## 5 KKT Conditions and the Fast Solution of Problem (2)
\right.\] which yield the next three cases: - \(\boxed{\varepsilon_{2}t-\varepsilon_{1}u>0}\) admitting other three cases, namely * \(\boxed{u>0}\), so that we set \[\beta_{1}=\max\biggl{\{}a_{2},-\frac{a_{1}t+y}{u},\frac{\varepsilon_{3}t+ \varepsilon_{1}y}{\varepsilon_{2}t-\varepsilon_{1}u}\biggr{\}}\leq\beta\leq \min\biggl{\{}b_{2},-\frac{b_{1}t+y}{u}\biggr{\}}=\beta_{2}\] * \(\boxed{u=0}\), so that \(b_{1}t+y\leq-u\beta\leq a_{1}t+y\) is always fulfilled and we set \[\beta_{1}=\max\biggl{\{}a_{2},\frac{\varepsilon_{3}t+\varepsilon_{1}y}{ \varepsilon_{2  
3. If \(\framebox{\varepsilon_{2}<0}\), then set \[\beta_{1}=\max\biggl{\{}a_{2},\frac{\varepsilon_{1}b_{1}-\varepsilon_{3}}{- \varepsilon_{2}}\biggr{\}}\leq\beta\leq\min\biggl{\{}b_{2},\frac{\varepsilon_ {1}a_{1}-\varepsilon_{3}}{-\varepsilon_{2}}\biggr{\}}=\beta_{2}.\] Set \(\bar{\beta}=(\beta_{1}+\beta_{2})/2\) and \(\bar{\alpha}=(-\varepsilon_{2}\bar{\beta}+\varepsilon_{3})/\varepsilon_{1}\); if \(a_{1}\leq\bar{\alpha}\leq b_{1}\), then set \[P_{11}=\left( \begin{array}{c}\bar{\alpha}\\ \bar{\beta}\end{array} \right),\qquad\varphi_{i}=\varphi(\bar{\alpha},\bar{ \beta}),\qquad y_{i}=P_{11},\qquad i=i+1,\] (26) otherwise, there is no solution for Case (XI);
* Case (XII): We set \(\bar{\beta}=a_{2}\) with \(\varepsilon_{1}\alpha+\varepsilon_{2}a_{2}=\varepsilon_{3}\), and we distinguish among three cases:
* if \(\framebox{\varepsilon_{1}=0\text{ AND }\varepsilon_{3}=\varepsilon_{2}a_{2}}\), then set \(\bar{\alpha}=(a_{1}+b_{1})/2\);
* if \(\framebox{\varepsilon_{1}=0\text{ AND }\varepsilon_{3}\
eq\varepsilon_{2}a_{2}}\), then there is no solution for Case (XII);
* if \(\framebox{\varepsilon_{1}\
eq 0}\), then set \[\alpha_{1}=\max\biggl{\{}a_{1},\frac{\varepsilon_{3}-\varepsilon_{2}a_{2}}{ \varepsilon_{1}}\biggr{\}}\leq\alpha\leq\min\biggl{\{}b_{1},\frac{\varepsilon _{3}-\varepsilon_{2}a_{2}}{\varepsilon_{1}}\biggr{\}}=\alpha_{2}.\] Set \(\bar{\alpha}=(\alpha_{1}+\alpha_{2})/2\) with \[P_{12}=\left( \begin{array}{c}\bar{\alpha}\\ a_{2}\end{array} \right),\qquad\varphi_{i}=\varphi(\bar{\alpha},a_{2}),\qquad y _{i}=P_{12},\qquad i=i+1.\] (27) The next lemma justifies the role of the last analysis for the computation of possible solutions of (14). Lemma 2: _Given the problem (14), and let the assumptions of Lemma 1 hold. Consider the sequence of the \(m\) points \(\{y_{i}\}\) and the sequence of the \(m\) values \(\{\varphi_{i}\}\), from (16)-(27), which are relabelled so that for any index \(i\geq 2\), we have_  
\[\varphi_{i-1}\leq\varphi_{i}\leq\varphi_{i+1}.\]  
_Then, if_  
\[\hat{\beta}\in\operatorname*{argmin}_{1\leq i\leq m}\{\varphi_{i}\}\]  
_then, the point \(y_{\hat{\beta}}\) is a global minimum for (14)._  
Proof of Lemma 2: The existence of a global minimum \(y^{*}\) and the corresponding value \(\varphi(y^{*})\) for (14) is ensured by Lemma 1. Moreover, each global minimum of (14) naturally fulfills KKT conditions, so that each global minimum must belong to the sequence \(\{y_{i}\}\). Now, assume by contradiction that there exists a point \(\bar{y}\in\{y_{i}\}\), with \(\bar{y}\in\operatorname*{argmin}_{1\leq i\leq m}\{\varphi_{i}\}\), but \(\bar{y}\) is _not_ a global minimum. This yields the contradictory fact that \(\varphi(y^{*})>\varphi(\bar{y})\).