Proof. Since f is not eventually $0$ , there exists the inverse function $1/f$ ; in particular, $f(t) \neq 0$ for t large enough. Now, the first part follows from continuity of f. The second part follows directly from the first part by considering $f'$ .

Proof. If g is eventually zero, the situation is trivial, so assume that this is not the case. Since $f/g$ is eventually monotone, the limit $\lim _{t \to \infty } |f(t)|/|g(t)| \in \mathbb {R} \cup \{\infty \}$ exists. If the limit is infinite then $f \succ g$ . If the limit is zero then $f \prec g$ . If the limit is finite and non-zero then $f \sim g$ .

Proof. Reasoning inductively, it is enough to consider the case where $\ell = 1$ . Suppose, for the sake of contradiction, that $|f'(t)| \gg t^{-n}$ for some $n \in \mathbb {N}$ . Since $f(t) \to 0$ as $t \to \infty $ and since f is eventually monotone, for sufficiently large t we have

$$ \begin{align*} |f(t)| = \int_{t}^{\infty} |f'(s)|\, ds \gg \int_{t}^{\infty} s^{-n} \,ds \gg t^{-n+1}, \end{align*} $$

contradicting the assumption on f.

Proof. Reasoning inductively, it is enough to consider the case where $\ell = 1$ . We consider the three possibilities in Lemma 2.4. If $f(t) \prec t^{-n}$ for all $n \in \mathbb {N}$ then the claim is trivially true by Lemma 2.5. If $f'(t) \ll f(t)/t$ then $f'(t) \ll t^{k-1}$ , as required. Finally, suppose that $f(t) \to c \neq 0$ as $n \to \infty $ . Clearly, in this case $k \geq 0$ . We may decompose ${f(t) = \overline f(t) + c}$ , where $\overline f(t) = f(t) - c$ and $\overline f(t) \prec 1$ . Repeating the reasoning with $\overline f$ in place of f, we conclude that $f'(t) = \overline f'(t) \ll t^{-1} \ll t^{k - 1}$ .

Proof. Combining (5) and Lemma 2.6. we have

$$ \begin{align*} y^{-\ell} R_{x,\ell}(y) \ll \sup_{\xi \in [0,y]} f^{(\ell)}(x+\xi) \ll \sup_{\xi \in [0,y]} (x+\xi)^{k-\ell} = \begin{cases} x^{k-\ell} & \text{if } k < \ell,\\ (x+y)^{k-\ell} & \text{if } k \geq \ell. \end{cases} \end{align*} $$

Assuming that $x \geq y$ , the two estimates are equivalent.

Proof. If $f^{(k)}(t)$ is constant zero for all $t \in I$ , then f is a polynomial of degree at most $k-1$ and the statement is trivially true. Thus, we assume without loss of generality that $f^{(k)}(t)> 0$ for all $t \in I$ . Let us assume for the sake of contradiction that f changes monotonicity at least k times. Thus, $f'$ has at least k zeros in I. It follows from the mean value theorem that $f"$ has at least $k-1$ zeros in I. Inductively applying this reasoning shows that $f^{(k)}$ has at least one zero in I, giving the desired contradiction.

Proof. We define $\varepsilon = H^{-\eta _0}$ for some $\eta _0> 0$ which only depends on $\ell $ and will be specified later. Let $N \in \mathbb {N}$ . Recall that by Proposition 2.9, we have

(7) $$ \begin{align} |R_{N,\ell}(h)| &\leq \varepsilon \quad\text{for all } 0 \leq h < H \end{align} $$

unless $N \ll \varepsilon ^{-{1}/( \ell -k ) }H^{{\ell }/( \ell - k )} = H^{( \ell + \eta _0 )/( \ell -k )}$ . Thus, the values of N such that (7) is false contribute only $O( H^{O(1)} )$ different sequences $e_N$ , and we may freely assume that N is large enough that (7) holds. In this case we have $e_N: [H] \to \{-1,0,1\}$ . Additionally, by Lemma 2.1 we may also assume that $f^{(\ell )}(x) \neq 0$ for all $x \geq N$ . As a consequence of (7), for each $0 \leq h < H$ , if

(8) $$ \begin{align} \varepsilon < \{P_{N,\ell}(h) \} < 1- \varepsilon \end{align} $$

then ⌊f(N + h)⌋ = ⌊P _{N, ℓ} (h)⌋ and hence $e_N(h) = 0$ .

