Proof. This is a direct consequence of Theorem 2.1 in view of

$$\begin{align*}|\mathbf{w}_1^{\top}\mathbf{v}|< 2^{\#S_t}\sigma, \qquad |(e^{i \widehat y_{t'} \omega^*}\mathbf{w}_1 -\mathbf{w}_2)^{\top}\mathbf{v}|< 2^{\#S_t+1}\sigma.\\[-38.5pt] \end{align*}$$

Proof. We only prove the theorem for $\omega ^* \leqslant \frac {\Omega }{\#S_t+1}$ . The case when $\omega ^* \leqslant \frac {\Omega }{\#S_t}$ for (7) is obvious afterwards. We fix $t\in \{1,\cdots , n\}$ in the following derivations. From (6), we can write

(13) $$ \begin{align} \widehat A \widehat a = A a +W, \end{align} $$

where $\widehat a = (\widehat a_1, \cdots , \widehat a_d)^{\top }$ , $a = (a_1, \cdots , a_n)^{\top }, W =\left (\mathbf{w}(0), \mathbf{w}(\omega ^*), \cdots , \mathbf{w}((\#S_t+1)\omega ^*)\right )^{\top }$ and

$$ \begin{align*} &\widehat A = \left( \phi_{0,\#S_t+1}(e^{i \widehat y_1 \omega^*}), \ \phi_{0, \#S_t+1}(e^{i \widehat y_2 \omega^*}), \ \cdots, \ \phi_{0,\#S_t+1}(e^{i \widehat y_d \omega^*})\right), \\ & A = \left( \phi_{0,\#S_t+1}(e^{i y_1 \omega^*}), \ \phi_{0,\#S_t+1}(e^{i y_2 \omega^*}), \ \cdots, \ \phi_{0,\#S_t+1}(e^{i y_n \omega^*})\right), \end{align*} $$

with $0< \omega ^*\leqslant \frac {\Omega }{\#S_t+1}$ . We further decompose (13) into the following two equations:

(14) $$ \begin{align} \widehat {a}_{t'} \phi_{0, \#S_t}(e^{i \widehat y_{t'}\omega^*}) = B_1 b + \mathbf{w}_1, \quad \widehat {a}_{t'} \phi_{1, \#S_t+1}(e^{i \widehat y_{t'}\omega^*}) = B_2 b + \mathbf{w}_2, \end{align} $$

where $\mathbf{w}_1 = (\mathbf{w}(0), \mathbf{w}(\omega ^*), \cdots , \mathbf{w}(\#S_t \omega ^*))^{\top }$ , $\mathbf{w}_2 = (\mathbf{w}(\omega ^*), \cdots , \mathbf{w}((\#S_t+1)\omega ^*))^{\top }$ and

$$ \begin{align*} B_1 &= \left( \phi_{0, \#S_t}(e^{i y_{1}\omega^*}),\ \cdots,\ \phi_{0, \#S_t}(e^{i y_{n}\omega^*}),\ \phi_{0, \#S_t}(e^{i\widehat y_{1}\omega^*}),\ \cdots, \phi_{0, \#S_t}(e^{i\widehat y_{t'-1}\omega^*}),\right. \\ & \qquad \qquad \ \left. \phi_{0, \#S_t}(e^{i\widehat y_{t'+1}\omega^*}),\ \cdots, \ \phi_{0, \#S_t}(e^{i \widehat y_{d}\omega^*}) \right),\\ B_2 &= \left( \phi_{1, \#S_t+1}(e^{i y_{1}\omega^*}),\ \cdots,\ \phi_{1, \#S_t+1}(e^{i y_{n}\omega^*}),\ \phi_{1, \#S_t+1}(e^{i\widehat y_{1}\omega^*}),\ \cdots, \phi_{1, \#S_t+1}(e^{i\widehat y_{t'-1}\omega^*}),\right. \\ & \qquad \qquad \ \left. \phi_{1, \#S_t+1}(e^{i\widehat y_{t'+1}\omega^*}),\ \cdots, \ \phi_{1, \#S_t+1}(e^{i \widehat y_{d}\omega^*}) \right). \end{align*} $$

We first consider the case when all the $e^{y_j \omega ^*}$ ’s and $e^{\widehat y_j\omega ^*}$ ’s are distinct. Thus, $b=(b_1, \cdots , b_{\#S_t+1})$ in (14) is such that

$$\begin{align*}b_{l} = \left\{\begin{array}{ll} a_l, & 1\leqslant l\leqslant n, \\ - \widehat a_{l-n}, & n<l\leqslant n+t'-1,\\ - \widehat a_{l-n+1}, & n+t'-1<l. \end{array}\right. \end{align*}$$

Observe that

$$ \begin{align*} &B_2 = B_1 \mathrm{diag}\left(e^{iy_1\omega^*} ,\cdots, e^{i y_n \omega^*}, e^{i \widehat y_{1} \omega^*}, \cdots, e^{i \widehat y_{t'-1} \omega^*}, e^{i \widehat y_{t'+1} \omega^*},\cdots, e^{i \widehat y_{d} \omega^*}\right),\\ & \phi_{1, \#S_t+1}(e^{i y_{l}\omega^*}) = e^{i y_{l}\omega^*}\phi_{0, \#S_t}(e^{i y_{l}\omega^*}). \end{align*} $$

We rewrite (14) as

(15) $$ \begin{align} \begin{aligned} &\widehat {a}_{t'} \phi_{0, \#S_t}(e^{i \widehat y_{t'}\omega^*}) = B_1 b + \mathbf{w}_1,\\ &e^{i \widehat y_{t'}\omega^*}\widehat {a}_{t'} \phi_{0, \#S_t}(e^{i \widehat y_{t'}\omega^*}) = B_1\mathrm{diag}\left(e^{i y_1\omega^*} ,\cdots, e^{i y_n \omega^*}, e^{i \widehat y_{1} \omega^*}, \cdots, e^{i \widehat y_{t'-1} \omega^*}, e^{i \widehat y_{t'+1} \omega^*},\cdots, e^{i \widehat y_{d} \omega^*}\right)b + \mathbf{w}_2. \end{aligned} \end{align} $$

Since all the $e^{iy_j\omega ^*}$ ’s and $e^{i\widehat y_j \omega ^*}$ ’s are pairwise distinct, $B_1$ is a regular matrix. We multiply both sides of the above equations by the inverse of $B_1$ to get from Lemma 2.4 that

(16) $$ \begin{align} \widehat {a}_{t'}\prod_{q\in S_t}\frac{e^{i\widehat y_{t'}\omega^*}-e^{i q\omega^*}}{e^{i y_t\omega^*}- e^{i q\omega^*}} &= a_t + (B_1^{-1})_{t} \mathbf{w}_1, \end{align} $$

(17) $$ \begin{align} e^{i \widehat y_{t'}\omega^*} \widehat {a}_{t'}\prod_{q\in S_t}\frac{e^{i\widehat y_{t'}\omega^*}-e^{i q\omega^*}}{e^{i y_t\omega^*}- e^{i q\omega^*}} &= e^{i y_{t}\omega^*} a_t + (B_1^{-1})_{t} \mathbf{w}_2, \end{align} $$

where $(B_1^{-1})_{t}$ is the t-th row of $B_1^{-1}$ . By Lemma 2.3, it follows that

(18) $$ \begin{align} (B_1^{-1})_{t}\mathbf{w}_1 = \frac{\sum_{p=0}^{\#S_t-1}\left(\mathbf{w}(p \omega^*) (-1)^{\#S_t-p}\sum_{\{q_1,\cdots, q_{\#S_{t}-p}\}\in S_{t, \#S_t-p}} e^{iq_1\omega^*}\cdots e^{iq_{\#S_t-p}\omega^*}\right)+ \mathbf{w}(\#S_t \omega^*)}{\prod_{q\in S_t}(e^{iy_t\omega^*} - e^{i q \omega^*})}. \end{align} $$

Thus, multiplying $\prod _{q\in S_t}\left (e^{i y_t\omega ^*}- e^{i q\omega ^*}\right )$ in both sides of (16) proves (7) in the case when all the $e^{iy_j\omega ^*}$ ’s and $e^{i\widehat y_j \omega ^*}$ ’s are pairwise distinct. Furthermore, equation (16) times $e^{i \widehat y_{t'} \omega ^*}$ minus (17) yields

$$\begin{align*}(e^{i y_{t}\omega^*} - e^{i \widehat y_{t'}\omega^*})a_t = (B_1^{-1})_{t}\left(e^{i \widehat y_{t'} \omega^*}\mathbf{w}_1-\mathbf{w}_2\right). \end{align*}$$

Similarly, further expanding $(B_1^{-1})_{t}\left (e^{i \widehat y_{t'} \omega ^*}\mathbf{w}_1-\mathbf{w}_2\right )$ explicitly by Lemma 2.3 and multiplying $\prod _{q\in S_t}\left (e^{i y_t\omega ^*}- e^{i q\omega ^*}\right )$ in both sides above yields (8).

