Proof of Corollary 2.5
By the ambiguity function relation and our assumptions, we find that
$$\begin{aligned} {\mathcal {A}}[x](m,n) = {\mathcal {A}}[y](m,n), \qquad \text{ for } m \in \{0,\dots ,\Delta +1\}, ~n\in \{0,\dots ,L-1\}. \end{aligned}$$
Therefore,
$$\begin{aligned} x(\ell ) \overline{x(\ell -m)} = y(\ell ) \overline{y(\ell -m)}, \qquad \text{ for } \ell \in \{0,\dots ,L-1\},~m \in \{0,\dots ,\Delta +1\},\nonumber \\ \end{aligned}$$
(5)
and in particular x and y have the same magnitudes. Let us now consider \(k \in \{1,\dots ,K\}\) as well as some \(\ell _0\) in \(V_k\). We have \(|x(\ell _0) |= |y(\ell _0) |\) and hence \(x(\ell _0) = \mathrm {e}^{\mathrm {i} \alpha _k} y(\ell _0)\), for some \(\alpha _k \in {\mathbb {R}}\). As \(V_k\) is the vertex set of a connected component of G, it follows that for all \(\ell \in V_k \setminus \{\ell _0\}\), there exists a (simple) path from \(\ell _0\) to \(\ell \). Therefore, we can consider \(\ell \in V_k \setminus \{\ell _0\}\) and let \((u_0,\dots ,u_n)\) be the vertex sequence of the path from \(u_0 = \ell _0\) to \(u_n = \ell \). For \(j \in \{0,\dots ,n-1\}\) one has, by definition of the edge set, that
$$\begin{aligned} |u_{j+1} - u_j |\in (0,\Delta +1] \cup [L-\Delta -1,L). \end{aligned}$$
Thus, there exists an \(m_j \in \{1,\dots ,\Delta +1\}\) such that \(u_{j+1} - u_j = m_j \mod L\) or \(u_j - u_{j+1} = m_j \mod L\). In either case, it follows from Eq. (5) that \(x(u_{j+1}) \overline{x(u_j)} = y(u_{j+1}) \overline{y(u_j)}\). By induction on j, we find that \(x(\ell ) = \mathrm {e}^{\mathrm {i} \alpha _k} y(\ell )\). \(\square \)
Proof of Theorem 3.6
The case \(4B \le L\) is similar to the case \(4B > L\) but simpler: Indeed consider \(4B \le L\). In this case, we have
$$\begin{aligned} \min _{\begin{array}{c} m \in \{0,1\}\\ n \in \{0,\dots ,L-1\} \end{array}} |{\mathcal {A}}[\phi ](m,n) |\ge \frac{1}{c} \end{aligned}$$
by assumption. Therefore, we can replace all sums over \(\{-2B,\dots ,2B\}\) by sums over \(\{0,\dots ,L-1\}\) in this proof. So let us consider \(4B > L\): Let \(\alpha \in {\mathbb {R}}\) be arbitrary. Employing Proposition A.1, we have
$$\begin{aligned} \left\Vert x - \mathrm {e}^{\mathrm {i} \alpha } y \right\Vert _{\ell ^2(V_\delta )} \le \left\Vert |x |- |y |\right\Vert _{\ell ^2(V_\delta )} + \left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
(6)
The magnitude difference is estimated as in Lemma 3.4. For the estimate of the phase difference, we develop inequalities in the following. Let \(\ell ,k \in S_\delta \). According to Proposition A.2, we have
$$\begin{aligned} \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|\le \left|\frac{x(k)}{|x(k) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(k)}{\vert y(k) |} \right|+ \frac{ 2 \left|x(\ell ) \overline{x(k)} - y(\ell ) \overline{y(k)} \right|}{ |x(\ell )x(k) |}. \end{aligned}$$
Using the above inequality recursively, one obtains that for all \(M \in {\mathbb {N}}\) and \(u_0,u_1,\dots ,u_M \in S_\delta \):