Let $\alpha _0,\dots ,\alpha _{\ell -1}$ denote the coefficients of $P_{N,\ell }$ :

$$ \begin{align*} P_{N,\ell}(h) = \alpha_0 + \alpha_1 h + \cdots + \alpha_{\ell-1}h^{\ell-1}. \end{align*} $$

By Proposition 2.8, we distinguish two cases.

1. (I) $(P_{N,\ell }(h))_{h \in [H]}$ has discrepancy at most $\varepsilon $ .

2. (II) There exists $1\leq q \ll \varepsilon ^{-O(1)}$ such that $\max _{0 \leq j < \ell } H^j\lVert q \alpha _j \rVert \ll \varepsilon ^{-O(1)}$ .

In the first case, it follows that the number of $h \in [H]$ such that (8) does not hold is at most ${3}\varepsilon H$ . Thus, $e_{N}$ is sparse, that is, it has at most ${3}\varepsilon H \ll H^{1-\eta _0}$ non-zero entries. It remains to estimate the number of the sequences $e_N$ of this type. Using a standard estimate $\binom {n}{k} \leq n^k/k! < (en)^k/k^k$ , we find

$$ \begin{align*} \log \left( \sum_{0\leq j \leq 3 \varepsilon H} \binom{H}{j} 2^j \right) & \ll \log ( 3\varepsilon H ) + \log \binom{H}{ 3 \varepsilon H} + 3 \varepsilon H \\ & \ll \log(3 H^{1-\eta_0}) + 3 \varepsilon H \log(e 3 H^{1-\eta_0}) + 3 H^{1-\eta_0}\\ &\ll_{\eta_0} H^{1-\eta_0/2}. \end{align*} $$

Thus the number of distinct sequences $e_N$ is bounded by $\exp (O(H^{1-\eta _0/2}))$ , which gives the desired result as long as $1-\eta _0/2\leq \eta $ .

In the second case we split $[H]$ into arithmetic progressions with common difference $q \ll \varepsilon ^{-O_{\ell }(1)}$ . This allows us to write (for $0\leq m < q$ )

$$ \begin{align*} P_{N,\ell}(q h + m) &= \alpha_{0} + (q h + m) \alpha_{1} + \cdots + (q h + m)^{\ell-1} \alpha_{\ell-1}\\ &= \beta_{0} + h \beta_{1} + \cdots + h^{\ell-1}\beta_{\ell-1}. \end{align*} $$

The defining property of q implies that

$$ \begin{align*} \max_{1\leq j < \ell} H^j \lVert \beta_{j} \rVert \ll \varepsilon^{-O_{\ell}(1)}. \end{align*} $$

In particular, we can write

$$ \begin{align*} \beta_{j} = z_{j} + s_{j}, \end{align*} $$

where $z_{j} \in \mathbb {Z}$ and $|s_{j}| \ll H^{-j} \cdot \varepsilon ^{-O_{\ell }(1)}$ for $ 0 \leq j < \ell $ . Putting everything together, we find

$$ \begin{align*} f(N+q h +m) = Q(h) + r(h) + R_{N,\ell}(q h + m), \end{align*} $$

where

$$ \begin{align*} Q(h) &= z_{0} + h z_{1} + \cdots + h^{\ell-1} z_{\ell-1},\\ r(h) &= s_{0} + h s_{1} + \cdots + h^{\ell-1} s_{\ell-1}. \end{align*} $$

In particular, Q is a polynomial of degree at most $\ell -1$ with integer coefficients and $P_{N, \ell }(q h +m) = Q(h) + r(h)$ . Moreover, $|r(h)| \ll \varepsilon ^{-O_{\ell }(1)}$ for all $h \in [0,H/q]$ . Since $|R_{N,\ell }(h)| \leq \varepsilon $ , we see that

$$ \begin{align*} \lfloor f(N+q h + m) \rfloor \neq \lfloor P_{N,\ell}(q h+m) \rfloor \end{align*} $$

holds exactly if either

(9) $$ \begin{align} \begin{split} \{r(h) \} \leq \varepsilon \quad &\text{and} \quad \{r(h) + R_{N,\ell}(q h + m) \} \geq 1-\varepsilon,\quad \text{or}\\\{r(h) \} \geq 1-\varepsilon \quad &\text{and} \quad \{r(h) + R_{N,\ell}(q h + m) \} \leq \varepsilon. \end{split} \end{align} $$

