2. The main results

Throughout the paper we use standard asymptotic notation. If f and g are complex-valued functions defined on some set, say $\mathcal D$ , then we write $f\ll g$ to signify that $|f(x)|\leqslant C|g(x)|$ for all $x\in\mathcal D$ for some implicit constant $C\in\mathbb R_+$ . The notation O(g) denotes a quantity that is $\ll g$ , and $f\asymp g$ means that both $f\ll g$ and $g\ll f$ . We write $f=o(g)$ if g never vanishes in $\mathcal D$ and $f(x)/g(x)\longrightarrow 0$ as $x\longrightarrow\infty$ . The notation $f=\Omega(g)$ means that $f\neq o(g)$ .

Our first main result shows the expected $\Omega$ -result for rationally additively twisted long sums when the denominator of the twist is sufficiently small. As far as the author is aware of, this is the first time $\Omega$ -results have been obtained for rationally additively twisted sums in a higher rank setting. Although it is not explicitly mentioned in what follows, we assume throughout the paper that the underlying Maass cusp form is a Hecke eigenform and normalised so that $A(1,1)=1$ . Our results should generalise for arbitrary $\mathrm{GL}_3$ Maass cusp forms, but we restrict ourselves to Hecke eigenforms to lighten the notation.

Theorem 2·1. Let $x\in[1,\infty[$ be sufficiently large and let k prime so that $k\ll x^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then

\begin{align*}\max_{h\in\mathbb Z_k^\times}\left|{\mathop{{\sum}^{\prime}}}_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\right|=\Omega\left(k^{1/2}x^{1/3}\right),\end{align*}

where the maximum is taken over all reduced residue classes modulo k.

Notice that $\ll_\varepsilon k^{1/2}x^{1/3+\varepsilon}$ is the conjectured upper bound in (1·2) for the range $k\ll x^{1/3}$ . Thus the result of Theorem 2·1 is essentially optimal in the sense of conjecture (1·2).

This follows immediately from the following mean square result.

Theorem 2·2. (see Theorem 10·2) Let $X\in[1,\infty[$ be sufficiently large and let k be a prime so that $k\ll X^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then

\begin{align*}\sum_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|{\mathop{{\sum}^{\prime}}}_{m\leq x}A(m,1)e\left(\frac{mh}k\right)\right|^2 \mathrm d x\gg X^{5/3}k^2.\end{align*}

As sketched in the introduction, the proof of the previous result relies on Voronoi identities for both $\widetilde A_1(x;\,h/k)$ and $\widetilde A_2(x;\,h/k)$ , but in Section 9 we shall also present an alternative proof for Theorem 2·1 that sidesteps the use of a Voronoi identity for $\widetilde A_1(x;\,h/k)$ .

The third main theorem establishes a non-trivial $\Omega$ -result for short sums of certain lengths.

Theorem 2·3. Let $x\in[1,\infty[$ be sufficiently large, let k prime and suppose that $k^{3/2}x^{1/2+\delta}\ll\Delta\ll kx^{2/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then

\begin{align*}\max_{h\in\mathbb Z_k^\times}\left|{\mathop{{\sum}^{\prime}}}_{x\leqslant m\leqslant x+\Delta}A(m,1)\,e\!\left(\frac{mh}k\right)\right|=\Omega\left(\Delta x^{-1/3}k^{-1/2}\right).\end{align*}

This is a direct consequence of the following mean square result.

Theorem 2·4 (see Theorem 13·1). Let $X\in[1,\infty[$ be sufficiently large and let k be a prime so that $k^{3/2}X^{1/2+\delta}\ll\Delta\ll kX^{2/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then

\begin{align*}\sum_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|{\mathop{{\sum}^{\prime}}}_{x\leqslant m\leqslant x+\Delta} A(m,1)\,e\!\left(\frac{mh}k\right)\right|^2\,\mathrm d x\gg\Delta^2X^{1/3}.\end{align*}

Notice that the result of Theorem 2·3 is of the right order of magnitude in the sense of the conjectural bound (1·2), up to $X^\delta$ , when $\Delta\asymp kX^{2/3-\delta}$ . However, for smaller $\Delta$ this $\Omega$ -result is farther away from the expected upper bound.

The restriction to prime denominators is made mainly for simplicity so we could handle the sum over the divisors of k arising from the Voronoi identities for Riesz weighted exponential sums efficiently. Since we are interested in $\Omega$ -results this is not such a serious restriction. Note also that theorems above hold in the case $k=1$ , which can be easily seen by making simple cosmetic modifications to the proofs.

Likewise, we are unable to obtain $\Omega$ -results for an arbitrary numerator h coprime to k. The averaging over the reduced residue classes modulo k has to be included in order to be able to exract main terms when evaluating moments of the sums $\widetilde A_a(x;\,h/k)$ . It seems that this cannot be achieved by our methods unless we perform such extra averaging. The presence of this is not so serious as we are mainly interested in limitations for obtaining bounds (1·2). We also point out that the primed sums can be replaced with ordinary sums if one assumes that $\vartheta\lt1/6$ (which is expected to be true, but currently we only know that $\vartheta\leq 5/14$ ).

Finally, as a by-product of our analysis we improve the best known upper bound for rationally additively twisted sums with a small denominator.

Theorem 2·5. Let $x\in[1,\infty[$ , and let h and k be coprime integers so that $1\leqslant k\ll x^{1/3}$ . Then

\begin{align*}\sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\ll_{\varepsilon} k^{3/4}x^{1/2+\vartheta/2+\varepsilon}.\end{align*}

Notice that this theorem does not require the assumption for the denominator k being a prime. Theorem 2·5 improves the best previously known upper bound [ Reference Jääsaari and Vesalainen22 , corollary 3], which gives for instance an upper bound $\ll_\varepsilon k^{3/4}x^{1/2+\vartheta/2+\varepsilon}+k^{9/8+3\vartheta/4}x^{1/4+3\vartheta^2/2+3\vartheta/4+\varepsilon}$ in the range $1\leq k\ll x^{1/3}$ under the assumption $\vartheta\leq 1/6$ .

This paper is organised as follows. In Sections 4 and 6 we gather basic facts concerning rationally additively twisted L-functions and linear exponential sums on $\mathrm{GL}_3$ , respectively. In Section 5 we introduce Meijer G-functions relevant for the present work and study their asymptotic behaviour. Then we proceed to derive Voronoi identities for the Riesz weighted sums $\widetilde A_a(x;\,h/k)$ in the next two Sections 7 (for $a\geq 2)$ and 8 (for $a=1$ ). In Section 9 we evaluate the mean square of $\widetilde A_2(x;\,h/k)$ and give the first proof for Theorem 2·1. In the next section the mean square of $\widetilde A_1(x;\,h/k)$ is evaluated and Theorem 2·2 (as well as Theorem 2·1) is deduced as a consequence. Theorem 2·5 is proved in Section 11. The final two sections are devoted to the proofs of Theorems 2·3 and 2·4.

3. Notation

The letter $\varepsilon$ denotes a positive real number, whose value can be fixed to be arbitrarily small, and whose value can be different in different instances in a proof. All implicit constants are allowed to depend on $\varepsilon$ , on the implicit constants appearing in the assumptions of theorem statements, and on anything that has been fixed. When necessary, we will use subscripts $\ll_{\alpha,\beta,...},O_{\alpha,\beta,...}$ , etc. to indicate when implicit constants are allowed to depend on quantities $\alpha,\beta,...$

As usual, complex variables are written in the form $s=\sigma+it$ with $\sigma$ and t real, and we write e(x) for $e^{2\pi ix}$ . The subscript in the integral $\int_{(\sigma)}$ means that we integrate over the vertical line $\Re(s)=\sigma$ . When $h\in\mathbb Z$ and $k\in\mathbb Z_+$ are coprime, then $\overline h$ denotes an integer such that $h\overline h\equiv 1\,(\text{mod }k)$ . We also write $\langle \cdot\rangle$ for $(1+|\cdot|^2)^{1/2}$ . The notation $\sum'_{m\leqslant x}$ means the sum up to x with the last term halved if x is an integer. Furthermore, ${\mathop{{\sum}^{\prime}}}_{x\leq m\leq x+\Delta}={\mathop{{\sum}^{\prime}}}_{m\leq x+\Delta}-{\mathop{{\sum}^{\prime}}}_{m\leq x-1}$ . The notation d(n) denotes the ordinary divisor function. Let us furthermore write

\begin{align*}1_{a\equiv b\,(\ell)}\,:\!=\,\begin{cases}1&\text{if }a\equiv b\pmod\ell\\0&\text{otherwise}.\end{cases}\end{align*}

Finally, for a positive integer k let us define the following averaging operator. Given a function $f\colon\mathbb Z_k^\times\longrightarrow\mathbb C$ , we set

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{x\in\mathbb Z_k^\times}f(x)\,:\!=\,\frac1{\varphi(k)}\sum_{x\in\mathbb Z_k^\times}f(x),\end{align*}

where $\varphi$ is Euler’s totient function, and $\mathbb Z_k^\times$ denotes the reduces residue classes modulo k, or a set of representatives of them.

4. Rationally additively twisted L-functions of $\mathrm{GL}_3$ Maass cusp forms

We shall derive our Voronoi identities from the additively twisted L-function of the Maass cusp form in question. To this end, we quote some of its properties from section 3 of [ Reference Goldfeld and Li12 ]. We fix a Hecke–Maass cusp form for $\mathrm{SL}_3(\mathbb Z)$ with Fourier coefficients $A(m_1,m_2)$ . Also, let us choose coprime integers h and k with k positive, as well as an index $j\in\left\{0,1\right\}$ . Then we have the rationally additively twisted version of the corresponding Godement–Jacquet L-function given by

\begin{equation*}L_j\!\left(s+j,\frac{h}{k}\right)\,:\!=\,\sum_{m=1}^\infty\frac{A(m,1)}{m^s}\left(e\!\left(\frac{mh}k\right)+\left({-}1\right)^je\!\left({-}\frac{mh}k\right)\right)\end{equation*}

for $s\in\mathbb C$ with $\Re(s)\gt1$ . This has an entire analytic extension and satisfies the functional equation

\begin{equation*}L_j\!\left(s+j,\frac{h}{k}\right)=i^{-j}\,k^{-3s+1}\,\pi^{3s-3/2}\,G_j(s+j)\,\widetilde L_j\!\left(1-s-j,\frac{\overline h}k\right)\!.\end{equation*}

The $\Gamma$ -factors are clumped together into the factor $G_j(s+j)$ given by

\begin{equation*}G_j(s+j)\,:\!=\,\frac{\displaystyle{\Gamma\!\left(\frac{1-s+j+\alpha}2\right)\Gamma\!\left(\frac{1-s+j+\beta}2\right)\Gamma\!\left(\frac{1-s+j+\gamma}2\right)}}{\displaystyle{\Gamma\!\left(\frac{s+j-\alpha}2\right)\Gamma\!\left(\frac{s+j-\beta}2\right)\Gamma\!\left(\frac{s+j-\gamma}2\right)}},\end{equation*}

where in turn the complex constants $\alpha$ , $\beta$ and $\gamma$ are the Langlands parameters of the underlying Maass cusp form. We know from [ Reference Goldfeld11 , propositions 6.3.1 and 12·1·9] and [ Reference Goldfeld and Li12 , equation (3·30)] that

\begin{equation*}\alpha+\beta+\gamma=0 \qquad\text{and}\qquad\max\left\{\left|\Re(\alpha)\right|,\left|\Re(\beta)\right|,\left|\Re(\gamma)\right|\right\}\leqslant\frac12.\end{equation*}

Finally, the Dirichlet series on the right-hand side of the functional equation is given by

\begin{equation*}\widetilde L_j\!\left(1-s-j,\frac{\overline h}k\right)\,:\!=\,\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{d^{1-2s}\,m^{1-s}}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\end{equation*}

for $\Re(s)\lt0$ , and by an entire analytic continuation elsewhere. Here $S(a,b;\,c)$ is the usual Kloosterman sum.

From Stirling’s formula, the functional equation above, and the Phragmén–Lindelöf principle, we obtain the convexity estimate

\begin{equation*}L_j\!\left(s+j,\frac{h}{k}\right)\ll_{\varepsilon} k^{3(1+\delta-\sigma)(1+\varepsilon)/2}\,\langle t\rangle^{3(1+\delta-\sigma)/2}\end{equation*}

in the strip $-\delta\leqslant\sigma\leqslant 1+\delta$ (see [ Reference Jääsaari and Vesalainen22 , p. 262]), where $\delta\in\mathbb R_+$ is arbitrary and fixed. Furthermore, it follows from this that, for each $j\in\left\{0,1\right\}$ and every $a\in\mathbb Z_+\cup\left\{0\right\}$ , as well as for any coprime integers h and k with k positive,

\begin{equation*}L_j\!\left({-}a+j,\frac{h}{k}\right)\ll_{a,{\varepsilon}} k^{3\left(1+2a\right)/2+\varepsilon}.\end{equation*}

In particular,

(4·1) \begin{align}L_j\!\left(0+j,\frac{h}{k}\right)\ll_{\varepsilon} k^{3/2+\varepsilon}, \quad L_j\!\left( -1+j,\frac{h}{k}\right)\ll_{\varepsilon} k^{9/2+\varepsilon},\quad L_j\!\left({-}2+j,\frac{h}{k}\right)\ll_{\varepsilon} k^{15/2+\varepsilon}\end{align}

and

\begin{equation*}L_j\!\left({-}3+j,\frac{h}{k}\right)\ll_{\varepsilon} k^{21/2+\varepsilon}.\end{equation*}

When summing by parts we will repeatedly use the following consequences of the Rankin–Selberg theory saying that

(4·2) \begin{align}\sum_{m\leqslant x}\left|A(d,m)\right|\ll_{\varepsilon} d^{\vartheta+\varepsilon}x\qquad\text{and}\qquad\sum_{m\leqslant x}\left|A(d,m)\right|^2\ll_{\varepsilon} d^{2\vartheta+\varepsilon}x,\end{align}

uniformly in $d\in\mathbb Z_+$ and $x\in\left[1,\infty\right[$ (see e.g. [ Reference Jääsaari and Vesalainen23 ]).

5. On certain Meijer G-functions and their asymptotics

Applying Perron’s formula to the additively twisted L-function, shifting the line of integration to the left, and then applying the additively twisted functional equation leads to integrals involving quotients of gamma functions. We could derive the asymptotics we need in the same way as similar integrals were treated in earlier works, e.g [ Reference Ernvall-Hytönen, Jääsaari and Vesalainen7 , Reference Ivić17 , Reference Jääsaari and Vesalainen22 , Reference Li30 , Reference Miller34 ]. However, here we prefer to follow the approach of [ Reference Czarnecki4 ], which observed that these types of integrals are actually Meijer G-functions, and thus one obtains the relevant asymptotics directly from the well-known asymptotic properties of the latter. Our main references for Meijer G-functions are [ Reference Luke31 , Reference Luke32 ].

Let $\alpha$ , $\beta$ and $\gamma$ be the complex numbers as in the functional equation of the additively twisted L-function, and let $y\in\mathbb Z_+$ , $a\in\mathbb Z_+$ and $j\in\left\{0,1\right\}$ . We define $\mathscr J_{a,\,j}(y)$ to be the specific Meijer G-function of interest to us as

\begin{align*}&\mathscr J_{a,\,j}(y)\\&\,:\!=\,G^{5,0}_{2,8}\!\left(\!y;\,\begin{array}{l}\displaystyle{1,\frac12}\\[9pt]\displaystyle{\frac{1+j+\alpha}2,\frac{1+j+\beta}2,\frac{1+j+\gamma}2,-\frac a2,\frac12-\frac a2,1-\frac{j-\alpha}2,1-\frac{j-\beta}2,1-\frac{j-\gamma}2}\end{array}\!\right)\\[2mm]&=\frac1{2\pi i}\int\limits_{\mathscr C}Q(s)\,y^s\,\mathrm ds,\end{align*}

where Q(s) denotes the $\Gamma$ -quotient

\begin{equation*}Q(s)\,:\!=\,\frac{\displaystyle{\Gamma\!\left(\frac{1+j+\alpha}2-s\right)\Gamma\!\left(\frac{1+j+\beta}2-s\right)\Gamma\!\left(\frac{1+j+\gamma}2-s\right)\Gamma\!\left({-}\frac a2-s\right)\Gamma\!\left(\frac12-\frac a2-s\right)}}{\displaystyle{\Gamma\!\left(s+\frac{j-\alpha}2\right)\Gamma\!\left(s+\frac{j-\beta}2\right)\Gamma\!\left(s+\frac{j-\gamma}2\right)\Gamma\!\left(1-s\right)\Gamma\!\left(\frac12-s\right)}}.\end{equation*}

The contour of integration $\mathscr C$ can be, say, a simple polygonal chain, which begins at $\sigma_0-i\infty$ with a vertical half-line with abscissa $\sigma_0$ satisfying $\sigma_0\gt1/4-a/6$ , ends at $\sigma_0+i\infty$ with another vertical half-line with the same abscissa $\sigma_0$ , and having the property that all the poles of the $\Gamma$ -factors in the numerator lie to the right of $\mathscr C$ [ Reference Luke32 , subsect. 5·3·1]. For definiteness, we may select $\mathscr C$ to be the contour $\mathscr C(\sigma_0,\sigma_1,\Lambda)$ consisting of line segments connecting the points $\sigma_0-i\infty$ , $\sigma_0-i\Lambda$ , $\sigma_1-i\Lambda$ , $\sigma_1+i\Lambda$ , $\sigma_0+i\Lambda$ and $\sigma_0+i\infty$ , in this order, where the second abscissa $\sigma_1$ is a real number satisfying $\sigma_1\lt-a/2$ and $\Lambda$ is a positive real number such that $\Lambda$ is larger than the imaginary parts of $\alpha/2$ , $\beta/2$ and $\gamma/2$ .

We will not repeat the above parameters and instead write the above expression as $G^{5,0}_{2,8}(y;\,\ldots)$ with the understanding that the parameters are exactly as above. The asymptotic we wish to use is a special case of the case (4) of [ Reference Luke31 , theorem 2, section 5·10]. In our special case the result reads as follows.