$$\begin{aligned}&\left|\frac{x(u_M)}{|x(u_M) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(u_M)}{|y(u_M) |} \right|\le \left|\frac{x(u_0)}{|x(u_0) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(u_0)}{|y(u_0) |} \right|\nonumber \\&\quad +\,2 \sum _{j=0}^{M-1} \frac{ \left|x(u_{j+1}) \overline{x(u_j)} - y(u_{j+1}) \overline{y(u_j)} \right|}{|x(u_{j+1}) x(u_j) |}. \end{aligned}$$
(7)
Suppose now that \(\ell _0\) is chosen such that any other vertex \(\ell \in S_\delta \) has graph distance (in the induced subgraph) at most \(|S_\delta |/2 \) from \(\ell _0\). Then, for any \(\ell \in S_\delta \setminus \{ \ell _0 \}\), there exists \(M(\ell ) \in {\mathbb {N}}\), with \(M(\ell ) \le |S_\delta |/ 2\), and a sequence \(u_0^\ell = \ell _0, u_1^\ell , \dots , u_{M(\ell )}^\ell = \ell \) in \(S_\delta \) such that (cf. Definition 3.1 with \(\Delta = 0\))
$$\begin{aligned} |u_{j+1}^\ell - u_j^\ell |\in \{1,L-1\}, \qquad \text{ for } j = 0,\dots ,M(\ell )-1. \end{aligned}$$
Therefore, there exists a sequence \(\sigma _1^\ell ,\dots ,\sigma _{M(\ell )}^\ell \) in \(\{ -1,1 \}\) such that
$$\begin{aligned} u_{j+1}^\ell - u_j^\ell = \sigma _{j+1}^\ell \mod L, \qquad \text{ for } j = 0,\dots ,M(\ell )-1. \end{aligned}$$
Now, let \(\alpha \in {\mathbb {R}}\) be such that
$$\begin{aligned} \left|\frac{x(\ell _0)}{|x(\ell _0) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell _0)}{|y(\ell _0) |} \right|= 0. \end{aligned}$$
Then, we have for any \(\ell \in S_\delta \), according to the above considerations (and inequality (7)), that
$$\begin{aligned} \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|\le 2 \sum _{j=1}^{M(\ell )} \frac{ \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell )} \right|}{|x(u_j^\ell ) x(u_j^\ell - \sigma _j^\ell ) |}. \end{aligned}$$
For the second term of the right-hand side of inequality (6) this yields
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2 \left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left( \sum _{j=1}^{M(\ell )} \frac{ \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell )} \right|}{ |x(u_j^\ell ) x(u_j^\ell - \sigma _j^\ell ) |} \right) ^2 \right) ^\frac{1}{2}. \end{aligned}$$
Applying Jensen’s inequality on the square of the inner sum and noting that \(M(\ell ) \le |S_\delta |/ 2\), we obtain
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \sqrt{2 |S_\delta |} \left( \sum _{\ell \in S_\delta } \sum _{j=1}^{M(\ell )} \frac{ \left|x(\ell ) \right|^2 }{ |x(u_j^\ell ) x(u_j^\ell - \sigma _j^\ell ) |^2 } \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell )} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
Since \(u^\ell _j \in S_\delta \), for \(j \in \{1,\dots ,M(\ell )\}\), we can further estimate
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{\sqrt{2 |S_\delta |}}{\delta ^2} \left( \sum _{\ell \in S_\delta } \sum _{j=1}^{M(\ell )} \left|x(\ell ) \right|^2 \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell )} \right|^2 \right) ^\frac{1}{2} \end{aligned}$$
and with \(\sigma ^\ell _j \in \{ \pm 1 \}\), for \(j \in \{1,\dots ,M(\ell )\}\), we also get
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{\sqrt{2 |S_\delta |}}{\delta ^2} \left( \sum _{\sigma \in \{-1,1\}} \sum _{\ell \in S_\delta } \sum _{j=1}^{M(\ell )} \left|x(\ell ) \right|^2 \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma )} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
