Proof. For any $g\in G$ denote

$$ \begin{align*} \Phi_0(g)=|\nu_{g(o)}|, \end{align*} $$

where o is the projection of $w_o$ onto ${\mathbb H}^n$ and for any $x\in {\mathbb H}^n$ , $\nu _x$ is the Patterson–Sullivan density defined in §3.1. Since $\Phi _0$ is $\Gamma $ -invariant, it can be considered as a smooth function on X. Moreover, by assuming B contains $K=\operatorname {Stab}_G(o)$ and using the Cauchy–Schwartz inequality, we get

$$ \begin{align*} m^{\operatorname{BR}}(B)&=\int_{B}\Phi_0(g)\,dm^{\operatorname{Haar}}(g)\\ &\leq\sqrt{dm^{\operatorname{Haar}}(B)}\Vert \Phi_0\Vert _2\\ &\ll\Vert \Phi_0\Vert _2. \end{align*} $$

According to [Reference Sullivan37, §7] and by the assumption $\delta>(n-1)/2$ , we have that $\phi _0\in L^2(X)$ .

Proof. For $s \in \mathbb R$ and $\textbf {r} \in \mathbb R^{n-1}$ , let $p = a_sv_{\textbf {r}}$ . Then

$$\begin{align*}p^{-1} = \begin{pmatrix} e^{-s} & & \\[3pt] -e^{-s} \textbf{r}^T & I & \\[3pt] \tfrac12e^{-s}\|\textbf{r}\|^2 & -\textbf{r} & e^{s} \end{pmatrix},\end{align*}$$

so

$$ \begin{align*}u_{\textbf{t}} p^{-1} = \begin{pmatrix} e^{-s}(1-(\textbf{t}\cdot \textbf{r})+\frac{1}{4}\|\textbf{r}\|^2\|\textbf{t}\|^2) & \textbf{t} - \frac{1}{2} \|\textbf{t}\|^2 \textbf{r} & \frac{1}{2}e^{s}\|\textbf{t}\|^2 \\[3pt] -e^{-s}\textbf{r}^T + \frac{1}{2}e^{-s}\|\textbf{r}\|^2\textbf{t}^T & I - \textbf{t}^T \textbf{r} & e^{s}\textbf{t}^T \\[3pt] \frac{1}{2}e^{-s}\|\textbf{r}\|^2 & -\textbf{r} & e^{s}\end{pmatrix}.\end{align*} $$

Now, if $p' = a_{s'}v_{\textbf {r}'},$ we obtain that

$$ \begin{align*}p'u_{\textbf{t}'} = \begin{pmatrix} e^{s'} & e^{s'}\textbf{t}' & \frac{1}{2}e^{s'}\|\textbf{t}'\|^2 \\[3pt] \textbf{r}^{\prime T} & \textbf{r}^{\prime T}\textbf{t}' + I & \frac{1}{2}\|\textbf{t}'\|^2\textbf{r}^{\prime T} + \textbf{t}^{\prime T} \\[3pt] \frac{1}{2}e^{-s'}\|\textbf{r}'\|^2 & \frac{1}{2}e^{-s'}\|\textbf{r}'\|^2\textbf{t}' + e^{-s'}\textbf{r}' & e^{-s'}(\frac{1}{4}\|\textbf{r}'\|^2\|\textbf{t}'\|^2 +(\textbf{r}'\cdot \textbf{t}')+1)\end{pmatrix}.\end{align*} $$

We wish to solve for $\textbf {t}'$ .

Setting entries equal yields

$$ \begin{align*}\textbf{t} - \tfrac{1}{2}\|\textbf{t}\|^2\textbf{r} = e^{s'}\textbf{t}'\end{align*} $$

and

(9) $$ \begin{align} e^{s'}=e^{-s}\big(1-(\textbf{t}\cdot\textbf{r})+\tfrac{1}{4}\|\textbf{r}\|^2\|\textbf{t}\|^2\big).\end{align} $$

Combining these implies that

$$ \begin{align*}\textbf{t}' = \frac{\textbf{t} - {1}/{2}\|\textbf{t}\|^2\textbf{r}}{e^s(1-(\textbf{t}\cdot\textbf{r})+{1}/{4}\|\textbf{r}\|^2\Vert \textbf{t}\Vert ^2)}.\end{align*} $$

We define $\rho _p(\textbf {t})$ to be this quantity. One can directly check that it satisfies the claim.

Proof. Since $\xi _i$ is assumed to be a bounded parabolic limit point, by definition, $(\Lambda (\Gamma )\setminus \{\xi _i\})/ \Gamma _{\xi _i}=(\Lambda (\Gamma )\setminus \{\xi _i\})/ \Gamma _{v_i}$ is compact, where

$$ \begin{align*}G_{v_i}=g_i^{-1}MUg_i\quad\text{and}\quad\Gamma_{v_i}=\Gamma\cap G_{v_i}.\end{align*} $$

If $\gamma \in \Gamma _{v_i}$ , then $\mathcal {H}_i(R_0)\gamma =\mathcal {H}_i(R_0)$ . Therefore, the visual map induces a homeomorphism between $\mathcal {H}_i(R_0)/\Gamma _{v_i}$ and $(\partial {\mathbb H}^n\setminus \{\xi _i\})/\Gamma _{v_i}$ . It follows that the quotient of $\{g^+\in \Lambda : g\in \mathcal {H}_i(R_0)\}$ by the action of $\Gamma _{v_i}$ is compact. Using Iwasawa decomposition, it follows that there exists a compact set $U_0\subset U$ such that for any $g=kaug_i\in \mathcal {H}_i(R_0)$ such that $g^+\in \Lambda (\Gamma )$ , $k\in K$ , $a\in A$ , and $u\in U$ , there exist $\gamma \in \Gamma _{v_i}$ , $k'\in K$ , $u'\in U_0$ so that $g\gamma =k'au'g_i$ .

Since $\xi _i$ is assumed to be a parabolic limit point, there exists a parabolic element $\gamma _0\in \Gamma _{\xi _i}$ , that is, $\gamma _0=g_i^{-1}mug_i$ .

Assume for contradiction that there exists an infinite sequence $\{\gamma _j\}\in \Gamma $ such that $\{\gamma _j v_i\}$ converges. Note that translating by an element of $\gamma $ allows us to assume, without loss of generality, that the limit of this sequence is $0$ . Using the Iwasawa decomposition, we get that for all j, there exist $a_{t_j}\in A$ , $k_j\in K$ , and uniformly bounded $u_j\in U$ such that $\gamma _j=k_ja_{t_j}u_jg_i$ . Since

$$ \begin{align*} \Vert \gamma_jv_i\Vert =\Vert k_ja_{t_j}u_je_1\Vert =e^{t_j}, \end{align*} $$

we may deduce that $t_j\rightarrow -\infty $ . In particular, $\gamma _j\in H_i(R_0)$ for all large enough j.

We have

$$ \begin{align*} \gamma_j\gamma_0\gamma_j^{-1} &=(k_ja_{t_j}u_jg_i)(g_i^{-1}mug_i)(g_i^{-1} u_j^{-1} a_{t_j}^{-1} k_{j}^{-1})\\ &=k_ja_{t_j}u_jm uu_j^{-1} a_{t_j}^{-1} k_j^{-1}. \end{align*} $$

Since $u_jmu_j^{-1}=m_ju^{\prime }_j\in MU$ , with $u^{\prime }_j$ uniformly bounded, and since M centralizes A, we have

$$ \begin{align*} \gamma_j\gamma_0\gamma_j^{-1} &=k_jm_j a_{t_j} u^{\prime}_j u a_{t_j}^{-1} k_j^{-1}. \end{align*} $$

Since $u^{\prime }_j u$ is in a bounded subset of U, we get that $a_{t_j} u^{\prime }_j u a_{t_j}^{-1}\rightarrow e$ as $t_j\rightarrow -\infty $ . Since K and M are compact, it then follows that the sequence $\gamma _j\gamma _0\gamma _j^{-1}$ has a convergent subsequence. This contradicts the discreteness of $\Gamma $ , since the $\gamma _j$ terms were assumed to be distinct.

Proof. Fix $T>T_0=\max \{s_0,\eta _0^{{1}/({\varepsilon -1})}\}$ . We will first show that

(36) $$ \begin{align} \inf_{\gamma\in\Gamma}\inf_{i=1,\ldots,q}\Vert a_{-\log T}g\gamma v_i\Vert>cT^{\varepsilon-1}, \end{align} $$

for some constant $1>c=c(\Gamma )>0$ .

There are two cases to consider. If $a_{-\log T}x\notin \mathcal {X}_i(R_0)$ , then (36) follows from (34) and the choice of T.

Otherwise, $a_{-\log T}x\in \mathcal {X}_i(R)$ for some maximal $R>R_0$ . According to Lemma 3.6, we have

$$ \begin{align*}d(x,\mathcal{C}_0)\ge R-R_0.\end{align*} $$

Then, because x is $(\varepsilon ,s_0)$ Diophantine and $T> s_0,$ by (32), we may deduce that

$$ \begin{align*}R-R_0<(1-\varepsilon)\log T.\end{align*} $$

Hence, $a_{-\log T}x \not \in \mathcal {X}_i((1-\varepsilon )\log T+R_0),$ so (33) implies (36).

