Note that
\({\textbf{y}}_j, {\hat{\textbf{y}}}_j\)’s are in
\([-\lambda , \lambda ]^2\) with
\(\lambda = \frac{(2n-1)\pi }{12\Omega }\) and
\(\hat{\mu }= \sum _{j=1}^n {\hat{a}}_j \delta _{{\hat{\textbf{y}}}_j}\) is a
\(\sigma \)-admissible measure of
\({\textbf{Y}}\). Let
\(\tau = \frac{\Omega }{2n-1}\). Similarly to the proof in the above section, we can construct
\({\textbf{x}}_j = {\textbf{y}}_j+{\textbf{v}}, {\hat{\textbf{x}}}_j = {\hat{\textbf{y}}}_j+{\textbf{v}}\) so that
\(\tau {\hat{\textbf{x}}}_j, \tau {\textbf{x}}_j \in [-\frac{\pi }{12},\frac{\pi }{12}]\times [\frac{5\pi }{12}, \frac{7\pi }{12}]\) and
$$\begin{aligned} \left| {\sum _{j=1}^n {\hat{a}}_j e^{i {\hat{\textbf{x}}}_j^\top \varvec{\omega }} - \sum _{j=1}^n a_j e^{i {\textbf{x}}_j^\top {\varvec{\omega }}}}\right| < 2\sigma , \quad \varvec{\omega } \in [0, \Omega ]^2. \end{aligned}$$
(3.7)
Thus we have
$$\begin{aligned} -\frac{\pi }{12}\le & {} \tau {\textbf{x}}_{j,1}\le \frac{\pi }{12},\quad \frac{5\pi }{12}\le \tau {\textbf{x}}_{j,2}\le \frac{7\pi }{12}, \quad \frac{\pi }{3}\le \tau {\textbf{x}}_{j,2}- \tau {\textbf{x}}_{j,1} \le \frac{2\pi }{3},\nonumber \\ \end{aligned}$$
(3.8)
$$\begin{aligned} -\frac{\pi }{12}\le & {} \tau {\hat{\textbf{x}}}_{j,1}\le \frac{\pi }{12},\quad \frac{5\pi }{12}\le \tau {\hat{\textbf{x}}}_{j,2}\le \frac{7\pi }{12}, \quad \frac{\pi }{3}\le \tau {\hat{\textbf{x}}}_{j,2}- \tau {\hat{\textbf{x}}}_{j,1} \le \frac{2\pi }{3}.\nonumber \\ \end{aligned}$$
(3.9)
Moreover, it follows that
$$\begin{aligned} -\frac{\pi }{12}\le & {} \tau {\textbf{x}}_{j,1}\le \frac{\pi }{12},\quad \frac{-7\pi }{12}\le \tau {\textbf{x}}_{j,2}-\pi \le \frac{-5\pi }{12}, \quad \frac{\pi }{3}\nonumber \\\le & {} \tau {\textbf{x}}_{j,1}- (\tau {\textbf{x}}_{j,2}-\pi ) \le \frac{2\pi }{3}, \end{aligned}$$
(3.10)
$$\begin{aligned} -\frac{\pi }{12}\le & {} \tau {\hat{\textbf{x}}}_{j,1}\le \frac{\pi }{12},\quad \frac{-7\pi }{12}\le \tau {\hat{\textbf{x}}}_{j,2}-\pi \le \frac{-5\pi }{12}, \quad \frac{\pi }{3}\nonumber \\\le & {} \tau {\hat{\textbf{x}}}_{j,1}- (\tau {\hat{\textbf{x}}}_{j,2}-\pi ) \le \frac{2\pi }{3}. \end{aligned}$$
(3.11)
Let
\({\hat{d}}_j = e^{i\tau {\hat{\textbf{x}}}_{j,1}}+e^{i\tau {\hat{\textbf{x}}}_{j,2}}, d_j = e^{i\tau {\textbf{x}}_{j,1}}+e^{i\tau {\textbf{x}}_{j,2}}\) and
\({\hat{g}}_j = e^{i\tau {\hat{\textbf{x}}}_{j,1}}+e^{i(\tau {\hat{\textbf{x}}}_{j,2}-\pi )}, g_j = e^{i\tau {\textbf{x}}_{j,1}}+e^{i(\tau {\textbf{x}}_{j,2}-\pi )}\). By (
3.7) and Lemma
3.1, we arrive at
