Higher-power moments of Fourier coefficients of holomorphic cusp forms for the congruence subgroup $\varGamma _0(N)$

Abstract

Let $S_k(N)$ be the space of all holomorphic cusp forms of even integral weight k for the congruence group $\varGamma _0(N).$ For any $f\in S_k(N)$ with $\Vert f\Vert _2=1,$ we study the higher-power moments of $\sum _{n\le x}a_f(n),$ where $a_f(n)$ is the nth normalized Fourier coefficient of f. Furthermore, as an application, we investigate the higher-power moments of Fourier coefficients in arithmetic progressions.

1 Introduction

Let $N\ge 1$ be a positive integer and $H_k(N)$ be the set of normalized primitive holomorphic cusp forms of even integral weight k for the congruence group $\varGamma _0(N).$ For any $f\in H_k(1),$ the summation

$$\begin{aligned} {S}_f(x):=\sum _{n\le x}\lambda _f(n) \end{aligned}$$

has attracted the interests of many mathematicians, where $\lambda _f(n)$ are Hecke eigenvalues with $\lambda _f(1)=1.$ The upper bound of ${S}_f(x)$ has been studied by many authors, for example, see [6, 7, 9, 10, 17, 22, 23, 25,26,27]. The best result to date is ${S}_f(x)\ll x^{1/3}(\log x)^{-0.118\ldots }$ proved by Wu [28]. In the opposite direction, Hafner and Ivić [9] also showed that there is a positive constant D such that

$$\begin{aligned} {S}_f(x)=\varOmega _{\pm }\left( x^{1/4}\exp \left\{ \frac{D(\log _2 x)^{1/4}}{(\log _2 x)^{3/4}}\right\} \right) . \end{aligned}$$

In fact, it is conjectured that ${S}_f(x)\ll x^{1/4+\varepsilon },$ which is supported by the mean square estimate (e.g., [12])

$$\begin{aligned} \int _{1}^{T}{S}_f(x)^2 \mathrm{d}x=B_fT^{3/2}+O\left( T^{1+\varepsilon }\right) , \end{aligned}$$

where $B_f$ is a constant depending on f. The third and fourth power moments of ${S}_f(x)$ are also studied by Cai [4]. In [30], Zhai obtained the asymptotic formula of $\int _1^{T}{S}_f(x)^{k}\mathrm{d}x,$ for $5\le k\le 7.$ Later, Zhai [31] studied the mean value of ${S}_f(x)$ in short interval for $2\le k\le 7$ and got the following asymptotic formula

$$\begin{aligned} \int _{T-H}^{T+H}{S}_f(x)^k \mathrm{d}x=D_k\int _{T-H}^{T+H}x^{k/4} \mathrm{d}x+O\left( HT^{k/4-\varepsilon }\right) \end{aligned}$$

holds for $T^{(k+4)/12+\sqrt{\varepsilon }}\le H\le T,$ where $D_k$ is a certain constant. One main ingredient of these mean value estimations is the truncated Voronoi summation formula proved by Jutia [16]

$$\begin{aligned} {S}_f(x)=\frac{x^{1/4}}{ \sqrt{2}\pi }\sum _{n\le M}\frac{\lambda _f(n)}{n^{3/4}}\cos \left( 4\pi \sqrt{nx}-\frac{\pi }{4}\right) +O\left( x^{1/2+\varepsilon }M^{-1/2}\right) , \end{aligned}$$

(1)

where $1\ll M\ll x.$

In this paper, we consider certain holomorphic cusp forms (may not be Hecke eigenforms) of higher levels and explore the explicit dependence on the level. Furthermore, as an application, we investigate the higher-power moments of Fourier coefficients in arithmetic progressions.

Put

$$\begin{aligned} \mathcal {F}=\big \{z=x+iy:0<x<1,\, |z|>1\ \mathrm {and}\ |z-1|>1\big \}. \end{aligned}$$

(2)

Then $\mathcal {F}$ is a fundamental domain for $\varGamma =SL_2(\mathbb {Z})$ and

$$\begin{aligned} \mathcal {F}_N=\bigcup \limits _{\sigma \in \varGamma _0(N)\backslash \varGamma }\sigma \mathcal {F} \end{aligned}$$

(3)

is a fundamental domain for $\varGamma _0(N).$ Let $S_k(N)$ denote the space of all holomorphic cusp forms of even integral weight k for the congruence group $\varGamma _0(N).$ For any $f\in S_k(N),$ we have the Fourier expansion

$$\begin{aligned} f(z)=\sum _{n=1}^\infty a_f(n)n^{(k-1)/2}\mathrm{e}^{2\pi i nz}, \end{aligned}$$

where $a_f(n)$ are the normalized Fourier coefficients. For any $f,\,g\in S_k(N),$ we define

$$\begin{aligned} \langle f,\,g\rangle =\int _{\mathcal {F}_N}f(z)\overline{g(z)}y^{k}\frac{\mathrm{d}x\,\mathrm{d}y}{y^2} \end{aligned}$$

and

$$\begin{aligned} \Vert f\Vert _2=\sqrt{\langle f,\,f\rangle }. \end{aligned}$$

For $N_1|N,$ denote the set of all $L^2(\varGamma _0(N)\backslash \mathbb {H})$-normalized [not $L^2(\varGamma _0(N_1)\backslash \mathbb {H})$-normalized] newforms of level $N_1$ by $\mathcal {B}^*(N_1,\, N).$ Then Deligne’s bound and [14, Corollary 5.45] imply that for any $f^*\in \mathcal {B}^*(N_1,\, N),$ we have

$$\begin{aligned} a_{f^*}(n)\ll N^{-1/2+\varepsilon }n^\varepsilon . \end{aligned}$$

(4)