There are no repetitions in the sequences \(u^\ell _1,u^\ell _2,\dots ,u^\ell _{M(\ell )}\) and hence
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{\sqrt{2 |S_\delta |}}{\delta ^2} \left( \sum _{\sigma \in \{-1,1\}} \sum _{\ell \in S_\delta } \sum _{u \in S_\delta } \left|x(\ell ) \right|^2 \left|x(u) \overline{x(u - \sigma )} - y(u) \overline{y(u - \sigma )} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
Therefore, we have
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{ \sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )} }{ \delta ^2 } \left( \sum _{\sigma \in \{-1,1\}} \sum _{u = 0}^{L-1} \left|x(u) \overline{x(u - \sigma )} - y(u) \overline{y(u - \sigma )} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
Suppose now that \(4B < L\). By Plancherel’s theorem, it holds that
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{ \sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )} }{ \delta ^2 } \left( \sum _{\sigma \in \{-1,1\}} \sum _{n = 0}^{L-1} \left|{\mathcal {A}}[x](\sigma ,n) - {\mathcal {A}}[y](\sigma ,n) \right|^2 \right) ^\frac{1}{2} \\&\quad = \frac{ \sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )} }{ \delta ^2 } \left( \sum _{\sigma \in \{-1,1\}} \sum _{n = -2B}^{2B} \left|{\mathcal {A}}[x](\sigma ,n) - {\mathcal {A}}[y](\sigma ,n) \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
It follows from the ambiguity function relation and the lower bound on the ambiguity function of the window on \(\{0,1\} \times \{-2B,\dots ,2B\}\) that
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le c \cdot \frac{ \sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )} }{ \delta ^2 } \left( \sum _{\sigma \in \{-1,1\}} \sum _{n = -2B}^{2B} \left|{\mathcal {F}}\left[ M_\phi [x] - M_\phi [y] \right] (n,-\sigma ) \right|^2 \right) ^\frac{1}{2} \\&\quad \le c \cdot \frac{ \sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )} }{ \delta ^2 } \cdot \left\Vert {\mathcal {F}}\left[ M_\phi [x] - M_\phi [y] \right] \right\Vert _{\mathrm {F}} \\&\quad = c \cdot \frac{ \sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )} }{ \delta ^2 } \cdot \left\Vert M_\phi [x] - M_\phi [y] \right\Vert _{\mathrm {F}}, \end{aligned}$$
where we have used Plancherel’s theorem in the last equality. \(\square \)
Proof of Theorem 3.11
Let \(k \in \{1,\dots ,K\}\) and \(\alpha _k \in {\mathbb {R}}\). As in the proof of Theorem 3.6, we start by splitting the estimate into a phase and a magnitude estimate using Proposition A.1:
$$\begin{aligned} \left\Vert x - \mathrm {e}^{\mathrm {i} \alpha _k} y \right\Vert _{\ell ^2(S_k)} \le \left\Vert |x |- |y |\right\Vert _{\ell ^2(S_k)} + \left( \sum _{\ell \in S_k} |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
As the connected components \(\{S_k\}_{k=1}^K\) are disjoint subsets of \(V_\delta \), we can use Jensen’s inequality to see that
$$\begin{aligned} \sum _{k=1}^K \left\Vert |x |- |y |\right\Vert _{\ell ^2(S_k)} \le \sqrt{K} \left( \sum _{k=1}^K \left\Vert |x |- |y |\right\Vert _{\ell ^2(S_k)}^2 \right) ^\frac{1}{2}&= \sqrt{K} \left\Vert |x |- |y |\right\Vert _{\ell ^2(\bigcup _{k=1}^K S_k)} \\&\le \sqrt{K} \left\Vert |x |- |y |\right\Vert _{\ell ^2(V_\delta )}. \end{aligned}$$
