## Appendix A  
### Appendix A  
In this subsection we give a proof of Proposition 6. First of all, notice that since \(X\
eq\emptyset\), the set \(\Delta X_{*}\) defined in (34) is not empty and from Proposition 5 it follows that \(T_{im}(X)=T_{im}(\mathcal{Z})\) and relations (35) hold true. Suppose that on the contrary to (40), there exists \(j_{0}\in J\) such that \(\tau^{*}(j_{0})\
ot\in W_{m_{*}}\). Since \(\tau^{*}(j_{0})\in T_{im}(X)\) and by construction relations (35) hold true, we conclude that  
\[\rho(\tau^{*}(j_{0}),\text{conv}\;W_{m_{*}})<\sigma(W_{m_{*}}).\]  
Hence, from Proposition 1 in [18], it follows that there exists \(i_{0}\in I_{m_{*}}\) such that  
\[P_{+}(\tau^{*}(j_{0}))\subset P_{+}(\tau(i_{0}))\text{ and }\tau^{*}(j_{0}) \
eq\tau(i_{0}).\]  
Then for all \(x\in X\), it holds that  
\[(\tau^{*}(j_{0}))^{\top}\mathcal{A}(x)\tau(i_{0})=\sum_{k\in P_{+}(\tau^{*}(j_ {0}))}\tau^{*}_{k}(j_{0})\epsilon^{\top}_{k}\mathcal{A}(x)\tau(i_{0})=0, \tag{10}\]  
where \(\tau^{*}_{k}(j_{0})\) is the \(k\)-th component of the vector \(\tau^{*}(j_{0})\). For a sufficiently small \(\theta>0\), consider a vector \(\tau:=(\tau(i_{0})-\theta\tau^{*}(j_{0}))/(1-\theta)\). It is evident that \(\tau\in T\). Taking into account (10), we conclude that for all \(x\in X\), it holds that  
\[ \begin{split}\tau^{\top}\mathcal{A}(x)\tau=&(1- \theta)^{-2}[(\tau(i_{0}))^{\top}\mathcal{A}(x)\tau(i_{0})\\ &-2\theta(\tau^{*}(j_{0}))^{\top}\mathcal{A}(x)\tau(i_{0})+ \theta^{2}(\tau^{*}(j_{0}))^{\top}\mathcal{A}(x)\tau^{*}(j_{0})]=0,\end{split} \]  
and therefore, \(\tau\in T_{im}(X)\). As a result, we have  
\[\tau^{*}(j_{0})=(1-\theta)\bar{\tau}+\theta\tau(i_{0})\text{ with }\bar{\tau}\in T_{im}(X),\;\tau(i_{0})\in T_{im}(X),\;\bar{\tau}\
eq\tau^{*}_{ i_{0}}\text{ and }\theta\in(0,1),\]  
which contradicts the assumption that \(\tau^{*}(j_{0})\) is a vertex of the set conv \(T_{im}(X)\). The proposition is proved. ### Appendix A  
Consider the algorithm RLCoP-3 and suppose that \(m_{*}>0\). Let us show that for all \(m=1,\dots,m_{*}\), it holds that  
\[P_{+}(\tau(i))\subset L^{m}(i)\;\forall i\in I_{m}. \tag{11}\]  
At the end of Iteration #0 we have data (25) satisfying relations (26) and (27) with \(\eta_{0}=0\). It follows from these relations that  
\[\sum_{i\in I_{1}}\gamma(i)(\tau(i))^{\top}\mathcal{B}(y,y_{0})\tau(i)=0\; \forall(y,y_{0})\in\mathbb{R}^{n+1}. \tag{12}\]  
Consider the set \(Z(1):=\{(y,y_{0})\in\mathbb{R}^{n+1}:\mathcal{B}(y,y_{0})\tau(i)\geq 0,\;i\in I _{1}\}\). It is evident that by construction,  
\[(\tau(i))^{\top}\mathcal{B}(y,y_{0})\tau(i)\geq 0\;\forall i\in I_{1},\; \forall(y,y_{0})\in Z(1).\]  
From the inequalities above and equality (12), it follows that:  
\[(\tau(i))^{\top}\mathcal{B}(y,y_{0})\tau(i)=0\;\forall i\in I_{1},\;\forall(y,y_{0})\in Z(1),\]  
wherefrom we get  
\[e^{\top}_{k}\mathcal{B}(y,y_{0})\tau(i)=0\;\forall k\in P_{+}(\tau(i)),\; \forall i\in I_{1},\;\forall(y,y_{0})\in Z(1).\]This implies that \(P_{+}(\tau(i))\subset L^{1}(i)\ \forall i\in I_{1}\). Suppose that at the beginning of Iteration \(\#m\), \(1\leq m<m_{*}\), we have the sets \(Z(m)\), \(S^{m}(i)\), \(L^{m}(i)\), \(i\in I_{m}\), constructed by the rules (36) and (37), such that  
\[\epsilon_{k}^{\top}\mathcal{B}(y,y_{0})\tau(i)=0\ \forall k\in L^{m}(i),\ \forall(y,y_{0})\in Z(m),\ P_{+}(\tau(i))\subset L^{m}(i),\ \forall i\in I_{m}. \tag{38}\]  
At the end of this iteration, we have the data presented in (38) which satisfy relations (29) and (30) with \(\eta_{m}=0\). These relations imply the equalities  
\[\sum_{i\in\Delta I_{m}}\gamma(i)(\tau(i))^{\top}\mathcal{B}(y,y_{0})\tau(i)+2 \sum_{i\in I_{m}}(\lambda^{m}(i))^{\top}\mathcal{B}(y,y_{0})\tau(i)=0\ \forall(y,y_{0})\in\mathbb{R}^{n+1}. \tag{39}\]  
Set \(I_{m+1}:=I_{m}\cup\Delta I_{m}\), and consider the set  
\[Z(m+1):=\{(y,y_{0})\in\mathbb{R}^{n+1}:\mathcal{B}(y,y_{0})\tau(i)\geq 0\ \forall i\in I_{m+1},y_{0}\geq 0\}.\]  
Notice that by construction, \(Z(m+1)\subset Z(m)\).