# A Graduated Non-Convexity Technique for Dealing Large Point Spread Functions
## Appendix A  
We prove here that \(\mathcal{B}_{n}=\mathcal{J}_{n,1}\). **Lemma A1**. : _The following inclusion holds:_  
\[\mathcal{B}_{n}\subset\mathcal{L}_{n,1}.\]  
Proof. : Let \(B\in\mathcal{B}_{n}\), \(B=(b_{k,l})_{k,l}\) and \(\Lambda^{(B)}=\text{diag}(\lambda^{(B)}_{0}\ \lambda^{(B)}_{1}\ \ldots\ \lambda^{(B)}_{n-1})\) be such that \(\lambda^{(B)}_{j}=-\lambda^{(B)}_{(n-j)\text{ mod }n}\) for every \(j\in\{0,1,\ldots,n-1\}\), and \(B=Q_{n}\Lambda^{(B)}\mathcal{Q}_{n}^{T}\). We have  
\[b_{k,l}=\sum_{j=0}^{n-1}q^{(n)}_{k,j}\,\lambda^{(B)}_{j}\,q^{(n)}_{l,j}.\]  
Observe that \(\lambda^{(B)}_{0}=0\) and \(\lambda^{(B)}_{n/2}=0\), if \(n\) is even. From this and (A1), we obtain  
\[b_{k,l}=\sum_{j=1}^{\lfloor(n-1)/2\rfloor}\lambda^{(B)}_{j}\cdot\big{(}q^{(n )}_{k,j}q^{(n)}_{l,j}-q^{(n)}_{k,n-j}q^{(n)}_{l,n-j}\big{)},\]  
both when \(n\) is even and when \(n\) is odd. From (5) and (A1), we deduce  
\[b_{k,l} = \frac{2}{n}\sum_{j=1}^{\lfloor(n-1)/2\rfloor}\lambda^{(B)}_{j} \Bigg{(}\cos\bigg{(}\frac{2\pi kj}{n}\bigg{)}\cos\bigg{(}\frac{2\pi lj}{n} \bigg{)}-\sin\bigg{(}\frac{2\pi kj}{n}\bigg{)}\sin\bigg{(}\frac{2\pi lj}{n} \bigg{)}\Bigg{)}\] \[= \frac{2}{n}\sum_{j=1}^{\lfloor(n-1)/2\rfloor}\lambda^{(B)}_{j} \cos\bigg{(}\frac{2\pi(k+l)j}{n}\bigg{)}.\]  
Let \(\mathbf{b}=(b_{0}\,b_{1}\,\cdots\,b_{n-1})^{T}\), where  
\[b_{t}=\frac{2}{n}\sum_{j=1}^{\lfloor(n-1)/2\rfloor}\lambda^{(B)}_{j}\cdot \cos\bigg{(}\frac{2\pi tj}{n}\bigg{)},\quad t\in\{0,1,\ldots,n-1\}.\] (A1)Thus, \(B=\text{circ}(\mathbf{b})\), because for each \(k,l\in\{0,1,\ldots,n-1\}\) we have \(b_{k,l}=b_{(k-l)\text{ mod }n}\). For any \(k,l\in\{0,1,\ldots,n-1\}\) it is \(b_{k,l}=b_{(k+l)\text{ mod }n}\). Hence, \(\mathcal{B}_{n}\subset\mathcal{L}_{n,1}\). One has  
\[\mathcal{B}_{n}\subset\mathcal{K}_{n,1}.\]  
Proof. : We recall that  
\[\mathcal{K}_{n,1}=\Big{\{}B\in\mathbb{R}^{n\times n}:\text{there is a symmetric }\mathbf{b}=(b_{0}\,\ldots\,b_{n-1})^{T}\in\mathbb{R}^{n}\text{ with }b_{k,j}=b_{(j+k)\text{ mod }n}\Big{\}}.\]  
By Lemma A, we obtain \(\mathcal{B}_{n}\subset\mathcal{L}_{n,1}\). Now, we prove the symmetry of \(\mathbf{b}\). Let \(B\in\mathcal{B}_{n}\) be such that there exists \(\Lambda^{(B)}\in\mathbb{R}^{n\times n}\), \(\Lambda^{(B)}=\text{diag}(\lambda_{0}^{(B)}\ \lambda_{1}^{(B)}\ \cdots\ \lambda_{n-1}^{(B)})\), such that \(C=Q_{n}\Lambda^{(B)}Q_{n}^{T}\) and \(\lambda_{j}^{(B)}=-\lambda_{(n-j)\text{ mod }n}^{(B)}\) for all \(j\in\{0,\,1,\ldots,n-1\}\). By Theorem A, \(b_{k,j}=b_{(j+k)\text{ mod }n}\). Moreover, by arguing as in Lemma A, we obtain (A), and hence  
\[b_{t} = \frac{2}{n}\sum_{j=1}^{\lfloor(n-1)/2\rfloor}\lambda_{j}^{(B)} \cdot\cos\left(\frac{2\pi tj}{n}\right)=\frac{2}{n}\sum_{j=1}^{\lfloor(n-1)/2 \rfloor}\lambda_{j}^{(B)}\cdot\cos\left(2\pi j-\frac{2\pi tj}{n}\right)\] \[= \frac{2}{n}\sum_{j=1}^{\lfloor(n-1)/2\rfloor}\lambda_{j}^{(B)} \cdot\cos\left(\frac{2\pi(n-t)j}{n}\right)=b_{n-t}\]  
for any \(t\in\{0,1,\ldots,n-1\}\). Thus, \(\mathbf{b}\) is symmetric. Now, we present the following:  
The following result holds:  
\[b_{k,l}=\sum_{j=0}^{n-1}q^{(n)}_{k,j}\,\lambda^{(B)}_{j}\,q^{(n)}_{l,j}.\]0  
Proof. : First of all, we recall that  
\[b_{k,l}=\sum_{j=0}^{n-1}q^{(n)}_{k,j}\,\lambda^{(B)}_{j}\,q^{(n)}_{l,j}.\]1 \[b_{k,l}=\sum_{j=0}^{n-1}q^{(n)}_{k,j}\,\lambda^{(B)}_{j}\,q^{(n)}_{l,j}.\]2  
We begin with proving that \(\mathcal{B}_{n}\subset\mathcal{J}_{n,1}\). Let \(B\in\mathcal{B}_{n}\). In Lemma A we proved that \(B\in\mathcal{K}_{n,1}\), that is, \(\mathbf{b}\) is symmetric and \(b_{k,j}=b_{(j+k)\text{ mod }n}\). Now we prove that  
\[b_{k,l}=\sum_{j=0}^{n-1}q^{(n)}_{k,j}\,\lambda^{(B)}_{j}\,q^{(n)}_{l,j}.\]3  
Since \(B\in\mathcal{B}_{n}\), the vector  
\[b_{k,l}=\sum_{j=0}^{n-1}q^{(n)}_{k,j}\,\lambda^{(B)}_{j}\,q^{(n)}_{l,j}.\]4  
is an eigenvector for the eigenvalue \(\lambda_{0}^{(B)}=0\). Hence, the formula (A) is a consequence of (12).