In the first case $e_N(q h +m) = 1$ and in the second case $e_N(q h + m) = -1$ . Since $r(h)$ is a polynomial of degree at most $\ell -1$ , it changes monotonicity at most $\ell - 2$ times. Since the $\ell $ th derivative of $r(h) + R_{N,\ell }(q h + m) = f(N + q h + m) - P_{N,\ell }(q h + m) + r(h)$ has constant sign, by Lemma 2.10 it changes monotonicity at most $\ell -1$ times on the interval $[0, H/q]$ . Hence, we can decompose $[0, H/q]$ into at most $2 \ell -2$ intervals $I_1, \ldots , I_p$ on which $r(h)$ and $r(h) + R_{N,\ell }(q h + m)$ are both monotone. As $|r(h)| \ll \varepsilon ^{-O_{\ell }(1)}$ , we can further subdivide each of the intervals $I_j$ into $O(\varepsilon ^{-O_{\ell }(1)})$ subintervals such that for each subinterval, each of the inequalities is either true on the entire subinterval or false on the entire subinterval. As a consequence, $e_N$ is structured, that is, $e_N$ is constant on each subinterval. Thus, we have found a decomposition of $[H]$ into $O(\varepsilon ^{-O_{\ell }(1)})$ arithmetic progressions on which $e_N$ is constant. We can write $O(\varepsilon ^{-O_{\ell }(1)}) = O(H^{C \eta _0})$ for some $C = C(\ell )>0$ . Using the rough estimate $H^3$ for the number of arithmetic sequences contained in $[H]$ , we can bound the number of sequences $e_N$ which arise this way by

$$ \begin{align*} (H^3)^{O(H^{C \eta_0})} = \exp( O( H^{C \eta_0} \log H) ) = \exp( O_{\eta_0}(H^{(C+1) \eta_0 } ). \end{align*} $$

It remains to choose $\eta _0 = (C+2)^{-1}$ and $\eta = 1-(2(C+2))^{-1}$ to finish the proof.

Proof. By Theorem 4.2, there exist a nilmanifold $G^{(0)}/\Gamma ^{(0)}$ together with a piecewise polynomial map $F^{(0)} \colon G^{(0)}/\Gamma ^{(0)} \to \mathbb {R}$ , and a group element $g_0 \in G^{(0)}$ such that ${f(n) = F^{(0)}(g_0^n \Gamma )}$ for all $n \in \mathbb {Z}$ . Following the strategy in [Reference FrantzikinakisFra09, Lemma 4.1], let $G := G^{(0)} \times \mathbb {R}$ and $\Gamma := \Gamma ^{(0)} \times \mathbb {Z}$ , and let $F \colon G/\Gamma \to \mathbb {R}$ be given by $F(t + \mathbb {Z},h \Gamma ^{(0)}) := F^{(0)}( g_0^{-\{ t \} } h \Gamma ^{(0)})$ for $t \in \mathbb {R}$ and $h \in G^{(0)}$ . This construction guarantees that F is piecewise polynomial and for all $t \in \mathbb {R}$ we have

$$ \begin{align*} F(t+\mathbb{Z}, g_0^t \Gamma) = F^{(0)}( g_0^{\lfloor t \rfloor } \Gamma ) = f(\lfloor t \rfloor ). \end{align*} $$

For $p \in \mathbb {R}[x]$ and $n \in \mathbb {Z}$ let g _p (n):= (p(n), g ₀ ^p(n)). Then $g_\alpha $ is polynomial with respect to the filtration $G_\bullet $ given by $G_{i} = G_{(\lfloor i/d \rfloor )}$ , where $( G_{(j)} )_j$ denotes the lower central series, and we have $f( \lfloor p(n) \rfloor ) = F(g_p(n)\Gamma )$ for all $n \in \mathbb {Z}$ .