Now, fix $\gamma \in \Gamma $ and $1\le i \le q$ , and let

$$ \begin{align*} \begin{pmatrix} x_1\\ \vdots\\ x_{n+1} \end{pmatrix}=a_{-\log T}g\gamma v_i.\end{align*} $$

According to (36), there exists $1\leq k\leq n$ such that $|x_k|>cT^{\varepsilon -1}$ . If $|x_1|>cT^{\varepsilon -1}$ , then it follows from the action of $a_{-\log T}$ on $\mathbb {R}^{n+1}$ that

$$ \begin{align*} \Vert g\gamma v_i\Vert \geq|cT x_1|>cT^{\varepsilon}. \end{align*} $$

Otherwise, there exists $2\leq k\leq n$ such that $|x_k|>cT^{\varepsilon -1}$ . Then, for any $\textbf {t}\in \mathbb {R}^{n-1}$ , the first coordinate of $u_{\textbf {t}}a_{-\log T}g\gamma v_i$ is

$$ \begin{align*} x_1+\textbf{t}\cdot \mathbf{x}'+\frac{1}{2}\Vert \textbf{t}\Vert ^2 x_{n+1}\quad\mbox{where } \mathbf{x}'=\begin{pmatrix} x_2\\ \vdots\\ x_n \end{pmatrix}.\end{align*} $$

In particular, by taking $t_k=\pm T$ (the kth entry in $\textbf {t}$ ) one can ensure that $\|a_{\log T}u_{\textbf {t}}a_{-\log T} g\gamma v_i\|>cT^{\varepsilon }$ .

Proof. First, note that if $\|f\|_{B_U(\eta )} < 2\varepsilon $ , then the result holds by assuming that the implied coefficient in (39) is bigger than $2^{\sigma }$ : in this case, the right-hand side is greater or equal to

$$ \begin{align*}2^{\sigma}\bigg(\frac{\varepsilon}{\Vert f\Vert _{B_U(\eta)}}\bigg)^{\sigma}\mu^{\operatorname{PS}}_y(B_U(\eta)) &\ge2^{\sigma}\bigg(\frac{\varepsilon}{2\varepsilon}\bigg)^{\sigma}\mu^{\operatorname{PS}}_y(B_U(\eta))\\ & \ge\mu^{\operatorname{PS}}_y(\{\textbf{t}\in B_U(\eta):\|f(\textbf{t})\|<\varepsilon\}), \end{align*} $$

as desired. Thus, we now assume that

(40) $$ \begin{align}\Vert f\Vert _{B_U(\eta)} \ge 2\varepsilon.\end{align} $$

If $\|f(\textbf {t})\|\ge \varepsilon $ for all $\textbf {t} \in B_U(\eta )$ such that $(u_{\textbf {t}} y)^+\notin \Lambda (\Gamma )$ , then there is nothing to prove. So assume that $\|f(\textbf {t})\|<\varepsilon $ and $(u_{\textbf {t}} y)^+\in \Lambda (\Gamma )$ . Since each $f_i$ is linear, for all $\textbf {t}'\in B_U(\eta )$ with $\|f(\textbf {t}')\|<\varepsilon $ , we get that for all $1\le i\le n-1$ ,

$$ \begin{align*} |f_i(\textbf{t}')|= |a_i+b\textbf{t}'| <\varepsilon, \end{align*} $$

$$ \begin{align*}|b(t^{\prime}_i-t_i)|=|f_i(\textbf{t}')-f_i(\textbf{t})|<2\varepsilon.\end{align*} $$

Therefore,

$$ \begin{align*}\|f(x)\|<\varepsilon {\implies} x \in B_U(2\varepsilon/b)z.\end{align*} $$

Thus, by (38), we have that there exists $\sigma = \sigma (\Gamma )>0$ so that

$$ \begin{align*} \mu^{\operatorname{PS}}_y(\{x \in B_U(\eta)y : \|f(x)\|<\varepsilon\})&\le\mu^{\operatorname{PS}}_z( B_U(2\varepsilon/b))\\ & \ll_{\Gamma}\bigg(\frac{2\varepsilon}{b\eta}\bigg)^{\sigma}\mu^{\operatorname{PS}}_z( B_U(\eta))\\ & \ll_{\Gamma}\bigg(\frac{2\varepsilon}{b\eta}\bigg)^{\sigma}\mu^{\operatorname{PS}}_y( B_U(3\eta))\\ & \ll_{\Gamma}\bigg(\frac{6\varepsilon}{b\eta}\bigg)^{\sigma}\mu^{\operatorname{PS}}_y( B_U(\eta)). \end{align*} $$

Assuming $\|f(\textbf {t})\|<\varepsilon $ for some $\textbf {t}\in B_U(\eta )$ (otherwise, as before, there is nothing to prove), for any $\textbf {t}"\in B_U(\eta )$ and $1\le i\le n-1$ , we have

$$ \begin{align*} |f_i(\textbf{t}")|\le |f_i(\textbf{t}")-f_i(\textbf{t})|+|f_i(\textbf{t})|<2b\eta+\varepsilon.\end{align*} $$

Thus, $\Vert f\Vert _{B_U(\eta )y}-\varepsilon \le 2b\eta $ , so by (40),

$$ \begin{align*}\tfrac{1}{2}\Vert f\Vert _{B_U(\eta)} \le 2b \eta,\end{align*} $$

which completes the proof.

Proof. For any $\textbf {t}\in B$ , denote

$$ \begin{align*} f_{\Lambda}(\textbf{t})=\min\{\Vert f(\textbf{t})v\Vert : v\in\Lambda\}. \end{align*} $$

Let

$$ \begin{align*} E=\{\textbf{t}\in B : f_{\Lambda}(\textbf{t})<\varrho\}\cap \operatorname{supp}\mu, \end{align*} $$

and for each $v \in \Lambda $ , define

$$ \begin{align*} E_v=\{\textbf{t} \in B : \|f(\textbf{t})v\|<\varrho\} \cap \operatorname{supp}\mu. \end{align*} $$

Observe that by assumption (3), the $E_v$ terms are a disjoint cover of E. For each $\textbf {t} \in E_v$ , define

$$ \begin{align*} r_{\textbf{t},v}=\sup\{r : \|f(\mathbf{s})v\|<\varrho \text{ for all } \textbf{s}\in B(\textbf{t},r)\}. \end{align*} $$

By assumption (2), we know that for every $\textbf {t}\in E$ , the set $B(\textbf {t},r_{\textbf {t},v})$ does not contain B. Thus, since $\textbf {t}\in B$ , we deduce that $r_{\textbf {t},v}<2r_0$ . For any fixed $r_{\textbf {t},v}<r^{\prime }_{\textbf {t},v}<2r_0$ , we have that

(42) $$ \begin{align} B(\textbf{t},r^{\prime}_{\textbf{t},v})\subset B(z,3r_0)=\tilde B, \end{align} $$

and by the definition of $r_{\textbf {t},v}$ , there exists $\textbf {s} \in B(\textbf {t},r^{\prime }_{\textbf {t},v})$ such that

$$ \begin{align*} \Vert f(\textbf{s})v\Vert \geq\varrho. \end{align*} $$

Note that $\{B(\textbf {t},r_{\textbf {t},v}) : \textbf {t}\in E, v\in \Lambda \}$ is a cover of E. According to the Besicovitch covering theorem, there exists a countable subset $I\subset E\times \Lambda $ such that $\{B(\textbf {t},r_{\textbf {t},v}) : (\textbf {t},v)\in ~I\}$ is a cover of E with a covering number bounded by a constant which only depends on m. Thus,

(43) $$ \begin{align} \sum_{(\textbf{t},v)\in I}\mu(B(\textbf{t},r_{\textbf{t},v})) \ll_{m} \mu\bigg(\bigcup_{(\textbf{t},v)\in I}B(\textbf{t},r_{\textbf{t},v})\bigg). \end{align} $$

By assumption (3) and the continuity of f, for any $(\textbf {t},v)\in I$ and $\mathbf{s}\in E\cap B(\textbf {t},r_{\textbf {t},v})$ ,

$$ \begin{align*} f_{\Lambda}(\textbf{s})=\Vert f(\textbf{s})v\Vert. \end{align*} $$

Thus,

$$ \begin{align*} \mu(\{\mathbf{s}\in B(\textbf{t},r_{\textbf{t},v}) : {f_{\Lambda}(\mathbf{s})}<\varepsilon\})&=\mu(\{\mathbf{s}\in B(\textbf{t},r_{\textbf{t},v}) : \Vert f(\mathbf{s})v\Vert <\varepsilon\})\\[3pt] &\le\mu(\{\mathbf{s}\in B(\textbf{t},r^{\prime}_{\textbf{t},v}) : \Vert f(\mathbf{s})v\Vert <\varepsilon\}).\end{align*} $$

Thus, assumption (1) and the assumption that $\mu $ is D-Federer together imply that

(44) $$ \begin{align} \mu(\{\mathbf{s}\in B(\textbf{t},r_{\textbf{t},v}) : {f_{\Lambda}(\mathbf{s})}<\varepsilon\})&\le\mu(\{\mathbf{s}\in B(\textbf{t},r^{\prime}_{\textbf{t},v}) : \Vert f(\mathbf{s})v\Vert <\varepsilon\})\nonumber\\[3pt] & \le C\bigg(\frac{\varepsilon}{\varrho}\bigg)^{\beta}\mu(B(\textbf{t},r^{\prime}_{\textbf{t},v}))\nonumber\\[3pt] & \le CD \bigg(\frac{\varepsilon}{\varrho}\bigg)^{\beta}\mu(B(\textbf{t},r_{\textbf{t},v})). \end{align} $$

Since E covers the set of points for which $f_{\Lambda }$ is less than $\varepsilon $ , we may now conclude

$$ \begin{align*} &\mu(\{\textbf{t}\in B : f_{\Lambda}(\textbf{t})<\varepsilon\})\\ &\quad\le \sum_{(\textbf{t},v)\in I}\mu(\{\mathbf{s}\in B(\textbf{t},r_{\textbf{t},v}) : f_{\Lambda}(\textbf{t})<\varepsilon\}) \\ & \quad\le CD \sum_{(\textbf{t},v)\in I}\bigg(\frac{\varepsilon}{\varrho}\bigg)^{\beta}\mu(B(\textbf{t},r_{\textbf{t},v})) \quad \text{by (44)}\\ & \quad\ll_{m} CD \bigg(\frac{\varepsilon}{\varrho}\bigg)^{\beta}\mu\bigg(\bigcup_{(\textbf{t},v)\in I}B(\textbf{t},r_{\textbf{t},v})\bigg)\quad \text{by (43)}\\ & \quad\ll_{m} CD \bigg(\frac{\varepsilon}{\varrho}\bigg)^{\beta}\mu(\tilde B ) \quad \text{by (42)} \\ & \quad\ll_{m} CD^2 \bigg(\frac{\varepsilon}{\varrho}\bigg)^{\beta}\mu(B ) \quad\mu\text{ is }D\text{-Federer.}\\[-3.6pc] \end{align*} $$

