Proof. We begin with the proof of $\textit {(I)}$. Let $\alpha \in \mathbb {C}$ and $M\geq 1$ an integer. We are going to apply lemma 4.1 with $\mathbf {b}_{{\scriptscriptstyle m}}(t)=\exp {(2\pi i\alpha \phi _{{\scriptscriptstyle 1,m}}(t))}$, and frequencies $\gamma _{{\scriptscriptstyle m}}=\sqrt {m}$. The elements of the set $\mathscr {B}=\{\sqrt {m}:|\mu (m)|=1\}$ are linearly independent over $\mathbb {Q}$, and since $\phi _{{\scriptscriptstyle 1,m}}(t)\equiv 0$ whenever $\sqrt {m}\notin \mathscr {B}$ (i.e. whenever $m$ is not square-free), we see that the conditions of lemma 4.1 are satisfied, and thus (4.2) holds.

Now, let us show that $\Phi _{{\scriptscriptstyle 1}}(\alpha )$ defines an entire function of $\alpha$. First, we note the following. By the definition of $\phi _{{\scriptscriptstyle 1,m}}(t)$, we have

(4.5)\begin{equation} \begin{aligned} \int\limits_{0}^{1}\phi_{{\scriptscriptstyle 1,m}}(t){\rm d}t & ={-}\frac{\sqrt{2}}{\pi}\frac{\mu^{2}(m)}{m}\sum_{k=1}^{\infty}\frac{r_{{\scriptscriptstyle2}}\big(mk^{2}\big)}{k^{2}}\int\limits_{0}^{1}\cos{(2\pi kt)}{\rm d}t\\ & =0, \end{aligned} \end{equation}

and we also have the uniform bound

(4.6)\begin{equation} \begin{aligned} \big|\phi_{{\scriptscriptstyle 1,m}}(t)\big| & \leq\frac{\sqrt{2}}{\pi}\frac{\mu^{2}(m)}{m}\sum_{k=1}^{\infty}\frac{r_{{\scriptscriptstyle2}}\big(mk^{2}\big)}{k^{2}}\\ & \leq2^{5/2}\pi^{{-}1}\mu^{2}(m)\frac{r_{{\scriptscriptstyle2}}(m)}{m}\sum_{k=1}^{\infty}\frac{r_{{\scriptscriptstyle2}}\big(k^{2}\big)}{k^{2}}\leq\mathfrak{a}\,\mu^{2}(m)\frac{r_{{\scriptscriptstyle2}}(m)}{m}, \end{aligned} \end{equation}

for some absolute constant $\mathfrak {a}>0$. By (4.5) and (4.6), we obtain

(4.7)\begin{equation} \begin{aligned} \Phi_{{\scriptscriptstyle1,m}}(\alpha) & =\int\limits_{0}^{1}\exp{\big(2\pi i\alpha\phi_{{\scriptscriptstyle 1,m}}(t)\big)}{\rm d}t\\ & =1+O\left(|\alpha|\frac{r_{{\scriptscriptstyle2}}(m)}{m}\right)^{2}, \end{aligned} \end{equation}

whenever, say, $m\geq |\alpha |^{2}$. Since $\sum _{m=1}^{\infty }r^{2}_{{\scriptscriptstyle 2}}(m)/m^{2}<\infty$, it follows from (4.7) that the infinite product $\prod _{m=1}^{\infty }\Phi _{{\scriptscriptstyle 1,m}}(\alpha )$ converges absolutely and uniformly on any compact subset of the plane, and so $\Phi _{{\scriptscriptstyle 1}}(\alpha )$ defines an entire function of $\alpha$.

We now estimate $\Phi _{{\scriptscriptstyle 1}}(\alpha )$ for large $|\alpha |$, $\alpha =\sigma +i\tau$. Let $c>0$ be an absolute constant such that (see [Reference Wigert15])

(4.8)\begin{equation} r_{{\scriptscriptstyle2}}(m)\leq m^{c/\log\log{m}}\quad:\quad m>2, \end{equation}

and for real $x>\textit {e}$ we write $\theta (x)=1-c/\log \log {x}$. Let $\epsilon >0$ be a small absolute constant which will be specified later, and set

\[ \ell=\ell(\alpha)=\Big[\Big(\epsilon^{{-}1}|\alpha|\Big)^{1/\theta(|\alpha|)}\Big]+1. \]

We are going to estimate the infinite product $\prod _{m=1}^{\infty }\Phi _{{\scriptscriptstyle 1,m}}(\alpha )$ separately for $m<\ell$ and $m\geq \ell$. In what follows, we assume that $|\alpha |$ is sufficiently large in terms of $c$ and $\epsilon$. The product over $m<\ell$ is estimated trivially by using the upper bound (4.6), leading to

(4.9)\begin{equation} \begin{aligned} \Big|\prod_{m<\ell}\Phi_{{\scriptscriptstyle1,m}}(\alpha)\Big| & \leq\exp{\left(2\pi\mathfrak{a}|\tau|\sum_{m<\ell}\mu^{2}(m)\frac{r_{{\scriptscriptstyle2}}(m)}{m}\right)}\\ & \leq\exp{\Big(\mathfrak{b}|\tau|\log{|\alpha|}\Big)}, \end{aligned} \end{equation}

for some absolute constant $\mathfrak {b}>0$.

Suppose now that $m\geq \ell$. By (4.8) we have

(4.10)\begin{equation} \frac{r_{{\scriptscriptstyle2}}(m)}{m}|\alpha|\leq m^{-\theta(m)}|\alpha|\leq\big(\epsilon^{{-}1}|\alpha|\big)^{-\frac{\theta(m)}{\theta(|\alpha|)}}|\alpha|\leq\epsilon \end{equation}

It follows from (4.5), (4.6) and (4.10) that

(4.11)\begin{equation} \Phi_{{\scriptscriptstyle1,m}}(\alpha)=1-\frac{(2\pi\alpha)^{2}}{2}\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t+\mathcal{R}_{{\scriptscriptstyle m}}(\alpha), \end{equation}

where the remainder term $\mathcal {R}_{{\scriptscriptstyle m}}(\alpha )$ satisfies the bound

(4.12)\begin{equation} \big|\mathcal{R}_{{\scriptscriptstyle m}}(\alpha)\big|\leq\bigg\{\frac{2}{3}\pi\mathfrak{a}\epsilon\exp{\big(2\pi\mathfrak{a}\epsilon\big)}\bigg\}\frac{(2\pi|\alpha|)^{2}}{2}\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t. \end{equation}

At this point, we specify $\epsilon$, by choosing $0<\epsilon <\frac {1}{64}$ such that

\[ 2\pi\mathfrak{a}\epsilon^{1/2}\exp{\big(2\pi\mathfrak{a}\epsilon\big)}\leq1. \]