Proof of Theorem 5.1 assuming Theorem 5.2

Applying a dyadic decomposition, it will suffice to show that

(15)

Fix a small $\varepsilon> 0$ . We will show that for all sufficiently large N we have

(16)

Splitting the average in (16) into intervals of length $\lceil (2N)^{0.7} \rceil $ , we see that (16) will follow once we show that for sufficiently large N and for H satisfying $N^{0.7} \leq H < N$ we have

(17)

Pick an integer $k \in \mathbb {N}$ such that $f(t) \ll t^k$ , and let $\ell = 10 k$ . By Theorem 2.11, we have

(18) $$ \begin{align} \lfloor f(N+h) \rfloor = \lfloor P_{N}(h) \rfloor + e_N(h), \end{align} $$

where $P_N$ is a polynomial of degree (at most) $\ell $ and one of the conditions (i)–(iii) in Theorem 2.11 holds. In the case (i) we have $N \ll _{\varepsilon } H^{10/9} \leq N^{7/9}$ , which implies that ${N = O_\varepsilon (1)}$ . Assuming that N is sufficiently large, we may disregard this case.

In case (ii) we have ${\mathbb {E}}_{h < H} |e_N(h)| < \varepsilon $ , and as a consequence

(19)

By Theorem 4.4, there exist a connected and simply connected nilpotent Lie group G, a lattice $\Gamma < G$ , a filtration $G_\bullet $ and a finite-valued piecewise polynomial map $F \colon G/\Gamma \to \mathbb {Z}$ such that for each polynomial P of degree at most $\ell $ there exists $g \in \mathrm {poly}(G_\bullet )$ such that $a(\lfloor P(h) \rfloor ) = F(g(h)\Gamma )$ ; it is crucial that we have the same $(G,\Gamma , G_{\bullet })$ for all polynomials P. In particular,

(20)

By Theorem 5.2 (which is uniform in g), for sufficiently large N the expression in (20) is bounded by $\varepsilon $ . Inserting this bound into (19) yields (17).

In case (iii), passing to an arithmetic progression, we may replace $e_N$ with a constant sequence:

(21)

(22)

To finish the argument, it suffices to apply Theorem 5.2 similarly to the previous case.

Proof. Item (i) follows from the fact that each pair of norms on the finite-dimensional vector space $\mathcal {V}$ are equivalent. For item (ii) we proceed by induction with respect to D. Multiplying P by a scalar, we may assume that $\lVert P \rVert _\infty = 1$ .

Suppose first that $D = 1$ . We proceed by induction on d. If $d = 1$ then P is an affine function $P(x) = a x + b$ , and the claim follows easily. Assume that $d \geq 2$ and that the claim has been proved for $d-1$ . By item (i), at least one of the coefficients of P has absolute value $\gg _{d,D} 1$ . In fact, we may assume that this coefficient is not the constant term, since otherwise for all $x \in [0,1)$ we would have $P(x) \in (\frac {99}{100}P(0),\frac {101}{100}P(0))$ and hence the set in (25) would be empty for sufficiently small $\delta $ . Thus, $\lVert P' \rVert _\infty \gg _{d,D} 1$ . By the inductive assumption,

(26) $$ \begin{align} \unicode{x3bb}( \{ x \in [0,1) \mid |P'(x)| < \delta^{d-1} \} ) \ll_{d} \delta. \end{align} $$

Thus, it will suffice to show that

(27) $$ \begin{align} \unicode{x3bb}( \{ x \in [0,1) \mid |P(x)| < \delta^d,\ |P'(x)|> \delta^{d-1} \} ) \ll_{d} \delta. \end{align} $$

For each interval $I \subseteq [0,1)$ such that $P'(x)$ has constant sign for $x \in I$ , we have

(28) $$ \begin{align} \unicode{x3bb}( \{ x \in I \mid |P(x)| < \delta^d,\ |P'(x)|> \delta^{d-1} \} ) \ll \delta. \end{align} $$

Since $[0,1)$ can be divided into $O(d)$ intervals where P is monotone, (27) follows.

