A Smoothing Inertial Neural Network for Sparse Signal Reconstruction with Noise Measurements via $L_p$-$L_1$ minimization

Abstract

In this paper, a smoothing inertial neural network (SINN) is proposed for the $L_{p}$-$L_{1}$ $(1\ge p>0)$ minimization problem, in which the objective function is non-smooth, non-convex, and non-Lipschitz. First, based on the smooth approximation technique, the objective function can be transformed into a smooth optimization problem, effectively solving the $L_{p}$-$L_{1}$ $(1\ge p>0)$ minimization model with non-smooth terms. Second, the Lipschitz property of the gradient that is smooth objective function is discussed. Then through theoretical analysis, the existence and uniqueness of the solution is discussed under the condition of restricted isometric property (RIP), and it is proved that the proposed SINN converges to the optimal solution of the minimization problem. Finally, the effectiveness and superiority of the proposed SINN are verified by the successful recovery performance under different pulse noise levels.

Appendix A

Proof of Theorem 2

We take $x^*\in \Omega$ and a stationary point is defined as (19), and consider the Lyapunov function $L(t)=\frac{1}{2}\Vert x(t)-x^*\Vert ^2$. Then, the following two equations hold:

$$\begin{aligned} \dot{L}(t)=(x(t)-x^*)^T\dot{x}(t),\ddot{L}(t)=(\dot{x}(t))^T\dot{x}(t)+(x(t)-x^*)^T\ddot{x}(t). \end{aligned}$$

(44)

and so we can obtain

$$\begin{aligned} \ddot{L}(t)+\lambda \dot{L}(t)=\Vert \dot{x}(t)\Vert ^2+\langle x(t)-x^*,\ddot{x}(t)+\lambda \dot{x}(t)\rangle . \end{aligned}$$

(45)

and calculate with (15), it is not hard to get

$$\begin{aligned} \begin{aligned} \ddot{x}(t)+\lambda \dot{x}(t)&=-\nabla \hat{f}(x(t),\varepsilon (t))\\&=-\mathbb {R}(x(t),\varepsilon (t)). \end{aligned} \end{aligned}$$

(46)

Then, (46) become

$$\begin{aligned} \Vert \dot{x}(t)\Vert ^2=\ddot{L}(t)+\lambda \dot{L}(t)+\langle x(t)-x^*,\mathbb {R}(x(t),\varepsilon (t))\rangle \end{aligned}$$

(47)

Since $A\mathbb {R}(x^*,\varepsilon (t))=0$, then

$$\begin{aligned} \begin{aligned} \Vert \dot{x}(t)\Vert ^2=\ddot{L}(t)+\lambda \dot{L}(t) +\langle x(t)-x^*,\mathbb {R}(x(t),\varepsilon (t)-A\mathbb {R}(x^*,\varepsilon (t)))\rangle . \end{aligned} \end{aligned}$$

(48)

From condition (ii), it follows that

$$\begin{aligned} \Vert \dot{x}(t)\Vert ^2\ge \ddot{L}(t)+\lambda \dot{L}(t)+\sigma \Vert \mathbb {R}(x(t),\varepsilon (t))\Vert ^2. \end{aligned}$$

(49)

Combining (48) and (49), we obtain

$$\begin{aligned} \Vert \dot{x}(t)\Vert ^2\ge \ddot{L}(t)+\lambda \dot{L}(t)+\sigma \Vert \ddot{x}(t)+\lambda \dot{x}(t)\Vert ^2. \end{aligned}$$

(50)

Therefore, (50) can be converted to

$$\begin{aligned} \begin{aligned} \ddot{L}(t)+\lambda \dot{L}(t)+\sigma \Vert \ddot{x}(t)\Vert ^2+\sigma \lambda \frac{d\Vert \dot{x}(t)\Vert ^2}{dt} +(\sigma \lambda ^2-1)\Vert y(t)\Vert ^2\le 0 \end{aligned} \end{aligned}$$

(51)

we can get it from (51)

$$\begin{aligned} \begin{aligned} V(t)&=\dot{L}(t)+\lambda \ L(t)\\&+\sigma \lambda \Vert \dot{x}(t)\Vert ^2+\int _{0}^t\sigma \Vert \ddot{x}(t)\Vert ^2ds+(\sigma \lambda ^2-1)\int _{0}^t\Vert \dot{x}(s)\Vert ^2ds \end{aligned} \end{aligned}$$

(52)

From (51), we know V(t) is a monotone nonincreasing function, so we obtain

$$\begin{aligned} \begin{aligned} V(t)&\le \dot{L}(0)+\lambda L(0)+\sigma \lambda \Vert \dot{x}(0)\Vert ^2\\&=\Vert x(0)-x^*\Vert \dot{x}(0)+\frac{\lambda }{2}\Vert x(0)-x^*\Vert +\sigma \lambda \Vert \dot{x}(0)\Vert ^2\\&=H_{h_{0}},t>0 \end{aligned} \end{aligned}$$

(53)