With this choice of $\epsilon$, we have

(4.13)\begin{equation} \big|\mathcal{R}_{{\scriptscriptstyle m}}(\alpha)\big|\leq\epsilon^{1/2}\frac{(2\pi|\alpha|)^{2}}{2}\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t, \end{equation}

and we also note that $\big |\Phi _{{\scriptscriptstyle 1,m}}(\alpha )-1\big |\leq \epsilon <\frac {1}{2}$. Thus, on rewriting (4.11) in the form

(4.14)\begin{equation} \Phi_{{\scriptscriptstyle1,m}}(\alpha)=\exp{\Bigg(-\frac{(2\pi\alpha)^{2}}{2}\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t+\widetilde{\mathcal{R}}_{{\scriptscriptstyle m}}(\alpha)\Bigg)}, \end{equation}

it follows from (4.13) that the remainder term $\widetilde {\mathcal {R}}_{{\scriptscriptstyle m}}(\alpha )$ satisfies the bound

(4.15)\begin{equation} \big|\widetilde{\mathcal{R}}_{{\scriptscriptstyle m}}(\alpha)\big|\leq\epsilon^{1/2}(2\pi|\alpha|)^{2}\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t. \end{equation}

From (4.14) and (4.15), we obtain

(4.16)\begin{equation} \prod_{m\geq\ell}\big|\Phi_{{\scriptscriptstyle1,m}}(\alpha)\big|\leq\exp{\Bigg(-\frac{\pi^{2}}{2}\big(3\sigma^{2}-5\tau^{2}\big)\sum_{m\,\geq\,\ell}\,\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t\Bigg)}. \end{equation}

Now, by (4.6) we have

(4.17)\begin{equation} \sum_{m\,\geq\,\ell}\,\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t\leq\mathfrak{a}^{2}\sum_{m\,\geq\,\ell}\mu^{2}(m)\frac{r^{2}_{{\scriptscriptstyle2}}(m)}{m^{2}}, \end{equation}

and since

(4.18)\begin{equation} \begin{aligned} \int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t & =\pi^{{-}2}\frac{\mu^{2}(m)}{m^{2}}\sum_{k=1}^{\infty}\frac{r^{2}_{{\scriptscriptstyle2}}\big(mk^{2}\big)}{k^{4}}\\ & \geq\pi^{{-}2}\mu^{2}(m)\frac{r^{2}_{{\scriptscriptstyle2}}(m)}{m^{2}}, \end{aligned} \end{equation}

we also have the lower bound

(4.19)\begin{equation} \sum_{m\,\geq\,\ell}\,\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t\geq\pi^{{-}2}\sum_{m\,\geq\,\ell}\mu^{2}(m)\frac{r^{2}_{{\scriptscriptstyle2}}(m)}{m^{2}}. \end{equation}

Since

(4.20)\begin{equation} \sum_{m\,\leq\,Y}\mu^{2}(m)r^{2}_{{\scriptscriptstyle2}}(m)\sim\mathfrak{h}Y\log{Y}, \end{equation}

as $Y\to \infty$ where $\mathfrak {h}>0$ is some constant, we obtain by partial summation

(4.21)\begin{equation} \big(3\sigma^{2}-5\tau^{2}\big)\sum_{m\,\geq\,\ell}\,\int\limits_{0}^{1}\phi^{2}_{{\scriptscriptstyle 1,m}}(t){\rm d}t\geq\big(A\sigma^{2}-B\tau^{2}\big)|\alpha|^{{-}1/\theta(|\alpha|)}\log{|\alpha|}, \end{equation}

for some absolute constants $A,\,B>0$. Inserting (4.21) into the RHS of (4.16) we arrive at

(4.22)\begin{equation} \prod_{m\geq\ell}\big|\Phi_{{\scriptscriptstyle1,m}}(\alpha)\big|\leq\exp{\left(-\frac{\pi^{2}}{2}\big(A\sigma^{2}-B\tau^{2}\big)|\alpha|^{{-}1/\theta(|\alpha|)}\log{|\alpha|}\right)}. \end{equation}

Finally, setting $C_{{\scriptscriptstyle 1}}=1+\text {max}\{A,\, A^{-1},\, B,\, \mathfrak {b}\}$, we deduce from (4.9) and (4.22)

(4.23)\begin{equation} \big|\Phi_{{\scriptscriptstyle1}}(\alpha)\big|\leq\exp{\left(-\frac{\pi^{2}}{2}\big(C_{{\scriptscriptstyle1}}^{{-}1}\sigma^{2}-C_{{\scriptscriptstyle1}}\tau^{2}\big)|\alpha|^{{-}1/\theta(|\alpha|)}\log{|\alpha|}+C_{{\scriptscriptstyle1}}|\tau|\log{|\alpha|}\right)}. \end{equation}

This completes the proof of $\textit {(I)}$.

We now turn to the proof of $\textit {(II)}$. By the definition of the Fourier transform, we have for real $x$

(4.24)\begin{equation} \mathcal{P}_{1}(x)=\int\limits_{-\infty}^{\infty}\Phi_{{\scriptscriptstyle1}}(\sigma)\exp{\big({-}2\pi ix\sigma\big)}{\rm d}\sigma, \end{equation}

and it follows from the decay estimates (4.23) that $\mathcal {P}_{1}(x)$ is of class $C^{\infty }$. Let us estimate $|\mathcal {P}^{(j)}_{1}(x)|$ for real $x$, $|x|$ sufficiently large in terms $j$. Let $\tau =\tau _{x}$, $|\tau |$ large, be a real number depending on $x$ to be determined later, which satisfies $\textit {sgn}(\tau )=-\text {sgn}(x)$. By Cauchy's theorem and the decay estimate (4.23), we have

(4.25)\begin{equation} \begin{aligned} \mathcal{P}^{(j)}_{1}(x) & =\big({-}2\pi i\big)^{j}\int\limits_{-\infty}^{\infty}\sigma^{j}\Phi_{{\scriptscriptstyle1}}(\sigma)\exp{\big({-}2\pi ix\sigma\big)}{\rm d}\sigma\\ & =\big({-}2\pi i\big)^{j}\exp{\big({-}2\pi |x||\tau|\big)}\int\limits_{-\infty}^{\infty}\big(\sigma+i\tau\big)^{j}\Phi_{{\scriptscriptstyle1}}(\sigma+i\tau)\exp{\big({-}2\pi ix\sigma\big)}{\rm d}\sigma. \end{aligned} \end{equation}

It follows that

(4.26)\begin{equation} \big|\mathcal{P}^{(j)}_{1}(x)\big|\leq\big(2\pi C_{{\scriptscriptstyle1}}|\tau|\big)^{j+1}\exp{\big({-}2\pi |x||\tau|\big)}\int\limits_{-\infty}^{\infty}\big(\sigma^{2}+1\big)^{j/2}\big|\Phi_{{\scriptscriptstyle1}}\big(C_{{\scriptscriptstyle1}}\tau\sigma+i\tau\big)\big|{\rm d}\sigma. \end{equation}