Lemma 5·1. Let $y\in\mathbb R_+$ be larger than some arbitrary fixed positive real constant, and let $j\in\left\{0,1\right\}$ and $a\in\mathbb Z_+$ be fixed. Then the above function $\mathscr J_{a,\,j}(y)$ has the asymptotic expansion

\begin{equation*}\mathscr J_{a,\,j}(y)\sim-\frac{1}{2\,\sqrt{3\pi}}\sum_\pm\exp\!\left(\pm6i\,y^{1/6}\pm\frac{\pi i}2\left(a+j\right)\right)y^{\left(1-a\right)/6}\sum_{\ell=0}^\infty M_\ell\,i^{\pm\ell}\,y^{-\ell/6},\end{equation*}

for some complex coefficients $M_0$ , $M_1$ , $M_2$ , for which $M_0=1$ , and which otherwise depend on $\alpha$ , $\beta$ , $\gamma$ , a, and j. In particular,

\begin{equation*}\mathscr J_{a,\,j}(y)=-\frac1{\sqrt{3\pi}}\,y^{(1-a)/6}\,\cos\!\left(6\,y^{1/6}+\frac\pi2\left(a+j\right)\right)+O(y^{-a/6}).\end{equation*}

Proof. We apply the asymptotics (4) of [ Reference Luke31 , theorem 2, section 5·10] which says that the Meijer G-function $G^{5,0}_{2,8}(y;\,\ldots)$ has the asymptotic expansion

\begin{equation*}G^{5,0}_{2,8}(y;\,\ldots)\sim A^{5,0}_8\,H_{2,8}(y\,e^{3\pi i})+\smash{\overline A}^{\,5,0}_8\,H_{2,8}(y\,e^{-3\pi i}),\end{equation*}

as $y\longrightarrow\infty$ . In the notation of [ Reference Luke31 , section 5·7], we have

\begin{equation*}\sigma=8-2=6\qquad\text{and}\qquad\nu=8-5-0=3.\end{equation*}

We observe that our parameters satisfy the conditions (A) and (B) of [ Reference Luke31 , section 5·7] vacuously as in our case $n=0$ , and they satisfy the condition (C) simply because $1-1/2=1/2\not\in\mathbb Z$ . In the notation of [ Reference Luke31 , theorem 5, section 5·7], we have

\begin{align*}\Xi_1 &=\frac{1+j+\alpha}2+\frac{1+j+\beta}2+\frac{1+j+\gamma}2-\frac a2+\frac12-\frac a2+1{-}\frac{j-\alpha}2+1{-}\frac{j-\beta}2+1{-}\frac{j-\gamma}2\\&=5-a,\end{align*}

and

\begin{equation*}\Lambda_1=1+\frac12=\frac32,\end{equation*}

so that

\begin{equation*}\theta=\frac1\sigma\left(\frac12\left(1-\sigma\right)+\Xi_1-\Lambda_1\right)=\frac{1-a}6.\end{equation*}

The asymptotic behaviour of $H_{2,8}$ is thus given by

\begin{equation*}H_{2,8}(y\,e^{\pm3\pi i})\sim i^{\pm\left(1-a\right)}\,\frac{\left(2\pi\right)^{5/2}}{\sqrt6}\,\exp(\mp6i\,y^{1/6})\,y^{(1-a)/6}\sum_{\ell=0}^\infty M_\ell\,i^{\mp\ell}\,y^{-\ell/6},\end{equation*}

as $y\longrightarrow\infty$ , for some complex constants $M_0$ , $M_1$ , $M_2,\dots$ . In particular, $M_0=1$ . Finally, in the notation of [ Reference Luke31 , section 5·9·2],

\begin{equation*}A^{5,0}_8=\left({-}1\right)^\nu\left(2\pi i\right)^{-\nu}\exp\!\left(i\pi\left(0-1+\frac{j-\alpha}2-1+\frac{j-\beta}2-1+\frac{j-\gamma}2\right)\right)=\left(2\pi\right)^{-3}i^{1-j},\end{equation*}

and similarly,

\begin{equation*}\smash{\overline A}^{\,5,0}_8=\left(2\pi\right)^{-3}i^{j-1}.\end{equation*}

The Meijer G-function appears from the integral in Perron’s formula. The following lemma connects the two.

Lemma 5·2. Let $a\in\mathbb Z_+$ , let $y\in\mathbb R_+$ , and let $\sigma_0$ and $\sigma_1$ be real numbers such that $\sigma_0\gt1/4-a/6$ and $\sigma_1\lt-a/2$ . Then

\begin{equation*}\frac1{2\pi i}\int\limits_{\mathscr C(2\sigma_0,2\sigma_1,2\Lambda)}\frac{G_j(s+j)\,y^s\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}=-\left({-}2\right)^{-a}\mathscr J_{a,\,j}(y^2).\end{equation*}

Proof. Observing

\begin{equation*}\frac1{s\left(s+1\right)\cdots\left(s+a\right)}=\frac{\displaystyle{\left({-}\frac12\right)^{a+1}\Gamma\!\left({-}\frac{s}{2}-\frac a2\right)\Gamma\!\left({-}\frac{s}{2}-\frac a2+\frac12\right)}}{\displaystyle{\Gamma\!\left({-}\frac{s}{2}+1\right)\Gamma\!\left({-}\frac s2+\frac12\right)}},\end{equation*}

we see that

\begin{align*}\frac1{2\pi i}\int\limits_{\mathscr C(2\sigma_0,2\sigma_1,2\Lambda)}\frac{G_j(s+j)\,y^s\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}&=-\left({-}2\right)^{-a}\frac1{2\pi i}\int\limits_{\mathscr C(2\sigma_0,2\sigma_1,2\Lambda)}Q\!\left(\frac s2\right)\big(y^2\big)^{s/2}\frac{\mathrm ds}2\\&=-\left({-}2\right)^{-a}\mathscr J_{a,\,j}(y^2),\end{align*}

as desired.

In order to derive Voronoi identities with Riesz weights in the $a=1$ case, we wish to differentiate the formula of case $a=2$ termwise. The following lemma allows this.

Lemma 5·3. Let $a\in\mathbb Z_+$ , let $j\in\left\{0,1\right\}$ , and let $y\in\mathbb R_+$ . Then

\begin{equation*}\frac{\mathrm d}{\mathrm dy}\left(y^{a+1}\mathscr J_{a+1,j}(y^2)\right)=-2\,y^a\mathscr J_{a,\,j}(y^2).\end{equation*}

Proof. Let $\sigma_0$ and $\sigma_1$ be real numbers such that $\sigma_0\gt1/4-a/6$ and $\sigma_1\lt-a/2-1/2$ . Using Lemma 5·2 twice we may compute

\begin{align*}&\frac{\mathrm d}{\mathrm dy}\left(y^{a+1}\mathscr J_{a+1,j}(y^2)\right)=\frac{\mathrm d}{\mathrm dy}\left({-}\left({-}2\right)^{a+1}\frac1{2\pi i}\int\limits_{\mathscr C(2\sigma_0,2\sigma_1,2\Lambda)}\frac{G_j(s+j)\,y^{s+a+1}\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)\left(s+a+1\right)}\right)\\&\qquad=\left({-}2\right)\cdot\left({-}1\right)\left({-}2\right)^{a}\frac1{2\pi i}\int\limits_{\mathscr C(2\sigma_0,2\sigma_1,2\Lambda)}\frac{G_j(s+j)\,y^{s+a}\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}=-2\,y^a\mathscr J_{a,\,j}(y^2),\end{align*}

which completes the proof.

6. Some known estimates for long linear exponential sums

We shall need estimates for long linear exponential sums involving Fourier coefficients of a $\mathrm{GL}_3$ Maass cusp form. The following is [ Reference Miller34 , theorem 1·1].

Lemma 6·1. Let $x\in\left[1,\infty\right[$ . Then

\begin{equation*}\sum_{m\leqslant x}A(m,1)\,e(m\alpha)\ll_{\varepsilon} x^{3/4+\varepsilon}\end{equation*}

uniformly in $\alpha\in\mathbb R$ .

On the other side of our Voronoi identities coefficients A(d, m), where d divides the denominator of the twist, will also appear. Fortunately, it is easy to extend Miller’s estimate for these coefficients with uniformity over d.

Lemma 6·2. Let $x\in\left[1,\infty\right[$ , $d\in\mathbb Z_+$ and $\alpha\in\mathbb R$ . Then

\begin{equation*}\sum_{m\leqslant x}A(m,d)\,e(m\alpha)\ll_{\varepsilon} d^{\vartheta+\varepsilon}\,x^{3/4+\varepsilon},\end{equation*}

uniformly in both $\alpha$ and d.

Proof. As the underlying form is a normalised Hecke eigenform we have

\begin{equation*}A(m,d)=\sum_{\ell\mid\left(d,m\right)}\mu\!\left(\ell\right)A\!\left(1,\frac d\ell\right)A\!\left(\frac{m}{\ell},1\right)\end{equation*}

for any $m,d\in\mathbb Z_+$ , where $\mu$ is the Möbius function. Using this, we may estimate

\begin{align*}\sum_{m\leqslant x}A(m,d)\,e(m\alpha)&=\sum_{m\leqslant x}\sum_{\ell\mid\left(d,m\right)}\mu\!\left(\ell\right)A\!\left(1,\frac d\ell\right)A\!\left(\frac{m}{\ell},1\right)e(m\alpha)\\&=\sum_{\ell\mid d}\mu\!\left(\ell\right)A\!\left(1,\frac d\ell\right)\sum_{m\leqslant x/\ell}A(m,1)\,e\!\left(m\ell\alpha\right)\\&\ll_\varepsilon\sum_{\ell\mid d}\left(\frac d\ell\right)^{\vartheta+\varepsilon}\left(\frac{x}{\ell}\right)^{3/4+\varepsilon}\ll_{\varepsilon} d^{\vartheta+\varepsilon}\,x^{3/4+\varepsilon},\end{align*}

completing the proof.

When $\alpha$ is a fraction with a small denominator, we have more precise estimates, which depend on $\vartheta$ . The following is [ Reference Jääsaari and Vesalainen22 , corollary 3].

Lemma 6·3. Let $x\in\left[1,\infty\right[$ , and let h and k be coprime integers such that $1\leqslant k\ll x^{2/3}$ . Then

\begin{equation*}\sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\ll_{\varepsilon} k^{1/2+\varepsilon}\,x^{2/3}+k\,x^{1/3+\vartheta+\varepsilon}.\end{equation*}

If in addition $\vartheta\leqslant1/3$ and $k\ll x^{2/3-2\vartheta}$ , then

\begin{equation*}\sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\ll_{\varepsilon} k^{3/4}\,x^{1/2+\vartheta/2+\varepsilon}+k^{9/8+3\vartheta/4}\,x^{1/4+3\vartheta^2/2+3\vartheta/4+\varepsilon}.\end{equation*}

In particular, if $\vartheta=0$ and $k\ll x^{2/3}$ , then

\begin{equation*}\sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\ll_{\varepsilon} k^{3/4}\,x^{1/2+\varepsilon}.\end{equation*}

7. Additively twisted Voronoi identities for Riesz means: the case $a\geqslant2$

In this section we study the modified Riesz weighted sums

\begin{align*}\widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right)&\,:\!=\,\frac1{a!}\mathop{{\sum}^{\prime}}\limits_{m\leqslant x}A(m,1)\left(e\!\left(\frac{mh}k\right)+\left({-}1\right)^je\!\left({-}\frac{mh}k\right)\right)\left(x-m\right)^a\\&\qquad\qquad\qquad-\sum_{0\leqslant\nu\leqslant a}\frac{\left({-}1\right)^\nu x^{a-\nu}}{\nu!\left(a-\nu\right)!}\,L_j\!\left({-}\nu+j,\frac{h}{k}\right),\end{align*}

where $x\in\left[0,\infty\right[$ , $j\in\left\{0,1\right\}$ , $a\in\mathbb Z_+\cup\left\{0\right\}$ , and h and k are coprime integers with k positive. Recall the key relation (see (1·5) above)

\begin{align*}\int\limits_x^t \widetilde A_{a,\,j}\left(u;\,\frac{h}{k}\right)\,\mathrm d u=\widetilde A_{a+1,j}\left(t;\,\frac{h}{k}\right)-\widetilde A_{a+1,j}\left(x;\,\frac{h}{k}\right)\end{align*}

for the Riesz means that we shall use repeatedly. Our goal in this section is to derive Voronoi-type identities for these sums when $a\geqslant2$ .

The classical Perron’s formula, which is e.g. [ Reference Brüdern2 , theorem 1·4·4], has the following generalisation for Riesz means (see e.g. [ Reference Montgomery and Vaughan36 , chapter 5]), which is the starting point for proving our Voronoi-type identities.

Lemma 7·1. Let $\sigma\in\mathbb R_+$ , and let $c\colon\mathbb Z_+\longrightarrow\mathbb C$ be a sequence such that the Dirichlet series $\sum_{n=1}^\infty c(n)/n^\sigma$ converges absolutely. Then, for $x\in\mathbb R_+$ , and for a non-negative integer a, we have

\begin{equation*}\frac1{a!}\mathop{{\sum}^{\prime}}\limits_{n\leqslant x}\,c(n)\left(x-n\right)^a=\frac1{2\pi i}\int\limits_{(\sigma)}\left(\sum_{n=1}^\infty\frac{c(n)}{n^s}\right)\frac{x^{s+a}\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)},\end{equation*}

where the integration is over the vertical line where the real part is $\sigma$ . When $a=0$ , the integral should be understood as the limit of $\int_{\sigma-iT}^{\sigma+iT}$ as $T\longrightarrow\infty$ .

The main result of this section is the following proposition.

Proposition 7·2. Let $x\in\mathbb R_+$ , $j\in\left\{0,1\right\}$ , $a\in\left\{2,3,\ldots\right\}$ , and let h and k be coprime integers with k positive. Then

\begin{align*}& \widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right)=i^{-j}\,\pi^{-3/2}\,k\,x^a\left({-}1\right)^{a+1}2^{-a}\\& \times\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\mathscr J_{a,\,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right)\!.\end{align*}

Proof. Let us first fix some $\delta\in\left]0,1/6\right[$ . Perron’s formula for Riesz means tells us that

\begin{align*}&\frac1{a!}{\mathop{{\sum}^{\prime}}}_{m\leqslant x}A(m,1)\left(e\!\left(\frac{mh}k\right)+\left({-}1\right)^je\!\left({-}\frac{mh}k\right)\right)\left(x-m\right)^a \\&=\frac1{2\pi i}\int\limits_{\left(1+\delta\right)}L_j\!\left(s+j,\frac{h}{k}\right)\frac{x^{s+a}\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}.\end{align*}

Let $\mathscr L$ be the polygonal chain $\mathscr C({-}\delta,-a-\delta,2\Lambda)$ connecting the points $-\delta-i\infty$ , $-\delta-2i\Lambda$ , $-a-\delta-2i\Lambda$ , $-a-\delta+2i\Lambda$ , $-\delta+2i\Lambda$ and $-\delta+i\infty$ , in this order, where $\Lambda$ is a fixed positive real number larger than the imaginary parts of $\alpha/2$ , $\beta/2$ and $\gamma/2$ . We shift the contour of integration from the vertical line $\Re(s)=1+\delta$ to $\mathscr L$ . This is justified by the convexity bound for $L_j(s+j,h/k)$ and the assumption $a\geqslant 2$ . This leads to residue terms from the simple poles at the points $-a$ , $-a+1$ , 0. For each $\nu\in\left\{0,1,\ldots,a\right\}$ we get a residue term

\begin{equation*}\frac{\left({-}1\right)^\nu\,x^{a-\nu}}{\nu!\left(a-\nu\right)!}\,L_j\!\left({-}\nu+j,\frac{h}{k}\right).\end{equation*}

Thus, we obtain

\begin{equation*}\widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right)=\frac1{2\pi i}\int\limits_{\mathscr L}L_j\!\left(s+j,\frac{h}{k}\right)\frac{x^{s+a}\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}.\end{equation*}

Applying the additively twisted functional equation gives

\begin{equation*}\widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right)=\frac1{2\pi i}\int\limits_{\mathscr L}i^{-j}\,k^{-3s+1}\,\pi^{3s-3/2}\,G_j(s+j)\,\widetilde L_j\!\left(1-s-j;\,\frac{\overline h}k\right)\,\frac{x^{s+a}\,\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}.\end{equation*}

Writing the L-function as a Dirichlet series and switching the order of summation and integration yields

\begin{align*}\widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right) & =i^{-j}\,k\,\pi^{-3/2}\,x^a\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\\& \times\frac1{2\pi i}\int\limits_{\mathscr L}G_j(s+j)\left(\frac{\pi^3\,d^2\,mx}{k^3}\right)^s\frac{\mathrm ds}{s\left(s+1\right)\cdots\left(s+a\right)}.\end{align*}

Invoking Lemma 5·2 finishes the proof.

Let us then study the Riesz means of order $a\in\mathbb Z_+\cup\{0\}$ defined as

(7·1) \begin{align}\widetilde A_a\!\left(x;\,\frac{h}{k}\right)&\,:\!=\,\frac12\left(\widetilde A_{a,0}\!\left(x;\,\frac{h}{k}\right)+\widetilde A_{a,1}\!\left(x;\,\frac{h}{k}\right)\right) \nonumber \\&=\frac1{a!}\mathop{{\sum}^{\prime}}\limits_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)(x-m)^a \nonumber \\&\qquad\qquad\qquad-\frac12\sum_{0\leqslant\nu\leqslant a}\frac{\left({-}1\right)^\nu x^{a-\nu}}{\nu!\left(a-\nu\right)!}\left(L_0\!\left({-}\nu,\frac{h}{k}\right)+L_1\!\left({-}\nu+1,\frac{h}{k}\right)\right),\end{align}

where $x\in\left[0,\infty\right[$ , and h and k are coprime integers with k positive. We observe at once that these sums have the pleasant property that

\begin{align*}\int\limits_0^x\!\widetilde A_a\!\left(t;\,\frac{h}{k}\right)\mathrm dt&=\widetilde A_{a+1}\!\left(x;\,\frac{h}{k}\right)-\widetilde A_{a+1}\!\left(0;\,\frac{h}{k}\right)\\&=\widetilde A_{a+1}\!\left(\!x;\,\frac{h}{k}\right)\!+\frac{\left({-}1\right)^{a+1}}{2\cdot\left(a+1\right)!}\!\left(L_0\!\left({-}\left(a+1\right);\,\frac{h}{k}\right)+L_1\!\left({-}\left(a+1\right)+1;\,\frac{h}{k}\right)\!\right)\!.\end{align*}

This allows us to later relate Riesz means of different orders, and in particular to turn our better control of higher order Riesz means to information about lower order Riesz means. Of course, similar identities hold for $\widetilde A_{a,\,j}$ with $j\in\left\{0,1\right\}$ .

The previous proposition, together with Lemma 5·1, immediately yields the following corollary using Weil’s boundFootnote ⁴ $|S(a,b;\,k)|\leq d(k)k^{1/2}(a,b,k)^{1/2}$ .

Corollary 7·3. Let $x\in\left[1,\infty\right[$ , let $a\geqslant 2$ be an integer, and let h and k be coprime integers with $1\leqslant k\ll x^{1/3}$ . Then

\begin{align*} & \widetilde A_a\!\left(x;\,\frac{h}{k}\right)\\& =\frac{\left({-}1\right)^ak^a\,x^{(2a+1)/3}}{(2\pi)^{a+1}\sqrt3}\sum_{d|k}\!\frac1{d^{(2a+1)/3}}\sum_{m=1}^\infty\!\frac{A(d,m)}{m^{(a+2)/3}}\sum_\pm\! i^{\pm a}\,S\!\left(\overline h,\pm m;\,\frac kd\right)e\!\left(\!\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}{k}\!\right)\\& \quad +O\left(k^{(2a+3)/2}\,d(k)\,x^{2a/3}\right).\end{align*}

We also record here the following upper bound.

Corollary 7·4. Let $x\in\left[1,\infty\right[$ , $j\in\left\{0,1\right\}$ , $a\in\left\{2,3,\ldots\right\}$ , and let h and k be coprime integers with $1\leqslant k\ll x^{1/3}$ . Then

\begin{equation*}\widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right)\ll k^{(2a+1)/2}\,d(k)\,x^{(2a+1)/3}.\end{equation*}

In particular,

\begin{equation*}\widetilde A_{2,j}\!\left(x;\,\frac{h}{k}\right)\ll k^{5/2}\,d(k)\,x^{5/3}\qquad\text{and}\qquad\widetilde A_{3,j}\!\left(x;\,\frac{h}{k}\right)\ll k^{7/2}\,d(k)\,x^{7/3}.\end{equation*}

Proof. Estimating the infinite series in Proposition 7·2 by absolute values using the asymptotics from Lemma 5·1 and Weil’s bound gives

\begin{align*}\widetilde A_{a,\,j}\!\left(x;\,\frac{h}{k}\right)&\ll k\,x^a\sum_{d\mid k}\sum_{m=1}^\infty\frac{\left|A(d,m)\right|}{dm}\left(\frac kd\right)^{1/2}d(k)\left(\frac{d^{2/3}\,m^{1/3}\,x^{1/3}}k\right)^{1-a}\\&\ll k^{a+1/2}\,d(k)\,x^{2a/3+1/3}\sum_{d\mid k}d^{\vartheta-5/6-2a/3}\ll k^{a+1/2}\,d(k)\,x^{2a/3+1/3},\end{align*}

as required.

8. Additively twisted Voronoi identities for Riesz means: the case $a=1$

In the process of deriving a truncated Voronoi identity for $\widetilde A_1(x;\,h/k)$ we need a reasonable upper bound for $\widetilde A_{1,j}(x;\,h/k)$ . After the Voronoi formula is derived, we will use it to deduce a better upper bound, which can then be fed back to the argument used to derive the Voronoi formula in the first place, leading to an improved error term in said formula.

