIX]{x1D6FE})+N(\unicode[STIX]{x1D6FE})^{-1}-2}{4}.\end{eqnarray}$$
Now
$\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FF}^{-1}$
is hyperbolic, so it is not an element of
$\unicode[STIX]{x1D6E4}_{a}$
. Assume that
$A_{\unicode[STIX]{x1D6E4}}$
is large enough and
$\unicode[STIX]{x1D70E}_{a}^{-1}\unicode[STIX]{x1D6FF}z\in P(A_{\unicode[STIX]{x1D6E4}}\sqrt{N(\unicode[STIX]{x1D6FE})})$
; then by Lemma 3.1 we have
$$\begin{eqnarray}u(\unicode[STIX]{x1D6FF}z,(\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D6FF}^{-1})\unicode[STIX]{x1D6FF}z)\geqslant D_{\unicode[STIX]{x1D6E4}}A_{\unicode[STIX]{x1D6E4}}^{2}N(\unicode[STIX]{x1D6FE}).\end{eqnarray}$$
If
$A_{\unicode[STIX]{x1D6E4}}$
is large enough, then this contradicts (6.1); the lemma is proved.◻
Lemma 6.2. For
$a\in A$
and any complex
$s$
, write
$$\begin{eqnarray}\displaystyle \begin{array}{@{}rcll@{}}A_{a}(z,s)\, & :=\, & (\operatorname{Im}\unicode[STIX]{x1D70E}_{a}^{-1}z)^{1-s} & \text{for }z\in F_{a}(Y_{\unicode[STIX]{x1D6E4}}),\\ A_{a}(z,s)\, & :=\, & 0 & \text{for }z\in F\setminus F_{a}(Y_{\unicode[STIX]{x1D6E4}})\end{array} & & \displaystyle \nonumber\end{eqnarray}$$
and finally let
$A_{a}(\unicode[STIX]{x1D6FE}z,s)=A_{a}(z,s)$
for
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
and
$z\in F$
. Let
$\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$
be hyperbolic; then, for
$0\leqslant \unicode[STIX]{x1D70E}:=\operatorname{Re}s\leqslant 1$
, one has (using the notation of Lemma 3.3) that
$$\begin{eqnarray}\int _{C_{\unicode[STIX]{x1D6FE}}}\bigg(1+\mathop{\sum }_{a\in A}|A_{a}(\ast ,s)|\bigg)\,dS\ll (N(\unicode[STIX]{x1D6FE}))^{(1-\unicode[STIX]{x1D70E})/2}\log N(\unicode[STIX]{x1D6FE}).\end{eqnarray}$$
Proof. It is well known that
$\int _{C_{\unicode[STIX]{x1D6FE}}}1\,dS\leqslant \log N(\unicode[STIX]{x1D6FE})$
, so it is enough to show that
$$\begin{eqnarray}|A_{a}(z,s)|\ll (N(\unicode[STIX]{x1D6FE}))^{(1-\unicode[STIX]{x1D70E})/2}\end{eqnarray}$$
for every
$a\in A$
and every
$z\in H$
lying on the non-euclidean line connecting the fixed points of
$\unicode[STIX]{x1D6FE}$
; this follows from Lemma 6.1. The lemma is proved.◻
Lemma 6.3. Let the function
$f$
be as in Theorem 1.1.
$\text{(i)}$
For integers
$j\geqslant 0$
, one has that
$$\begin{eqnarray}\sup _{z\in F}|u_{j}^{Y_{\unicode[STIX]{x1D6E4}}}(z)|\ll (1+|t_{j}|)^{C}\end{eqnarray}$$
and, for any
$a\in A$
and
$R>0$
, one has that
$$\begin{eqnarray}\sup _{z\in F}\int _{-R}^{R}\bigg|E_{a}^{Y_{\unicode[STIX]{x1D6E4}}}\bigg(z,\frac{1}{2}+\text{i}r\bigg)\bigg|^{2}\,dr\ll (1+R)^{C}\end{eqnarray}$$
with some absolute constant
$C$
.