We decompose the range of integration in (4.26) as follows:

(4.27)\begin{equation} \begin{aligned} \int\limits_{-\infty}^{\infty}\big(\sigma^{2}+1\big)^{j/2}\big|\Phi_{{\scriptscriptstyle1}}\big(C_{{\scriptscriptstyle1}}\tau\sigma+i\tau\big)\big|{\rm d}\sigma & =\int\limits_{|\sigma|\leq\sqrt{2}}\ldots{\rm d}\sigma+\int\limits_{|\sigma|>\sqrt{2}}\ldots{\rm d}\sigma\\ & ={\textrm L}_{1}+{\textrm L}_{2}. \end{aligned} \end{equation}

In what follows, we assume that $|\tau |$ is sufficiently large in terms of $j$. In the range $|\sigma |\leq \sqrt {2}$, we have by (4.23)

(4.28)\begin{equation} \big|\Phi_{{\scriptscriptstyle1}}\big(C_{{\scriptscriptstyle1}}\tau\sigma+i\tau\big)\big|\leq\exp{\big(4C_{{\scriptscriptstyle1}}|\tau|\log{|\tau|}\big)}. \end{equation}

From (4.28), it follows that ${\textrm L}_{1}$ satisfies the bound

(4.29)\begin{equation} {\textrm L}_{1}\leq\exp{\big(5C_{{\scriptscriptstyle1}}|\tau|\log{|\tau|}\big)}. \end{equation}

Referring to (4.23) once again, we have in the range $|\sigma |>\sqrt {2}$

(4.30)\begin{align} & \big|\Phi_{{\scriptscriptstyle1}}\big(C_{{\scriptscriptstyle1}}\tau\sigma+i\tau\big)\big|\nonumber\\ & \quad \leq\exp{\left({-}C_{{\scriptscriptstyle1}}|\tau|\bigg\{\,\bigg|\left(\frac{\pi^{2}}{16C^{2}_{{\scriptscriptstyle1}}}\right)^{\frac{\theta(|\tau|)}{2\theta(|\tau|)-1}}|\tau|^{\frac{\theta(|\tau|)-1}{2\theta(|\tau|)-1}}\sigma\bigg|^{\frac{2\theta(|\tau|)-1}{\theta(|\tau|)}}-1\bigg\}\log{|C_{{\scriptscriptstyle1}}\tau\sigma+i\tau|}\right)} \end{align}

From (4.30), it follows that

(4.31)\begin{equation} {\textrm L}_{2}\leq2^{j/2}{\tilde \tau}^{j+1}\int\limits_{-\infty}^{\infty}|\sigma|^{j}\exp\left({-}C_{{\scriptscriptstyle1}}|\tau|\Big\{|\sigma|^{\frac{2\theta(|\tau|)-1}{\theta(|\tau|)}}-1\Big\}\log{|C_{{\scriptscriptstyle1}}\tau{\tilde \tau}\sigma+i\tau|}\right){\rm d}\sigma, \end{equation}

where ${\tilde \tau}$ is given by

\[ {\tilde \tau}=\left(\frac{16C^{2}_{{\scriptscriptstyle1}}}{\pi^{2}}\right)^{\frac{\theta(|\tau|)}{2\theta(|\tau|)-1}}|\tau|^{\frac{1-\theta(|\tau|)}{2\theta(|\tau|)-1}}. \]

We decompose the range of integration in (4.31) as follows:

(4.32)\begin{equation} \begin{aligned} & \int\limits_{-\infty}^{\infty}|\sigma|^{j}\exp\left({-}C_{{\scriptscriptstyle1}}|\tau|\Big\{|\sigma|^{\frac{2\theta(|\tau|)-1}{\theta(|\tau|)}}-1\Big\}\log{|C_{{\scriptscriptstyle1}}\tau{\tilde \tau}\sigma+i\tau|}\right){\rm d}\sigma\\ & \quad =\int\limits_{|\sigma|\leq u}\ldots{\rm d}\sigma+\int\limits_{|\sigma|>u}\ldots{\rm d}\sigma ={\textrm L}_{3}+{\textrm L}_{4}, \end{aligned} \end{equation}

where $u=2^{\frac {\theta (|\tau |)}{2\theta (|\tau |)-1}}$. In the range $|\sigma |\leq u$ we estimate trivially, obtaining

(4.33)\begin{equation} {\textrm L}_{3}\leq\exp{\big(3C_{{\scriptscriptstyle1}}|\tau|\log{|\tau|}\big)}. \end{equation}

In the range $|\sigma |>u$ we have $|\sigma |^{\frac {2\theta (|\tau |)-1}{\theta (|\tau |)}}-1\geq \frac {1}{2}|\sigma |^{\frac {2\theta (|\tau |)-1}{\theta (|\tau |)}}\geq \frac {1}{2}|\sigma |^{1/2}$. It follows that

(4.34)\begin{equation} {\textrm L}_{4}\leq\int\limits_{-\infty}^{\infty}|\sigma|^{j}\exp\big(-|\sigma|^{1/2}\,\big){\rm d}\sigma. \end{equation}

Combining (4.33) and (4.34), we see that ${\textrm L}_{2}$ satisfies the bound

(4.35)\begin{equation} {\textrm L}_{2}\leq2^{j/2+1}{\tilde \tau}^{j+1}\exp{\big(3C_{{\scriptscriptstyle1}}|\tau|\log{|\tau|}\big)}. \end{equation}

From (4.29) and (4.35) we find that ${\textrm L}_{1}$ dominates. By (4.26) and (4.27) we arrive at

(4.36)\begin{equation} \big|\mathcal{P}^{(j)}_{1}(x)\big|\leq\big(4\pi C_{{\scriptscriptstyle1}}|\tau|\big)^{j+1}\exp{\left({-}2\pi|\tau|\bigg\{|x|-\frac{5C_{{\scriptscriptstyle1}}}{2\pi}\log{|\tau|}\bigg\}\right)\,} \end{equation}

Finally, we specify $\tau$. We choose

\[ \tau={-}\text{sgn}(x)\exp{\big(\rho|x|\big)}\quad;\quad\rho=\frac{\pi}{5C_{{\scriptscriptstyle1}}}. \]

With this choice, we have the bound (recall that $|x|$ is assumed to be large in terms of $j$)

(4.37)\begin{equation} \begin{aligned} \big|\mathcal{P}^{(j)}_{1}(x)\big| & \leq\big(4\pi C_{{\scriptscriptstyle1}}\big)^{j+1}\exp{\left(-\pi|x|\exp{\big(\rho|x|\big)}+(j+1)\rho|x|\right)}\\ & \leq\exp{\left(-\frac{\pi}{2}|x|\exp{\big(\rho|x|\big)}\right)}. \end{aligned} \end{equation}

It remains to show that $\mathcal {P}_{1}(x)$ defines a probability density. This will be a consequence of the proof of theorem 1.1. This concludes the proof of proposition 4.2.