Lemma 8·1. Let $x\in\left[1,\infty\right[$ , let $j\in\left\{0,1\right\}$ , and let h and k be coprime integers so that $1\leqslant k\ll x^{1/3}$ . Then

\begin{equation*}\widetilde A_{1,j}\!\left(x;\,\frac{h}{k}\right)\ll_{\varepsilon} k^{5/4}\,x^{29/24+\varepsilon}.\end{equation*}

Proof. We choose $H=k^{5/4}\,x^{11/24}$ . Then $1\leqslant H\ll x$ , and using (1·6), Lemma 6·1 and Corollary 7·4 gives

\begin{align*}\widetilde A_{1,j}\!\left(x;\,\frac{h}{k}\right)&=\frac1H\int\limits_x^{x+H}\widetilde A_{1,j}\!\left(t;\,\frac{h}{k}\right)\mathrm dt-\frac1H\int\limits_x^{x+H}\int\limits_x^t\widetilde A_{0,j}\!\left(u;\,\frac{h}{k}\right)\mathrm du\,\mathrm dt\\&=\frac1H\left(\widetilde A_{2,j}\!\left(x+H;\,\frac{h}{k}\right)-\widetilde A_{2,j}\!\left(x;\,\frac{h}{k}\right)\right)-\frac1H\int\limits_x^{x+H}\int\limits_x^t\widetilde A_{0,j}\!\left(u;\,\frac{h}{k}\right)\mathrm du\,\mathrm dt\\&\ll_{\varepsilon}\frac1H\,k^{5/2}d(k)\,x^{5/3}+H\,x^{3/4+\varepsilon}\ll_{\varepsilon} k^{5/4}\,x^{29/24+\varepsilon},\end{align*}

as desired.

The proof of the case $a=1$ gives rise to sums of the kind treated in the above lemma, but with coefficients A(m, d) instead of A(m, 1). The following lemma deals with this minor complication.

Lemma 8·2. Let $x\in\left[1,\infty\right[$ , let $j\in\left\{0,1\right\}$ , let h and k be coprime integers satisfying $1\leqslant k\ll x^{1/3}$ , and let $d\in\mathbb Z_+$ be so that $d|k$ . Then

\begin{equation*}\sum_{m\leqslant x}A(m,d)\left(e\!\left(\frac{mhd}k\right)+\left({-}1\right)^je\!\left({-}\frac{mhd}k\right)\right)\left(x-m\right)\ll_\varepsilon d^{\vartheta+\varepsilon}\,k^{5/4}\,x^{29/24+\varepsilon}.\end{equation*}

Proof. We may compute using Möbius inversion, Lemma 8·1, and (4·1),

\begin{align*}&\sum_{m\leqslant x}A(m,d)\left(e\!\left(\frac{mhd}k\right)+\left({-}1\right)^je\!\left({-}\frac{mhd}k\right)\right)\left(x-m\right)\\&=\sum_{m\leqslant x}\sum_{\ell\mid\left(d,m\right)}\mu\!\left(\ell\right)A\!\left(1,\frac d\ell\right)A\!\left(\frac{m}{\ell},1\right)\left(e\!\left(\frac{mhd}k\right)+\left({-}1\right)^je\!\left({-}\frac{mhd}k\right)\right)\left(x-m\right)\\&=\sum_{\ell\mid d}\mu\!\left(\ell\right)A\!\left(1,\frac d\ell\right)\ell\sum_{m\leqslant x/\ell}A\!\left(m,1\right)\left(e\!\left(\frac{\ell mhd}{k}\right)+\left({-}1\right)^je\!\left({-}\frac{\ell mhd}{k}\right)\right)\left(\frac x\ell-m\right)\end{align*}

\begin{align*}&=\sum_{\ell\mid d}\mu\!\left(\ell\right)A\!\left(1,\frac d\ell\right)\ell\left(\widetilde A_{1,j}\!\left(\frac x\ell;\,\frac{\ell h d}{k}\right)+O_{\varepsilon}\!\left(\left(\frac k{\ell}\right)^{3/2+\varepsilon}\cdot\frac x\ell\right)+O_{\varepsilon}\!\left(\left(\frac k{\ell}\right)^{9/2+\varepsilon}\right)\right)\\&\ll_{\varepsilon}\sum_{\ell\mid d}\left(\frac d\ell\right)^{\vartheta+\varepsilon}\ell\left(k^{5/4}\left(\frac{x}{\ell}\right)^{29/24+\varepsilon}+\left(\frac k\ell\right)^{3/2+\varepsilon}\cdot\frac x\ell+\left(\frac{k}\ell\right)^{9/2+\varepsilon}\right),\end{align*}

where we have used the fact that since $k\ll x^{1/3}$ , also $k/\ell\ll\left(x/\ell\right)^{1/3}$ , so that Lemma 8·1 is applicable. Finally, since $k\ll x^{1/3}$ , we have

\begin{equation*}k^{3/2+\varepsilon}\,x+k^{9/2+\varepsilon}\ll_{\varepsilon} k^{5/4}\,x^{29/24+\varepsilon},\end{equation*}

and the total contribution is $\ll_{\varepsilon} d^{\vartheta+\varepsilon}\,k^{5/4}\,x^{29/24+\varepsilon}$ .

We will need to handle various sums involving both Fourier coefficients and Kloosterman sums. Taking discrete Fourier transforms of the Kloosterman sums allows us to reduce things back to exponential sums weighted by just the Fourier coefficients.

Lemma 8·3. Let $h\in\mathbb Z$ and $k\in\mathbb Z_+$ be coprime, and define

\begin{equation*}\widehat S(h,\xi;\,k)\,:\!=\,\frac1k\sum_{\ell=1}^kS(h,\ell;\,k)\,e\!\left({-}\frac{\ell\xi}k\right)\end{equation*}

for every $\xi\in\mathbb Z$ . Then, for any $m,\xi\in\mathbb Z$ , we have

\begin{equation*}S(h,m;\,k)=\sum_{\xi=1}^k\widehat S(h,\xi;\,k)\,e\!\left(\frac{m\xi}k\right)\end{equation*}

and

\begin{equation*}\widehat S(h,\xi;\,k)\ll k^{1/2}\,d(k),\end{equation*}

as well as

\begin{equation*}\sum_{\xi=1}^k\bigl|\widehat S(h,\xi;\,k)\bigr|\ll k\,d(k).\end{equation*}

Proof. The first claim follows directly from the discrete Fourier inversion formula, and the second follows directly from Weil’s bound. The third follows from the Cauchy–Schwarz inequality, the discrete Parseval identity and Weil’s bound since

\begin{align*}\sum_{\xi=1}^k\bigl|\widehat S(h,\xi;\,k)\bigr|&\leqslant k^{1/2}\sqrt{\sum_{\xi=1}^k\bigl|\widehat S(h,\xi;\,k)\bigr|^2}\\&=k^{1/2}\sqrt{\frac1k\sum_{\ell=1}^k\bigl|S(h,\ell;\,k)\bigr|^2}\\&\ll k^{1/2}\,\sqrt{\frac1k\cdot k\left(k^{1/2}\,d(k)\right)^2}\ll k\,d(k). \end{align*}

The argument used in the proof of Proposition 7·2 runs into difficulties in the case $a=1$ due to convergence issues, and so we need to proceed differently. Our next goal is to establish the case $a=1$ essentially by differentiating the formula for $a=2$ termwise. The requisite convergence considerations give a practical truncated version of the summation formula as a by-product.

Proposition 8·4. Let $x\in\left[1,\infty\right[$ , $j\in\left\{0,1\right\}$ , and let h and k be positive integers with $1\leqslant k\ll x^{1/3}$ . Then

\begin{align*}&\widetilde A_{1,j}\!\left(x;\,\frac{h}{k}\right)=i^{-j}\,2^{-1}\,\pi^{-3/2}\,kx\\&\quad\times\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\mathscr J_{1,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right).\end{align*}

Here the infinite series converges both boundedly and uniformly when x is restricted to a compact interval in $\mathbb R_+$ . Furthermore, if $N\in\left[1,\infty\right[$ , and $N\gg k^{3}$ , then

\begin{align*}&\widetilde A_{1,j}\!\left(x;\,\frac{h}{k}\right)=i^{-j}\,2^{-1}\,\pi^{-3/2}\,kx\\&\times\sum_{d\mid k}\sum_{m\leqslant N}\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\mathscr J_{1,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right)+\text{error},\end{align*}

where the error is

\begin{equation*}\ll_{\varepsilon} k^{5/4+{\varepsilon}}\,x^{5/3}\,N^{\varepsilon-1/8}.\end{equation*}

Proof. Our starting point is the formula

\begin{align*}&\widetilde A_{2,j}\!\left(x;\,\frac{h}{k}\right)=-i^{-j}\,2^{-2}\,\pi^{-3/2}\,k\,x^2\\&\times\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\mathscr J_{2,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right),\end{align*}

and our plan is to differentiate this identity. The derivative of the left-hand side is of course simply $\widetilde A_{1,j}(x;\,h/k)$ . On the other hand, Lemma 5·3 tells us that

\begin{equation*}\frac{\mathrm d}{\mathrm dx}\left(x^2\mathscr J_{2,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right)\right)=-2x\mathscr J_{1,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right),\end{equation*}

and using this it is straightforward to check that, formally, the derivative of the right-hand side coincides with the right-hand side of the desired identity for $\widetilde A_{1,j}(x;\,h/k)$ . From the analysis of the previous section it is clear that the above series for $\widetilde A_{2,j}(x;\,h/k)$ converges for any x. Thus, by standard arguments, it only remains to prove that the series for $\widetilde A_{1,j}(x;\,h/k)$ converges boundedly and uniformly when x is restricted to a bounded interval.

We take arbitrary numbers $a,b\in\left[1,\infty\right[$ satisfying $a\lt b$ and $a\gg k^{3}$ , and consider the partial sum

\begin{align*}&\Sigma(a,b)\\[-5pt]&\,:\!=\,kx\sum_{d\mid k}\sum_{a\leqslant m\leqslant b}\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\mathscr J_{1,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right).\end{align*}

By Lemma 5·1, we have

\begin{align*}&\mathscr J_{1,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right)=-\frac{1}{2\sqrt{3\pi}}\sum_\pm\exp\!\left(\pm\frac{6i\pi\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\pm\frac{\pi i}2\left(1+j\right)\right)\\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad+O\!\left(\left(d^{2/3}\,m^{1/3}\,x^{1/3}\,k^{-1}\right)^{-1}\right).\end{align*}

We may estimate the contribution of the O-term in $\Sigma(a,b)$ by

\begin{align*}\ll_{\varepsilon} kx\sum_{d\mid k}\sum_{a\leqslant m\leqslant b}\frac{\left|A(d,m)\right|}{dm}\left(\frac kd\right)^{1/2+\varepsilon}\left(d^{2/3}\,m^{1/3}\,x^{1/3}\,k^{-1}\right)^{-1}\ll_{\varepsilon} k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3}.\end{align*}

We also write in each term of $\Sigma(a,b)$

\begin{equation*}S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)=\sum_{\xi=1}^{k/d}\widehat S\!\left(\overline h,\xi;\,\frac kd\right)\left(e\!\left(\frac{m\xi d}k\right)+\left({-}1\right)^je\!\left({-}\frac{m\xi d}k\right)\right).\end{equation*}

Combining all these observations gives

\begin{align*}&\Sigma(a,b)\\&=-\frac{1}{2\sqrt{3\pi}}\sum_\pm e\!\left(\pm\frac{\pi i }2\left(1+j\right)\right)\sum_{d\mid k}\frac1d\sum_{\xi=1}^{k/d}\widehat S\!\left(\overline h,\xi;\,\frac kd\right)kx\,\Sigma_\pm(a,b)+O(k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3}),\end{align*}

where

\begin{equation*}\Sigma_\pm(a,b)\,:\!=\,\sum_{a\leqslant m\leqslant b}\frac{A(d,m)}m\left(e\!\left(\frac{m\xi d}k\right)+\left({-}1\right)^je\!\left({-}\frac{m\xi d}k\right)\right) e\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\right).\end{equation*}

To simplify the formulae below, we shall write, for each $\ell\in\left\{0,1\right\}$ , and all $x\in\left[1,\infty\right[$ ,

\begin{equation*}A_{\ell,j,d}\!\left(x;\,\frac \xi k\right)\,:\!=\,\sum_{m\leqslant x}A(d,m)\left(e\!\left(\frac{m\xi d}k\right)+\left({-}1\right)^je\!\left({-}\frac{m\xi d}k\right)\right)\left(x-m\right)^\ell.\end{equation*}

We start the estimation of $\Sigma_\pm(a,b)$ by summing by parts getting

\begin{align*}\Sigma_\pm(a,b)&=\left.\frac1t\,e\!\left(\pm\frac{3\,d^{2/3}\,t^{1/3}\,x^{1/3}}k\right)A_{0,j,d}\!\left(t;\,\frac\xi k\right)\right]_a^{t=b}\\&\quad-\int\limits_a^bA_{0,j,d}\!\left(t;\,\frac\xi k\right)\left({-}\frac1{t^2}\pm\frac{2\pi i\,d^{2/3}\,x^{1/3}}{k\,t^{5/3}}\right)e\!\left(\pm\frac{3\,d^{2/3}\,t^{1/3}\,x^{1/3}}k\right)\mathrm dt.\end{align*}

By Lemma 6·2, the substitution term is $\ll_{\varepsilon} d^{\vartheta+\varepsilon}\,a^{\varepsilon-1/4}$ , as is the contribution from the term involving $1/t^2$ , and it only remains to estimate the term involving $t^{-5/3}$ . We do this by integration by parts to get

\begin{align*}&\int\limits_a^bA_{0,j,d}\!\left(t;\,\frac\xi k\right)\frac{2\pi i\,d^{2/3}\,x^{1/3}}{k\,t^{5/3}}\,e\!\left(\pm\frac{3\,d^{2/3}\,t^{1/3}\,x^{1/3}}k\right)\mathrm dt\\&=\left.A_{1,j,d}\!\left(t;\,\frac\xi k\right)\frac{2\pi i\,d^{2/3}\,x^{1/3}}{k\,t^{5/3}}\,e\!\left(\pm\frac{3\,d^{2/3}\,t^{1/3}\,x^{1/3}}k\right)\right]_a^{t=b}\\&\qquad-\int\limits_a^bA_{1,j,d}\!\left(t;\,\frac\xi k\right)\left({-}\frac{10\,\pi i\,d^{2/3}\,x^{1/3}}{3\,k\,t^{8/3}}\pm\frac{4\,\pi^2\,d^{4/3}\,x^{2/3}}{k^2\,t^{7/3}}\right)e\!\left(\pm\frac{d^{2/3}\,t^{1/3}\,x^{1/3}}k\right)\mathrm dt.\end{align*}

By Lemma 8·2, the substitution terms and the term involving $t^{-8/3}$ contribute

\begin{equation*}\ll_{\varepsilon} d^{2/3+\vartheta+\varepsilon}\,x^{1/3}\,k^{1/4}\,a^{\varepsilon-11/24},\end{equation*}

whereas the contribution from the term involving $t^{-7/3}$ is

\begin{equation*}\ll_{\varepsilon} d^{4/3+\vartheta+\varepsilon}\,k^{-3/4}\,x^{2/3}\,a^{\varepsilon-1/8}.\end{equation*}

Altogether, our estimate for $\Sigma_\pm(a,b)$ reads

\begin{equation*}\Sigma_\pm(a,b)\ll_\varepsilon d^{\vartheta+\varepsilon}\,a^{\varepsilon-1/4}+d^{2/3+\vartheta+\varepsilon}\,x^{1/3}\,k^{1/4}\,a^{\varepsilon-11/24}+d^{4/3+\vartheta+\varepsilon}\,k^{-3/4}\,x^{2/3}\,a^{\varepsilon-1/8}.\end{equation*}

Plugging this into our previous expression for $\Sigma(a,b)$ gives

\begin{align*}\Sigma(a,b)&\ll_\varepsilon\sum_{d\mid k}\frac1d\sum_{\xi=1}^{k/d}\left|\widehat S\!\left(\overline h,\xi;\,\frac kd\right)\right|kx\\&\qquad\times\left(d^{\vartheta+\varepsilon}\,a^{\varepsilon-1/4}+d^{2/3+\vartheta+\varepsilon}\,x^{1/3}\,k^{1/4}\,a^{\varepsilon-11/24}+d^{4/3+\vartheta+\varepsilon}\,k^{-3/4}\,x^{2/3}\,a^{\varepsilon-1/8}\right)\\&\qquad+k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3}\\&\ll_\varepsilon k^{2+\varepsilon}\,x\left(a^{\varepsilon-1/4}+x^{1/3}\,k^{1/4}\,a^{\varepsilon-11/24}+k^{-3/4}\,x^{2/3}\,a^{\varepsilon-1/8}\right)\\&\qquad+k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3}\\&\ll_{\varepsilon} k^{2+\varepsilon}\,x\,a^{\varepsilon-1/4}+k^{9/4+\varepsilon}\,x^{4/3}\,a^{\varepsilon-11/24}+k^{5/4+\varepsilon}\,x^{5/3}\,a^{\varepsilon-1/8}+k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3},\end{align*}

and we are done with the first assertion. For the truncated formula we observe that

\begin{equation*}k^{2+\varepsilon}\,x\,a^{\varepsilon-1/4}\ll_{\varepsilon} k^{5/4+\varepsilon}\,x^{5/3}\,a^{\varepsilon-1/8},\end{equation*}

and that

\begin{equation*}k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3}\ll_{\varepsilon} k^{5/4+\varepsilon}\,x^{5/3}\,a^{\varepsilon-1/8},\end{equation*}

as well as

\begin{equation*}k^{9/4+\varepsilon}\,x^{4/3}\,a^{\varepsilon-11/24}\ll_{\varepsilon} k^{5/4+\varepsilon}\,x^{5/3}\,a^{\varepsilon-1/8},\end{equation*}

since $k\ll x^{1/3}$ . Thus we have obtained

\begin{equation*}\Sigma(a,b)\ll_{\varepsilon} k^{5/4+\varepsilon}x^{5/3}a^{{\varepsilon}-1/8},\end{equation*}

which immediately gives the truncated identity.

From the above formula we get directly a better upper bound for $\widetilde A_{1,j}(x;\,h/k)$ compared to Lemma 8·1.

Corollary 8·5. Let $x\in\left[1,\infty\right[$ , let $j\in\left\{0,1\right\}$ , and let h and k be coprime integers with $1\leqslant k\ll x^{1/3}$ . Then

\begin{equation*}\widetilde A_{1,j}\!\left(x;\,\frac{h}{k}\right)\ll_\varepsilon k^{3/2}\,x^{1+\varepsilon}.\end{equation*}

Furthermore, if $d\in\mathbb Z_+$ and $d\mid k$ , then also

\begin{equation*}\sum_{m\leqslant x}A\!\left(m,d\right)\left(e\!\left(\frac{mhd}k\right)+\left({-}1\right)^je\!\left({-}\frac{mhd}k\right)\right)\left(x-m\right)\ll_\varepsilon d^{\vartheta+\varepsilon}\,k^{3/2}\,x^{1+\varepsilon}.\end{equation*}

We can now feed this back into the estimation of $\Sigma(a,b)$ in the proof of Proposition 8·4, where we used the weaker estimate from Lemma 8·2 (which in turn was based on Lemma 8·1). The resulting estimate for $\Sigma_\pm(a,b)$ is

\begin{equation*}\Sigma_\pm(a,b)\ll_{\varepsilon} d^{\vartheta+\varepsilon}\,a^{\varepsilon-1/4}+d^{2/3+\vartheta+\varepsilon}\,k^{1/2}\,x^{1/3}\,a^{\varepsilon-2/3}+d^{4/3+\vartheta+\varepsilon}\,k^{-1/2}\,x^{2/3}\,a^{\varepsilon-1/3},\end{equation*}

and the end result is

\begin{align*}\Sigma(a,b)&\ll_{\varepsilon} k^{2+\varepsilon}\,x\,a^{\varepsilon-1/4}+k^{5/2+\varepsilon}\,x^{4/3}\,a^{\varepsilon-2/3}+k^{3/2+\varepsilon}\,x^{5/3}\,a^{\varepsilon-1/3}+k^{5/2+\varepsilon}\,x^{2/3}\,a^{-1/3}\\&\ll_{\varepsilon} k^{2+\varepsilon}\,x\,a^{\varepsilon-1/4} +k^{3/2+\varepsilon}\,x^{5/3}\,a^{\varepsilon-1/3}.\end{align*}

Thus we have obtained the following.