By [3, Lemma 9] (see also [21, Theorem 2.3.6]), we know that

$$\begin{aligned} \bigcup _{N_1|N}\bigcup _{f^*\in \mathcal {B}^*(N_1,\, N)}\bigg \{f^{(g)}:=\sum _{d|g}\xi _g(d)f^*|_d:g|\frac{N}{N_1}\bigg \} \end{aligned}$$

is an orthonormal basis of $S_k(N),$ where $\xi _g(d)$ is a parameter defined by [3, (5.6)] and $f^*|_d(z)=d^{k/2}f^*(dz).$ For any $f^{(g)}$ in this basis, by the estimate $\xi _g(d)\ll g^{\varepsilon }(d/g)^{1/2}$ in [3, (5.6)] and (4), we have

$$\begin{aligned} a_{f^{(g)}}(n)\ll \sum _{d|g}\left| \xi _g(d)\right| d^{1/2}\left| a_{f^*}\left( \frac{n}{d}\right) \right| \ll (Nn)^\varepsilon N^{-1/2}\sum _{d|g}d^{1/2}. \end{aligned}$$

(5)

Then for any $h\in S_k(N)$ with $\Vert h\Vert _2=1,$ we have

$$\begin{aligned} h=\sum _{N_1|N}\sum _{f^*\in \mathcal {B}^*(N_1,\, N)}\sum _{g\big |\frac{N}{N_1}}\alpha _{f^{(g)}}f^{(g)}. \end{aligned}$$

Moreover, by (5), we have

$$\begin{aligned}&a_h(n)= \sum _{N_1|N}\sum _{f^*\in \mathcal {B}^*(N_1,\, N)}\sum _{g\big |\frac{N}{N_1}}\alpha _{f^{(g)}}a_{f^{(g)}}(n)\\&\quad \ll (Nn)^\varepsilon N^{-1/2}\sqrt{\left( \sum _{N_1|N}\sum _{f^*\in \mathcal {B}^*(N_1,\, N)}\sum _{g\big |\frac{N}{N_1}}\left| \alpha _{f^{(g)}}\right| ^2\right) \left( \sum _{N_1|N}\sum _{f^*\in \mathcal {B}^*(N_1,\, N)}\sum _{g\big |\frac{N}{N_1}}\sum _{d|g}d\right) }. \end{aligned}$$

Since the basis $\{f^{(g)}\}$ is orthonormal and $\Vert h\Vert _2=1,$ the first bracket inside the root sign is 1. Inasmuch as $|\mathcal {B}^*(N_1,\, N)|\ll N_1^{1+\varepsilon },$ the second bracket is bounded by

$$\begin{aligned} N^\varepsilon \sum _{N_1|N}\sum _{g\big |\frac{N}{N_1}}\sum _{d|g}dN_1\ll N^\varepsilon \sum _{g'g'' dN_1=N}dN_1 \ll N^{1+\varepsilon }. \end{aligned}$$

It follows

$$\begin{aligned} a_h(n)\ll (Nn)^\varepsilon N^{-1/2}\sqrt{1\cdot N^{1+\varepsilon }}\ll (Nn)^\varepsilon . \end{aligned}$$

(6)

For any $f\in S_k(N),$ define

$$\begin{aligned} \mathcal {S}_f(x):=\sum _{n\le x}a_f(n). \end{aligned}$$

We first study the moments of $\mathcal {S}_f(x).$ We start with the large value of $\mathcal {S}_f(x)$ (see Theorem 4) by Halasz–Montgomery inequality, and the estimates of exponential sums. Then we divide the interval $[T,\,2T]$ into subintervals $[T+(j-1)V,\, T+jV],\, j=1,\,2, \ldots ,$ and pick the maximal $|\mathcal {S}_f(x)|$ in $[T+(j-1)V,\, T+jV].$ Finally, we can get the higher-power moments of $\mathcal {S}_f(x)$ by Theorem 4. More precisely, we have the following theorem.

Theorem 1

Let $f\in S_k(N)$ and $\Vert f\Vert _2=1.$ For $2\le A\le 11$ and $N\le T,$

$$\begin{aligned}&\int _T^{2T} \left| \mathcal {S}_f(x)\right| ^A\mathrm{d}x \ll T^{1+\varepsilon }(TN)^{A/4}+T^{\varepsilon }(TN)^{(A+1)/3}. \end{aligned}$$

Remark 1

If $T\gg N^{\frac{A+4}{8-A}},$ and $A<8,$ we know that the first term dominates the upper bound.

On the other hand, for $f\in S_k(N)$ with $\Vert f\Vert _2=1,$ the mean value of $a_f(n)$ in arithmetic progressions also attracts the attention of mathematicians, see [2, 8, 19, 20, 29]. Let a and r be two positive integers. It is well known that $\{a_f(n)|n\equiv a\; \mathrm {mod}\, r\}$ determines a cusp form of higher level. More precisely, (e.g., see [1] and [33, Lemma 3.1])

$$\begin{aligned} g(z)&:=\mathop {\mathop {\sum }\limits _{n\ge 1}}\limits _{n\equiv a\; \mathrm {mod }\, r}a_f(n)n^{(k-1)/2}e( nz)\\&=\frac{1}{r}\sum _{\ell =1}^re\left( \frac{-a\ell }{r}\right) \sum _{n\ge 1}a_f(n)n^{(k-1)/2}e\left( nz+\frac{n\ell }{r}\right) \end{aligned}$$

is a cusp form on $\varGamma _0(Nr^2).$ Set $S(a)= \left( \begin{array}{cc} 1 &{} a\\ 0&{} 1 \end{array} \right) .$ By Cauchy–Schwarz’s inequality, we have

$$\begin{aligned} |g(z)|^2\le \frac{1}{r}\sum _{l=1}^r\left| f|_{S({\ell }/{r})}(z)\right| ^2, \end{aligned}$$

where $f|_{\gamma }(z)=(cz+d)^{-k}f(\gamma z), \ \gamma =\left( \begin{array}{cc} a &{}b\\ c &{}d \end{array} \right) \in SL_2(\mathbb {R}).$ For the same reason as the norm reduction of $[\varGamma _0(N_1):\varGamma _0(N)]^{-1/2}$ under the map $\mathcal {B}^*(N_1,\, N)\rightarrow \mathcal {B}^*(N_1,\, N_1),$ the following holds:

$$\begin{aligned} \Vert g\Vert _2^2\ll \frac{1}{r}\sum _{l=1}^r\int _{\varGamma _0(Nr^2)\setminus \mathbb {H}}y^k|f(z)|^2\frac{\mathrm{d}x\mathrm{d}\,y}{y^2}=\left[ \varGamma _0(N):\varGamma _0\left( Nr^2\right) \right] \ll (Nr)^{\varepsilon }r^2. \end{aligned}$$

