The $\ell _1$-Analysis in Phase Retrieval with Redundant Dictionary

Abstract

This article presents new results concerning the recovery of a signal from the magnitude only measurements where the signal is not sparse in an orthonormal basis but in a redundant dictionary, which we call it phase retrieval with redundant dictionary for short. To solve this phaseless problem, we analyze the $\ell _1$-analysis model. Firstly we investigate the noiseless case with presenting a null space property of the measurement matrix under which the $\ell _1$-analysis model provides an exact recovery. Secondly we introduce a new property (S-DRIP) of the measurement matrix. By solving the $\ell _1$-analysis model, we prove that this property can guarantee a stable recovery of real signals that are nearly sparse in overcomplete dictionaries.

Appendix

The following two lemmas are useful in the proof of Lemma 4.1.

Lemma 6.1

(Sparse representation of a polytope [6, 21]): Suppose $\alpha >0$ is a constant and $s>0$ is an integer. Set

$$\begin{aligned} T(\alpha ,s):=\{v\in \mathbb {R}^n: \Vert v\Vert _\infty \le \alpha ,\ \Vert v\Vert _1\le s\alpha \}. \end{aligned}$$

For any $v\in \mathbb {R}^n$, set

$$\begin{aligned} U (\alpha ,s,v):=\{u\in \mathbb {R}^n:supp (u)\subseteq supp (v),\Vert u\Vert _0\le s,\Vert u\Vert _1=\Vert v\Vert _1,\Vert u\Vert _\infty \le \alpha \}. \end{aligned}$$

Then $v\in T(\alpha ,s)$ if and only if v is in the convex hull of $U (\alpha ,s,v)$. In particular, any $v\in T(\alpha ,s)$ can be expressed as

$$\begin{aligned} v=&\sum _{i=1}^{M}\lambda _iu_i\quad \text {and}\quad 0\le \lambda _i\le 1,\, \sum _{i=1}^{M}\lambda _i=1,\\&u_i\in U (\alpha ,s,v). \end{aligned}$$

Lemma 6.2

(Lemma 5.3 in [5]): Suppose $m\ge r$, $a_1\ge a_2\ge \cdots \ge a_m\ge 0$ and $\sum _{i=1}^{r}a_i\ge \sum _{i=r+1}^{m}a_i$. Then for all $\alpha \ge 1$, we have

$$\begin{aligned} \sum _{j=r+1}^{m}a_j^\alpha \le \sum _{i=1}^{r}a_i^\alpha . \end{aligned}$$

Now we are ready to prove Lemma 4.1.

Proof of the Lemma 4.1

We assume that the tight frame $D\in {\mathbb R}^{n\times N}$ is normalized, i.e., $DD^*=I$ and $\Vert y\Vert _2=\Vert D^*y\Vert _2$ for all $y\in \mathbb {R}^n$. For a subset $T\subseteq \{1,2,\ldots ,N\}$, we denote $D_T$ as the matrix D restricted to the columns indexed by T (replacing other columns by zero vectors).

Set $h:=\hat{x}-x_0$. Let $T_0$ denote the index set of the largest k coefficients of $D^*x_0$ in magnitude. Then

$$\begin{aligned} \Vert D^*x_0\Vert _1+\rho \ge \Vert D^*\hat{x}\Vert _1&=\Vert D^*x_0+D^*h\Vert _1\\ {}&=\Vert D^*_{T_0}x_0+D^*_{T_0}h+D^*_{T_0^c}x_0+D^*_{T_0^c}h\Vert _1\\ {}&\ge \Vert D^*_{T_0}x_0\Vert _1-\Vert D^*_{T_0}h\Vert _1-\Vert D^*_{T_0^c}x_0\Vert _1+\Vert D^*_{T_0^c}h\Vert _1, \end{aligned}$$

which implies

$$\begin{aligned} \Vert D^*_{T_0^c}h\Vert _1&\le \Vert D^*_{T_0}h\Vert _1+2\Vert D^*_{T_0^c}x_0\Vert _1+\rho \\&=\Vert D^*_{T_0}h\Vert _1+2\sigma _k(D^*x_0)_1+\rho . \end{aligned}$$