Proof. We split the proof into three cases, depending on whether $j=1$, $j=2$ or $j\geq 3$.

Case 1. $j=1$. Since by definition, $\Xi (m,\,1)=0$ for any integer $m\geq 1$, we need to show that $\frac {1}{X}\int _{X}^{2X}\widehat {\mathcal {E}}_{1}(x){\rm d}x\to 0$ as $X\to \infty$. By proposition 2.1 and lemma 3.2, we have

(5.2)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}_{1}(x){\rm d}x={-}\frac{\sqrt{2}}{\pi}\sum_{m\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\frac{1}{X}\int\limits_{ X}^{2X}\cos{\big(2\pi\sqrt{m}x\big)}{\rm d}x+O_{\epsilon}\big(X^{{-}1+\epsilon}\big)\,. \end{equation}

It follows that

(5.3)\begin{equation} \begin{aligned} \bigg|\frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}_{1}(x){\rm d}x\bigg| & \ll_{\epsilon}\frac{1}{X}\sum_{m=1}^{\infty}\frac{r_{{\scriptscriptstyle 2}}(m)}{m^{3/2}}+X^{{-}1+\epsilon}\\ & \ll_{\epsilon} X^{{-}1+\epsilon}. \end{aligned} \end{equation}

This proves (5.1) in the case where $j=1$.

Case 2. $j=2$. By proposition 2.1 and lemma 3.2, we have

(5.4)\begin{align} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{2}_{1}(x){\rm d}x& =\frac{2}{\pi^{2}}\sum_{m,\,n\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\frac{r_{{\scriptscriptstyle 2}}(n)}{n}\frac{1}{X}\nonumber\\ & \quad\times\int\limits_{ X}^{2X}\cos{\big(2\pi\sqrt{m}x\big)}\cos{\big(2\pi\sqrt{n}x\big)}{\rm d}x+O_{\epsilon}\big(X^{{-}1+\epsilon}\big), \end{align}

Now, we have

(5.5)\begin{equation} \begin{aligned} & \frac{1}{X}\int\limits_{ X}^{2X}\cos{\big(2\pi\sqrt{m}x\big)}\cos{\big(2\pi\sqrt{n}x\big)}{\rm d}x\nonumber\\ & \quad =\frac{1}{2}{\mathbb 1}_{m=n}+\frac{1}{4\pi X}{\mathbb 1}_{m\neq n}\frac{\sin{\big(2\pi(\sqrt{m}-\sqrt{n}\,)x\big)}}{\sqrt{m}-\sqrt{n}}\Bigg|^{x=2X}_{x=X}\\ & \qquad+ O\left(\frac{1}{X(mn)^{1/4}}\right)\\ & \quad =\frac{1}{2}{\mathbb 1}_{m=n}+\frac{1}{4\pi X}{\mathbb 1}_{m\neq n}{\textrm K}_{(\sqrt{m},\sqrt{n}\,)}+O\left(\frac{1}{X(mn)^{1/4}}\right), \end{aligned} \end{equation}

say. Inserting (5.5) into the RHS of (5.4), we obtain

(5.6)\begin{align} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{2}_{1}(x){\rm d}x& =\frac{1}{\pi^{2}}\sum_{m\,\leq\,X^{2}}\frac{r^{2}_{{\scriptscriptstyle 2}}(m)}{m^{2}}\nonumber\\ & \quad +\frac{1}{2\pi^{3}X}\sum_{m\neq n\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\frac{r_{{\scriptscriptstyle 2}}(n)}{n}{\textrm K}_{(\sqrt{m},\sqrt{n}\,)}+O_{\epsilon}\big(X^{{-}1+\epsilon}\big), \end{align}

To estimate the off-diagonal terms, we use the identity $\sin {t}=\frac {1}{2\pi i}(\exp {(it)}-\exp {(-it}))$ and then apply Hilbert's inequality, obtaining

(5.7)\begin{equation} \bigg|\sum_{m\neq n\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\frac{r_{{\scriptscriptstyle 2}}(n)}{n}{\textrm K}_{(\sqrt{m},\sqrt{n}\,)}\bigg|\ll\sum_{m=1}^{\infty}r^{2}_{{\scriptscriptstyle2}}(m)m^{{-}3/2}. \end{equation}

Inserting (5.7) into the RHS of (5.6), we arrive at

(5.8)\begin{equation} \begin{aligned} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{2}_{1}(x){\rm d}x & =\frac{1}{\pi^{2}}\sum_{m\,\leq\,X^{2}}\frac{r^{2}_{{\scriptscriptstyle 2}}(m)}{m^{2}}+O_{\epsilon}\big(X^{{-}1+\epsilon}\big)\\ & =\frac{1}{\pi^{2}}\sum_{m=1}^{\infty}\frac{r^{2}_{{\scriptscriptstyle 2}}(m)}{m^{2}}+O_{\epsilon}\big(X^{{-}1+\epsilon}\big). \end{aligned} \end{equation}

Recalling that $\Xi (m,\,1)=0$, it follows that the RHS of (5.1) in the case where $j=2$ is given by

(5.9)\begin{equation} \begin{aligned} \sum_{m=1}^{\infty}\Xi(m,2) & =\frac{1}{2\pi^{2}}\sum_{m=1}^{\infty}\frac{\mu^{2}(m)}{m^{2}}\sum_{\textit{e}_{1},\textit{e}_{2}=\pm1}\underset{\textit{e}_{1}k_{1}+\textit{e}_{2}k_{2}=0}{\sum_{k_{1}, k_{2}=1}^{\infty}}\prod_{i=1}^{2}\frac{r_{{\scriptscriptstyle2}}\big(mk^{2}_{i}\big)}{k_{i}^{2}}\\ & =\frac{1}{\pi^{2}}\sum_{m,k=1}^{\infty}\mu^{2}(m)\frac{r^{2}_{{\scriptscriptstyle2}}\big(mk^{2}\big)}{(mk^{2})^{2}}=\frac{1}{\pi^{2}}\sum_{m=1}^{\infty}\frac{r^{2}_{{\scriptscriptstyle 2}}(m)}{m^{2}}. \end{aligned} \end{equation}

This proves (5.1) in the case where $j=2$.