Therefore, we could apply Theorem 1 to $g/ \Vert g\Vert _2,$ whose Fourier coefficient is $a_f(n)/ \Vert g\Vert _2$ if $n\equiv a\mod r$ or 0 else. For $f\in S_k(N),$ define

$$\begin{aligned} \mathcal {S}_f(x;\,a,\,r):=\mathop {\mathop {\sum }\limits _{n\le x}}\limits _{n\equiv a\; \mathrm {mod}\, r}a_f(n). \end{aligned}$$

We have the following result.

Theorem 2

Let $f\in S_k(N)$ and $\Vert f\Vert _2=1.$ For $2\le A\le 11$ and $Nr^2\le T,$

$$\begin{aligned} \int _T^{2T} \left| \mathcal {S}_f(x;\,a,\,r)\right| ^A\mathrm{d}x \ll T^{1+\varepsilon }\left( T^{\frac{1}{4}}N^{\frac{1}{4}}r^{\frac{3}{2}}\right) ^A+ T^{\varepsilon }\left( TNr^2\right) ^{\frac{1}{3}}\left( T^{\frac{1}{3}}N^{\frac{1}{3}}r^{\frac{5}{3}}\right) ^A. \end{aligned}$$

2 Preliminary lemmas

In this section we introduce some properties of holomorphic cusp forms and their corresponding L-functions.

For any $f\in S_k(N),$ its corresponding L-function is defined by

$$\begin{aligned} L(s,\,f)=\sum _{n=1}^\infty a_f(n)n^{-s}. \end{aligned}$$

By (6), this L-function is holomorphic in the region $\mathfrak {R}s>1.$ Moreover, it can be analytically extended to the whole complex plane and has the functional equation

$$\begin{aligned}&\bigg (\frac{\sqrt{N}}{2\pi }\bigg )^{s}\varGamma \bigg (s+\frac{k-1}{2}\bigg )L(s,\,f)\nonumber \\&\quad =\epsilon _f\bigg (\frac{\sqrt{N}}{2\pi }\bigg )^{1-s}\varGamma \bigg (1-s+\frac{k-1}{2}\bigg )L(1-s,\,\mathrm{Wf}). \end{aligned}$$

(7)

Here $\epsilon _f={\pm }1$ and

$$\begin{aligned} \mathrm{Wf}(z)=N^{-k/2}z^{-k}f\left( \frac{-1}{Nz}\right) =\sum _{n=1}a_{\mathrm{Wf}}(n)n^{(k-1)/2}\mathrm{e}^{2\pi i nz} \end{aligned}$$

is also a holomorphic cusp form of weight k for $\varGamma _0(N).$ Furthermore, we know that $\Vert \mathrm{Wf}\Vert _2=\Vert f\Vert _2$ (see [13, §6.7]). Suppose $\Vert f\Vert _2=1$ from now on. Then by (6), we have

$$\begin{aligned} a_f(n)\ll (Nn)^{\varepsilon }, \quad a_{\mathrm{Wf}}(n)\ll (Nn)^{\varepsilon }. \end{aligned}$$

(8)

By standard arguments, we get the following convexity bound.

Lemma 1

For any $f\in S_k(N)$ with $\Vert f\Vert _2=1$ and $\varepsilon >0,$ we have

$$\begin{aligned} L(s,\,f)\ll _\varepsilon N^{\frac{1}{2}{(1-\sigma )+\varepsilon }}(1+|\tau |)^{1-\sigma +\varepsilon } \end{aligned}$$

(9)

in the strip $-\varepsilon<\mathfrak {R}(s)=\sigma <1+\varepsilon ,$ where $\tau =\mathfrak {I}(s).$

Proof

By (6), we know that $L(s,\,f)\ll _\varepsilon 1$ on the line $\mathfrak {R}(s)=1+\varepsilon .$ For $\mathfrak {R}(s)=-\varepsilon ,$ by the functional equation (7), we have

$$\begin{aligned} L(s,\,f) =\frac{\epsilon _f\bigg (\frac{\sqrt{N}}{2\pi }\bigg )^{1-s} \varGamma \bigg (1-s+\frac{k-1}{2}\bigg )}{\bigg (\frac{\sqrt{N}}{2\pi }\bigg )^{s}\varGamma \bigg (s+\frac{k-1}{2}\bigg )} L(1-s,\,\mathrm{Wf}). \end{aligned}$$

Then, by [18, Lemma 3.1] and (8), we obtain that for $\mathfrak {R}(s)=-\varepsilon$

$$\begin{aligned} L(s,\,f)\ll N^{1/2+2\varepsilon }(1+|\tau |)^{1+\varepsilon } \end{aligned}$$

and the lemma follows by Phragmén–Lindelöf principle. $\square$

At the end of this section, we introduce two well-known lemmas for self-containedness.

Lemma 2

Let F(x) be a real differentiable function such that $F'(x)$ is monotonic and $F'(x)\ge m>0$ or $F'(x)\le -m<0$ for $a\le x\le b.$ Then

$$\begin{aligned} \left| \int _a^b \cos F(x)\mathrm{d}x\right| \le 2m^{-1},\quad \left| \int _a^b \sin F(x)\mathrm{d}x\right| \le 2m^{-1}. \end{aligned}$$

Proof

See [11, Lemma 2.1].