Employing Lemma 3.4, we obtain for the magnitude retrieval estimate
$$\begin{aligned} \left\Vert |x |- |y |\right\Vert _{\ell ^2(V_\delta )} \le \frac{c}{\delta } \left\Vert M_\phi [x]-M_\phi [y] \right\Vert _\mathrm {F}. \end{aligned}$$
The phase difference is estimated just like in Theorem 3.6: First, observe that there must exist a vertex \(\ell _0 \in S_k\) such that any other vertex \(\ell \in S_k\) has graph distance \(M(\ell )\) at most \(\tfrac{L+\Delta }{2+\Delta }\) from \(\ell _0\). Indeed, consider the following argument: The worst case which could happen is that we need to connect the vertex 0 to the vertex \(\lfloor L/2 \rfloor \). By definition of the graph, it will take us exactly one step to go from 0 to any \(\ell \in \{1,\dots ,\Delta +1\} \cap S_k\); it will take us exactly two steps to go from 0 to \(\Delta + 2\), if the latter is in \(S_k\); and it will take us at most three steps to go from 0 to \(\ell \in \{\Delta +3,2\Delta +3\} \cap S_k\). Following this logic, it is not too hard to see that it will take us at most 2n steps to go from 0 to \(n(\Delta +2)\), if the latter is in \(S_k\), and it will take us at most \(2n+1\) steps to go from 0 to \(\ell \in \{n(\Delta +2)+1,\dots ,(n+1)(\Delta +2)-1\} \cap S_k\). So, if there exists an element \(n \in {\mathbb {N}}\) such that \(n(\Delta +2) = \lfloor L/2 \rfloor \), then it will take us at most
$$\begin{aligned} 2n = \frac{2 \lfloor L/2 \rfloor }{\Delta + 2} \le \frac{L}{\Delta +2} \le \frac{L+\Delta }{2 + \Delta } \end{aligned}$$
steps to connect 0 and \(\lfloor L/2 \rfloor \). Similarly, if there is an element \(n \in {\mathbb {N}}_0\) such that \(\lfloor L/2 \rfloor \in \{n(\Delta +2)+1,\dots , (n+1)(\Delta +2)-1\}\), then there is a \(\beta \in \{1,\dots ,\Delta +1\}\) such that
$$\begin{aligned} 2n+1 = \frac{2 \lfloor L/2 \rfloor - 2\beta }{\Delta + 2} + 1 \le \frac{L - 2}{\Delta + 2} + 1 = \frac{L + \Delta }{\Delta +2}. \end{aligned}$$
So for any \(\ell \in S_k \setminus \{\ell _0\}\), there exists a path \(u_0^\ell = \ell _0,u_1^\ell ,\dots ,u_{M(\ell )}^\ell = \ell \) from \(\ell _0\) to \(\ell \). By definition, this path satisfies
$$\begin{aligned} \left|u_{j+1}^\ell - u_j^\ell \right|\in (0,\Delta +1] \cup [L-\Delta -1,L), \qquad \text{ for } j = 0,\dots ,M(\ell )-1. \end{aligned}$$
Therefore, there exist sequences \(\sigma _1^\ell ,\dots ,\sigma _{M(\ell )}^\ell \in \{\pm 1\}\) and \(\Delta _1^\ell ,\dots ,\Delta _{M(\ell )}^\ell \in \{1,\dots ,\Delta +1\}\) such that
$$\begin{aligned} u_{j+1}^\ell - u_j^\ell = \sigma _{j+1}^\ell \Delta _{j+1}^\ell \mod L, \qquad \text{ for } j = 0, \dots ,M(\ell )-1. \end{aligned}$$
We let \(\alpha _k \in {\mathbb {R}}\) be chosen in such a way that
$$\begin{aligned} \left|\frac{x(\ell _0)}{|x(\ell _0) |} - \mathrm {e}^{\mathrm {i} \alpha _k} \frac{y(\ell _0)}{|y(\ell _0) |} \right|= 0. \end{aligned}$$
Proceeding as in the proof of Theorem 3.6 (now with \(M(\ell ) \le (L+\Delta )/(2+\Delta )\)), we derive