Proof of Theorem 4.1

Let $x_0=a_{-\log s}x$ and fix $g\in G$ such that $x=g\Gamma $ . By Lemma 4.3, for all $T \gg _{\Gamma ,\varepsilon } s_0,$ we have (35), that is, that

$$ \begin{align*}\sup_{\Vert \textbf{t}\Vert \leq T}\inf_{\gamma\in\Gamma}\inf_{i=1,\ldots,q}\Vert u_{\textbf{t}} g\gamma v_i\Vert>T^{\varepsilon}.\end{align*} $$

Let $f:\mathbb R^{n-1}\rightarrow \operatorname {SL}_{n+1}(\mathbb R)$ be defined by

$$ \begin{align*}f(\textbf{t})=u_{\textbf{t}} a_{-\log s}g.\end{align*} $$

We first show that parts (1), (2), and (3) of Proposition 4.5 for $\mu =\mu ^{\operatorname {PS}}_{x_0}$ , f, $\varrho =1$ , $z=x_0$ , $r=T/s$ , $\eta =e^{-R_0}$ , and

$$ \begin{align*}\Lambda=\Gamma \{v_1,\ldots,v_q\}.\end{align*} $$

Note that $0\notin \Lambda $ .

It follows from the action of $u_{\textbf {t}}$ on $\mathbb {R}^{n+1}$ that for any $v\in \mathbb R^{n+1}$ , there exists $v'=(v^{\prime }_1,\ldots ,v^{\prime }_{n+1})^T\in \mathbb R^{n+1}$ such that

(45) $$ \begin{align} f(\textbf{t})v=\big(v^{\prime}_1+\textbf{t}\cdot v"+\tfrac{1}{2}\Vert \textbf{t}\Vert ^2v^{\prime}_{n+1},v^{\prime}_2-t_1v^{\prime}_{n+1},\ldots,v^{\prime}_{n}-t_{n-1}v^{\prime}_{n+1},v^{\prime}_{n+1}\big)^T,\end{align} $$

where $v"=(v^{\prime }_2,\ldots ,v^{\prime }_{n})^T$ . Thus, if $v^{\prime }_{n+1}\ne 0$ , then $\textbf {t} \mapsto f(\textbf {t})v$ is bounded from below by a function which satisfies the assumption of Lemma 4.4. Therefore, by (41), for any $v\in \Lambda $ , the function $\textbf {t}\mapsto \Vert f(\textbf {t})v\Vert $ is $(C,\beta )$ -good on $\tilde B$ with respect to $\mu ^{\operatorname {PS}}_{x_0}$ for some $C=C(\Gamma )\ge 1$ , $\beta =\beta (\Gamma )>0$ , which proves (1) of Proposition 4.5. Note that these constants are uniform across all v so that $v_{n+1}'\ne 0.$

However, if $v\neq 0$ and

$$ \begin{align*}v_{n+1}' = 0,\end{align*} $$

then $\textbf {t} \mapsto f(\textbf {t})v$ is bounded below by some positive constant, and since positive constant functions are $(C,\beta )$ -good for any $C\ge 1,\beta>0,$ we conclude that so is this function by (41).

By (6), we have

$$ \begin{align*} u_{\textbf{t}}a_{-\log s}=a_{-\log s} u_{s\textbf{t}}.\end{align*} $$

Since multiplication by $a_{-\log s}$ only changes the matrix entries by scaling, using (35), for $i=1,\ldots ,q$ , we get

$$ \begin{align*} \sup_{\Vert \textbf{t}\Vert \leq T/s}\Vert a_{-\log s}u_{s\textbf{t}} g\gamma v_i\Vert>s^{-1}\sup_{\Vert \textbf{t}\Vert \leq T}\Vert u_{\textbf{t}} g\gamma v_i\Vert >s^{-1}T^{\varepsilon}.\end{align*} $$

Thus, for any $s\le T^{\varepsilon }$ , $\|\textbf {t}\|<T/s$ , and $v\in \Lambda $ ,

$$ \begin{align*} \Vert f(\textbf{t})v\Vert \ge 1, \end{align*} $$

which establishes (2) of Proposition 4.5.

Since $\eta =e^{-R_0}$ and $\mathcal {H}_i(R_0)$ terms are pairwise disjoint, part (3) of Proposition 4.5 follows from the uniqueness of $\gamma $ in (33) and (34).

According to 37, the measure $\mu ^{\operatorname {PS}}_{x_0}$ is D-Federer for any $D>0$ . Thus, we may now use (33) and Proposition 4.5 to deduce

$$ \begin{align*} &\mu^{\operatorname{PS}}_{x_0}(B_U(T/s)\cap\mathcal{H}_i(R))\\[3pt] &\quad=\mu^{\operatorname{PS}}_{x_0}(\{\textbf{t}\in B_U(T/s) : \text{ there exists } \gamma\in\Gamma,1\le i\le q\text{ such that }\Vert f(\textbf{t})\gamma v_i\Vert <e^{-R}\})\\[3pt] &\quad\ll e^{-R \beta} \mu^{\operatorname{PS}}_{x_0}(B_U(T/s)x_0), \end{align*} $$

where the implied constant depends on n and $\Gamma $ .

Proof of Theorem 5.2 assuming Proposition 5.3

Using the product structure of $\nu $ , we can write

$$ \begin{align*}\nu(P_{\eta} g) = \int_{A_{\eta}}\int_{M_{\eta}} \mu^{\operatorname{PS}-}_g(B_{\tilde{U}} (\eta))\,dm\,ds.\end{align*} $$

Then, by an analogous statement to Proposition 5.3 for $\mu ^{\operatorname {PS}-}$ , there exists a constant $c_0 = c_0(\Gamma )>0$ such that

$$ \begin{align*} \nu(P_{\eta+\xi}g) &= \int_{A_{\xi+\eta}}\int_{M_{\xi+\eta}}\mu^{\operatorname{PS}-}_g(B_{\tilde U}(\xi + \eta))\,dm\,ds \\[3pt] &\le\int_{A_{\xi+\eta}}\int_{M_{\xi+\eta}}\mu^{\operatorname{PS}-}_g(B_{\tilde U}(\eta))\bigg[1+c_0\frac{\xi^{\alpha}}{\eta^{\theta}} e^{\omega\operatorname{height}(g\Gamma)}\bigg]\, dm\, ds\\ &=\bigg[1+c_0\frac{\xi^{\alpha}}{\eta^{\theta}} e^{\omega\operatorname{height}(g\Gamma)}\bigg]\bigg[ \frac{(\xi+\eta)^{{1}/{2}(n-1)(n-2)+1}}{\eta^{{1}/{2}(n-1)(n-2)+1}}\nu(P_{\eta} g)\bigg]\\[4pt] &\le\bigg[1+c_0\frac{\xi^{\alpha}}{\eta^{\theta}} e^{\omega\operatorname{height}(g\Gamma)}\bigg]\bigg[1+c_1\frac{\xi}{\eta}\bigg]\nu(P_{\eta} g), \end{align*} $$

where $c_1>0$ is an absolute constant (which depends only on n) arising from the binomial theorem. Therefore,

(46) $$ \begin{align} \frac{\nu(P_{\xi+\eta}g)}{\nu(P_{\eta} g)}-1\ll_{\Gamma} e^{\omega\operatorname{height}(g\Gamma)}\frac{\xi^{\alpha}}{\eta^{\theta}}\cdot\frac{\xi}{\eta}+\frac{\xi}{\eta}+e^{\omega\operatorname{height}(g\Gamma)}\frac{\xi^{\alpha}}{\eta^{\theta}}. \end{align} $$

Since $\xi <\eta $ , the first term on the left-hand side of (46) is dominated by the last term, and so

$$ \begin{align*}\frac{\nu(P_{\xi+\eta}g)}{\nu(P_{\eta} g)}-1\ll_{\Gamma}\frac{\xi}{\eta}+e^{\omega\operatorname{height}(g\Gamma)}\frac{\xi^{\alpha}}{\eta^{\theta}}.\end{align*} $$

Since $e^{\omega \operatorname {height}(g\Gamma )}\ge 1$ , if we define

$$ \begin{align*}\alpha'=\min\{1,\alpha\},\quad \theta'=\max\{1,\theta\},\end{align*} $$

then both terms are dominated by

$$ \begin{align*}e^{\omega\operatorname{height}(g\Gamma)}\frac{\xi^{\alpha'}}{\eta^{\theta'}},\end{align*} $$

which completes the proof.

Proof of Proposition 5.3

It follows from the geometry of $B_U(\xi +\eta )x-B_U(\eta )x$ that there exist hyperplanes $L_1,\ldots ,L_m$ , where m only depends on n, such that

$$ \begin{align*} B_U(\xi+\eta)x-B_U(\eta)x\subseteq\bigcup_{i=1}^{m} \mathcal{N}_U(L_i,2\xi).\end{align*} $$

For any $0<\xi <\eta \ll _{\Gamma } e^{-\operatorname {height}(x)},$ we have that

By (37), $\mu ^{\operatorname {PS}}_x$ is D-Federer (see Corollary A.8 for more detail), in particular,

$$ \begin{align*}\mu^{\operatorname{PS}}_x(B_U(2\eta)) \ll_{\Gamma} \mu^{\operatorname{PS}}_x(B_U(\eta)).\end{align*} $$

Thus, we obtain

$$ \begin{align*} \frac{\mu^{\operatorname{PS}}_x(B_U(\xi+\eta))}{\mu^{\operatorname{PS}}_x(B_U(\eta))} -1 \ll_{\Gamma}e^{{\omega}\operatorname{height}(x)}\frac{ \xi^{\alpha}}{\eta^{\theta}},\end{align*} $$

and relabeling the constants completes the proof.