Case 3. $j\geq 3$. In what follows, all implied constants in the Big $O$ notation are allowed to depend on $j$. By proposition 2.1 and lemma 3.2, together with the trivial bound $|x^{-2}T(x^{2})|\ll 1$, we have

(5.10)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{j}_{1}(x){\rm d}x=\left(-\frac{\sqrt{2}}{\pi}\right)^{j}\frac{1}{X}\int\limits_{ X}^{2X}\left(\,\sum_{m\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\cos{\big(2\pi\sqrt{m}x\big)}\right)^{j}{\rm d}x\!+O_{\epsilon}\big(X^{{-}1\!+\!\epsilon}\big). \end{equation}

Let $1\leq Y\leq X^{2}$ be a large parameter to be determined later. In the range $X< x<2X$, we have

(5.11)\begin{equation} \begin{aligned} & \Bigg(\sum_{m\leq X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\cos{\big(2\pi\sqrt{m}x\big)}\Bigg)^{j}=\Bigg(\,\sum_{m\,\leq\,Y}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\cos{\big(2\pi\sqrt{m}x\big)}\Bigg)^{j}\\ & +O\left((\log{X})^{j-1}\bigg|\sum_{Y < m\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\exp{\big(2\pi i\sqrt{m}x\big)}\bigg|\,\right). \end{aligned} \end{equation}

We first estimate the mean-square of the second summand appearing on the RHS of (5.11). We have

(5.12)\begin{equation} \begin{aligned} & \frac{1}{X}\int\limits_{ X}^{2X}\bigg|\sum_{Y < m\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\exp{\big(2\pi i\sqrt{m}x\big)}\bigg|^{2}{\rm d}x =\sum_{Y < m\,\leq\,X^{2}}\frac{r^{2}_{{\scriptscriptstyle 2}}(m)}{m^{2}}\\ & \quad +\frac{1}{2\pi iX}\sum_{Y < m\neq n\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\frac{r_{{\scriptscriptstyle 2}}(n)}{n}\frac{\exp{\big(2\pi i(\sqrt{m}-\sqrt{n}x)\big)}}{\sqrt{m}-\sqrt{n}}\bigg|^{x=2X}_{x=X}. \end{aligned} \end{equation}

Applying Hilbert's inequality, we find that

(5.13)\begin{equation} \begin{aligned} \Bigg|\,\sum_{Y < m\neq n\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\frac{r_{{\scriptscriptstyle 2}}(n)}{n}\frac{\exp{\big(2\pi i(\sqrt{m}-\sqrt{n}x)\big)}}{\sqrt{m}-\sqrt{n}}\bigg|^{x=2X}_{x=X}\,\Bigg| & \ll\sum_{m > Y}r^{2}_{{\scriptscriptstyle2}}(m)m^{{-}3/2}\\ & \ll Y^{{-}1/2}\log{Y}. \end{aligned} \end{equation}

Inserting (5.13) into the RHS of (5.12), we obtain

(5.14)\begin{equation} \begin{aligned} \frac{1}{X}\int\limits_{ X}^{2X}\bigg|\sum_{Y < m\,\leq\,X^{2}}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\exp{\big(2\pi i\sqrt{m}x\big)}\bigg|^{2}{\rm d}x & =\sum_{Y < m\,\leq\,X^{2}}\frac{r^{2}_{{\scriptscriptstyle 2}}(m)}{m^{2}}+O\big(X^{{-}1}Y^{{-}1/2}\log{Y}\big)\\ & \ll Y^{{-}1}\log{Y}+X^{{-}1}Y^{{-}1/2}\log{Y}\\ & \ll Y^{{-}1}\log{Y}. \end{aligned} \end{equation}

Integrating both sides of (5.11), we have by (5.10), (5.14) and Cauchy–Schwarz inequality

(5.15)\begin{equation} \begin{aligned} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{j}_{1}(x){\rm d}x & =\left(-\frac{\sqrt{2}}{\pi}\right)^{j}\frac{1}{X}\int\limits_{ X}^{2X}\left(\,\sum_{m\,\leq\,Y}\frac{r_{{\scriptscriptstyle 2}}(m)}{m}\cos{\big(2\pi\sqrt{m}x\big)}\right)^{j}{\rm d}x+O_{\epsilon}\big(X^{\epsilon}Y^{{-}1/2}\big)\\ & =\left(\frac{-1}{\sqrt{2}\pi}\right)^{j}\sum_{\textit{e}_{1},\ldots,\textit{e}_{j}=\pm1}\sum_{m_{1},\ldots,m_{j}\leq Y}\prod_{i=1}^{j}\frac{r_{{\scriptscriptstyle 2}}(m_{i})}{m_{i}}\frac{1}{X}\\ & \quad\times\int\limits_{ X}^{2X}\cos{\left(2\pi\Big(\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}\Big)x\right)}{\rm d}x + O_{\epsilon}\big(X^{\epsilon}Y^{{-}1/2}\big). \end{aligned} \end{equation}

Before we proceed to evaluate the RHS of (5.15), we need the following result (see [Reference Wenguang14], § 2 lemma $2.2$). Let $\textit {e}_{1},\,\ldots,\,\textit {e}_{j}=\pm 1$, and suppose that $m_{1},\,\ldots,\,m_{j}\leq Y$ are integers. Then it holds

(5.16)\begin{equation} \sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}\neq0\Longrightarrow\bigg|\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}\bigg|\gg Y^{1/2-2^{j-2}}, \end{equation}

where the implied constant depends only on $j$. It follows from (5.16) that

(5.17)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\cos{\left(2\pi\Big(\,\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}\Big)x\right)}{\rm d}x=\left\{ \begin{array}{ll} 1 & ;\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}=0 \\ O\big(X^{{-}1}Y^{2^{j-2}-1/2}\big) & ; \sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}\neq0 . \end{array} \right. \end{equation}

Estimate (5.17) in the case where $\sum _{i=1}^{j}\textit {e}_{i}\sqrt {m_{i}}\neq 0$ is somewhat wasteful, but nevertheless it will suffice for us. Inserting (5.17) into the RHS of (5.15), we obtain

(5.18)\begin{align} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{j}_{1}(x){\rm d}x& =\left(\frac{-1}{\sqrt{2}\pi}\right)^{j}\sum_{\textit{e}_{1},\ldots,\textit{e}_{j}=\pm1}\,\,\underset{\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}=0}{\sum_{m_{1},\ldots,m_{j}\,\leq\,Y}}\prod_{i=1}^{j}\frac{r_{{\scriptscriptstyle 2}}(m_{i})}{m_{i}}\nonumber\\ & \quad +O_{\epsilon}\Big(X^{\epsilon}Y^{{-}1/2}\Big\{1+X^{{-}1}Y^{2^{j-2}}\Big\}\Big). \end{align}

It remains to estimate the first summand appearing on the RHS of (5.18). Let $\textit {e}_{1},\,\ldots,\,\textit {e}_{j}=\pm 1$. We have