$$\begin{aligned}&\left| {\sum _{j=1}^n {\hat{a}}_j {\hat{d}}_j^{t} - \sum _{j=1}^n a_j d_j^{t}}\right| < 2^{t+1} \sigma , \quad t=0,1, \ldots , 2n-1, \end{aligned}$$
(3.12)
$$\begin{aligned}&\left| {\sum _{j=1}^n {\hat{a}}_j {\hat{g}}_j^{t} - \sum _{j=1}^n a_j g_j^{t}}\right| < 2^{t+1} \sigma , \quad t =0, 1, \ldots , 2n-1. \end{aligned}$$
(3.13)
Let
$$\begin{aligned} {\textbf{d}}= & {} \Bigg (\sum _{j=1}^n {\hat{a}}_j {\hat{d}}_j^{0} - \sum _{j=1}^n a_j d_j^{0}, \quad \sum _{j=1}^n {\hat{a}}_j {\hat{d}}_j^{1} - \sum _{j=1}^n a_j d_j^{1}, \quad \ldots ,\\{} & {} \quad \sum _{j=1}^n {\hat{a}}_j {\hat{d}}_j^{2n-1} - \sum _{j=1}^n a_j d_j^{2n-1}\Bigg )^\top , \end{aligned}$$
and
$$\begin{aligned} {\textbf{g}}= & {} \Bigg (\sum _{j=1}^n {\hat{a}}_j {\hat{g}}_j^{0} - \sum _{j=1}^n a_j g_j^{0}, \quad \sum _{j=1}^n {\hat{a}}_j {\hat{g}}_j^{1} - \sum _{j=1}^n a_j g_j^{1}, \quad \ldots ,\\{} & {} \quad \sum _{j=1}^n {\hat{a}}_j {\hat{g}}_j^{2n-1} - \sum _{j=1}^n a_j g_j^{2n-1}\Bigg )^\top . \end{aligned}$$
Equations (
3.12) and (
3.13) imply respectively
$$\begin{aligned} ||{\textbf{d}}||_2< \frac{2^{2n+1} \sigma }{\sqrt{3}}, \quad ||{\textbf{g}}||_2 < \frac{2^{2n+1} \sigma }{\sqrt{3}}. \end{aligned}$$
Note also that by (
3.8), (
3.9), (
3.10), and (
3.11), we get
$$\begin{aligned} |{\hat{d}}_j|, |d_j|, |{\hat{g}}_j|, |g_j|\le \sqrt{3},\quad j=1, \ldots , n. \end{aligned}$$
Define
\(d_{\min }:=\min _{p\ne q}\left| {d_p - d_q}\right| \) and
\(g_{\min }:=\min _{p\ne q}\left| {g_p - g_q}\right| \). Applying Theorem
6.2, we thus have that
$$\begin{aligned} \Big |\Big |\eta _{n,n}(d_1,\ldots , d_n, {\hat{d}}_1, \ldots , {\hat{d}}_n)\Big |\Big |_{\infty }<\frac{(1+\sqrt{3})^{2n-1}}{ d_{\min }^{n-1}} \frac{2^{2n+1} }{\sqrt{3}}\frac{\sigma }{m_{\min }}, \end{aligned}$$
(3.14)
and
$$\begin{aligned} \Big |\Big |\eta _{n,n}(g_1,\ldots , g_n, {\hat{g}}_1, \ldots , {\hat{g}}_n)\Big |\Big |_{\infty }<\frac{(1+\sqrt{3})^{2n-1}}{ g_{\min }^{n-1}} \frac{2^{2n+1} }{\sqrt{3}}\frac{\sigma }{m_{\min }}, \end{aligned}$$
(3.15)
where
\(\eta _{n,n}(\ldots )\)’s are vectors defined as in (
6.10). We now demonstrate that we can reorder
\({\hat{d}}_j, {\hat{g}}_j\) to have
\(|{\hat{d}}_j -d_j|< \frac{d_{\min }}{2}\) and
\(|{\hat{g}}_j -g_j|< \frac{g_{\min }}{2}, j=1, \ldots , n\). First, since (
3.8) and (
3.10) hold, by Lemma
3.2 we have
$$\begin{aligned} d_{\min }\ge \frac{3}{2\pi } \min _{p\ne q}\tau \Big |\Big |{\textbf{y}}_p - {\textbf{y}}_q\Big |\Big |_1 \ge 11.475 \Big (\frac{\sigma }{m_{\min }}\Big )^{\frac{1}{2n-1}}> 2^{3/2}(1+\sqrt{3})\Big (\frac{2^{5/2}}{\sqrt{3}} \frac{\sigma }{m_{\min }}\Big )^{\frac{1}{2n-1}}, \end{aligned}$$
