An Optimal Recovery Condition for Sparse Signals with Partial Support Information via OMP

Abstract

This paper considers the orthogonal matching pursuit (OMP) algorithm for sparse recovery in both noiseless and noisy cases when the partial prior information is available. The prior information is included in an estimated subset of the support of the sparse signal. First, we show that if $\varvec{A}$ satisfies $\delta _{k+b+1}<\frac{1}{\sqrt{k-g+1}}$, then the OMP algorithm can perfectly recover any k-sparse signal $\varvec{x}$ from $\varvec{y}=\varvec{Ax}$ in $k-g$ iterations when the prior support of $\varvec{x}$ includes g true indices and b wrong indices. Furthermore, we show that the condition $\delta _{k+b+1}<\frac{1}{\sqrt{k-g+1}}$ is optimal. Second, we achieve the exact recovery of the remainder support (i.e., it is composed of indices in the true support of $\varvec{x}$ but not in the prior support) from $\varvec{y}=\varvec{Ax}+\varvec{v}$ under appropriate conditions. On the other hand, for the remainder support recovery, we also obtain a necessary condition based on the minimum magnitude of nonzero elements in the remainder support of $\varvec{x}$. Compared to the OMP algorithm, numerical experiments demonstrate that the OMP algorithm with the partial prior information has better recovery performance.

Appendices

Appendix A: The Proof of Lemma 1

Proof

For simplicity, let

$$\begin{aligned} \alpha _1^{(t)}=\max _{i\in T{\setminus } \Lambda _t}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| =\left\| \varvec{A}_{T{\setminus }\Lambda _t}^{\top }\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\| _\infty \end{aligned}$$

(13)

and

$$\begin{aligned} \beta _1^{(t)}=\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| =\left| \left\langle \varvec{Ae}_{i_t}, \varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| \end{aligned}$$

(14)

where $i_t=\arg \max \nolimits _{i\in (T\cup \Lambda _t)^c}|\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\rangle |$. By $|T\cap T_0|=g$, $|T|=k$, $\Lambda _t\subseteq (T\cup T_0)$, $T_0\subseteq \Lambda _t$ and $|\Lambda _t|=t$, it is clear that the fact $|T\backslash \Lambda _t|=k-g-t$ holds. Then by (13), one obtains that

$$\begin{aligned}&\sqrt{k-g-t}\left\| \varvec{z}_{T\cup \Lambda _t}\right\| _2\alpha _1^{(t)}\nonumber \\&\quad =\sqrt{k-g-t}\left\| \varvec{z}_{T\cup \Lambda _t}\right\| _2\left\| \varvec{A}_{T {\setminus } \Lambda _t}^{\top }\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\| _\infty \nonumber \\&\quad {\mathop {\ge }\limits ^{(a)}}\left\| \varvec{z}_{T\cup \Lambda _t}\right\| _2\left\| \varvec{A}_{T {\setminus } \Lambda _t}^{\top }\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\| _2\nonumber \\&\quad {\mathop {=}\limits ^{(b)}}\left\| \varvec{z}_{T\cup \Lambda _t}\right\| _2\left\| \varvec{A}_{T\cup \Lambda _t}^{\top }\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\| _2\nonumber \\&\quad {\mathop {\ge }\limits ^{(c)}} \left\langle \varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _k}, \varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t} \right\rangle \nonumber \\&\quad =\left\langle \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t} \right\rangle \end{aligned}$$

(15)

where (a) and (c), respectively, follow from $|T\backslash \Lambda _t|=k-g-t$ and the Hölder inequality, and (b) is due to the fact that $\varvec{A}_{\Lambda _t}^{\top }\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}=\varvec{0}$.

Let $s=-\frac{\sqrt{k-g-t+1}-1}{\sqrt{k-g-t}}$ and

$$\begin{aligned} {\hat{s}}_{i_t}=\left\{ \begin{array}{ll} +\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2s, &{} \langle \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}, \varvec{Ae}_{i_t}\rangle \ge 0, \\ -\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2s, &{} \langle \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t},\varvec{Ae}_{i_t}\rangle <0, \end{array} \right. \end{aligned}$$

then ${\hat{s}}_{i_t}^2=\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2^2s^2$ and

$$\begin{aligned} s^2=\frac{\sqrt{k-g-t+1}-1}{\sqrt{k-g-t+1}+1}<1. \end{aligned}$$

Further, by (15), (14), $s^2<1$ and some simple calculations, we derive that

$$\begin{aligned}&\left( 1-s^4\right) \sqrt{k-g-t}\left\| \varvec{z}_{T\cup \Lambda _t}\right\| _2\left( \alpha _1^{(t)}-\beta _1^{(t)}\right) \nonumber \\&\quad \ge \left( 1-s^4\right) \left( \left\langle \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t} \right\rangle -\sqrt{k-g-t}\left\| \varvec{z}_{T\cup \Lambda _t}\right\| _2\left| \left\langle \varvec{Ae}_{i_t},\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| \right) \nonumber \\&\quad =\left\| \varvec{A}\left( \tilde{\varvec{z}}_{T\cup \Lambda _t}+{\hat{s}}_{i_t}\varvec{e}_{i_t}\right) \right\| _2^2-\left\| \varvec{A}\left( s^2\tilde{\varvec{z}}_{T\cup \Lambda _t}-{\hat{s}}_{i_t}\varvec{e}_{i_t}\right) \right\| _2^2, \end{aligned}$$

(16)

where the last equality is from the definitions of s and ${\hat{s}}_{i_t}$ and

$$\begin{aligned} \frac{2{\hat{s}}_{i_t}}{1-s^2} =\left\{ \begin{array}{ll} -\sqrt{k-g-t}\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2,&{}\langle \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}, \varvec{Ae}_{i_t}\rangle \ge 0; \\ \sqrt{k-g-t}\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2,&{}\langle \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}, \varvec{Ae}_{i_t}\rangle <0. \end{array} \right. \end{aligned}$$