Proof. By Lemma 3.8, for some $T_0\gg _{\Gamma ,\varepsilon } s_0$ , there exists

$$ \begin{align*}y \in B_U(T_0)x\cap\operatorname{supp}m^{\operatorname{BMS}}.\end{align*} $$

Then for $T \ge T_0$ , we have

$$ \begin{align*}B_U(T-T_0)y\subseteq B_U(T)x\subseteq B_U(T+T_0)y.\end{align*} $$

Since $c \ge 1,$ we therefore have that for $T \ge 2T_0,$

Proof. Let $T_0\gg _{\Gamma ,\varepsilon } s_0$ satisfy the conclusion of Lemma 3.8. For $T \ge T_0$ , let

$$ \begin{align*}E_T:=\mathcal{N}(L,\xi)\cap B_U(T) \cap H_R\cap \operatorname{supp}m^{\operatorname{BMS}},\end{align*} $$

and observe that $\mu ^{\operatorname {PS}}_x(E_T)=\mu ^{\operatorname {PS}}_x(\mathcal {N}(L,\xi )\cap B_U(T)\cap H_R).$

Let $c_1=c_1(\Gamma )>0$ be the implied constant in Proposition 5.4. Fix $r=c_1e^{-R}$ and let $\{u_1,\ldots ,u_k\}$ be a maximal ${r}/{2}$ -separated set in $E_{T-{r}/{4}}$ . Then,

$$ \begin{align*}E_T\subseteq\bigcup_{i=1}^k B_U(r)u_i.\end{align*} $$

Note also that by (37), we have that there exists a constant $c_2=c_2(\Gamma )>0$ such that for all $u_i$ ,

(47) $$ \begin{align} \mu^{\operatorname{PS}}_{u_i}(B_U(r))=\mu^{\operatorname{PS}}_{u_i}(B_U(8(r/8)) \le c_2\mu^{\operatorname{PS}}_{u_i}(B_U(r/8)). \end{align} $$

Therefore,

By Corollary 5.5, there exists $\sigma =\sigma (\Gamma ) \ge \delta _{\Gamma }$ so that

$$ \begin{align*}\mu^{\operatorname{PS}}_x(B_U(T+1)) \subseteq \mu^{\operatorname{PS}}_x(B_U(2T)) \ll_{\Gamma} 2^{\sigma} \mu^{\operatorname{PS}}_x(B_U(T)).\end{align*} $$

Let

$$ \begin{align*}\omega' = \omega+\theta.\end{align*} $$

It follows from the geometry of $B_U(\xi +T)x-B_U(T)x$ that there exist $L_1,\ldots ,L_m$ , where m only depends on n, such that

$$ \begin{align*} B_U(\xi+T)x-B_U(T)x\subseteq\bigcup_{i=1}^{m} \mathcal{N}_U(L_i,2\xi).\end{align*} $$

Thus, we also have

$$ \begin{align*}(B_U(\xi + T)x - B_U(T)x) \cap H_R\subseteq\bigcup_{i=1}^{m} \mathcal{N}_U(L_i,2\xi).\end{align*} $$

We arrive at

$$ \begin{align*} \frac{\mu^{\operatorname{PS}}_x((B_U(\xi+T) \cap H_R)-(B_U(T)\cap H_R) )}{\mu^{\operatorname{PS}}_x(B_U(T))} & \le \sum_{i=1}^m\frac{\mu^{\operatorname{PS}}_x(\mathcal{N}(L_i,2\xi)\cap B_U(\xi+T))}{\mu^{\operatorname{PS}}_x(B_U(T))}\\[4pt] & \ll_{\Gamma}m e^{\omega'R}{\xi^{\alpha}} \frac{\mu^{\operatorname{PS}}_x( B_U(\xi+T))}{\mu^{\operatorname{PS}}_x(B_U(T))}. \end{align*} $$

By Corollary 5.5 again, we conclude that

$$ \begin{align*} \frac{\mu^{\operatorname{PS}}_x((B_U(\xi+T) \cap H_R)-(B_U(T)\cap H_R) )}{\mu^{\operatorname{PS}}_x(B_U(T))} \ll_{\Gamma}e^{\omega'R}\xi^{\alpha},\end{align*} $$

which completes the proof.

Proof of Theorem 5.1

Observe that by Lemma 3.6, for any $R>R_0$ ,

(48) $$ \begin{align} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T))&=\mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T)\cap H_{R-R_0}) + \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T)\cap \mathcal{X}(R)),\nonumber\\[4pt] \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi))&=\mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi)\cap H_{R-R_0}) + \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi)\cap \mathcal{X}(R)). \end{align} $$

By Theorem 4.1, for $T \gg _{\Gamma ,\varepsilon } s_0$ , $0<s\le T^{{\varepsilon }/({1-\varepsilon })}$ , and any $R \ge R_0,$

$$ \begin{align*} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi)\cap \mathcal{X}(R))&= \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U((s(T+\xi)/s) \cap \mathcal{X}(R)) \\[4pt] &\ll_{\Gamma} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi))e^{-\beta R}\\[4pt] &\ll_{\Gamma} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T))e^{-\beta R} \quad\text{by Corollary 5.5}. \end{align*} $$

Observe that use of Corollary 5.5 is justified if $T \gg _{\Gamma ,\varepsilon } s_0$ by Remark 5.6. Similarly, by Proposition 5.7 and the same reasoning as in Remark 5.6, for $T \gg _{\Gamma ,\varepsilon } s_0$ , we have

$$ \begin{align*} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi)\cap H_{R-R_0})-\mu^{\operatorname{PS}}_{a_{-s}x}(B_U(T)\cap H_{R-R_0})\ll_{\Gamma} e^{\omega R}\xi^{\alpha} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T)). \end{align*} $$

Putting this together with (48), we conclude

$$ \begin{align*} &\mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T+\xi))-\mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T))\\[4pt] &\quad\ll_{\Gamma} e^{\omega R}\xi^{\alpha} \mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T)) + 2e^{-\beta R}\mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T))\\[4pt] &\quad\ll_{\Gamma} (e^{\omega R}\xi^{\alpha} +2e^{-\beta R})\mu^{\operatorname{PS}}_{a_{-\log s}x}(B_U(T)). \end{align*} $$

Taking $R=-{\alpha }/({\omega +\beta })\log \xi $ implies the result, provided that $\xi $ is sufficiently small so that this is larger than $R_0.$ Note that since $\alpha ,\omega ,\beta , R_0$ are all constants depending only on $\Gamma $ , this is equivalent to requiring $\xi \ll _{\Gamma } 1.$

Proof. Without loss of generality, assume that f and $\psi $ are non-negative functions. Step 1: Setup and approximations.

Let $\kappa ',\ell '$ satisfy the conclusion of Assumption 1.1, and let $\ell>\ell '$ satisfy the conclusion of Lemma 3.12. Observe that $\ell $ can be increased if necessary while maintaining this property.

Because $\psi $ is supported on an admissible box, there exists $0<\eta _0<1/2$ (depending on $\operatorname {supp}\psi $ ) such that $G_{3\eta _0}\operatorname {supp}\psi $ is still an admissible box. For $0<\eta <\eta _0$ , let $\psi _{\eta ,\pm }$ satisfy the conclusion of Lemma 3.12 for $G,3\eta $ , and $\psi $ . In particular, for all small $\eta>0$ ,

(49) $$ \begin{align} S_{\ell'}(\psi_{\eta,\pm})\ll_{\operatorname{supp}\psi}\eta^{-2\ell}S_{\ell}(\psi). \end{align} $$

Since $\psi $ is uniformly continuous and the BMS measure is finite, we may deduce from Lemma 3.12(2) that

(50) $$ \begin{align} |\tilde{m}^{\operatorname{BMS}}(\psi_{\eta,\pm})-\tilde{m}^{\operatorname{BMS}}(\psi)|\ll_{\operatorname{supp}\psi,\Gamma}\eta S_{\ell}(\psi). \end{align} $$

According to Lemma 3.1, for any $p \in P_{\eta }$ , there exists $\rho _p : B_U(1) \to B_U(1+O(\eta ))$ that is a diffeomorphism onto its image and a constant $D=D(\eta )<3\eta $ such that

(51) $$ \begin{align}u_{\textbf{t}} p^{-1} \in P_D u_{\rho_p(\textbf{t})}.\end{align} $$

Step 2: Assuming that f is supported on a small ball.

We start by proving that there exists $\kappa> 0$ such that if $f\in C_c^{\infty }(B_U(r_1))$ , where $r_1\le \operatorname {inj}(g)$ , then for $s>0$ ,

(52) $$ \begin{align} & \sum_{\gamma\in\Gamma}\int_U\psi(a_s u_{\textbf{t}} g\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_g (\textbf{t})-\tilde{m}^{\operatorname{BMS}}(\psi)\mu^{\operatorname{PS}}_g(f)\nonumber\\[3pt] &\quad\ll_{\Gamma,\operatorname{supp}\psi}S_{\ell}(\psi)S_{\ell}(f)e^{-2\kappa s}\mu^{\operatorname{PS}}_g(B_U(1)). \end{align} $$

For any $s>0$ and $\gamma \in \Gamma $ , from (51), we have that

$$ \begin{align*} &\int_{B_U(r_1)} \psi(a_s u_{\textbf{t}} g\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_g(\textbf{t}) \\[3pt] &\quad= \frac{1}{\nu(P_{\eta} g)}\int_{P_{\eta} g}\int_{B_U(r_1)}\psi(a_{s}u_{\textbf{t}} p^{-1} pg\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\,d\nu(pg)\\[3pt] &\quad\le \frac{1}{\nu(P_{\eta} g)} \int_{P_{\eta} g}\int_{B_U(r_1)} \psi_{\eta,+}(a_{s}u_{\rho_p(\textbf{t})}pg\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\,d\nu(pg), \end{align*} $$

where the last inequality follows since $a_s P_{3\eta }a_{-s}\subset P_{3\eta }$ for any positive s.