Suppose $S_0$ is the index set of the k largest entries in absolute value of $D^*h$. We get

$$\begin{aligned} \Vert D^*_{S_0^c}h\Vert _1\le \Vert D^*_{T_0^c}h\Vert _1&\le \Vert D^*_{T_0}h\Vert _1+2\sigma _k(D^*x_0)_1+\rho \\&\le \Vert D^*_{S_0}h\Vert _1+2\sigma _k(D^*x_0)_1+\rho . \end{aligned}$$

Set

$$\begin{aligned} \alpha :=\frac{\Vert D^*_{S_0}h\Vert _1+2\sigma _k(D^*x_0)_1+\rho }{k}. \end{aligned}$$

We divide $D^*_{S_0^c }h$ into two parts $D^*_{S_0^c }h=h^{(1)}+h^{(2)}$, where

$$\begin{aligned} h^{(1)}:=D^*_{S_0^c}h\cdot I_{\{i:\,\,|D^*_{S_0^c}h(i)|>\alpha /(t-1)\}}, \quad h^{(2)}:=D^*_{S_0^c}h\cdot I_{\{i:\,\,|D^*_{S_0^c}h(i)|\le \alpha /(t-1)\}} . \end{aligned}$$

Then a simple observation is that $\Vert h^{(1)}\Vert _1\le \Vert D^*_{S_0^c}h\Vert _1\le \alpha k$. Set

$$\begin{aligned} \ell := |supp (h^{(1)})|=\Vert h^{(1)}\Vert _0 . \end{aligned}$$

Since all non-zero entries of $h^{(1)}$ have magnitude larger than $\alpha /(t-1)$, we have

$$\begin{aligned} \alpha k\ge \Vert h^{(1)}\Vert _1=\sum _{i\in supp (h^{(1)})}|h^{(1)}(i)|\ge \sum _{i\in supp (h^{(1)})}\frac{\alpha }{t-1}=\ell \cdot \frac{\alpha }{t-1}, \end{aligned}$$

which implies $\ell \le (t-1)k$.

Note that

$$\begin{aligned} \Vert h^{(2)}\Vert _1&=\Vert D^*_{S_0^c}h\Vert _1-\Vert h^{(1)}\Vert _1\le k\alpha -\ell \cdot \frac{ \alpha }{t-1}=(k(t-1)-\ell )\frac{\alpha }{t-1},\\ \Vert h^{(2)}\Vert _\infty&\le \frac{\alpha }{t-1}. \end{aligned}$$

Then in Lemma 6.1, by setting $s:=k(t-1)-\ell$, we can express $h^{(2)}$ as a weighted mean:

$$\begin{aligned} h^{(2)}=\sum _{i=1}^{M}\lambda _iu_i, \end{aligned}$$

where $0\le \lambda _i\le 1$, $\sum _{i=1}^{M}\lambda _i=1$, $\Vert u_i\Vert _0\le k(t-1)-\ell$, $\Vert u_i\Vert _\infty \le \alpha /(t-1)$ and $supp (u_i)\subseteq supp (h^{(2)})$. Thus

$$\begin{aligned} \Vert u_i\Vert _2\le \sqrt{\Vert u_i\Vert _0}\cdot \Vert u_i\Vert _\infty&=\sqrt{k(t-1)-\ell }\cdot \Vert u_i\Vert _\infty \\&\le \sqrt{k(t-1)}\cdot \Vert u_i\Vert _\infty \\&\le \alpha \sqrt{k/(t-1)}. \end{aligned}$$

Recall that $\alpha =\frac{\Vert D^*_{S_0}h\Vert _1+2\sigma _k(D^*x_0)_1+\rho }{k}$. Then

$$\begin{aligned} \Vert u_i\Vert _2&\le \alpha \sqrt{k/(t-1)} \nonumber \\&\le \frac{\Vert D^*_{S_0}h\Vert _2}{\sqrt{t-1}}+\frac{2\sigma _k(D^*x_0)_1+\rho }{\sqrt{k(t-1)}}\nonumber \\&\le \frac{\Vert D^*_{S_0}h+h^{(1)}\Vert _2}{\sqrt{t-1}}+\frac{2\sigma _k(D^*x_0)_1+\rho }{\sqrt{k(t-1)}}\nonumber \\&=\frac{z+R}{\sqrt{t-1}}, \end{aligned}$$

(6.26)

where $z:=\Vert D^*_{S_0}h+h^{(1)}\Vert _2, \,\, R:=\frac{2\sigma _k(D^*x_0)_1+\rho }{\sqrt{k}}$.

