Generalized Vandermonde matrices

One of the primary difficulties which we will encounter in § 6 is the fact that taking derivatives of exponentials (which arise as the inverse of the $\log$s defining our sequence) increases in complexity as we take derivatives. To handle this we appeal to a recent result of Khare and Tao [Reference Khare and TaoKT21] which requires us to set up some notation. Given an $M$-tuple $\boldsymbol {\mathbf {u}} \in \mathbb {R}^M$, let

\[ V(\boldsymbol{\mathbf{u}}) := \prod_{1\le i < j \le M}(u_j-u_i) \]

denote the Vandermonde determinant. Furthermore, given two tuples $\boldsymbol {\mathbf {u}}$ and $\boldsymbol {\mathbf {n}}$, we define

\[ \boldsymbol{\mathbf{u}}^{\boldsymbol{\mathbf{n}}} := u_1^{n_1} \cdots u_M^{n_M}\quad \mbox{and}\quad \boldsymbol{\mathbf{u}}^{\circ\boldsymbol{\mathbf{n}}} := \begin{pmatrix} u_1^{n_1} & u_1^{n_2} & \cdots & u_1^{n_M}\\ u_2^{n_1} & u_2^{n_2} & \cdots & u_2^{n_M}\\ \vdots & \vdots & \vdots & \vdots \\ u_M^{n_1} & u_M^{n_2} & \cdots & u_M^{n_M} \end{pmatrix}, \]

the latter being a generalized Vandermonde matrix. Finally, let $\boldsymbol {\mathbf {n}}_{\min } :=(0,1, \ldots, M-1)$. Then Khare and Tao established the following lemma.

Lemma 2.4 [Reference Khare and TaoKT21, Lemma 5.3]

Let $K$ be a compact subset of the cone

\[ \{(n_1, \ldots, n_{M})\in \mathbb{R}^M \ : \ n_1 < \cdots < n_{M}\}. \]

Then there exist constants $C,c>0$ such that

(2.3)\begin{equation} cV(\boldsymbol{\mathbf{u}}) \boldsymbol{\mathbf{u}}^{\boldsymbol{\mathbf{n}}-\boldsymbol{\mathbf{n}}_{\min}} \le \det(\boldsymbol{\mathbf{u}}^{\circ \boldsymbol{\mathbf{n}}}) \le C V(\boldsymbol{\mathbf{u}}) \boldsymbol{\mathbf{u}}^{\boldsymbol{\mathbf{n}}-\boldsymbol{\mathbf{n}}_{\min}} \end{equation}

for all $\boldsymbol {\mathbf {u}} \in (0,\infty )^M$ with $u_1 \le \dots \le u_M$ and all $\boldsymbol {\mathbf {n}} \in K$.

3.1 Combinatorial target

We will need the following combinatorial definitions to explain how to prove convergence of the $m$-point correlation from (3.1). Given a partition $\mathcal {P}$ of $[ m]$, we say that $j\in [m]$ is isolated if $j$ belongs to a partition element of size $1$. A partition is called non-isolating if no element is isolated (and otherwise we say it is isolating). For our example $\mathcal {P} = \{\{1,3\}, \{4\}, \{2,5,6\}\}$ we have that $4$ is isolated, and thus $\mathcal {P}$ is isolating.

Now consider the middle line of (3.1). We apply Poisson summation to each of the $k_i$ sums. That is, we insert

(3.2)\begin{equation} \sum_{k \in \mathbb{Z}} f(N(\omega(n)+k + s)) = \frac{1}{N}\sum_{k\in \mathbb{Z}} e(k(\omega(n)+s)) \widehat{f}\bigg(\frac{k}{N}\bigg) \end{equation}

yielding

(3.3)\begin{equation} \mathcal{M}^{(m)}(N) = \frac{1}{N^m} \int_0^1 \sum_{\boldsymbol{\mathbf{n}} \in [N]^m} \sum_{\boldsymbol{\mathbf{k}} \in \mathbb{Z}^m} \widehat{f}\bigg(\frac{\boldsymbol{\mathbf{k}}}{N}\bigg) e(\boldsymbol{\mathbf{k}}\cdot \omega(\boldsymbol{\mathbf{n}}) + \boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{1}}s)\,{d} s, \end{equation}

where $\omega (\boldsymbol {\mathbf {n}}) := (\omega (n_1), \omega (n_2), \ldots, \omega (n_m))$ and where $\widehat {f}(\frac {\boldsymbol {\mathbf {k}}}{N}) = \prod _{i=1}^m \widehat {f}(\frac {k_i}{N})$.

In [Reference Lutsko and TechnauLT21, § 3] we showed that, if

\[ \mathcal{E}(N):= \frac{1}{N^m} \int_0^1 \sum_{\boldsymbol{\mathbf{n}} \in [N]^m} \sum_{\substack{\boldsymbol{\mathbf{k}} \in (\mathbb{Z}^{\ast})^m}} \widehat{f}\bigg(\frac{\boldsymbol{\mathbf{k}}}{N}\bigg) e( \boldsymbol{\mathbf{k}}\cdot \omega(\boldsymbol{\mathbf{n}}) + \boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{1}}s)\,{d} s, \]

then for fixed $m$, and assuming the inductive hypothesis, Theorem 1.3 reduces to the following lemma.

Lemma 3.1 Let $\mathscr {P}_m$ denote the set of non-isolating partitions of $[m]$. We have that

(3.4)\begin{equation} \lim_{N \to \infty}\mathcal{E}(N) = \sum_{\mathcal{P}\in \mathscr{P}_m} {\mathbf{E}(f^{|P_1|})}\cdots {\mathbf{E}(f^{|P_d|})}.\end{equation}

where the partition $\mathcal {P} = (P_1, P_2, \ldots, P_d)$, and $|P_i|$ is the size of $P_i$.

3.2 Dyadic decomposition

It is convenient to decompose the sums over $n$ and $k$ within $S_N(s,f)$ into (nearly) dyadic ranges in a smooth manner. Given $N$, we let $Q>1$ be the unique integer with $e^{Q}\leq N < e^{Q+1}$. Now, we describe a smooth partition of unity which approximates the indicator function of $[1,N]$. Strictly speaking, these partitions depend on $Q$, but we suppress this from the notation. Furthermore, since we want asymptotics of $\mathcal {E}(N)$, we need to take a bit of care at the right end point of $[1,N]$, and so a tighter than dyadic decomposition is needed. Let us make this precise, and point out that a detailed construction can be found in the Appendix. For $0\le q < Q$ we let $\mathfrak {N}_q: \mathbb {R}\to [0,1]$ denote a smooth function for which

\[ \operatorname{supp}(\mathfrak{N}_q) \subset [e^{q}/2, 3 e^q), \quad \mathrm{for}\ 0\leq q < Q, \]

and such that $\mathfrak {N}_{q}(x) + \mathfrak {N}_{q+1}(x) = 1$ for $x\in [e^q,e^{q+1})$. Now for $q \ge Q$ we let $\mathfrak {N}_q$ form a smooth partition of unity for which

(3.5)\begin{equation} \sum_{q=0}^{2Q-1} \mathfrak{N}_q (x) = \begin{cases} 1 & \mbox{if } 1< x < e^{Q},\\ 0 & \mbox{if } x< 1/2 \mbox{ or } x > N + \dfrac{3N}{\log(N)} \end{cases} \end{equation}

and

(3.6)\begin{equation} \operatorname{supp}(\mathfrak{N}_q) \subset \bigg[e^{Q} + (q-Q-1.1)\frac{e^{Q}}{Q}, e^{Q} + (3+q-Q)\frac{e^{Q}}{Q}\bigg) \quad \mathrm{for}\ Q< q\leq 2Q-1, \end{equation}

while $\operatorname {supp}(\mathfrak {N}_Q) \subset (0.9\cdot e^{Q-1}, 1.1\cdot e^{Q})$. Let $\Vert g \Vert _{\infty }$ denote the sup norm of a function $g : \mathbb {R} \to \mathbb {R}$. We impose the following condition on the derivatives:

(3.7)\begin{equation} \Vert \mathfrak{N}_{q}^{(j)}\Vert_{\infty} \ll \begin{cases} e^{-qj} & \mbox{for } q < Q,\\ (e^{Q}/Q)^{-j} & \mbox{for } Q< q, \end{cases} \end{equation}

for $j \ge 1$. For technical reasons, assume $\mathfrak {N}_q^{(1)}$ changes sign only once.