Finally, we consider the case when the $e^{iy_j\omega ^*}$ ’s and $e^{i\widehat y_j \omega ^*}$ ’s are not pairwise distinct. Since it is a limiting case of the above cases, (7) and (8) still hold in this case. This completes the proof.

Proof. Note that

(23) $$ \begin{align} \left|e^{i\theta_j}-e^{i\theta_p} \right| \geqslant \frac{2}{\pi} \left|\theta_j-\theta_p\right|, \quad \text{for all}\,\, \theta_j, \theta_p \in \Big[-\frac{\pi}{2}, \frac{\pi}{2}\Big]. \end{align} $$

Then we have

$$ \begin{align*} \prod_{1\leqslant p\leqslant k,p\neq j}{\left|e^{i\theta_j}-e^{i\theta_p}\right|}\ \geqslant \left(\frac{2}{\pi}\right)^{k-1} \prod_{1\leqslant p\leqslant k,p\neq j}{\left|\theta_j-\theta_p\right|} \geqslant \zeta(k)\left(\frac{2\theta_{\min}}{\pi}\right)^{k-1}.\\[-40pt] \end{align*} $$

Proof. See Corollary 7 in [Reference Liu and Zhang26].

Proof. We are now ready to prove Theorem 3.3. Suppose that $\widehat \mu =\sum _{j=1}^{d}\widehat a_j \delta _{\widehat y_j}, d\leqslant n-1$ is a $\sigma $ -admissible measure of $\mathbf{Y}$ . By the Definition 3.1 and the model (1), we have

$$\begin{align*}\mathscr F[\widehat \mu](\omega)= \mathscr F[\mu](\omega)+\mathbf{W}_1(\omega),\quad \omega \in [-\Omega,\Omega] \end{align*}$$

for some $\mathbf{W}_1$ with $|\mathbf{W}_1(\omega )|<2\sigma $ . Note that by adding some point sources in $\widehat \mu $ , from above we can actually construct $\widehat \mu =\sum _{j=1}^{n-1}\widehat a_j \delta _{\widehat y_j}$ such that

$$\begin{align*}\mathscr F[\widehat \mu](\omega)= \mathscr F[\mu](\omega)+\mathbf{W}_2(\omega),\quad \omega \in [-\Omega,\Omega], \end{align*}$$

for some $\mathbf{W}_2$ with $|\mathbf{W}_2(\omega )|<2\sigma $ . For ease of exposition, we consider in the following that the measure $\widehat \mu $ is with $n-1$ point sources. The above equation implies that $\widehat \rho = \sum _{j=1}^{n-1}e^{-\widehat y_j \Omega }\widehat a_{j} \delta _{\widehat y_j}$ and $\rho = \sum _{j=1}^ne^{-y_j \Omega }a_{j} \delta _{y_j}$ satisfy

(24) $$ \begin{align} \mathscr F[\widehat \rho](\omega)= \mathscr F[\rho](\omega)+\mathbf{W}_3(\omega),\quad \omega \in [0,2\Omega], \end{align} $$

for some $\mathbf{W}_3$ with $|\mathbf{W}_3(\omega )|<2\sigma , \omega \in [0,2\Omega ]$ . For any $y_t$ and $\widehat y_{t'}$ , define $S_t$ as

$$\begin{align*}S_t:= \left\{y_1,\cdots, y_{t-1}, y_{t+1}, \cdots ,y_n, \widehat y_1,\cdots, \widehat y_{t'-1}, \widehat y_{t'+1},\cdots, \widehat y_{n-1} \right\}. \end{align*}$$

Then $\#S_t = 2n-3$ . Let $\omega ^*=\frac {2\Omega }{2n-2}$ . Applying (11) to (24), we obtain that

$$\begin{align*}\prod_{q=1,\cdots, n, q\neq t} \left|e^{i y_{t}\omega^*} - e^{i y_{q}\omega^*} \right|\prod_{q=1,\cdots, n-1}\left|e^{iy_{t}\omega^*} - e^{i\widehat y_q\omega^*}\right|\left|a_t \right|< 2^{\#S_t+2}\sigma, \ t=1,\cdots, n. \end{align*}$$

This gives

$$\begin{align*}\prod_{q=1,\cdots, n, q\neq t} \left|e^{i y_{t}\omega^*} - e^{i y_{q}\omega^*} \right|\prod_{q=1,\cdots, n-1}\left|e^{iy_{t}\omega^*} - e^{i\widehat y_q\omega^*}\right|< 2^{2n-1}\frac{\sigma}{m_{\min}}, \ t=1,\cdots, n. \end{align*}$$

Therefore, it follows that

$$\begin{align*}\min_{t=1,\cdots, n}\prod_{q=1,\cdots, n, q\neq t} \left|e^{i y_{t}\omega^*} - e^{i y_{q}\omega^*} \right|\max_{t=1,\cdots, n}\prod_{q=1,\cdots, n-1}\left|e^{iy_{t}\omega^*} - e^{i\widehat y_q\omega^*}\right|< 2^{2n-1}\frac{\sigma}{m_{\min}}. \end{align*}$$

Denote $y_q \omega ^*$ by $\theta _q$ and $\theta _{\min } = \min _{p\neq q}|\theta _p-\theta _q|$ . Since the $y_j$ ’s are in $B_{\frac {(n-1) \pi }{2\Omega }}^1(0)$ and $\omega ^* = \frac {2\Omega }{2n-2}$ , we have that the $\theta _j$ ’s are in $\left [-\frac {\pi }{2}, \frac {\pi }{2}\right ]$ . Thus, by Lemma 3.4, we get

(25) $$ \begin{align} \min_{t=1,\cdots, n}\prod_{q=1,\cdots, n, q\neq t} \left|e^{i y_{t}\omega^*} - e^{i y_{q}\omega^*} \right|\geqslant \zeta(n)\left(\frac{2\theta_{\min}}{\pi}\right)^{n-1}. \end{align} $$

Moreover, using Lemma 3.5 yields

(26) $$ \begin{align} \max_{t=1, \cdots, n}\prod_{q=1,\cdots,n-1}\left|e^{iy_{t}\omega^*} - e^{i\widehat y_q\omega^*}\right|\geqslant \xi(n-1) \left(\frac{2\theta_{\min}}{\pi}\right)^{n-1}. \end{align} $$

Combining the above estimates, it follows that

(27) $$ \begin{align} \zeta(n)\xi(n-1) \left(\frac{2\theta_{\min}}{\pi}\right)^{2n-2}< \frac{2^{2n-1}\sigma}{m_{\min}}. \end{align} $$

Thus,

$$\begin{align*}\theta_{\min} < \pi\left( \frac{2}{\zeta(n)\xi(n-1)}\right)^{\frac{1}{2n-2}}\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2n-2}}, \end{align*}$$

and consequently,

(28) $$ \begin{align} d_{\min}= \frac{\theta_{\min}}{\omega^*}< \frac{(2n-2)\pi}{2\Omega}\left( \frac{2}{\zeta(n)\xi(n-1)}\right)^{\frac{1}{2n-2}}\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2n-2}}\leqslant \frac{2\pi e}{\Omega}\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2n-2}}, \end{align} $$

where $d_{\min }:=\min _{p\neq q}|y_p-y_q|$ and the last inequality is from Lemma A.1. Therefore, if $d_{\min }\geqslant \frac {2\pi e}{\Omega }\left (\frac {\sigma }{m_{\min }}\right )^{\frac {1}{2n-2}}$ , then there is no $\sigma $ -admissible measure of $\mathbf{Y}$ with less than n supports.

The last part consists in proving the cases when $n=2,3$ . When $n=3$ , the result is enhanced by noting that $\frac {2}{\zeta (n)\xi (n-1)}=8$ in (28). When $n=2$ , the result is enhanced by improving the estimates (25) and (26). For (25), we now have

$$\begin{align*}\left|{e^{iy_1\omega^*} - e^{iy_2\omega^*}}\right|\geqslant 2\sin \left(\frac{\theta_{\min}}{2}\right), \end{align*}$$

where $\theta _{\min } = |y_1\omega ^*-y_2\omega ^*|$ and $\omega ^*= \Omega $ . For (26), we have

$$\begin{align*}\max_{j=1,2}\left|e^{iy_{j}\omega^*} - e^{i\widehat y_1\omega^*}\right|\geqslant 2\sin\left(\frac{\theta_{\min}}{4}\right). \end{align*}$$

Thus, similarly to (27), we obtain that

$$\begin{align*}2\sin \left(\frac{\theta_{\min}}{2}\right)2\sin\left(\frac{\theta_{\min}}{4}\right)<\frac{2^{3}\sigma}{m_{\min}}, \end{align*}$$

which gives

$$\begin{align*}\sin^2 \left(\frac{\theta_{\min}}{2}\right)<\frac{4\sigma}{m_{\min}}. \end{align*}$$

It then follows that

$$\begin{align*}d_{\min} = \frac{\theta_{\min}}{\omega^*} < \frac{2\arcsin\left(2\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}, \end{align*}$$

which completes the proof.