Now we suppose $0\le \mu \le 1$, $d\ge 0$ are two constants to be determined. Set

$$\begin{aligned} \beta _j:=D^*_{S_0}h+h^{(1)}+\mu \cdot u_j,\,\, j=1,\ldots ,M. \end{aligned}$$

Then for any fixed $i\in [1:M]$,

$$\begin{aligned} \sum _{j=1}^{M}\lambda _j\beta _j-d\beta _i&=D^*_{S_0}h+h^{(1)}+\mu \cdot h^{(2)}-d\beta _i\\\nonumber&=(1-\mu -d)(D^*_{S_0}h+h^{(1)})-d\mu u_i+\mu D^*h. \end{aligned}$$

(6.27)

For $\sum _{i=1}^M\lambda _i=1$, we have the following identity

$$\begin{aligned}&(2d-1)\sum _{1\le i<j\le M}\lambda _i\lambda _j\Vert AD(\beta _i-\beta _j)\Vert _2^2\nonumber \\&\quad = \sum _{i=1}^{M}\lambda _i\Vert AD(\sum _{j=1}^{M}\lambda _j\beta _j-d\beta _i)\Vert _2^2 -\sum _{i=1}^{M}\lambda _i(1-d)^2\Vert AD\beta _i\Vert _2^2. \end{aligned}$$

(6.28)

In (6.27), we chose $d=1/2$ and $\mu =\sqrt{t(t-1)}-(t-1)< 1/2$. Then (6.28) implies

$$\begin{aligned} 0&=\sum _{i=1}^{M}\lambda _i\Vert AD(\sum _{j=1}^{M}\lambda _j\beta _j-d\beta _i)\Vert _2^2-\sum _{i=1}^{M}\frac{\lambda _i}{4}\Vert AD\beta _i\Vert _2^2\nonumber \nonumber \\&\overset{(2.27)}{=}\sum _{i=1}^{M}\lambda _i\Vert AD\left( (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}u_i+\mu D^*h\right) \Vert _2^2-\sum _{i=1}^{M}\frac{\lambda _i}{4}\Vert AD\beta _i\Vert _2^2\nonumber \\&=\sum _{i=1}^{M}\lambda _i\Vert AD\left( (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}u_i\right) \Vert _2^2\nonumber \\&\qquad +2\left\langle AD\left( (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}h^{(2)}\right) , \mu ADD^*h\right\rangle +\mu ^2\Vert ADD^*h\Vert _2^2\nonumber \\&\qquad -\sum _{i=1}^{M}\frac{\lambda _i}{4}\Vert AD\beta _i\Vert _2^2 \nonumber \\&=\sum _{i=1}^{M}\lambda _i\Vert AD\left( (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}u_i\right) \Vert _2^2 \\&\qquad +\mu (1-\mu )\left\langle AD(D^*_{S_0}h+h^{(1)}),ADD^*h\right\rangle -\sum _{i=1}^{M}\frac{\lambda _i}{4}\Vert AD\beta _i\Vert _2^2.\nonumber \end{aligned}$$

(6.29)

We next estimate the three terms in (6.29). First we give the following useful relation:

$$\begin{aligned}&\left\langle D(D^*_{S_0}h+h^{(1)}), Dh^{(2)}\right\rangle \nonumber \\&\quad =\left\langle D(D^*_{S_0}h+h^{(1)}), D(D^*h-D^*_{S_0}h-h^{(1)})\right\rangle \nonumber \\&\quad =\left\langle D(D^*_{S_0}h+h^{(1)}), h\right\rangle -\left\langle D(D^*_{S_0}h+h^{(1)}),D(D^*_{S_0}h+h^{(1)})\right\rangle \nonumber \\&\quad =\left\langle D^*_{S_0}h+h^{(1)}, D^*h\right\rangle -\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2\nonumber \\&\quad =\Vert D^*_{S_0}h+h^{(1)}\Vert _2-\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2. \end{aligned}$$

(6.30)

Noting that $\Vert D^*_{S_0}h\Vert _0\le k$, $\Vert h^{(1)}\Vert _0= \ell \le (t-1)k$ and $\Vert u_i\Vert _0\le s =k(t-1)-\ell$, we obtain

$$\begin{aligned} \Vert D^*_{S_0}h+h^{(1)}\Vert _0\le \ell +k\le t\cdot k,\quad \Vert \beta _i\Vert _0\le \Vert D^*_{S_0}h\Vert _0 + \Vert h^{(1)}\Vert _0+ \Vert u_i\Vert _0\le t\cdot k, \end{aligned}$$

and

$$\begin{aligned} \Vert (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}u_i\Vert _0\le t\cdot k. \end{aligned}$$