(5.19)\begin{equation} \underset{\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}=0}{\sum_{m_{1},\ldots,m_{j}\,\leq\,Y}}\prod_{i=1}^{j}\frac{r_{{\scriptscriptstyle 2}}(m_{i})}{m_{i}}=\sum_{m_{1},\ldots,m_{j}\,\leq\,Y}\prod_{i=1}^{j}\frac{\mu^{2}(m_{i})}{m_{i}}\,\,\underset{\sum_{i=1}^{j}\textit{e}_{i}k_{i}\sqrt{m_{i}}=0}{\sum_{k_{1}\,\leq\,\sqrt{Y/m_{1}},\ldots,k_{j}\,\leq\,\sqrt{Y/m_{j}}}}\,\,\prod_{i=1}^{j}\frac{r_{{\scriptscriptstyle 2}}\big(m_{i}k^{2}_{i}\big)}{k^{2}_{i}}\,. \end{equation}

Now, since the elements of the set $\mathscr {B}=\{\sqrt {m}:|\mu (m)|=1\}$ are linearly independent over $\mathbb {Q}$, it follows that the relation $\sum _{i=1}^{j}\textit {e}_{i}k_{i}\sqrt {m_{i}}=0$ with $m_{1},\,\ldots,\,m_{j}$ square-free is equivalent to the relation $\sum _{i\in S_{\ell }}\textit {e}_{i}k_{i}=0$ for $1\leq \ell \leq s$, where $\biguplus _{\ell =1}^{s} S_{\ell }=\{1,\,,\,\ldots,\,j\}$ is a partition. Multiplying (5.19) by $(-1/\sqrt {2}\pi )^{j}$, summing over all $\textit {e}_{1},\,\ldots,\,\textit {e}_{j}=\pm 1$ and rearranging the terms in ascending order, it follows that

(5.20)\begin{equation} \begin{aligned} \left(\frac{-1}{\sqrt{2}\pi}\right)^{j}\sum_{\textit{e}_{1},\ldots,\textit{e}_{j}=\pm1} & \underset{\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}=0}{\sum_{m_{1},\ldots,m_{j}\,\leq\,Y}}\prod_{i=1}^{j}\frac{r_{{\scriptscriptstyle 2}}(m_{i})}{m_{i}}\\ & =\sum_{s=1}^{j}\underset{\ell_{1},\ldots,\ell_{s}\geq1}{\sum_{\ell_{1}+\cdots+\ell_{s}=j}}\frac{j!}{\ell_{1}!\cdots\ell_{s}!}\sum_{m_{1}<\cdots< m_{s}\leq Y}\prod_{i=1}^{s}\Xi\big(m_{i},\ell_{i};\sqrt{Y/m_{i}}\,\big), \end{aligned} \end{equation}

where for $y\geq 1$, and integers $m,\,\ell \geq 1$, the term $\Xi (m,\,\ell ;y)$ is given

(5.21)\begin{equation} \Xi(m,\ell;y)=({-}1)^{\ell}\left(\frac{1}{\sqrt{2}\pi}\right)^{\ell}\frac{\mu^{2}(m)}{m^{\ell}}\sum_{\textit{e}_{1},\ldots,\textit{e}_{\ell}=\pm1}\underset{\textit{e}_{1}k_{1}+\cdots+\textit{e}_{\ell}k_{\ell}=0}{\sum_{k_{1},\ldots,k_{\ell}\leq y}}\prod_{i=1}^{\ell}\frac{r_{{\scriptscriptstyle2}}\big(mk^{2}_{i}\big)}{k_{i}^{2}}. \end{equation}

Denoting by $\tau (\cdot )$ the divisor function, we have for $m$ square-free

(5.22)\begin{equation} \sum_{k > y}\frac{r_{{\scriptscriptstyle2}}\big(mk^{2}\big)}{k^{2}}\ll r_{{\scriptscriptstyle2}}(m)\sum_{k > y}\frac{\tau^{2}(k)}{k^{2}}\ll r_{{\scriptscriptstyle2}}(m)y^{{-}1}\big(\log{2y}\big)^{3}. \end{equation}

Using (5.22) repeatedly, it follows that

(5.23)\begin{equation} \Xi(m,\ell;y)=\Xi(m,\ell)+O\left(\frac{\mu^{2}(m)r^{\ell}_{{\scriptscriptstyle2}}(m)}{m^{\ell}}y^{{-}1}\big(\log{2y}\big)^{3}\right), \end{equation}

where $\Xi (m,\,\ell )$ is given as in (1.4). Using (5.23) repeatedly, and noting that $\Xi (m,\,1;y)=0$, we have by (5.20)

(5.24)\begin{equation} \begin{aligned} & \left(\frac{-1}{\sqrt{2}\pi}\right)^{j}\sum_{\textit{e}_{1},\ldots,\textit{e}_{j}=\pm1}\underset{\sum_{i=1}^{j}\textit{e}_{i}\sqrt{m_{i}}=0}{\sum_{m_{1},\ldots,m_{j}\,\leq\,Y}}\prod_{i=1}^{j}\frac{r_{{\scriptscriptstyle 2}}(m_{i})}{m_{i}}\\ & =\sum_{s=1}^{j}\underset{\ell_{1},\ldots,\ell_{s}\geq1}{\sum_{\ell_{1}+\cdots+\ell_{s}=j}}\frac{j!}{\ell_{1}!\cdots\ell_{s}!}\sum_{m_{1}<\cdots< m_{s}\leq Y}\prod_{i=1}^{s}\Xi(m_{i},\ell_{i})+O\big(Y^{{-}1/2}(\log{Y})^{3}\big)\\ & =\sum_{s=1}^{j}\underset{\ell_{1},\ldots,\ell_{s}\geq1}{\sum_{\ell_{1}+\cdots+\ell_{s}=j}}\frac{j!}{\ell_{1}!\cdots\ell_{s}!}\underset{m_{s}>\cdots>m_{1}}{\sum_{m_{1},\ldots\,,m_{s}=1}^{\infty}}\prod_{i=1}^{s}\Xi(m_{i},\ell_{i})+O\big(Y^{{-}1/2}(\log{Y})^{3}\big), \end{aligned} \end{equation}

where the series appearing on the RHS of (5.24) converges absolutely. Finally, inserting (5.24) into the RHS of (5.18) and making the choice $Y=X^{2^{2-j}}$, we arrive at

(5.25)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{j}_{1}(x){\rm d}x=\sum_{s=1}^{j}\underset{\ell_{1},\ldots\,,\ell_{s}\geq1}{\sum_{\ell_{1}+\cdots+\ell_{s}=j}}\,\,\frac{j!}{\ell_{1}!\cdots\ell_{s}!}\underset{m_{s}>\cdots>m_{1}}{\sum_{m_{1},\ldots,m_{s}=1}^{\infty}}\prod_{i=1}^{s}\Xi(m_{i},\ell_{i})+O_{\epsilon}\Big(X^{-\eta_{j}+\epsilon}\Big), \end{equation}

where $\eta _{j}=2^{1-j}$. This settles the proof in the case where $j\geq 3$. The proof of proposition 5.1 is therefore complete.