Proof. See Corollary 9 in [Reference Liu and Zhang26].

Proof. Since $\widehat \mu =\sum _{j=1}^{n}\widehat a_j \delta _{\widehat y_j}, \widehat y_j \in B_{\frac {(n-1) \pi }{2\Omega }}^1(0)$ is a $\sigma $ -admissible measure of $\mathbf{Y}$ , from the model (1), we have

$$\begin{align*}\mathscr F[\widehat \mu](\omega)= \mathscr F[\mu](\omega)+\mathbf{W}_1(\omega),\quad \omega \in [-\Omega,\Omega], \end{align*}$$

for some $\mathbf{W}_1$ with $|\mathbf{W}_1(\omega )|<2\sigma , \omega \in [-\Omega ,\Omega ] $ . This implies that $\widehat \rho = \sum _{j=1}^{n}e^{-\widehat y_j \Omega }\widehat a_{j} \delta _{\widehat y_j}$ and $\rho = \sum _{j=1}^ne^{-y_j \Omega } a_{j} \delta _{y_j}$ satisfy

(33) $$ \begin{align} \mathscr F[\widehat \rho](\omega)= \mathscr F[\rho](\omega)+\mathbf{W}_2(\omega),\quad \omega \in [0,2\Omega], \end{align} $$

for some $\mathbf{W}_2$ with $|\mathbf{W}_2(\omega )|<2\sigma , \omega \in [0,2\Omega ]$ . For any $y_t$ , let $\widehat y_{t'}$ be the one in $\widehat y_j$ ’s that is the closest to $y_{t}$ and define $S_t$ as

$$\begin{align*}S_t:= \left\{y_1,\cdots, y_{t-1}, y_{t+1}, \cdots ,y_n, \widehat y_1,\cdots, \widehat y_{t'-1}, \widehat y_{t'+1},\cdots, \widehat y_{n} \right\}. \end{align*}$$

Then $\#S_t = 2n-2$ . Let $\omega ^*=\frac {2\Omega }{2n-1}$ . Since (33) holds, by (11), we have

(34) $$ \begin{align} \prod_{q=1,\cdots, n, q\neq t} \left|e^{i y_{t}\omega^*} - e^{i y_{q}\omega^*} \right|\prod_{q=1,\cdots, n}\left|e^{iy_{t}\omega^*} - e^{i\widehat y_q\omega^*}\right|< \frac{2^{2n}\sigma}{m_{\min}}, \ t=1,\cdots, n. \end{align} $$

Therefore, it follows that

$$\begin{align*}\min_{t=1,\cdots, n}\prod_{q=1,\cdots, n, q\neq t} \left|e^{i y_{t}\omega^*} - e^{i y_{q}\omega^*} \right|\max_{t=1,\cdots, n}\prod_{q=1,\cdots, n}\left|e^{iy_{t}\omega^*} - e^{i\widehat y_q\omega^*}\right|< \frac{2^{2n}\sigma}{m_{\min}}. \end{align*}$$

Denote $y_q \omega ^*, \widehat y_q\omega ^*$ by, respectively, $\theta _q, \widehat \theta _q$ and $\theta _{\min } = \min _{p\neq q}|\theta _p-\theta _q|$ . Since $y_j$ ’s in $B_{\frac {(n-1) \pi }{2\Omega }}^1(0)$ and $\omega ^* = \frac {2\Omega }{2n-1}$ , we have $\theta _j, \widehat \theta _{j}$ ’s in $\left [-\frac {\pi }{2}, \frac {\pi }{2}\right ]$ . By Lemma 3.4, we further get

(35) $$ \begin{align} \max_{j=1, \cdots, n}\prod_{q=1,\cdots,n}\left|e^{iy_{j}\omega^*} - e^{i \widehat y_q\omega^*}\right|<\frac{1}{\zeta(n)}\left(\frac{\pi}{2\theta_{\min}}\right)^{n-1} \frac{2^{2n}\sigma}{m_{\min}}. \end{align} $$

We then utilize Lemma 3.9 to estimate the recovery of the locations. For this purpose, let $\epsilon = \frac {\pi ^{2n-1}}{\zeta (n)(\theta _{\min })^{n-1}} \frac {2\sigma }{m_{\min }}$ . Then (35) is equivalent to

$$ \begin{align*}\left|\left|{\eta_{n ,n}(e^{i \theta_1},\cdots,e^{i \theta_n}, e^{i \widehat \theta_1},\cdots, e^{i \widehat \theta_n})}\right|\right|{}_{\infty}<\left(\frac{2}{\pi}\right)^n\epsilon.\end{align*} $$

We thus only need to check the following condition:

(36) $$ \begin{align} \theta_{\min}\geqslant \Big(\frac{4\epsilon}{\lambda(n)}\Big)^{\frac{1}{n}}, \quad \text{or equivalently}\,\,\, (\theta_{\min})^n \geqslant \frac{4\epsilon}{\lambda(n)}. \end{align} $$

Indeed, by the separation condition (29),

(37) $$ \begin{align} \theta_{\min}= d_{\min}\omega^*\geqslant \frac{2.36\pi e}{n-\frac{1}{2}}\Big(\frac{\sigma}{m_{\min}}\Big)^{\frac{1}{2n-1}}\geqslant \pi \Big(\frac{4}{\lambda(n)\zeta(n)}\frac{2\sigma}{m_{\min}}\Big)^{\frac{1}{2n-1}}, \end{align} $$

where we have used Lemma A.2 in the last inequality. Then

$$\begin{align*}(\theta_{\min})^{2n-1}\geqslant \frac{4\pi^{2n-1}}{\lambda(n)\zeta(n)}\frac{2\sigma}{m_{\min}}, \end{align*}$$

whence we get (36). Therefore, we can apply Lemma 3.9 to get that, after reordering $\widehat \theta _j$ ’s,

(38) $$ \begin{align} \begin{aligned} &\left|\widehat \theta_{j}-\theta_j\right|< \frac{\theta_{\min}}{2},\\ &\left|\widehat \theta_{j}-\theta_j\right|< \frac{2^{n}\pi^{2n-1}}{\zeta(n)(n-2)!(\theta_{\min})^{2n-2}}\frac{\sigma}{m_{\min}} , 1\leqslant j\leqslant n. \end{aligned} \end{align} $$

Finally, we estimate $\left |\widehat y_j - y_j\right |$ . Since $\left |\widehat \theta _{j}-\theta _j\right |< \frac {\theta _{\min }}{2}$ , it is clear that $\left |\widehat y_j-y_j\right |< \frac {d_{\min }}{2}.$ Thus, $\widehat \mu $ is within the $\frac {d_{\min }}{2}$ -neighborhood of $\mu $ . Moreover,

$$\begin{align*}\left|\widehat y_j-y_j\right|= \frac{2n-1}{2\Omega}\left|\widehat \theta_j -\theta_j\right|, \quad j =1, \cdots, n. \end{align*}$$

Using (38) and Lemma A.3, a direct calculation shows that

(39) $$ \begin{align} \left|\widehat y_j-y_j\right|< \frac{C(n)}{\Omega} \left(\frac{\pi}{\Omega d_{\min}}\right)^{2n-2} \frac{\sigma}{m_{\min}}, \end{align} $$

where $C(n)=\sqrt {n-0.5}\ 2^{2n-\frac {3}{2}}e^{2n}4.5^{-1}$ .