Notice that (3.5) implies

\[ \sum_{n \in \mathbb{Z}}\sum_{q=0}^{2Q-2} \mathfrak{N}_q (n) \sum_{k \in \mathbb{Z}} f(N(\omega(n) +k +s)) \leq S_N(s,f) \leq\sum_{n\in \mathbb{Z}}\sum_{q=0}^{2Q-1} \mathfrak{N}_q (n) \sum_{k \in \mathbb{Z}} f(N(\omega(n) +k +s)). \]

Ignoring the lower bound, which can be treated similarly, applying Poisson summation for every fixed $n$ and using the positivity of the inner sum, we then have

\[ S_N(s,f)\leq \frac{1}{N}\sum_{n \in \mathbb{Z}}\sum_{q=0}^{2Q-1} \mathfrak{N}_q(n) \sum_{k \in \mathbb{Z}} \widehat{f} \bigg(\frac{k}{N}\bigg) e(k(\omega(n) +s)). \]

Next, by positivity, we have that

(3.8)\begin{equation} \mathcal{M}^{(m)}(N) \le \int_0^1 \bigg(\frac{1}{N}\sum_{q=0}^{2Q-1} \sum_{n \in \mathbb{Z}} \mathfrak{N}_{q}(n)\sum_{k \in \mathbb{Z}} \widehat{f}\bigg(\frac{k}{N}\bigg) e( k \omega(n) + ks)\bigg)^m {d} s. \end{equation}

All frequencies $\boldsymbol {\mathbf {k}}$ for which $k_j=0$ for at least some index $1\leq j\leq m-1$ contribute to $\mathcal {M}^{(m)}(N)$ exactly

\[ \sum_{i=1}^{m-1} \begin{pmatrix} m-1\\ i \end{pmatrix}\widehat{f}(0)^{i}\int_0^1 \bigg(\frac{1}{N}\sum_{n \in [N]}\sum_{k \neq 0} \widehat{f}\bigg(\frac{k}{N}\bigg) e( k \omega(n) + ks)\bigg)^{m-i} {d} s. \]

Subtracting exactly the above term from both sides of (3.8), while using our inductive assumption that the $\mathcal {M}^{m-i}(N)$ converge (for $1\leq i\leq m-2)$, then yields

(3.9)\begin{equation} \mathcal{E}(N) \le \int_0^1\bigg(\frac{1}{N}\sum_{q=0}^{2Q-1} \sum_{n \in \mathbb{Z}} \mathfrak{N}_{q}(n)\sum_{k\neq 0}\widehat{f}\bigg(\frac{k}{N}\bigg) e(k \omega(n) + ks)\bigg)^m {d} s +o(1). \end{equation}

The same argument can be used to yield

\[ \mathcal{E}(N)+o(1) \ge \int_0^1\bigg(\frac{1}{N}\sum_{q=0}^{2Q-2} \sum_{n \in \mathbb{Z}} \mathfrak{N}_{q}(n)\sum_{k\neq 0}\widehat{f}\bigg(\frac{k}{N}\bigg) e( k \omega(n) + ks)\bigg)^m {d} s . \]

We similarly decompose the $k$ sums, although thanks to the compact support of $\widehat {f}$ we do not need to worry about $k\ge N$. Let $\mathfrak {K}_u:\mathbb {R} \to [0,1]$ be a smooth function such that, for $U :=\lceil \log N \rceil$,

\[ \sum_{u=-U}^{U} \mathfrak{K}_u(k) =\begin{cases} 1 & \mbox{if } \vert k\vert \in [1, N),\\ 0 & \mbox{if } \vert k\vert < 1/2 \mbox{ or } \vert k\vert > 2N, \end{cases} \]

and the symmetry $\mathfrak {K}_{-u}(k) = \mathfrak {K}_{u}(-k)$ holds true for all $u,k> 0$. Similarly,

\begin{align*} & \operatorname{supp}(\mathfrak{K}_u) \subset[e^{u }/3, 3e^{u})\quad \mbox{if } u \ge 0,\quad \mbox{and } \\ & \Vert \mathfrak{K}_{u}^{(j)} \Vert_{\infty} \ll e^{-|u|j},\quad \mbox{for all } j\ge 1. \end{align*}

As for $\mathfrak {N}_q$, we also assume $\mathfrak {K}_u^{(1)}$ changes sign exactly once.

Therefore, a central role is played by the smoothed exponential sums

(3.10)\begin{equation} \mathcal{E}_{q,u}(s):=\frac{1}{N}\sum_{k\in \mathbb{Z}} \mathfrak{K}_{u}(k) \widehat{f}\bigg(\frac{k}{N}\bigg)e( ks)\sum_{n\in \mathbb{Z}} \mathfrak{N}_{q}(n)e( k\omega(n) ). \end{equation}

Notice that (3.9) and the compact support of $\widehat {f}$ imply

\[ \mathcal{E}(N) \ll \bigg\Vert \sum_{u=-U}^{U}\sum_{q=0}^{2Q-1} \mathcal{E}_{q,u}\bigg\Vert_{L^m(\mathbb{R})}^m. \]

Now write

\[ \mathcal{F}(N) := \frac{1}{N^m}\sum_{\boldsymbol{\mathbf{q}}\in [0,2Q-1]^m} \sum_{\boldsymbol{\mathbf{u}} = [-U,U]^m}\sum_{\boldsymbol{\mathbf{k}},\boldsymbol{\mathbf{n}} \in \mathbb{Z}^m} \mathfrak{K}_{\boldsymbol{\mathbf{u}}}(\boldsymbol{\mathbf{k}})\mathfrak{N}_{\boldsymbol{\mathbf{q}}}(\boldsymbol{\mathbf{n}}) \int_0^1\widehat{f}\bigg(\frac{\boldsymbol{\mathbf{k}}}{N}\bigg) e( \boldsymbol{\mathbf{k}}\cdot \omega(\boldsymbol{\mathbf{n}}) + \boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{1}} s)\, {d} s, \]

where $\mathfrak {N}_{\boldsymbol {\mathbf {q}}}(\boldsymbol {\mathbf {n}}) := \mathfrak {N}_{q_1}(n_1)\mathfrak {N}_{q_2}(n_2)\cdots \mathfrak {N}_{q_m}(n_m)$ and $\mathfrak {K}_{\boldsymbol {\mathbf {u}}}(\boldsymbol {\mathbf {k}}) := \mathfrak {K}_{u_1}(k_1)\mathfrak {K}_{u_2}(k_2)\cdots \mathfrak {K}_{u_m}(k_m)$. Our goal will be to establish that $\mathcal {F}(N)$ is equal to the right-hand side of (3.4) up to an $o(1)$ term. Then, since we can establish the same asymptotic for the lower bound, we may conclude the asymptotic for $\mathcal {E}(N)$. Since the details are identical, we will only focus on $\mathcal {F}(N)$.

Fixing $\boldsymbol {\mathbf {q}}$ and $\boldsymbol {\mathbf {u}}$, we let

\[ \mathcal{F}_{\boldsymbol{\mathbf{q}},\boldsymbol{\mathbf{u}}}(N)=\frac{1}{N^m}\int_0^1\sum_{\substack{\boldsymbol{\mathbf{n}},\boldsymbol{\mathbf{k}} \in \mathbb{Z}^m}} \mathfrak{N}_{\boldsymbol{\mathbf{q}}}(\boldsymbol{\mathbf{n}})\mathfrak{K}_{\boldsymbol{\mathbf{u}}}(\boldsymbol{\mathbf{k}}) \widehat{f}\bigg(\frac{\boldsymbol{\mathbf{k}}}{N}\bigg) e(\boldsymbol{\mathbf{k}}\cdot \omega(\boldsymbol{\mathbf{n}}) + \boldsymbol{\mathbf{k}}\cdot \boldsymbol{\mathbf{1}}s) \,{d} s. \]

4.1 Degenerate regimes

Fix $\delta = {1/(m + 1)}$. We say that $(q,u)\in [2 Q] \times [-U,U]$ is degenerate if either

\[ |u| < q^{{(A-1)/2}}\quad \mathrm{or}\quad q \leq \delta Q \]

holds. Otherwise $(q,u)$ is called non-degenerate. Let $\mathscr {G}(N)$ denote the set of all non-degenerate pairs $(q,u)$. In this section it is enough to suppose that $u>0$ (and therefore $k>0$). Next, we show that degenerate $(q,u)$ contribute a negligible amount to $\mathcal {F}(N)$.

First, assume $q \le \delta Q$. Expanding the $m$th power, evaluating the $s$-integral and trivial estimation yield

\begin{align*} \Vert \mathcal{E}_{q,u} \Vert_{L^m}^m &= \int_0^1\bigg(\frac{1}{N} \sum_{k \in \mathbb{Z}} \mathfrak{K}_u(k)\widehat{f}\bigg(\frac{k}{N}\bigg) e(ks) \sum_{n \in \mathbb{Z}} \mathfrak{N}_{q}(n)e(k\omega(n))\bigg)^m {d}s\\ &\ll \frac{1}{N^m}\sum_{\substack{k_i \asymp e^u,\\ i =1,\ldots m}} \max_x\bigg(\widehat{f}\bigg(\frac{x}{N}\bigg)\bigg)^m\sum_{\substack{n_i \asymp e^q,\\ i =1,\ldots m}} \bigg|\int_0^1 e((k_1+ \cdots + k_m)s)\,{d}s\bigg|\\ &\ll \frac{1}{N^m} \#\{k_1,\ldots,k_m\asymp e^{u}: k_1 +\cdots +k_m = 0 \}N^{m\delta}\ll N^{m\delta-1}. \end{align*}

If $u< q^{(A-1)/2}$ and $q>\delta Q$, then we can apply the Euler summation formula, followed by van der Corput's lemma with $j=1$, to conclude that

\[ \sum_{n\in \mathbb{Z}}\mathfrak{N}_{q}(n) e( k \omega(n)) \ll \frac{e^q}{k q^{A-1}}, \]

where the numerator is the size of the support of $\mathfrak {N}_q$ and the denominator is the maximum value of $k \omega '(x)$ for $x$ in that support. Hence,

\[ \Vert \mathcal{E}_{q,u} \Vert_\infty\ll \frac{1}{N}\sum_{k\asymp e^{u}} \frac{e^q}{k q^{A-1}}\ll \frac{1}{N}\frac{e^q}{ q^{A-1}}. \]

Note

\[ \sum_{q\leq Q}\frac{e^{q}}{q^{A-1}}\ll\int_{1}^{Q}\frac{e^{q}}{q^{A-1}}{d}q= \int_{1}^{Q/2}\frac{e^{q}}{q^{A-1}}{d}q+\int_{Q/2}^{Q} \frac{e^{q}}{q^{A-1}}{d}q\ll e^{Q/2}+\frac{1}{Q^{A-1}} \int_{Q/2}^{Q}e^{Q}\,{d}q\ll\frac{e^{Q}}{Q^{A-1}}. \]

Thus,

\[ \bigg\Vert \sum_{\delta Q\leq q\leq Q} \sum_{u\leq q^{(A-1)/2}} \mathcal{E}_{q,u}\bigg\Vert_\infty \ll\frac{1}{N} \sum_{q\leq Q} \sum_{u\leq q^{(A-1)/2}}\sum_{k\asymp e^{u}}\frac{1}{k} \frac{e^{q}}{q^{A-1}}\ll\frac{1}{N} \frac{e^{Q}}{Q^{A-1}} \sum_{u\leq Q^{(A-1)/2}}1\leq\frac{1}{Q^{{(A-1)/2}}}. \]

Taking the $L^m$-norm then yields

\[ \bigg\Vert \sum_{(q,u)\in [2Q]\times[-U,U] \setminus \mathscr{G}(N)}\mathcal{E}_{q,u}\bigg\Vert_{L^m}^m \ll_{\delta} \log(N)^{-\rho}, \]

for some $\rho >0$. Hence, the triangle inequality implies

(4.1)\begin{equation} \mathcal{F}(N) =\bigg\Vert \sum_{(u,q) \in \mathscr{G}(N)}\mathcal{E}_{q,u} \bigg\Vert_{L^m}^m + O(N^{-\rho}). \end{equation}

Next, to dismiss the degenerate regimes, let $w,W$ denote strictly positive numbers satisfying $w< W$. Consider

\[ g_{w,W}(x):=\min\bigg(\frac{1}{\Vert xw\Vert},\frac{1}{W}\bigg), \]

where $\| \cdot \|$ denotes the distance to the nearest integer. We shall need (as in [Reference Lutsko and TechnauLT21, Proof of Lemma 4.1]) the following lemma.