Proof Proof of theorem 1.1

We shall first prove (1.1) in the particular case where $\mathcal {F}\in C^{\infty }_{0}(\mathbb {R})$, that is, $\mathcal {F}$ is an infinitely differentiable function having compact support.

To that end, let $\mathcal {F}$ be test function as above, and note that the assumptions on $\mathcal {F}$ imply that $|\mathcal {F}(w)-\mathcal {F}(y)|\leq c_{\mathcal {F}}|w-y|$ for all $w,\,y$, where $c_{\mathcal {F}}>0$ is some constant which depends on $\mathcal {F}$. It follows that for any integer $M\geq 1$ and any $X>0$, we have

(6.1)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x=\frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\Big(\sum_{m\leq M}\phi_{{\scriptscriptstyle 1,m}}\big(\sqrt{m}x\big)\Big){\rm d}x+\mathscr{E}_{\mathcal{F}}\big(X,M\big), \end{equation}

where $\phi _{{\scriptscriptstyle 1,m}}(t)$ is defined as in proposition 3.1, and the remainder term $\mathscr {E}_{\mathcal {F}}(X,\,M)$ satisfies the bound

(6.2)\begin{equation} \big|\mathscr{E}_{\mathcal{F}}\big(X,M\big)\big|\leq c_{\mathcal{F}}\,\frac{1}{X}\int\limits_{X}^{2X}\Big|\widehat{\mathcal{E}}_{1}(x)-\sum_{m\leq M}\phi_{{\scriptscriptstyle 1,m}}\big(\sqrt{m}x\big)\Big|{\rm d}x. \end{equation}

By proposition 3.1 it follows that

(6.3)\begin{equation} \lim_{M\to\infty}\limsup_{X\to\infty}\big|\mathscr{E}_{\mathcal{F}}\big(X,M\big)\big|=0. \end{equation}

Denoting by $\widehat {\mathcal {F}}$ the Fourier transform of $\mathcal {F}$, the assumptions on the test function $\mathcal {F}$ allows us to write

(6.4)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\Big(\sum_{m\,\leq\,M}\phi_{{\scriptscriptstyle 1,m}}\big(\sqrt{m}x\big)\Big){\rm d}x=\int\limits_{-\infty}^{\infty}\widehat{\mathcal{F}}(\alpha)\mathfrak{M}_{X}(\alpha;M){\rm d}\alpha, \end{equation}

where $\mathfrak {M}_{X}(\alpha ;M)$ is defined at the beginning of § 4. Letting $X\to \infty$ in (6.4), we have by (4.2) in proposition 4.2 together with an application of Lebesgue's dominated convergence theorem

(6.5)\begin{equation} \lim_{X\to\infty}\frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\Big(\sum_{m\,\leq\,M}\phi_{{\scriptscriptstyle 1,m}}\big(\sqrt{m}x\big)\Big){\rm d}x=\int\limits_{-\infty}^{\infty}\widehat{\mathcal{F}}(\alpha)\mathfrak{M}(\alpha;M){\rm d}\alpha, \end{equation}

where $\mathfrak {M}(\alpha ;M)=\prod _{m\,\leq \,M}\Phi _{{\scriptscriptstyle 1,m}}(\alpha )$, and $\Phi _{{\scriptscriptstyle 1,m}}(\alpha )$ is defined at the beginning of § 4. Letting $M\to \infty$ in (6.5), and recalling that $\mathcal {P}_{1}(\alpha )=\widehat {\Phi }_{{\scriptscriptstyle 1}}(\alpha )$ where $\Phi _{{\scriptscriptstyle 1}}(\alpha )=\prod _{m=1}^{\infty }\Phi _{{\scriptscriptstyle 1,m}}(\alpha )$, we have by Lebesgue's dominated convergence theorem

(6.6)\begin{equation} \begin{aligned} \lim_{M\to\infty}\lim_{X\to\infty}\frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\Big(\sum_{m\,\leq\,M}\phi_{{\scriptscriptstyle 1,m}}\big(\sqrt{m}x\big)\Big){\rm d}x & =\int\limits_{-\infty}^{\infty}\widehat{\mathcal{F}}(\alpha)\Phi_{{\scriptscriptstyle1}}(\alpha){\rm d}\alpha\\ & =\int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha, \end{aligned} \end{equation}

where in the second equality we made use of Parseval's theorem which is justified by the decay estimate (4.4) for $\mathcal {P}^{(j)}_{1}(\alpha )$ with $\alpha$ real stated in proposition 4.2. It follows from (6.4), (6.5) and (6.6) that

(6.7)\begin{equation} \frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\Big(\sum_{m\,\leq\,M}\phi_{{\scriptscriptstyle 1,m}}\big(\sqrt{m}x\big)\Big){\rm d}x=\int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha+\mathscr{E}^{\flat}_{\mathcal{F}}\big(X,M\big), \end{equation}

where the remainder term $\mathscr {E}^{\flat }_{\mathcal {F}}(X,\,M)$ satisfies

(6.8)\begin{equation} \lim_{M\to\infty}\lim_{X\to\infty}\big|\mathscr{E}^{\flat}_{\mathcal{F}}\big(X,M\big)\big|=0. \end{equation}

Inserting (6.7) into the RHS of (6.1), we deduce from (6.3) and (6.8) that

(6.9)\begin{equation} \begin{aligned} & \limsup_{X\to\infty}\bigg|\frac{1}{X}\int\limits_{ X}^{2X}\mathcal{F}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x-\int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha\bigg|\\ & \quad\leq\lim_{M\to\infty}\lim_{X\to\infty}\big|\mathscr{E}^{\flat}_{\mathcal{F}}\big(X,M\big)\big|+\lim_{M\to\infty}\limsup_{X\to\infty}\big|\mathscr{E}_{\mathcal{F}}\big(X,M\big)\big|=0. \end{aligned} \end{equation}

We conclude that

(6.10)\begin{equation} \lim\limits_{X\to\infty}\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x=\int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha, \end{equation}

whenever $\mathcal {F}\in C^{\infty }_{0}(\mathbb {R})$.

The result (6.10) extends easily to include the class $C_{0}(\mathbb {R})$ of continuous functions with compact support. To see this, fix a smooth bump function $\varphi (y)\geq 0$ supported in $[-1,\,1]$ having total mass $1$, and for an integer $n\geq 1$ let $\varphi _{n}(y)=n\varphi (ny)$. Given $\mathcal {F}\in C_{0}(\mathbb {R})$, let $\mathcal {F}_{n}=\mathcal {F}\star \varphi _{n}\in C^{\infty }_{0}(\mathbb {R})$, where $\star$ denotes the Euclidean convolution operator. We then have

(6.11)\begin{equation} \begin{aligned} & \bigg|\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x-\int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha\bigg|\\ & \quad\leq\bigg|\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}_{n}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x-\int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha\bigg|+\underset{y\in\mathbb{R}}{\textit{max}}\big|\mathcal{F}(y)-\mathcal{F}_{n}(y)\big|. \end{aligned} \end{equation}

Since $\underset {y\in \mathbb {R}}{\textit {max}}\,\big |\mathcal {F}(y)-\mathcal {F}_{n}(y)\big |\to 0$ as $n\to \infty$, and

(6.12)\begin{equation} \begin{aligned} \int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha & =\lim_{n\to\infty}\int\limits_{-\infty}^{\infty}\mathcal{F}_{n}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha\\ & =\lim_{n\to\infty}\lim\limits_{X\to\infty}\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}_{n}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x, \end{aligned} \end{equation}

it follows that (6.10) holds whenever $\mathcal {F}\in C_{0}(\mathbb {R})$.