The last part is to prove the case when $n=2$ . When $n=2$ , by (34), we have

(40) $$ \begin{align} \left|e^{i y_{1}\omega^*} - e^{i \widehat y_{1}\omega^*} \right| \left|e^{iy_{1}\omega^*} - e^{i\widehat y_2\omega^*}\right| \left|e^{iy_{1}\omega^*} - e^{iy_2\omega^*}\right| < \frac{2^{4}\sigma}{m_{\min}}. \end{align} $$

Denote $\omega ^*|y_1-y_2|=\theta _{\min }$ . Reordering $\hat y_j$ ’s so that $|\widehat y_1- y_1|\leqslant |\widehat y_2- y_1|$ . Thus, if $|\widehat y_{1}-y_1|\omega ^*\geqslant \frac {\theta _{\min }}{2}$ , we have $|\widehat y_{2}-y_1|\omega ^*\geqslant \frac {\theta _{\min }}{2}$ . Recall also that $\widehat y_j\in B_{\frac {(n-1) \pi }{2\Omega }}^1(0), j=1,2$ . Then (40) yields

$$\begin{align*}\left(2\sin\left(\frac{\theta_{\min}}{4}\right)\right)^2 2\sin\left(\frac{\theta_{\min}}{2}\right) < \frac{2^4\sigma}{m_{\min}}, \end{align*}$$

which gives

$$\begin{align*}\sin^3\left(\frac{\theta_{\min}}{2}\right)< \frac{2^3\sigma}{m_{\min}}. \end{align*}$$

It then follows that

$$\begin{align*}\theta_{\min}< 2\arcsin\left(\frac{8\sigma}{m_{\min}}\right)^{\frac{1}{3}} \end{align*}$$

and

$$\begin{align*}d_{\min}=\frac{\theta_{\min}}{\omega^*}< \frac{3}{\Omega}\arcsin\left(2\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{3}}\right), \end{align*}$$

where we have set $\omega ^* = \frac {2\Omega }{3}$ . Therefore, if

$$\begin{align*}d_{\min}\geqslant \frac{3}{\Omega}\arcsin\left(2\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{3}}\right), \end{align*}$$

then we must have $|\widehat y_{1}-y_1|\omega ^*<\frac {\theta _{\min }}{2}$ and consequently, $|\widehat y_{1}-y_1|< \frac {d_{\min }}{2}$ . In the same manner, we also have $|\widehat y_{2}-y_2|< \frac {d_{\min }}{2}$ . This completes the proof.

Proof. By Theorem 3.8, the separation condition (41) implies

(44) $$ \begin{align} \Big|\widehat y_j-y_j\Big|<\frac{\sqrt{n-0.5}\ 2^{2n-\frac{3}{2}}e^{2n}4.5^{-1}}{\Omega}\left(\frac{\pi}{\Omega d_{\min}}\right)^{2n-2}\frac{\sigma}{m_{\min}}, \quad 1\leqslant j\leqslant n. \end{align} $$

Together with (41), this gives $|\widehat y_j-y_j|<\frac {d_{\min }}{9}, j=1,\cdots , n$ . Hence, among $\{\widehat y_q\}_{q=1}^n$ , $\widehat y_j$ is the closest point to $y_j$ . We write $y_{j} = \widehat y_j+\epsilon _j$ with $0\leqslant \left |{\epsilon _{j}}\right |< \frac {d_{\min }}{9}$ . Since $\widehat \mu =\sum _{j=1}^{n}\widehat a_j \delta _{\widehat y_j}, \widehat y_j \in B_{\frac {(n-1) \pi }{2\Omega }}^1(0)$ is a $\sigma $ -admissible measure of $\mathbf{Y}$ , from the model (1), we have

$$\begin{align*}\mathscr F[\widehat \mu](\omega)= \mathscr F[\mu](\omega)+\mathbf{W}_1(\omega),\quad \omega \in [0,\Omega], \end{align*}$$

for some $\mathbf{W}_1$ with $|\mathbf{W}_1(\omega )|<2\sigma , \omega \in [0,\Omega ] $ . For each $j\in \{1,\cdots , n\}$ , denote by $S_j$ the set containing all $y_p$ ’s and $\widehat y_p$ ’s except $y_j, \widehat y_{j}$ . By $\left |{\widehat y_p - y_p}\right |<\frac {d_{\min }}{9}, 1\leqslant p\leqslant n$ , for all $q\in S_j$ we have $q\neq y_j$ and $q\neq \widehat y_j$ . Let $\omega ^* = \frac {\Omega }{2n-1}$ . As $y_p$ ’s and $\widehat y_p$ ’s are in $B_{\frac {(n-1) \pi }{2\Omega }}^1(0)$ , we also have $e^{iq\omega ^*}\neq e^{iy_j\omega ^*}, e^{iq\omega ^*}\neq e^{i\widehat y_j\omega ^*}, q\in S_j$ . Thus, by (10), we can have

(45) $$ \begin{align} \left|\widehat {a}_{j}\prod_{q\in S_j}\frac{e^{i\widehat y_{j}\omega^*}-e^{i q\omega^*}}{e^{i y_j\omega^*}- e^{i q\omega^*}}-a_j\right| < \frac{2^{2n-1}\sigma}{\prod_{q\in S_j}\left|e^{iy_j\omega^*} - e^{iq\omega^*}\right|}. \end{align} $$

Equivalently, we have

(46) $$ \begin{align} \left|\widehat {a}_{j}-a_j\prod_{q\in S_j}\frac{e^{iy_{j}\omega^*}-e^{i q\omega^*}}{e^{i \widehat y_j\omega^*}- e^{i q\omega^*}}\right| < \frac{2^{2n-1}\sigma}{\prod_{q\in S_j}\left|e^{i\widehat y_j\omega^*} - e^{iq\omega^*}\right|}. \end{align} $$

We rewrite its LHS as

(47) $$ \begin{align} \left|\widehat {a}_{j}-a_j\prod_{q\in S_j}\left(\frac{e^{i y_{j}\omega^*}- e^{i \widehat y_{j}\omega^*}}{e^{i \widehat y_j\omega^*}- e^{i q\omega^*}}+1\right)\right|, \end{align} $$

and expand that

(48) $$ \begin{align} \prod_{q\in S_j}\left(\frac{e^{i y_{j}\omega^*}- e^{i \widehat y_{j}\omega^*}}{e^{i \widehat y_j\omega^*}- e^{i q\omega^*}}+1\right) = 1+ (e^{i y_{j}\omega^*}- e^{i \widehat y_{j}\omega^*})g(\epsilon_j, S_j, \widehat y_j, y_j, \omega^*) \end{align} $$

where

(49) $$ \begin{align} \left|{g(\epsilon_j, S_j, \widehat y_j, y_j, \omega^*)}\right| \leqslant \frac{C_3(n)}{d_{\min}} \end{align} $$

for a certain constant $C_{3}(n)$ . Thus, combining (46), (47) and (48) yields that

(50) $$ \begin{align} \left|\widehat {a}_{j}-a_j\right| < \left|a_j (e^{i y_{j}\omega^*}- e^{i \widehat y_{j}\omega^*}) g(\epsilon_j, S_j, \widehat y_j, y_j, \omega^*)\right| + \frac{2^{2n-1}\sigma}{\prod_{q\in S_j}\left|e^{i\widehat y_j\omega^*} - e^{iq\omega^*}\right|}. \end{align} $$

Now we estimate the two terms in the RHS of the above equation. First, by (11), we have

(51) $$ \begin{align} \left|a_j (e^{i y_{j}\omega^*}- e^{i \widehat y_{j}\omega^*})\right| < \frac{2^{2n-1}\sigma}{\prod_{q\in S_j}\left|e^{iy_j\omega^*} - e^{iq\omega^*}\right|}. \end{align} $$

Second, based on the estimate $|\widehat y_p -y_p|< \frac {d_{\min }}{9}, p=1,\cdots , n$ , it is easy to prove that

(52) $$ \begin{align} \frac{\sigma}{\prod_{q\in S_j}\left|e^{i\widehat y_j\omega^*} - e^{iq\omega^*}\right|}\leqslant \frac{C_4(n)\sigma}{d_{\min}^{2n-2}} \end{align} $$

holds for some constant $C_4(n)$ . Combining estimates (49), (50), (51) and (52) yields

$$\begin{align*}\left|\widehat {a}_{j}-a_j\right|< \frac{C_5(n)}{d_{\min}^{2n-1}}\sigma \end{align*}$$

for some constant $C_5(n)$ . Now we consider the case when $\widehat y_j = y_j$ . This time, by (45), we have

$$\begin{align*}\left|{\widehat a_j - a_j}\right|<\frac{2^{\#S_j}\sigma}{\prod_{q\in S_j}\left|e^{iy_j\omega^*} - e^{iq\omega^*}\right|}. \end{align*}$$

Together with (52), we demonstrate (43) and complete the proof.

Proof. Step 1. We first prove the one-dimensional case. Let $\mu = \sum _{j=1}^2a_j \delta _{y_j}$ and $\widehat \mu = a \delta _{\widehat y}$ . A crucial relation is

(57) $$ \begin{align} \left|{\mathscr F[\widehat \mu](\omega)- \mathscr F[\mu](\omega) }\right|< 2\sigma, \ \omega \in [-\Omega, \Omega]. \end{align} $$

Note that if (57) holds, $\widehat \mu $ can be a $\sigma $ -admissible measure of some $\mathbf{Y}$ generated by model (1). This time, resolving two point sources is impossible. Conversely, if (57) does not hold, $\widehat \mu $ cannot be any $\sigma $ -admissible measure of some $\mathbf{Y}$ generated by $\mu $ as in model (1). Let $\mathscr R$ be the constant such that (57) holds when $|y_1-y_2|< \mathscr R$ and fails to hold in the opposite case. Based on the above discussions, proving Lemma 4.1 is to show $\mathscr R = \frac {4\arcsin \left (\left (\frac {\sigma }{m_{\min }}\right )^{\frac {1}{2}}\right )}{\Omega }$ .