$\text{(ii)}$
For every positive integer
$K$
, one has for
$j\geqslant 0$
that
$$\begin{eqnarray}|(f,u_{j})|\ll _{f,K}(1+|t_{j}|)^{-K}.\end{eqnarray}$$
$\text{(iii)}$
For any
$a\in A$
, the function
$$\begin{eqnarray}\int _{F}f(z)E_{a}(z,s)\,d\unicode[STIX]{x1D707}_{z}\end{eqnarray}$$
is meromorphic for
${\textstyle \frac{1}{2}}\leqslant \operatorname{Re}s\leqslant 2$
having poles only at the points
$\{S_{l}:l\in L\}$
and, for every positive integer
$K$
and every
${\textstyle \frac{1}{2}}\leqslant \unicode[STIX]{x1D70E}\leqslant 2$
, one has
$$\begin{eqnarray}\int _{-\infty }^{-1}\bigg|\int _{F}f(z)E_{a}(z,\unicode[STIX]{x1D70E}+\text{i}r)\,d\unicode[STIX]{x1D707}_{z}\bigg|^{2}r^{2K}\,dr\ll _{f,K}1\end{eqnarray}$$
and
$$\begin{eqnarray}\int _{1}^{\infty }\bigg|\int _{F}f(z)E_{a}(z,\unicode[STIX]{x1D70E}+\text{i}r)\,d\unicode[STIX]{x1D707}_{z}\bigg|^{2}r^{2K}\,dr\ll _{f,K}1.\end{eqnarray}$$
Proof. Part (i) follows e.g. from [Reference Iwaniec9, Proposition 7.2 and formulas (9.13), (8.1), (8.2), (8.5) and (8.6)]. For parts (ii) and (iii), we use that the Laplace operator is self-adjoint (see [Reference Iwaniec9, (4.2)]) and we apply repeatedly [Reference Iwaniec9, (4.2)]. Part (ii) follows at once in this way. Part (iii) also follows in this way if we can show that there is an absolute constant
$K_{0}>0$
such that
$$\begin{eqnarray}\int _{-\infty }^{-1}\frac{\int _{F}|E_{a}^{Y}(z,\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\,d\unicode[STIX]{x1D707}_{z}}{r^{2K_{0}}}\,dr+\int _{1}^{\infty }\frac{\int _{F}|E_{a}^{Y}(z,\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\,d\unicode[STIX]{x1D707}_{z}}{r^{2K_{0}}}\,dr\ll _{Y}1\end{eqnarray}$$
for
${\textstyle \frac{1}{2}}<\unicode[STIX]{x1D70E}\leqslant 2$
and
$Y\geqslant 1$
. (We can assume indeed that
$\unicode[STIX]{x1D70E}>{\textstyle \frac{1}{2}}$
, since the case
$\unicode[STIX]{x1D70E}={\textstyle \frac{1}{2}}$
of the last statement of (iii) follows from the
${\textstyle \frac{1}{2}}<\unicode[STIX]{x1D70E}\leqslant 2$
case of that statement by continuity, since the upper bound is uniform in
$\unicode[STIX]{x1D70E}$
.)
Statement (6.2) can be deduced from the Maass–Selberg relations in the following way. One first shows by [Reference Iwaniec9, (6.31)] (using the notation of that book) that
$$\begin{eqnarray}|\unicode[STIX]{x1D719}_{a,a}(\unicode[STIX]{x1D70E}+\text{i}r)|\ll 1\end{eqnarray}$$
for
${\textstyle \frac{1}{2}}<\unicode[STIX]{x1D70E}\leqslant 2$
,
$|r|\geqslant 1$
and for any cusp
$a$
; then the same estimate follows for
$\unicode[STIX]{x1D719}_{a,b}(\unicode[STIX]{x1D70E}+\text{i}r)$
for any two cusps
$a$
,
$b$
. Finally, still by [Reference Iwaniec9, (6.31)], we get
$$\begin{eqnarray}\mathop{\sum }_{b}|\unicode[STIX]{x1D719}_{a,b}(\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\leqslant 1+O\bigg(\unicode[STIX]{x1D70E}-\frac{1}{2}\bigg)\end{eqnarray}$$
for
$1/2<\unicode[STIX]{x1D70E}\leqslant 2$
,
$|r|\geqslant 1$
and for any cusp
$a$
. By the Hadamard inequality (see e.g. [Reference Horn and Johnson7, Corollary 7.8.2]), we then see that for the determinant
$\unicode[STIX]{x1D719}=\det (\unicode[STIX]{x1D719}_{a,b})$
we have that
$$\begin{eqnarray}|\unicode[STIX]{x1D719}(\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\leqslant \mathop{\prod }_{a}\bigg(\mathop{\sum }_{b}|\unicode[STIX]{x1D719}_{a,b}(\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\bigg)\leqslant \bigg(1+O\bigg(\unicode[STIX]{x1D70E}-\frac{1}{2}\bigg)\bigg)\mathop{\sum }_{b}|\unicode[STIX]{x1D719}_{a_{0},b}(\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\end{eqnarray}$$
for any fixed cusp
$a_{0}$
and, combining it with [Reference Hejhal6, Propositions 12.7 and 12.8], we get that
$$\begin{eqnarray}\mathop{\sum }_{b}|\unicode[STIX]{x1D719}_{a_{0},b}(\unicode[STIX]{x1D70E}+\text{i}r)|^{2}\geqslant 1-O\bigg(\bigg(\unicode[STIX]{x1D70E}-\frac{1}{2}\bigg)\unicode[STIX]{x1D714}(r)\bigg)\end{eqnarray}$$
for
${\textstyle \frac{1}{2}}<\unicode[STIX]{x1D70E}\leqslant 2$
,
$|r|\geqslant 1$
and for any cusp
$a_{0}$
with
$\unicode[STIX]{x1D714}$
defined in [Reference Hejhal6, Proposition 12.7]. Then, using again [Reference Iwaniec9, (6.31)] and [Reference Hejhal6, Proposition 12.7], we get (6.2). The lemma is proved.◻
Lemma 6.4. For any
$a\in A$
and
$l\in L$
such that
$\frac{1}{2}<S_{l}<1$
, we have that
$$\begin{eqnarray}\mathop{\sum }_{j\geqslant 0,\,s_{j}=S_{l}}(f,u_{j})\unicode[STIX]{x1D6FD}_{a,j}(0)=\int _{F}f(z)\text{res}_{s=S_{l}}E_{a}(z,s)\,d\unicode[STIX]{x1D707}_{z}.\end{eqnarray}$$
Remark 6.5. Note that since
$f$
could be any function satisfying the conditions of Theorem 1.1, so we could easily get
$$\begin{eqnarray}\mathop{\sum }_{j\geqslant 0,s_{j}=S_{l}}\overline{u_{j}(z)}\unicode[STIX]{x1D6FD}_{a,j}(0)=\text{res}_{s=S_{l}}E_{a}(z,s)\end{eqnarray}$$
for every
$z\in H$
, but we will use (6.3) during the proof of Theorem 1.1, so it is enough for our purposes.