Suppose now that $D \geq 2$ and the claim has been proved for all $D' < D$ . Reasoning as above, we infer from item (i) that P has a coefficient with absolute value $\gg _{d,D} 1$ other than the constant. We may expand P as

$$ \begin{align*} P(y,t) &= \sum_{i=0}^{d} t^{i} Q_i(y), \quad y \in [0,1)^{D-1},\ t \in [0,1), \end{align*} $$

where the $Q_i$ are polynomials in $D-1$ variables of degree $d-i$ . Changing the order of variables if necessary, we may assume that there exists j with $1 \leq j \leq d$ such that $Q_j$ has a coefficient $\gg _{d,D} 1$ , and hence $\lVert Q_j \rVert _{\infty } \gg _{d,D} 1$ . For $k \in \mathbb {N}$ , let us consider the set

$$ \begin{align*} E_k := \{ (y,t) \in [0,1)^D \mid |P(y,t)| < \delta^d,\ 2^{-k} \leq |Q_j(y)| < 2^{-k+1} \}. \end{align*} $$

The set in (28) is the disjoint union $\bigcup _{k=1}^\infty E_i$ , so our goal is to show that

(29) $$ \begin{align} \sum_{k=1}^\infty \unicode{x3bb}(E_k) \ll_{d,D} \delta. \end{align} $$

Fix a value of k. By the inductive assumption, as long as $j \neq d$ , we have

(30) $$ \begin{align} \unicode{x3bb}( \{ y \in [0,1)^{D-1} \mid |Q_j(y)| < 2^{-k+1} \} ) \ll_{d,D} 2^{-k/(d-j)}. \end{align} $$

(If $j = d$ , the set in (30) is empty for all sufficiently large k, and the reasoning simplifies.) For each $y \in [0,1)^{D-1}$ such that $2^{-k} \leq |Q_j(y)| < 2^{-k+1}$ , by the inductive assumption (for $D=1$ ) we have

(31) $$ \begin{align} \unicode{x3bb}( \{ t \in [0,1) \mid |P(y,t)| < \delta^d \} ) \ll_{d,D} 2^{k/d} \delta. \end{align} $$

Combining (30) and (31) yields

(32) $$ \begin{align} \unicode{x3bb}( E_k ) \ll_{d,D} 2^{-k j/d(d-j)} \delta \leq 2^{-k/d^2} \delta. \end{align} $$

Summing (32) gives (29) and finishes the argument.

Proof. We may freely assume that $\delta \geq 1/\log ^A N$ , since otherwise there is nothing to prove. In particular, $\delta = \log ^{O(A)}N$ and $1/\delta = O(\log ^A N)$ . Decomposing S into a bounded number of pieces, we may assume that S is a basic semialgebraic set. We will assume that $\operatorname {int} S \neq \emptyset $ ; the case where $\operatorname {int} S = \emptyset $ can be handled using similar methods and is somewhat simpler. Thus, S takes the form

(34) $$ \begin{align} S = \{ (t,x) \in (\mathbb{R}/\mathbb{Z}) \times (G/\Gamma) \mid P_1(t,x)> 0,\ P_2(t,x) > 0,\ \dots, P_r(t,x) > 0 \} , \end{align} $$

where $r = O(1)$ and $P_i$ are polynomial maps (under identification of $(\mathbb {R}/\mathbb {Z}) \times (G/\Gamma )$ with $[0,1)^{1+D}$ ) with $\deg P_i = O(1)$ for $1 \leq i \leq r$ . Scaling, we may assume that ${\lVert P_i \rVert _\infty = 1}$ for all $1 \leq i \leq r$ . Let $\tau _1$ denote Mal’cev coordinates on $(\mathbb {R}/\mathbb {Z}) \times (G/\Gamma )$ , given by $\tau _1(t ,x) = (t,\tau (x))$ , where we identify $[0,1)$ with $\mathbb {R}/\mathbb {Z}$ in the standard way. Furthermore, splitting S further and applying a translation if necessary, we may assume that $\tau _1(S) \subseteq ( \frac {1}{10},\frac {9}{10} )^{1+D}$ , implying in particular that $\tau _1$ is continuous in a neighbourhood of S.