Here we assume $t\cdot k$ as an integer first. Since A satisfies the DRIP of order $t\cdot k$ with constant $\delta$, we can obtain

$$\begin{aligned}&\quad \sum _{i=1}^{M}\lambda _i \Vert AD\left( (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}u_i\right) \Vert _2^2\\&\le \sum _{i=1}^{M}\lambda _i(1+\delta )\Vert D\left( (\frac{1}{2}-\mu )(D^*_{S_0}h+h^{(1)})-\frac{\mu }{2}u_i\right) \Vert _2^2\\&=(1+\delta )\left( (\frac{1}{2}-\mu )^2 \Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2+\frac{\mu ^2}{4}\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2\right. \nonumber \\&\qquad \left. -\,\mu (\frac{1}{2}-\mu )\left\langle D(D^*_{S_0}h+h^{(1)}), Dh^{(2)}\right\rangle \right) \\&\overset{(7.30)}{=}(1+\delta )\left( \frac{1}{2}(\frac{1}{2}-\mu ) \Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2+\frac{\mu ^2}{4}\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2\right. \nonumber \\&\qquad \left. -\,\mu (\frac{1}{2}-\mu ) \Vert D^*_{S_0}h+h^{(1)}\Vert _2^2\right) , \end{aligned}$$

$$\begin{aligned} \left\langle AD(D^*_{S_0}h+h^{(1)}), ADD^*h\right\rangle= & {} \left\langle AD(D^*_{S_0}h+h^{(1)}), Ah\right\rangle \\\le & {} \sqrt{1+\delta }\cdot \Vert D(D^*_{S_0}h+h^{(1)})\Vert _2\cdot \epsilon \end{aligned}$$

and

$$\begin{aligned}&\quad \sum _{i=1}^{M}\lambda _i\Vert AD\beta _i\Vert _2^2\\&=\sum _{i=1}^{M}\lambda _i\Vert AD(D^*_{S_0}h+h^{(1)}+\mu \cdot u_i)\Vert _2^2\\&\ge (1-\delta )\sum _{i=1}^{M}\lambda _i\Vert D(D^*_{S_0}h+h^{(1)}+\mu \cdot u_i)\Vert _2^2\\&=(1-\delta )\left( \Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2+\mu ^2\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2+2\mu \left\langle D(D^*_{S_0}h+h^{(1)}), Dh^{(2)}\right\rangle \right) \\&\overset{(7.30)}{=}(1-\delta )\left( (1-2\mu )\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2+\mu ^2\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2+2\mu \Vert D^*_{S_0}h+h^{(1)}\Vert _2^2 \right) . \end{aligned}$$