Lemma 4.1 If $W<1/10$, then

\[ \sum_{e^{u}\leq |k|< e^{u+1}}g_{w,W}(k)\ll\bigg(e^{u}+\frac{1}{w}\bigg)\log(1/W) \]

where the implied constant is absolute.

With the previous lemma at hand, we can show that an additional degenerate regime is negligible. Specifically, when we apply the $B$-process, the first step is to apply Poisson summation. Depending on the new summation variable there may, or may not, be a stationary point. The following lemma allows us to dismiss the contribution when there is no stationary point. Fix $k\asymp e^{u}$ and let $[a,b]:=\mathrm {supp}(\mathfrak {N}_{q})$. Consider

\[ \mathrm{Err}(k):=\sum_{\underset{m_{q}(r)>0}{r\in\mathbb{Z}}} \int_{\mathbb{R}} e(\Phi_r(x)) \mathfrak{N}_{q}(x)\,{d}x, \]

where

\begin{align*} \Phi_r(x):=k \omega(x)-rx,\quad m_{q}(r):=\min_{x\in[a,b]}\vert\Phi_{r}(x)\vert. \end{align*}

Our next aim is to show that the smooth exponential sum

\[ \mathrm{Err}_{u}(s):=\sum_{k\in\mathbb{Z}}e(ks)\mathrm{Err}(k)\mathfrak{K}_{u}(k) \widehat{f}\bigg(\frac{k}{N}\bigg) \]

is small on average over $s$.

Lemma 4.2 Fix any constant $C>0$. Then the bound

(4.2)\begin{equation} I_{\boldsymbol{\mathbf{u}}}:=\int_0^1\prod_{i=1}^m \operatorname{Err}_{u_i}(s)\,{d}s\ll Q^{-C}N^{m} \end{equation}

holds uniformly in $Q^{{(A-1)/2}}\leq \boldsymbol {\mathbf {u}}\ll Q$.

Proof. Let $\mathcal {L}_{\boldsymbol {\mathbf {u}}}$ denote the truncated sub-lattice of $\mathbb {Z}^{m}$ defined by gathering all $\boldsymbol {\mathbf {k}}\in \mathbb {Z}^{m}$ so that $k_{1}+\cdots +k_{m}=0$ and $\vert k_{i}\vert \asymp e^{u_i}$ for all $i\leq m$. The quantity $\mathcal {L}_{\boldsymbol {\mathbf {u}}}$ arises from

(4.3)\begin{equation} I_{\boldsymbol{\mathbf{u}}}=\sum_{\underset{i\leq m}{|k_{i}|\asymp e^{u}}}\bigg(\bigg(\prod_{i\leq m} \mathrm{Err}(k_{i})\mathfrak{K}_{u}(k_{i})\widehat{f}\bigg(\frac{k_i}{N}\bigg)\bigg) \int_{0}^{1}e((k_{1}+\cdots+k_{m})s)\,{d}s\bigg)\ll \sum_{\boldsymbol{\mathbf{k}}\in\mathcal{L}_{\boldsymbol{\mathbf{u}}}} \bigg(\prod_{i\leq m}\mathrm{Err}(k_{i})\bigg).\end{equation}

Partial integration and the dyadic decomposition allow one to show that the contribution of $\vert r\vert \geq Q^{O(1)}$ to $\operatorname {Err}(k_i)$ can be bounded by $O(Q^{-C})$ for any $C>0$. Hence, from van der Corput's lemma (Lemma 2.3) with $j=2$ and the assumption $m_{q}(r)>0$, we infer

\[ \mathrm{Err}(k)\ll Q^{O(1)}\min\bigg(\frac{1}{\Vert k\omega'(a)-r\Vert}, \frac{1}{(k\omega''(a))^{1/2}}\bigg)= Q^{O(1)} \min\bigg(\frac{1}{\Vert k\omega'(a)\Vert}, \frac{1}{(k\omega''(a))^{1/2}}\bigg) \]

where the implied constant is absolute. Notice that $\omega '(a)\asymp q^{A-1}e^{-q}=:w$ and

\[ k\omega''(a)\asymp(e^{u-2q}q^{A-1})^{1/2}=:W. \]

Thus, $\mathrm {Err}(k)\ll g_{w,W}(k) Q^{O(1)}.$ Using $\mathrm {Err}(k_{i})\ll g_{w,W}(k_{i}) Q^{O(1)}$ for $i< m$ and $\mathrm {Err}(k_{m})\ll Q^{O(1)}/W$ in (4.3) produces the estimate

(4.4)\begin{equation} I_{\boldsymbol{\mathbf{u}}}\ll\frac{Q^{O(1)}}{W}\sum_{\underset{i< m}{\vert k_{i}\vert\asymp e^{u}}} \bigg(\prod_{i< m}g_{w,W}(k_{i})\bigg)=\frac{Q^{O(1)}}{W} \bigg(\sum_{\vert k \vert \asymp e^{u}}g_{w,W}(k)\bigg)^{m-1}.\end{equation}

Suppose $W\geq N^{-\varepsilon }$. Then $g_{w,W}(k)\leq N^{\varepsilon }$ and we obtain that

\[ I_{\boldsymbol{\mathbf{u}}}\ll Q^{O(1)} N^{\varepsilon m}e^{u_1+\dots+u_{m-1}}\ll N^{m-1+\varepsilon m}\ll Q^{-C}N^{m}. \]

Now suppose $W< N^{-\varepsilon }\leq 1/10$. Then Lemma 4.1 is applicable and yields

\[ \sum_{\vert k\vert\asymp e^{u}}g_{w,W}(k)\ll(e^{u}+1/w)\log(1/W)\ll(e^{u}+e^{q})\log(1/W)\ll NQ. \]

Plugging this into (4.4) and using $1/W\ll e^{q-u/2}q^{(1-A)/2}\ll Ne^{-\frac {u}{2}}$ shows that

\[ I_{\boldsymbol{\mathbf{u}}}\ll Q^{O(1)}\frac{(NQ)^{m-1}}{W}\ll Q^{O(1)} (NQ)^{m}e^{-\frac{u}{2}}. \]

Because $u\geq Q^{{(A-1)/2}}$, we certainly have $e^{-\frac {u}{2}}\ll Q^{-C}$ for any $C>0$ and thus the proof is complete.

4.2 First application of the $B$-process

First, following the lead set out in [Reference Lutsko, Sourmelidis and TechnauLST24], we apply the $B$-process in the $n$-variable. Assume without loss of generality that $k >0$ (if $k <0$ we take complex conjugates and the without loss of generality assumption that $f$ is even).

Then, after applying the $B$-process, the phase function $x\mapsto k\omega (x) - rx$ will be transformed to

\[ \phi(k,r): = k\omega(x_{k,r}) - r x_{k,r}\quad \mathrm{where}\ x_{k,r} := \begin{cases} \widetilde{\omega}\bigg(\dfrac{r}{k}\bigg) & \text{if } \dfrac{r}{k}>e^{A-1},\\ 3 & \text{otherwise,} \end{cases} \]

and $\widetilde {\omega }(x):= (\omega ^{\prime })^{-1}(x)$ is the inverse of the derivative of $\omega$. Importantly, the next proposition states that $\mathcal {E}_{q,u}$ is well approximated by

(4.5)\begin{equation} \mathcal{E}_{q,u}^{(B)}(s):=\frac{e(-1/8)}{N}\sum_{k > 0} \mathfrak{K}_{ u}(k)\widehat{f} \bigg(\frac{ k}{N}\bigg) e(ks)\sum_{r\geq 0} \frac{\mathfrak{N}_{q}(x_{k,r})}{\sqrt{|k\omega^{\prime\prime}(x_{k,r})|}}e(\phi(k,r)). \end{equation}

We comment on the definition of $x_{k,r}$ and $\mathcal {E}_{q,u}^{(B)}$ in the next remark.

Proposition 4.3 If $u\geq Q^{(A-1)/2}$, then

(4.6)\begin{equation} \Vert \mathcal{E}_{q,u} - \mathcal{E}^{(B)}_{q,u}\Vert_{L^m}^m \ll Q^{-100m} \end{equation}

uniformly for all non-degenerate $(u,q)\in \mathscr {G}(N)$.