Instead of considering all the $\omega \in [-\Omega ,\Omega ]$ directly, we consider

(58) $$ \begin{align} \left|{\mathscr F[\widehat \mu](\omega)-\mathscr F[\mu](\omega)}\right|< 2\sigma, \ \omega \in [0, \Omega]. \end{align} $$

In the sequel, we intend to find $\mathscr R$ so that (58) holds when $|y_1-y_2|< \mathscr R$ and does not hold in the opposite case. Afterward, we will show that (57) holds as well under some circumstances when $|y_1-y_2|< \mathscr R$ .

Step 2. Note that for the general source locations $y_1, y_2$ , shifting them by x and getting that

$$\begin{align*}\left|{\mathscr F[\widehat \mu](\omega)e^{ix\omega}- \mathscr F[\mu](\omega)e^{ix\omega} }\right|< 2\sigma, \ \omega \in [-\Omega, \Omega], \end{align*}$$

we can transform the problem into the case when $y_1=-y_2$ . Thus, we consider that the true source is $\mu = m_{\min } \delta _{y_1}+ m_{\min } \delta _{y_2}$ with $y_1>0, y_1 = -y_2$ . The measure $\widehat \mu $ is $a\delta _{\widehat y}$ with a and $\widehat y$ to be determined.

Then the existence of $\widehat \mu $ satisfying (58) is equivalent to solving the condition on $y_1$ so that

$$\begin{align*}\min_{a\in \mathbb C, \hat y\in \mathbb R}\left|{ae^{i\widehat y\omega}-2m_{\min}\cos\left(y_1\omega\right)}\right|<2\sigma, \ \omega \in [0, \Omega], \end{align*}$$

where we have used

$$ \begin{align*} ae^{i \widehat y\omega}-m_{\min}(e^{iy_1\omega}+e^{i y_2\omega})= ae^{i\widehat y\omega}-2m_{\min}\cos\left(y_1\omega\right). \end{align*} $$

We denote $d_{\min }:=|y_1-y_2|$ and first consider the case when $0<d_{\min }<\frac {\pi }{\Omega }$ . Note that for two nonnegative values $x,y$ , we have

$$\begin{align*}\left|xe^{i\theta}-y\right|{}^2 = (x\cos\theta-y)^2+x^2\sin^2\theta = x^2+y^2-2xy\cos\theta \geqslant (x-y)^2, \end{align*}$$

and the equality is attained when $\theta =0$ . Since $0<d_{\min }\leqslant \frac {\pi }{\Omega } \left (0<y_1\leqslant \frac {\pi }{2\Omega }\right )$ , we have $\cos (y_1\omega )\geqslant 0, \omega \in [0,\Omega ]$ . Thus, for every $\omega $ ,

$$\begin{align*}\min_{a\in \mathbb C, \hat y\in \mathbb R}\left|{ae^{i\widehat y\omega}-2m_{\min}\cos\left(y_1\omega\right)}\right|\geqslant \left||a|-2m_{\min}\cos\left(y_1\omega\right)\right|, \end{align*}$$

and the minimum is attained when $\widehat y=0$ and a is a positive number. We now try to find the condition on $y_1$ so that there exists $a\in \mathbb R^+$ satisfying

$$\begin{align*}\left|a-2m_{\min}\cos\left(y_1\omega\right)\right|<2\sigma, \quad \omega\in [0,\Omega]. \end{align*}$$

This is equivalent to

(59) $$ \begin{align} \max_{ \omega, \omega' \in [0,\Omega]}\left|2m_{\min}\left(\cos\left(y_1\omega\right)- \cos\left(y_1\omega'\right)\right)\right|<4\sigma, \end{align} $$

since the existence of $\omega _1, \omega _2 \in [0,\Omega ]$ so that

$$\begin{align*}\left|2m_{\min}\left(\cos\left(y_1\omega_1\right)- \cos\left(y_1\omega_2\right)\right)\right|\geqslant 4\sigma \end{align*}$$

results in

$$\begin{align*}\max_{\omega \in [0,\Omega]} \left|a-2m_{\min}\cos\left(y_1\omega\right)\right|\geqslant 2\sigma,\quad \forall a\in \mathbb R^+. \end{align*}$$

If $d_{\min }=2y_1\leqslant \frac {\pi }{\Omega }$ , then $0\leqslant y_1\omega \leqslant \frac {\pi }{2}, \omega \in [0,\Omega ]$ . Then problem (59) becomes

$$\begin{align*}2m_{\min}\left|1-\cos\left(\frac{d_{\min}}{2}\Omega\right)\right|< 4\sigma. \end{align*}$$

Thus, $4\sin ^2\left (\frac {d_{\min }\Omega }{4}\right )<\frac {4\sigma }{m_{\min }}$ , and equivalently,

$$\begin{align*}d_{\min} < \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}. \end{align*}$$

Note that when $d_{\min } < \frac {4\arcsin \left (\left (\frac {\sigma }{m_{\min }}\right )^{\frac {1}{2}}\right )}{\Omega }$ , choosing $a= \frac {m_{\min }+m_{\min }\cos (y_1\Omega )}{2}$ and $\widehat y=0$ makes

$$\begin{align*}\left|{\mathscr F[\widehat \mu](\omega)-\mathscr F[\mu](\omega)}\right|< 2\sigma, \ \omega \in [0, \Omega]. \end{align*}$$

As $\mathscr F[\widehat \mu ](-\omega )-\mathscr F[\mu ](-\omega ) = \mathscr F[\widehat \mu ](\omega )-\mathscr F[\mu ](\omega )$ this time, the solution makes

$$\begin{align*}\left|{\mathscr F[\widehat \mu](\omega)-\mathscr F[\mu](\omega)}\right|< 2\sigma, \ \omega \in [-\Omega, \Omega]. \end{align*}$$

Thus, this is exactly the resolution limit $\mathscr R$ when $0<d_{\min }\leqslant \frac {\pi }{\Omega }$ for the case when $\frac {\sigma }{m_{\min }}\leqslant \frac {1}{2}$ . For the case when $d_{\min }> \frac {\pi }{\Omega }$ , by $\frac {4\arcsin \left (\left (\frac {\sigma }{m_{\min }}\right )^{\frac {1}{2}}\right )}{\Omega }\leqslant \frac {\pi }{\Omega }$ when $\frac {\sigma }{m_{\min }}\leqslant \frac {1}{2}$ and above discussions, we have

$$\begin{align*}\min_{a\in \mathbb C, \hat y\in \mathbb R} \max_{\omega\in[0,\Omega]}\left|{ae^{i\widehat y\omega}-2m_{\min}\cos\left(y_1\omega\right)}\right|\geqslant 2\sigma. \end{align*}$$

Thus, there does not exist any $\sigma $ -admissible measure of $\mathbf Y$ with less than two supports. This proves the statements for the case when $\frac {\sigma }{m_{\min }}\leqslant \frac {1}{2}$ in the lemma.

Now, we consider the case when $\frac {\sigma }{m_{\min }}>\frac {1}{2}$ . We choose a specific case where $a=m_{\min }$ and $\widehat y =y_1$ . Then

$$\begin{align*}\left|{m_{\min}e^{iy_1\omega}-m_{\min}(e^{iy_1\omega}+e^{i y_2\omega})}\right| = \left|{m_{\min}e^{iy_2\omega}}\right| < 2\sigma \end{align*}$$

gives

$$\begin{align*}\left|{\mathscr F[\widehat \mu](\omega)-\mathscr F[\mu](\omega)}\right|<2\sigma, \ \omega \in \mathbb R. \end{align*}$$

Thus, the case when $\frac {\sigma }{m_{\min }}>\frac {1}{2}$ is meaningless. There are always some $\sigma $ -admissible measures for some images with only one point source.

