## 2 Analytical Properties of 5-Point-Window \(L_{1}\) Splines
\right.\]  
_When \(c>0\),_  
* _If_ \(q>\frac{\sqrt{10}+1}{3}c\)_, then_ \(p(q)=c\) _and_ \[\frac{4\sqrt{10}-8}{3}<\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=\frac{10q^{2}+20 cq+ 6c^{2}}{6(c+q)^{2}}<\frac{5}{3}\,.\]_Algorithms_ **2010**, \(3\)  
* _If_ \(0<q\leq\frac{\sqrt{10}+1}{3}c\)_, then_ \(p(q)=\frac{\sqrt{10}-1}{3}q\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=\frac{4\sqrt{10}-8}{3}\,.\]
* _If_ \(\frac{2-\sqrt{10}}{\sqrt{10}}c\leq q<0\)_, then_ \(p(q)=\frac{\sqrt{10}}{2-\sqrt{10}}q\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=0\,.\]
* _If_ \(-\frac{1}{2}c<q<\frac{2-\sqrt{10}}{\sqrt{10}}c\)_, then_ \(p(q)=c\) _and_ \[-1<\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=\frac{10q^{2}+20 cq+6c^{2}}{6(c+q)^{2} }<0\,.\]
* _If_ \(-2c\leq q\leq-\frac{1}{2}c\)_, then_ \(p(q)=c\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=-1\,.\]
* _If_ \(q<-2c\)_, then_ \(p(q)=c\) _and_ \[-\frac{5}{3}<\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=-\frac{10q^{2}+20 cq+6c^{2}}{ 6(c+q)^{2}}<-1\,.\]  
_When \(c<0\),_  
* _If_ \(q>-2c\)_, then_ \(p(q)=c\) _and_ \[1<\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=\frac{10q^{2}+20 cq+6c^{2}}{6(c+q)^{2}}< \frac{5}{3}\,.\]
* _If_ \(-\frac{1}{2}c\leq q\leq-2c\)_, then_ \(p(q)=c\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=1\,.\]
* _If_ \(\frac{2-\sqrt{10}}{\sqrt{10}}c<q<-\frac{1}{2}c\)_, then_ \(p(q)=c\) _and_ \[0<\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=-\frac{10q^{2}+20 cq+6c^{2}}{6(c+q)^{2} }<1\,.\]
* _If_ \(0<q\leq\frac{2-\sqrt{10}}{\sqrt{10}}c\)_, then_ \(p(q)=\frac{\sqrt{10}}{2-\sqrt{10}}q\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=0\,.\]
* _If_ \(\frac{\sqrt{10}+1}{3}c\leq q<0\)_, then_ \(p(q)=\frac{\sqrt{10}-1}{3}q\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=\frac{4\sqrt{10}-8}{3}\,.\]0* _If_ \(q<\frac{\sqrt{10}+1}{3}\)_c, then_ \(p(q)=c\) _and_ \[\frac{\mathrm{d}G(q;c)}{\mathrm{d}q}=\frac{4\sqrt{10}-8}{3}\,.\]1  
Proof. : The function \(\phi(p,q;c)\) is the sum of two convex functions, so it is also convex. The convexity of \(G(q;c)\) comes from the fact that it is the partial minimization of \(\phi(p,q;c)\) (see [23]). If \(q>0\), we calculate using Lemma 2  
\[\frac{\partial\theta(p,q)}{\partial p}=\left\{ \begin{array}{ll}- \frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}},&\quad\text{if }p<-2q,\\ -1,&\quad\text{if }-2q\leq p\leq-\frac{1}{2}q,\\ \frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}},&\quad\text{if }p>-\frac{1}{2}q, \end{array} \right.\]  
which is a nondecreasing function of \(p\) for any fixed \(q\). Moreover, when \(p<\frac{\sqrt{10}}{2-\sqrt{10}}q\),  
\[\frac{\partial\theta(p,q)}{\partial p}=-\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}} <-\frac{2(\sqrt{10}-1)}{3}\,.\]  
When \(p>\frac{\sqrt{10}-1}{3}q\),  
\[\frac{\partial\theta(p,q)}{\partial p}=\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}} >\frac{2(\sqrt{10}-1)}{3}\,.\]  
Analogously, if \(q<0\), we calculate from Lemma 2  
\[\frac{\partial\theta(p,q)}{\partial p}=\left\{ \begin{array}{ll}- \frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}},&\quad\text{if }p<-\frac{1}{2}q,\\ 1,&\quad\text{if }-\frac{1}{2}q\leq p\leq-2q,\\ \frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}},&\quad\text{if }p>-2q, \end{array} \right.\]  
which is a nondecreasing function of \(p\) for any fixed \(q\). When \(p<\frac{\sqrt{10}-1}{3}q\),  
\[\frac{\partial\theta(p,q)}{\partial p}=-\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2} }<-\frac{2(\sqrt{10}-1)}{3}\,.\]  
When \(p>\frac{\sqrt{10}}{2-\sqrt{10}}q\),  
\[\frac{\partial\theta(p,q)}{\partial p}=\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}} >\frac{2(\sqrt{10}-1)}{3}\,.\]  
Note that  
\[\phi(p,q;c)=\frac{2(\sqrt{10}-1)}{3}|p-c|+\theta(p,q).\]  
When \(c=0\) and \(q>0\),  
\[\frac{\partial\phi(p,q;c)}{\partial p}=\left\{ \begin{array}{ll}\frac{2( \sqrt{10}-1)}{3}+\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}}>0,&\quad\text{if }p>0,\\ -\frac{2(\sqrt{10}-1)}{3}+\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}}<0,&\quad\text{ if }-\frac{1}{2}q<p<0,\\ -\frac{2(\sqrt{10}-1)}{3}-1<0,&\quad\text{if }-2q\leq p\leq-\frac{1}{2}q, \\ -\frac{2(\sqrt{10}-1)}{3}-\frac{10p^{2}+20 pq+6q^{2}}{6(p+q)^{2}}<0,&\quad\text{ if }p<-2q.\end{array}