Proof. Let $[a,b]:= \operatorname {supp}(\mathfrak {N}_q)$, let $\Phi _r(x):= k\omega (x) - rx$, and let $m(r):= \min \{|\Phi _r^\prime (x)|:\ x \in [a,b]\}$. As usual when applying the $B$-process we first apply Poisson summation and integration by parts:

\[ \sum_{n \in \mathbb{Z}} \mathfrak{N}_q(n) e(k\omega(n)) = \sum_{r \in \mathbb{Z}} \int_{-\infty}^\infty \mathfrak{N}_q(x) e(\Phi_r(x)) \,{d}x = M(k) + \operatorname{Err}(k), \]

where $M(k)$ gathers the contributions when $r\in \mathbb {Z}$ with $m(r)=0$ (i.e. with a stationary point) and $\operatorname {Err}(k)$ gathers the contribution of $0 < m(r)$.

In the notation of Lemma 2.2, let $w(x):= \mathfrak {N}_q(x)$, $\Lambda _{\psi } := \omega (e^q)e^{u} = q^{A}e^{u}$, and $\Omega _\psi = \Omega _{w} := e^q$. Since $(u,q)$ is non-degenerate we have that $\Lambda _\psi /\Omega _\psi \gg q$, and hence

(4.7)\begin{equation} M(k) = e(-1/8)\sum_{r\geq 0} \frac{\mathfrak{N}_{q}(x_{k,r})}{\sqrt{|k\omega^{\prime\prime}(x_{k,r})|}} e(\phi(k,r)) + O((q \Lambda_\psi^{1/2+O(\varepsilon)})^{-1}). \end{equation}

Summing (4.7) against $N^{-1} \mathfrak {K}_{u}(k) \widehat {f}( k/N) e(ks)$ for $k\geq 0$ gives rise to $\mathcal {E}_{q,u}^{\mathrm {(B)}}$. The term coming from

\[ \operatorname{Err}(k) N^{-1} \mathfrak{K}_{u}(k) \widehat{f}( k/N) e(ks)=\frac{1}{N}\mathrm{Err}_{u}(s) \]

can be bounded sufficiently by Lemma 4.2 and the triangle inequality.

Since $x_{k,r}$ is roughly of size $e^{q}$, if we stop here and apply the triangle inequality to (4.5) we would get

\begin{align*} |\mathcal{E}_{q,u}^{(B)}(s)|\ll\frac{1}{N}\sum_{k > 0} \mathfrak{K}_{u}(k) e^q \frac{1}{\sqrt{k}} \frac{k}{e^q} \ll\frac{1}{N} e^{3u/2} \ll N^{1/2}. \end{align*}

Hence, we still need to find a saving of $O(N^{1/2})$. To achieve most of this, we now apply the $B$-process in the $k$ variable. This will require the following a priori bounds.

4.3 Amplitude bounds

Before proceeding with the second application of the $B$-process, we require bounds on the amplitude function

\[ \Psi_{q,u}(k,r,s) = \Psi_{q,u} :=\frac{\mathfrak{N}_q(x_{k,r}) \mathfrak{K}_u(k)}{\sqrt{|k \omega^{\prime\prime}(x_{k,r})|}} \widehat{f}\bigg(\frac{k}{N}\bigg), \]

and its derivatives; for which we have the following lemma.

Lemma 4.4 For any pair $q,u$ as above, and any $j \ge 1$, we have the bounds

(4.8)\begin{equation} \|\partial_k^j \Psi_{q,u}(k,r,\cdot)\|_\infty \ll e^{-uj} Q^{O(1)}\Vert \Psi_{q,u} \Vert_{\infty}, \end{equation}

where the implicit constant in the exponent depends on $j$, but not $q,u$. Moreover,

\[ \|\Psi_{q,u}\|_\infty \ll e^{q-u/2} q^{-(1/2)(A-1)}. \]

Proof. First, note that since $\Psi _{q,u}$ is a product of functions of $k$, if we can establish (4.8) for each of these functions, then the overall bound will hold for $\Psi _{q,u}(k,r,s)$ by the product rule. Moreover, the bound is obvious for $\mathfrak {K}_u(k)$, $\widehat {f}(k/N)$, and $k^{-1/2}$.

Thus, consider first $\partial _k \mathfrak {N}_q(x_{k,r}) = \mathfrak {N}_q^\prime (x_{k,r}) \partial _k(x_{k,r})$. By assumption, since $x_{k,r} \asymp e^{q}$, we have that $\mathfrak {N}_q^\prime (x_{k,r}) \ll e^{-q}$. Again, by repeated application of the product rule, it suffices to show that $\partial _k^j x_{k,r} \ll e^{q-uj}Q^{O(1)}$. To that end, begin with the equation

\[ 1 = \partial_x(x) =\partial_x( \widetilde{\omega}(\omega^{\prime}(x)))= \widetilde{\omega}^\prime (\omega^\prime(x)) \omega^{\prime\prime}(x). \]

Hence, $\widetilde {\omega }^\prime (\omega ^\prime (x)) = \frac {1}{\omega ^{\prime \prime }(x)}$ which we can write as

\[ \widetilde{\omega}^\prime (\omega^\prime(x)) = x^2 f_1(\log(x)), \]

where $f_1$ is a rational function. Now we take $j-1$ derivatives of each side. Inductively, one sees that there exist rational functions $f_j$ such that

\[ \widetilde{\omega}^{(j)} (\omega^\prime(x)) = x^{j+1} f_j(\log(x)). \]

Setting $x= x_{k,r}=\widetilde \omega (r/k)$ then gives

(4.9)\begin{equation} \widetilde{\omega}^{(j)} (r/k) = x_{k,r}^{j+1} f_j(\log(x_{k,r})). \end{equation}

With (4.9), we can use repeated application of the product rule to bound

\begin{align*} \partial_k^{j} x_{k,r} &= \partial_k^{j} \widetilde{\omega}(r/k)\\ &= -\partial_k^{j-1} \widetilde{\omega}^\prime (r/k)\bigg(\frac{r}{k^2}\bigg)\\ &\ll \widetilde{\omega}^{(j)} (r/k)\bigg(\frac{r}{k^2}\bigg)^{j} + \widetilde{\omega}^\prime (r/k)\bigg(\frac{r}{k^{1+j}}\bigg)\\ &\ll x_{k,r}^{j+1} f_j(\log(x_{k,r}))\bigg(\frac{r}{k^2}\bigg)^{j} + x_{k,r}^2 f_1(\log(x_{k,r}))\bigg(\frac{r}{k^{1+j}}\bigg). \end{align*}

Now recall that $k \asymp e^{u}$, $x_{k,r} \asymp e^{q}$, and $r \asymp e^{u-q}q^{A-1}$, thus

\begin{align*} \partial_k^{j} x_{k,r} &\ll \bigg(e^{q(j+1)} \bigg(\frac{e^{u-q}}{e^{2u}}\bigg)^{j} + e^{2q}\bigg(\frac{e^{u-q}}{e^{(1+j)u}}\bigg)\bigg)Q^{O(1)}\\ &\ll e^{q-ju}Q^{O(1)}. \end{align*}

Hence, $\partial _k^{(j)} \mathfrak {N}_q(x_{k,r}) \ll e^{-ju} Q^{O(1)}$.

The same argument suffices to prove that $\partial _k^{j} (1/\sqrt {|\omega ^{\prime \prime }(x_{k,r})|}) \ll e^{q-ju} Q^{O(1)}$.

4.4 Second application of the $B$-process

We now apply the $B$-process in the $k$-variable. At the present stage, the phase function is $\phi (k,r)+ks$. Thus, for $h \in \mathbb {Z}$, let $\mu = \mu _{h,r,s}$ be the unique stationary point of $k \mapsto \phi (k,r) - (h-s)k$. Namely,

\[ (\partial_\mu\phi)(\mu,r) = h-s. \]

After the second application of the $B$-process, the phase will be transformed to

\[ \Phi(h,r,s) = \phi(\mu,r) - (h-s)\mu. \]

With that, let (again for $u >0)$

(4.10)\begin{equation} \mathcal{E}^{(\mathrm{BB})}_{q,u}(s):= \frac{1}{N} \sum_{r\ge 0}\sum_{h \ge 0} \widehat{f} \bigg(\frac{\mu}{N}\bigg)\mathfrak{K}_u(\mu)\mathfrak{N}_{q}(x_{\mu,r}) \frac{1}{\sqrt{|\mu \omega^{\prime\prime}(x_{\mu,r}) \cdot (\partial_{\mu\mu}\phi)(\mu,r) |}} e(\Phi(h,r,s)). \end{equation}

We can now apply the $B$-process for a second time and conclude the following proposition.

Proposition 4.5 We have

(4.11)\begin{equation} \Vert \mathcal{E}_{q,u}^{\mathrm{(BB)}}-\mathcal{E}_{q,u}^{\mathrm{(B)}} \Vert_{L^{m}([0,1])}=O(N^{-\frac{1}{2m} +\varepsilon}) \end{equation}

uniformly for any non-degenerate $(q,u)\in \mathscr {G}(N)$.