Let $\eta \in (0,\delta )$ be a small positive quantity, to be specified in the course of the argument, and let $\Psi ,\Psi ' \colon \mathbb {R} \to [0,1]$ be given by

$$ \begin{align*} \Psi(t) = \begin{cases} 0 & \text{if } t < 0,\\ t/\eta & \text{if } t \in [0,\eta],\\ 1 & \text{if } t> \eta, \end{cases} \quad \Psi'(t) = \begin{cases} 0 & \text{if } |t| > 2\eta,\\ 2- |t|/\eta & \text{if } |t| \in [\eta,2\eta],\\ 1 & \text{if } |t| < \eta. \end{cases} \end{align*} $$

It is clear that $\lVert \Psi \rVert _{\mathrm {Lip}} = \lVert \Psi ' \rVert _{\mathrm {Lip}} = 1/\eta $ . Let $\Psi _{\square } \colon [0,1)^{1+D} \to [0,1]$ be an $O(1)$ -Lipschitz function with $\Psi _{\square }(t,u) = 1$ if $(t,u) \in ( \frac {1}{10},\frac {9}{10} )^{1+D}$ and $\Psi _{\square }(t,u) = 0$ if $(t,u) \not \in ( \frac {1}{20},\frac {19}{20} )^{1+D}$ . For $1 \leq i \leq r$ , put

$$ \begin{align*} F_i(t,x) &= \Psi(P_i(t,x)), \quad F_i'(t,x) = \Psi_{\square}(\tau_1(t,x))\Psi'(P_i(t,x)),\\[-1pt] F(t,x) &= \prod_{i=1}^r F_i(t,x), \quad F'(t,x) = \min\bigg( \sum_{i=1}^r F_i'(t,x),1 \bigg). \end{align*} $$

It is routine (although tedious) to verify that F and $F'$ are $1/\eta ^{O(1)}$ -Lipschitz (cf. [Reference Green and TaoGT12b, Lemma A.4]); this follows from the aforementioned bounds on the Lipschitz norms of $\Psi $ and $\Psi '$ and the fact that the derivatives of the polynomials $P_i$ are bounded by $O(1)$ on $[0,1)^{D+1}$ , which follows, for example, from Lemma 5.5. Directly from the definitions, we see that for each $t \in \mathbb {R}/\mathbb {Z}$ and $x \in G/\Gamma $ we have $F(t,x) = 1_S(t,x)$ or $F'(t,x) = 1$ . It follows that

(35)

In order to estimate either of the summands in (35), we begin by dividing the interval $[H]$ into $O(1/\alpha )$ subintervals with lengths between $\alpha H$ and $2\alpha H$ , where

(36) $$ \begin{align} \alpha := ( \log^A N \max( \lVert F \rVert_{\mathrm{Lip}},\lVert F' \rVert_{\mathrm{Lip}},1 ) )^{-1} = \eta^{O(1)}/\log^A N \end{align} $$

To estimate the first summand, we note that for each such subinterval $[k ,k +H')$ (where $\alpha H \leq H' < 2\alpha H < H$ ), for each $h \in [k ,k +H')$ , we have

(37) $$ \begin{align} F\bigg( \frac{h}{H},g(h)\Gamma \bigg) &= F\bigg( \frac{k}{H},g(h)\Gamma \bigg) + O\bigg( \frac{H'}{H} \lVert F \rVert_{\mathrm{Lip}} \bigg)\nonumber\\& = F\bigg( \frac{k}{H},g(h)\Gamma \bigg) + O\bigg( \frac{1}{\log^A N} \bigg). \end{align} $$

Applying Theorem 5.3 to each subinterval, for each constant $C \geq 1$ we obtain

(38)

Let us now consider the second summand. We have, similarly to (37),

$$ \begin{align*} F'\bigg( \frac{h}{H},g(h)\Gamma \bigg) = F'\bigg( \frac{k}{H},g(h)\Gamma \bigg) + O\bigg( \frac{1}{\log^A N} \bigg). \end{align*} $$

For now, let us assume that $\alpha> \widetilde \delta $ , which we will verify at the end of the argument. We conclude from the fact that $(g(h)\Gamma )_{h =0}^{H-1}$ is totally $\widetilde \delta $ -equidistributed that

(39)

where we use $dx$ as a shorthand for $d \mu _{G/\Gamma }(x)$ . Taking the weighted average of (39) over all subintervals, we conclude that

(40)

Applying Lemma 5.5(ii) to estimate the measure of the support of $F_i'$ for each $1 \leq i \leq r$ , we conclude that

(41) $$ \begin{align} \int_{[0,1)} \int_{G/\Gamma} F'(t,x) \,dx dt \ll \eta^{1/O(1)}. \end{align} $$

Thus, we may choose $\eta = 1/\log ^{O(A)} N$ such that

(42) $$ \begin{align} \int_{[0,1)} \int_{G/\Gamma}F'(t,x) \,dx dt \leq \frac{1}{\log^A N}, \end{align} $$

which allows us to simplify (40) to

(43)

Combining (38) and (43) with (35), we conclude that

(44)

Letting C and B be sufficiently large multiples of A, we conclude that

(45)

as needed. Note that choosing B as a large multiple of A also guarantees that ${\alpha = 1{/}\kern1pt{\log} ^{O(A)}N> \widetilde \delta = 1{/}\kern1pt{\log} ^B N}$ .

