## Appendix A  
**Proof.**: **Positivity of Theorem 1**  
\[J_{\lambda}(x)=\lambda\|x\|_{2}^{2}=\sum\limits_{k=1}^{p}\lambda_{k}x_{(k)}^{ 2}=\sum\limits_{k=1}^{p}(\sqrt{\lambda_{k}}x_{(k)})^{2}\]  
Take the square root on both side for the following:  
\[\|\sqrt{\lambda}x\|_{2}=\sqrt{\sum\limits_{k=1}^{p}(\sqrt{\lambda_{k}}x_{(k)} )^{2}}\]  
For \(\forall x\in\mathbb{R}^{p}\), hold Equation (2); thus, \((\lambda_{1..p})\) is positive and \((\lambda_{1..p})\
eq 0\). \(\|x\|_{2}\) will never be negative because of the square of \(x\).  
\[\|\sqrt{\lambda}x\|_{2}=\sqrt{\sum\limits_{k=1}^{p}(\sqrt{\lambda_{k}}x_{(k)} )^{2}}\geq 0\]  
If \(x=0\), then we have the following:  
\[\|\sqrt{\lambda}x\|_{2}=\sqrt{\sum\limits_{k=1}^{p}(\sqrt{\lambda_{k}}.0)^{2} }=0\] (A1)  
If \(\|x\|_{2}=0\), then we have the following:  
\[\|\sqrt{\lambda}x\|_{2}=\sqrt{\sum\limits_{k=1}^{p}(\sqrt{\lambda_{k}}x_{(k)} )^{2}}=0\Rightarrow x_{(k)}=0\forall k\Rightarrow x=0\]  
Since \((\lambda_{k})\
eq 0\), \(\|x\|_{2}\) will be only zero if and only if \(x=0\).  
**Proof.**: **Homogeneity of Theorem 1**  
The first two steps are the same as that of the proof of positivity of Theorem 1; then, we have the following:  
\[\|\sqrt{\lambda}x\|_{2}=\sqrt{\sum\limits_{k=1}^{p}(\sqrt{\lambda_{k}})^{2}x_ {(k)}^{2}}=|\sqrt{\lambda}|\sqrt{\sum\limits_{k=1}^{p}x_{(k)}^{2}}\]  
\[\|\sqrt{\lambda}x\|_{2}=|\sqrt{\lambda}|\|x\|_{2}\]  
where (\(c=\sqrt{\lambda}\)). Then, we have the following:  
\[\|cx\|_{2}=|c|\|x\|_{2}\]  
**Proof.**: **Triangle inequality of Theorem 1**The first two steps are the same as that of the proof of positivity of Theorem 1. Now, with \(x+y\) in place of \(x\), we have the following:  
\[\|\sqrt{\lambda}(x+y)\|_{2} =\sqrt{\frac{p}{k}}(\sqrt{\lambda_{k}}x_{(k)}+\sqrt{\lambda_{k}}y_ {(k)})^{2}\] \[\|\sqrt{\lambda}(x+y)\|_{2} =\sqrt{\frac{p}{k}}(\sqrt{\lambda_{k}}x_{(k)})^{2}+\frac{p}{k}( \sqrt{\lambda_{k}}y_{(k)})^{2}+2\frac{p}{k}\lambda_{k}x_{(k)}y_{(k)}\]  
From Cauchy-Schwarz inequality, \(x.y\leq\|x\|.\|y\|\), we have the following:  
\[\|\sqrt{\lambda}x+\sqrt{\lambda}y\|_{2} \leq\sqrt{\|\sqrt{\lambda}x\|^{2}+\|\sqrt{\lambda}y\|^{2}+2\| \sqrt{\lambda}x\|.\|\sqrt{\lambda}y\|}\] \[\|\sqrt{\lambda}x+\sqrt{\lambda}y\|_{2} \leq\sqrt{(\|\sqrt{\lambda}x\|+\|\sqrt{\lambda}y\|)^{2}}\] \[\|\sqrt{\lambda}x+\sqrt{\lambda}y\|_{2} \leq(\|\sqrt{\lambda}x\|+\|\sqrt{\lambda}y\|)\]  
where \(x=\sqrt{\lambda}x\) and \(y=\sqrt{\lambda}y\). Then, we have the following:  
\[\|x+y\|\leq\|x\|+\|y\|\]  
**Proof. Corollary 1**  
From Equation (3):  
\[J_{\lambda}(x) =\lambda_{1}x_{(1)}^{2}+\lambda_{2}x_{(2)}^{2}+\cdots=+\lambda_{p }x_{(p)}^{2}\] All \(\lambda_{k}\) take on an equal positive value, i.e., \(\lambda_{1}=\lambda_{2}=\cdots=\lambda_{p}\)  
\[J_{\lambda}(x) =\lambda x_{(1)}^{2}+\lambda x_{(2)}^{2}+\cdots=+\lambda x_{(p)}^ {2}\] \[J_{\lambda}(x) =\lambda(x_{(1)}^{2}+x_{(2)}^{2}+\cdots+x_{(p)}^{2})\] \[J_{\lambda}(x) =\lambda\sum\limits_{k=1}^{p}x_{(k)}^{2}\] \[J_{\lambda}(x) =\lambda\|x\|_{2}^{2}\]  
where \(\lambda\) is positive scalar.  
**Proof. Corollary 2**  
The first two steps are the same as that of the proof of positivity of Theorem 1. When \(\lambda_{2}=\ldots=\lambda_{p}\) = 0, we get the first term only and \(p=1\). The remaining terms are zero when \(p>1\).  
\[\|\sqrt{\lambda}x\|_{2} =\sqrt{\sum\limits_{k=1}^{p=1}(\sqrt{\lambda_{k}}x_{(k)})^{2}}\] \[\|\sqrt{\lambda}x\|_{2} =\sqrt{(\sqrt{\lambda_{1}}x_{(1)})^{2}}\] \[\|\sqrt{\lambda}x\|_{2} =|\sqrt{\lambda_{1}}x_{(1)}|\]  
where \(x_{1}=\sqrt{\lambda_{1}}x_{(1)}\) and \(\|x\|_{\infty}=max|x_{k}|\). Then, we have the following:  
\[\|x\|_{2} =\|x\|_{\infty}\]