Step 3. Now we consider the case when the sources $\mathbf{y}_j$ ’s are in $\mathbb R^k$ . We still consider the crucial relation that

(60) $$ \begin{align} \left|{\mathscr F[\widehat \mu](\boldsymbol{{\omega}})- \mathscr F[\mu](\boldsymbol{{\omega}}) }\right|< 2\sigma, \ \left|\left|{\boldsymbol{{\omega}}}\right|\right|{}_{2}\leqslant \Omega. \end{align} $$

By a similar argument as the one in step 1, we need to compute constant $\mathscr R$ such that (60) holds when $\left |\left |{\mathbf{y}_1-\mathbf{y}_2}\right |\right |{}_2< \mathscr R$ and fails to hold in the opposite case. Note that by choosing suitable axes or transforming the problem, we can make $\mathbf{y}_1= (y_1, 0, \cdots , 0)^\top , \mathbf{y}_2= (y_2, 0, \cdots , 0)^\top $ . Consider $\widehat \mu = a \delta _{\mathbf {\widehat y}}, \mathbf {\widehat y} \in \mathbb R^k$ with a and $\mathbf {\widehat y}$ to be determined. We now have

$$\begin{align*}\mathscr F[\widehat \mu](\boldsymbol{{\omega}})- \mathscr F[\mu](\boldsymbol{{\omega}}) = ae^{i \mathbf{\widehat y} \cdot \boldsymbol{{\omega}}} - \sum_{j=1}^{2}a_j e^{i \mathbf{y}_j \cdot \boldsymbol{{\omega}}}= ae^{i \mathbf{\widehat y}_{2:k}\cdot \boldsymbol{{\omega}}_{2:k}} e^{\mathbf{\widehat y}_1 \boldsymbol{{\omega}}_1} - \sum_{j=1}^{2}a_j e^{i {y}_j \boldsymbol{{\omega}}_1}, \end{align*}$$

where $\mathbf{x}_1$ is the first vector element and $\mathbf{x}_{2:k}$ is the vector consisting of the $2$ -nd to the k-th elements of $\mathbf{x}$ . Thus, analyzing when (60) holds can be reduced to the one-dimensional case, and it is not hard to see that the result for the one-dimensional space still holds for multi-dimensional spaces.

Proof. Step 1. We only need to consider the case when $\frac {\sigma }{m_{\min }} \leqslant \frac {1}{2}$ , as the case when $\frac {\sigma }{m_{\min }}>\frac {1}{2}$ is trivial. Also, we only consider the one-dimensional case since the treatment for multi-dimensional spaces is similar to the one in the proof of Theorem 4.1.

Let $\mu = \sum _{j=1}^2a_j \delta _{y_j}$ and $\widehat \mu = a \delta _{\widehat y}$ . Similarly to step 1 in the proof of Theorem 4.1, the computational resolution limit $\mathscr D_{num}^+(k, 2)$ should be the constant such that

(61) $$ \begin{align} \left|{\mathscr F[\widehat \mu](\omega)- \mathscr F[\mu](\omega)}\right|< 2\sigma, \ \omega \in [-\Omega, \Omega] \end{align} $$

holds when $|y_1-y_2|< \mathscr D_{num}^+(k, 2)$ and fails to hold in the opposite case. Choosing a suitable axis, we assume that the true source is $\mu = m_{\min }\alpha \delta _{y_1}+ m_{\min } \delta _{y_2}$ with $y_1 = -y_2$ , $\alpha \geqslant 1$ . We consider $\widehat \mu = a\delta _{\widehat y}$ with $a>0$ and $\widehat y$ to be determined. We shall prove that if

$$\begin{align*}\left|y_1-y_2\right|\geqslant \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}, \end{align*}$$

then (61) does not hold for any $\widehat \mu $ consisting of only one positive source. In the opposite case, Theorem 4.1 already ensures the existence of such $\widehat \mu , \mu $ , making (61) hold. By the above two results, we prove the theorem.

In the following proof, we will find a necessary condition for

$$\begin{align*}\min_{a>0, \alpha\geqslant 1, \widehat y\in \mathbb R}\left|{\mathscr F[\widehat \mu](\omega)- \mathscr F[\mu](\omega)}\right|<2\sigma, \omega \in [-\Omega,\Omega]. \end{align*}$$

By the definition of $\mathscr D_{num}^+(k, 2)$ in Definition 3.2, we only consider the case when $d_{\min }\leqslant \frac {\pi }{\Omega }$ .

Step 2. We analyze a necessary condition – that is,

(62) $$ \begin{align} \min_{a>0, \alpha\geqslant 1, \widehat y\in \mathbb R}\left|{\mathscr F[\widehat \mu](\Omega)- \mathscr F[\mu](\Omega)}\right|+\left|{\mathscr F[\widehat \mu](0)- \mathscr F[\mu](0)}\right|<4\sigma. \end{align} $$

Since from (61) and the assumption for $\mu $ ,

(63) $$ \begin{align} \mathscr F[\widehat \mu](\omega)- \mathscr F[\mu](\omega) = ae^{i \widehat y\omega}-m_{\min}\left(\alpha e^{iy_1\omega}+ e^{-iy_1\omega}\right), \end{align} $$

we thus consider

(64) $$ \begin{align} \min_{a>0, \alpha\geqslant 1, \widehat y\in \mathbb R}\left|ae^{i\widehat y \Omega} -m_{\min}\left(\alpha e^{iy_1 \Omega}+ e^{-i y_1\Omega}\right)\right|+\left|a -m_{\min}\left(\alpha+1\right)\right|. \end{align} $$

Let $\alpha = 1+h, h\geqslant 0$ , and rewrite the above formula as

$$\begin{align*}\min_{a>0, h\geqslant 0, \widehat y\in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|. \end{align*}$$

A key observation is that if $2m_{\min }\cos (y_1\Omega )+hm_{\min }\leqslant a \leqslant 2m_{\min }+hm_{\min }$ , then we have

(65) $$ \begin{align} & \min_{a>0, h\geqslant 0, \widehat y\in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\nonumber \\\geqslant &\min_{a>0, h\geqslant 0, \widehat y\in \mathbb R, \widehat x\in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{i\widehat x \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\nonumber \\= &\ \min_{a> 0, h\geqslant 0}\left|a -h m_{\min}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\nonumber \\= &\ \min_{2m_{\min}\cos(y_1\Omega)\leqslant b \leqslant 2 m_{\min}}\left|b- 2m_{\min} \cos(y_1 \Omega)\right|+|2m_{\min}-b| \nonumber\\= &\ 2m_{\min} - 2m_{\min} \cos(y_1 \Omega), \end{align} $$

where the second equality is because $a\geqslant 2m_{\min }\cos (y_1\Omega )+hm_{\min }$ , $hm_{\min }\geqslant 0$ and $2m_{\min }\cos (y_1\Omega )= 2m_{\min }\cos (\frac {d_{\min }}{2}\Omega )\geqslant 0$ for $d_{\min }\leqslant \frac {\pi }{\Omega }$ .

However, letting $h=0, \widehat y =0$ , we obtain that

$$ \begin{align*} &\min_{a>0}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\\=&\ \min_{a>0}\left|a - 2m_{\min} \cos(y_1 \Omega)\right|+|a-2m_{\min}|\\=&\ \min_{a>0}(a - 2m_{\min} \cos(y_1 \Omega))+2m_{\min}-a \quad \big(\text{choose } 2m_{\min} \cos(y_1 \Omega)\leqslant a\leqslant 2m_{\min}\big)\\=&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega). \end{align*} $$

Together with (65), this yields

$$ \begin{align*} &\min_{a>0, h\geqslant 0, \widehat y\in \mathbb R}\left|a e^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\\ =&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega) \end{align*} $$

in the case when $2m_{\min }\cos (y_1\Omega )+hm_{\min }\leqslant a \leqslant 2m_{\min }+hm_{\min }$ .

Now we consider the case when $a< 2m_{\min }\cos (y_1\Omega )+hm_{\min }$ . In this case, we have

$$ \begin{align*} &\min_{a>0, h\geqslant 0, a< 2m_{\min}\cos(y_1\Omega)+hm_{\min}, \widehat y\in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\\\geqslant & \min_{a>0, h\geqslant 0, a< 2m_{\min}\cos(y_1\Omega)+hm_{\min}}|a-(2m_{\min}+hm_{\min})|\\=&\ \min_{a>0, h\geqslant 0, a< 2m_{\min}\cos(y_1\Omega)+hm_{\min}}2m_{\min}+hm_{\min}-a\\>&\ 2m_{\min} -2m_{\min}\cos(y_1 \Omega). \end{align*} $$

Last, we consider the case when $a>2m_{\min }+hm_{\min }$ . This time, we have

$$ \begin{align*} &\min_{a>0, h\geqslant 0, a>2m_{\min}+hm_{\min}, \widehat y\in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\\\geqslant &\ \min_{a>0, h\geqslant 0, a>2m_{\min}+hm_{\min}, \widehat y \in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\\geqslant &\ \min_{a>0, h\geqslant 0, a>2m_{\min}+hm_{\min}}a -h m_{\min}- 2m_{\min} \cos(y_1 \Omega)\\>&\ 2m_{\min} -2m_{\min}\cos(y_1 \Omega). \end{align*} $$

Therefore, combining the above discussions yields

$$ \begin{align*} &\min_{a\geqslant 0, h\geqslant 0, \widehat y\in \mathbb R}\left|ae^{i\widehat y\Omega} -h m_{\min} e^{iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\\ =&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega), \end{align*} $$

and that is,

$$ \begin{align*} &\min_{a\geqslant 0, \alpha\geqslant 1, \widehat y\in \mathbb R}\left|a e^{\widehat y\Omega}-m_{\min}\left(\alpha e^{iy_1\Omega}+ e^{-i y_1 \Omega}\right)\right|+\left|a -m_{\min}\left(\alpha+1\right)\right|\\ =&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega). \end{align*} $$

Thus, (62) is equivalent to

$$\begin{align*}2m_{\min}-2m_{\min} \cos(y_1 \Omega)<4\sigma. \end{align*}$$

Similarly to the proof of Theorem 4.1, this shows that

$$\begin{align*}d_{\min} < \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega} \end{align*}$$

and completes the proof.