Proof of Theorem 5.4

The argument is very similar to the proof of Theorem 1.1, assuming [Reference Green and TaoGT12a, Proposition 2.1]. As the first step, we apply the factorization theorem [Reference Green and TaoGT12b, Theorem 1.19], Theorem 4.1 above, with $M_0 = \log N$ and parameter C to be determined in the course of the argument. We conclude that there exists an integer M with $\log N \leq M \ll \log ^{O_C(1)}N$ such that g admits a factorization of the form

(46) $$ \begin{align} g(h) = \varepsilon(h) g'(h) \gamma(h), \end{align} $$

where $\varepsilon $ is $(M,N)$ -smooth, $\gamma $ is M-rational, and $g'$ takes values in a rational subgroup $G' < G$ which admits a Mal’cev basis where each element is an M-rational combination of elements of , and $(g'(h)\Gamma )_{h =0}^{H-1}$ is totally $1/M^{C}$ -equidistributed in $G'/(\Gamma \cap G')$ (with respect to the metric induced by ).

With the same reasoning as in [Reference Green and TaoGT12a], we conclude that $(\gamma (h)\Gamma )_h$ is a periodic sequence with some period $q \leq M$ , and for each $0 \leq j < q$ and $h \equiv j \bmod q$ we have $\gamma (h) \Gamma = \gamma _j \Gamma $ for some $\gamma _j \in G$ with coordinates $\tau (\gamma _j)$ that are rationals with height much less than $ M^{O(1)}$ . Splitting the average in (24) into subprogressions, it will suffice to show that for each residue $0 \leq j < q$ modulo q, and for each arithmetic progression $Q \subseteq q \mathbb {Z} + j$ with diameter at most $N/M$ , we have

(47)

The key difference between our current work and the corresponding argument in [Reference Green and TaoGT12a] is that $1_S$ is not continuous and hence in (47) we cannot replace $\varepsilon (h)$ with a constant and hope that the value of the average will remain approximately unchanged. Instead, we will use an argument of a more algebraic type. We note that, as a consequence of invariance of the metric on G under multiplication on the right, for each $h,h' \in Q$ we have

$$ \begin{align*} d( \varepsilon(h)g'(h)\gamma_j, \varepsilon(h')g'(h)\gamma_j ) = d( \varepsilon(h), \varepsilon(h') ) = O(1). \end{align*} $$

Let us fix $k \in Q$ and put $\varepsilon '(h) = \varepsilon (h)\varepsilon (k )^{-1}$ . Then $d(\varepsilon '(h),e_G) = O(1)$ and $g(h)\Gamma = \varepsilon (h) g'(h) \gamma _j \Gamma = \varepsilon '(h) \varepsilon (k )g'(h)\gamma _j\Gamma $ .

Let $\Omega \subseteq G$ be a bounded semialgebraic set such that $\varepsilon '(h) \in \Omega $ for all $h \in Q$ . For instance, we may take $\Omega $ to be the preimage of a certain ball with radius $1/\delta ^{O(1)}$ under $\widetilde \tau $ . Let also $\Pi := \widetilde \tau ^{-1}( [0,1)^D )$ denote the standard fundamental domain for $G/\Gamma $ . Consider the set

$$ \begin{align*} R = \{ (g_1,g_2) \in \Omega \times \Pi \mid g_1g_2\Gamma \in S \}. \end{align*} $$

We may decompose R as

(48) $$ \begin{align} R = \bigcup_{\gamma \in \Gamma} R_\gamma \quad\text{where } R_\gamma = \{ (g_1,g_2) \in \Omega \times \Pi \mid g_1g_2 \Gamma \in S,\ g_1g_2\gamma \in \Pi \}. \end{align} $$