Let us now consider the general case in which $\mathcal {F}$ is a continuous function of polynomial growth, say $|\mathcal {F}(\alpha )|\ll |\alpha |^{j}$ for all sufficiently large $|\alpha |$, where $j\geq 1$ is some integer. Let $\psi \in C^{\infty }_{0}(\mathbb {R})$ satisfy $0\leq \psi (y)\leq 1$, $\psi (y)=1$ for $|y|\leq 1$, and set $\psi _{n}(y)=\psi (y/n)$. Define $\mathcal {F}_{n}(y)=\mathcal {F}(y)\psi _{n}(y)\in C_{0}(\mathbb {R})$. For $n$ sufficiently large we have by proposition 5.1

(6.13)\begin{equation} \begin{aligned} \frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x & =\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}_{n}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x+O\Bigg(\frac{1}{n^{j}}\frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{2j}_{1}(x){\rm d}x\Bigg)\\ & =\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}_{n}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x+O\left(\frac{1}{n^{j}}\right), \end{aligned} \end{equation}

where the implied constant depends only on $j$ and the implicit constant appearing in the relation $|\mathcal {F}(\alpha )|\ll |\alpha |^{j}$. Since $\mathcal {F}_{n}\to \mathcal {F}$ pointwise as $n\to \infty$, it follows from the rapid decay of $\mathcal {P}_{1}(\alpha )$ that

(6.14)\begin{equation} \begin{aligned} \int\limits_{-\infty}^{\infty}\mathcal{F}(\alpha)\mathcal{P}_{1}(\alpha){\rm d}\alpha & =\lim_{n\to\infty}\int\limits_{-\infty}^{\infty}\mathcal{F}_{n}(\alpha)\mathcal{P}_{}(\alpha){\rm d}\alpha\\ & =\lim_{n\to\infty}\lim\limits_{X\to\infty}\frac{1}{X}\int\limits_{X}^{2X}\mathcal{F}_{n}\big(\widehat{\mathcal{E}}_{1}(x)\big){\rm d}x\,. \end{aligned} \end{equation}

We conclude from (6.13) and (6.14) that $\frac {1}{X}\int _{X}^{2X}\mathcal {F}(\widehat {\mathcal {E}}_{1}(x)){\rm d}x\to \int _{-\infty }^{\infty }\mathcal {F}(\alpha )\mathcal {P}_{1}(\alpha ){\rm d}\alpha$ as $X\to \infty$. It follows that (6.10) holds for all continuous functions of polynomial growth. The extension to include the class of (piecewise)-continuous functions of polynomial growth is now straightforward, and so (1.1) is proved.

The decay estimates (1.2) stated in theorem 1.1 have already been proved in proposition 4.2, and it remains to show that $\mathcal {P}_{1}(\alpha )$ defines a probability density. To that end, we note that the LHS of (6.10) is real and non-negative whenever $\mathcal {F}$ is. Since $\mathcal {P}_{1}(\alpha )$ is continuous, by choosing a suitable test function $\mathcal {F}$ in (6.10), we conclude that $\mathcal {P}_{q}(\alpha )\geq 0$ for real $\alpha$. Taking $\mathcal {F}\equiv 1$ in (6.10) we find $\int _{-\infty }^{\infty }\mathcal {P}_{1}(\alpha ){\rm d}\alpha =1$. The proof of theorem 1.1 is therefore complete.

Proof Proof of theorem 1.3

In the particular case where $\mathcal {F}(\alpha )=\alpha ^{j}$ with $j\geq 1$ an integer, we have by theorem 1.1

(6.15)\begin{equation} \lim\limits_{X\to\infty}\frac{1}{X}\int\limits_{ X}^{2X}\widehat{\mathcal{E}}^{j}_{1}(x){\rm d}x=\int\limits_{-\infty}^{\infty}\alpha^{j}\mathcal{P}_{1}(\alpha){\rm d}\alpha. \end{equation}

It immediately follows from (5.1) in proposition 5.1 that

(6.16)\begin{equation} \int\limits_{-\infty}^{\infty}\alpha^{j}\mathcal{P}_{1}(\alpha){\rm d}\alpha=\sum_{s=1}^{j}\underset{\,\,\ell_{1},\,\ldots\,,\ell_{s}\geq1}{\sum_{\ell_{1}+\cdots+\ell_{s}=j}}\,\,\frac{j!}{\ell_{1}!\cdots\ell_{s}!}\underset{\,\,\,m_{s}>\cdots>m_{1}}{\sum_{m_{1},\,\ldots\,,m_{s}=1}^{\infty}}\prod_{i=1}^{s}\Xi(m_{i},\ell_{i}), \end{equation}

where the series on the RHS of (6.16) converges absolutely, and for integers $m,\,\ell \geq 1$, the term $\Xi (m,\,\ell )$ is given as in (1.4).

By definition $\Xi (m,\,1)=0$, and so it follows from (6.16) that ${\int _{-\infty }^{\infty }\alpha \mathcal {P}_{1}(\alpha ){\rm d}\alpha =0}$. Suppose now that $j\geq 3$ is an integer which satisfies $j\equiv 1\,(2)$, and consider a summand

(6.17)\begin{equation} \underset{m_{s}>\cdots>m_{1}}{\sum_{m_{1},\ldots,m_{s}=1}^{\infty}}\prod_{i=1}^{s}\Xi(m_{i},\ell_{i}), \end{equation}

with $\ell _{1}+\cdots +\ell _{s}=j$. We may clearly assume that $\ell _{1},\,\ldots,\,\ell _{s}\geq 2$, for otherwise (6.17) vanishes. As $j$ is odd, it follows that $|\{1\leq i\leq s: \ell _{i}\equiv 1\,(2)\}|$ is also odd. Since $\Xi (m,\,\ell )<0$ for $\ell >1$ odd, and $\Xi (m,\,\ell )>0$ for $\ell$ even, it follows that (6.17) is strictly negative. We conclude that $\int _{-\infty }^{\infty }\alpha ^{j}\mathcal {P}_{1}(\alpha ){\rm d}\alpha <0$. The proof of theorem 1.3 is therefore complete.