Corollary 8·6. Let $x\in\left[1,\infty\right[$ , $j\in\left\{0,1\right\}$ , let h and k be positive integers with $1\leqslant k\ll x^{1/3}$ , and let $N\in\left[1,\infty\right[$ with $N\gg k^3$ . Then

\begin{align*}&\widetilde A_{1,j}\!\left(x;\,\frac{h}{k}\right)=i^{-j}\,2^{-1}\,\pi^{-3/2}\,kx\\&\times\sum_{d\mid k}\sum_{m\leqslant N}\frac{A(d,m)}{dm}\left(S\!\left(\overline h,m;\,\frac kd\right)+\left({-}1\right)^jS\!\left(\overline h,-m;\,\frac kd\right)\right)\mathscr J_{1,j}\!\left(\frac{\pi^6\,d^4\,m^2\,x^2}{k^6}\right)+\text{error},\end{align*}

where the error is

\begin{equation*}\ll_{\varepsilon} k^{2+\varepsilon}\,x\,N^{\varepsilon-1/4} +k^{3/2+\varepsilon}\,x^{5/3}\,N^{\varepsilon-1/3}.\end{equation*}

Proposition 8·4 and Corollary 8·6 yield the following result when combined with averaging over $j\in\left\{0,1\right\}$ and Lemma 5·1.

Corollary 8·7. Let $x\in\left[1,\infty\right[$ , $j\in\left\{0,1\right\}$ , and let h and k be positive integers with $1\leqslant k\ll x^{1/3}$ . Then

\begin{align*}\widetilde A_{1}\!\left(x;\,\frac{h}{k}\right)&=\frac{k\,x}{4\,\pi^2\,\sqrt3}\sum_{d\mid k}\frac1{d}\sum_{m=1}^\infty\frac{A(d,m)}{m}\sum_\pm \left(\mp i\right)\,S\!\left(\overline h,\pm m;\,\frac kd\right)e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}{k}\right)\\&\qquad\qquad\qquad+O_{\varepsilon}\big(k^{5/2+{\varepsilon}}\,x^{2/3}\big).\end{align*}

Here the infinite series converges both boundedly and uniformly when x is restricted to a compact interval in $\mathbb R_+$ . Furthermore, if $N\in\left[1,\infty\right[$ , and $N\ll k^{-3}\,x^3$ , then

\begin{align*}\widetilde A_{1}\!\left(x;\,\frac{h}{k}\right)&=\frac{k\,x}{4\,\pi^2\,\sqrt3}\sum_{d\mid k}\frac1{d}\sum_{m\leqslant N}\frac{A(d,m)}{m}\sum_\pm \left(\mp i\right)\,S\!\left(\overline h,\pm m;\,\frac kd\right)e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}{k}\right)\\[4pt]&\qquad\qquad\qquad+O_{\varepsilon}\left(k^{3/2+{\varepsilon}}\,x^{5/3+{\varepsilon}}N^{{\varepsilon}-1/3}\right).\end{align*}

9. Pointwise $\Omega$ -result for long sums on $\mathrm{GL}_3$

Let us first gather some auxiliary results which will be useful in what follows. We start with results concerning Kloosterman sums.

Lemma 9·1. Let k be a prime. Suppose also that m and n are integers. Then

\begin{equation*}\sum_{a\in\mathbb Z_k^\times}S(a,m;\,k)=\begin{cases}1&\text{if $m\not\equiv0\pmod k$}\\-k+1&\text{if $m\equiv0\pmod k$}\end{cases}\end{equation*}

as well as

\begin{align*}\Sigma(m,n,k)\,:\!=\,\sum_{a\in\mathbb Z_k^\times} S(a,m;\,k)\,S(a,n;\,k)=\begin{cases}k^2-k-1&\text{if $m\equiv n\not\equiv0\pmod k$}\\k-1&\text{if $m\equiv n\equiv0\pmod k$}\\-1&\text{if $m\equiv0$ and $n\not\equiv0\pmod k$}\\-1&\text{if $m\not\equiv0$ and $n\equiv0\pmod k$}\\-k-1&\text{if $0\not\equiv m\not\equiv n\not\equiv0\pmod k$}.\end{cases}\end{align*}

We will use the first derivative test repeatedly and record one of its formulations here. Results such as this are discussed for instance in [ Reference Huxley15 , section 5·1].

Lemma 9·2. Let $a,b\in\mathbb R$ with $a\lt b$ , let $\lambda\in\mathbb R_+$ , and let f be a real-valued continuously differentiable function on $\left]a,b\right[$ such that $\left|f'(x)\right|\geqslant\lambda$ for $x\in\left]a,b\right[$ . Also, let g be a complex-valued continuously differentiable function on the interval $\left[a,b\right]$ , and let $G\in\mathbb R_+$ be such that $g(x)\ll G$ for $x\in\left[a,b\right]$ . Then

\begin{align*}\int\limits_a^bg(x)\,e\!\left(f(x)\right)\,\mathrm dx\ll\frac G{\lambda}+\frac1{\lambda}\int\limits_a^b\left|g'(x)\right|\mathrm d x.\end{align*}

Now we are in a position to prove an asymptotic formula for the averaged mean square of the long second order Riesz sums.

Theorem 9·3. Suppose that $X\in\left[1,\infty\right[$ and that k is a prime such that $k\ll X^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then we have

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx=B(k)\,k^5\,X^{13/3}+O\big(k^{6}\,X^4\big),\end{align*}

where $B(k)\in\mathbb R_+$ and $B(k)\asymp1$ .

In particular, if $x\in\left[1,\infty\right[$ and $k\in\mathbb Z_+$ is a prime so that $k\ll x^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ , then we have

\begin{equation*}\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_2\!\left(x;\,\frac{h}{k}\right)\right|=\Omega\big(k^{5/2}\,x^{5/3}\big).\end{equation*}

Proof. Note that the $\Omega$ -result follows immediately from the moment result. For the first statement, using Corollary 7·3 the second moment is

\begin{align*}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\\&=\frac{k^4}{192\,\pi^6}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}x^{10/3}\left|\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{d^{5/3}\,m^{4/3}}\sum_\pm S\!\left(\overline h,\pm m;\,\frac kd\right)\,e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\right)\right|^2\,\mathrm dx\\&\qquad-\frac{k^2}{8\pi^3\sqrt 3}\,\Re\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}x^{5/3}\sum_{d\mid k}\sum_{m=1}^\infty\frac{A(d,m)}{d^{5/3}\,m^{4/3}}\\&\qquad\times\sum_\pm S\!\left(\overline h,\pm m;\,\frac kd\right)e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\right)\cdot O(k^{7/2}\,x^{4/3})\,\mathrm dx+O(k^{7}\,X^{11/3}).\end{align*}

Once we have proved that the first term on the right-hand side is $\asymp k^{5}\,X^{13/3}$ , it immediately follows from the Cauchy–Schwarz inequality that the mixed term integral is

\begin{equation*}\ll\sqrt{k^{5}\,X^{13/3}}\,\sqrt{k^{7}\,X^{11/3}}\ll k^{6}\,X^{4},\end{equation*}

as required. Also, since $k\ll X^{1/3}$ , we have $k^{7}\,X^{11/3}\ll k^{6}\,X^4$ .

In the first integral, we expand $\left|\Sigma\right|^2$ as $\Sigma\,\overline\Sigma$ . The diagonal terms, i.e. those terms where $d_1^2\,m_1=d_2^2\,m_2$ and where the signs in sums $\sum_\pm$ are chosen to be equal, give the contribution

\begin{align*}&\frac{k^4}{192\pi^6}\smash{\sum_{d_1\mid k}\sum_{d_2\mid k}\underset{d_1^2m_1=d_2^2m_2}{\sum_{m_1=1}^\infty\sum_{m_2=1}^\infty}}\frac{A(d_1,m_1)\,\overline{A(d_2,m_2)}}{d_1^{5/3}\,d_2^{5/3}\,m_1^{4/3}\,m_2^{4/3}}\\[9pt]&\qquad\times\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\sum_\pm S\!\left(\overline h,\pm m_1;\,\frac k{d_1}\right)S\!\left(\overline h,\pm m_2;\,\frac k{d_2}\right)\int\limits_X^{2X}x^{10/3}\,\mathrm dx.\end{align*}

We will examine the contribution of the different pairs $(d_1,d_2)$ separately at first. The terms where $d_1=d_2=1$ contribute

(9·1) \begin{align}\frac{k^4}{96\pi^6}\sum_{m=1}^\infty\frac{\left|A(1,m)\right|^2}{m^{8/3}}\left(\frac{k^2-k-1}{k-1}1_{m\not\equiv 0\,(k)}+1_{m\equiv 0\,(k)}\right)\int\limits_X^{2X}x^{10/3}\,\mathrm dx\end{align}

by Lemma 9·1.

Similarly, the terms with $d_1=d_2=k$ contribute

(9·2) \begin{align}\frac{k^4}{96\pi^6}\sum_{m=1}^\infty\frac{\left|A(k,m)\right|^2}{k^{10/3}m^{8/3}}\int\limits_X^{2X}x^{10/3}\,\mathrm d x\end{align}

and the terms where $d_1d_2=k$ contribute

(9·3) \begin{align}-\frac{k^4}{48\pi^6}\Re\left(\sum_{m=1}^\infty\frac{A(1,k^2m)\overline{A(k,m)}}{k^{13/3}m^{8/3}}\right)\int\limits_X^{2X}x^{10/3}\,\mathrm d x.\end{align}

Our aim is to show that these three terms sum up to $B(k)k^5X^{13/3}$ , where $B(k)\in\mathbb R$ with $B(k)\asymp 1$ . The upper bound is straightforward to establish just by using the known estimates for the Fourier coefficients ( $A(m,n)\ll_{\varepsilon} (mn)^{\vartheta+{\varepsilon}}$ and (4·2)), but the lower bound requires some work. Towards this, observe first that by the arithmetic-geometric mean inequality, the absolute value of (9·3) is bounded from above by

(9·4) \begin{align}&\frac{k^4}{96\pi^6}\sum_{m=1}^\infty\frac{\left|A(1,k^2m)\right|^2}{k^{16/3}m^{8/3}}\int\limits_X^{2X}x^{10/3}\,\mathrm dx+\frac{k^4}{96\pi^6}\sum_{m=1}^\infty\frac{\left|A(k,m)\right|^2}{k^{10/3}m^{8/3}}\int\limits_X^{2X}x^{10/3}\,\mathrm d x \nonumber\\[7pt]&\leq\frac{k^4}{96\pi^6}\sum_{m=1}^\infty\frac{\left|A(1,m)\right|^2}{m^{8/3}}\int\limits_X^{2X}x^{10/3}\,\mathrm dx+\frac{k^4}{96\pi^6}\sum_{m=1}^\infty\frac{\left|A(k,m)\right|^2}{k^{10/3}m^{8/3}}\int\limits_X^{2X}x^{10/3}\,\mathrm d x.\end{align}

Notice that the latter term in (9·4) is exactly (9·2). Thus, using the identity $1_{m\not\equiv 0(k)}=1-1_{m\equiv 0(k)}$ , a simple computation shows that the sum of (9·1), (9·2) and (9·3) is bounded from below by

\begin{equation*}\frac{k^4}{96\pi^6}\cdot\frac{k^2-2k}{k-1}\sum_{m=1}^\infty\frac{|A(1,m)|^2}{m^{8/3}}\left(1-1_{m\equiv 0\,(k)}\right)\int\limits_X^{2X}x^{10/3}\,\mathrm d x.\end{equation*}

This is bounded from below by the required lower bound $\gg k^5X^{13/3}$ just by dropping all the terms except the one corresponding to $m=1$ (recall that $A(1,1)=1$ ) as each of the terms involved is non-negative.

The contribution coming from the terms $d_1^2m=d_2^2n$ , where the signs in $\sum_\pm$ are chosen to be distinct, can be estimated by using the first derivative test together with Weil’s bound as

\begin{align*}&\ll k^4\sum_\pm\sum_{d_1|k}\sum_{d_2|k}\sum_{m=1}^\infty\sum_{\substack{n=1\\d_1^2m=d_2^2n}}^\infty\frac{\left|A(d_1,m)\overline{A(d_2,n)}\right|}{d_1^{5/3}d_2^{5/3}m^{4/3}n^{4/3}}\left|S\left(\overline h,\pm m;\,\frac k{d_1}\right)S\left(\overline h,\mp n;\,\frac k{d_2}\right)\right|\\&\qquad\qquad\times\int\limits_X^{2X}x^{10/3}\,e\!\left(\pm\frac{6(d_1^2mx)^{1/3}}k\right)\mathrm dx\\&\ll k^4\cdot k\,X^{10/3}X^{2/3}k\sum_{d_1|k}\sum_{d_2|k}\sum_{m=1}^\infty\sum_{\substack{n=1\\d_1^2m=d_2^2n}}^\infty\frac{\left|A(d_1,m)\overline{A(d_2,n)}\right|}{d_1^{17/6}d_2^{13/6}m^{5/3}n^{4/3}}\\&\ll X^{4}k^{6}.\end{align*}

For the off-diagonal terms we restrict to the case where the signs in $\sum_\pm$ are chosen to be the same as the case of distinct signs can be treated similarly with easier arguments. The prior terms contribute, using again the first derivative test and Weil’s bound,

\begin{align*}&\ll k^4\sum_\pm\sum_{d_1|k}\sum_{d_2|k}\sum_{m=1}^\infty\frac{\left|A(d_1,m)\right|}{m^{4/3}d_1^{5/3}}\sum_{\substack{n=1\\d_1^2m\neq d_2^2n}}^\infty\frac{\left|A(d_2,n)\right|}{n^{4/3}d_2^{5/3}}\left|S\left(\overline h,\pm m;\frac k{d_1}\right)S\left(\overline h,\pm n;\,\frac k{d_2}\right)\right|\\&\qquad\times\int\limits_X^{2X}x^{10/3}\,e\left(\frac{\pm 3(mx d_1^2)^{1/3}\pm 3(nxd_2^2)^{1/3}}k\right)\mathrm dx\\&\ll k^4\sum_{d_1|k}\sum_{d_2|k}\sum_{m=1}^\infty\frac{\left|A(d_1,m)\right|}{m^{4/3}d_1^{5/3}}\sum_{\substack{n=1\\d_1^2m\neq d_2^2n}}^\infty\frac{\left|A(d_2,n)\right|}{n^{4/3}d_2^{5/3}}\cdot\frac{X^{4}k}{\left|d_1^{2/3}m^{1/3}-d_2^{2/3}n^{1/3}\right|}\\&\qquad\qquad\times\left|S\left(\overline h,\pm m;\,\frac k{d_1}\right)S\left(\overline h,\pm n;\,\frac k{d_2}\right)\right|\\&\ll X^{4}k^6\sum_{d_1|k}\sum_{d_2|k}\sum_{n=1}^\infty\frac{\left|A(d_2,n)\right|}{n^{4/3}d_2^{5/3}}\sum_{\substack{m=1\\d_1^2m\neq d_2^2n}}^\infty\frac{\left|A(d_1,m)\right|}{m^{4/3}d_1^{5/3}}\cdot\frac{1}{\left|d_1^{2/3}m^{1/3}-d_2^{2/3}n^{1/3}\right|}.\end{align*}

By symmetry it is enough to treat the terms in the m-sum with $d_1^2m\lt d_2^2n$ (if this is not the case, one would consider the terms with $d_2^2n\lt d_1^2m$ in the n-sum). For these we split the m-sum into two parts according to whether $d_1^2m\lt d_2^2n/2$ or $d_2^2n/2\leq d_1^2m\lt d_2^2n$ . Note that for $d_1^2m\ll d_2^2n$ we have

(9·5) \begin{align}\frac1{d_2^{5/3}}\cdot\frac1{|d_2^{2/3}n^{1/3}-d_1^{2/3}m^{1/3}|}\ll\frac1{d_2^{1/3}}\cdot\frac{n^{2/3}}{|d_2^2n-d_1^2m|}.\end{align}

Using (9·5) together with the pointwise bound $A(d_2,n)\ll_\varepsilon (d_2n)^{\vartheta+\varepsilon}$ the terms with $d_1^2m\lt d_2^2n/2$ contribute

\begin{align*}&\ll X^{4}k^6\sum_{d_1|k}\sum_{d_2|k}\sum_{n=1}^\infty\frac{\left|A(d_2,n)\right|}{n^{4/3}d_2^{5/3}}\sum_{\substack{m=1\\d_1^2m\lt d_2^2n/2}}^\infty\frac{\left|A(d_1,m)\right|}{m^{4/3}d_1^{5/3}}\cdot\frac{1}{\left|d_1^{2/3}m^{1/3}-d_2^{2/3}n^{1/3}\right|}\\&\ll_\varepsilon X^4k^6\sum_{d_1|k}\sum_{d_2|k}\sum_{n=1}^\infty\sum_{\substack{m=1\\d_1^2m\lt d_2^2n/2}}^\infty\frac{|A(d_1,m)|}{m^{4/3}}\cdot\frac{d_2^{\vartheta+\varepsilon}}{n^{2/3-\vartheta-\varepsilon}d_1^{5/3}d_2^{1/3}(d_2^2n-d_1^2m)}\\&\ll_\varepsilon X^4k^6\sum_{d_1|k}\sum_{d_2|k}\sum_{n=1}^\infty\sum_{\substack{m=1\\d_1^2m\lt d_2^2n/2}}^\infty\frac{|A(d_1,m)|}{m^{4/3}n^{5/3-\vartheta-\varepsilon}d_1^{5/3}d_2^{7/3-\vartheta-\varepsilon}}\\&\ll X^4k^6,\end{align*}

where the last estimate follows from (4·2) using partial summation, and the estimate $\vartheta\leq 5/14$ .

Recall that $\delta\gt0$ is sufficiently small, but fixed. For the terms with $d_2^2n/2\leq d_1^2m\lt d_2^2n$ we again use (9·5) and the pointwise estimate $A(d,n)\ll (dn)^{\vartheta+\delta}$ to see that these terms contribute

\begin{equation*}\ll X^4k^6\sum_{d_1|k}\sum_{d_2|k}\sum_{n=1}^\infty\frac{n^{\vartheta+2\delta}}{d_2^{1/3-\vartheta-\delta}n^{2/3+\delta}}\sum_{\substack{m=1\\ d_2^2n/2\leq d_1^2m\lt d_2^2n}}^\infty\frac{|A(d_1,m)|}{m^{4/3}d_1^{5/3}}\cdot\frac1{|d_2^2n-d_1^2m|}.\end{equation*}

Observe that for $d_2^2n\gt d_1^2m$ we have

\begin{equation*}n^{2/3}=n^{1/6}n^{1/2}\gg\left(\frac{d_1^2m}{d_2^2}\right)^{1/6}n^{1/2}.\end{equation*}

Combining this with the estimate, which holds as $d_2^2n/2\leq d_1^2m$ ,

\begin{equation*}n^{\vartheta+2\delta}\leq\left(\frac{2d_1^2m}{d_2^2}\right)^{5/14+2\delta}\end{equation*}

shows that the terms with $d_2^2n/2\leq d_1^2m\lt d_2^2n$ contribute

(9·6) \begin{align}\ll X^4k^6\sum_{d_1|k}\sum_{d_2|k}\sum_{n=1}^\infty\sum_{\substack{m=1\\d_2^2n/2\leq d_1^2m\lt d_2^2n}}^\infty\frac{|A(d_1,m)|}{m^{8/7-2\delta}d_1^{9/7-4\delta}n^{1/2+\delta}d_2^{5/7-\vartheta+3\delta}(d_2^2n-d_1^2m)}.\end{align}

Next, we exchange the order of m- and n-sums. After that we apply the Cauchy–Schwarz inequality to the n-sum to see it is

\begin{align*}&\sum_{\substack{n=1\\d_2^2n/2\leq d_1^2m\lt d_2^2n}}^\infty\frac1{n^{1/2+\delta}(d_2^2n-d_1^2m)}\\&\leq\left(\sum_{\substack{n=1\\d_2^2n/2\leq d_1^2m\lt d_2^2n}}^\infty\frac1{n^{1+2\delta}}\right)^{1/2}\left(\sum_{\substack{n=1\\d_2^2n/2\leq d_1^2m\lt d_2^2n}}^\infty\frac1{(d_2^2n-d_1^2m)^2}\right)^{1/2}\ll 1,\end{align*}

where the last estimate is uniform in $m,\,d_1,$ and $d_2$ . Now using this in (9·6), partial summation and (4·2) show that the terms with $d_1^2m\geq d_2^2n/2$ contribute $\ll X^4k^6$ . This finishes the proof.