(3.16)
and
$$\begin{aligned} g_{\min }\ge \frac{3}{2\pi } \min _{p\ne q}\tau \Big |\Big |{\textbf{y}}_p - {\textbf{y}}_q\Big |\Big |_1 \ge 11.475 \Big (\frac{\sigma }{m_{\min }}\Big )^{\frac{1}{2n-1}}> 2^{3/2}(1+\sqrt{3})\Big (\frac{2^{5/2}}{\sqrt{3}} \frac{\sigma }{m_{\min }}\Big )^{\frac{1}{2n-1}}, \end{aligned}$$
(3.17)
where we also use separation condition (
2.6) in the above derivation. Let
$$\begin{aligned} \epsilon _d = \frac{(1+\sqrt{3})^{2n-1}}{ d_{\min }^{n-1}} \frac{2^{2n+1}}{\sqrt{3}}\frac{\sigma }{m_{\min }},\quad \epsilon _g = \frac{(1+\sqrt{3})^{2n-1}}{ g_{\min }^{n-1}}\frac{2^{2n+1} }{\sqrt{3}}\frac{\sigma }{m_{\min }}. \end{aligned}$$
By (
3.16),
$$\begin{aligned} d_{\min }^{2n-1}\ge \frac{(1+\sqrt{3})^{2n-1}2^{3n+1}}{\sqrt{3}} \frac{\sigma }{m_{\min }}, \ \text {or equivalently}, d_{\min }^{n}\ge 2^n \epsilon _d, \end{aligned}$$
and by (
3.17),
$$\begin{aligned} g_{\min }^{2n-1}\ge \frac{(1+\sqrt{3})^{2n-1}2^{3n+1}}{\sqrt{3}} \frac{\sigma }{m_{\min }}, \ \text {or equivalently}, g_{\min }^{n}\ge 2^n \epsilon _g. \end{aligned}$$
Thus the conditions of Lemma
6.8 are satisfied. By Lemma
6.8, we have that after reordering
\({\hat{d}}_j, {\hat{g}}_j\),
$$\begin{aligned} \Big |{\hat{d}}_j- d_j\Big |< \frac{d_{\min }}{2},\quad \Big |{\hat{g}}_j- g_j\Big |< \frac{g_{\min }}{2}, \end{aligned}$$
and
$$\begin{aligned}&\left| {{\hat{d}}_j -d_j}\right| \le \Big (\frac{2}{d_{\min }}\Big )^{n-1}\epsilon _d = \Big (\frac{1}{d_{\min }}\Big )^{2n-2}\frac{(1+\sqrt{3})^{2n-1}2^{3n}}{\sqrt{3}}\frac{\sigma }{m_{\min }},\\&\left| {{\hat{g}}_j -g_j}\right| \le \Big (\frac{2}{g_{\min }}\Big )^{n-1}\epsilon _g= \Big (\frac{1}{g_{\min }}\Big )^{2n-2} \frac{(1+\sqrt{3})^{2n-1}2^{3n}}{\sqrt{3}}\frac{\sigma }{m_{\min }}. \end{aligned}$$
Observing
$$\begin{aligned} e^{i\tau {\hat{\textbf{x}}}_{j,1}}- e^{i\tau {\textbf{x}}_{j, 1}}= & {} \frac{1}{2} \Big ({\hat{d}}_j -d_j+ {\hat{g}}_j -g_j\Big ),\nonumber \\ e^{i\tau {\hat{\textbf{x}}}_{j,2}}- e^{i \tau {\textbf{x}}_{j, 2}}= & {} \frac{1}{2} \Big ({\hat{d}}_j -d_j- ({\hat{g}}_j -g_j)\Big ), \end{aligned}$$
(3.18)
we conclude that
$$\begin{aligned} \Big |e^{i\tau {\hat{\textbf{x}}}_{j,1}}- e^{i\tau {\textbf{x}}_{j, 1}}\Big |+\Big |e^{i\tau {\hat{\textbf{x}}}_{j,2}}- e^{i\tau {\textbf{x}}_{j, 2}}\Big |\le & {} \Big (\Big (\frac{1}{d_{\min }}\Big )^{2n-2} + \Big (\frac{1}{g_{\min }}\Big )^{2n-2}\Big )\nonumber \\{} & {} \frac{(1+\sqrt{3})^{2n-1}2^{3n}}{\sqrt{3}}\frac{\sigma }{m_{\min }}. \end{aligned}$$