Lemma 3

Let S be an inner-product vector space over $\mathbb {C},$ $(\mathbf {a},\,\mathbf {b})$ denote the inner product in S and $\Vert \mathbf {a}\Vert ^2=(\mathbf {a},\,\mathbf {a}).$ Suppose that $\mathbf {a},\,\mathbf {b}_1,\ldots ,\mathbf {b}_R$ are arbitrary vectors in S. Then

$$\begin{aligned} \sum _{\ell \le R}\left| \left( \mathbf {a},\,\mathbf {b}_\ell \right) \right| ^2\le \Vert \mathbf {a}\Vert ^2\max _{s\le R}\sum _{t\le R}\left| \left( \mathbf {b}_{s},\,\mathbf {b}_{t}\right) \right| . \end{aligned}$$

Proof

This is the well-known Halasz–Montgomery inequality. See [11, (A.40)]. $\square$

3 A truncated Voronoi formula

The following theorem is our main tool, an analogue of [15, Theorem 3] with a very similar proof.

Theorem 3

For any $f\in S_k(N)$ and any $\varepsilon >0,$ we have

$$\begin{aligned} \mathcal {S}_f(x)&=\frac{\epsilon _f(xN)^{1/4}}{\pi \sqrt{2}}\sum _{n\le M}\frac{a_{\mathrm{Wf}}(n) }{n^{3/4}}\cos \left( 4\pi \sqrt{\frac{nx}{N}}-\frac{k+\frac{1}{2}}{2}\pi \right) \nonumber \\&\quad +O_{\varepsilon }\left( \left( (xN)^{1/2}M^{-1/2}+N^{1/2}\right) (xN)^\varepsilon \right) \end{aligned}$$

(10)

uniformly for $2\le M\le x$ and $1\le N\le x.$

In particular, we have

$$\begin{aligned} \mathcal {S}_f(x)\ll _{\varepsilon } (xN)^{1/3+\varepsilon }, \quad \text {when} \ x\ge N^{1/2}. \end{aligned}$$

(11)

Proof

Let $1\le N\le x,\, 2\le M \le x$ and $T>1$ be chosen as

$$\begin{aligned} T^2=N^{-1}4\pi ^2(M+1/2)x. \end{aligned}$$

By the Perron formula (see [24, Theorem II.2]) with $\kappa =1+\varepsilon$ and (6), we have

$$\begin{aligned} \mathcal {S}_f(x) =\frac{1}{2\pi i}\int _{\kappa -iT}^{\kappa +iT}L(s,\,f)\frac{x^s}{s}\mathrm{d}s +O_{\varepsilon }\left( x^\kappa \sum _{n\ge 1}\frac{|a_f(n)|}{n^{\kappa }(1+T|\log (x/n)|)}\right) . \end{aligned}$$

(12)

The O-term in (12) contributes

$$\begin{aligned} \ll (Nx)^{\varepsilon }x\left( \sum _{n\le \frac{x}{2}}+\sum _{\frac{x}{2}<n<\frac{3x}{2}}+\sum _{n\ge \frac{3x}{2}}\right) \frac{1}{n^{1+\varepsilon }(1+T| \log (x/n)|)}. \end{aligned}$$

It is easy to see that

$$\begin{aligned} \left( \sum _{n\le \frac{x}{2}}+\sum _{n\ge \frac{3x}{2}}\right) \frac{1}{n^{1+\varepsilon }(1+T| \log (x/n)|)}\ll T^{-1}N^{\varepsilon }. \end{aligned}$$

For the middle sum, we have

$$\begin{aligned} \sum _{\frac{x}{2}<n<\frac{3x}{2}}\frac{1}{n^{1+\varepsilon }(1+T| \log (x/n)|)}&\ll \sum _{\frac{x}{2}<n<\frac{3x}{2}}\frac{1}{x+T|x-n|}\\&\ll x^{-1}+T^{-1}(Nx)^{\varepsilon }. \end{aligned}$$

Combining the above estimates and noting that $T\ll x,$ we have

$$\begin{aligned} \mathcal {S}_f(x) =\frac{1}{2\pi i}\int _{\kappa -iT}^{\kappa +iT}L(s,\,f)\frac{x^s}{s}\mathrm{d}s +O_{\varepsilon }\left( (Nx)^{\varepsilon }xT^{-1}\right) . \end{aligned}$$

(13)

We deform the line of integration to the contour $\mathfrak {L}$ joining the points $\kappa -iT,$ $-\varepsilon -iT,$ $-\varepsilon +iT$ and $\kappa +iT.$ Let $\mathfrak {L}_\mathrm{v}:=[-\varepsilon -iT,\,-\varepsilon +iT].$ By (9), the integral over the horizontal segments of $\mathfrak {L}$ is

$$\begin{aligned} \ll _{\varepsilon }(Nx)^\varepsilon \left( \frac{x}{T} +N^{1/2}\right) . \end{aligned}$$

The residue of the integrand at $s=0$ gives

$$\begin{aligned} {L}(0,\,f)\ll _{\varepsilon } N^{1/2+\varepsilon }. \end{aligned}$$

By (7), the integral over $\mathfrak {L}_\mathrm{v}$ equals

$$\begin{aligned}&\frac{1}{2\pi i}\int _{\mathfrak {L}_\mathrm{v}}{L}(s,\,f)\frac{x^s}{s}\mathrm{d}s\\&\quad = \frac{1}{2\pi i}\int _{\mathfrak {L}_\mathrm{v}}\frac{\epsilon _f\left( \frac{\sqrt{N}}{2\pi }\right) ^{1-s} \varGamma \left( 1-s+\frac{k-1}{2}\right) }{\left( \frac{\sqrt{N}}{2\pi }\right) ^s\varGamma \left( s+\frac{k-1}{2}\right) } \sum _{n\ge 1}a_{\mathrm{Wf}}(n) n^{s-1}\frac{x^s}{s}\mathrm{d}s\\&\quad =\epsilon _f\frac{\sqrt{N}}{2\pi }\sum _{n\ge 1}\frac{a_{\mathrm{Wf}}(n) }{n} \frac{1}{2\pi i}\int _{\mathfrak {L}_\mathrm{v}}\frac{\varGamma \left( 1-s+\frac{k-1}{2}\right) }{\varGamma \left( s+\frac{k-1}{2}\right) }\left( 2\pi {\sqrt{\frac{nx}{N}}}\right) ^{2s}\frac{\mathrm{d}s}{s}\\&\quad =\epsilon _f\frac{\sqrt{N}}{2\pi }\sum _{n\ge 1}\frac{a_{\mathrm{Wf}}(n) }{n}I_{\mathfrak {L}_\mathrm{v}}\left( 2\pi {\sqrt{\frac{nx}{N}}}\right) , \end{aligned}$$

where

$$\begin{aligned} I_{\mathfrak {L}_\mathrm{v}}(y):=\frac{1}{2\pi i}\int _{\mathfrak {L}_\mathrm{v}}\frac{\varGamma \left( 1-s+\frac{k-1}{2}\right) }{\varGamma \left( s+\frac{k-1}{2}\right) }y^{2s}\frac{\mathrm{d}s}{s}. \end{aligned}$$