Next we present the first proof for Theorem 2·1. The idea is to relate upper bounds of Riesz weighted sums with different orders. We first connect the sizes of $\widetilde A_1(x;\,h/k)$ and $\widetilde A_2(x;\,h/k)$ .

Proposition 9·4. Let k be a prime number. Let $\gamma\in\left[1,2\right]$ and $\eta\in\left[1,2\right]$ be such that $\gamma/2+\eta/6\gt 3/4$ and

\begin{equation*}\widetilde A_1\left(x;\,\frac{h}{k}\right)\ll x^\gamma k^\eta\end{equation*}

for $k\ll x^{1/3-\delta}$ with $\delta\gt0$ sufficiently small and fixed. Then we have

\begin{equation*}\widetilde A_2\left(x;\,\frac{h}{k}\right)\ll x^{7/6+\gamma/2}k^{7/4+\eta/2}.\end{equation*}

Furthermore, we have

\begin{equation*}\underset{h\in\mathbb Z_k^\times}{\max}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|=\Omega\big(xk^{3/2}\big)\end{equation*}

for any prime $k\ll x^{1/3-\delta}$ .

Proof. For any $H\gt0$ we have by (1·6)

(9·7) \begin{align}\widetilde A_2\left(x;\,\frac{h}{k}\right)=\frac1H\int\limits_x^{x+H}\widetilde A_2\left(t;\,\frac{h}{k}\right)\,\mathrm dt-\frac1H\int\limits_x^{x+H}\int\limits_x^t\widetilde A_1\left(u;\,\frac{h}{k}\right)\,\mathrm du\,\mathrm dt.\end{align}

Substituting the Voronoi identity of Corollary 7·3 into the first term, we may estimate it by the first derivative test as

(9·8) \begin{align}\frac1H\int\limits_x^{x+H}\widetilde A_2\left(t;\,\frac{h}{k}\right)\,\mathrm dt\ll\frac1H\,x^{7/3}k^{7/2}+k^{7/2}x^{4/3}.\end{align}

Now, if $\widetilde A_1(x,h/k)\ll x^\gamma k^\eta$ , then the second term on the right-hand side of (9·7) is $\ll H\,x^\gamma\,k^\eta$ , and choosing $H=x^{7/6-\gamma/2}k^{7/4-\eta/2}$ immediately gives

\begin{equation*}\widetilde A_2\left(x;\,\frac{h}{k}\right)\ll x^{7/6+\gamma/2}k^{7/4+\eta/2}\end{equation*}

using (9·8) and the assumption $\gamma/2+\eta/6\gt3/4$ .

To prove the second statement, note that from Theorem 9·3, (9·8) and (9·7) it follows that

\begin{align*}\Omega\left(x^{5/3}k^{5/2}\right)&=\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_2\left(x;\frac{h}{k}\right)\right|\\&\ll \frac1H\,x^{7/3}k^{7/2}+k^{7/2}x^{4/3}+H\max_{u\in[x,x+H]}\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_1\left(u;\,\frac{h}{k}\right)\right|.\end{align*}

Thus for any $H\leq x$ with $x^{2/3}k=o(H)$ we have

\begin{equation*}\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|=\Omega\left(H^{-1}x^{5/3}k^{5/2}\right),\end{equation*}

from which the claim follows.

Lemma 9·5. Let $x\in\left[1,\infty\right[$ , $H\in\left[1,x\right]$ , and $k\ll x^{1/3-\delta}$ be a prime with $\delta\gt0$ sufficiently small and fixed. Then

\begin{equation*}\frac1H\int\limits_x^{x+H}\widetilde A_1\left(t;\,\frac{h}{k}\right)\,\mathrm dt\ll H^{-1}\,x^{5/3}k^{5/2}.\end{equation*}

Proof. We have

\begin{equation*}\frac1H\int\limits_x^{x+H}\widetilde A_1\left(t;\,\frac{h}{k}\right)\,\mathrm dt=\frac1H\left(\widetilde A_2\left(x+H;\,\frac{h}{k}\right)-\widetilde A_2\left(x;\,\frac{h}{k}\right)\right),\end{equation*}

and by Corollary 7·4, this is $\ll H^{-1}\,x^{5/3}k^{5/2}$ , where there is no ${\varepsilon}$ in the exponent of k as the denominator is a prime.

Proposition 9·4 together with the first part of the following result proves Theorem 2·1, a result which is formulated as the latter part.

Theorem 9·6. Let $\alpha\in\left]0,1\right]$ and $\beta\in]0,1]$ be such that $3\alpha+\beta\gt3/2$ , $\alpha\gt\vartheta$ , and

(9·9) \begin{align}\sum_{m\leqslant x}A(m,1)e\left(\frac{mh}k\right)\ll x^\alpha k^\beta\end{align}

for $x\in\left[1,\infty\right[$ . Suppose that k is a prime so that $k\ll x^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then

\begin{equation*}\widetilde A_1\left(x;\,\frac{h}{k}\right)\ll x^{5/6+\alpha/2}k^{5/4+\beta/2}.\end{equation*}

Furthermore, when $x\longrightarrow\infty$ , we have

\begin{equation*}\underset{h\in\mathbb Z_k^\times}{\max}\left|{\mathop{{\sum}^{\prime}}}_{m\leqslant x}A(m,1)e\!\left(\frac{mh}k\right)\right|=\Omega\big(x^{1/3}k^{1/2}\big)\end{equation*}

for any prime $k\ll x^{1/3-\delta}$ .

Proof. By (1·6) we have

(9·10) \begin{align}\widetilde A_1\left(x;\,\frac{h}{k}\right)=\frac1H\int\limits_x^{x+H}\widetilde A_1\left(t;\,\frac{h}{k}\right)\,\mathrm dt-\frac1H\int\limits_x^{x+H}\int\limits_x^t \widetilde A_0\left(u;\,\frac{h}{k}\right)\,\mathrm du\,\mathrm dt.\end{align}

For the first term, we already know by Lemma 9·5 that

\begin{equation*}\frac1H\int\limits_x^{x+H}\widetilde A_1\left(t;\,\frac{h}{k}\right)\,\mathrm dt\ll H^{-1}\,x^{5/3}k^{5/2}.\end{equation*}

On the other hand, recalling that (by (4·1))

(9·11) \begin{align}\sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)=\widetilde A_0\!\left(x;\,\frac{h}{k}\right)+O(k^{3/2+\varepsilon})+O(x^{\vartheta+\varepsilon})\end{align}

it follows from (9·9) and the assumptions on $\alpha$ and $\beta$ that we certainly have $\widetilde A_0(x;\,h/k)\ll x^\alpha k^\beta$ and so

\begin{equation*}\frac1H\int\limits_x^{x+H}\int\limits_x^t\widetilde A_0\left(u;\,\frac{h}{k}\right)\,\mathrm du\,\mathrm dt\ll H\,x^\alpha k^\beta.\end{equation*}

Choosing $H=x^{5/6-\alpha/2}k^{5/4-\beta/2}$ gives

\begin{equation*}\widetilde A_1\left(x;\,\frac{h}{k}\right)\ll x^{5/6+\alpha/2}k^{5/4+\beta/2}.\end{equation*}

To prove the second statement, note that from Proposition 9·4 and (9·10) it follows that

\begin{equation*}\Omega\big(xk^{3/2}\big)=\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|\ll \frac1H\,x^{5/3}k^{5/2}+H\max_{u\in[x,x+H]}\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_0\left(u;\,\frac{h}{k}\right)\right|.\end{equation*}

Thus for any $H\leq x$ with $x^{2/3}k=o(H)$ we have

\begin{equation*}\max_{h\in\mathbb Z_k^\times}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|=\Omega\left(H^{-1}xk^{3/2}\right),\end{equation*}

from which the claim follows as the residue term in $\widetilde A_0(x;\,h/k)$ has size $O(k^{3/2+{\varepsilon}})$ by (4·1).

Corollary 9·7. The exponent pairs $(\alpha,\beta)=(1,0)$ , $(\alpha,\beta)=(3/4+\varepsilon,0)$ , $(\alpha,\beta)=$ $(1/3+\varepsilon,1/2)$ and $(\alpha,\beta)=(1/2+\varepsilon,3/4)$ , each of which either holds or holds under some assumptions, in the previous estimate lead to

\begin{equation*}\widetilde A_1\left(x;\,\frac{h}{k}\right)\ll x^{4/3}k^{5/4},\quad\ll_{\varepsilon} x^{29/24+\varepsilon}k^{5/4},\quad\ll_{\varepsilon} x^{1+\varepsilon}k^{3/2}\quad\text{and}\quad\ll_{\varepsilon} x^{13/12+\varepsilon}k^{13/8},\end{equation*}

respectively. Here the choice $(\alpha,\beta)=(1,0)$ is admissible by the Rankin–Selberg theory, the second one by Lemma 6·1, the third under the conjectural bounds (1·2), and the final possibility by Theorem 2·5 (assuming $\vartheta=0$ ).

10. Second moments of long sums

Using the truncated Voronoi identity for $\widetilde A_1(x;\,h/k)$ we deduce the following mean square result.

Theorem 10·1. Let $x\in[1,\infty[$ and suppose that k is a prime so that $k\ll X^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then we have

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx=C(k)\cdot X^3k^3+O_{\varepsilon}\left(k^{4}\,X^{8/3+{\varepsilon}}\right),\end{align*}

where $C(k)\in\mathbb R_+$ with $C(k)\asymp 1$ .

Proof. We apply Corollary 8·7 with the choice $N= \lfloor k^{-3}X^3\rfloor$ to get

\begin{align*}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\\[4pt]&=\frac{k^2}{48\,\pi^4}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}x^{2}\left|\sum_{d\mid k}\sum_{m\leq N}\frac{A(d,m)}{d\,m}\sum_\pm (\mp i)S\!\left(\overline h,\pm m;\,\frac kd\right)\,e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\right)\right|^2\,\mathrm dx\\[4pt]&\qquad+\frac{k}{4\pi^2\sqrt 3}\,\Re\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}x\sum_{d\mid k}\sum_{m\leq N}\frac{A(d,m)}{d\,m}\\[4pt]&\qquad\quad\times\sum_\pm (\mp i) S\!\left(\overline h,\pm m;\,\frac kd\right)e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\right)\cdot O(k^{5/2}\,x^{2/3})\,\mathrm dx+O_{\varepsilon}(k^{5}\,X^{7/3+2{\varepsilon}}).\end{align*}

Once we have proved that the first term on the right-hand side is $\asymp k^{3}\,X^{3}$ , it immediately follows from the Cauchy–Schwarz inequality that the mixed term integral is

\begin{equation*}\ll_{\varepsilon}\sqrt{k^{3}\,X^{3}}\,\sqrt{k^{5}\,X^{7/3+2{\varepsilon}}}\ll_{\varepsilon} k^{4}\,X^{8/3+{\varepsilon}}.\end{equation*}

Also, as $k\ll X^{1/3-\delta}$ , we have $k^{5}X^{7/3+2{\varepsilon}}\ll_{\varepsilon} k^{4}X^{8/3+{\varepsilon}}$ , as required.

In the first integral, we expand $\left|\Sigma\right|^2$ as $\Sigma\,\overline\Sigma$ . The diagonal terms, i.e. those terms where $d_1^2\,m=d_2^2\,n$ and where the signs in sums $\sum_\pm$ are chosen to be equal, give the contribution

\begin{align*}&\frac{k^2}{48\pi^4}\sum_{d_1\mid k}\sum_{d_2\mid k}\underset{d_1^2m=d_2^2n}{\sum_{m\leq N}\sum_{n\leq N}}\frac{A(d_1,m)\,\overline{A(d_2,n)}}{d_1\,d_2\,m\,n}\\&\qquad\qquad\qquad\times\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\sum_\pm S\!\left(\overline h,\pm m;\,\frac k{d_1}\right)S\!\left(\overline h,\pm n;\,\frac k{d_2}\right)\int\limits_X^{2X}x^{2}\,\mathrm dx.\end{align*}

We will examine the contribution of the different pairs $(d_1,d_2)$ separately at first. The terms where $d_1=d_2=1$ contribute

(10·1) \begin{align}\frac{k^2}{24\pi^4}\sum_{m\leq N}\frac{\left|A(1,m)\right|^2}{m^{2}}\left(\frac{k^2-k-1}{k-1}1_{m\not\equiv 0\,(k)}+1_{m\equiv 0\,(k)}\right)\int\limits_X^{2X}x^2\,\mathrm dx\end{align}

by Lemma 9·1.

Similarly, the terms with $d_1=d_2=k$ contribute

(10·2) \begin{align}\frac{k^2}{24\pi^4}\sum_{m\leq N}\frac{\left|A(k,m)\right|^2}{k^{2}m^{2}}\int\limits_X^{2X}x^{2}\,\mathrm d x\end{align}

and the terms where $d_1d_2=k$ contribute

(10·3) \begin{align}-\frac{k^2}{12\pi^4}\Re\left(\sum_{m\leq N/k^2}\frac{A(1,k^2m)\overline{A(k,m)}}{k^{3}m^{2}}\right)\int\limits_X^{2X}x^{2}\,\mathrm d x.\end{align}

Our aim is to show that these three terms (10·1), (10·2) and (10·3), sum up to $C(k)k^3X^{3}$ , where $C(k)\in\mathbb R$ with $C(k)\asymp 1$ . The argument is similar as in the proof of Theorem 9·3.

The upper bound of the right order of magnitude follows easily from the averaged estimates for the Fourier coefficients (4·2). For the lower bound, observe first that by using the arithmetic-geometric mean inequality the absolute value of (10·3) is bounded from above by

(10·4) \begin{align}&\frac{k^2}{24\pi^4}\sum_{m\leq N/k^2}\frac{\left|A(1,k^2m)\right|^2}{k^{4}m^{2}}\int\limits_X^{2X}x^{2}\,\mathrm dx+\frac{k^2}{24\pi^4}\sum_{m\leq N/k^2}\frac{\left|A(k,m)\right|^2}{k^{2}m^{2}}\int\limits_X^{2X}x^{2}\,\mathrm d x \nonumber\\[8pt]&\leq \frac{k^2}{24\pi^4}\sum_{m\leq N}\frac{\left|A(1,m)\right|^2}{m^{2}}\int\limits_X^{2X}x^{2}\,\mathrm dx+\frac{k^2}{24\pi^4}\sum_{m\leq N}\frac{\left|A(k,m)\right|^2}{k^{2}m^{2}}\int\limits_X^{2X}x^{2}\,\mathrm d x .\end{align}

Notice that the latter term in (10·4) is exactly (10·2). Hence the sum of (10·1), (10·2) and (10·3) is bounded from below by

\begin{equation*}\frac{k^2}{24\pi^4}\cdot\frac{k^2-2k}{k-1}\sum_{m\leq N}\frac{|A(1,m)|^2}{m^2}\left(1-1_{m\equiv 0\,(k)}\right)\int\limits_X^{2X}x^2\,\mathrm d x.\end{equation*}

This gives the lower bound $\gg k^3X^3$ as in the proof of Theorem 9·3.

The contribution arising from the terms $d_1^2m=d_2^2n$ where the signs in the sum $\sum_\pm$ are chosen to be distinct is by the first derivative test and Weil’s bound given by

\begin{align*}&\ll k^2\sum_\pm\sum_{d_1|k}\sum_{d_2|k}\sum_{m\leq N}\sum_{\substack{n\leq N\\d_1^2m=d_2^2n}}\frac{|A(d_1,m)\overline{A(d_2,n)|}}{d_1d_2mn}\left|S\left(\overline h,\pm m;\,\frac k{d_1}\right)S\left(\overline h,\mp n;\,\frac k{d_2}\right)\right|\\&\qquad\times\int\limits_X^{2X}x^{2}\,e\!\left(\frac{\pm 6(d_1^2mx)^{1/3}}k\right)\mathrm dx\\&\ll k^2\cdot k\,X^{2}\,X^{2/3}k\sum_{m\leq N}\frac{\left|A(1,m)\right|^2}{m^{7/3}}\ll X^{8/3}k^{4}\end{align*}

as in the proof of Theorem 9·3.

Next we study the off-diagonal contribution in the same sum, which is given, up to a constant, by

\begin{align*}&k^2\sum_\pm\sum_\pm\sum_{d_1|k}\sum_{d_2|k}\underset{d_1^2m\neq d_2^2n}{\sum_{m\leq N}\sum_{n\leq N}}\frac{A(m,d_1)\overline{A(n,d_2)}}{mnd_1d_2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}S\left(\overline h,\pm m;\,\frac k{d_1}\right)S\left(\overline h,\pm n;\,\frac k{d_2}\right)\\&\qquad\qquad\times\int\limits_X^{2X}x^2e\,\left(\pm\frac{3(mxd_1^2)^{1/3}}k\mp\frac{3(nxd_2^2)^{1/3}}k\right)\,\mathrm d x.\end{align*}

We restrict to the case where the signs in $\sum_\pm$ are chosen to be the same as the case of distinct signs can be treated similarly with easier arguments. The prior terms contribute, using again the first derivative test and Weil’s bound,

\begin{align*}\ll k^3\sum_{d_1|k}\sum_{d_2|k}\underset{d_1^2m\neq d_2^2n}{\sum_{m\leq N}\sum_{n\leq N}}\frac{|A(m,d_1)\overline{A(n,d_2)}|}{mn(d_1d_2)^{3/2}}\cdot\frac{X^{8/3}k}{\bigl|d_1^{2/3}m^{1/3}-d_2^{2/3}n^{1/3}\bigr|}.\end{align*}

By symmetry it is enough to treat the terms in the m-sum with $d_1^2m\lt d_2^2n$ . For these we split the m-sum into two parts according to whether $d_1^2m\lt d_2^2n/2$ or $d_2^2n/2\leq d_1^2m\lt d_2^2n$ . Note that in any case we have

(10·5) \begin{align}\frac1{d_2^{3/2}}\cdot\frac1{|d_2^{2/3}n^{1/3}-d_1^{2/3}m^{1/3}|}\ll\frac{n^{2/3}}{d_2^{1/6}|d_2^2n-d_1^2m|}.\end{align}

Using this, the terms with $d_1^2m\lt d_2^2n/2$ contribute

\begin{align*}&\ll X^{8/3}k^{4}\sum_{d_1|k}\sum_{d_2|k}\sum_{m\leq N}\frac{|A(d_1,m)|}{md_1^{3/2}}\underset{d_1^2m\lt d_2^2n/2}{\sum_{n\leqslant N}}\frac{|A(d_2,n)|}{nd_2^{1/6}}\cdot\frac{n^{2/3}}{|d_2^2n-d_1^2m|}\\[6pt]&\ll X^{8/3}k^{4}\sum_{d_1|k}\sum_{d_2|k}\sum_{m\leq N}\frac{|A(d_1,m)|}{md_1^{3/2}}\underset{d_1^2m\lt d_2^2n/2}{\sum_{n\leqslant N}}\frac{|A(d_2,n)|}{d_2^{13/6}n^{4/3}}\ll_{\varepsilon} X^{8/3+\varepsilon}k^{4}.\end{align*}

In the last step we have used partial summation together with the averaged bounds (4·2) and the fact that $\vartheta\leq 5/14$ .