$$\begin{aligned}&\left( \sum _{\ell \in S_k} |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha _k} \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{1}{\delta ^2} \left( \sum _{\ell \in S_k} \sum _{j=1}^{M(\ell )} \left|x(\ell ) \right|^2 \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell \Delta _j^\ell )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell \Delta _j^\ell )} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
We first treat the case \(2\Delta < L-2\) and use that \(\sigma _j^\ell \Delta _j^\ell \in \{-\Delta -1,\dots ,\Delta +1\}\) to estimate
$$\begin{aligned}&\left( \sum _{\ell \in S_k} |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha _k} \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{1}{\delta ^2} \left( \sum _{m = -\Delta -1}^{\Delta +1} \sum _{\ell \in S_k} \sum _{j=1}^{M(\ell )} \left|x(\ell ) \right|^2 \left|x(u_j^\ell ) \overline{x(u_j^\ell - m)} - y(u_j^\ell ) \overline{y(u_j^\ell - m)} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
We may use that there are no repetitions in the sequences \(u_1^\ell ,\dots , u_{M(\ell )}^\ell \) to obtain
$$\begin{aligned}&\left( \sum _{\ell \in S_k} |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha _k} \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{1}{\delta ^2} \left( \sum _{m = -\Delta -1}^{\Delta +1} \sum _{\ell \in S_k} \sum _{u = 0}^{L-1} \left|x(\ell ) \right|^2 \left|x(u) \overline{x(u - m)} - y(u) \overline{y(u - m)} \right|^2 \right) ^\frac{1}{2} \\&\quad = 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _{\ell ^2(S_k)}}{\delta ^2} \left( \sum _{m = -\Delta -1}^{\Delta +1} \sum _{u = 0}^{L-1} \left|x(u) \overline{x(u - m)} - y(u) \overline{y(u - m)} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
Suppose furthermore that \(4B > L\). According to Plancherel’s theorem, we find that
$$\begin{aligned}&\left( \sum _{\ell \in S_k} |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha _k} \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _{\ell ^2(S_k)}}{\delta ^2} \left( \sum _{m = -\Delta -1}^{\Delta +1} \sum _{n = 0}^{L-1} \left|{\mathcal {A}}[x](m,n) - {\mathcal {A}}[y](m,n) \right|^2 \right) ^\frac{1}{2} \\&\quad = 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _{\ell ^2(S_k)}}{\delta ^2} \left( \sum _{m = -\Delta -1}^{\Delta +1} \sum _{n = -2B}^{2B} \left|{\mathcal {A}}[x](m,n) - {\mathcal {A}}[y](m,n) \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
Next, we use the ambiguity function relation, inequality (4) as well as the symmetry of the ambiguity function of the window around the origin to derive
$$\begin{aligned}&\left( \sum _{\ell \in S_k} |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha _k} \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2c \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _{\ell ^2(S_k)}}{\delta ^2} \left( \sum _{m = -\Delta -1}^{\Delta +1} \sum _{n = -2B}^{2B} \left|{\mathcal {F}}\left[ M_\phi [x] - M_\phi [y] \right] (n,m) \right|^2 \right) ^\frac{1}{2} \\&\quad \le 2c \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _{\ell ^2(S_k)}}{\delta ^2} \left\Vert {\mathcal {F}}\left[ M_\phi [x] - M_\phi [y] \right] \right\Vert _2 \\&\qquad \qquad \qquad \qquad = 2c \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _{\ell ^2(S_k)}}{\delta ^2} \left\Vert M_\phi [x] - M_\phi [y] \right\Vert _2. \end{aligned}$$