Proof. Step 1. We only need to analyze the case when $\frac {\sigma }{m_{\min }} \leqslant \frac {1}{2}$ , as the case when $\frac {\sigma }{m_{\min }}>\frac {1}{2}$ is trivial. Also, we only consider the one-dimensional case since the treatment for multi-dimensional spaces is similar to the one in the proof of Theorem 4.1.

Let $\mu = \sum _{j=1}^2a_j \delta _{y_j}$ and $\widehat \mu = a \delta _{\widehat y}$ . Similarly to step 1 in the proof of Theorem 4.1, the resolution limit $\mathscr D_{num}(k, 2)$ should be the constant such that the estimate

(67) $$ \begin{align} \left|{\mathscr F[\widehat \mu](\omega)- \mathscr F[\mu](\omega) }\right|< 2\sigma, \ \omega \in [-\Omega, \Omega] \end{align} $$

holds when $|y_1-y_2|< \mathscr D_{num}(k,2)$ and fails to hold in the opposite case.

Step 2. Without loss of generality, we assume the true source is

$$ \begin{align*}\mu = m_{\min}\alpha e^{-i\beta} \delta_{y_1}+ m_{\min}e^{i\beta} \delta_{y_2}\end{align*} $$

with $y_1 = -y_2$ , $0< y_1\leqslant \frac {\pi }{2\Omega }$ , $\alpha \geqslant 1$ and $0\leqslant \beta \leqslant \frac {\pi }{2}$ . It is not hard to see that the other cases can all be transformed to the above setting. We consider $\widehat \mu = ae^{i\gamma }\delta _{\widehat y}$ with $a>0$ , $\gamma $ and $\widehat y$ to be determined. We shall prove that if

$$\begin{align*}\left|y_1-y_2\right|\geqslant \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}, \end{align*}$$

then (67) does not hold for any $\widehat \mu $ consisting of only one source. In the opposite case, Theorem 4.1 already ensures the existence of such $\widehat \mu , \mu $ satisfying (67). By the two results, we prove the theorem.

From (67), we have

$$ \begin{align*} \mathscr F[\widehat \mu](\omega)- \mathscr F[\mu](\omega) &= ae^{i\gamma}e^{i \widehat y \omega}-m_{\min}\left(\alpha e^{-i\beta}e^{iy_1\omega}+ e^{i\beta}e^{-iy_1\omega}\right). \end{align*} $$

We rewrite it as

(68) $$ \begin{align} ae^{i\gamma}e^{i \widehat y \omega}-m_{\min}\left(\alpha e^{i(y_1 \omega-\beta)}+ e^{i(\beta-y_1 \omega)}\right) \end{align} $$

and analyze it by considering the two cases: (1) $y_1\Omega \geqslant \beta $ ; (2) $y_1\Omega <\beta $ .

Part 1: ( $y_1\Omega \geqslant \beta $ )

In the first case, when $y_1\Omega \geqslant \beta $ , we define $\omega ^*=\frac {\beta }{y_1}\in [0,\Omega ]$ . Considering

(69) $$ \begin{align} \mathscr F[\widehat \mu](\omega+\omega^*)- \mathscr F[\mu](\omega+\omega^*)&= ae^{i\gamma}e^{i \widehat y (\omega+\omega^*)}-m_{\min}\left(\alpha e^{i(y_1 \omega-\beta+y_1\omega^*)}+ e^{i(\beta-y_1\omega^*-y_1 \omega)}\right)\nonumber \\&= ae^{i\gamma+\widehat y\omega^*}e^{i \widehat y \omega}-m_{\min}\left(\alpha e^{i(y_1 \omega)}+ e^{i(-y_1 \omega)}\right), \end{align} $$

(67) is equivalent to

$$\begin{align*}\left|ae^{i\gamma+\widehat y\omega^*}e^{i \widehat y \omega}-m_{\min}\left(\alpha e^{i(y_1 \omega)}+ e^{i(-y_1 \omega)}\right)\right|<2\sigma, \quad \omega\in [-\Omega-\omega^*, \Omega-\omega^*]. \end{align*}$$

Note that this reduces the problem to a case similar to the one for positive sources. Since the interval $[-\Omega -\omega ^*, \Omega -\omega ^*]$ includes the interval $[-\Omega , 0]$ , in the same fashion as the proof for positive sources, we consider the necessary condition that

(70) $$ \begin{align} \min_{a>0, \alpha\geqslant 1, \gamma \in \mathbb R, \widehat y\in \mathbb R, 0\leqslant \beta\leqslant y_1\Omega}\left|{ \mathscr F[\widehat \mu](-\Omega+\omega^*)- \mathscr F[\mu](-\Omega+\omega^*)}\right|+\left|{\mathscr F[\widehat \mu](\omega^*)- \mathscr F[\mu](\omega^*)}\right|<4\sigma. \end{align} $$

Note that minimizing over $0\leqslant \beta \leqslant y_1\Omega $ is now equivalent to minimizing over $0\leqslant \omega ^*\leqslant \Omega $ . We thus consider

$$ \begin{align*} \min_{a>0, \alpha\geqslant 1, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -m_{\min}\left(\alpha e^{-iy_1 \Omega}+ e^{i y_1\Omega}\right)\right|+\left|ae^{i\gamma+\widehat y\omega^*} -m_{\min}\left(\alpha+1\right)\right|. \end{align*} $$

Let $\alpha = 1+h, h\geqslant 0$ , and rewrite the above formula as

$$ \begin{align*} &\min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\ &\qquad \qquad \qquad \qquad \qquad+\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|. \end{align*} $$

A key observation is that if $2m_{\min }\cos (y_1\Omega )+hm_{\min }\leqslant a\leqslant 2m_{\min }+hm_{\min }$ , we have

(71) $$ \begin{align} &\min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y\Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\nonumber \\&\qquad \qquad \qquad \qquad \qquad +\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|\nonumber \\\geqslant &\ \min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \widehat x\in \mathbb R}\left|ae^{-i\widehat y\Omega} -h m_{\min} e^{-i\widehat x \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+|ae^{i\gamma}-(2m_{\min}+hm_{\min})|\nonumber \\= &\ \min_{a> 0, h\geqslant 0}\left|a -h m_{\min}- 2m_{\min} \cos(y_1 \Omega)\right|+\left|a-(2m_{\min}+hm_{\min})\right|\nonumber \\= &\ \min_{2m_{\min}\cos(y_1\Omega)\leqslant b\leqslant 2m_{\min}}\left|b- 2m_{\min} \cos(y_1 \Omega)\right|+|2m_{\min}-b|\nonumber \\=&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega), \end{align} $$

where the second equality is because $2m_{\min }\cos (y_1\Omega )+hm_{\min }\leqslant a\leqslant 2m_{\min }+hm_{\min }$ and $2m_{\min }\cos (y_1\Omega )= 2m_{\min }\cos (\frac {d_{\min }}{2}\Omega )\geqslant 0$ for $d_{\min }\leqslant \frac {\pi }{\Omega }$ .

However, letting $h=0, \widehat y =0, \gamma =0, \omega ^*=0$ , we have

$$ \begin{align*} &\min_{a>0}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|+\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|.\\=&\ \min_{a>0}\left|a - 2m_{\min} \cos(y_1 \Omega)\right|+|a-(2m_{\min}+hm_{\min})|\\=&\ \min_{a>0}(a - 2m_{\min} \cos(y_1 \Omega))+2m_{\min}-a \quad \big(\text{choose} 2m_{\min} \cos(y_1 \Omega)\leqslant a\leqslant 2m_{\min}\big)\\=&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega). \end{align*} $$

Together with (71), this yields

$$ \begin{align*} &\min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\&\qquad \qquad \qquad \qquad \qquad+\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|\\=&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega), \end{align*} $$

in the case when $2m_{\min }\cos (y_1\Omega )+hm_{\min }\leqslant a\leqslant 2m_{\min }+hm_{\min }$ .