Proof. For
$z\in H$
, we have that
$$\begin{eqnarray}f(z)=\mathop{\sum }_{j=0}^{\infty }(f,u_{j})u_{j}(z)+\mathop{\sum }_{a}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)E_{a}\bigg(z,\frac{1}{2}+\text{i}r\bigg)\,dr\end{eqnarray}$$
and this expression is uniformly and absolutely convergent on compact subsets of
$H$
. Formula (6.4) follows from the spectral theorem [Reference Iwaniec9, Theorems 4.7 and 7.3].
Using Lemma 6.3 and that
$f(\unicode[STIX]{x1D70E}_{a}z)$
is also bounded, (6.4) implies that for any
$a\in A$
, the sum
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=0}^{\infty }(f,u_{j})\unicode[STIX]{x1D6FD}_{a,j}(0)y^{1-s_{j}}+\mathop{\sum }_{c}\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(f,E_{c}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)\nonumber\\ \displaystyle & & \displaystyle \quad \times \,\bigg(\unicode[STIX]{x1D6FF}_{ca}y^{1/2+\text{i}r}+\unicode[STIX]{x1D719}_{c,a}\bigg(\frac{1}{2}+\text{i}r\bigg)y^{1/2-\text{i}r}\bigg)\,dr\nonumber\end{eqnarray}$$
is bounded for
$z\in P(Y_{\unicode[STIX]{x1D6E4}})$
. Since, by [Reference Iwaniec9, (6.22) and (6.27)], we have
$$\begin{eqnarray}\mathop{\sum }_{c}\bigg(f,E_{c}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)\unicode[STIX]{x1D719}_{c,a}\bigg(\frac{1}{2}+\text{i}r\bigg)=\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}-\text{i}r\bigg)\bigg)\end{eqnarray}$$
for any real
$r$
, so for any
$a\in A$
the sum
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathop{\sum }_{j=0}^{\infty }(f,u_{j})\unicode[STIX]{x1D6FD}_{a,j}(0)y^{1-s_{j}}+\frac{1}{4\unicode[STIX]{x1D70B}}\int _{-\infty }^{\infty }\bigg(\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}+\text{i}r\bigg)\bigg)y^{1/2+\text{i}r}\nonumber\\ \displaystyle & & \displaystyle \quad +\,\bigg(f,E_{a}\bigg(\ast ,\frac{1}{2}-\text{i}r\bigg)\bigg)y^{1/2-\text{i}r}\bigg)\,dr\nonumber\end{eqnarray}$$
is bounded for
$z\in P(Y_{\unicode[STIX]{x1D6E4}})$
, i.e.
$$\begin{eqnarray}\mathop{\sum }_{j=0}^{\infty }(f,u_{j})\unicode[STIX]{x1D6FD}_{a,j}(0)y^{1-s_{j}}+\frac{1}{2\unicode[STIX]{x1D70B}\text{i}}\int _{1/2}\bigg(\int _{F}f(z)E_{a}(z,s)\,d\unicode[STIX]{x1D707}_{z}\bigg)y^{1-s}\,ds\end{eqnarray}$$
is bounded for
$z\in P(Y_{\unicode[STIX]{x1D6E4}})$
. We now shift the integration to the right, to
$\operatorname{Re}s=1-\unicode[STIX]{x1D6FF}$
with a small
$\unicode[STIX]{x1D6FF}>0$
, and we see that every residue must be
$0$
because of the boundedness. The lemma follows.◻
Acknowledgement
The author is grateful to the referee for giving useful remarks on the first version of this paper.