Note that we may once again use Jensen’s inequality to show that
$$\begin{aligned} \sum _{k=1}^K \Vert x \Vert _{\ell ^2(S_k)} \le \sqrt{K} \Vert x \Vert _2. \end{aligned}$$
Thus, combining the phase and the magnitude estimates yields
$$\begin{aligned} \inf _{\alpha _1,\dots ,\alpha _K} \sum _{k=1}^K \left\Vert x - \mathrm {e}^{\mathrm {i} \alpha _k} y \right\Vert _{\ell ^2(S_k)} \le \frac{c \sqrt{K}}{\delta } \left( 1 + 2 \sqrt{\frac{L+\Delta }{2+\Delta }} \frac{\Vert x \Vert _2}{\delta } \right) \left\Vert M_\phi [x] - M_\phi [y] \right\Vert _2. \end{aligned}$$
The cases in which \(2\Delta \ge L-2\) or \(4B \ge L\) are dealt with similarly. \(\square \)
Proof of Theorem 4.1
Let \(\alpha \in {\mathbb {R}}\). As in the proof of Theorem 3.6, we start by splitting the estimate into a phase and a magnitude estimate
$$\begin{aligned} \left\Vert x - \mathrm {e}^{\mathrm {i} \alpha _k} y \right\Vert _{\ell ^2(V_\delta )} \le \left\Vert |x |- |y |\right\Vert _{\ell ^2(V_\delta )} + \left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2}. \end{aligned}$$
First, noting that
$$\begin{aligned} {\mathcal {F}}\left[ |\phi |^2 \right] (k) = {\mathcal {A}}[\phi ](0,k), \qquad \text{ for } k = 0,\dots ,L-1, \end{aligned}$$
we can apply Lemma 3.4 to obtain the estimate
$$\begin{aligned} \left\Vert |x |- |y |\right\Vert _{\ell ^2(V_\delta )} \le \frac{c}{\delta } \left\Vert M_\phi [x]-M_\phi [y] \right\Vert _\mathrm {F}. \end{aligned}$$
For the estimate on the phase retrieval part, we need to consider a new strategy based on Eq. (1). First, we find that there must exist a vertex \(\ell _0 \in S_\delta \) such that any other vertex \(\ell \in S_\delta \) has graph distance \(M(\ell )\) at most \(|S_\delta |/2\) from \(\ell _0\). So for any \(\ell \in S_\delta \setminus \{\ell _0\}\), there exists a path \(u_0^\ell = \ell _0,u_1^\ell ,\dots ,u_{M(\ell )}^\ell = \ell \) from \(\ell _0\) to \(\ell \). By definition, this path satisfies
$$\begin{aligned} \left|u_{j+1}^\ell - u_j^\ell \right|\in \{\ell _\phi , L-\ell _\phi \}, \qquad \text{ for } j = 0,\dots ,M(\ell )-1. \end{aligned}$$
Therefore, there exists a sequence \(\sigma _1^\ell ,\dots ,\sigma _{M(\ell )}^\ell \in \{\pm 1\}\) such that
$$\begin{aligned} u_{j+1}^\ell - u_j^\ell = \sigma _{j+1}^\ell \ell _\phi \mod L, \qquad \text{ for } j = 0, \dots ,M(\ell )-1. \end{aligned}$$
With this at hand, we proceed similarly to the proof of Theorem 3.6. We let \(\alpha \in {\mathbb {R}}\) be chosen in such a way that
$$\begin{aligned} \left|\frac{x(\ell _0)}{|x(\ell _0) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell _0)}{|y(\ell _0) |} \right|= 0. \end{aligned}$$
Then, we have that for any \(\ell \in S_\delta \)
$$\begin{aligned} \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|\le 2 \sum _{j=1}^{M(\ell )} \frac{ \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell \ell _\phi )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell \ell _\phi )} \right|}{|x(u_j^\ell ) x(u_j^\ell - \sigma _j^\ell \ell _\phi ) |} \end{aligned}$$
and