Because $0\le t< k-g$, $\varvec{A}$ satisfies the RIP of order $k+b+1$ with $\delta _{k+b+1}$, $|T\cup \Lambda _t|=k+b$ and $i_t\in (T\cup \Lambda _t)^c$, one obtains that

$$\begin{aligned}&\left\| \varvec{A}\left( \tilde{\varvec{z}}_{T\cup \Lambda _t}+{\hat{s}}_{i_t}\varvec{e}_{i_t}\right) \right\| _2^2-\left\| \varvec{A}\left( s^2\tilde{\varvec{z}}_{T\cup \Lambda _t}-{\hat{s}}_{i_t}\varvec{e}_{i_t}\right) \right\| _2^2\nonumber \\&\quad \ge \left( 1-\delta _{k+b+1}\right) \left( \left\| \tilde{\varvec{z}}_{T\cup \Lambda _t}+{\hat{s}}_{i_t}\varvec{e}_{i_t}\right\| _2^2\right) -\left( 1+\delta _{k+b+1}\right) \left( \left\| s^2\tilde{\varvec{z}}_{T\cup \Lambda _t}-{\hat{s}}_{i_t}\varvec{e}_{i_t}\right\| _2^2\right) \nonumber \\&\quad =\left\| \tilde{\varvec{z}}_{T\cup \Lambda _t}\right\| _2^2\left( 1+s^2\right) ^2\bigg (\frac{1-s^2}{1+s^2}-\delta _{k+b+1}\bigg ), \end{aligned}$$

(17)

where the equality is from $i_t\in (T\cup \Lambda _t)^c$, $\Vert \varvec{e}_{i_t}\Vert _2=1$ and the fact ${\hat{s}}_{i_t}^2=\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2^2s^2.$ Combining with (16), we have that

$$\begin{aligned} \alpha _1^{(t)}-\beta _1^{(t)} \ge&\frac{\left( 1+s^2\right) ^2\bigg (\frac{1-s^2}{1+s^2}-\delta _{k+b+1}\bigg )}{\left( 1-s^4\right) \sqrt{k-g-t}}\Vert \tilde{\varvec{z}}_{T\cup \Lambda _t}\Vert _2\\ =&\bigg (1-\sqrt{k-g-t+1}\delta _{k+b+1}\bigg )\frac{\Vert \tilde{\varvec{z}}_{T\cup \Lambda _t}\Vert _2}{\sqrt{k-g-t}}, \end{aligned}$$

where the equality is due to the definition of s meaning the fact that

$$\begin{aligned} \frac{1-s^2}{1+s^2}=\frac{1-\frac{\sqrt{k-g-t+1}-1}{\sqrt{k-g-t+1}+1}}{1+\frac{\sqrt{k-g-t+1}-1}{\sqrt{k-g-t+1}+1}}=\frac{1}{\sqrt{k-g-t+1}}. \end{aligned}$$

$\square$

Appendix B: The Proof of Theorem 1

Proof

By the inductive method, we first prove that under the condition (4), i.e., $\delta _{k+b+1}<\frac{1}{\sqrt{k-g+1}}$, the $\mathrm {OMP}_{T_0}$ algorithm succeeds in the sense of Definition 2.

For the first iteration, $\Lambda _0=T_0$ and $\varvec{r}^{(0)}=\varvec{A}_{T\cup T_0}\varvec{z}_{T\cup T_0}$. By Lemma 1 with $t=0$ and (4), i.e., $\delta _{k+b+1}<\frac{1}{\sqrt{k-g+1}}$, we have that

$$\begin{aligned}&\max _{i\in T{\setminus } T_0}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup T_0}\varvec{z}_{T\cup T_0}\right\rangle \right| -\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup T_0}\varvec{z}_{T\cup T_0}\right\rangle \right| \\&\quad \ge \left( 1-\sqrt{k-g+1}\delta _{k+b+1}\right) \frac{\Vert \tilde{\varvec{z}}_{T\cup T_0}\Vert _2}{\sqrt{k-g}}>0. \end{aligned}$$

That is,

$$\begin{aligned} \max \limits _{i\in T{\setminus } T_0}\left| \left\langle \varvec{Ae}_i, \varvec{r}^{(0)}\right\rangle \right| >\max \limits _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i,\varvec{r}^{(0)}\right\rangle \right| . \end{aligned}$$

Then the $\mathrm {OMP}_{T_0}$ algorithm selects a correct index in the first iteration, i.e., $j_1\in T{\setminus } T_0$.

Suppose that the $\mathrm {OMP}_{T_0}$ algorithm has performed t ($1\le t <k-g$) iterations successfully, that is, $\Lambda _{t}{\setminus } T_0 \subseteq T{\setminus } T_0$. For the $(t+1)$th iteration, by the equality (7) with $\varvec{v}=\varvec{0}$, one has that

$$\begin{aligned}&\max _{i\in T{\setminus } \Lambda _t}\left| \left\langle \varvec{Ae}_i,\varvec{r}^{(t)}\right\rangle \right| -\max _{i\in (T\cup \Lambda _t)^c}\left| \left\langle \varvec{Ae}_i,\varvec{r}^{(t)}\right\rangle \right| \\&\quad =\max _{i\in T{\setminus } \Lambda _t}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| -\max _{i\in (T\cup \Lambda _t)^c}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| \\&\quad \overset{(a)}{\ge }\frac{1}{\sqrt{k-g-t}}\bigg (1-\sqrt{k-g-t+1}\delta _{k+b+1}\bigg )\\&\quad \ge \frac{1}{\sqrt{k-g}}\bigg (1-\sqrt{k-g+1}\delta _{k+b+1}\bigg )\overset{(b)}{>}0, \end{aligned}$$

where (a) and (b) follow from Lemma 1 and (4), respectively. Then the $\mathrm {OMP}_{T_0}$ algorithm makes a success in the $(t+1)$th iteration, i.e., $j_{t+1}\in T{\setminus }\Lambda _t\subseteq T{\setminus } T_0$. Therefore, if $\delta _{k+b+1}<\frac{1}{\sqrt{k-g+1}}$, then the $\mathrm {OMP}_{T_0}$ algorithm succeeds by Definition 2.