Step 2.1: Use the product structure of the BMS measure. For any $p\in P_{\eta }$ , $(u_{\textbf {t}} g)^+= (u_{\rho _p (\textbf {t})}pg)^+$ , the measures $d\mu ^{\operatorname {PS}}_{g}(\textbf {t})$ and $d({\rho _p}_{\ast }\mu ^{\operatorname {PS}}_{pg}(\textbf {t}))=d\mu ^{\operatorname {PS}}_{pg}(\rho _p (\textbf {t}))$ are absolutely continuous with each other, and the Radon–Nikodym derivative at $\textbf {t}$ is given by

(53) $$ \begin{align} \frac{d\mu^{\operatorname{PS}}_{g}(\textbf{t})}{d\mu^{\operatorname{PS}}_{pg}(\rho_p (\textbf{t}))}=e^{\delta_{\Gamma}\beta_{(u_{\textbf{t}} g)^+}(u_{\textbf{t}} g(o),u_{\rho_p(\textbf{t})}pg(o))}. \end{align} $$

Let $0<\xi <\eta $ . Let $\chi _{\eta ,\xi }$ satisfy the conclusion of Lemma 3.10 for $H=P$ , $\xi _1=\eta -\xi $ , $\xi _2=\xi $ , and g. Let $\varphi _{\eta ,g}$ be the function defined on $B_U (1) P_{\eta }g$ given by

$$ \begin{align*} \varphi_{\eta,g}(u_{\rho_p(\textbf{t})}pg):=\frac{f(\textbf{t})\chi_{\eta,\xi}(pg)}{\nu(P_{\eta} g)e^{\delta_{\Gamma}\beta_{(u_{\textbf{t}} g)^+}(u_{\textbf{t}} g(o),u_{\rho_p(\textbf{t})}pg(o))}}.\end{align*} $$

We will need a bound on $S_{\ell }(\varphi _{\eta ,g}).$ To that end, note that

$$ \begin{align*} |\beta_{(u_{\textbf{t}} g)^+}(u_{\textbf{t}} g(o), u_{\rho_p(\textbf{t})} pg(o))| &\le d(u_{\textbf{t}} g(o), u_{\rho_p(\textbf{t})}pg(o))\\[3pt] &= d(g(o), u_{-\textbf{t}} u_{\rho_p(\textbf{t})}pg(o)). \end{align*} $$

Since $u_{-\textbf {t}} u_{\rho _p(\textbf {t})}p\in G_{5\eta }$ , the above is bounded by some absolute constant (depending only on $\Gamma $ ) for all $\eta <\tfrac 12$ . Observe that this bound holds on the support of $f(\textbf {t})\chi _{\eta ,\xi }(pg).$ Moreover, because the Busemann function is Lipschitz, all of its derivatives are bounded.

It then follows immediately from the product rule and the definition of the Sobolev norm that

(54) $$ \begin{align}S_{\ell}(f(\textbf{t})\chi_{\eta,\xi}(pg)\exp(-\delta_{\Gamma} \beta_{(u_{\textbf{t}} g)^+}(u_{\textbf{t}} g(o),u_{\rho_p(\textbf{t})}pg(o)))) \ll_{\Gamma,\ell} S_{\ell}(f(\textbf{t})\chi_{\eta,\xi}(pg)).\end{align} $$

By [Reference Kleinbock and Margulis17, Lemma 2.4.7(a)], Lemma 3.10, (54), and Lemma 3.9, we have

(55) $$ \begin{align} S_{\ell}(\varphi_{\eta,g})&\ll_{\Gamma,\ell} \nu(P_{\eta} g)^{-1} S_{\ell}(f)S_{\ell}(\chi_{\eta,\xi})\nonumber\\[3pt] &\ll_{\Gamma,\ell} \eta^{-(\delta_{\Gamma} + {1}/{2}(n-1)(n-2)+1)}e^{(\delta_{\Gamma} - k_2(x,\eta))d(\pi(\mathcal{C}_0),\pi(a_{\log \eta}g))}S_{\ell}(\chi_{\eta,\xi})S_{\ell}(f)\nonumber \\[3pt] &\ll_{\Gamma,\ell} e^{\delta_{\Gamma}(|\log \eta|+\operatorname{height}(g\Gamma))} \eta^{-(\delta_{\Gamma} + {1}/{2}(n-1)(n-2)+1)}\eta^{n-1} {\xi}^{-\ell-(n-1)/2} S_{\ell}(f)\nonumber\\[3pt] &\ll_{\Gamma,\ell} e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)}\eta^{-(2\delta_{\Gamma} + {1}/{2}(n-1)(n-2)+1)}\eta^{n-1} {\xi}^{-\ell-(n-1)/2} S_{\ell}(f)\nonumber\\[3pt] &\ll_{\Gamma,\ell} e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)} \eta^{4n-({1}/{2})n^2-3-2\delta_{\Gamma}}\xi^{-\ell-(n-1)/2}S_{\ell}(f). \end{align} $$

Note that the dependence on $\ell $ arises from the exponential of the Busemann function in the denominator.

Also, using the product structure of $\tilde {m}^{\operatorname {BMS}}$ in (25), we get

$$ \begin{align*} & \frac{1}{\nu(P_{\eta}g)} \int_{P_{\eta} g}\int_{B_U(r_1)} \psi_{\eta,+}(a_{s}u_{\rho_p(\textbf{t})}pg\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\,d\nu(pg)\\[3pt] &\quad= \frac{1}{\nu(P_{\eta}g)}\int_{P_{\eta} g}\int_{B_U(r_1)} \psi_{\eta,+}(a_{s}u_{\rho_p(\textbf{t})}pg\gamma)f(\textbf{t})\frac{d\mu^{\operatorname{PS}}_{g}(\textbf{t})}{d\mu^{\operatorname{PS}}_{pg}(\rho_p (\textbf{t}))}\,d\mu^{\operatorname{PS}}_{pg}(\rho_p (\textbf{t}))\,d\nu(pg)\\[3pt] &\quad\leq\int_G \psi_{\eta,+}(a_{s}h\gamma)\varphi_{\eta,g}(h)\,d\tilde{m}^{\operatorname{BMS}}(h). \end{align*} $$

Step 2.2: Use the exponential mixing assumption.

By defining $\Psi _{\eta ,+}(h\Gamma ) = \sum \nolimits _{\gamma \in \Gamma } \psi _{\eta ,+}(h\gamma )$ and $\Phi _{\eta ,g}(h\Gamma ):=\sum \nolimits _{\gamma \in \Gamma }\varphi _{\eta ,g}(h\gamma ),$ we obtain

$$ \begin{align*} \sum_{\gamma\in\Gamma}\int_G\psi_{\eta,+}(a_sh\gamma)\varphi_{\eta,g}(h)\,d\tilde{m}^{\operatorname{BMS}}(h)\leq\int_{X} \Psi_{\eta,+}(a_s x)\Phi_{\eta,g}(x)\,dm^{\operatorname{BMS}}(x)\end{align*} $$

for any positive s. Note that

(56) $$ \begin{align} S_{\ell'}(\Psi_{\eta,+})=S_{\ell'}(\psi_{\eta,+})\quad\text{and}\quad S_{\ell'}(\Phi_{\eta,g})=S_{\ell'}(\varphi_{\eta,g}). \end{align} $$

In particular, (49) and (55) imply

(57) $$ \begin{align} &S_{\ell'}(\Psi_{\eta,+})\ll_{\operatorname{supp}\psi}\eta^{-2\ell}S_{\ell}(\psi)\quad\text{and}\nonumber\\[3pt] &S_{\ell'}(\Phi_{\eta,g}) \ll_{\Gamma} e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)} \eta^{4n-({1}/{2})n^2-3-2\delta_{\Gamma}}\xi^{-\ell-(n-1)/2}S_{\ell}(f). \end{align} $$

By Assumption 1.1,

$$ \begin{align*} &\int\Psi_{\eta,+}(a_{s}x)\Phi_{\eta,g}(x)\,dm^{\operatorname{BMS}}(x)-m^{\operatorname{BMS}}(\Psi_{\eta,+})m^{\operatorname{BMS}}(\Phi_{\eta,g})\\ &\quad\ll_{\Gamma} S_{\ell'}(\Psi_{\eta,+})S_{\ell'}(\Phi_{\eta,g})e^{-\kappa' s}. \end{align*} $$

Then, by (57), there exists $c_1=c_1(\Gamma ,\operatorname {supp}\psi )$ such that

$$ \begin{align*} &\sum_{\gamma\in\Gamma}\int_{B_U(r_1)} \psi(a_su_{\textbf{t}} g\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_g(\textbf{t})\\[3pt] &\quad<m^{\operatorname{BMS}}(\Psi_{\eta,+})m^{\operatorname{BMS}}(\Phi_{\eta,g})\\[3pt] &\qquad+c_1e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)} \eta^{4n-({1}/{2})n^2-3-2\delta_{\Gamma}}\xi^{-\ell-(n-1)/2}S_{\ell}(\psi)S_{\ell}(f)e^{-\kappa' s}. \end{align*} $$

Step 2.3: Rewrite in terms of $\psi $ and f.

