## 3 Proposed Method
After the accelerated step, the convergence rate increases from \(\mathbf{O}(1/k)\) to \(\mathbf{O}(1/k^{2})\). This algorithm, which is referred to as SRR using the fast plug-and-play ADMM, is summarized in Algorithm 2. Regarding the sub-problem \(\mathbf{Z}\), Equation (22) can be converted into the following constraint problem:  
\[\mathbf{Z}=\arg\min_{\mathbf{Z}}\frac{1}{2}\|\mathbf{Z}-\widehat{\mathbf{Z}}\|_{2}^{2}+\lambda \sum_{i=1}^{4}\|\mathbf{T}_{i}\|_{p}^{p},\quad\mathbf{s.t.}\quad\mathbf{T}_{i}=\mathbf{K}_{i} *\mathbf{Z},\ (i=1,2,3,4). \tag{27}\]  
Accordingly, the augmented Lagrangian function is  
\[\mathbf{L}_{\rho}(\mathbf{Z},\mathbf{T}_{1},\mathbf{T}_{2},\mathbf{T}_{3},\mathbf{T}_{4}, \mathbf{V}_{1},\mathbf{V}_{2},\mathbf{V}_{3},\mathbf{V}_{4},) =\frac{1}{2}\|\mathbf{Z}-\widehat{\mathbf{Z}}\|_{2}^{2}+\lambda\sum_{i=1} ^{4}\|\mathbf{T}_{i}\|_{p}^{p} \tag{28}\] \[+\sum_{i=1}^{4}\langle\mathbf{V}_{i},(\mathbf{T}_{i}-\mathbf{K}_{i}*\mathbf{Z})\rangle\] \[+\frac{\eta}{2}\sum_{i=1}^{4}\|\mathbf{T}_{i}-\mathbf{K}_{i}*\mathbf{Z}\|_{2}^ {2},\]  
where \(\mathbf{V}_{i}\ (i=1,2,3,4)\) is the Lagrange multiplier, and \(\eta>0\) is a penalty parameter. Furthermore,based on the ADMM algorithm, the sub-problem \(\mathbf{Z}\) can be solved as follows:  
\[ \begin{split}\mathbf{Z}^{(n+1)}&=\arg\min_{\mathbf{Z}}\frac{1} {2}\|\mathbf{Z}-\widehat{\mathbf{Z}}\|_{2}^{2}+\sum_{i=1}^{4}\langle\mathbf{V}_{i}^{(n)},( \mathbf{T}_{i}^{(n)}-\mathbf{K}_{i}*\mathbf{Z})\rangle\\ &+\frac{\eta}{2}\sum_{i=1}^{4}\|\mathbf{T}_{i}^{(n)}-\mathbf{K}_{i}*\mathbf{ Z}\|_{2}^{2}\\ &=\arg\min_{\mathbf{Z}}\frac{1}{2}\|\mathbf{Z}-\widehat{\mathbf{Z}}\|_{2}^{2 }+\frac{\eta}{2}\sum_{i=1}^{4}\|\mathbf{T}_{i}^{(n)}-\mathbf{K}_{i}*\mathbf{Z}+\frac{\mathbf{V }_{i}^{(n)}}{\eta}\|_{2}^{2}.\end{split}  \tag{29}\]  
```
Initialize:\(\rho,\mu,tol,\tau\in(0,1)\) While\(\|\mathrm{PSNR}(F^{(k+1)},\mathbf{G})-\mathrm{PSNR}(\mathbf{F}^{(k)},\mathbf{G})\|_{2}/\| \mathrm{PSNR}(\mathbf{F}^{(k)},\mathbf{G})\|_{2}>tol\)do
1. \(\mathbf{F}^{(k+1)}=\arg\min_{\mathbf{F}}\frac{1}{2}\|\mathbf{SHF}-\mathbf{G}\|_{2}^{2}+\frac{ \rho}{2}\|\mathbf{F}-(\widehat{\mathbf{Z}}^{(k)}-\frac{\hat{\mathbf{Y}}^{(k)}}{\rho})\|_{2 }^{2}\)
2. \(\mathbf{Z}^{(k+1)}=\mathbf{D}_{\lambda}(\widehat{\mathbf{Z}}^{(k+1)})\), where \(\widehat{\mathbf{Z}}=\mathbf{F}^{(k+1)}+\frac{\hat{\mathbf{Y}}^{(k)}}{\rho}\), \(\lambda=\frac{\mu}{\rho}\)
3. \(\mathbf{Y}^{(k+1)}=\widehat{\mathbf{Y}}^{(k)}+\rho(\mathbf{F}^{(k+1)}-\mathbf{Z}^{(k+1)})\)
4. \(c^{(k+1)}=\rho^{-1}\|\mathbf{Y}^{(k+1)}-\widehat{\mathbf{Y}}^{(k+1)}\|^{2}+\rho\|\mathbf{Z }^{(k+1)}-\widehat{\mathbf{Z}}^{(k+1)}\|^{2}\)
5. if\(c^{(k+1)}\leq\tau c^{(k)}\)then
6. \(\alpha^{(k+1)}=\frac{1+\sqrt{1+4(\alpha^{(k)})^{2}}}{2}\)
7. \(\widehat{\mathbf{Z}}^{(k+1)}=\mathbf{Z}^{(k+1)}+\frac{\alpha^{(k)}-1}{\alpha^{(k+1)}} (\mathbf{Z}^{(k+1)}-\mathbf{Z}^{(k)})\)
8. \(\widehat{\mathbf{Y}}^{(k+1)}=\mathbf{Y}^{(k+1)}+\frac{\alpha^{(k)}-1}{\alpha^{(k+1)}} (\mathbf{Y}^{(k+1)}-\mathbf{Y}^{(k)})\)
9. else
10. \(\alpha^{(k+1)}=1\), \(\widehat{\mathbf{Z}}^{(k+1)}=\mathbf{Z}^{(k+1)}\), \(\widehat{\mathbf{Y}}^{(k+1)}=\mathbf{Y}^{(k+1)}\)
11. \(c^{(k+1)}=\tau^{-1}c^{(k)}\)
12. endif
13. \(k=k+1\)
```  
**Algorithm 2** Super-resolution using the fast plug-and-play ADMM  
By employing the convolution theorem [48], the 2D Fourier transform of \(\mathbf{Z}\) can be obtained as follows:  
\[ \begin{split}\mathcal{F}(\mathbf{Z}^{(n+1)})&=\arg \min_{\mathbf{Z}}\frac{1}{2}\|\mathcal{F}(\mathbf{Z})-\mathcal{F}(\widehat{\mathbf{Z}})\|_ {2}^{2}\\ &+\frac{\eta}{2}\sum_{i=1}^{4}\|\mathcal{F}(\mathbf{T}_{i}^{(n)})- \mathcal{F}(\mathbf{K}_{i})\circ\mathcal{F}(\mathbf{Z})+\frac{\mathcal{F}(\mathbf{V}_{i}^ {(n)})}{\eta}\|_{2}^{2},\end{split}  \tag{30}\]  
where the symbol \(\circ\) represents component-wise multiplication.