It remains to prove $\varvec{x}=\hat{\varvec{x}}$, where $\varvec{{\hat{x}}}$ is the output by the $\mathrm {OMP}_{T_0}$ algorithm after $k-g$ successful iterations. As the $\mathrm {OMP}_{T_0}$ algorithm has performed $k-g$ iterations successfully, which implies the fact $\Lambda _{k-g}=T\cup T_0$, one has that

$$\begin{aligned} \hat{\varvec{x}}_{\Lambda _{k-g}}&=\varvec{x}^{(k-g)}=\varvec{A}^{\dagger }_{\Lambda _{k-g}}\varvec{y}\\&{\mathop {=}\limits ^{(a)}}\left( \varvec{A}^{\top }_{\Lambda _{k-g}}\varvec{A}_{\Lambda _{k-g}}\right) ^{-1}\varvec{A}^{\top }_{\Lambda _{k-g}}\varvec{A}_T\varvec{x}_T\\&=\left( \varvec{A}^{\top }_{\Lambda _{k-g}}\varvec{A}_{\Lambda _{k-g}}\right) ^{-1}\varvec{A}^{\top }_{\Lambda _{k-g}}\varvec{A}_{\Lambda _{k-g}}\varvec{x}_{\Lambda _{k-g}}\\&\quad -\left( \varvec{A}^{\top }_{\Lambda _{k-g}}\varvec{A}_{\Lambda _{k-g}}\right) ^{-1}\varvec{A}^{\top }_{\Lambda _{k-g}}\varvec{A}_{\Lambda _{k-g}{\setminus } T}\varvec{x}_{\Lambda _{k-g}{\setminus } T}\\&{\mathop {=}\limits ^{(b)}}\varvec{x}_{\Lambda _{k-g}} \end{aligned}$$

where (a) and (b), respectively, follow from the fact that $\varvec{A}$ satisfies the RIP of order $k+b+1$, which means that $\varvec{A}_{\Lambda _{k-g}}$ is full column rank, and $\Lambda _{k-g}=T\cup T_0$. We have completed the proof of the theorem. $\square$

Appendix C: The Proof of Theorem 2

Proof

For given integers $k>0$, $b\ge 0$ and $0 \le g<k$, let $\varvec{A}_1\in {\mathbb {R}}^{(k+b+1)\times (k+b+1)}$ be

$$\begin{aligned} \varvec{A}_1= \begin{bmatrix}&&0&\cdots&0&\beta \\&\sqrt{\frac{k-g}{k-g+1}}\varvec{I}_{k-g}&\vdots&\vdots&\vdots \\&&0&\cdots&0&\beta \\ 0&\cdots&0&\\ \vdots&\vdots&&\varvec{I}_{g+b+1}&\\ 0&\cdots&0&&\\ \end{bmatrix}, \end{aligned}$$

(18)

where $\beta =\frac{1}{\sqrt{(k-g+1)(k-g)}}$. Then

$$\begin{aligned} \varvec{A}_1^{\top }\varvec{A}_1= \begin{bmatrix}&&0&\cdots&0&\gamma \\&\frac{k-g}{k-g+1}\varvec{I}_{k-g}&\vdots&\vdots&\vdots \\&&0&\cdots&0&\gamma \\ 0&\cdots&0&&0\\ \vdots&\vdots&\varvec{I}_{g+b}&\vdots \\ 0&\cdots&0&&0 \\ \gamma&\cdots&\gamma&0&\cdots&0&1+\gamma \\ \end{bmatrix}, \end{aligned}$$

where $\gamma =(k-g)\beta ^2=\frac{1}{k-g+1}$. By elementary transformation of determinant, one can verify that

$$\begin{aligned} \begin{vmatrix} \varvec{A}_1^{\top }\varvec{A}_1-\lambda \varvec{I}_{k+b+1} \end{vmatrix}&= \begin{vmatrix}&&0&\cdots&0&\lambda \\&(\frac{k-g}{k-g+1}-\lambda )\varvec{I}_{k-g}&\vdots&\vdots&0\\&&0&\cdots&0&\vdots \\ 0&\cdots&0&&\vdots \\ \vdots&\vdots&\mu \varvec{I}_{g+b}&\vdots \\ 0&\cdots&0&&0 \\ (k-g)\gamma&\cdots&\gamma&0&\cdots&0&\mu +\gamma \\ \end{vmatrix}\\&=(-1)^{1+1}\left( \frac{k-g}{k-g+1}-\lambda \right) ^{k-g}\mu ^{g+b}\left( \mu +\gamma \right) \\&\ \ +(-1)^{2(k+b+1)+1}\frac{k-g}{k-g+1}\gamma \left( \frac{k-g}{k-g+1}-\lambda \right) ^{k-g-1}\mu ^{g+b}\\&=(1-\lambda )^{g+b}\left( \frac{k-g}{k-g+1}-\lambda \right) ^{k-g-1}\left( \lambda ^2{-}2\lambda +\frac{k{-}g}{k{-}g{+}1}\right) , \end{aligned}$$

where $\mu =1-\lambda$. Then the eigenvalues $\{\lambda _i(\varvec{A}_1^{\top }\varvec{A}_1)\}_{i=1}^{k+b+1}$ of $\varvec{A}_1^{\top }\varvec{A}_1$ are

$$\begin{aligned}&\displaystyle \lambda _1\left( \varvec{A}_1^{\top }\varvec{A}_1\right) =\cdots =\lambda _{k-g-1}\left( \varvec{A}_1^{\top }\varvec{A}_1\right) =\frac{k-g}{k-g+1},\\&\displaystyle \lambda _{k-g}\left( \varvec{A}_1^{\top }\varvec{A}_1\right) =\cdots =\lambda _{k+b-1}\left( \varvec{A}_1^{\top }\varvec{A}_1\right) =1, \lambda _{k+b}\left( \varvec{A}_1^{\top }\varvec{A}_1\right) =1-\frac{1}{\sqrt{k-g+1}},\\&\displaystyle \lambda _{k+b+1}\left( \varvec{A}_1^{\top }\varvec{A}_1\right) =1+\frac{1}{\sqrt{k-g+1}}. \end{aligned}$$

Now, taking the matrix

$$\begin{aligned} \varvec{A}=\begin{pmatrix} \varvec{A}_1 &{} \varvec{0} \\ \varvec{0} &{} \varvec{A}_2 \\ \end{pmatrix}\in {\mathbb {R}}^{m\times n} \end{aligned}$$

(19)

where $\varvec{A}_2\in {\mathbb {R}}^{(m-k-b-1)\times (n-k-b-1)}$ satisfies $1-\frac{1}{\sqrt{k-g+1}}\le \lambda ((\varvec{A}^{\top }\varvec{A})_{|S_1|\times |S_2|}) \le 1+\frac{1}{\sqrt{k-g+1}}$ with $S_1\subseteq [m]$, $S_2\subseteq [n]$ and $|S_1|=|S_2|=k+b+1$.