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{\sqrt{2 |S_\delta |}}{\delta ^2} \left( \sum _{\ell \in S_\delta } \sum _{j=1}^{M(\ell )} \left|x(\ell ) \right|^2 \left|x(u_j^\ell ) \overline{x(u_j^\ell - \sigma _j^\ell \ell _\phi )} - y(u_j^\ell ) \overline{y(u_j^\ell - \sigma _j^\ell \ell _\phi )} \right|^2 \right) ^\frac{1}{2}, \end{aligned}$$
by Jensen’s inequality, \(M(\ell ) \le |S_\delta |/ 2\) and the fact that \(u_j^\ell , u_j^\ell - \sigma _j^\ell \ell _\phi \in S_\delta \), for all \(j \in \{1,\dots ,M(\ell )\}\). As before, we can further employ a crude estimate to derive
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad = \frac{\sqrt{2 |S_\delta |} \Vert x \Vert _{\ell ^2(S_\delta )}}{\delta ^2} \left( \sum _{\sigma \in \{\pm 1\}} \sum _{u \in S_\delta } \left|x(u) \overline{x(u - \sigma \ell _\phi )} - y(u) \overline{y(u - \sigma \ell _\phi )} \right|^2 \right) ^\frac{1}{2}, \end{aligned}$$
since \(\sigma _j^\ell \in \{\pm 1\}\), for all \(j \in \{1,\dots ,M(\ell )\}\) and all \(\ell \in S_\delta \), and because for fixed \(\ell \in S_\delta \), the \(u_j^\ell \) are all distinct. Next, we note that due to \({\text {supp}} \phi = \{n_0,\dots ,n_0 + \ell _\phi \} \mod L\), \(2\ell _\phi < L\), and the autocorrelation relation, we have for \(u \in S_\delta \)
$$\begin{aligned}&x(u)\overline{x(u-\ell _\phi )} \phi (n_0) \overline{\phi (n_0 + \ell _\phi )}\\&\quad = \sum _{\ell = 0}^{L-1} x(\ell ) \overline{x(\ell -\ell _\phi )} \phi (\ell + n_0 - u) \overline{ \phi (\ell + n_0 + \ell _\phi - u) } \\&\quad = \sqrt{L} {\mathcal {F}}^{-1} [M_\phi [x](u-n_0-\ell _\phi ,\cdot )](\ell _\phi ), \end{aligned}$$
as well as
$$\begin{aligned}&x(u)\overline{x(u+\ell _\phi )} \phi (n_0+\ell _\phi ) \overline{\phi (n_0)}\\&\quad = \sum _{\ell = 0}^{L-1} x(\ell ) \overline{x(\ell +\ell _\phi )} \phi (\ell + \ell _\phi + n_0 - u) \overline{ \phi (\ell + n_0 - u) } \\&\quad = \sqrt{L} {\mathcal {F}}^{-1} [M_\phi [x](u-n_0,\cdot )](-\ell _\phi ). \end{aligned}$$
This implies
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{\sqrt{2 |S_\delta |L} \Vert x \Vert _{\ell ^2(S_\delta )}}{\delta ^2 |\phi (n_0) \phi (n_0 + \ell _\phi ) |}\\&\qquad \Bigg ( \sum _{u \in S_\delta } \Big ( \left|{\mathcal {F}}^{-1} [M_\phi [x](u-n_0-\ell _\phi ,\cdot ) - M_\phi [y](u-n_0-\ell _\phi ,\cdot )](\ell _\phi ) \right|^2 \\&\qquad + \left|{\mathcal {F}}^{-1} [M_\phi [x](u-n_0,\cdot ) - M_\phi [y](u-n_0,\cdot )](-\ell _\phi ) \right|^2 \Big ) \Bigg )^\frac{1}{2}. \end{aligned}$$
Yet another crude estimate results in
$$\begin{aligned}&\left( \sum _{\ell \in S_\delta } |x(\ell ) |^2 \left|\frac{x(\ell )}{|x(\ell ) |} - \mathrm {e}^{\mathrm {i} \alpha } \frac{y(\ell )}{|y(\ell ) |} \right|^2 \right) ^\frac{1}{2} \\&\quad \le \frac{2 \sqrt{|S_\delta |L} \Vert x \Vert _{\ell ^2(S_\delta )}}{\delta ^2 |\phi (n_0) \phi (n_0 + \ell _\phi ) |} \left( \sum _{m,n = 0}^{L-1} \left|{\mathcal {F}}^{-1} [M_\phi [x](m,\cdot ) - M_\phi [y](m,\cdot )](n) \right|^2 \right) ^\frac{1}{2} \\&\quad = \frac{2 \sqrt{|S_\delta |L} \Vert x \Vert _{\ell ^2(S_\delta )}}{\delta ^2 |\phi (n_0) \phi (n_0 + \ell _\phi ) |} \left\Vert M_\phi [x] - M_\phi [y] \right\Vert _{\mathrm {F}}. \end{aligned}$$
\(\square \)