For the terms with $d_2^2n/2\leq d_1^2m\lt d_2^2n$ we note that

\begin{equation*}m\geq m^{1/2}\left(\frac{d_2^2n}{2d_1^2}\right)^{1/2}.\end{equation*}

Combining this with (10·5) shows that the terms with $d_2^2n/2\leq d_1^2m\lt d_2^2n$ contribute

\begin{align*}& \ll X^{8/3}k^4\sum_{d_1|k}\sum_{d_2|k}\sum_{m\leq N}\underset{d_2^2n/2\leq d_1^2m \lt d_2^2n}{\sum_{n\leqslant N}}\frac{|A(d_1,m)A(d_2,n)|}{m^{1/2}n^{5/6}d_1^{1/2}d_2^{7/6}}\cdot\frac1{|d_2^2n-d_1^2m|}.\end{align*}

Applying the Cauchy–Schwarz inequality shows that this is bounded by

(10·6) \begin{align}&\ll k^4X^{8/3}\sum_{d_1|k}\sum_{d_2|k}\frac1{d_1^{1/2}d_2^{7/6}}\left(\sum_{m\leq N}\underset{d_2^2n/2\leq d_1^2m \lt d_2^2n}{\sum_{n\leqslant N}}\frac{|A(m,d_1)A(n,d_2)|^2}{mn^{5/3}}\right)^{1/2} \nonumber\\[6pt]&\qquad\times\left(\sum_{m\leq N}\underset{d_2^2n/2\leq d_1^2m \lt d_2^2n}{\sum_{n\leqslant N}}\frac1{(d_2^2n-d_1^2m)^2}\right)^{1/2}\\[6pt]&\ll_{\varepsilon} k^4X^{8/3+\varepsilon}\sum_{d_1|k}\sum_{d_2|k}\frac1{d_1^{1/2-\vartheta}d_2^{7/6-\vartheta}} \nonumber\\[3pt]&\ll_{\varepsilon} k^4X^{8/3+\varepsilon} \nonumber.\end{align}

Above we have used partial summation together with (4·2) and the observation that the final double sum in (10·6) is uniformly bounded with respect to $d_1$ and $d_2$ . This finishes the proof.

As a consequence of the above theorem we obtain lower bounds for certain averaged mean squares. Theorem 2·2 follows immediately from the latter bound by (1·4).

Theorem 10·2. Suppose that k is a prime so that $k\ll X^{1/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then the lower bounds

\begin{equation*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg X^3k^3\quad\text{and}\quad\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg X^{5/3}k\end{equation*}

hold for sufficiently large X.

Indeed, for any $0\lt H\leq X$ to be chosen later, we compute

(10·7) \begin{align}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx \nonumber\\&=\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\int\limits_x^{x+H}\widetilde A_1\left(x;\,\frac{h}{k}\right)\,\mathrm d t\right|^2\,\mathrm d x \nonumber\\&=\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\int\limits_x^{x+H}\left(\widetilde A_1\left(t;\,\frac{h}{k}\right)-\int\limits_x^t\widetilde A_0\left(u;\,\frac{h}{k}\right)\,\mathrm d u\right)\,\mathrm d t\right|^2\,\mathrm d x \nonumber\\&\ll\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\int\limits_x^{x+H}\widetilde A_1\left(t;\,\frac{h}{k}\right)\,\mathrm d t\right|^2\,\mathrm d x+\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\int\limits_x^{x+H}\int\limits_x^t\widetilde A_0\left(u;\,\frac{h}{k}\right)\,\mathrm d u\,\mathrm d t\right|^2\,\mathrm d x.\end{align}

By simply integrating, the first term on the previous line is clearly

(10·8) \begin{align}\ll\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X+H}\left|\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx.\end{align}

Using the upper bound $\widetilde A_2(x;\,h/k)\ll x^{5/3}k^{5/2}$ from Corollary 7·4 we see that (10·8) is

\begin{equation*}\ll\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H X^{10/3}k^5.\end{equation*}

The latter term in (10·7) can be estimated as follows. First note that

\begin{align*}\int\limits_x^{x+H}\int\limits_x^t\widetilde A_0\left(u;\,\frac{h}{k}\right)\,\mathrm d u\,\mathrm d t&=\int\limits_x^{x+H}\widetilde A_0\left(u;\,\frac{h}{k}\right)(u-x)\,\mathrm d u\ll H\int\limits_x^{x+H}\left|\widetilde A_0\left(u;\,\frac{h}{k}\right)\right|\,\mathrm d u.\end{align*}

Using this together with the Cauchy–Schwarz inequality yields

\begin{align*}\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\int\limits_x^{x+H}\int\limits_x^t\widetilde A_0\left(u;\,\frac{h}{k}\right)\,\mathrm d u\,\mathrm d t\right|^2\,\mathrm d x &\ll\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left(\int\limits_x^{x+H}\left|\widetilde A_0\left(u;\,\frac{h}{k}\right)\right|\mathrm d u\right)^2\,\mathrm d x\\&\ll H\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X+H}\int\limits_x^{x+H}\left|\widetilde A_0\left(u;\,\frac{h}{k}\right)\right|^2\mathrm d u\,\mathrm d x \\&\ll H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X+H}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm d x\\&\ll H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm d x.\end{align*}

Thus we have shown that

\begin{align*}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\\&\ll\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H X^{10/3}k^5.\end{align*}

As the second moment of $\widetilde A_2(x;\,h/k)$ averaged over h modulo k is $\asymp X^{13/3}k^5$ when $k\ll X^{1/3-\delta}$ by Theorem 9·3, this implies

(10·9) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx&\ll\frac{X^{13/3}k^5}{H^2}+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H X^{10/3}k^5 \nonumber \\&\ll\frac{X^{13/3}k^5}{H^2}+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx,\end{align}

where the last estimate holds as $H\leq X$ .

Next we shall optimise the choice of the parameter H. Actually, before that we need a preliminary lower bound for the moment

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\end{align*}

we are interested in. This is to guarantee that our optimal choice for H satisfies the constraint $H\leq X$ .

We make the initial choice $H=X$ in (10·9). This gives

(10·10) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg Xk^3\end{align}

by Theorem 10·1 as $k\ll X^{1/3-\delta}$ . We now make the final optimal choice for H by choosing

\begin{equation*}H=\frac{X^{13/12}k^{5/4}}{\sqrt[4]{\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx}}.\end{equation*}

This satisfies the constraint $H\leq X$ for sufficiently large X as by (10·10) and $k\ll X^{1/3-\delta}$ we have

\begin{equation*}H\ll\frac{X^{13/12}k^{5/4}}{X^{1/4}k^{3/4}}=X^{5/6}k^{1/2}\ll X^{1-\delta/2}.\end{equation*}

Using this choice for H in (10·9) gives

\begin{equation*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg X^{5/3}k\end{equation*}

as, by the previous theorem, the mean-square of $\widetilde A_1(x;\,h/k)$ averaged over h modulo k is $\asymp X^3k^3$ when $k\ll X^{1/3-\delta}$ . Hence, the second bound is also established.

11. An improved pointwise bound for long sums with rational twists

In this section we improve the best known upper bound for rationally additively twisted sums with sufficiently small denominators.

Theorem 11·1. Let $x\in\left[1,\infty\right[$ , and let h and k be coprime integers with $1\leqslant k\ll x^{1/3}$ . Then

\begin{equation*}\sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\ll_{\varepsilon} k^{3/4}\,x^{1/2+\vartheta/2+\varepsilon}.\end{equation*}

Proof. By the estimate (9·11) it is enough to prove the desired bound for $\widetilde A_0(x;\,h/k)$ . Let $H\in\left[1,\infty\right[$ satisfy $H\ll x$ , which will be chosen at the end. Estimating a short exponential sum by absolute values yields

\begin{equation*}\int\limits_x^{x+H}\widetilde A_0\!\left(t;\,\frac{h}{k}\right)\,\mathrm dt-H\,\widetilde A_0\!\left(x;\,\frac{h}{k}\right)=\int\limits_x^{x+H}\left(\widetilde A_0\!\left(t;\,\frac{h}{k}\right)-\widetilde A_0\!\left(x;\,\frac{h}{k}\right)\right)\mathrm dt\ll_\varepsilon H^2\,x^{\vartheta+\varepsilon},\end{equation*}

so that

\begin{equation*}H\,\widetilde A_0\!\left(x;\,\frac{h}{k}\right)=\int\limits_x^{x+H}\widetilde A_0\!\left(t;\,\frac{h}{k}\right)\,\mathrm dt+O(H^2\,x^{\vartheta+\varepsilon}).\end{equation*}

We know from Corollary 8·5 that

\begin{equation*}\widetilde A_1\!\left(x;\,\frac{h}{k}\right)\ll_{\varepsilon} k^{3/2}\,x^{1+\varepsilon}.\end{equation*}

This bound gives

\begin{align*}\widetilde A_0\!\left(x;\,\frac{h}{k}\right)&=\frac1{H}\int\limits_{x}^{x+H}\widetilde A_0\!\left(t;\,\frac{h}{k}\right)\,\mathrm dt+O(H\,x^{\vartheta+\varepsilon})\\&=\frac1{H}\left(\widetilde A_1\!\left(x+H;\,\frac{h}{k}\right)-\widetilde A_1\!\left(x;\,\frac{h}{k}\right)\right)+O(H\,x^{\vartheta+\varepsilon})\\&\ll_{\varepsilon}\frac1H\,k^{3/2}\,x^{1+\varepsilon}+H\,x^{\vartheta+\varepsilon}.\end{align*}

Choosing $H=k^{3/4}\,x^{(1-\vartheta)/2}$ yields

\begin{equation*}\widetilde A_0\!\left(x;\,\frac{h}{k}\right)\ll_{\varepsilon} k^{3/4}\,x^{1/2+\vartheta/2+\varepsilon},\end{equation*}

as desired.

12. Short second moment for Riesz weighted sums

The goal of this section is to compute the short second moment of Riesz weighted sums for $a=2$ and $a=3$ . Let us start by listing some consequences of the first derivative test.

Lemma 12·1. Let $\Xi,U,T\in\left[0,X\right]$ and $k,m,n,d_1,d_2\in\mathbb Z_+$ . Then

\begin{align*}&\int\limits_X^{X+\Xi}x^{(4a+2)/3}\,e\!\left(\pm\frac{3\!\left(d_1^{2/3}m^{1/3}\left(x+T\right)^{1/3}+d_2^{2/3}n^{1/3}\left(x+U\right)^{1/3}\right)}k\right)\mathrm dx\\& \ll\frac{X^{(4a+4)/3}k}{d_1^{2/3}m^{1/3}+d_2^{2/3}n^{1/3}}.\end{align*}

Lemma 12·2. Let $\Xi,U,T\in\left[0,X\right]$ and $k,m,n,d_1,d_2\in\mathbb Z_+$ with $U\leqslant T$ and $d_1^2m\lt d_2^2n$ . Then

\begin{align*}&\int\limits_X^{X+\Xi}x^{(4a+2)/3}\,e\!\left(\pm\frac{3\!\left(d_1^{2/3}m^{1/3}\left(x+T\right)^{1/3}-d_2^{2/3}n^{1/3}\left(x+U\right)^{1/3}\right)}k\right)\mathrm dx\\& \ll\frac{X^{(4a+4)/3}k}{d_2^{2/3}n^{1/3}-d_1^{2/3}m^{1/3}}.\end{align*}

Lemma 12·3. Let $\Xi,T\in\left]0,X\right]$ and $k,m,n,d_1,d_2\in\mathbb Z_+$ with $d_1^2m\lt d_2^2n$ , $d_1\geqslant d_2$ , as well as $d_1^2 m\leqslant X/6T$ . Then

(12·1) \begin{align}&\int\limits_X^{X+\Xi}x^{(4a+2)/3}\,e\!\left(\pm\frac{3\!\left(d_1^{2/3}m^{1/3}\,x^{1/3}-d_2^{2/3}n^{1/3}\left(x+T\right)^{1/3}\right)}k\right)\mathrm dx \nonumber \\&\qquad\qquad\ll\frac{X^{(4a+4)/3}\,k\,n^{2/3}\,d_2^{-2/3}}{n-m}.\end{align}

The first two lemmas are straightforward to prove, but the third one requires some explanation. The derivative of the phase function is, up to a sign, given by

\begin{align*}\frac{d_1^{2/3}m^{1/3}}{x^{2/3}k}-\frac{d_2^{2/3}n^{1/3}}{(x+T)^{2/3}k}=\frac{d_1^{2/3}m^{1/3}\left(x+T\right)^{2/3}-d_2^{2/3}n^{1/3}\,x^{2/3}}{\left(x+T\right)^{2/3}\,x^{2/3}\,k}.\end{align*}

The denominator is of course $\asymp X^{4/3}k$ , and the numerator is, by using the identity $x^3-y^3=(x-y)(x^2+xy+y^2)$ , given by

\begin{align*}\frac{\left(d_1^2m-d_2^2n\right)x^2+2\,x\,Td_1^2m+d_1^2m\,T^2}{\left(d_1^{2/3}m^{1/3}\left(x+T\right)^{2/3}\right)^2+d_1^{2/3}m^{1/3}\,d_2^{2/3}n^{1/3}\,x^{2/3}\left(x+T\right)^{2/3}+\left(d_2^{2/3}n^{1/3}\,x^{2/3}\right)^2}.\end{align*}

In the numerator, the denominator is given by $\asymp d_2^{4/3}n^{2/3}\,X^{4/3}$ and the numerator is

\begin{equation*}\asymp\left(d_2^2n-d_1^2m\right)X^2.\end{equation*}

Now (12·1) follows immediately from Lemma 9·2.

Let us then proceed to the proof of the short mean square result for Riesz weighted exponential sums related to Hecke–Maass cusp forms.

Theorem 12·4. Let $X\in\left[1,\infty\right[$ , k be a prime, and let $\Delta,\Xi\in\left[1,X\right]$ satisfyFootnote ⁵ $X^{1/2+\delta}k^{3/2}\ll\Delta\ll X^{2/3-\delta}k$ , and $X^{2+\delta}k^3\ll\Xi\,\Delta^2$ for any sufficiently small fixed $\delta\gt0$ . Then, for X large enough,

\begin{equation*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{X+\Xi}\left|\widetilde A_2\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\mathrm dx\asymp\Xi\,\Delta^2\,X^2k^3\end{equation*}

and

\begin{equation*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_3\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_3\left(x;\,\frac{h}{k}\right)\right|^2\mathrm dx\asymp\Xi\,\Delta^2\,X^{10/3}k^5.\end{equation*}

Proof. We shall actually prove a more general result concerning $\widetilde A_a(x;\,h/k)$ for any integer $a\geqslant2$ satisfying $a\equiv0\pmod3$ or $a\equiv2\pmod3$ . Our starting point is the formula

\begin{align*}&\widetilde A_a\left(\!x+\Delta;\,\frac{h}{k}\right)-\widetilde A_a\left(x;\,\frac{h}{k}\right)=\frac{({-}1)^ak^a}{(2\pi)^{a+1}\sqrt3}\sum_\pm i^{\pm a}\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}S\!\left(\overline h,\pm m;\,\frac kd\right)\\&\qquad\times\left(\left(x+\Delta\right)^{(2a+1)/3}e\!\left(\pm\frac{3(md^2(x+\Delta))^{1/3}}k\right)-x^{(2a+1)/3}\,e\!\left(\pm\frac{3(md^2x)^{1/3}}k\right)\right)\\&\qquad\qquad+O\,\left(X^{2a/3}k^{(2a+3)/2}\right),\end{align*}

which follows from Corollary 7·3. Note that d(k) does not appear in the error term as we assume k to be a prime.

Exchanging here the factor $(x+\Delta)^{(2a+1)/3}$ to $x^{(2a+1)/3}$ costs $\ll\Delta\,X^{(2a-2)/3}k^{(2a+1)/2}$ , which is $\ll X^{2a/3}k^{(2a+3)/2}$ provided that $\Delta\ll kX^{2/3-\delta}$ . Thus,

\begin{align*}&\widetilde A_a\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_a\left(x;\,\frac{h}{k}\right)\\&=\frac{({-}1)^ax^{(2a+1)/3}k^a}{(2\pi)^{a+1}\sqrt3}\sum_\pm i^{\pm a}\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}S\left(\overline h,\pm m;\,\frac kd\right)\\&\qquad\times\left(e\!\left(\pm\frac{3(md^2(x+\Delta))^{1/3}}k\right)-e\!\left(\pm\frac{3(md^2x)^{1/3}}k\right)\right)+O\,\left(X^{2a/3}k^{(2a+3)/2}\right).\end{align*}

Now the second moment may be expanded as

\begin{align}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{X+\Xi}\left|\widetilde A_a\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_a\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx \nonumber\\&=\frac{k^{2a}}{(2\pi)^{2a+2}\cdot3} \nonumber\end{align}

(12·2) \begin{align}&\!\!\!\!\times\int\limits_X^{X+\Xi}\!x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|\sum_\pm i^{\pm a}\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}S\!\left(\overline h,\pm m;\,\frac kd\right)\left(e\!\left(\ldots\right)-e\!\left(\ldots\right)\right)\right|^2\!\mathrm dx \nonumber\\[5pt]&\!\!-\frac{({-}1)^ak^a}{2^a\pi^{a+1}\sqrt3}\Re\int\limits_X^{X+\Xi}x^{(2a+1)/3}\sum_\pm i^{\pm a}\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}S\left(\overline h,\pm m;\,\frac kd\right) \nonumber \\[3pt]&\qquad\qquad\qquad\times\left(e\!\left(\ldots\right)-e\!\left(\ldots\right)\right)\cdot O\big(X^{2a/3}k^{(2a+3)/2}\big)\,\mathrm dx+O\big(\Xi\,X^{4a/3}k^{2a+3}\big).\end{align}

Once we have proved that the first term is $\asymp\Xi\,\Delta^2\,X^{(4a-2)/3}k^{2a-1}$ , then the middle term is by the Cauchy–Schwarz inequality

(12·3) \begin{align}\ll\sqrt{\Xi\,\Delta^2\,X^{(4a-2)/3}k^{2a-1}}\sqrt{\Xi\,X^{4a/3}k^{2a+3}}\ll\Xi\,\Delta\,X^{(4a-1)/3}\,k^{2a+1}.\end{align}

Thus, we can focus on the first term on the right-hand side of (12·2). In the course of examining this term various subterms arise. We will derive an upper bound for each of them and at the end of the proof show that these terms contribute less than the main term that will turn out to be $\asymp\Xi\,\Delta^2\,X^{(4a-2)/3}\,k^{2a-1}$ .

For $\lambda\in\{0,\Delta\}$ we write

\begin{equation*}s_\lambda^+\,:\!=\,i^a\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}e\!\left(\frac{3(md^2(x+\lambda))^{1/3}}k\right)S\left(\overline h,m;\,\frac kd\right)\end{equation*}

and

\begin{equation*}s_\lambda^-\,:\!=\,i^{-a}\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}e\!\left({-}\frac{3(md^2(x+\lambda))^{1/3}}k\right)S\left(\overline h,-m;\,\frac kd\right),\end{equation*}

so that, up to a constant, the integral under study takes the shape

(12·4) \begin{align} & k^{2a}\!\int\limits_X^{X+\Xi}\!\!x^{(4a+2)/3}\!\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|\sum_\pm i^{\pm a}\sum_{d|k}\sum_{m=1}^\infty\frac{A(d,m)}{m^{(a+2)/3}d^{(2a+1)/3}}S\!\left(\overline h,\pm m;\,\frac kd\right)\left(e\!\left(\ldots\right)-e\!\left(\ldots\right)\right)\right|^2\mathrm dx \nonumber\\[3pt]&=k^{2a}\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|\left(s_\Delta^+-s_0^+\right)+\left(s_\Delta^{-\,-}s_0^-\right)\right|^2\,\mathrm d x \nonumber\\[3pt]&= k^{2a}\sum_{\pm}\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|s_\Delta^\pm-s_0^\pm\right|^2\mathrm dx \nonumber \\[3pt]&\qquad\qquad\quad+2k^{2a}\Re\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left(s_\Delta^+-s_0^+\right)\overline{\left(s_\Delta^{-\,-}s_0^-\right)}\,\mathrm d x,\end{align}

where we sum over both choices of the sign $\pm$ . Furthermore, we split each of the sums $s_\lambda^\pm$ into parts

\begin{equation*}s_\lambda^{\pm,\leqslant}\,:\!=\,i^{\pm a}\sum_{d|k}\sum_{m\leqslant N_0}\ldots\quad\text{and}\quad s_\lambda^{\pm,\gt}\,:\!=\,i^{\pm a}\sum_{d|k}\sum_{m\gt N_0}\ldots,\end{equation*}

where we set $N_0\,:\!=\,X/6\Delta$ .