Moreover, by definition of the RIP and [7, Remark 1], the matrix $\varvec{A}$ satisfies the RIP with

$$\begin{aligned} \delta _{k+b+1}=&\max \left\{ 1-\lambda _{\min }\left( \left( \varvec{A}^{\top }\varvec{A}\right) _{|S_1|\times |S_2|}\right) , \lambda _{\max }\left( \left( \varvec{A}^{\top }\varvec{A}\right) _{|S_1|\times |S_2|}\right) -1\right\} \\ =&\max \left\{ 1-\lambda _{\min }\left( \varvec{A}_1^{\top }\varvec{A}_1\right) , \lambda _{\max }\left( \varvec{A}_1^{\top }\varvec{A}_1\right) -1\right\} \\ =&\frac{1}{\sqrt{k-g+1}}. \end{aligned}$$

Let k-sparse signal

$$\begin{aligned} \bar{\varvec{x}}=(\underbrace{1,\ldots ,1,}_{k}0,\ldots ,0)^{\top }\in {\mathbb {R}}^{n} \end{aligned}$$

and the prior support

$$\begin{aligned} T_0=\{k-g+1,\ldots ,k,k+1,\ldots ,k+b\}, \end{aligned}$$

then

$$\begin{aligned} T=supp(\bar{\varvec{x}})=\{1,\ldots ,k\},\ \ T{\setminus } T_0=\{1,\ldots , k-g\}. \end{aligned}$$

For the first iteration, from (7) with $t=0$ and $\varvec{y}=\varvec{A}\bar{\varvec{x}}$ [$\varvec{A}$ is defined in (19)] it follows that

$$\begin{aligned} \varvec{r}^{(0)}&=\varvec{A}_{T{\setminus } T_0}\bar{\varvec{x}}_{T{\setminus } T_0} -\varvec{A}_{T_0}\varvec{A}_{T{\setminus } T_0}^{\dagger }\varvec{A}_{T{\setminus } T_0}\bar{\varvec{x}}_{T{\setminus } T_0}\\&=\left( \underbrace{\sqrt{\frac{k-g}{k-g+1}},\ldots ,\sqrt{\frac{k-g}{k-g+1}}}_{k-g},0,\ldots ,0\right) ^{\top }, \end{aligned}$$

where the last equality is from (19) and the fact

$$\begin{aligned} \varvec{A}_{T{\setminus } T_0}\bar{\varvec{x}}_{T{\setminus } T_0} =\left( \underbrace{\sqrt{\frac{k-g}{k-g+1}},\ldots ,\sqrt{\frac{k-g}{k-g+1}}}_{k-g},0,\ldots ,0\right) ^{\top }, \end{aligned}$$

and $\varvec{A}_{T_0}^{\top }\varvec{A}_{T{\setminus } T_0}\bar{\varvec{x}}_{T{\setminus } T_0}=\varvec{0}$.

On the other hand, for $i\in T{\setminus } T_0$, one has

$$\begin{aligned} \left| \left\langle \varvec{Ae}_i, \varvec{r}^{(0)}\right\rangle \right| =\frac{k-g}{k-g+1}. \end{aligned}$$

And for $i\in (T\cup T_0)^c=\{k+b+1\}$, it follows immediately that

$$\begin{aligned} \left| \left\langle \varvec{Ae}_i, \varvec{r}^{(0)}\right\rangle \right| =\frac{k-g}{k-g+1}. \end{aligned}$$

Clearly,

$$\begin{aligned} \max \limits _{i\in T{\setminus } T_0}\left\langle \varvec{Ae}_i,\varvec{r}^{(0)}\right\rangle =\max \limits _{i\in (T\cup T_0)^c}\left\langle \varvec{Ae}_i,\varvec{r}^{(0)}\right\rangle , \end{aligned}$$

which implies the $\mathrm {OMP}_{T_0}$ algorithm may fail to identify one index of the subset $T{\setminus } T_0$ in the first iteration. So the $\mathrm {OMP}_{T_0}$ algorithm may fail for the given matrix $\varvec{A}$ in (19), the k-sparse signal $\bar{\varvec{x}}$ and the prior support $T_0$. $\square$

Appendix D: The Proof of Theorem 3

Proof

The proof consists of two parts. In the first part, we show that the $\mathrm {OMP}_{T_0}$ algorithm selects indices from the remainder support $T{\setminus } T_0$ in each iteration under the conditions (4) and (5). In the second part, we prove that the $\mathrm {OMP}_{T_0}$ algorithm exactly performs $|T{\setminus } T_0|=k-g$ iterations under the stopping rule $\Vert \varvec{r}^{(t)}\Vert _2\le \varepsilon$.

Part I: By induction method, suppose first that the $\mathrm {OMP}_{T_0}$ algorithm performed t ($1 \le t < k-g$) iterations successfully, that is, $\Lambda _t\subseteq T\cup T_0$ and $j_1,\ldots , j_t \in T{\setminus } T_0$. Then by the $\mathrm {OMP}_{T_0}$ algorithm in Table 1, we need to show $j_{t+1} \in T{\setminus } \Lambda _t$ which means the $\mathrm {OMP}_{T_0}$ algorithm makes a success in the $(t+1)$th iteration. By the fact that $\varvec{r}^{(t)}$ is orthogonal to each column of $\varvec{A}_{\Lambda _t}$, we only need to prove that

$$\begin{aligned} \max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{r}^{(t)}\right\rangle \right| >\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{r}^{(t)}\right\rangle \right| \end{aligned}$$

(20)

for the $(t+1)$th iteration.

From (7), one has that

$$\begin{aligned} \max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{r}^{(t)}\right\rangle \right|&=\max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}+\varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| \nonumber \\&\ge \max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}\right\rangle \right| -\max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| \end{aligned}$$

(21)

and

$$\begin{aligned} \max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{r}^{(t)}\right\rangle \right| =&\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}+\varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| \nonumber \\ \le&\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}\right\rangle \right| +\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| . \end{aligned}$$

(22)

Therefore, to show (20), by (21) and (22), it suffices to prove that

$$\begin{aligned}&\max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}\right\rangle \right| -\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{A}\tilde{\varvec{z}}_{T\cup \Lambda _t}\right\rangle \right| \nonumber \\&>\max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| +\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| . \end{aligned}$$

(23)

One first gives a lower bound on the left-hand side of (23). From Lemma 1, it follows that

$$\begin{aligned}&\max _{i\in T{\setminus } \Lambda _t}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| -\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i,\varvec{A}_{T\cup \Lambda _t}\varvec{z}_{T\cup \Lambda _t}\right\rangle \right| \nonumber \\&\quad \ge \left( 1-\sqrt{k-g-t+1}\delta _{k+b+1}\right) \frac{\Vert \tilde{\varvec{z}}_{T\cup \Lambda _t}\Vert _2}{\sqrt{k-g-t}}\nonumber \\&\quad \ge \left( 1-\sqrt{k-g-t+1}\delta _{k+b+1}\right) \frac{\Vert \varvec{x}_{T{\setminus }\Lambda _t}\Vert _2}{\sqrt{k-g-t}}\nonumber \\&\quad \ge \left( 1-\sqrt{k-g+1}\delta _{k+b+1}\right) \min _{i\in T{\setminus } T_0}|x_i|, \end{aligned}$$