Combining the above formulas with (13), we have

$$\begin{aligned} \mathcal {S}_f(x)&=\epsilon _f\frac{\sqrt{N}}{2\pi }\sum _{n\ge 1}\frac{a_{\mathrm{Wf}}(n) }{n}I_{\mathfrak {L}_\mathrm{v}}\left( 2\pi {\sqrt{\frac{nx}{N}}}\right) \nonumber \\&\quad +O_{\varepsilon }(Nx)^\varepsilon \left( \left( \frac{x}{T}+N^{1/2}\right) \right) . \end{aligned}$$

(14)

Next, we apply the stationary phrase method to bound $I_{\mathfrak {L}_\mathrm{v}}(y)$ for large y and give an asymptotic expansion in terms of trigonometric functions for small y. By Stirling’s formula, for $\tau \ge 1,$ the integrand equals

$$\begin{aligned} \mathrm{e}^{-i\pi (k-1)/2}y^{2\sigma }\tau ^{-2\sigma }\mathrm{e}^{2i\tau \log (ey/\tau )}\left\{ 1+c_1\tau ^{-1}+O\left( \tau ^{-2}\right) \right\} \end{aligned}$$

for $|\sigma |\le A,$ where $c_1$ and $A>0$ are some suitable constants and the implied constant is independent of $\tau$ and y. Set $g(\tau ):=2\tau \log (ey/\tau ).$ Then $g^\prime (\tau )=2\log (y/\tau ).$ With the second mean value theorem for integrals (cf. [24, Theorem I.0.3]), we obtain for $y>T$ and $\sigma =-\varepsilon ,$

$$\begin{aligned}&\int _1^Ty^{2\sigma }\tau ^{-2\sigma }\mathrm{e}^{ig(\tau )}\left\{ 1+c_1\tau ^{-1}+O\left( \tau ^{-2}\right) \right\} \mathrm{d}\tau \nonumber \\&\quad \ll T^{2\varepsilon }y^{-2\varepsilon }\left| \log \frac{y}{T}\right| ^{-1}+T^{2\varepsilon -1}y^{-2\varepsilon }, \end{aligned}$$

(15)

and for $y<T$ and $\sigma =\frac{1}{2}+\varepsilon ,$

$$\begin{aligned}&\int _T^\infty y^{2\sigma }\tau ^{-2\sigma }\mathrm{e}^{ig(\tau )}\left\{ 1+c_1\tau ^{-1}+O\left( \tau ^{-2}\right) \right\} \mathrm{d}\tau \nonumber \\&\quad \ll T^{-1-2\varepsilon }y^{1+2\varepsilon }\left| \log \frac{y}{T}\right| ^{-1}+T^{-2-2\varepsilon }y^{1+2\varepsilon }. \end{aligned}$$

(16)

For $n>M,$ we infer by (15) that

$$\begin{aligned} I_{\mathfrak {L}_\mathrm{v}}\left( 2\pi {\sqrt{\frac{nx}{N}}}\right) \ll _{\varepsilon } \left( \frac{x}{{n}}\right) ^{\varepsilon }\left( \left| \log \frac{n}{M+1/2}\right| ^{-1}+\sqrt{\frac{N}{xM}}\right) . \end{aligned}$$

By (8), we obtain that

$$\begin{aligned}&\epsilon _f\frac{\sqrt{N}}{2\pi }\sum _{n>M}\frac{a_{\mathrm{Wf}}(n) }{n}I_{\mathfrak {L}_{rm v}}\left( 2\pi {\sqrt{\frac{nx}{N}}}\right) \nonumber \\&\quad \ll _{\varepsilon } (xN)^\varepsilon {{N^{1/2}}}\left( \sum _{M<n<3M/2}\left| \frac{1}{M+\frac{1}{2}-n}\right| + \sum _{n>M}\frac{1}{n^{1+\varepsilon }}\right) \nonumber \\&\quad \ll _{\varepsilon } (xN)^\varepsilon {{N^{1/2}}}, \end{aligned}$$

(17)

if we notice that $\sqrt{\frac{N}{xM}}\ll 1.$

For $n\le M,$ we complete the path $\mathfrak {L}_\mathrm{v}$ to the contour $\mathfrak {L}_\mathrm{v}^*$ in order to apply [5, Lemma 1], where $\mathfrak {L}_\mathrm{v}^*$ is the positively oriented contour consisting of $\mathfrak {L}_\mathrm{v},$ $\mathfrak {L}_\mathrm{v}^\pm$ and $\mathfrak {L}_\mathrm{h}^\pm$ with

$$\begin{aligned} \mathfrak {L}_\mathrm{v}^\pm :=\left[ \frac{1}{2}+\varepsilon \pm iT,\,\frac{1}{2}+\varepsilon \pm i\infty \right] ,\quad \mathfrak {L}_\mathrm{h}^\pm :=\left[ -\varepsilon \pm iT,\,\frac{1}{2}+\varepsilon \pm iT\right] . \end{aligned}$$