Now the first integral on the right-hand side of (12·4) is

(12·5) \begin{align}&k^{2a}\sum_\pm\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|s_\Delta^{\pm,\leqslant}-s_0^{\pm,\leqslant}\right|^2\mathrm dx\\&\qquad+k^{2a}\sum_\pm\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|s_\Delta^{\pm,\gt}-s_0^{\pm,\gt}\right|^2\mathrm dx \nonumber\\&\qquad+2k^{2a}\Re\sum_\pm\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left(s_\Delta^{\pm,\leqslant}-s_0^{\pm,\leqslant}\right)\overline{\left(s_\Delta^{\pm,\gt}-s_0^{\pm,\gt}\right)}\mathrm dx. \nonumber\end{align}

Of the three new integrals we treat the last one first. It is given by

\begin{align*}&2k^{2a}\Re\sum_\pm\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left(s_\Delta^{\pm,\leqslant}-s_0^{\pm,\leqslant}\right)\overline{\left(s_\Delta^{\pm,\gt}-s_0^{\pm,\gt}\right)}\,\mathrm dx\\&=k^{2a}\sum_\pm\!\int\limits_X^{X+\Xi}\!x^{(4a+2)/3}\!\left(\overline{\left(s_\Delta^{\pm,\leqslant}-s_0^{\pm,\leqslant}\right)}\left(s_\Delta^{\pm,\gt}-s_0^{\pm,\gt}\right)+\left(s_\Delta^{\pm,\leqslant}-s_0^{\pm,\leqslant}\right)\overline{\left(s_\Delta^{\pm,\gt}-s_0^{\pm,\gt}\right)}\right)\!\mathrm dx\\&\ll k^{2a}\sum_\pm\sum_{d_1|k}\sum_{m\leq N_0}\frac{A\left(d_1,m\right)}{m^{(a+2)/3}d_1^{(2a+1)/3}}S\left(\overline h,\pm m;\,\frac k{d_1}\right)\\&\qquad\qquad\times\sum_{d_2|k}\sum_{n\gt N_0}\frac{\overline{A\left(d_2,n\right)}}{n^{(a+2)/3}d_2^{(2a+1)/3}}S\left(\overline h,\pm n;\,\frac k{d_2}\right)\\&\quad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left(e\!\left(\mp\frac{3\left(x+\Delta\right)^{1/3}\left(m^{1/3}d_1^{2/3}-n^{1/3}d_2^{2/3}\right)}k\right) \right. \\&\left. \qquad\qquad\qquad-e\!\left(\mp\frac{3\left(\left(x+\Delta\right)^{1/3}m^{1/3}d_1^{2/3}-x^{1/3}n^{1/3}d_2^{2/3}\right)}k\right)\right. \\&\left.\qquad\qquad\qquad\qquad \qquad\quad-e\!\left(\mp\frac{3\left(x^{1/3}m^{1/3}d_1^{2/3}-\left(x+\Delta\right)^{1/3}n^{1/3}d_2^{2/3}\right)}k\right) \right. \\&\left. \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad+e\!\left(\mp\frac{3x^{1/3}\left(m^{1/3}d_1^{2/3}-n^{1/3}d_2^{2/3}\right)}k\right)\right)\mathrm d x.\end{align*}

As $n\gt m$ and k is a prime, the diagonal contribution from $d_1^2m=d_2^2n$ can only arise when $d_1=k$ and $d_2=1$ in the integrals corresponding to the first and fourth exponential term on the right-hand side of the previous display. This in particular means that $N_0\lt n=k^2m$ and so we have $m\gt N_0/k^2$ in the diagonal terms corresponding to the first and fourth exponential term. Hence, their total contribution is, for any sufficiently small fixed $\varepsilon\gt0$ ,

(12·6) \begin{align}&\ll k^{2a}\sum_{N_0/k^2\lt m\leq N_0}\frac{\left|A\left(k,m\right)\right|}{m^{(a+2)/3}k^{(2a+1)/3}}\cdot\frac{\left|A\left(1,k^2m\right)\right|}{\left(k^2m\right)^{(a+2)/3}}\cdot k^{1/2}\Xi X^{(4a+2)/3} \nonumber\\&\ll_{\varepsilon} k^{3\vartheta+2a/3-7/6}\Xi X^{(4a+2)/3}\sum_{N_0/k^2\lt m\leq N_0}\frac1{m^{(2a+4)/3-2\vartheta-2\varepsilon}} \nonumber\\&\ll_{\varepsilon} k^{3\vartheta+2a/3-7/6}\Xi X^{(4a+2)/3}\left(\frac{N_0}{k^2}\right)^{-2/3-\varepsilon}\sum_{N_0/k^2\lt m\leq N_0}\frac1{m^{(2a+2)/3-2\vartheta-3\varepsilon}} \nonumber\\&\ll_{\varepsilon} k^{3\vartheta+2a/3+1/6+2\varepsilon}\Delta^{2/3+\varepsilon}\Xi X^{4a/3-{\varepsilon}},\end{align}

where we have applied the pointwise bound $\left|A\left(m,n\right)\right|\ll (mn)^{\vartheta+\varepsilon}$ in the second step and in the final step we recall the definition of $N_0=X/6\Delta$ .

The diagonal contribution in the integrals corresponding to the two other exponentials are estimated similarly, noting that by the first derivative test

\begin{align*}\int\limits_X^{X+\Xi}x^{(4a+2)/3}e\!\left(\mp\frac{3m^{1/3}d^{2/3}\left(\left(x+\Delta\right)^{1/3}-x^{1/3}\right)}k\right)\,\mathrm d x\ll\frac{X^{(4a+7)/3}\,k}{\Delta\,m^{1/3}d^{2/3}}.\end{align*}

We may estimate the off-diagonal terms with $d_1^2m\neq d_2^2n$ by Lemma 12·2 or Lemma 12·3 depending on the term in question. By symmetry it is enough to consider the terms with $d_1^2m\lt d_2^2n$ . As a model case, we consider the sum

\begin{align*}&k^{2a}\sum_\pm\sum_{d_1|k}\sum_{m\leq N_0}\frac{A\left(d_1,m\right)}{m^{(a+2)/3}d_1^{(2a+1)/3}}S\left(\overline h,\pm m;\,\frac k{d_1}\right)\sum_{d_2|k}\sum_{n\gt N_0}\frac{\overline{A\left(d_2,n\right)}}{n^{(a+2)/3}d_2^{(2a+1)/3}}S\left(\overline h,\pm n;\,\frac k{d_2}\right)\\[5pt]&\qquad\qquad\quad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}e\!\left(\mp\frac{3\left(x^{1/3}m^{1/3}d_1^{2/3}-\left(x+\Delta\right)^{1/3}n^{1/3}d_2^{2/3}\right)}k\right)\,\mathrm d x.\end{align*}

For simplicity, we further restrict to terms with $d_1=k$ , $d_2=1$ (in which case $n=d_2^2n\gt d_1^2m=k^2m$ ) as the other terms are easier to deal with.

Note that when $N_0/k^2\lt m\leq N_0$ we may argue as in the diagonal case to see that these terms are by Weil’s bound

(12·7) \begin{align}&\ll k^{2a+1/2}\sum_{N_0/k^2\leq m\leq N_0}\frac{|A(k,m)|}{m^{(a+2)/3}k^{(2a+1)/3}}\sum_{n\gt N_0}\frac{|A(1,n)|}{n^{(a+2)/3}}\cdot X^{(4a+2)/3}\Xi \nonumber\\&\ll k^{4a/3+1/6+\vartheta}X^{(4a+2)/3}\Xi N_0^{-a/3+1/3}\left(\frac{N_0}{k^2}\right)^{-a/3+1/3} \nonumber\\&\ll k^{2a-1/2+\vartheta}X^{(2a+4)/3}\Xi\Delta^{(2a-2)/3}.\end{align}

For the terms with $m\leq N_0/k^2$ we use Lemma 12·3 and Weil’s bound to estimate the sum in question as

(12·8) \begin{align}&\ll k^{2a+1/2}\sum_{m\leqslant N_0/k^2}\frac{\left|A(k,m)\right|}{m^{(a+2)/3}k^{(2a+1)/3}}\sum_{n\gt N_0}\frac{\left|A(1,n)\right|}{n^{(a+2)/3}}\cdot\frac{X^{(4a+4)/3}\,n^{2/3}k}{|n-m|} \nonumber\\&\ll k^{4a/3+7/6}X^{(4a+4)/3}\sum_{m\leq N_0/k^2}\frac{|A(k,m)|}{m^{(a+2)/3}}\sum_{n\gt N_0}\frac{|A(1,n)|}{n^{a/3}}\cdot\frac1{|n-m|}.\end{align}

To the inner sum we apply the Cauchy–Schwarz inequality and partial summation (together with the estimate (4·2)) to see that

\begin{align*}\sum_{n\gt N_0}\frac{|A(1,n)|}{n^{a/3}}\cdot\frac1{|n-m|}&\leq\left(\sum_{n\gt N_0}\frac{|A(1,n)|^2}{n^{2a/3}}\right)^{1/2}\left(\sum_{n\gt N_0}\frac1{(n-m)^2}\right)^{1/2} \\&\ll N_0^{-a/3+1/2},\end{align*}

where the last estimate holds uniformly in $m\leq N_0$ . Using this in (12·8) gives a contribution

(12·9) \begin{align}&\ll k^{4a/3+7/6}X^{(4a+4)/3}\sum_{m\leq N_0/k^2}\frac{|A(k,m)|^2}{m^{(a+2)/3}}\left(\frac X\Delta\right)^{-a/3+1/2} \nonumber \\&\ll_\varepsilon k^{4a/3+7/6+2\vartheta+\varepsilon}X^{a+11/6}\Delta^{a/3-1/2},\end{align}

where we have in addition used partial summation together with (4·2).

Note also that the second integral on the right-hand side of (12·4) can be treated using even simpler arguments. Indeed, applying Lemma 12·1 to bound the exponential integrals, and estimating then trivially using the Weil bound shows that these terms contribute

(12·10) \begin{align}&\ll k^{2a+1}\sum_{d_1|k}\sum_{m=1}^\infty\frac{|A(d_1,m)|}{m^{(a+2)/3}d_1^{(2a+1)/3}}\sum_{d_2|k}\sum_{n=1}^\infty\frac{|A(d_2,n)|}{n^{(a+2)/3}d_1^{(2a+1)/3}}\cdot\frac1{\sqrt{d_1d_2}} \nonumber\\&\qquad\qquad\qquad\qquad\times\frac{X^{(4a+4)/3}k}{d_1^{2/3}m^{1/3}+d_2^{2/3}n^{1/3}} \nonumber \\&\ll X^{(4a+4)/3}k^{2a+2}.\end{align}

Let us next treat the middle term in (12·5). Using the triangle inequality it is clearly enough to estimate the integrals

\begin{equation*}k^{2a}\int\limits_X^{X+\Xi}x^{(4a+2)/3}\,\bigl|s_\lambda^{\pm,\gt}\bigr|^2\,\mathrm dx.\end{equation*}

We expand the square to see that the integral is bounded from above by the expression

\begin{align*}&k^{2a}\sum_{d_1|k}\sum_{d_2|k}\sum_{m\gt N_0}\sum_{n\gt N_0}\frac{\left|A\left(d_1,m\right)A\left(d_2,n\right)\right|}{m^{(a+2)/3}n^{(a+2)/3}d_1^{(2a+1)/3}d_2^{(2a+1)/3}}S\left(\overline h,\pm m;\,\frac k{d_1}\right)S\left(\overline h,\pm n;\,\frac k{d_2}\right)\\&\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}e\left(\pm\frac{3\left(x+\lambda\right)^{1/3}\left(m^{1/3}d_1^{2/3}-n^{1/3}d_2^{2/3}\right)}k\right)\mathrm d x.\end{align*}

We treat the diagonal and off-diagonal terms separately. Let us first concentrate on the diagonal, which correspond to the terms where $d_1^2m_1=d_2^2m_2$ . These contribute

(12·11) \begin{align}&\ll k^{2a+1}\sum_{m\gt N_0}\frac{\left|A(1,m)\right|^2}{m^{(2a+4)/3}}\cdot\Xi\,X^{(4a+2)/3} \nonumber\\&\ll k^{2a+1}\,\Xi\,X^{(4a+2)/3}\,N_0^{-(2a+1)/3} \nonumber\\&\ll k^{2a+1}\Xi\,X^{(2a+1)/3}\,\Delta^{(2a+1)/3}\end{align}

by partial summation coupled with (4·2).

Similarly, the off-diagonal terms contribute, using Lemma 12·2 and Weil’s bound,

\begin{align*}&\ll k^{2a+1}\sum_{d_1|k}\sum_{d_2|k}\sum_{m\gt N_0}\frac{\left|A(d_1,m)\right|}{m^{(a+2)/3}d_1^{(2a+1)/3}}\sum_{n\gt m}\frac{\left|A(d_2,n)\right|}{n^{(a+2)/3}d_2^{(2a+1)/3}}\cdot\frac{X^{(4a+4)/3}k}{|d_1^{2/3}m^{1/3}-d_2^{2/3}n^{1/3}|}.\end{align*}

By symmetry it is enough to treat the terms in the m-sum with $d_1^2m\lt d_2^2n$ . For these we split the m-sum into two parts according to whether $d_1^2m\lt d_2^2n/2$ or $d_2^2n/2\leq d_1^2m\lt d_2^2n$ . Let $\eta\gt0$ be arbitrarily small, but fixed. Note that for $d_2^2n/2\leq d_1^2m$ we have, using (9·5) and the pointwise bound $A(m,d)\ll (md)^{\vartheta+\eta}$ , that

(12·12) \begin{align}&\frac{|A(d_2,n)|}{m^{(a+2)/3}d_1^{(2a+1)/3}n^{(a+2)/3}d_2^{(2a+1)/3}}\cdot\frac1{|d_2^{2/3}n^{1/3}-d_1^{2/3}m^{1/3}|} \nonumber\\&\ll\frac{n^{\vartheta+\eta}}{m^{(a+2)/3}d_1^{(2a+1)/3}n^{(a+2)/3}d_2^{2a/3-1-\vartheta-\eta}|d_2^2n-d_1^2m|}.\end{align}

Observe that

\begin{equation*}n^{a/3}=n^{1/6}n^{a/3-1/6}\gg\left(\frac{d_1^2m}{d_2^2}\right)^{1/6}n^{a/3-1/6}.\end{equation*}

Using this together with an easy estimate

\begin{equation*}n^\vartheta\ll\left(\frac{d_1^2m}{d_2^2}\right)^\vartheta\leq\left(\frac{d_1^2m}{d_2^2}\right)^{5/14}\end{equation*}

gives that (12·12) is

\begin{align*}\ll \frac1{m^{a/3+10/21}d_1^{2a/3-1/21}n^{a/3-1/6-\eta}d_2^{2a/3-13/21-\vartheta-\eta}|d_2^2n-d_1^2m|}.\end{align*}

This means that the terms with $d_2^2n/2\leq d_1^2m\lt d_2^2n$ contribute

\begin{align*}&\ll k^{2a+2}X^{(4a+4)/3}\\&\quad\times\sum_{d_1|k}\sum_{d_2|k}\sum_{n\gt N_0}\\&\quad\sum_{\substack{m\gt N_0\\d_2^2n/2\leq d_1^2m\lt d_2^2n}}\frac{\left|A(d_1,m)\right|}{m^{a/3+10/21}d_1^{2a/3-1/21}n^{a/3-1/6-\eta}d_2^{2a/3-13/21-\vartheta-\eta}(d_2^2n-d_1^2m)}\\&\ll k^{2a+2}X^{(4a+4)/3^{\vphantom{(^{(}}}}\\&\quad\times\sum_{d_1|k}\sum_{d_2|k}\sum_{n\gt N_0}\\&\quad\sum_{m\gt N_0}\left(\frac{d_1^2m}{d_2^2}\right)^{2\eta}\frac{\left|A(d_1,m)\right|}{m^{a/3+10/21}d_1^{2a/3-1/21}n^{a/3-1/6+\eta}d_2^{2a/3-13/21-\vartheta-\eta}(d_2^2n-d_1^2m)},\end{align*}

where in the last step we have multiplied both the numerator and denominator by $n^{2\eta}$ , used the estimate $n\ll d_1^2m/d_2^2$ , and finally dropped the constraint $d_2^2n/2\leq d_1^2m\lt d_2^2n$ in the m-sum. Next, observe that for a fixed m the n-sum can be bounded by the Cauchy–Schwarz inequality as

\begin{equation*}\sum_{n\gt N_0}\frac1{n^{a/3-1/6+\eta}(d_2^2n-d_1^2m)}\leq\left(\sum_{n\gt N_0}\frac1{n^{2a/3-1/3+2\eta}}\right)^{1/2}\left(\sum_{n\gt N_0}\frac1{(d_2^2n-d_1^2m)^2}\right)^{1/2}\ll 1,\end{equation*}

where the final bound is uniform in m, $d_1$ , and $d_2$ .

Combining everything we have shown that in total the terms with $d_2^2n/2\leq d_1^2m\lt d_2^2n$ contribute, using partial summation together with (4·2),

(12·13) \begin{align}&\ll k^{2a+2}X^{(4a+4)/3}\sum_{d_1|k}\sum_{d_2|k}\sum_{m\gt N_0}\frac{1}{m^{a/3+10/21-2\eta}d_1^{2a/3-1/21-4\eta-\vartheta}d_2^{2a/3-13/21-\vartheta+3\eta}} \nonumber\\&\ll k^{2a+2}X^{(4a+4)/3},\end{align}

where the inner triple sum is $\ll 1$ by the estimate $\vartheta\leq 5/14$ .

For the terms with $d_1^2m\lt d_2^2n/2$ we note that by (12·12) one has

\begin{align*}&\frac{|A(d_2,n)|}{m^{(a+2)/3}d_1^{(2a+1)/3}n^{(a+2)/3}d_2^{(2a+1)/3}}\cdot\frac1{|d_2^{2/3}n^{1/3}-d_1^{2/3}m^{1/3}|} \nonumber\\&\ll_{\varepsilon}\frac{n^{\vartheta+\varepsilon}}{m^{(a+2)/3}d_1^{(2a+1)/3}n^{(a+2)/3}d_2^{2a/3-1-\vartheta-\varepsilon}(d_2^2n-d_1^2m)}\\&\ll_{\varepsilon} \frac1{m^{(a+2)/3}d_1^{(2a+1)/3}n^{(a+5)/3-\vartheta-\varepsilon}d_2^{(2a+7)/6-\vartheta-\varepsilon}}.\end{align*}

Hence, the terms with $d_1^2m\lt d_2^2n/2$ contribute

\begin{align*}\ll_{\varepsilon} k^{2a+2}X^{(4a+4)/3}\sum_{d_1|k}\sum_{d_2|k}\sum_{n\gt N_0}\sum_{m\gt N_0}\frac{|A(d_1,m)|}{m^{(a+2)/3}d_1^{(2a+1)/3}n^{(a+5)/3-\vartheta-\varepsilon}d_2^{(2a+7)/6-\vartheta-\varepsilon}},\end{align*}

which is again

(12·14) \begin{align}\ll k^{2a+2}X^{(4a+4)/3}\end{align}

using partial summation and (4·2).