(24)

where the second inequality is due to the definition of $\varvec{z}_{T\cup T_0}$ in (8) and the last inequality is from $0\le t<k-g$ and the induction assumption $j_1,\ldots ,j_t\in T{\setminus } T_0$ (i.e., $|T{\setminus }\Lambda _t|=k-g-t$) implying

$$\begin{aligned} \Vert \varvec{x}_{T{\setminus }\Lambda _t}\Vert _2\ge \sqrt{k-g-t+1}\min _{i\in T{\setminus } T_0}|x_i|. \end{aligned}$$

One now gives an upper bound on the right-hand side of (23). There exist the indices $i^{(t)}\in T{\setminus } \Lambda _t$ and $i_1^{(t)}\in (T\cup T_0)^c$ satisfying

$$\begin{aligned} \max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| =\left| \left\langle \varvec{Ae}_{i^{(t)}}, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| \end{aligned}$$

and

$$\begin{aligned} \max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| =\left| \left\langle \varvec{Ae}_{i_1^{(t)}}, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| , \end{aligned}$$

respectively. Therefore,

$$\begin{aligned}&\max _{i\in T{\setminus }\Lambda _t}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| +\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{Ae}_i, \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\rangle \right| \nonumber \\&\quad =\Vert \varvec{A}_{\{i^{(t)},i_1^{(t)}\}}^{\top } \varvec{P}_{\Lambda _t}^\perp \varvec{v}\Vert _1\nonumber \\&\quad \le \sqrt{2}\Vert \varvec{A}_{\{i^{(t)},i_1^{(t)}\}}^{\top } \varvec{P}_{\Lambda _t}^\perp \varvec{v}\Vert _2\nonumber \\&\quad {\mathop {\le }\limits ^{(a)}}\sqrt{2(1+\delta _{k-g+1})}\Vert \varvec{P}_{\Lambda _t}^\perp \varvec{v}\Vert _2\nonumber \\&\quad {\mathop {\le }\limits ^{(b)}}\sqrt{2(1+\delta _{k-g+1})}\varepsilon \end{aligned}$$

(25)

where (a) follows from $\varvec{A}$ fulfilling the RIP with order $k-g+1$ ($g<k$) and (b) is because the fact

$$\begin{aligned} \left\| \varvec{P}_{\Lambda _t}^\perp \varvec{v}\right\| _2\le \left\| \varvec{P}_{\Lambda _t}^\perp \right\| _2\left\| \varvec{v}\right\| _2\le \left\| \varvec{v}\right\| _2\le \varepsilon . \end{aligned}$$

By (4) and (5), there is

$$\begin{aligned} \bigg (1-\sqrt{k-g+1}\delta _{k+b+1}\bigg )\min _{i\in T{\setminus } T_0}|x_i| >\sqrt{2(1+\delta _{k-g+1})}\varepsilon . \end{aligned}$$

Combining with (24), (25), it is obvious that (23) holds. Then the $\mathrm {OMP}_{T_0}$ algorithm selects one index from the subset $T{\setminus } \Lambda _t$ in the $(t+1)$th iteration. In conclusion, we have shown that the $\mathrm {OMP}_{T_0}$ algorithm selects one index from $T{\setminus } T_0$ in each iteration under the conditions (4) and (5).

Part II: We prove that the $\mathrm {OMP}_{T_0}$ algorithm performs exactly $k-g$ iterations. That is, we show that $\Vert \varvec{r}^{(k-g)}\Vert _2\le \varepsilon$ and $\Vert \varvec{r}^{(t)}\Vert _2>\varepsilon$ for $0\le t <k-g$.

Since the $\mathrm {OMP}_{T_0}$ algorithm selects an index of $T{\setminus } T_0$ in each iteration under the conditions (4) and (5), $\Lambda _t\subseteq T\cup T_0$ and $|\Lambda _t|=k-g+t$ with $0\le t<k-g$ and $\Lambda _{k-g}=T\cup T_0$ meaning $\varvec{P}^{\perp }_{\Lambda _{k-g}}\varvec{A}_T\varvec{x}_T=\varvec{0}$. Moreover,

$$\begin{aligned} \left\| \varvec{r}^{(k-g)}\right\| _2=&\left\| \varvec{P}^{\perp }_{\Lambda _{k-g}}\varvec{A}_T\varvec{x}_T+\varvec{P}^{\perp }_{\Lambda _{k-g}}\varvec{v}\right\| _2 =\left\| \varvec{P}^{\perp }_{\Lambda _{k-g}}\varvec{v}\right\| _2\le \Vert \varvec{v}\Vert _2\le \varepsilon . \end{aligned}$$

For $0\le t<k-g$, by (7) we have

$$\begin{aligned} \left\| \varvec{r}^{(t)}\right\| _2&=\Vert \varvec{A}_{T\cup \Lambda _{t}}\varvec{z}_{T\cup \Lambda _t}+(\varvec{I}-\varvec{P}_{\Lambda _t})\varvec{v}\Vert _2\\&\ge \Vert \varvec{A}_{T\cup \Lambda _{t}}\varvec{z}_{T\cup \Lambda _t}\Vert _2-\Vert \varvec{P}^{\perp }_{\Lambda _{t}}\varvec{v}\Vert _2\\&{\mathop {\ge }\limits ^{(a)}}\sqrt{1-\delta _{k+b}}\Vert \varvec{z}_{T\cup \Lambda _t}\Vert _2-\varepsilon \\&\ge \sqrt{1-\delta _{k+b+1}}\Vert \varvec{x}_{T{\setminus }\Lambda _t}\Vert _2-\varepsilon \\&\ge \sqrt{1-\delta _{k+b+1}}\sqrt{k-g-t}\min _{T{\setminus }\Lambda _t}|x_i|-\varepsilon \\&\ge \sqrt{1-\delta _{k+b+1}}\min _{T{\setminus } T_0}|x_i|-\varepsilon {\mathop {>}\limits ^{(b)}}\varepsilon \end{aligned}$$

where (a) is because $\varvec{A}$ satisfies the RIP with order $k+b+1$, $|T\cup \Lambda _{t}|=k+b$ and $\Vert \varvec{P}_{\Lambda _t}^\perp \varvec{e}\Vert _2\le \varepsilon$ and (b) is because of (5). We have completed the proof. $\square$

