## Appendix  
_Remark 1: the Penalty is Logarithmic Function._  
_Proof for Proposition 3:_ Let the \(t\)-th element \(x_{t}=0\) be activated as \(\hat{x}_{t}=\alpha=|\alpha|e^{j\arg\{\alpha\}}\), and the updated solution vector is denoted as \(\hat{\mathbf{x}}\). The update of the cost function is then given by  
\[\Delta J=J\{\hat{\mathbf{x}}\}-J\{\mathbf{x}\}=\frac{1}{2}|\alpha|^{2}\;R_{{}_ {\alpha}}-\Re\{\alpha^{\prime}b_{{}_{t}}\}+\tau\frac{1}{a}\log\left(1+a|\alpha|\right) \tag{24}\]  
The cost function is reduced if \(\Delta J<0\). For a fixed \(|\alpha|\), \(\Delta J\) achieves a minimum if \(\arg\{\alpha\}=\arg\{b_{t}\}\), and in this case  
\[\Delta J=\frac{1}{2}|\alpha|^{2}\;R_{{}_{\alpha}}-|\alpha|b_{{}_{t}}|+\tau \frac{1}{a}\log\left(1+a|\alpha|\right) \tag{25}\]  
As we can see, if \(|\alpha|=0\), then \(\Delta J=0\). So if \(\left.\frac{\partial\Delta J}{\partial|\alpha|}\right|_{\alpha|=0}<0\), when \(|\alpha|>0\), there exists a \(|\alpha|\) that makes \(\Delta J<0\). \[\left.\frac{\partial\Delta J}{\partial|\alpha|}\right|=R_{{}_{\alpha}}|\alpha| -|b_{{}_{t}}|+\frac{\tau}{1+a|\alpha|} \tag{26}\]  
If \(\left.\frac{\partial\Delta J}{\partial|\alpha|}\right|_{|\alpha|=0}<0\), we can get  
\[\tau-|b_{{}_{t}}|<0 \tag{27}\]  
Therefore, if \(|b_{{}_{t}}|>\tau\), there exists a \(|\alpha|\) that decreases the cost function. _Proof for Proposition 4:_ Let the \(t\)-th element \(x_{t}\
eq 0\) be activated as \(\hat{x}_{t}=0\), and the updated solution vector is denoted as \(\hat{\mathbf{x}}\). The update of the cost function is then given by  
\[\Delta J=J\left[\hat{\mathbf{x}}\right]-J\left[\mathbf{x}\right]=\frac{1}{2} \left|x_{t}\right|^{2}R_{t,x}+\Re\left\{x_{t}^{*}b_{t}\right\}-\tau\frac{1}{a} \log\left(1+a\left|x_{t}\right|\right) \tag{28}\]  
Thus, if \(\frac{1}{2}\left|x_{t}\right|^{2}R_{t,t}+\Re\{x_{t}^{*}b_{t}\}-\tau\frac{1}{a} \log\left(1+a\left|x_{t}\right|\right)<0\), then \(\Delta J<0\), there exists a nonzero value of the \(t\)-th element that decreases the cost function that should be removed from the support. _Remark 2: The Penalty is \(l_{p}\) Norm Function._  
_Proof for Proposition 5:_ Let the \(t\)-th element \(x_{t}=0\) be activated as \(\hat{x}_{t}=\alpha=|\alpha|e^{j\arg\{\alpha\}}\), and the updated solution vector is denoted as \(\hat{\mathbf{x}}\). The update of the cost function is then given by  
\[\Delta J=J\left[\hat{\mathbf{x}}\right]-J\left[\mathbf{x}\right]=\frac{1}{2} \left|\alpha\right|^{2}R_{t,x}-\Re\left\{\alpha^{*}b_{t}\right\}+\tau|\alpha| ^{p} \tag{29}\]  
The cost function is reduced if \(\Delta J<0\). For a fixed \(|\alpha|\), \(\Delta J\) achieves a minimum if \(\arg\{\alpha\}=\arg\{b_{t}\}\), and in this case  
\[\Delta J=\frac{1}{2}\left|\alpha\right|^{2}R_{t,x}-|\alpha|\left|b_{t}\right| +\tau|\alpha|^{p}=\frac{1}{2}R_{t,x}\left(|\alpha|-\frac{\left|b_{t}\right|}{R _{t,x}}\right)^{2}+\tau|\alpha|^{p}-\frac{\left|b_{t}\right|^{2}}{2R_{t,x}} \tag{30}\]  
From Equation (30) we can see that if \(|\alpha|=\frac{\left|b_{t}\right|}{R_{t,t}}\) and \(\tau|\alpha|^{p}-\frac{\left|b_{t}\right|^{2}}{2R_{t,t}}<0\), then \(\Delta J<0\). Thus, the value of \(|\alpha|\) results in the decrement of the cost function is given by  
\[\alpha=\frac{\left|b_{t}\right|}{R_{t,x}}e^{j\arg\{b_{t}\}} \tag{31}\]  
And we can get the proposition that if \(|b_{t}|^{2-p}>2\tau R_{t,t}^{1-p}\), the \(t\)-th element would be added into the support. _Proof for Proposition 6:_ Let the \(t\)-th element \(x_{t}\
eq 0\) be activated as \(\hat{x}_{t}=0\), and the updated solution vector is denoted as \(\hat{\mathbf{x}}\). The update of the cost function is then given by  
\[\Delta J=J\left[\hat{\mathbf{x}}\right]-J\left[\mathbf{x}\right]=\frac{1}{2} \left|x_{t}\right|^{2}R_{t,x}+\Re\left\{x_{t}^{*}b_{t}\right\}-\tau\left|x_{t} \right|^{p} \tag{32}\]  
Thus, if \(\frac{1}{2}\left|x_{t}\right|^{2}R_{t,t}+\Re\{x_{t}^{*}b_{t}\}-\tau|x_{t}|^{p}<0\), then \(\Delta J<0\), there exists a nonzero value of the \(t\)-th element that decreases the cost function that should be removed from the support.