Using Lemma 3.10 and (53), one can calculate

$$ \begin{align*} m^{\operatorname{BMS}}(\Phi_{\eta,g})&=\int_{G}\varphi_{\eta,g}(h)\,d\tilde{m}^{\operatorname{BMS}}(h)\\[3pt] &=\frac{1}{\nu(P_{\eta} g)}\int_{Pg}\int_U\frac{f(\textbf{t})\chi_{\eta,\xi}(p)}{e^{\delta_{\Gamma}\beta_{(u_{\textbf{t}} g)^+}(u_{\textbf{t}} g(o),u_{\rho_p(\textbf{t})}pg(o))}}\,d\mu^{\operatorname{PS}}_{pg}(\rho_p (\textbf{t}))\,d\nu(pg)\\[3pt] &=\frac{1}{\nu(P_{\eta} g)}\int_{Pg}\int_U f(\textbf{t})\chi_{\eta,\xi}(p)\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\,d\nu(pg)\\[3pt] &\leq\frac{\nu(P_{\eta+\xi} g)}{\nu(P_{\eta} g)}\int_{B_U(r_1)}f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t}). \end{align*} $$

Thus, by Theorem 5.2, there exist $\alpha ,\theta ,\omega ,c_0>0$ depending only on $\Gamma $ such that for any $0<\xi <\eta \ll _{\Gamma } e^{-\operatorname {height}(g\Gamma )}$ ,

$$ \begin{align*} m^{\operatorname{BMS}}(\Phi_{\eta,g})&\leq \bigg(1+c_2e^{\omega\operatorname{height}(g\Gamma)} \frac{\xi^{\alpha}}{\eta^{\theta}}\bigg)\int_{B_U(r_1)}f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\\[3pt] &=\bigg(1+c_2e^{\omega\operatorname{height}(g\Gamma)} \frac{\xi^{\alpha}}{\eta^{\theta}}\bigg)\int_{B_U(r_1)}f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\\[3pt] &=\bigg(1+c_2e^{\omega\operatorname{height}(g\Gamma)} \frac{\xi^{\alpha}}{\eta^{\theta}}\bigg)\mu^{\operatorname{PS}}_g(f). \end{align*} $$

Using (50), we get that there exists $c_3=c_3(\Gamma ,\operatorname {supp}\psi )$ such that

$$ \begin{align*} m^{\operatorname{BMS}}(\Psi_{\eta,+})&\leq\int_G\psi_{\eta,+}(g)\,d\tilde{m}^{\operatorname{BMS}}(g)\\[4pt] &<\tilde{m}^{\operatorname{BMS}}(\psi)+c_3\eta S_{\ell}(\psi). \end{align*} $$

To summarize, we have

$$ \begin{align*} &\sum_{\gamma\in\Gamma}\int_{B_U(r_1)}\psi(a_s u_{\textbf{t}} g\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_g(\textbf{t})\\[3pt] &\quad\le \frac{1}{\nu(P_{\eta} g)}\sum_{\gamma\in\Gamma}\int_{P_{\eta} g}\int_{B_U(r_1)} \psi_{\eta,+}(a_{s}u_{\rho_p(\textbf{t})}pg\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_{g}(\textbf{t})\,d\nu(pg)\\[3pt] &\quad\leq\sum_{\gamma\in\Gamma}\int_G \psi_{\eta,+}(a_{s}h\gamma)\varphi_{\eta,g}(h)\,d\tilde{m}^{\operatorname{BMS}}(h)\\ &\quad\leq\int_{X} \Psi_{\eta,+}(a_s x)\Phi_{\eta,g}(x)\,dm^{\operatorname{BMS}}(x)\\[3pt] &\quad<m^{\operatorname{BMS}}(\Psi_{\eta,+})m^{\operatorname{BMS}}(\Phi_{\eta,g})+c_1{\eta}^{4n-({1}/{2})n^2-3-\delta_{\Gamma}-2\ell}{\xi}^{-\ell -(n-1)/2}S_{\ell}(\psi)S_{\ell}(f)e^{-\kappa' s}\\[3pt] &\quad<(\tilde{m}^{\operatorname{BMS}}(\psi)+c_3\eta S_{\ell}(\psi))\bigg(\bigg(1+c_2e^{\omega\operatorname{height}(g\Gamma)} \frac{\xi^{\alpha}}{\eta^{\theta}}\bigg)\mu^{\operatorname{PS}}_g(f)\bigg)\\[3pt] & \qquad+c_1e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)} \eta^{4n-({1}/{2})n^2-3-2\delta_{\Gamma}}\xi^{-\ell-(n-1)/2}S_{\ell}(\psi)S_{\ell}(f)e^{-\kappa' s}. \end{align*} $$

It follows from the proof of Lemma 3.12 that $\tilde {m}^{\operatorname {BMS}}(\psi )\ll _{\operatorname {supp}\psi }S_{\ell }(\psi )$ and $\mu ^{\operatorname {PS}}_g(f)\ll S_{\ell }(f)\mu ^{\operatorname {PS}}_g(B_U(1))$ . Then, using Proposition 3.4, we arrive at

$$ \begin{align*} &\sum_{\gamma\in\Gamma}\int_{B_U(r_1)}\psi(a_s u_{\textbf{t}} g\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_g(\textbf{t})- \mu^{\operatorname{PS}}_g(f)\tilde{m}^{\operatorname{BMS}}(\psi)\\ &\quad\ll_{\Gamma}\bigg(e^{\omega\operatorname{height}(g\Gamma)} \frac{\xi^{\alpha}}{\eta^{\theta}} +e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)} \eta^{4n-({1}/{2})n^2-3-2\delta_{\Gamma}}\xi^{-\ell-(n-1)/2}e^{-\kappa' s}\bigg)\\ &\qquad\cdot S_{\ell}(\psi)S_{\ell}(f)\mu^{\operatorname{PS}}_g(B_U(1)). \end{align*} $$

Define

$$ \begin{align*}\kappa = \frac{3\alpha\theta\kappa'}{2\theta(2\ell+n-1)+9\alpha(2\delta_{\Gamma}+3+n^2/2-4n)},\end{align*} $$

and note that by making $\ell $ larger if necessary, we guarantee $\kappa>0$ .

Recall from (22) that

$$ \begin{align*}e^{-\operatorname{height}(g\Gamma)}\ll_{\Gamma}\operatorname{inj}(g).\end{align*} $$

For $s \ge \max \{\theta ,\omega \}\operatorname {height}(g\Gamma ) /\kappa $ , choose

(58) $$ \begin{align} \eta=e^{-\kappa s/\theta},\quad\xi=e^{-{4\kappa s}/{\alpha} }. \end{align} $$

Note that $\eta < \operatorname {inj}(g\Gamma )$ by choice of s, $\omega \operatorname {height}(g\Gamma )\le \kappa s$ , and $\xi <\eta $ since by Proposition 5.2, $\alpha <\theta $ . By Proposition 5.2, we have $\omega>\delta _{\Gamma }$ ; therefore, $\delta _{\Gamma } \operatorname {height}(g\Gamma )\le \kappa s$ . Note also that $\max \{\theta ,\omega \}\operatorname {height}(g\Gamma ) /\kappa \ll _{\Gamma } \operatorname {height}(g\Gamma )$ .

With these choices, we obtain

(59) $$ \begin{align}e^{\omega\operatorname{height}(g\Gamma)}\bigg(\frac{\xi}{\eta^{\theta'}}\bigg)^{\alpha'} +e^{\delta_{\Gamma} \operatorname{height}(g\Gamma)} \eta^{4n-({1}/{2})n^2-3-2\delta_{\Gamma}}\xi^{-\ell-(n-1)/2}e^{-\kappa' s} \le 2e^{-2\kappa s}. \end{align} $$

In a similar way, using $\psi _{\eta ,-}$ , one can show a lower bound, proving (52).

Step 3: Covering argument for general f.

We now deduce the claim by decomposing f into a sum of functions, each defined on a ball of radius $r_1$ in U.

Let $u_1,\ldots ,u_k$ and $\sigma _1,\ldots ,\sigma _k\in C_c^{\infty }(B_U(r))$ satisfy the conclusion of Lemma 3.11 for $E=B_U(r)$ and $r_1$ . For $1\leq i\leq k$ , let

$$ \begin{align*} f_i:=f\sigma_i.\end{align*} $$

Then, $f\le \sum _{i-1}^k f_i$ , and by Lemma 3.11 and [Reference Kleinbock and Margulis17, Lemma 2.4.7(a)],

(60) $$ \begin{align} S_{\ell}(f_i)\ll_{\Gamma} S_{\ell}(f)S_{\ell}(\sigma_i)\ll_{\Gamma} r_1^{-\ell+n-1} S_{\ell}(f). \end{align} $$

Since each $f_i$ is supported on $B_U(r_1)u_i$ for some $u_i\in B_U(1)$ , by (52), we have

$$ \begin{align*} & \sum_{\gamma\in\Gamma}\int_{B_U(r_1)}\psi(a_s u_{\textbf{t}} g\gamma)f_i(\textbf{t})\,d\mu^{\operatorname{PS}}_g(\textbf{t})-\tilde{m}^{\operatorname{BMS}}(\psi)\mu^{\operatorname{PS}}_g(f_i)\\ &\quad \ll_{\Gamma,\operatorname{supp}\psi}\mu^{\operatorname{PS}}_g(B_U(1)) S_{\ell}(\psi)S_{\ell}(f_i)e^{-2\kappa s}. \end{align*} $$

Summing the above expressions for $i=1,\ldots ,k$ , we get

$$ \begin{align*} &\sum_{\gamma\in\Gamma}\int_{B_U(r)}\psi(a_s u_{\textbf{t}} g\gamma)f(\textbf{t})\,d\mu^{\operatorname{PS}}_g(\textbf{t})-\tilde{m}^{\operatorname{BMS}}(\psi)\mu^{\operatorname{PS}}_g(f)\\[3pt] &\quad \ll k r_1^{-\ell+n-1}S_{\ell}(\psi)S_{\ell}(f)e^{-2\kappa s}\mu^{\operatorname{PS}}_g(B_U(1))\\[3pt] &\quad \ll\bigg(\frac{r}{r_1}\bigg)^{n-1} r_1^{-\ell+n-1}S_{\ell}(\psi)S_{\ell}(f)e^{-2\kappa s}\mu^{\operatorname{PS}}_g(B_U(1))\\[3pt] &\quad \ll r_1^{-\ell}S_{\ell}(\psi)S_{\ell}(f)e^{-2\kappa s}\mu^{\operatorname{PS}}_g(B_U(1))\\[3pt] &\quad \ll S_{\ell}(\psi)S_{\ell}(f)e^{-\kappa s}\mu^{\operatorname{PS}}_g(B_U(1)), \end{align*} $$

where the first inequality is by Lemma 3.11, the second inequality follows from $r_1=\operatorname {inj}(g)>e^{-\kappa s/\ell }$ , the third is by (58) and because $r<1$ , and the implied constants depend on $\Gamma $ and $\operatorname {supp}\psi $ .

