# Convergence Gain in Compressive Deconvolution: Application to Medical Ultrasound Imaging
## 2 Method
### Preliminaries
#### 2.1.2 Notations  
We denote the 2-norm of a vector by \(\|\cdot\|\) and let \(\|z\|_{G}^{2}=z^{\mathsf{T}}Gz\) for \(z\in\mathbb{R}^{N}\) and \(G\in\mathbb{R}^{N\times N}\). For a real symmetric matrix \(S\), \(S\succeq 0\) (\(S\succ 0\)) represents \(S\), which is positive semidefinite (positive definite). For ease of the analysis, we define the following matrices as  
\[H=\left( \begin{array}{cc}R&0&0\\ 0&\frac{\rho+\rho-\rho\varrho}{\rho+\varrho}\beta B^{\mathsf{T}}B&-\frac{\rho} {\rho+\varrho}B^{\mathsf{T}}\\ 0&-\frac{\rho}{\rho+\varrho}B&\frac{1}{(\rho+\varrho)\beta}I_{N}\end{array} \right) \tag{14}\]  
\[M=\left( \begin{array}{ccc}I_{n}&0&0\\ 0&I_{N}&0\\ 0&-\varrho\beta B&(\rho+\varrho)I_{N}\end{array} \right) \tag{15}\]  
and  
\[Q=\left( \begin{array}{ccc}R&0&0\\ 0&\beta I_{N}&-\rho B^{\mathsf{T}}\\ 0&-B&\frac{1}{\rho}I_{N}\end{array} \right). \tag{16}\]  
Below we prove three assertions regarding the matrices just defined. These assertions make it possible to present our convergence analysis for the new algorithm compactly with alleviated notation. **Lemma 1**. : _Given \(R\succeq 0\), let \(\beta>0\), \(\mu,\rho\in(0,1)\), and \(\varrho\in(0,1]\). The matrices \(H,M\), and \(Q\) defined, respectively, in Equations (15) and (16) have the relationships as follows:_  
\[H\succ 0,\quad HM=Q \tag{17}\]\[G:=Q^{\mathsf{T}}+Q-M^{\mathsf{T}}HM\succeq 0. \tag{18}\]  
**Proof. ** We consider two cases. (I). \(\rho\in(0,1)\), \(\varrho\in(0,1)\). We only need to check that  
\[\bar{H}=\left( \begin{array}{ccc}R&0&0\\ 0&\rho+\varrho-\rho\varrho&\rho\\ 0&\rho&1\end{array} \right)\succ 0. \tag{19}\]  
Note that  
\[\left\{ \begin{array}{l}\rho+\varrho-\rho\varrho=\rho+\varrho(1-\rho)>0,\\ \rho+\varrho-\rho\varrho-\rho^{2}=(\varrho+\varrho)(1-\rho)>0.\end{array} \right.\]  
Then we have  
\[\bar{H}\succ 0. \tag{20}\]  
For any \(w=(x_{1},z_{2},\lambda)\
eq 0\), since \(\rho,\varrho\in(0,1)\), the assertion \(H\succ 0\) is verified. With the matrices \(H,M\), and \(Q\) at hand, we easily obtain  
\[HM =\left( \begin{array}{ccc}R&0&0\\ 0&\frac{\varrho+\varrho-\rho\varrho}{\rho+\varrho}\beta B^{\mathsf{T}}B&- \frac{\rho}{\varrho+\varrho}B^{\mathsf{T}}\\ 0&-\frac{\rho}{\varrho+\varrho}B&\frac{1}{(\varrho+\varrho)\varrho}I_{N} \end{array} \right)\left( \begin{array}{ccc}I_{N}&0&0\\ 0&I_{N}&0\\ 0&-\varrho\beta B&(\varrho+\varrho)I_{N}\end{array} \right)\] \[=\left( \begin{array}{ccc}R&0&0\\ 0&\beta I_{N}&-\rho B^{\mathsf{T}}\\ 0&-B&\frac{1}{\beta}I_{N}\end{array} \right)=Q. \tag{21}\]  
The second assertion \(HM=Q\) is proved. Consequently, we have  
\[M^{\mathsf{T}}HM =M^{\mathsf{T}}Q=\left( \begin{array}{ccc}I_{N}&0&0\\ 0&I_{N}&-\varrho\beta B^{\mathsf{T}}\\ 0&0&(\varrho+\varrho)I_{N}\end{array} \right)\left( \begin{array}{ccc}R&0&0 \\ 0&\beta I_{N}&-\rho B^{\mathsf{T}}\\ 0&-B&\frac{1}{\beta}I_{N}\end{array} \right) \tag{22}\] \[=\left( \begin{array}{ccc}R&0&0\\ 0&(1+\varrho)\beta B^{\mathsf{T}}B&-(\varrho+\varrho)B^{\mathsf{T}}\\ 0&-(\varrho+\varrho)B&\frac{\varrho+\varrho}{\beta}I_{N}\end{array} \right). \tag{23}\]  
Using Equations (15) and (16), and the above equation, we have  
\[G =Q^{\mathsf{T}}+Q-M^{\mathsf{T}}HM\] \[=\left( \begin{array}{ccc}2R&0&0\\ 0&2\beta I_{N}&-(1+\varrho)B^{\mathsf{T}}\\ 0&-(1+\varrho)B&\frac{2}{\beta}I_{N}\end{array} \right)-\left( \begin{array}[] {ccc}R&0&0\\ 0&(1+\varrho)\beta B^{\mathsf{T}}B&-(\varrho+\varrho)B^{\mathsf{T}}\\ 0&-(\varrho+\varrho)B&\frac{\varrho+\varrho}{\beta}I_{N}\end{array} \right)\] \[=\left( \begin{array}{ccc}R&0&0\\ 0&(1-\varrho)\beta B^{\mathsf{T}}B&(\varrho-1)B^{\mathsf{T}}\\ 0&(\varrho-1)B&\frac{2-\rho-\varrho}{\beta}I_{N}\end{array} \right).\]  
Note that \(\beta>0\) and \(\rho,\varrho\in(0,1)\). Thus, for any \(w=(z_{1},z_{2},\lambda)\
eq 0\), we have  
\[\left\{ \begin{array}{l}1-\varrho>0\\ (1-\varrho)(2-\rho-\varrho)-(1-\varrho)^{2}=(1-\varrho)\{(1-\varrho)+(1-\rho) \}-(1-\varrho)^{2}>0\end{array} \right..\]Therefore, the matrix \(G\) is positive definite. (II). \(\rho\in(0,1)\) and \(\varrho=1\). Note that  
\[H=\left(