Before we can prove the above proposition, we need some preparations. Note that

\[ k\omega'(n)=Ak \frac{(\log n)^{A-1}}{n}\leq 10 A e^{u-q}q^{A-1}. \]

If $u-q+(A-1)\log q<-10$ then $10 A e^{u-q}q^{A-1}=10 Ae^{-10 A}\leq 0.6$. Hence, there is no stationary point in the first application of the $B$-process. Thus, the contribution from this regime is disposed of by the first $B$-process. Therefore, from now on we assume that

(4.12) \begin{equation} u\geq q-(A-1)\log q-10A,\quad\text{in particular}, e^{u}\gg e^{q}q^{1-A}.\end{equation}

4.5 Non-essential regimes

In this section we estimate the contribution from regimes where $u$ is smaller by a power of a logarithm than the top scale $Q$. We shall see that this regime can be disposed of. More precisely, let

\[ \mathcal{T}(N):=\{(q,u)\in\mathscr{G}(N):\ u\leq\log N- 10 A\log\log N\}. \]

We shall see that contribution $\mathcal {T}(N)$ is negligible by showing that the function

(4.13)\begin{equation} T_{N}(s):=\sum_{(q,u)\in\mathcal{T}(N)}\mathcal{E}_{q,u}^{(\mathrm{B})}(s)\end{equation}

has a small $\Vert \cdot \Vert _{\infty }$-norm (in the $s\in [0,1]$ variable). To prove this, we need to ensure that in

\[ \mathcal{E}_{q,u}^{(\mathrm{B})}(s)=\frac{e(-1/8)}{N}\sum_{r\geq0} \sum_{k\geq0}\Psi_{q,u}(k,r,s)e(\phi(k,r)-ks) \]

the amplitude function

\[ \Psi_{q,u}(k,r,s):=\frac{\mathfrak{N}_{q}(x_{k,r}) \mathfrak{K}_{u}(k)}{\sqrt{|k\omega''(x_{k,r})|}}\widehat{f}\bigg(\frac{k}{N}\bigg) \]

has a suitably good decay in $k$.

Lemma 4.6 If (4.12) holds, then

\[ \Vert k\mapsto\partial_{k}\Psi_{q,u}(k,r,s)\Vert_{L^{1}(\mathbb{R})} \ll e^{u/2} q^{-(1/2)(A-1)} \]

uniformly for $r$ and $s$ in the prescribed ranges.

Proof. First, use the triangle inequality to bound

\begin{align*} \Vert k\mapsto\partial_{k}\Psi_{q,u}(k,r,s)\Vert_{L^{1}(\mathbb{R})} &\ll \bigg\Vert \partial_k\bigg\{\frac{\mathfrak{N}_{q}(x_{k,r}) \mathfrak{K}_{u}(k)}{\sqrt{|k\omega''(x_{k,r})|}}\bigg\} \widehat{f}\bigg(\frac{k}{N}\bigg)\bigg\Vert_{L^{1}(\mathbb{R})} \\ &\quad +\bigg\Vert \frac{\mathfrak{N}_{q}(x_{k,r}) \mathfrak{K}_{u}(k)}{\sqrt{|k\omega''(x_{k,r})|}} \partial_k\widehat{f}\bigg(\frac{k}{N}\bigg)\bigg\Vert_{L^{1}(\mathbb{R})}. \end{align*}

Since $\widehat {f}$ has bounded derivative, the term on the right can be bounded by $1/N$ times the sup norm times $e^{u}$. Since $\widehat {f}(\frac {k}{N})$ is bounded, and $\frac {\mathfrak {N}_{q}(x_{k,r}) \mathfrak {K}_{u}(k)}{\sqrt {|k\omega ''(x_{k,r})|}}$ changes sign finitely many times, we can apply the fundamental theorem of calculus and bound the whole by

\[ \Vert k\mapsto\partial_{k}\Psi_{q,u}(k,r,s)\Vert_{L^{1}(\mathbb{R})} \ll \bigg\Vert k \mapsto \frac{1}{\sqrt{|k\omega''(x_{k,r})|}}\bigg\Vert_{L^{\infty}(\mathbb{R})}. \]

Now we are in the position to prove that the contribution from (4.13) is negligible thanks to a second derivative test. This is one of the places where, in contrast to the monomial case, we only win by a logarithmic factor. Moreover, this logarithmic saving goes to $0$ as $A$ approaches $1$.

Lemma 4.7 The oscillatory integral

(4.14)\begin{equation} I_{q,u}(h,r):=\int_{-\infty}^{\infty}\Psi_{q,u}(t,r,s)e(\phi(t,r)-t(h-s))\,{d}t \end{equation}

satisfies the bound

(4.15)\begin{equation} I_{q,u}(h,r)\ll e^{q} q^{1-A}\end{equation}

uniformly in $h,$ and $r$ in ranges prescribed by $\Psi$.

Proof. We aim to apply van der Corput's lemma (Lemma 2.3) for a second derivative bound. For that, first note that $\partial _{t}\phi (t,r)=\omega (x_{t,r})+t\partial _{t}(\omega (x_{t,r}))-r\partial _{t}(x_{t,r})$. Now, since

(4.16)\begin{equation} \partial_{t}(\omega(x_{t,r}))=\omega'(x_{t,r})\partial_{t}(x_{t,r})= \frac{r}{t}\partial_{t}(x_{t,r}),\end{equation}

it follows that

(4.17)\begin{equation} \partial_{t}\phi(t,r)=\omega(x_{t,r}).\end{equation}

Now we bound the second derivative of $\phi (t,r)-t(s+h)$. By (4.16) and (4.17), we have

\[ \partial_{t}^{2}[\phi(t,r)] =\partial_{t}[\omega(x_{t,r})]=\frac{r}{t}\partial_{t}[x_{t,r}]. \]

Thus,

\[ \partial_{t}^{2}[\phi(t,r)]=-\frac{1}{\omega''(x_{t,r})}\frac{r^{2}}{t^{3}}. \]

Taking $x_{t,r}\asymp e^{q}$ into account gives

(4.18)\begin{equation} \partial_{t}^{2}[\phi(t,r)]\asymp\frac{1}{e^{-2q}q^{A-1}}\frac{(e^{u-q}q^{A-1})^{2}}{e^{3u}}=e^{-u}q^{A-1}.\end{equation}

The upshot, by van der Corput's lemma (Lemma 2.3), is that

\[ I_{q,u}(h,r)\ll \|\Psi\|_\infty (e^{-u}q^{A-1})^{-1/2} \ll e^{q} q^{1-A}. \]

Now we are in the position to prove the following lemma.

Proof. Note that

(4.19)\begin{equation} \mathcal{E}_{q,u}^{(\mathrm{B})}(s)\ll\frac{1}{N}\sum_{r\asymp e^{u-q}q^{A-1}} \vert\Xi(r)\vert, \quad\mathrm{where}\ \Xi(r):=\sum_{k > 0}\Psi_{q,u}(k,r,s)e(\phi(k,r)-ks).\end{equation}

By Poisson summation,

\begin{align*} \Xi(r)=\sum_{h\in\mathbb{Z}}I_{q,u}(h,r). \end{align*}

We decompose the right-hand side into the contribution $\Xi _{1}(r)$ coming from $\vert h\vert >(4Q)^{A}$, and a contribution $\Xi _{2}(r)$ from the regime $\vert h\vert \leq (4Q)^{A}$. Next, we argue that $\Xi _{1}(r)$ can be disposed of by partial integration. Because $x_{k,r}\leq 2N$, we have

\[ \omega(x_{k,r})=(\log x_{k,r})^{A}\leq(3Q)^{A}. \]

Note for $\vert h\vert > (4Q)^{A}$, by (4.17), that the inequality

\[ \partial_{k}[\phi(k,r)-k(s+h)]\gg h \]

holds true uniformly in $r$ and $s$. As a result, partial integration yields, for any constant $C>0$, the bound

\begin{align*} I_{q,u}(h,r)\ll \Vert k\mapsto\partial_{k}\Psi_{q,u}(k,r,s)\Vert _{L^{1}(\mathbb{R})}h^{-C}. \end{align*}

Therefore,

\[ \Xi_{1}(r)\ll \Vert k\mapsto\partial_{k}\Psi_{q,u}(k,r,s) \Vert_{L^{1}(\mathbb{R})}\sum_{h\geq(4Q)^{A}}h^{-C}. \]

Recall that we have $q\geq {1/(m + 1)} Q$. Thus, taking $C$ to be large and using Lemma 4.6, we deduce that

(4.20)\begin{equation} \Xi_{1}(r)\ll_{C_{1}}e^{\frac{u}{2}}Q^{-C_{1}}\end{equation}

for any constant $C_{1}>0$. All in all, we have shown so far that

\[ \Xi(r)\ll\Xi_{2}(r)+e^{\frac{u}{2}}Q^{-C_{1}}. \]

In $\Xi _{2}(r)$ there are $O(Q^{A})$ choices of $h$. By using Lemma 4.7 we conclude

(4.21)\begin{equation} \Xi_{2}(r)\ll e^{q}q^{1-A}Q^{A}.\end{equation}

By combining (4.20) and (4.21), we deduce from (4.19) that

\begin{align*} \Vert \mathcal{E}_{q,u}^{(\mathrm{B})}(\cdot)\Vert _{\infty}\ll \frac{1}{N}\sum_{r\asymp e^{u-q}q^{A-1}}e^{q}q^{1-A} Q^{A}\ll \frac{1}{N}e^{u} Q^{A}. \end{align*}

As a result,

\begin{align*} \Vert T_{N}(\cdot)\Vert_\infty &\ll\frac{1}{N}\sum_{(u,q)\in\mathcal{T}(N)}e^{u}Q^{A}\ll \frac{1}{N}\sum_{u\leq\log N-10A\log\log N}e^{u}Q^{A+1}\\ &\ll\frac{1}{(\log N)^{10A}}(\log N)^{A+1}\ll\frac{1}{(\log N)^{8A}}. \end{align*}

4.6 Essential regimes

We are now ready to apply our stationary phase expansion (Proposition 2.1), and thus effectively apply the $B$-process a second time. Recall that after applying Poisson summation, the phase will be $\psi _{r,h}(t) = \psi (t): = \phi (t,r) - t(h-s)$. Let

\[ W_{q,u}(t):=\frac{\mathfrak{N}_{q}(x_{t,r})\mathfrak{K}_{u}(t)}{\sqrt{|t\omega''(x_{t,r})|}} \widehat{f}\bigg(\frac{t}{N}\bigg)e\bigg(\psi(t)- \psi(\mu)-\frac{1}{2}(t-\mu)^{2} \psi^{\prime\prime}(\mu)\bigg). \]

Furthermore, define

\[ p_{j}(\mu):=c_j \bigg(\frac{1}{|\psi''(\mu)|}\bigg)^{j}W_{q,u}^{(2j)}(\mu), \]

where $p_0(\mu ) = e(1/8) W_{q,u}(\mu )$. Note that, by (4.18), one can bound

(4.22)\begin{equation} p_{j}(\mu) \ll p_1(\mu) \ll N^{\varepsilon}\frac{1}{|\psi^{\prime\prime}(\mu)|} \frac{1}{\mu^{1/2}} \frac{\omega^{\prime\prime\prime\prime}(x_{\mu,r}) (\partial_t x_{t,r}|_{t=\mu})^2}{|\omega^{\prime\prime}(x_{\mu,r})|^{3/2}} \ll e^{u/2-q} N^\varepsilon,\quad j \ge 1. \end{equation}

Hence, let

\[ P_{q,u}(h,r,s):=\frac{e(\psi(\mu))}{\sqrt{|\psi^{\prime\prime}(\mu)|}}(p_{0}(\mu)+p_1(\mu)), \]

and set

\[ E_{q,u}^{\mathrm{(BB)}}(s):=\frac{e(-1/8)}{N}\sum_{r\geq0}\sum_{h\geq0}P_{q,u}(h,r,s). \]

Before proving Proposition 4.5 we need the following lemma.