Correspondingly, we denote by $I_{\mathfrak {L}_\mathrm{v}^\pm }$ and $I_{\mathfrak {L}_\mathrm{h}^\pm }$ the integrals over these segments. By (16), the integral over the vertical line segments $\mathfrak {L}_\mathrm{v}^\pm$ is

$$\begin{aligned} I_{\mathfrak {L}_\mathrm{v}^\pm }\ll _{\varepsilon } x^\varepsilon \left( \frac{n}{M}\right) ^{1/2}\left| \log \frac{n}{M+1/2}\right| ^{-1}+x^\varepsilon \left( \frac{n}{M}\right) ^{1/2}, \end{aligned}$$

while for the horizontal segments, $I_{\mathfrak {L}_\mathrm{h}^\pm }$ contributes at most $O((M/n)^\varepsilon ).$ Therefore, by (8), we have

$$\begin{aligned}&\epsilon _f\frac{\sqrt{N}}{2\pi }\sum _{n\le M}\frac{a_{\mathrm{Wf}}(n) }{n}\left( I_{\mathfrak {L}_\mathrm{v}^\pm }+I_{\mathfrak {L}_\mathrm{h}^\pm }\right) \nonumber \\&\quad \ll _{\varepsilon } (xN)^\varepsilon {{N^{1/2}}}\left( \sum _{M/2<n\le M}\left| \frac{1}{M+\frac{1}{2}-n}\right| + \sum _{n\le M}\frac{1}{n^{1+\varepsilon }}\right) \nonumber \\&\quad \ll _{\varepsilon } (xN)^\varepsilon {{N^{1/2}}}, \end{aligned}$$

(18)

if we notice that $\frac{n}{M}\ll 1$ when $n\le M.$ Inserting (18) and (17) into (14), we get from our choice of T, 

$$\begin{aligned} \mathcal {S}_f(x)&=\epsilon _f\frac{\sqrt{N}}{2\pi }\sum _{1\le n\le M}\frac{a_{\mathrm{Wf}}(n) }{n}I_{\mathfrak {L}_\mathrm{v}^*}\left( 2\pi {\sqrt{\frac{nx}{N}}}\right) \nonumber \\&\quad +O_{\varepsilon }\left( \left( (xN)^{1/2}M^{-1/2}+N^{1/2}\right) (xN)^\varepsilon \right) . \end{aligned}$$

(19)

Now all the poles of the integrand in

$$\begin{aligned} I_{\mathfrak {L}_\mathrm{v}^*}(y):=\frac{1}{2\pi i}\int _{\mathfrak {L}_\mathrm{v}^*}\frac{\varGamma \left( 1-s+\frac{k-1}{2}\right) \varGamma (s)}{\varGamma \left( s+\frac{k-1}{2}\right) \varGamma (s+1)}y^{2s}{\mathrm{d}s} \end{aligned}$$

lie on the right of the contour $\mathfrak {L}_\mathrm{v}^*.$ After a change of variable s into $1-s,$ we have

$$\begin{aligned} I_{\mathfrak {L}_\mathrm{v}^*}(y)=\frac{1}{\pi }I_0\left( y^2\right) , \end{aligned}$$

where

$$\begin{aligned} I_0(y):=\frac{1}{2\pi i}\int _{\mathcal {L}_\varepsilon }\frac{\varGamma \left( s+\frac{k-1}{2}\right) \varGamma (1-s)}{\varGamma \left( 1-s+\frac{k-1}{2}\right) \varGamma (2-s)}y^{1-s}{\mathrm{d}s}. \end{aligned}$$

Here $\mathcal {L}_\varepsilon$ consists of the line $s=\frac{1}{2}-\varepsilon +i\tau$ with $|\tau |\ge T,$ together with three sides of the rectangle whose vertices are $\frac{1}{2}-\varepsilon - iT,$ $1+\varepsilon - iT,$ $1+\varepsilon + iT$, and $\frac{1}{2}-\varepsilon + iT.$ Clearly, our $I_0$ is a particular case of $I_\rho$ defined in [5, Lemma 1], corresponding to the choice of parameters $A=\delta =N=\omega =\alpha _1=1,$ $\beta _1=\mu =(k-1)/2,$ $\rho =m=0,$ $a=-\frac{3}{4},$ $c_0=\frac{1}{2},$ $h=2$, and $k_0=-\frac{3}{4}-\frac{k-1}{2}.$ It hence follows that

$$\begin{aligned} I_{\mathfrak {L}_\mathrm{v}^*}\left( 2\pi {\sqrt{\frac{nx}{N}}}\right)&=e_0'\sqrt{{2\pi }}(nx/N)^{1/4}\cos \left( 4\pi \sqrt{\frac{nx}{N}}+k_0\pi \right) \nonumber \\&\quad +O\left( N^{1/4}(nx)^{-1/4}\right) . \end{aligned}$$

(20)

Here the value of $e_0'$ in [5, Lemma 1] is $1/\sqrt{\pi }.$ The main term in (10) follows from (20) and (19). With a simple checking, the contribution of the error term in (20) will be absorbed in the error term in (19).

Finally, we set $M=(xN)^{1/3}$ and note that

$$\begin{aligned} \sum _{n\le M}\frac{|a_{\mathrm{Wf}}(n) |}{n^{3/4}}\ll _{\varepsilon } (NM)^{\varepsilon }M^{1/4}. \end{aligned}$$

Then (11) follows plainly. $\square$

4 A large value estimate of $\mathcal {S}_f(x)$

Theorem 4

Let $H\ge N$ and $V\gg (HN)^{1/4}.$ Suppose $H\le x_1<x_2<\cdots <x_R\le 2 H$ satisfy $|x_j-x_i|\gg V \ \text {for}\ (i\ne j)$ and $|\mathcal {S}_f(x_\ell )|\gg VH^{2\varepsilon }.$ Then we have

$$\begin{aligned} R\ll H^\varepsilon \left( V^{-3}HN+V^{-12}H(HN)^{11/4}\right) . \end{aligned}$$

Proof

Suppose that $H_0>V$ is a parameter to be determined later. Let I be any subinterval of $[H,\,2H]$ of length not exceeding $H_0$ and let $G=I\cap \{x_1,\,x_2,\ldots ,x_R\}.$ Without loss of generality, we may assume that $G=\{x_1,\,x_2,\ldots ,x_{R_0}\}.$