(3.19)
On the other hand, by (
3.8) and (
3.9),
$$\begin{aligned} |{\hat{\textbf{x}}}_{j,1}- {\textbf{x}}_{j, 1}|\le \frac{\pi }{6} \quad \text{ and } \quad |{\hat{\textbf{x}}}_{j,2}- {\textbf{x}}_{j, 2}|\le \frac{\pi }{6}. \end{aligned}$$
We further have
$$\begin{aligned}&\tau \Big |{\hat{\textbf{x}}}_{j,1}- {\textbf{x}}_{j, 1}\Big |+ \tau \Big |{\hat{\textbf{x}}}_{j,2}- {\textbf{x}}_{j, 2}\Big | \le \frac{\pi }{3}\Big (\Big |e^{i{\hat{\textbf{x}}}_{j,1}}- e^{i{\textbf{x}}_{j, 1}}\Big |+\Big |e^{i{\hat{\textbf{x}}}_{j,2}}- e^{i{\textbf{x}}_{j, 2}}\Big |\Big )\\&\quad \le \Big (\Big (\frac{1}{d_{\min }}\Big )^{2n-2} + \Big (\frac{1}{g_{\min }}\Big )^{2n-2}\Big ) \frac{(1+\sqrt{3})^{2n-1}2^{3n}\pi }{3\sqrt{3}}\frac{\sigma }{m_{\min }}. \end{aligned}$$
Recalling that
\(\tau =\frac{\Omega }{2n-1}\), we have
$$\begin{aligned} \Big |{\hat{\textbf{x}}}_{j,1}- {\textbf{x}}_{j, 1}\Big |+ \Big |{\hat{\textbf{x}}}_{j,2}- {\textbf{x}}_{j, 2}\Big |&\le \frac{2n-1}{\Omega } \Big (\Big (\frac{1}{d_{\min }}\Big )^{2n-2} + \Big (\frac{1}{g_{\min }}\Big )^{2n-2}\Big )\\&\quad \frac{(1+\sqrt{3})^{2n-1}2^{3n}\pi }{3\sqrt{3}}\frac{\sigma }{m_{\min }}. \end{aligned}$$
Note that by (
3.16), we obtain that
$$\begin{aligned} D_{\min }\le \frac{2\pi (2n-1)}{3\Omega }d_{\min } \quad \text{ and } \quad D_{\min }\le \frac{2\pi (2n-1)}{3\Omega }g_{\min }. \end{aligned}$$
Thus
$$\begin{aligned} \left| \left| {{\hat{\textbf{x}}}_j-{\textbf{x}}_j}\right| \right| _1 \le&\frac{(1+\sqrt{3})^{2n-1}2^{3n+1}\pi (2n-1)}{3\sqrt{3}\Omega } \Big (\frac{2(2n-1)}{3}\Big )^{2n-2} \Big (\frac{\pi }{\Omega D_{\min }}\Big )^{2n-2} \frac{\sigma }{m_{\min }}\\ =&\frac{(1+\sqrt{3})^{2n-1}2^{5n-1}(2n-1)^{2n-1}\pi }{3^{2n-0.5}\Omega } \Big (\frac{\pi }{\Omega D_{\min }}\Big )^{2n-2} \frac{\sigma }{m_{\min }}. \end{aligned}$$
Since
\(||{\hat{\textbf{y}}}_{j}- {\textbf{y}}_{j}||_1 = ||{\hat{\textbf{x}}}_{j}- {\textbf{x}}_{j}||_1 \), we further get
$$\begin{aligned} \left| \left| {{\hat{\textbf{y}}}_j-{\textbf{y}}_j}\right| \right| _1 \le \frac{(1+\sqrt{3})^{2n-1}2^{5n-1}(2n-1)^{2n-1}\pi }{3^{2n-0.5}\Omega } \Big (\frac{\pi }{\Omega D_{\min }}\Big )^{2n-2} \frac{\sigma }{m_{\min }}. \end{aligned}$$
Since
\(D_{\min }\ge \frac{15.3\pi (n-0.5)}{\Omega } \Big ( \frac{\sigma }{m_{\min }}\Big )^{\frac{1}{2n-1}}\), together with the above estimate, we can also show that
$$\begin{aligned} \left| \left| {{\hat{\textbf{y}}}_j-{\textbf{y}}_j}\right| \right| _1< \frac{D_{\min }}{2}. \end{aligned}$$
This completes the proof.
\(\square \)