Lemma 4.9 For any $c\in [0,1]$ and any $M>10$, we have the bound

\[ \int_{0}^{1}\min (\Vert c+s\Vert ^{-1},M )\,{d}s \leq 2 \log M. \]

We can now prove Proposition 4.5.

Proof Proof of Proposition 4.5

Fix $s \in [0,1]$ and recall the definition of $I_{q,u}(h,r)$ from (4.14). Then, by Poisson summation,

\[ \mathcal{E}_{q,u}^{\mathrm{(B)}}(s)=\frac{e(-1/8)}{N}\sum_{r\geq0}\sum_{h\in\mathbb{Z}}I_{q,u}(h,r). \]

Let $[a,b]:=\operatorname {supp}(\mathfrak {K}_u)$, and

\[ m_{r}(h):=\min_{k\in[a,b]}\vert \psi_{r,h}^\prime(k)\vert. \]

We decompose the $h$-summation into three different ranges:

\[ \sum_{h \in \mathbb{Z}} I_{q,u}(h,r) = \mathfrak{C}_1+\mathfrak{C}_2+\mathfrak{C}_3, \]

where the first contribution $\mathfrak {C}_{1}(r,s)$ is where $m_{r}(h)=0$, the second contribution $\mathfrak {C}_{2}(r,s)$ is where $0< m_{r}(k)\leq N^{\varepsilon }$, and the third contribution $\mathfrak {C}_{3}(r,s)$ is where $m_{r}(h)\geq N^{\varepsilon }$. Integration by parts shows that

\[ \mathfrak{C}_{3}(r,s)\ll N^{-100}. \]

Next, we handle $\mathfrak {C}_{1}(r,s)$. To this end, we shall apply Proposition 2.1, in the notation of which we have

\[ \Omega_w :=e^{u},\quad \Lambda_w :=e^{q-u/2+\varepsilon},\quad \Lambda_{\psi}:=e^{u}q^{A-1},\quad \Omega_{\psi}:=e^{u}. \]

The decay of the amplitude function was shown in Lemma 4.4; the decay of the phase function follows from a short calculation we omit. Next, since we have disposed of the inessential regimes, we see

\[ Z:=\Omega_{\psi}+\Lambda_w+\Lambda_{\psi}+\Omega_w+1\asymp e^{u}q^{A-1}\asymp N^{1+o(1)}. \]

Furthermore,

\[ \frac{\Omega_{\psi}}{\Lambda_{\psi}^{1/2}}Z^{\frac{\delta}{2}}= \frac{e^{\frac{u}{2}}}{q^{(1/2)A}}Z^{\frac{\delta}{2}} \asymp\frac{e^{u(\frac{1}{2}+\delta)}}{q^{(1/2)A + (\delta/2)(A-1)}}.\]

Hence, taking $\delta :=1/11$ is compatible with the assumption (2.1). Thus,

\[ \mathfrak{C}_{1}(r,s)=\sum_{h\geq0} P_{q,u}(h,r,s) + O(N^{-1/11}). \]

Now we bound $\mathfrak {C}_{2}(r,s)$. First note that $\omega (x_{t,r})$ is monotonic in $t$. To see this, set the derivative equal to $0$:

\[ A\frac{\log(x_{t,r})^{A-1}}{x_{t,r}} \partial_t x_{t,r} = A \frac{\log(x_{t,r})^{A-1}}{x_{t,r}} \omega^{\prime}(r/t)(-r/t^2) = 0. \]

However, since $x_{t,r}\asymp e^q$, this implies $\omega ^\prime (r/k) =0$, which is a contradiction. Thus, by van der Corput's lemma (Lemma 2.3) for the first derivative, and monotonicity, we have

\[ \mathfrak{C}_{2}(r,s)\ll N^{\frac{1}{2}+\varepsilon}\min (\Vert \omega(x_{a,r})+s\Vert ^{-1},N^{\frac{1}{2}+o(1)}), \]

where we used (4.18) and the fact that $\partial _t \phi (t,r) = \omega (x_{t,r})$. Notice that

\[ \bigg\Vert \frac{1}{N}\sum_{r\geq0}\mathfrak{C}_{2}(r,\cdot)\bigg\Vert_{L^{m}}^{m} \ll\bigg\Vert \frac{1}{N}\sum_{r\geq0}\mathfrak{C}_{2}(r,\cdot) \bigg\Vert _{\infty}^{m-1}\bigg\Vert \frac{1}{N} \sum_{r\geq0}\mathfrak{C}_{2}(r,\cdot)\bigg\Vert_{L^{1}}. \]

By (4.15) we see that

\[ \bigg\Vert\frac{1}{N}\sum_{r\geq0}\mathfrak{C}_{2}(r,s)\bigg\Vert_{\infty}^{m-1}\ll N^{O(\varepsilon)}. \]

Hence, it remains to estimate

\[ N^{O(\varepsilon)}\sum_{r\asymp e^{u-q}q^{A-1}} \frac{1}{\sqrt{N}}\int_{0}^{1}\min(\Vert \omega(x_{a,r})+s\Vert^{-1},N^{\frac{1}{2}+o(1)})\,{d}s. \]

By exploiting Lemma 4.9 we see that

\[ \bigg\Vert \frac{1}{N}\sum_{r\geq0}\mathfrak{C}_{2}(r,\cdot)\bigg\Vert _{L^{m}}^m \ll \sum_{r\asymp e^{u-q}q^{A-1}}\frac{N^{o(1)}}{\sqrt{N}}\ll N^{o(1)-\frac{1}{2}} \]

which implies the claim.

Finally, it remains to show that

\[ \|\mathcal{E}^{(\mathrm{BB})}_{q,u}(\cdot) - E^{(\mathrm{BB})}_{q,u}(\cdot)\|_{L^m}^m =O(N^{-1/2 +\varepsilon}) \]

from which we can apply the triangle inequality to conclude Proposition 4.5. For this, recall the bounds (4.22). Since $\mathcal {E}_{q,u}^{(\mathrm {BB})}$ is simply the term arising from $p_0(\mu )$, we have that

\[ \|\mathcal{E}^{(\mathrm{BB})}_{q,u}(\cdot) - E^{(\mathrm{BB})}_{q,u}(\cdot)\|_{L^m}^m \ll \frac{1}{N}\sum_{r\in \mathbb{Z}} p_1 \ll \frac{e^{u-q/2}N^\varepsilon}{N}. \]

From here the bound follows from the ranges of $q$ and $u$.

Before proceeding, we note that (4.10) can be simplified. In particular, we have the following lemma.

Lemma 4.10 Given, $h$, $r$, and $s$ as above, we have

\[ \mu_{h,r,s} = \frac{r}{\omega^\prime(\omega^{-1}(h-s))},\quad \Phi(h,r,s) = -r \omega^{-1}(h-s) \]

and

(4.23)\begin{equation} \mu \omega^{\prime\prime}(x_{\mu,r}) \cdot (\partial_{\mu\mu}\phi)(\mu,r) =-\frac{r^2}{\mu^2}. \end{equation}

Proof. Recall $x_{k,r}=\tilde {\omega }(\frac {r}{k})$. Now, to compute $\mu$, we have

\begin{align*} 0 &= \partial_\mu (\mu \omega(x_{\mu,r})-rx_{\mu,r}) -( h-s)\\ &= \omega(x_{\mu,r}) + \mu \omega^\prime(x_{\mu,r}) (\partial_\mu x_{\mu,r}) - r \partial_\mu x_{\mu,r} - (h-s). \end{align*}

Consider first

\[ \partial_\mu x_{\mu,r} = \partial_\mu\bigg(\widetilde{\omega}\bigg(\frac{r}{\mu}\bigg)\bigg) = \widetilde{\omega}^\prime \bigg(\frac{r}{\mu}\bigg) \bigg(-\frac{r}{\mu^2}\bigg). \]

Furthermore, since $\mu = \widetilde {\omega }(\omega ^\prime (\mu ))$, we may differentiate both sides and then change variables to see that

(4.24)\begin{equation} \widetilde{\omega}^\prime(r/\mu) = \frac{1}{\omega^{\prime\prime}(\widetilde{\omega}(r/\mu))}. \end{equation}

Hence,

\[ \partial_\mu (x_{\mu, r}) = - \omega\bigg(\widetilde{\omega}\bigg(\frac{r}{\mu}\bigg)\bigg) \frac{r}{\mu^2 \omega^{\prime\prime}(\widetilde{\omega}(r/\mu))}. \]