As before, using similar arguments, one can show a lower bound, proving the claim.

Proof. According to [Reference Oh and Shah26, Lemma 2.17], there exists an admissible box $B_y$ around y for any $y\in X$ . Then, $\{B_y : y\in \operatorname {supp}\psi \}$ is an open cover of the compact set $\operatorname {supp}\psi $ . Hence, there exists a minimal sub-cover $B_{y_1},\ldots ,B_{y_k}$ . Using a similar construction to that in Lemma 3.11, there exist $\sigma _1,\ldots , \sigma _k$ , a partition of unity for $\operatorname {supp}\psi $ , such that for $i=1,\ldots ,k$ , we have $\sigma _i \in C_c^{\infty }(B_{y_i})$ and for $i=1,\ldots ,k$ and $m=1,\ldots ,\ell $ ,

(61) $$ \begin{align} |\sigma_i^{(m)}|\ll_{\operatorname{supp}\psi,\Gamma} 1 \end{align} $$

(the implied constant depends on the chosen sub-cover).

Define $\psi _i=\psi \sigma _i$ . Then

(62) $$ \begin{align} \psi=\sum_{i=1}^k\psi_i, \end{align} $$

and by (61) and the product rule, we have

(63) $$ \begin{align} S_{\ell}(\psi_i)\ll_{\operatorname{supp}\psi,\Gamma}S_{\ell}(\psi). \end{align} $$

According to Proposition 6.1 and Proposition 3.4, there exist $c=c(\Gamma ,\operatorname {supp}\psi )>0$ , $\unicode{x3bb} =\unicode{x3bb} (\Gamma )>1$ such that for $s\gg _{\Gamma } \operatorname {height}(x)$ ,

$$ \begin{align*} &\int_{B_U(r)} \psi(a_s u_{\textbf{t}} x)f(\textbf{t})d\textbf{t}\\ &\quad= \sum\limits_{i=1}^k \int_{B_U(r)} \psi_i(a_s u_{\textbf{t}} x)f(\textbf{t})\,d\textbf{t} \\ &\quad \le \sum\limits_{i=1}^k m^{\operatorname{BMS}}(\psi_i) \mu^{\operatorname{PS}}_x(f) + c S_{\ell}(\psi_i)S_{\ell}(f)e^{-\kappa s}\mu^{\operatorname{PS}}_x(B_U(1))\\ &\quad \le \sum\limits_{i=1}^k m^{\operatorname{BMS}}(\psi_i) \mu^{\operatorname{PS}}_x(f) + c \unicode{x3bb} S_{\ell}(\psi_i)S_{\ell}(f)e^{-\kappa s+(n-1-\delta_{\Gamma})d(\pi(\mathcal{C}_0),\pi(x))}\\ &\quad\ll_{\Gamma,\operatorname{supp}\psi} m^{\operatorname{BMS}}(\psi) \mu^{\operatorname{PS}}_x(f) + c \unicode{x3bb} S_{\ell}(\psi)S_{\ell}(f)e^{-\kappa s+(n-1-\delta_{\Gamma})\operatorname{height}(x)}, \end{align*} $$

where the last line follows by the definition of $\operatorname {height}(x)$ , and (62) and (63). Moreover, we may assume that $s \ge {2(n-1-\delta _{\Gamma })}/{\kappa }\operatorname {height}(x)$ without changing the assumption $s\gg _{\Gamma } \operatorname {height}(x)$ . Then,

$$ \begin{align*}e^{-\kappa s+(n-1-\delta_{\Gamma})\operatorname{height}(x)} \ll_{\Gamma} e^{-\kappa s/2 },\end{align*} $$

as desired.

Proof. Step 1: Setup and approximations. Assume $s \gg _{\Gamma } \operatorname {height}(x)$ , and let $\kappa ,\ell '$ satisfy the conclusion of Theorem 1.5 and $\ell>\ell '$ satisfy the conclusion of Lemma 3.12.

Since $\psi $ is assumed to be supported on an admissible box, there exist $r_0,\eta ,\varepsilon _0,\varepsilon _1>0$ (depending only on $\operatorname {supp} \psi $ ) and $z\in X$ such that

$$ \begin{align*}\operatorname{supp}{\psi}=B_U(r_0) P_{\eta}z,\end{align*} $$

and

$$ \begin{align*}G_{\varepsilon_0}\operatorname{supp}\psi\subset B_U(r_0+\varepsilon_1) P_{\eta+\varepsilon_1}z,\end{align*} $$

where $B_U(r+\varepsilon _1) P_{\eta +\varepsilon _1}z$ is also an admissible box. Denote $\eta '=\eta +\varepsilon _1$ and $r^{\prime }_0=r_0+\varepsilon _1$ .

Without loss of generality, assume that f is a non-negative function. Continuously extend $\psi $ to $P_{\eta '}$ by defining $\psi =0$ on $P_{\eta '}\setminus P_{\eta }$ .

For $0<\varepsilon <\varepsilon _0$ , let $\psi _{\varepsilon ,\pm }$ and $f_{\varepsilon ,\pm }$ for Lemma 3.12 for $G,\varepsilon ,\psi $ and $U,\varepsilon ,f$ respectively. By Lemma 3.12,

(64) $$ \begin{align} S_{\ell'}(\psi_{\varepsilon,\pm})\ll_{\Gamma,\operatorname{supp}(\psi)} \varepsilon^{-2\ell}S_{\ell}(\psi)\quad\text{and}\quad S_{\ell'}(f_{\varepsilon,\pm})\ll_{\Gamma}\varepsilon^{-2\ell}S_{\ell}(f). \end{align} $$

Moreover, by Lemma 3.12(2),

(65) $$ \begin{align} \|f_{\varepsilon,\pm}-f\|_{\infty} \le \varepsilon S_{\ell}(f).\end{align} $$

For $p \in P_{\eta '}$ , define

(66) $$ \begin{align} \varphi(p):= \mu^{\operatorname{PS}}_{pz}(B_U(r^{\prime}_0)pz). \end{align} $$

Step 1.1: Construct a smooth approximation to $1/\varphi $ . Since the Busemann function is smooth and $\varphi $ is bounded below by a positive quantity on $P_{\eta '}$ by Corollary 3.3, the mean value theorem implies that for any $0<\varepsilon <\varepsilon _0$ and all $p, p' \in P_{\varepsilon },$ there exists a constant $d = d(\Gamma ,\operatorname {supp}\psi )$ such that

(67) $$ \begin{align} \bigg|\frac{1}{\varphi(p)} - \frac{1}{\varphi(p')}\bigg| \le \frac{d\varepsilon}{\varphi(p)}.\end{align} $$

By Lemma 3.10, for any $\xi>0$ , there exists a non-negative smooth function $\chi _{\xi }$ with

(68) $$ \begin{align} 1_{P_{\varepsilon}-\xi}\le \chi_{\xi} \le {\mathbf{1}}_{P_{\varepsilon}}\end{align} $$

and $S_{\ell '}(\chi _{\xi }) \ll _{\Gamma ,n} (\varepsilon -\xi /2)^{n-1}(\xi /2)^{-\ell '-(n-1)/2}.$ Define

(69) $$ \begin{align}\sigma(p):= \frac{1}{\varphi}\ast \frac{\chi_{\xi}}{m(P_{\varepsilon-\xi})},\end{align} $$

where m denotes the probability Haar measure on P. Then, assuming $\varepsilon _0<1/2$ and $\xi \le \varepsilon ^2$ , by (67), (68), and (69), we have that

(70) $$ \begin{align} \frac{1-d\varepsilon}{\varphi(p)} & \le\frac{1}{m(P_{\varepsilon-\xi})}\int_{pP_{\varepsilon - \xi}}\frac{1}{\varphi(p')}\,dp' \end{align} $$

(71) $$ \begin{align} & \kern30pt \le \sigma(p) \nonumber\\ & \kern30pt \le \frac{1}{m(P_{\varepsilon-\xi})}\int_{pP_{\varepsilon}}\frac{1+d\varepsilon}{\varphi(p')}\,dp' \nonumber \\ & \le \bigg(\frac{\varepsilon}{\varepsilon-\xi}\bigg)^n\frac{1+d\varepsilon}{\varphi(p)}\nonumber\\[3pt] & \le\frac{1+d'\varepsilon}{\varphi(p)}, \end{align} $$

for some absolute constant $d'>0$ .