Then, it can be obtained from condition (c)

$$\begin{aligned} \dot{L}(t)+\lambda L(t)\le H_{h_{0}} \end{aligned}$$

(54)

Next, multiply the two sides of the equation by $\mathrm{exp}(\lambda t)$ and integrate from 0 to t

$$\begin{aligned} \begin{aligned} \left( \dot{L}(t)+\lambda L(t)\right) \mathrm{exp}(\lambda t)\le H_{h_{0}}\mathrm{exp}(\lambda t)\\ \frac{d}{dt}\left( L(t)\mathrm{exp}(\lambda t)\right) \le H_{h_{0}}\mathrm{exp}(\lambda t)\\ L(t)\le \frac{H_{h_{0}}}{\lambda }+L(0)\mathrm{exp}(-\lambda t) \end{aligned} \end{aligned}$$

(55)

where $h_{0}=0$, so the trajectory of the algorithm and L(t) are bounded.

And then from the inequality (53), we can get

$$\begin{aligned} \dot{L}(t)+\sigma \lambda \Vert \dot{x}(t)\Vert ^2=\langle x(t)-x^*,\dot{x}(t)\rangle +\sigma \lambda \Vert \dot{x}(t)\Vert ^2\le H_{h_{0}} \end{aligned}$$

(56)

That is, we can get that $\Vert \dot{x}(t)\Vert$ is also bounded, then the following conditions can be obtained by (49) and (53) : $\int _{0}^{+\infty }\Vert \dot{x}(r)\Vert ^2dr=A<+\infty ,\ \Vert \ddot{x}(r)\Vert dr\le B$ where t is at $[0,+\infty ),\ A,\ B>0$, therefore

$$\begin{aligned} \begin{aligned} \int _{0}^{+\infty }\Vert \dot{x}(r)\Vert ^2\Vert \ddot{x}(r)\Vert dr&=\frac{1}{3}\lim \limits _{t\rightarrow \infty }\left( \Vert \dot{x}(t)^3\Vert -\Vert \dot{x}(0)\Vert \right) \\&\le A\int _{0}^{+\infty }\Vert \ddot{x}(r)\Vert dr<+\infty \end{aligned} \end{aligned}$$

(57)

Thus, the limit of $\Vert \dot{x}(t)\Vert ^3$ exists. Furthermore, it can be obtained that the limit of $\lim \limits _{t\rightarrow \infty }\Vert \dot{x}(t)\Vert$ also exists, since $\int _{0}^{+\infty }\Vert \dot{x}(r)\Vert ^2dr=A<+\infty$ and L(t) are bounded, we can obtain $\lim \limits _{t\rightarrow \infty }\Vert \dot{x}(t)\Vert =0$, and further obtain $\lim \limits _{t\rightarrow \infty }\dot{x}(t)=0$.

Afterward, we give the proof of $\lim \limits _{t\rightarrow \infty }\Vert \ddot{x}(t)\Vert =0$.

Define $h_{\upsilon }(t)=(\frac{1}{\upsilon })(\dot{x}(t+\upsilon )-\dot{x}(t))$, and available by simple calculations:

$$\begin{aligned} \begin{array}{lr} \dot{h}_{\upsilon }(t)+\lambda \dot{h}_{\upsilon }(t)= -\frac{1}{\upsilon }(\mathbb {R}(x(t+\upsilon ),\varepsilon (t+\upsilon ))-\mathbb {R}(x(t),\varepsilon (t))) \end{array} \end{aligned}$$

(58)

Then, according to the condition (a), we can get

$$\begin{aligned} \begin{array}{lr} \Vert \mathbb {R}(x(t+\upsilon ),\varepsilon (t+\upsilon ))-\mathbb {R}(x(t),\varepsilon (t))\Vert &{}\\ \le \Vert \mathbb {R}(x(t+\upsilon ),\varepsilon (t+\upsilon ))-\mathbb {R}(x(t),\varepsilon (t+\upsilon ))\Vert &{}\\ \quad +\Vert \mathbb {R}(x(t),\varepsilon (t+\upsilon ))-\mathbb {R}(x(t),\varepsilon (t))\Vert &{}\\ \le \frac{\upsilon }{\zeta } \sup _{r \in [t,+\infty )}\Vert \dot{x}(r)\Vert +k_{j}(1-\exp (-\upsilon ))\Vert u(t)\Vert &{} \end{array} \end{aligned}$$

(59)

Integrate (58), we can easily get $\lim \limits _{t\rightarrow +\infty } \sup \Vert h_{\upsilon }(t)\Vert =0$, then we can get condition $\lim \limits _{t\rightarrow \infty }\Vert \ddot{x}(t)\Vert =0$ from $\Vert \ddot{x}(r)\Vert \le \sup \Vert h_{\upsilon }(t)\Vert$. Thence, we have

$$\begin{aligned} \lim \limits _{t\rightarrow \infty }\nabla _{x}\hat{f}(x(t),\varepsilon (t))=0 \end{aligned}$$

(60)

Therefore, we demonstrate that $x^*$ is a stable point defined by (19).