Suppose $J=\left[ \frac{(1+4\varepsilon )\log H-2\log V}{\log 2}\right] .$ For any $x_\ell \in \{x_1,\,x_2,\ldots ,x_R\},$ we apply Theorem 3 with $x=x_\ell$ and $M=2^{J+1}\asymp HNV^{-2}$ to get

$$\begin{aligned}&\mathcal {S}_f\left( x_\ell \right) \nonumber \\&\quad \ll (HN)^{1/4}\left| \sum _{n\le M}\frac{a_{\mathrm{Wf}}(n) }{n^{3/4}}\cos \left( 4\pi \sqrt{\frac{nx}{N}}-\frac{k+\frac{1}{2}}{2}\pi \right) \right| +(HN)^{\varepsilon }\left( V+N^{1/2}\right) \nonumber \\&\quad \ll (HN)^{\frac{1}{4}}\sum _{j=0}^J\left| \sum _{n\sim 2^j}\xi _n e\left( 2\sqrt{\frac{nx_\ell }{N}}\right) \right| +(HN)^{\varepsilon }\left( V+N^{1/2}\right) , \end{aligned}$$

(21)

where

$$\begin{aligned} \xi _n=a_{\mathrm{Wf}}(n)n^{-3/4}\, \text {or} \, \overline{a_{\mathrm{Wf}}(n)}n^{-3/4}, \end{aligned}$$

and $n\sim 2^j$ means $2^j\le n<2^{j+1}.$ It is easy to see that in (21), $N^{1/2}\le V$ when $H \ge N$ and $V\gg (HN)^{1/4}.$ Squaring, summing over the set G and then using the Cauchy’s inequality, we get that

$$\begin{aligned}&\sum _{\ell \le R_0} \mathcal {S}_f^2\left( x_\ell \right) \nonumber \\&\quad \ll (HN)^{\frac{1}{2}}\sum _{\ell \le R_0}\left| \sum _{j=0}^J\left| \sum _{n\sim 2^j}\xi _n e\left( 2\sqrt{\frac{nx_\ell }{N}}\right) \right| \right| ^2+(HN)^{2\varepsilon }V^2R_0\nonumber \\&\quad \ll (HN)^{\frac{1}{2}}\log H\sum _{\ell \le R_0}\sum _{j=0}^J\left| \sum _{n\sim 2^j}\xi _n e\left( 2\sqrt{\frac{nx_\ell }{N}}\right) \right| ^2+(HN)^{2\varepsilon }V^2R_0\nonumber \\&\quad \ll (HN)^{\frac{1}{2}}\log ^2 H\max _{0\le j\le J}\sum _{\ell \le R_0}\left| \sum _{n\sim 2^j}\xi _n\varphi _n(\ell ) \right| ^2+(HN)^{2\varepsilon }V^2R_0\nonumber \\&\quad \ll (HN)^{\frac{1}{2}}\log ^2 H\sum _{\ell \le R_0}\left| \sum _{n\sim L}\xi _n\varphi _n(\ell ) \right| ^2+(HN)^{2\varepsilon }V^2R_0 \end{aligned}$$

(22)

for some $L\le M,$ where $\varphi _n(\ell )=e\left( 2\sqrt{\frac{nx_\ell }{N}}\right) .$

Take $\mathbf {a}=\{a_n\}_{n=1}^\infty$ with $a_n=\xi _n$ for $n\sim L$ and zero otherwise. Take $\mathbf {b}_\ell =\{{b}_{\ell ,n}\}_{n=1}^\infty$ with ${b}_{\ell ,n}=\overline{\varphi _n(\ell )}$ for $n\sim L$ and zero otherwise. Then

$$\begin{aligned} \left( \mathbf {a},\,\mathbf {b}_\ell \right)&=\sum _{n\sim L}\xi _n\varphi _n(\ell ) ,\\ \left( \mathbf {b}_s,\,\mathbf {b}_t\right)&=\sum _{n\sim L}e\left( 2\sqrt{\frac{n}{N}}\left( \sqrt{x_{t}}-\sqrt{x_{s}}\right) \right) ,\\ \Vert \mathbf {a}\Vert ^2&=\sum _{n\sim L}\frac{|a_{\mathrm{Wf}}(n)|^2}{n^{3/2}}\ll L^{-1/2+\varepsilon }N^{\varepsilon }, \end{aligned}$$

where we have used (8) in the last formula. Then, by Lemma 3 and (22), we get

$$\begin{aligned} R_0V^2H^{4\varepsilon }&\ll (HN)^{\frac{1}{2} +\varepsilon }L^{-\frac{1}{2}}(\log H)^2\max _{s\le R_0}\sum _{t\le R_0}\left| \sum _{n\sim L} e\left( 2\sqrt{\frac{n}{N}}\left( \sqrt{x_t}-\sqrt{x_s}\right) \right) \right| \nonumber \\&\quad +(HN)^{2\varepsilon }V^2R_0. \end{aligned}$$

As $H^{4\varepsilon }\gg (HN)^{2\varepsilon },$ we have

$$\begin{aligned}&R_0V^2H^{4\varepsilon }\nonumber \\&\quad \ll (HN)^{\frac{1}{2}+\varepsilon }L^{-\frac{1}{2}}(\log H)^2\max _{s\le R_0}\sum _{t\le R_0}\left| \sum _{n\sim L} e\left( 2\sqrt{\frac{n}{N}}\left( \sqrt{x_t}-\sqrt{x_s}\right) \right) \right| . \end{aligned}$$

(23)