Now, we consider the case when $a< 2m_{\min }\cos (y_1\Omega )+hm_{\min }$ . In this case, we have

$$ \begin{align*} &\min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\&\qquad \qquad \qquad \qquad \qquad+\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|\\\geqslant &\ \min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \omega^*\in [0,\Omega], a< 2m_{\min}\cos(y_1\Omega)+hm_{\min}}|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})|\\\geqslant&\ \min_{a>0, h\geqslant 0, a< 2m_{\min}\cos(y_1\Omega)+hm_{\min}}|a-(2m_{\min}+hm_{\min})|\\=&\ \min_{a>0, h\geqslant 0, a< 2m_{\min}\cos(y_1\Omega)+hm_{\min}}2m_{\min}+hm_{\min}-a\\>&\ 2m_{\min} -2m_{\min}\cos(y_1 \Omega). \end{align*} $$

Finally, we consider the case when $a> 2m_{\min }+hm_{\min }$ . In this case, we have

$$ \begin{align*} &\min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\&\qquad \qquad \qquad \qquad \qquad+\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|\\\geqslant &\ \min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega], a> 2m_{\min}+hm_{\min}}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\\geqslant&\ \min_{a>0, h\geqslant 0, a> 2m_{\min}+hm_{\min}}a-(2m_{\min}\cos(y_1 \Omega)+hm_{\min})\\ >& 2m_{\min} -2m_{\min}\cos(y_1 \Omega). \end{align*} $$

Therefore, combining all the above discussions, we arrive at

$$ \begin{align*} &\min_{a>0, h\geqslant 0, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -h m_{\min} e^{-iy_1 \Omega}- 2m_{\min} \cos(y_1 \Omega)\right|\\&\qquad \qquad \qquad \qquad \qquad+\left|ae^{i\gamma+\widehat y\omega^*}-(2m_{\min}+hm_{\min})\right|\\=&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega), \end{align*} $$

or equivalently,

$$ \begin{align*} &\min_{a>0, \alpha\geqslant 1, \gamma \in \mathbb R, \widehat y\in \mathbb R, \omega^*\in [0,\Omega]}\left|ae^{i\gamma+\widehat y\omega^*}e^{-i\widehat y \Omega} -m_{\min}\left(\alpha e^{-iy_1 \Omega}+ e^{i y_1\Omega}\right)\right|+\left|ae^{i\gamma+\widehat y\omega^*} -m_{\min}\left(\alpha+1\right)\right|\\=&\ 2m_{\min}-2m_{\min} \cos(y_1 \Omega). \end{align*} $$

Thus, (70) is equivalent to

$$\begin{align*}2m_{\min}-2m_{\min} \cos(y_1 \Omega)<4\sigma. \end{align*}$$

Similar to the proof of Theorem 4.1, this yields

$$\begin{align*}d_{\min} < \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}. \end{align*}$$

Part 2: ( $y_1\Omega <\beta $ )

In part 2, because $y_1\Omega <\beta $ , the trick used in the former proof does not work now. We utilize another finding for the proof. Suppose $d_{\min }\geqslant \frac {4\arcsin \left (\left (\frac {\sigma }{m_{\min }}\right )^{\frac {1}{2}}\right )}{\Omega }$ and there exists some measure $\widehat \mu =a\delta _{\widehat y}$ so that

$$\begin{align*}|\mathscr F[\widehat \mu](\omega)-\mathbf{Y}(\omega)|<\sigma, \qquad \omega \in [-\Omega, \Omega]. \end{align*}$$

Then, this is in contradiction with (76) and (77) in Theorem 5.1. Thus, we have proved that

$$\begin{align*}d_{\min}< \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}. \end{align*}$$

Note that this new finding can also be used to prove the first part, but we keep the first part for a stronger understanding of the optimization problem and its underlying difficulty. The new finding comes from an optimal algorithm described in the next section. Now we have completed the proof.

Proof. Observe that $\mathbf H$ has the decomposition

(78) $$ \begin{align} \mathbf H = DAD^{\top}+\Delta, \end{align} $$

where $A=\text {diag}(e^{-iy_1\Omega }a_1, e^{-iy_2\Omega }a_2)$ and $D=\big (\phi _{1}(e^{i y_1 \Omega }), \phi _{1}(e^{i y_2\Omega })\big )$ with $\phi _{1}(\omega )$ being defined as $(1, \omega )^\top $ and

$$ \begin{align*} \Delta = \left(\begin{array}{cc} \mathbf {W}(-\Omega)& \mathbf {W}(0)\\ \mathbf {W}(0)&\mathbf {W}(\Omega) \end{array} \right). \end{align*} $$

We denote the singular values of $DAD^{\top }$ by $\sigma _1, \sigma _2$ .

We first estimate $\left |\left |{\Delta }\right |\right |{}_2$ . We have

$$ \begin{align*} &\max_{x_1^2+x_2^2=1}\left|\left|{\Delta(x_1, x_2)^{\top}}\right|\right|{}_2\\ = &\max_{x_1^2+x_2^2=1}\sqrt{(x_1\mathbf{W}(-\Omega) +x_2 \mathbf{W}(0))^2+ (x_1\mathbf{W}(0) +x_2 \mathbf{W}(\Omega))^2}\\ = &\max_{x_1^2+x_2^2=1}\sqrt{\mathbf{W}(0)^2+ 2x_1x_2\mathbf{W}(0)(\mathbf{W}(-\Omega)+\mathbf{W}(\Omega))+x_1^2\mathbf{W}(-\Omega)^2+x_2^2\mathbf{W}(\Omega)^2}\\ <&\max_{x_1^2+x_2^2=1}\sqrt{\sigma^2+ 4 \sigma^2 x_1x_2+ (x_1^2+x_2^2)\sigma^2} \quad \Big(\text{by the condition on the noise}\Big)\\ =&2\sigma. \end{align*} $$

Thus, we have $||\Delta ||_2<2\sigma $ . By Weyl’s theorem, we have

(79) $$ \begin{align} \left|{\widehat \sigma_j-\sigma_j}\right|\leqslant ||\Delta||_2<2\sigma, j=1,2. \end{align} $$

Now we estimate the minimum singular value of $DAD^{\top }$ in the presence of two sources. Denote $\sigma _{\min }(M)$ and $\lambda _{\min }(M)$ as, respectively, the minimum singular value and eigenvalue of matrix M. We have

$$ \begin{align*} \sigma_{\min}(DAD^{\top})\geqslant m_{\min}\sigma_{\min}(D)^2=m_{\min}\lambda_{\min}(DD^*)= 4m_{\min}\sin^2\left(\left|{\frac{y_1-y_2}{4}}\right|\Omega\right). \end{align*} $$

Therefore, when $y_j\in B_{\frac {(n-1) \pi }{2\Omega }}^1(0),j=1,2,$ and (75) holds, $\sigma _{\min }(DAD^{\top })\geqslant 4\sigma $ . This is $\sigma _2 \geqslant 4\sigma $ . Similarly, by Weyl’s theorem, $|\widehat \sigma _2-\sigma _2|\leqslant ||\Delta ||_2$ . Thus, $\widehat \sigma _2\geqslant 4\sigma -||\Delta ||_2>2\sigma $ . Conclusion (76) follows.

However, note that if there exists $\widehat \mu = \widehat a_1\delta _{\widehat y_1}$ consisting of one source being a $\sigma $ -admissible measure of $\mathbf{Y}$ , we can substitute the D in (78) by $\big (\phi _{1}(e^{i \widehat y_1 \Omega })\big )$ with the $\mathbf{W}$ and $\Delta $ being modified. Now we have $\sigma _2=0$ and also $||\Delta ||_2< 2\sigma $ . Thus, by (79), we get $|\widehat \sigma _2|\leqslant ||\Delta ||_2< 2\sigma $ and prove (77).

Proof. By (81), there exists $\mathbf{v}_{q^*}$ such that

$$\begin{align*}\left|{\mathbf{y}_1\cdot \mathbf{v}_{q^*}- \mathbf{y}_2\cdot \mathbf{v}_{q^*}}\right|= \cos\theta_{\min}\left|\left|{\mathbf{y}_1-\mathbf{y}_2}\right|\right|{}_2\geqslant \frac{4\arcsin\left(\left(\frac{\sigma}{m_{\min}}\right)^{\frac{1}{2}}\right)}{\Omega}. \end{align*}$$

Hence, similar to the proof of Theorem 5.1, we can show that $\widehat \sigma _{q^*, 2}> 2\sigma $ . This proves (82). Also, we can show (84) in the same way as the one in the proof of Theorem 5.1.