For $upz \in B_U(r^{\prime }_0) P_{\eta '} z$ and $0<\varepsilon <\varepsilon _0$ , let

$$ \begin{align*} \Psi_{\varepsilon,\pm}(upz)=\sigma(p)\int_{Upz}\psi_{c_1\varepsilon,\pm}(u_{\textbf{t}}pz)\,d\textbf{t}.\end{align*} $$

Then, by (67),

(72) $$ \begin{align} \sup_{w\in G_{\varepsilon}}\Psi_{\varepsilon,\pm}(wupz) &=\sup_{w\in P_{\varepsilon}}\sigma(wp)\int_{Uwpz}\psi_{c_1\varepsilon,+}(u_{\textbf{t}}wpz)\,d\textbf{t}\nonumber\\[3pt] &\le(1+d'\varepsilon)\Psi_{2\varepsilon,\pm}. \end{align} $$

Step 2: Bounding with PS measure.

Let

$$ \begin{align*}P(f,\psi, x; s) = \{p \in P_{\eta} : a_s\operatorname{supp}(f)x \cap B_U(r_0)pz \ne \emptyset\}.\end{align*} $$

By [Reference Mohammadi and Oh23, Lemma 6.2], there exists an absolute constant $c_1>0$ such that

(73) $$ \begin{align} &e^{(n-1)s}\int_{B_U(r)}\psi(a_s u_{\textbf{t}} x)f(\textbf{t})\,d\textbf{t}\\[3pt] &\quad\le(1+c_1\varepsilon)\sum_{p\in P(f,\psi,x;s)}f_{c_1e^{-s}\eta}(a_{-s}pz)\int_{Upz}\psi_{c_1\varepsilon,+}(u_{\textbf{t}}pz)\,d\textbf{t}.\nonumber \end{align} $$

It now follows from [Reference Mohammadi and Oh23, Lemma 6.5], (70), and (72) that there exists an absolute constant $c_2>0$ such that

$$ \begin{align*} &e^{-\delta_{\Gamma} s}\sum_{p\in P(f,\psi,x;s)}f_{c_1e^{-s}\eta}(a_{-s}pz)\int_{Upz}\psi_{c_1\varepsilon,+}(u_{\textbf{t}}pz)\,d\textbf{t}\\[3pt] &\quad\le\frac{(1+c_2\varepsilon)(1+d'\varepsilon)}{1-d\varepsilon}\int_U\Psi_{2c_2\varepsilon,+}(a_{s}u_{\textbf{t}}x)f_{(c_1+c_2)e^{-s}\varepsilon_0,+}(\textbf{t})\,d\mu^{\operatorname{PS}}_{x}(\textbf{t}). \end{align*} $$

Note that (70) is needed because our definition of $\Psi _{\varepsilon , +}$ is not identical to $\Psi $ as defined in [Reference Mohammadi and Oh23, Lemma 6.5]. The latter is bounded above by ${1}/({1-d\varepsilon }) \Psi _{\varepsilon ,+}$ by (70).

Combining the above with (73), we get that there exist constants $c_3,c_4=c_4(\Gamma ,\operatorname {supp}\psi )>0$ such that

$$ \begin{align*} &e^{(n-1-\delta_{\Gamma})s}\int_{B_U(r)}\psi(a_s u_{\textbf{t}} x)f(\textbf{t})\,d\textbf{t}\\[3pt] &\quad\leq(1+c_4\varepsilon)\int_U\Psi_{c_3\varepsilon,+}(a_{s}u_{\textbf{t}}x)f_{c_3e^{-s}\varepsilon_0,+}(\textbf{t})\,d\mu^{\operatorname{PS}}_{x}(\textbf{t}). \end{align*} $$

It follows from Theorem 1.5 that for some constant $c_5=c_5(\Gamma ,\operatorname {supp}\psi )>0$ ,

(74) $$ \begin{align} &e^{(n-1-\delta_{\Gamma})s}\int_{B_U(r)}\psi(a_s u_{\textbf{t}} x)f(\textbf{t})\,d\textbf{t}\nonumber\\[3pt] &\quad\leq(1+c_4\varepsilon)(\mu^{\operatorname{PS}}_x(f_{c_3e^{-s}\varepsilon_0,+})m^{\operatorname{BMS}}(\Psi_{c_3\varepsilon,+})+c_5S_{\ell'}(\Psi_{c_3\varepsilon,+})S_{\ell'}(f_{c_3e^{-s}\varepsilon_0,+})e^{-\kappa s}). \end{align} $$

Step 3: Bounding the error terms. We now show how to bound the various error terms to obtain the desired conclusion.

To compute $m^{\operatorname {BMS}}(\Psi )$ , we use (25), (65), and (71) to deduce that for some $c_6=c_6(\Gamma ,\operatorname {supp}\psi )$ , if $\xi = \varepsilon ^2$ ,

(75) $$ \begin{align} &m^{\operatorname{BMS}}(\Psi_{c_3\varepsilon,+})\nonumber\\[3pt] &\quad\le (1+d'\varepsilon)\!\int_{P_{\eta'} z} \int_{B_U(r^{\prime}_0)} \frac{1}{\mu^{\operatorname{PS}}_{pz}(B_U(r^{\prime}_0)pz)} \int_{B_U(r^{\prime}_0)pz}\psi_{c_1\varepsilon,\pm}(u_{\textbf{t}}pz)\,d\textbf{t} \,d\mu^{\operatorname{PS}}_{pz}(\textbf{t})\,d\nu(pz)\nonumber\\[3pt] &\quad\le (1+d'\varepsilon)\int_{P_{\eta'} z} \int_{B_U(r^{\prime}_0)pz}\psi_{c_1\varepsilon,\pm}(u_{\textbf{t}}pz)\,d\textbf{t} \,d\nu(pz)\nonumber\\[3pt] &\quad\le (1+d'\varepsilon)(m^{\operatorname{BR}}(\psi)+c_6\varepsilon S_{\ell}(\psi)). \end{align} $$

By Proposition 3.4, if s is sufficiently large so that $r+c_3e^{-s}\varepsilon _0\le 1$ (note that this requirement on s depends only on $\Gamma $ and $\operatorname {supp}\psi $ ), we have that

(76) $$ \begin{align} \mu^{\operatorname{PS}}_x(B_U(r+c_3e^{-s}\varepsilon_0))\leq\mu^{\operatorname{PS}}_x(B_U(1))\ll_{\Gamma} e^{(n-1-\delta_{\Gamma})d(\pi(\mathcal{C}_0),\pi(x))}. \end{align} $$

Hence, by (65) and (76), we have

(77) $$ \begin{align} \mu^{\operatorname{PS}}_x(f_{c_3e^{-s}\varepsilon_0,+})-\mu^{\operatorname{PS}}_x(f)&\ll_{\Gamma}e^{- s}\varepsilon_0 S_{\ell}(f)\mu^{\operatorname{PS}}_x(B_U(r+c_3e^{-s}\varepsilon_0)) \nonumber \\[3pt] &\ll_{\Gamma} e^{-s}\varepsilon_0 S_{\ell}(f) e^{(n-1-\delta_{\Gamma})d(\pi(\mathcal{C}_0),\pi(x))}. \end{align} $$

According to [Reference Kleinbock and Margulis17, Lemma 2.4.7(a)] and (64), if $\xi = \varepsilon ^2$ and

(78) $$ \begin{align}\varepsilon = e^{-{\kappa s}/{2(n+4\ell)}},\end{align} $$

then

(79) $$ \begin{align} S_{\ell'}(\Psi_{c_3\varepsilon,+}) &\ll_{\Gamma} S_{\ell'}(\psi_{c_3\varepsilon,+})S_{\ell'}(\sigma)\nonumber\\[3pt] &\ll_{\Gamma} (m(P_{\varepsilon-\xi}))^{-1} (\varepsilon-\xi/2)^{n-1}\xi^{-{\ell'}-(n-1)/2}\varepsilon^{-2\ell}S_{\ell}(\psi)\nonumber\\[3pt] &\ll \varepsilon^{-1-2\ell}\xi^{-{\ell'}-(n-1)/2}S_{\ell}(\psi) \nonumber\\[3pt] &\le e^{\kappa s/2} S_{\ell}(\psi). \end{align} $$

Using (74), (75), (77), and (79), we obtain

(80) $$ \begin{align} &e^{(n-1-\delta_{\Gamma})s}\int_{B_U(r)}\psi(a_su_{\textbf{t}} x)f(\textbf{t})\,d\textbf{t} - \mu^{\operatorname{PS}}_x(f)m^{\operatorname{BR}}(\psi)\nonumber\\[3pt] &\quad\le (1+c_4\varepsilon) [d'\varepsilon \mu^{\operatorname{PS}}_x(f)m^{\operatorname{BR}}(\psi) + (1+d'\varepsilon)\{ c_6 \varepsilon\mu^{\operatorname{PS}}_x(f)S_{\ell}(\psi)\nonumber\\[3pt] &\qquad+(e^{- s}\varepsilon_0 m^{\operatorname{BR}}(\psi)S_{\ell}(f) + c_6 e^{- s}\varepsilon_0 \varepsilon S_{\ell}(f) S_{\ell}(\psi))e^{(n-1-\delta_{\Gamma})d(\pi(\mathcal{C}_0),\pi(x))}\}\nonumber\\[3pt] &\qquad+ c_8 S_{\ell}(\psi)S_{\ell}(f)e^{-\kappa s/2} ]. \end{align} $$

These remaining error terms can be controlled as follows. Using (76), we can deduce

(81) $$ \begin{align} \mu^{\operatorname{PS}}_x(f)\le\Vert f\Vert _{\infty}\mu^{\operatorname{PS}}_x(B_U(r))\ll_{\Gamma}S_{\ell}(f)e^{(n-1-\delta_{\Gamma})(\pi(\mathcal{C}_0),\pi(x))}. \end{align} $$

We also have that

(82) $$ \begin{align}m^{\operatorname{BR}}(\psi) \ll_{\Gamma, \operatorname{supp}\psi} S_{\ell}(\psi).\end{align} $$

Combining (80), (81), and (82) implies