Hence,

\begin{align*} 0 &= \omega(x_{\mu,r}) - r \bigg(\omega\bigg(\widetilde{\omega}\bigg(\frac{r}{\mu}\bigg)\bigg) \frac{r}{\mu^2 \omega^{\prime\prime}(\widetilde{\omega}(r/\mu))}\bigg) + \omega\bigg(\widetilde{\omega}\bigg(\frac{r}{\mu}\bigg)\bigg) \frac{r^2}{\mu^2 \omega^{\prime\prime}(\widetilde{\omega}(r/\mu))} - (h-s)\\ &= \omega(x_{\mu,r}) - (h-s). \end{align*}

Hence, $\omega (\widetilde {\omega }(r/\mu )) = h-s$. Solving for $\mu$ gives

\[ \mu = \frac{r}{\omega^\prime(\omega^{-1}(h-s))}. \]

Moreover, we can simplify the phase as follows:

\begin{align*} \Phi(h,r,s) &= \phi(\mu,r) - (h-s)\mu\\ &= \mu \omega(x_{\mu,r}) - r x_{\mu,r} - (h-s)\mu\\ &= \mu \omega(\widetilde{\omega}(r/\mu)) - r \widetilde{\omega}(r/\mu) - (h-s)\mu\\ &= \frac{r(h-s)}{\omega^\prime(\omega^{-1}(h-s))} - r \omega^{-1}(h-s) - (h-s) \frac{r}{\omega^\prime(\omega^{-1}(h-s))}\\ &= - r \omega^{-1}(h-s). \end{align*}

Turning now to (4.23), we note that since, by the definition of $\mu$, we have that $\partial _\mu \phi (\mu,r) = h-s$ and $h-s = \omega (\widetilde {\omega }(r/\mu ))$, we may differentiate both sides of the former to deduce that

\begin{align*} \partial_{\mu\mu}\phi(\mu,r) &= \partial_\mu(\omega (\widetilde{\omega}(r/\mu)))\\ &= \omega^\prime(\widetilde{\omega}(r/\mu)) \widetilde{\omega}^\prime(r/\mu) (- r/\mu^2)\\ &= - (r^2/\mu^3) \widetilde{\omega}^\prime(r/\mu). \end{align*}

Now, using (4.24), we conclude that

\begin{align*} \mu \omega^{\prime\prime}(x_{r,\mu}) \cdot (\partial_{\mu\mu}\phi)(\mu,r) &= -\mu \omega^{\prime\prime}(\widetilde{\omega}(r/\mu)) (r^2/\mu^3) \widetilde{\omega}^\prime(r/\mu)\\ &= -(r^2/\mu^2) \omega^{\prime\prime}(\widetilde{\omega}(r/\mu)) \frac{1}{\omega^{\prime\prime}(\widetilde{\omega}(r/\mu))} = -\frac{r^2}{\mu^2}. \end{align*}

Applying Lemma 4.10 and inserting some definitions allows us to write

(4.25)\begin{equation} \mathcal{E}^{(\mathrm{BB})}_{q,u}(s)=\frac{1}{N}\sum_{r\ge 0}\sum_{h \ge 0} \widehat{f} \bigg(\frac{\mu}{N}\bigg)\mathfrak{K}_u(\mu)\mathfrak{N}_{q}(\widetilde{\omega}(r/\mu)) \frac{\mu}{r} e(-r\omega^{-1}(h-s)). \end{equation}

Returning now to the full $L^m$ norm, let $\sigma _i := \sigma (u_i) := \frac {u_i}{|u_i|}$. Applying Propositions 4.3 and 4.5 and expanding the $m$th power yields

(4.26)\begin{equation} \mathcal{F}(N) = \sum_{\sigma_1,\ldots,\sigma_m\in \{\pm 1\} } \sum_{\substack{(u_i,q_i) \in \mathscr{G}(N)\\ u_i>0}} \int_{0}^{1}\prod_{\substack{i\leq m\\ \sigma_i >0}} \mathcal{E}^{\mathrm{(BB)}}_{q_i, u_i}(s)\prod_{\substack{i\leq m\\ \sigma_i <0}} \overline{\mathcal{E}^{\mathrm{(BB)}}_{q_i, u_i}(s)}\,d s+ O(N^{-\varepsilon/2}). \end{equation}

To simplify this expression, for a fixed $\boldsymbol {\mathbf {u}}$ and $\boldsymbol {\mathbf {q}}$, and $\boldsymbol {\mathbf {\mu }}=(\mu _1,\ldots,\mu _m)$, let $\mathfrak {K}_{\boldsymbol {\mathbf {u}}}(\boldsymbol {\mathbf {\mu }}):= \prod _{i\leq m}\mathfrak {K}_{u_i}(\mu _i)$. The functions $\mathfrak {N}_{\boldsymbol {\mathbf {q}}}(\boldsymbol {\mathbf {\mu }},s)$ and $\widehat {f}(\boldsymbol {\mathbf {\mu }}/N)$ are defined similarly. Aside from the error term, the right-hand side of (4.26) splits into a sum over

\[ \mathcal{F}_{\boldsymbol{\mathbf{q}},\boldsymbol{\mathbf{u}}} := \frac{1}{N^m}\sum_{\boldsymbol{\mathbf{r}} \in \mathbb{Z}^m}\frac{1}{r_1r_2\cdots r_m} \int_0^1\sum_{\boldsymbol{\mathbf{h}}\in \mathbb{Z}^m}\mathfrak{K}_{\boldsymbol{\mathbf{u}}}(\boldsymbol{\mathbf{\mu}}) \mathfrak{N}_{\boldsymbol{\mathbf{q}}}(\boldsymbol{\mathbf{\mu}},s)A_{\boldsymbol{\mathbf{h}},\boldsymbol{\mathbf{r}}}(s) e(\varphi_{\boldsymbol{\mathbf{h}},\boldsymbol{\mathbf{r}}}(s))\,{d}s, \]

where the phase function is given by

\[ \varphi_{\boldsymbol{\mathbf{h}},\boldsymbol{\mathbf{r}}}(s) := -(r_1\omega^{-1}(h_1-s)+r_2\omega^{-1}(h_2-s) + \cdots + r_m\omega^{-1}(h_m-s)), \]

and where

\[ A_{\boldsymbol{\mathbf{h}},\boldsymbol{\mathbf{r}}}(s) : = \widehat{f}\bigg(\frac{\boldsymbol{\mathbf{\mu}}}{N}\bigg) \mu_1\mu_2 \cdots \mu_m. \]

Now we distinguish between two cases. First, the set of all $(\boldsymbol {\mathbf {r}},\boldsymbol {\mathbf {h}})$ where the phase $\varphi _{\boldsymbol {\mathbf {h}},\boldsymbol {\mathbf {r}}}(s)$ vanishes identically, which we call the diagonal; and its complement, the off-diagonal. Let

\[ \mathscr{A} := \{(\boldsymbol{\mathbf{r}}, \boldsymbol{\mathbf{h}}) \in \mathbb{N}\times \mathbb{N} : \varphi_{\boldsymbol{\mathbf{h}},\boldsymbol{\mathbf{r}}}(s) = 0,\ \forall s \in [0,1]\}, \]

and let

\[ \eta(\boldsymbol{\mathbf{r}},\boldsymbol{\mathbf{h}}):=\begin{cases} 1 & \mbox{if } (\boldsymbol{\mathbf{r}},\boldsymbol{\mathbf{h}}) \not\in \mathscr{A}, \\ 0 & \mbox{if } (\boldsymbol{\mathbf{r}},\boldsymbol{\mathbf{h}}) \in \mathscr{A}. \end{cases} \]

The diagonal, as we show, contributes the main term, while the off-diagonal contribution is negligible (see § 6).

8.1 The case $m=2$

Suppose we knew that

(8.2)\begin{equation} \lim_{N\rightarrow \infty} R^{(2)}(N,f)= {\mathbf{E}(f)},\quad \text{for any positive}\ f\in C_c^\infty. \end{equation}

Then, by the linearity of $f\mapsto R^{(2)}(N,f)$ and $f\mapsto \int _\mathbb {R} f(x) \,d x$, the convergence of $R^{(2)}(N,f) \rightarrow {\mathbf {E}(f)}$ follows for all $f\in C_c^\infty$. To verify (8.2) we show, for any positive $f\in C_c^\infty$, that

(8.3)\begin{equation} \limsup_{N\rightarrow \infty} R^{(2)}(N,f)\leq {\mathbf{E}(f)}\leq \liminf_{N\rightarrow \infty} R^{(2)}(N,f). \end{equation}

We first argue the estimate for the limit superior. Given $\varepsilon >0$, Lemma 8.1 guarantees that there exists a pointwise majorant $F$ of $f$ satisfying (8.1). Observe that $R^{(2)}(N,f) - R^{(2)}(N,F)= R^{(2)}(N,f-F)$ and $R^{(2)}(N,f-F)=-R^{(2)}(N,F-f)$. As a result,

(8.4)\begin{equation} \vert R^{(2)}(N,f)- R^{(2)}(N,F)\vert \leq R^{(2)}(N,F-f). \end{equation}

Let $g:=F-f$. Combining (8.4), Theorem 1.3, and (8.1), we see that

(8.5)\begin{equation} \limsup_{N \to \infty} R^{(2)}(N,f)\leq \limsup_{N \to \infty} R^{(2)}(N,F) + \limsup_{N \to \infty}R^{(2)}(N,g) \leq {\mathbf{E}(f)} + \varepsilon + \limsup_{N \to \infty}R^{(2)}(N,g). \end{equation}

Using Lemma 8.1, we find a pointwise majorant $G$ of $g$ with (8.1).