Now only one integral remains, namely

\begin{equation*}k^{2a}\sum_\pm\int\limits_X^{X+\Xi}x^{(4a+2)/3}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left|s_\Delta^{\pm,\leqslant}-s_0^{\pm,\leqslant}\right|^2\mathrm dx,\end{equation*}

where we have for simplicity dropped the constant $(2\pi)^{-(2a+2)}/3$ in front. When we again multiply out $\left|\Sigma\right|^2=\Sigma\,\overline\Sigma$ , we obtain from the off-diagonal terms by Lemma 12·2 or Lemma 12·3 depending on the term in question, and Weil’s bound a contribution

\begin{align*}&\ll k^{2a+2}\sum_{d_1|k}\sum_{d_2|k}\sum_{m\leqslant N_0}\sum_{m\lt n\leqslant N_0}\frac{\left|A(d_1,m)\right|\cdot\left|A(d_2,n)\right|}{m^{(a+2)/3}\,n^{(a+2)/3}d_1^{2a/3+5/6}d_2^{2a/3+5/6}}\cdot\frac{X^{(4a+4)/3}}{|d_1^{2/3}m^{1/3}-d_2^{2/3}n^{1/3}|}.\end{align*}

Estimating this is completely analogous to the computation above and leads to an upper bound

(12·15) \begin{align}\ll X^{(4a+4)/3}k^{2a+2}.\end{align}

To evaluate the diagonal contribution we set $N_1\,:\!=\,\ell\,X^2k^3\,\Delta^{-3}$ with a constant $\ell$ so small that

\begin{equation*}\frac{m^{1/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\leqslant\frac1{12},\end{equation*}

say, for each $m\leqslant N_1$ . Clearly $N_1\lt N_0$ for X sufficiently large, as $\Delta\gg X^{1/2+\delta}k^{3/2}$ .

By estimating trivially, the higher frequencies $m\geq N_1$ in the diagonal terms contribute

(12·16) \begin{align}&\ll k^{2a+1}\sum_{N_1\lt m\leqslant N_0}\frac{\left|A(1,m)\right|^2}{m^{(2a+4)/3}}\,\Xi\,X^{(4a+2)/3} \nonumber\\[4pt]&\ll k^{2a+1} N_1^{-2a/3-1/3}\,\Xi\,X^{(4a+2)/3}\ll\Xi\,\Delta^{2a+1}.\end{align}

In the remaining smaller frequency diagonal terms, we see that these terms are given by

(12·17) \begin{align}&k^{2a}\sum_\pm\sum_{d_1|k}\sum_{d_2|k}\sum_{m\leqslant N_1}\sum_{\substack{n\leqslant N_1\\[4pt]d_1^2m=d_2^2n}}\frac{A(d_1,m)\overline{A(d_2,n)}}{m^{(a+2)/3}n^{(a+2)/3}d_1^{(2a+1)/3}d_2^{(2a+1)/3}}\nonumber \\[4pt]&\qquad\qquad\times\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}S\left(\overline h,\pm m;\frac k{d_1}\right)S\left(\overline h,\pm n;\,\frac k{d_2}\right) \nonumber\\&\qquad\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\!\left(\pm\frac{3m^{1/3}d_1^{2/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\mathrm dx.\end{align}

Using Lemma 9·1 we note that this equals $S_1+S_2+S_3+S_4$ , where

\begin{align*}&S_1\,:\!=\,k^{2a}\left(k-\frac1{k-1}\right)\sum_\pm\sum_{\substack{m\leq N_1\\m\not\equiv 0\,(k)}}\frac{|A(1,m)|^2}{m^{(2a+4)/3}}\\[4pt]&\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\,\mathrm d x\end{align*}

\begin{align*}&S_2\,:\!=\,k^{2a}\sum_\pm\sum_{m\leq N_1}\frac{|A(k,m)|^2}{m^{(2a+4)/3}k^{(4a+2)/3}}\\[4pt]&\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}k^{2/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\,\mathrm d x\\[4pt]&S_3\,:\!=\,-2k^{2a}\sum_\pm\Re\sum_{m\leq N_1/k^2}\frac{A(1,k^2m)\overline{A(k,m)}}{(k^2m)^{(a+2)/3}m^{(a+2)/3}k^{(2a+1)/3}}\\[4pt]&\qquad\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}k^{2/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\,\mathrm d x \\[4pt]&S_4\,:\!=\,k^{2a}\sum_\pm\sum_{\substack{m\leq N_1\\m\equiv 0\,(k)}}\frac{|A(1,m)|^2}{m^{(2a+4)/3}}\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\,\mathrm d x.\end{align*}

Our next objective is to show that

(12·18) \begin{align}S_1+S_2+S_3+S_4\asymp \Xi\Delta^2 X^{(4a-2)/3}k^{2a-1}.\end{align}

The upper bound follows simply by estimating trivially using known average growth properties of the Fourier coefficients. Towards the lower bound, note that by the arithmetic-geometric mean inequality we have

\begin{align*}2\cdot\frac{|A(1,k^2m)\overline{A(k,m)|}}{(k^2m)^{(a+2)/3}m^{(a+2)/3}k^{(2a+1)/3}}\leq\frac{|A(1,k^2m)|^2}{(k^2m)^{(2a+4)/3}}+\frac{|A(k,m)|^2}{k^{ (4a+2)/3}m^{(2a+4)/3}}\end{align*}

and so

\begin{align*}|S_3|&\leq k^{2a}\sum_\pm\sum_{m\leq N_1/k^2}\frac{|A(1,k^2m)|^2}{(k^2m)^{(2a+4)/3}}\\[4pt]&\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}k^{2/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\,\mathrm d x\\[4pt]&\quad+k^{2a}\sum_\pm\sum_{m\leq N_1}\frac{|A(k,m)|^2}{k^{(4a+2)/3}m^{(2a+4)/3}}\\[4pt]&\qquad\qquad\times\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}k^{2/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}k\right)-1\right|^2\,\mathrm d x\\[4pt]&\leq |S_4|+|S_2|.\end{align*}

From this it follows that

\begin{align*}S_1+S_2+S_3+S_4 &\gg k^{2a+1}\sum_\pm\sum_{m\leq N_1}\frac{|A(1,m)|^2}{m^{(2a+4)/3}}\left(1-1_{m\equiv 0\,(k)}\right)\\&\qquad\qquad\times\int\limits_X^{X+\Xi} x^{(4a+2)/3}\left|e\left(\pm\frac{3m^{1/3}\left((x+\Delta)^{1/3}-x^{1/3}\right)}{k}\right)-1\right|^2\,\mathrm d x.\end{align*}

Using the non-negativity of the terms involved and dropping all but the first term in the sum, this is

\begin{equation*}\geq k^{2a+1}\sum_{\pm}\int\limits_X^{X+\Xi}x^{(4a+2)/3}\left|e\left(\pm\frac{3\left((x+\Delta)^{1/3}-x^{1/3}\right)}{k}\right)-1\right|^2\,\mathrm d x.\end{equation*}

Using the estimate $\left|e(\alpha)-1\right|\asymp\left|\alpha\right|$ , which holds for small $|\alpha|$ (recall that $\Delta\ll X^{2/3-\delta}k$ ), we can bound this from below by

\begin{align*}&\gg k^{2a+1}\Xi\,X^{(4a+2)/3}\left(\frac{\Delta}{X^{2/3}k}\right)^2\\&\asymp\Xi\,\Delta^2\,X^{(4a-2)/3}\,k^{2a-1}.\end{align*}

We conclude that the expression (12·17) is $\asymp \Xi\,\Delta^2\,X^{(4a-2)/3}\,k^{2a-1}$ .

Combining estimates (12·2)–(12·18) we have shown that

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{X+\Xi}\left|\widetilde A_a\left(x+\Delta;\frac{h}{k}\right)-\widetilde A_a\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx=S+O(E),\end{align*}

where $S\asymp\Xi\,\Delta^2\,X^{(4a-2)/3}\,k^{2a-1}$ and

\begin{align*}E\,:\!=\,&\Xi X^{4a/3}k^{2a+3}+\Xi\Delta X^{(4a-1)/3}k^{2a+1}+k^{3\vartheta+2a/3+1/6+2{\varepsilon}}\Delta^{2/3+{\varepsilon}}\Xi X^{4a/3-\varepsilon} \\&+k^{2a+\vartheta-1/2}X^{(2a+4)/3}\Xi\Delta^{(2a-2)/3}+k^{4a/3+7/6+\vartheta+\varepsilon}X^{a+11/6}\Delta^{a/3-1/2}\\&+X^{(4a+4)/3}k^{2a+2}+k^{2a+1}\Xi X^{(2a+1)/3}\Delta^{(2a+1)/3}+\Xi \Delta^{2a+1}.\end{align*}

Now we note that $E=o(S)$ by considering the terms in E individually. Below we list which assumptions are needed to bound each of the terms. These claims can be easily verified by straightforward computations:

1. (i) $\Xi X^{4a/3}k^{2a+3}=o(S)$ holds as $\Delta\gg X^{1/2+\delta}k^{3/2}$ and $k\ll X^{1/3-\delta}$ ;

2. (ii) $\Xi\Delta X^{(4a-1)/3}k^{2a+1}=o(S)$ holds as $\Delta\gg X^{1/2+\delta}k^{3/2}$ and $k\ll X^{1/3-\delta}$ ;

3. (iii) $k^{3\vartheta+2a/3+1/6+2\varepsilon}\Delta^{2/3+\varepsilon}\Xi X^{4a/3-{\varepsilon}}=o(S)$ holds as $\Delta\gg X^{1/2+\delta}k^{3/2}$ and $\vartheta\leq 5/14$ ;

4. (iv) $k^{2a+\vartheta-1/2}X^{(2a+4)/3}\Xi\Delta^{(2a-2)/3}=o(S)$ holds as $\Delta\ll X^{2/3-\delta}k$ , $k\ll X^{1/3-\delta}$ , and $\vartheta\lt1/2$ (when $a\geq 4)$ and as $\Delta\gg X^{1/2+\delta}k^{3/2}$ , $k\ll X^{1/3-\delta}$ , and $\vartheta\lt1/2$ (when $a\lt4)$ ;

5. (v) $k^{4a/3+7/6+2\vartheta+\varepsilon}X^{a+11/6}\Delta^{a/3-1/2}=o(S)$ holds as $\Xi\Delta^2\gg X^{2+\delta}k^3$ , $\Delta\ll X$ , and $\vartheta\leq 5/14$ ;

6. (vi) $X^{(4a+4)/3}k^{2a+2}=o(S)$ holds as $X^{2+\delta}k^3\ll\Xi\Delta^2$ ;

7. (vii) $k^{2a+1}\Xi X^{(2a+1)/3}\Delta^{(2a+1)/3}=o(S)$ holds as $\Delta\gg X^{1/2+\delta}k^{3/2}$ and $k\ll X^{1/3-\delta}$ ;

8. (viii) $\Xi \Delta^{2a+1}=o(S)$ holds as $\Delta\ll X^{2/3-\delta}k$ .

These yield the claimed estimate

\begin{equation*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{X+\Xi}\left|\widetilde A_a\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_a\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\asymp\Xi\,\Delta^2\,X^{(4a-2)/3}\,k^{2a-1}. \end{equation*}

The statements of the theorem follow by specialising to the cases $a=2$ and $a=3$ .

13. $\Omega$ -result for short sums of Fourier coefficients

Choosing $\Xi=X$ (which clearly satisfies the required assumptions) in Theorem 12·4 leads to the bounds

(13·1) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_2\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\asymp\Delta^2\,X^3k^3\end{align}

and

(13·2) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_3\left(x+\Delta;\frac{h}{k}\right)-\widetilde A_3\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\asymp\Delta^2\,X^{13/3}k^5\end{align}

in the range $k^{3/2}X^{1/2+\delta}\ll\Delta\ll X^{2/3-\delta}k$ . Using similar ideas as before, these can be used to deduce lower bounds for twisted short sums of the Hecke eigenvalues on average when one also averages over the numerator of the exponential twist, which will in turn yield an $\Omega$ -result for short sums.

Theorem 13·1. Let $X\in[1,\infty[$ , let k be a prime and suppose that $k^{3/2}\,X^{1/2+\delta}\ll\Delta\ll kX^{2/3-\delta}$ for any sufficiently small fixed $\delta\gt0$ . Then, for sufficiently large X, we have

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg\Delta^2\,X^{1/3}k^{-1}.\end{align*}

Notice that this result immediately implies Theorem 2·4 (and consequently Theorem 2·3) as the residue terms in the sums $\widetilde A_0(x;\,h/k)$ cancel each other.

Proof. We begin by proving a lower bound for the mean square of a short first order Riesz sum $\widetilde A_1(x+\Delta;\,h/k)-\widetilde A_1(x;\,h/k)$ . As in the proof of Theorem 10·2, this time using the pointwise bound $\widetilde A_3(x;\,h/k)\ll k^{7/2}x^{7/3}$ from Corollary 7·4, we have for any $0\lt H\leq X$ that

(13·3) \begin{align}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_2\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx \nonumber\\&\ll \frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{2X}\left|\widetilde A_3\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_3\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx \nonumber\\&\qquad\qquad+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H k^7X^{14/3} \nonumber\\&\ll\frac1{H^2}\,\Delta^2\,X^{13/3}k^5+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H k^7X^{14/3},\end{align}

where the last estimate follows from (13·2).

Next we shall optimise the choice of the parameter H. Actually, before that we need a preliminary lower bound for the moment

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\end{align*}

we are interested in. This is to guarantee that our optimal choice for H satisfies the constraint $H\ll \Delta^2X^{-1/3}k^{-2}$ (and in particular $H\leq X$ when X is large enough). This is needed so that $H^{-1}X^{14/3}k^7\ll H^{-2}\Delta^2X^{13/3}k^5$ .

We make the initial choice $H=\Delta^2X^{-1/3}k^{-2}$ (which satisfies the constraint $H\leq X$ for sufficiently large X by the assumption $\Delta\ll X^{2/3-\delta}k$ ) in (13·3). This gives

(13·4) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg \Delta^{-2}k^7X^{11/3}\end{align}

using (13·1) as $\Delta\gg k^{3/2}X^{1/2+\delta}$ . We now make the final choice for H by choosing

\begin{equation*}H=\frac{\Delta^{1/2}X^{13/12}k^{5/4}}{\sqrt[4]{\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx}}.\end{equation*}

This satisfies the constraint $H\ll \Delta^2X^{-1/3}k^{-2}$ thanks to (13·4) and the assumption $\Delta\gg X^{1/2+\delta}k^{3/2}$ . Using this choice in (13·3) gives

(13·5) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg\left(\frac{\Delta^2X^3k^3}{\Delta X^{13/6}k^{5/2}}\right)^2=\Delta^2X^{5/3}k\end{align}

by (13·1), giving a lower bound for the short first order Riesz sum.

We now turn into estimating the sum appearing in the statement of the theorem. Again, arguing as in the proof of Theorem 10·2 and using (13·1) together with the bound $\widetilde A_2(x;\,h/k)\ll x^{5/3}k^{5/2}$ gives

(13·6) \begin{align}&\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx \nonumber\\&\ll\frac1{H^2}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\left(\int\limits_X^{2X}+\int\limits_{3X/2}^{3X}\right)\left|\widetilde A_2\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_2\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx \nonumber \\&\qquad+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H X^{10/3}k^5 \nonumber\\&\ll\frac1{H^2}\,\Delta^2\,X^3\,k^3+H^2\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx+\frac1H X^{10/3}k^5\end{align}

for any $0\lt H\leq X$ .

Next we shall optimise the choice of the parameter H. Actually, before that we need a preliminary lower bound for the moment

\begin{align*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\end{align*}

we are interested in. This is to guarantee that our optimal choice for H satisfies the constraint $H\ll \Delta^2X^{-1/3}k^{-2}$ (and in particular $H\leq X$ when X is large enough in view of the assumption $\Delta\ll kX^{2/3-\delta}$ ). This is needed so that $H^{-1}X^{10/3}k^5\ll H^{-2}\Delta^2X^3k^3$ .

We make the initial choice $H=\Delta^2X^{-1/3}k^{-2}$ (which again satisfies the constraint $H\leq X$ for sufficiently large X) in (13·6). This gives

(13·7) \begin{align}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg \Delta^{-2}X^{7/3}k^5\end{align}

using (13·5) as $\Delta\gg X^{1/2+\delta}k^{3/2}$ . We now make the final optimal choice for H by choosing

\begin{equation*}H=\frac{\Delta^{1/2}X^{3/4}k^{3/4}}{\sqrt[4]{\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx}}.\end{equation*}

This satisfies the constraint $H\ll \Delta^2X^{-1/3}k^{-2}$ thanks to (13·7) and the assumption $\Delta\gg X^{1/2+\delta}k^{3/2}$ . Using this choice in (13·6) gives

\begin{align*}k\,\Delta^2\,X^{5/3}&\ll \mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{3X}\left|\widetilde A_1\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_1\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\\&\ll k^{3/2}\,X^{3/2}\Delta\sqrt{\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx},\end{align*}

and hence

\begin{equation*}\mathop{\LARGE\unicode{x1D53C}}\limits_{h\in\mathbb Z_k^\times}\int\limits_X^{4X}\left|\widetilde A_0\left(x+\Delta;\,\frac{h}{k}\right)-\widetilde A_0\left(x;\,\frac{h}{k}\right)\right|^2\,\mathrm dx\gg\left(\frac{\Delta^2 X^{5/3}k}{\Delta X^{3/2}k^{3/2}}\right)^2=\Delta^2\,X^{1/3}\,k^{-1}.\end{equation*}

This completes the proof.

Appendix: corrections to [ Reference Jääsaari and Vesalainen22 ]

In [ Reference Jääsaari and Vesalainen22 ] there was a wrong factor d instead of the correct factor $d^{1/3}$ in Theorem 1. This was corrected in [ Reference Jääsaari and Vesalainen23 ], which also described how the proofs of Theorem 2 and Corollary 3 need to be modified in order to obtain the same conclusions. Alas, the computation of the asymptotics for the Meijer G-function revealed another mistake in the proof of Theorem 1 of [ Reference Jääsaari and Vesalainen22 ]. Namely, when $j=1$ , the $\Gamma$ -quotient expression used to approximate the $\Gamma$ -quotient coming from the functional equation of the twisted L-function of the underlying Maass cusp form is correct only in the upper half-plane; in the lower half-plane the factor $i^j$ should read $({-}i)^j$ , thereby changing Theorem 1, though fortunately not really affecting the proofs of Theorem 2 and Corollary 3 apart from obvious cosmetic changes. The corrected formulation of [ Reference Jääsaari and Vesalainen22 , theorem 1] reads as follows.

Theorem A·1. Let $x,N\in\left[2,\infty\right[$ with $N\ll x$ , and let h and k be coprime integers with $1\leqslant k\leqslant x$ and $k\ll\left(Nx\right)^{1/3}$ , the latter having a sufficiently small implicit constant depending on the underlying Maass cusp form. Then we have

\begin{align*}& \sum_{m\leqslant x}A(m,1)\,e\!\left(\frac{mh}k\right)\\& =\frac{x^{1/3}}{2\pi\sqrt3}\sum_{d\mid k}\frac1{d^{1/3}}\sum_{d^2\,m\leqslant N_k}\frac{A(d,m)}{m^{2/3}}\sum_\pm S\!\left(\overline h,\pm m;\,\frac kd\right)e\!\left(\pm\frac{3\,d^{2/3}\,m^{1/3}\,x^{1/3}}k\right)\\& \quad +O(k\,x^{2/3+\vartheta+\varepsilon}\,N^{-1/3})+O(k\,x^{1/6+\varepsilon}\,N^{1/6+\vartheta}).\end{align*}