Appendix E: The Proof of Theorem 5

Proof

By (6), the condition (5) holds. Then, from Theorem 3, the conditions (4) (i.e., $\delta _{k+b+1}<\frac{1}{\sqrt{k-g+1}}$) and (5) ensure the $\mathrm {OMP}_{T_0}$ algorithm with the stopping rule $\Vert \varvec{r}^{(t)}\Vert _2\le \varepsilon$ exactly stops after performing $k-g$ iterations successfully. Clearly, $\Lambda _{k-g}=T\cup T_0$. For the $\mathrm {OMP}_{T_0}$ algorithm in Table 1, by $\varvec{A}$ satisfying the RIP with order $k+b+1$ (i.e., $\varvec{A}_{T\cup T_0}$ is column full rank, which implies $\varvec{A}^{\dag }_{T\cup T_0}=(\varvec{A}_{T\cup T_0}^{\top }\varvec{A}_{T\cup T_0})^{-1}$), one has

$$\begin{aligned} \varvec{x}^{(k-g)}&=\arg \min _{\varvec{u}}\Vert \varvec{y}-\varvec{A}_{\Lambda _{k-g}}\varvec{u}\Vert =\varvec{A}^{\dag }_{T\cup T_0}\varvec{y} =\varvec{A}^{\dag }_{T\cup T_0}\left( \varvec{A}_{T\cup T_0}\varvec{x}_{T\cup T_0}+\varvec{v}\right) \nonumber \\&= \varvec{x}_{T\cup T_0}+\varvec{\omega }, \end{aligned}$$

(26)

where

$$\begin{aligned} \varvec{\omega }=\left( \varvec{A}_{T\cup T_0}^{\top }\varvec{A}_{T\cup T_0}\right) ^{-1}\varvec{A}_{T\cup T_0}^{\top }\varvec{v}. \end{aligned}$$

(27)

Furthermore,

$$\begin{aligned} x^{(k-g)}_i=\left\{ \begin{array}{ll} x_i+\omega _i, &{}\quad i\in T, \\ \omega _i, &{}\quad i\in T_0{\setminus } T,\\ 0,&{}\quad i\in (T_0\cup T)^c, \end{array} \right. \end{aligned}$$

(28)

and

$$\begin{aligned} \sqrt{1-\delta _{k+b+1}}\Vert \varvec{\omega }\Vert _2\le&\Vert \varvec{A}_{T\cup T_0}\varvec{\omega }\Vert _2 =\Vert \varvec{P}_{T\cup T_0}\varvec{v}\Vert _2\le \Vert \varvec{v}\Vert \le \varepsilon . \end{aligned}$$

(29)

Then, by (28), one obtains

$$\begin{aligned}&\min _{i\in T\cap T_0}|x^{(k-g)}_i| \ge \min _{i\in T}(|x_i|-|\omega _i|) \overset{(a)}{>}\frac{\varepsilon }{\sqrt{1-\delta _{k+b+1}}}, \\&\max _{i\in T_0{\setminus } T}|x^{(k-g)}_i|=\max _{i\in T_0{\setminus } T}|\omega _i| \le \Vert \varvec{\omega }\Vert _2\overset{(b)}{\le }\frac{\varepsilon }{\sqrt{1-\delta _{k+b+1}}}, \end{aligned}$$

where (a) is from (29) and (5), and (b) follows from (29). In addition, by the definition of RIP, $\mathrm {supp}(\varvec{x})=T$ and (28) meaning $\mathrm {supp}(\varvec{x}^{(k-g)})=T\cup T_0$, one has

$$\begin{aligned} \Vert \varvec{x}-\varvec{x}^{(k-g)}\Vert _2&\le \frac{1}{\sqrt{1-\delta _{T\cup T_0}}}\left\| \varvec{A}\left( \varvec{x}-\varvec{x}^{(k-g)}\right) \right\| _2\\&\overset{(a)}{=}\frac{1}{\sqrt{1-\delta _{T\cup T_0}}}\left\| \varvec{A}_{T\cup T_0}\varvec{\omega }\right\| _2\\&\overset{(b)}{\le } \frac{\varepsilon }{\sqrt{1-\delta _{k+b+1}}}, \end{aligned}$$

where (a) is due to (26), and (b) is from (5), (27) and (29). $\square$

Appendix F: The Proof of Theorem 4

Proof

The proof below roots in [31]. However, some essential modifications are necessary in order to adapt the results to the sparse signal $\varvec{x}$ with the prior support $T_0$. Using proofs by contradiction, we show our result in the theorem. Construct a linear model: $\varvec{y}=\varvec{Ax}+\varvec{v}$, where $\varvec{A}$ and $\varvec{v}$, respectively, satisfy the RIP of order $k+b+1$ with $0\le \delta _{k+b+1}(\varvec{A})=\delta _{k+b+1}<1$ and $\Vert \varvec{v}\Vert _2\le \varepsilon$, and $\varvec{x}$ is a k-sparse signal with the prior support $T_0$ and satisfies

$$\begin{aligned} \min _{T{\setminus } T_0}|x_i|\le \theta :=\frac{\sqrt{1-\delta _{k+b+1}}\varepsilon }{1-\sqrt{k-g+1}\delta _{k+b+1}}, \end{aligned}$$

(30)

such that the $\mathrm {OMP}_{T_0}$ algorithm may fail to exactly recover the remainder support $T{\setminus } T_0$ of the signal $\varvec{x}$ within $k-g$ iterations.

It is well known that there exist the unit vectors $\varvec{\xi }^{(1)},\varvec{\xi }^{(2)},\ldots ,\varvec{\xi }^{(k-g-1)}\in {\mathbb {R}}^{k-g}$ such that the matrix

$$\begin{aligned} \begin{bmatrix}&\varvec{\xi }^{(1)}&\varvec{\xi }^{(2)}&\cdots&\frac{1}{\sqrt{k-g}}\varvec{1}_{k-g} \\ \end{bmatrix} \end{aligned}$$

is orthogonal, which implies $\langle \varvec{\xi }^{(i)}, \varvec{\xi }^{(j)}\rangle =0$ and $\langle \varvec{\xi }^{(i)}, \varvec{1}_{k-g}\rangle =0$ for $i,j=1,\ldots ,k-g-1$ and $i\ne j$, where $\varvec{1}_{k-g}=(1,\ldots ,1)^{\top }\in {\mathbb {R}}^{k-g}$. Let the matrix

$$\begin{aligned} \varvec{U}^{\top }=\begin{bmatrix}&\varvec{\Xi }&\varvec{a}&\varvec{0}_{(k-g)\times (g+b)}&\varvec{a} \\ 0&\cdots&0&0&0 \\ \vdots&\vdots&\vdots&\varvec{I}_{g+b}&\vdots \\ 0&\cdots&0&0&0\\ 0&\cdots&0&\eta \tau&\varvec{0}_{g+b}^{\top }&-\tau \\ \end{bmatrix}, \end{aligned}$$