Applying Theorem 1.3 once more,

(8.6)\begin{equation} \limsup_{N \to \infty} R^{(2)}(N,g) \leq \limsup_{N \to \infty} R^{(2)}(N,G)= {\mathbf{E}(G)} \leq {\mathbf{E}(g)} + \varepsilon \leq 2 \varepsilon . \end{equation}

By inserting this into (8.5) we obtain

(8.7)\begin{equation} \limsup_{N \to \infty} R^{(2)}(N,f) \leq {\mathbf{E}(f)} + 3 \varepsilon , \end{equation}

for any choice of $\varepsilon >0$. Hence, $\limsup _{N\rightarrow \infty } R^{(2)}(N,f)\leq {\mathbf{E}(f)}$. Next, we show the right-hand side of (8.3). We notice that (8.4) and (8.6) yield

\[ \liminf_{N \to \infty} R^{(2)}(N,f)\geq \liminf_{N \to \infty} R^{(2)}(N,F) - \limsup_{N \to \infty}R^{(2)}(N,g)\geq {\mathbf{E}(f)} - 2 \varepsilon . \]

Since $\varepsilon >0$ was arbitrary, $\liminf _{N\rightarrow \infty } R^{(2)}(N,f)\geq {\mathbf {E}(f)}$ follows.

8.2 linearity-based $m\geq 2$

Let $f \in C_c^\infty (\mathbb {R}^{m-1})$ be positive, which we may assume by a linearity-based argument. Let $\varepsilon >0$. It suffices to demonstrate that

(8.8)\begin{equation} \limsup_{N\rightarrow \infty} R^{(m)}(N,f) \leq {\mathbf{E}(f)} + 2\varepsilon\quad \mathrm{and}\quad {\mathbf{E}(f)} - 2\varepsilon \leq \liminf_{N\rightarrow \infty} R^{(m)}(N,f). \end{equation}

Up to an error of at most $\varepsilon$, we can approximate $f$ with respect to the supremum norm on $\mathbb {R}^{m-1}$ by a finite linear combination of functions $\widetilde {f}$ each of which is the form $\widetilde {f}(\boldsymbol {\mathbf {x}}):=f_1(x_1)\cdots f_{m-1}(x_{m-1})$, where all $f_i\in C_c^\infty (\mathbb {R})$. Thus, it suffices to establish (8.8) for $\widetilde {f}$ in place of $f$. Given $1\leq i< m$, Lemma 8.1 implies that there exists a pointwise majorant $F_i$ of $f_i$ satisfying (8.1). As a result, $F(\boldsymbol {\mathbf {x}}):= F_1(x_1)\cdots F_{m-1}(x_{m-1})$ is a pointwise majorant of $\widetilde {f}$. By using the identity $ab-a'b'=a(b-b')+(a-a')b'$ inductively, we conclude that there exists a constant $C_{\widetilde {f}}>0$ so that

\[ \int_{\mathbb{R}^{m-1}} F(\boldsymbol{\mathbf{x}}) - \widetilde{f}(\boldsymbol{\mathbf{x}})\,d \boldsymbol{\mathbf{x}} < C_{\widetilde{f}} \cdot \varepsilon \]

no matter what our initial choice of the small parameter $\varepsilon$ was. Since $\widehat {F}(\boldsymbol {\mathbf {x}})=\widehat {F_1}(x_1)\cdots \widehat {F_{m-1}}(x_{m-1})$ is the product of compactly supported functions, $\widehat {F}\in C_c^{\infty }(\mathbb {R}^{m-1})$. Notice that

\[ \vert R^{(m)}(N,f)- R^{(m)}(N,F)\vert \leq R^{(m)}(N,F-f). \]

From here we reason exactly as we did in the case $m=2$ and thus we infer (8.8).

8.3 Proof of Lemma 8.1

The purpose of this subsection is to establish Lemma 8.1. We do so using arguments inspired by Marklof [Reference MarklofMar03, Section 8.6]. First, chose a $\delta >0$ small enough to ensure that

\[ \int_\mathbb{R} \frac{\delta}{1+x^2} d x < \frac{\varepsilon}{2}. \]

Next, we perturb $f$ slightly by considering an auxiliary function $h$ defined via its Fourier transform

\[ \widehat{h}(x) := f(x) + \frac{\delta}{1+x^2}. \]

The reason for working with this perturbation is to create the leeway needed to ensure that the final approximation $F$ is indeed positive. Let $\chi : \mathbb {R}\rightarrow [0,1]$ be an even, smooth function with $\operatorname {supp}(\chi )\subseteq [-2,2]$ and $\chi (x) = 1$ for $x \in [-1,1]$. We consider

\[ h_{T}(x):= h(x) \chi\bigg(\frac{x}{T}\bigg), \]

and proceed to argue that there exists a cut-off $T>0$ so that

(8.9)\begin{equation} |\widehat{h_T}(x) - \widehat{h}(x)| < \frac{\delta}{1+x^2} \end{equation}

for all $x\in \mathbb {R}$. Observe that (8.9) is trivially true at $x=0$. From now on let $x\neq 0$. By using partial integration twice, we conclude

\begin{align*} \widehat{h_T}(x) - \widehat{h}(x) &= \int_{\mathbb{R}}\bigg[h(y) \chi\bigg(\frac{y}{T}\bigg)- h(y)\bigg] \bigg(\frac{d^2}{{{d}y}^2}\frac{e(-yx)}{(-2\pi i x)^2}\bigg)d y\\ &= \frac{1}{(-2\pi i x)^2}\int_{\mathbb{R}} \bigg(\frac{d^2}{{{d}y}^2}\bigg[h(y) \chi\bigg(\frac{y}{T}\bigg)- h(y)\bigg]\bigg) e(-yx) \,d y. \end{align*}

Furthermore,

(8.10)\begin{equation} \frac{d^2}{{{d}y}^2}\bigg(h(y) \bigg(1-\chi\bigg(\frac{y}{T}\bigg)\bigg)\bigg) = h''(y) \bigg(1-\chi\bigg(\frac{y}{T}\bigg)\bigg) + 2\frac{1}{T}h'(y) \chi'\bigg(\frac{y}{T}\bigg)+ \frac{1}{T^2}h(y) \chi''\bigg(\frac{y}{T}\bigg). \end{equation}

Let $j\geq 0$ be an integer. To estimate the right-hand side of (8.10) efficiently, our next goal is to control the decay of the $j$th derivative of $h$ uniformly. To this end, thanks to Fourier inversion, we see that

\[ h^{(j)}(y)=\int_{\mathbb{R}}\widehat{h}(\xi)(2\pi i \xi)^j e(y\xi)\,d \xi. \]

Let $y\neq 0$. By using partial integration $t\geq j$ many times,

\begin{align*} \vert h^{(j)}(y)\vert=\bigg\vert \int_{\mathbb{R}}\widehat{h}(\xi)(2\pi i \xi)^j \frac{d^t}{d\xi^t}\bigg(\frac{e(y\xi)}{(2\pi i y)^t}\bigg) d \xi\bigg\vert &\leq \frac{(2\pi)^{j}}{\vert 2\pi y\vert^{t}}\cdot \int_{\mathbb{R}}\bigg\vert \frac{d^t}{d\xi^t} (\xi^j \widehat{h}(\xi)) \bigg\vert d \xi\\ &\leq \frac{1}{\vert y\vert^{t}} \int_{\mathbb{R}}\bigg\vert\frac{d^t}{d\xi^t} \bigg(\xi^j f(\xi)+\frac{\delta \xi^j}{1+\xi^2}\bigg)\bigg\vert d \xi. \end{align*}

By applying Leibniz's rule we conclude that, for any integers $t\geq j\geq 0$, there exists a constant $C(j,t,f)>0$ so that $\vert h^{(j)}(y)\vert \leq C(j,t,f)\vert y\vert ^{-t}$ for any $y\neq 0$. Taking this information in (8.10) into account yields

\[ \frac{d^2}{{{d}y}^2}\bigg(h(y) \bigg(1-\chi\bigg(\frac{y}{T}\bigg)\bigg)\bigg)\ll \frac{1}{y^2}\bigg[1-\chi\bigg(\frac{y}{T}\bigg) +\frac{1}{T}\chi'\bigg(\frac{y}{T}\bigg)+ \frac{1}{T^2}\chi''\bigg(\frac{y}{T}\bigg)\bigg]\ll \frac{1}{y^2} \]

whenever $y\neq 0$. Here the implied constants depend only on $f$ and $\chi$. Because the left-hand side of (8.10) is supported on $\vert y\vert \geq T$, it follows that

\[ \vert \widehat{h_T}(x) - \widehat{h}(x)\vert\ll\frac{1}{\vert x\vert^2} \int_{\mathbb{R}}\big\vert\frac{d^2}{{{d}y}^2}\bigg(h(y) \bigg(1-\chi\bigg(\frac{y}{T}\bigg)\bigg)\bigg)\bigg \vert d y\ll \frac{1}{\vert x\vert^2}\int_{\vert y \vert>T} \frac{1}{y^2} d y \ll \frac{1}{\vert x \vert^2 T}. \]

Upon choosing $T$ large enough to overwhelm the implicit constant, we deduce that (8.9) holds true.

Now put $F(x):=\widehat {h_T}(x)$. By the convolution theorem, $F(x)=\int _\mathbb {R} \widehat {h}(x-y) \widehat {\chi }(y) \,d y$. Notice that $\widehat {h}$ and $\widehat {\chi }$ are real-valued (the latter being the Fourier transform of an even, real-valued function). Therefore, $F$ is real-valued. To proceed, we notice that the triangle inequality implies

\[ f(x)= \widehat{h}(x) - \frac{\delta}{1+x^2}< \widehat{h}(x) + (\widehat{h_T}(x)-\widehat{h}(x))=F(x). \]

Hence, $F$ is a pointwise majorant of $f$ and, in particular, positive. Furthermore,

\[ \int_\mathbb{R} F(x)- f(x) \,d x < \int_\mathbb{R} |\widehat{h_T}(x) - \widehat{h}(x)| \,d x+ \int_\mathbb{R} \frac{\delta}{1+x^2} d x < \varepsilon. \]