Following the calculation on [32, p. 236], we get, by the Kuzmin–Landau inequality and the exponent pair $(4/18,\,11/18),$

$$\begin{aligned} \sum _{n\sim L}e\left( 2\sqrt{\frac{n}{N}}\left( \sqrt{x_t}-\sqrt{x_s}\right) \right)&\ll \frac{(NL)^{\frac{1}{2}}}{|\sqrt{x_t}-\sqrt{x_s}|}+\left( \frac{|\sqrt{x_t}-\sqrt{x_s}|}{(NL)^{1/2}}\right) ^{\frac{4}{18}}L^{\frac{11}{18}}\\&\ll \frac{(HNL)^{\frac{1}{2}}}{|{x_t}-{x_s}|}+\left( \frac{|\sqrt{x_t}-\sqrt{x_s}|}{(HNL)^{1/2}}\right) ^{\frac{4}{18}}L^{\frac{11}{18}}\\&\ll \frac{(HNL)^{\frac{1}{2}}}{|{x_t}-{x_s}|}+H^{-\frac{1}{9}}H_0^{\frac{2}{9}}L^{\frac{1}{2}}N^{-\frac{1}{9}}, \end{aligned}$$

where we have used the mean value theorem and the estimate $|\sqrt{x_t}-\sqrt{x_s}|\le H_0.$ Then we have

$$\begin{aligned}&\max _{s\le R_0}\sum _{t\le R_0}\left| \sum _{n\sim L}e\left( 2\sqrt{\frac{n}{N}}\left( \sqrt{x_t}-\sqrt{x_s}\right) \right) \right| \\&\quad \ll L+\max _{s\le R_0}\mathop {\mathop {\sum }\limits _{t\le R_0}}\limits _{t\ne s}\left| \sum _{n\sim L}e\left( 2\sqrt{\frac{n}{N}}\left( \sqrt{x_t}-\sqrt{x_s}\right) \right) \right| \\&\quad \ll L+\frac{(HNL)^{\frac{1}{2}}\log H}{V}+\frac{H_0^{\frac{2}{9}}L^{\frac{1}{2}}}{(HN)^{\frac{1}{9}}}R_0, \end{aligned}$$

where we use the fact that $\{x_s\}$ is V-spaced. Substituting this estimate into (23) and noting that $L\le M\ll HNV^{-2},$ we obtain

$$\begin{aligned} R_0V^2H^{4\varepsilon } \ll (HN)^\varepsilon (\log H)^2\left( HNV^{-1}(\log H)+H_0^{\frac{2}{9}}(HN)^{\frac{7}{18}}R_0\right) . \end{aligned}$$

Set $H_0=V^9(HN)^{-7/4}.$ Then

$$\begin{aligned} R_0\ll H^{1+\varepsilon }NV^{-3}. \end{aligned}$$

Now we divide the interval $[H,\,2H]$ into $O(1+H/H_0)$ subintervals of length not exceeding $H_0.$ In each interval of this type, the number of $x_\ell$’s is at most $O(H^{1+\varepsilon }NV^{-3}).$ So we have

$$\begin{aligned} R&\ll H^{1+\varepsilon }NV^{-3}\left( 1+H/H_0\right) \\&\ll H^\varepsilon \left( V^{-3}HN+V^{-12}H(HN)^{11/4}\right) . \end{aligned}$$

$\square$

5 Proof of Theorem 1

Put $H=T$ and

$$\begin{aligned} (HN)^{1/4+\varepsilon }\le 2^m=V\ll (HN)^{\frac{1}{3}+\varepsilon }. \end{aligned}$$

Obviously, there are $O(\log H)=O(\log T)$ choices for V. We denote by $\tau _j(V)$ the point with

$$\begin{aligned} \left| \mathcal {S}_f\left( \tau _j(V)\right) \right| =\sup \limits _{x\in [H+(j-1)V,\,H+jV]}\left| \mathcal {S}_f(x)\right| , \quad (j=1,\,2,\ldots ,[{H}/{V}]+1). \end{aligned}$$

By (11), we have

$$\begin{aligned} \left| \mathcal {S}_f\left( \tau _j(V)\right) \right| \ll (HN)^{\frac{1}{3}+\varepsilon }, \quad \text {for}\ x\sim H, \quad H\ge N^{1/2}. \end{aligned}$$

We consider those $\tau _{j_\ell }(V)$ with

$$\begin{aligned} H^{2\varepsilon }V\le \left| \mathcal {S}_f\left( \tau _{j_\ell }(V)\right) \right| <2H^{2\varepsilon }V. \end{aligned}$$

According to the parity of subindex $\ell ,$ we divide these points $\tau _{j_\ell }(V)$ into two groups of points $x_1^\pm (V),\,x_2^\pm (V),\ldots ,x^\pm _{R^\pm }(V),$ which satisfy

$$\begin{aligned} H^{2\varepsilon }V\le \left| \mathcal {S}_f\left( x_j^\pm (V)\right) \right| <2H^{2\varepsilon }V, \quad \left| x_i^\pm (V)-x_j^\pm (V)\right| \ge V \end{aligned}$$

for $i\ne j\le R^\pm =R^\pm (V).$ Then we have

$$\begin{aligned} \int _{H}^{2H}\left| \mathcal {S}_f(x)\right| ^A\mathrm{d}x\ll H (HN)^{\left( \frac{1}{4}+\varepsilon \right) A}+\sum _{V}V\sum _{j\le R^\pm }\left| \mathcal {S}_f\left( x_j^\pm (V)\right) \right| ^A. \end{aligned}$$

Therefore, by Theorem 4, we have

$$\begin{aligned} \int _H^{2H}\left| \mathcal {S}_f(x)\right| ^A\mathrm{d}x\ll H (HN)^{\left( \frac{1}{4}{+}\varepsilon \right) A}{+}H^{\varepsilon }\max _{V}\left( HNV^{A-2}{+}H(HN)^{11/4}V^{A-11}\right) , \end{aligned}$$

where the maximum is taken over

$$\begin{aligned} (HN)^{1/4+\varepsilon }\ll V\ll (HN)^{1/3+\varepsilon }. \end{aligned}$$

Since $2\le A\le 11$ and $N\le T=H,$ we have

$$\begin{aligned}&\int _T^{2T} \left| \mathcal {S}_f(x)\right| ^A\mathrm{d}x \ll T^{1+\varepsilon }(TN)^{A/4}+T^{\varepsilon }(TN)^{(A+1)/3}. \end{aligned}$$