(31)

where the submatrix $\varvec{\Xi }=[\varvec{\xi }^{(1)},\varvec{\xi }^{(2)},\ldots ,\varvec{\xi }^{(k-g-1)}]$,

$$\begin{aligned} \eta =\frac{\sqrt{k-g+1}-1}{\sqrt{k-g}},\varvec{a}=\frac{\varvec{1}_{k-g}}{\sqrt{(k-g)(\eta ^2+1)}} \end{aligned}$$

and $\tau =\frac{1}{\sqrt{\eta ^2+1}}$. Then $\varvec{U}$ is also an orthogonal matrix.

Let $\varvec{D}\in {\mathbb {R}}^{(k+b+1)\times (k+b+1)}$ be a diagonal matrix with

$$\begin{aligned} d_{ii}=\left\{ \begin{array}{ll} \sqrt{1-\delta _{k+b+1}}, &{}\quad i=k-g, \\ \sqrt{1+\delta _{k+b+1}}, &{}\quad i\ne k-g, \end{array} \right. \end{aligned}$$

(32)

and $\varvec{A}=\varvec{DU}$, then $\varvec{A}^{\top }\varvec{A}=\varvec{U}^{\top }\varvec{D}^2\varvec{U}$. In the following, we show that $\delta _{k+b+1}(\varvec{A})=\delta _{k+b+1}$. For any $\varvec{x}\in {\mathbb {R}}^{k+b+1}$, setting $\hat{\varvec{\nu }}=\varvec{Ux}$, we have that

$$\begin{aligned} \Vert \varvec{Ax}\Vert _2^2&=\langle \varvec{Ax}, \varvec{Ax} \rangle =\varvec{x}^{\top }\varvec{A}^{\top }\varvec{Ax}\\&=(\varvec{Ux})^{\top }\varvec{D}^{\top }\varvec{D}(\varvec{Ux}) =\hat{\varvec{\nu }}^{\top }\varvec{D}^2\hat{\varvec{\nu }}\\&=(1+\delta _{k+b+1})\Vert \hat{\varvec{\nu }}\Vert _2^2-2\delta _{k+b+1}{\hat{\nu }}_{k-g}^2\\&\le (1+\delta _{k+b+1})\Vert \hat{\varvec{\nu }}\Vert _2^2 {\mathop {=}\limits ^{(a)}}(1+\delta _{k+b+1})\Vert \varvec{x}\Vert _2^2 \end{aligned}$$

and

$$\begin{aligned} \Vert \varvec{Ax}\Vert _2^2=&\langle \varvec{Ax}, \varvec{Ax} \rangle =\left( 1-\delta _{k+b+1}\right) \Vert \hat{\varvec{\nu }}\Vert _2^2 +2\delta _{k+b+1}\sum _{1\le i\le k+b+1,\ i\ne k-g}{\hat{\nu }}_i^2\\ \ge&\left( 1-\delta _{k+b+1}\right) \Vert \hat{\varvec{\nu }}\Vert _2^2 {\mathop {=}\limits ^{(b)}}\left( 1-\delta _{k+b+1}\right) \Vert \varvec{x}\Vert _2^2, \end{aligned}$$

where (a) and (b) result of the fact that $\varvec{U}$ is an orthogonal matrix. Then, based on the definition 1, we have $\delta _{k+b+1}(\varvec{A})\le \delta _{k+b+1}$. It remains to prove that $\varvec{A}=\varvec{DU}$ satisfies $\delta _{k+b+1}(\varvec{A})\ge \delta _{k+b+1}.$ Let the vector

$$\begin{aligned} \hat{\varvec{x}}=\left( \left( \varvec{\xi }^{(1)}\right) ^{\top },0,\ldots ,0\right) ^{\top }\in {\mathbb {R}}^{k+b+1}, \end{aligned}$$

then $\hat{\varvec{x}}$ is $(k+b+1)$-sparse and $\Vert \hat{\varvec{x}}\Vert _2^2=1$. By the definitions of $\varvec{D}$ and $\varvec{A}$, we obtain that

$$\begin{aligned} \Vert \varvec{A}\hat{\varvec{x}}\Vert _2^2=&(\varvec{U}\hat{\varvec{x}})^{\top }\varvec{D}^2\varvec{U}\hat{\varvec{x}} =\varvec{e}_1^{\top }\varvec{D}^2\varvec{e}_1 =1+\delta _{k+b+1}=(1+\delta _{k+b+1})\Vert \hat{\varvec{x}}\Vert _2^2. \end{aligned}$$

Then, $\delta _{k+b+1}(\varvec{A})\ge \delta _{k+b+1}$. In conclusion, $\delta _{k+b+1}(\varvec{A})=\delta _{k+b+1}$.

Let the original signal

$$\begin{aligned} \varvec{x}=(\overbrace{\theta ,\ldots , \theta ,}^{T{\setminus } T_0}\overbrace{1,\ldots ,1,}^{T_0\cap T}\overbrace{0,\ldots ,0}^{T_0{\setminus } T},0)^{\top }\in {\mathbb {R}}^{k+b+1}, \end{aligned}$$

(33)

where $\theta$ is defined in (30). Then $\varvec{x}$ in (33) is k-sparse with the support $T=\{1,2,\ldots ,k\}$, the prior support $T_0=\{k-g+1,\ldots ,k+b\}$ and satisfies (30). It is not hard to prove that $\varvec{A}_{T{\setminus } T_0}=\varvec{DU}_{T{\setminus } T_0}$. Moreover, by some simple calculations we derive that

$$\begin{aligned} \varvec{A}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0}&=\varvec{DU}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0}\\&\quad =\varvec{D}\sqrt{\frac{k-g}{\eta ^2+1}}\bigg (\underbrace{0,\ldots ,0}_{k-g-1},\theta ,\underbrace{0,\ldots , 0}_{g+b}, \eta \theta \bigg )^{\top }\\&\quad =\sqrt{\frac{k-g}{\eta ^2+1}}\bigg (\underbrace{0,\ldots , 0}_{k-g-1}, \sqrt{1-\delta _{k+b+1}}\theta , \underbrace{0, \cdots ,0}_{g+b}, \sqrt{1+\delta _{k+b+1}}\eta \theta \bigg )^{\top } \end{aligned}$$

and

$$\begin{aligned} \varvec{A}^{\top }\varvec{A}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0} =(\underbrace{\mu ,\ldots ,\mu ,}_{k-g}\underbrace{0,\ldots ,0,}_{g+b} -\frac{2\eta }{\eta ^2+1}\sqrt{k-g}\delta _{k+b+1}\theta )^{\top }, \end{aligned}$$