Using the quantitative bounds in [Reference Green and TaoGT12b, Lemmas A.2 and A.3], we see that for each $\gamma \in \Gamma $ such that $R_\gamma \neq \emptyset $ we have $|\widetilde \tau (\gamma )| = O(1/\delta ^{O(1)})$ . Hence, the union in (48) involves $O(1/\delta ^{O(1)})$ non-empty terms, and in particular is finite. Each of the sets $R_\gamma $ is semialgebraic with complexity $O(1)$ . Moreover, since $\varepsilon '$ is a polynomial map of bounded degree, for each $\gamma \in \Gamma $ the set

$$ \begin{align*} T_\gamma = \{ (t,x) \in [0,1) \times \Pi \mid ( \varepsilon'(tH),x ) \in R_\gamma \} \end{align*} $$

is also semialgebraic with complexity $O(1)$ . Hence, (47) will follow once we show that for each semialgebraic set $T \subseteq [0,1) \times G/\Gamma $ with bounded complexity we have

(49)

Following [Reference Green and TaoGT12a], we put $\widetilde{G}' := \gamma _j^{-1}G'\gamma _j$ , $\Lambda := \Gamma \cap \widetilde{G}'$ and $ \widetilde{g}'(n) := \gamma _j^{-1}g'(n)\gamma _j$ . Let also $D' = \dim G'$ , let $\sigma $ and $\widetilde \sigma $ denote the coordinate maps on $\widetilde{G}'/\Lambda $ and $\widetilde{G}'$ respectively, and let $\Delta = \widetilde \sigma ^{-1}( [0,1)^{D'} )$ denote the fundamental domain. Then $\widetilde{g}'$ is a polynomial sequence with respect to the filtration $\widetilde{G}'_{\bullet }$ given by $\widetilde{G}'_{i} = \gamma _j^{-1}G^\prime _i\gamma _j$ . We have a well-defined map $\iota \colon \widetilde{G}'/\Lambda \to G/\Gamma $ given by

$$ \begin{align*} \iota(x\Lambda) = \varepsilon(k)\gamma_j x \Gamma. \end{align*} $$

Thus, for all $h \in [H]$ we have

$$ \begin{align*} \varepsilon(k) g'(h) \gamma_j \Gamma = \iota( \widetilde{g}'(h) \Lambda). \end{align*} $$

As discussed in [Reference Green and TaoGT12b], the Lipschitz norm of the map $\iota $ is $O(M^{O(1)})$ and the sequence $(\widetilde{G}'(h)\Lambda )_{h = 0}^{H-1}$ is $1/M^{\unicode{x3bb} C+O(1)}$ -equidistributed, where $\unicode{x3bb}> 0$ is a constant dependent only on d and D.

For each $\gamma \in \Gamma $ , the map $\iota $ is a polynomial on the semialgebraic set $\Delta \cap \iota ^{-1}(\Pi \gamma )$ . The estimate on the Lipschitz norm of $\iota $ implies that $\Delta $ can be partitioned into $M^{O(1)}$ semialgebraic sets with complexity $O(1)$ such that on each of the pieces $\iota $ is a polynomial of degree $O(1)$ (using the coordinates $\widetilde \tau $ and $\widetilde \sigma $ ). Applying the corresponding partition in (49), we see that it will suffice to show that for each semialgebraic set $T \subseteq (\mathbb {R}/\mathbb {Z}) \times (\widetilde{G}'/\Lambda )$ with bounded complexity and for each constant $A'> 0$ we have

(50)

Bearing in mind that $M \geq \log N$ , it will suffice to show that

(51)

We are now in position to apply Proposition 5.6 on $\widetilde{G}'/\Lambda $ . The complexity of $(\widetilde{G}',\Lambda ,\widetilde{G}'_\bullet )$ is $1/\delta '$ , where $\delta ' = 1/M^{O(1)}$ . The largest exponent $A'$ with which Proposition 5.6 is applicable to $(\widetilde{g}'(h))_{h=0}^{H-1}$ satisfies $\log ^{A'}N \gg M^{\mu C}$ for a constant $\mu \gg 1$ , leading to

(52)

In order to derive (51) it is enough to let C be a sufficiently large multiple of A.