Combining the above results with (6.26) and (6.29), we get

$$\begin{aligned} 0&\le \frac{1}{2}(1+\delta )\left( \frac{1}{2}-\mu \right) \Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2+\frac{1+\delta }{4}\mu ^2\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2\nonumber \\&\qquad -(1+\delta )\mu \left( \frac{1}{2}-\mu \right) \Vert D^*_{S_0}h+h^{(1)}\Vert _2^2 \\&\qquad +\mu (1-\mu )\sqrt{1+\delta }\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2\cdot \epsilon \\&\qquad -\frac{1}{4}(1-\delta )(1-2\mu )\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2-\frac{1-\delta }{4}\mu ^2\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2\nonumber \\&\qquad -\frac{1-\delta }{2}\mu \Vert D^*_{S_0}h+h^{(1)}\Vert _2^2 \\&=\delta (\frac{1}{2}-\mu )\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2^2+(\mu ^2(1+\delta )-\mu )\Vert D^*_{S_0}h+h^{(1)}\Vert _2^2\nonumber \\&\qquad +\frac{\delta }{2}\mu ^2\sum _{i=1}^{M}\lambda _i\Vert Du_i\Vert _2^2+\mu (1-\mu )\sqrt{1+\delta }\Vert D(D^*_{S_0}h+h^{(1)})\Vert _2\cdot \epsilon \\&\le (\delta (\frac{1}{2}-\mu )+\mu ^2(1+\delta )-\mu )z^2+\frac{\delta }{2}\mu ^2\sum _{i=1}^{M}\lambda _i\Vert u_i\Vert _2^2 +\mu (1-\mu )\sqrt{1+\delta }\cdot z\cdot \epsilon \\&\overset{(7.26)}{\le } \left( (1+\delta )(\frac{1}{2}-\mu )^2-\frac{1-\delta }{4} \right) z^2+\frac{\delta }{2}\mu ^2\frac{(z+R)^2}{t-1} +\mu (1-\mu )\sqrt{1+\delta }\cdot z\cdot \epsilon \\&=\left( (\mu ^2-\mu )+\delta \left( \frac{1}{2}-\mu +(1+\frac{1}{2(t-1)})\mu ^2\right) \right) z^2\nonumber \\&\qquad +\left( \mu (1-\mu )\sqrt{1+\delta }\cdot \epsilon +\frac{\delta \mu ^2R}{t-1}\right) z+\frac{\delta \mu ^2R^2}{2(t-1)} \\&=-t\left( (2t-1)-2\sqrt{t(t-1)} \right) (\sqrt{\frac{t-1}{t}}-\delta )z^2\nonumber \\&\qquad +\left( \mu ^2\sqrt{\frac{t}{t-1}}\sqrt{1+\delta }\cdot \epsilon +\frac{\delta \mu ^2R}{t-1}\right) z+\frac{\delta \mu ^2R^2}{2(t-1)}\\&=\frac{\mu ^2}{t-1}\left( -t(\sqrt{\frac{t-1}{t}}-\delta )z^2+(\sqrt{t(t-1)(1+\delta )}\epsilon +\delta R)z+\frac{\delta R^2}{2} \right) , \end{aligned}$$

which is a quadratic inequality for z. Recall that $\delta <\sqrt{(t-1)/t}$. So by solving the above inequality, we get

$$\begin{aligned} z&\le \frac{(\sqrt{t(t-1)(1+\delta )}\epsilon +\delta R)+\left( (\sqrt{t(t-1)(1+\delta )}\epsilon +\delta R)^2+2t(\sqrt{(t-1)/t}-\delta )\delta R^2 \right) ^{1/2} }{2t(\sqrt{(t-1/t)}-\delta ) }\\&\le \frac{\sqrt{t(t-1)(1+\delta )}}{t(\sqrt{(t-1)/t}-\delta )}\epsilon +\frac{2\delta +\sqrt{2t(\sqrt{(t-1)/t}-\delta )\delta }}{2t(\sqrt{(t-1)/t}-\delta )}R. \end{aligned}$$

We know $\Vert D^*_{S_0^c}h\Vert _1\le \Vert D^*_{S_0}h\Vert _1+R\sqrt{k}$. In the Lemma 6.2, if we set $m=N$, $r=k$, $\lambda =R\sqrt{k}\ge 0$ and $\alpha =2$, we can obtain

$$\begin{aligned} \Vert D^*_{S_0^c}h\Vert _2\le \Vert D^*_{S_0}h\Vert _2+R. \end{aligned}$$

So

$$\begin{aligned} \Vert h\Vert _2&=\Vert D^*h\Vert _2\\&=\sqrt{\Vert D^*_{S_0}h\Vert _2^2+\Vert D^*_{S_0^c}h\Vert _2^2}\\&\le \sqrt{\Vert D^*_{S_0}h\Vert _2^2+(\Vert D^*_{S_0}h\Vert _2+R)^2}\\&\le \sqrt{2\Vert D^*_{S_0}h\Vert _2^2}+R\le \sqrt{2}z+R\\&\le \frac{\sqrt{2(1+\delta )}}{1-\sqrt{t/(t-1)}\delta }\epsilon +\left( \frac{\sqrt{2}\delta +\sqrt{t(\sqrt{(t-1)/t}-\delta )\delta }}{t(\sqrt{(t-1)/t}-\delta )}+1 \right) R. \end{aligned}$$

Substituting R into this inequality, we can get the conclusion. For the case where $t\cdot k$ is not an integer, we set $t^*:=\lceil tk\rceil / k$, then $t^*>t$ and $\delta _{t^*k}=\delta _{tk}<\sqrt{\frac{t-1}{t}}<\sqrt{\frac{t^*-1}{t^*}}$. We can prove the result by working on $\delta _{t^*k}$. $\square$