(34)

where $\mu =\frac{(1-\delta _{k+b+1})+(1+\delta _{k+b+1})\eta ^2}{\eta ^2+1}\theta$. Similarly, let the error vector

$$\begin{aligned} \varvec{v}&=\varvec{D}^{-1}\varvec{U}(\underbrace{0,\ldots ,0}_{k+b},-\sqrt{1-\delta _{k+b+1}}\varepsilon )^{\top }\\&=\varvec{D}^{-1}\sqrt{\frac{1-\delta _{k+b+1}}{\eta ^2+1}}(\underbrace{0, \cdots ,0}_{k-g-1},-\eta \varepsilon ,\underbrace{0,\ldots , 0}_{g+b},\varepsilon )^{\top }\\&=\frac{1}{\sqrt{\eta ^2+1}}\bigg (\underbrace{0,\ldots ,0}_{k-g-1},-\eta \varepsilon , \underbrace{0,\ldots ,0}_{g+b}, \sqrt{\frac{1-\delta _{k+b+1}}{1+\delta _{k+b+1}}}\varepsilon \bigg )^{\top } \end{aligned}$$

then $\Vert \varvec{v}\Vert _2\le \varepsilon$ and

$$\begin{aligned} \varvec{A}^{\top }\varvec{v}&=\varvec{U}^{\top }\varvec{DD}^{-1}\varvec{U}(\underbrace{0,\ldots ,0}_{k+b}, -\sqrt{1-\delta _{k+b+1}}\varepsilon )^{\top }\nonumber \\&=(\underbrace{0,\ldots ,0}_{k+b}, -\sqrt{1-\delta _{k+b+1}}\varepsilon )^{\top }. \end{aligned}$$

(35)

By (34) and (35), it is clear that

$$\begin{aligned} \varvec{A}_{T_0}^{\top }\varvec{A}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0}=\varvec{0},\ \ \ \ \varvec{A}_{T_0}^{\top }\varvec{v}=\varvec{0}. \end{aligned}$$

Therefore, using (7) and the above equality, we obtain that

$$\begin{aligned} \varvec{r}^{(0)}&=\varvec{A}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0}-\varvec{A}_{T_0}(\varvec{A}_{T_0}^{\top }{\varvec{A}_{T_0}})^{-1}\varvec{A}_{T_0}^{\top }\varvec{v}\\&\quad -\varvec{A}_{T_0}(\varvec{A}_{T_0}^{\top }{\varvec{A}_{T_0}})^{-1}\varvec{A}_{T_0}^{\top }\varvec{A}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0}+\varvec{v}\\&=\varvec{A}_{T{\setminus } T_0}\varvec{x}_{T{\setminus } T_0}+\varvec{v}. \end{aligned}$$

Therefore, we have that

$$\begin{aligned} \langle \varvec{Ae}_i, \varvec{r}^{(0)} \rangle&=\left\{ \begin{array}{ll} \frac{(1-\delta _{k+b+1})+(1+\delta _{k+b+1})\eta ^2}{\eta ^2+1}\theta , \ \ \ \ &{}\quad i\in T{\setminus } T_0 \\ -\frac{2\eta \sqrt{k-g}\delta _{k+b+1}\theta }{\eta ^2+1} -\sqrt{1-\delta _{k+b+1}}\varepsilon , &{}\quad i=k+b+1 \end{array} \right. \\&=\left\{ \begin{array}{ll} (1-\frac{\delta _{k+b+1}}{\sqrt{k-g+1}})\theta ,&{}\quad i\in T{\setminus } T_0 \\ -\frac{(k-g)\delta _{k+b+1}\theta }{\sqrt{k-g+1}}-\sqrt{1-\delta _{k+b+1}}\varepsilon ,&{}\quad i=k+b+1. \end{array} \right. \end{aligned}$$

From (30) and the above inequality, it follows that

$$\begin{aligned} \max _{i\in T{\setminus } T_0}\left| \left\langle \varvec{Ae}_i, \varvec{r}^{(0)} \right\rangle \right| =\max _{i\in (T\cup T_0)^c}\left| \left\langle \varvec{A e}_i, \varvec{r}^{(0)} \right\rangle \right| , \end{aligned}$$

which means the $\mathrm {OMP}_{T_0}$ algorithm may choose a wrong index $k+b+1$ in the first iteration. That is, the remainder support $T{\setminus } T_0$ of the signal $\varvec{x}$ may not be exactly recovered in $k-g$ iterations by the $\mathrm {OMP}_{T_0}$ algorithm. We completed the proof. $\square$

Appendix G: The Proof of Theorem 6

Proof

By $g\ge (1-\frac{1}{c^2})(k+1)$, we derive that

$$\begin{aligned} \frac{c}{\sqrt{k+1}}\le \frac{1}{\sqrt{k-g+1}}. \end{aligned}$$

(36)

Since $1 \le b\le (c-2)\lceil \frac{k}{2}\rceil$, we have $k+b+1\le c\lceil \frac{k}{2}\rceil$. Then, $\delta _{k+b+1}\le \delta _{c\lceil \frac{k}{2}\rceil }$. Therefore, from $\delta _{cr}<c\cdot \delta _{2r}$ for any positive integers c and r (seeing [25, Corollary 3.4]), the fact $k+1\ge 2\lceil \frac{k}{2}\rceil$ with $k\ge 2$, $\delta _{k+1}<\frac{1}{\sqrt{k+1}}$ and the inequality (36), it follows that

$$\begin{aligned} \delta _{k+b+1}\le \delta _{c\lceil \frac{k}{2}\rceil }<c\delta _{2\lceil \frac{k}{2}\rceil } \le c\delta _{k+1} <\frac{c}{\sqrt{k+1}}\le \frac{1}{\sqrt{k-g+1}}, \end{aligned}$$

which implies the condition $\delta _{k+b+1}$ in this paper is weaker than the sufficient condition $\delta _{k+1}<\frac{1}{\sqrt{k+1}}$. We complete the proof of the theorem. $\square$