## Appendix A
### _Appendix A.3_  
The aim of this subsection is to prove Lemma 2. The proof is based on Proposition 6 and the following one.  
**Proposition A1**.: _Suppose that \(X\
eq\emptyset\). Let \(W_{*}:=\{\tau^{*}(j),j\in J\}\) be the set of all vertices of the set \(\operatorname{conv}T_{im}(X)\). Consider the sets_  
\[Z_{*}:=\{x\in\mathbb{R}^{n}:\mathcal{A}(x)\tau^{*}(j)\geq 0,\ j\in J\},\] \[\overline{M}(j):=\{k\in P:e_{k}^{\top}\mathcal{A}(x)\tau^{*}(j)=0 \ \forall x\in Z_{*}\},\ j\in J.\]  
_Then the set_  
\[F_{*}:=\{D\in\mathcal{COP}^{p}:e_{k}^{\top}D\tau^{*}(j)=0,\ k\in\overline{M} (j),\ j\in J\} \tag{41}\]  
_is the minimal face containing the set \(\mathcal{D}\) defined in (6)._  
Proof.: In our previous works, it has been shown (see Theorem 4 in [18]) that the minimal face \(\mathbf{F_{min}}\) containing the set \(\mathcal{D}\) can be presented as follows:  
\[\mathbf{F_{min}}=\{D\in\mathcal{COP}^{p}:e_{k}^{\top}D\tau^{*}(j)=0,\ k\in M(j ),\ j\in J\}, \tag{42}\]  
where \(M(j)=\{k\in P:e_{k}^{\top}\mathcal{A}(x)\tau^{*}(j)=0\ \forall x\in X\},\ j\in J\).  
Since \(X\subset Z_{*}\), \(\overline{M}(j)\subset M(j)\) for all \(j\in J\). Let us prove that  
\[\overline{M}(j)=M(j)\ \forall j\in J.\] (A9)  
Suppose the contrary: there exist \(j_{0}\in J\) and \(k_{0}\in M(j_{0})\) such that \(k_{0}\
ot\in\overline{M}(j_{0})\). In this case, there exists \(\tilde{x}\in Z_{*}\) such that  
\[\mathcal{A}(\tilde{x})\tau^{*}(j)\geq 0\ \forall j\in J,\ e_{k_{0}}^{\top} \mathcal{A}(\tilde{x})\tau^{*}(j_{0})>0.\] (A10)  
It was shown in [18] (see Theorem 2 there) that there exists \(\tilde{x}\in X\) such that  
\[t^{\top}\mathcal{A}(\tilde{x})t>0\ \forall t\in\Omega(W_{*}),\ \mathcal{A}( \tilde{x})\tau^{*}(j)\geq 0\ \forall j\in J.\] (A11)  
For a sufficiently small \(\theta>0\), let us consider a vector \(x(\theta):=(1-\theta)\tilde{x}+\theta\tilde{x}\). From (A10) and (A11) we get  
\[t^{\top}\mathcal{A}(x(\theta))t\geq 0\ \forall t\in\Omega(W_{*}),\ \mathcal{A}(x( \theta))\tau^{*}(j)\geq 0\ \forall j\in J,\] (A12)  
\[e_{k_{0}}^{\top}\mathcal{A}(x(\theta))\tau^{*}(j_{0})>0.\] (A13)  
From (A12) and Theorem 1 in [18], it follows that \(x(\theta)\in X\). Then from (A13), taking into account the definition of the set \(M(j_{0})\), we conclude that \(k_{0}\
ot\in M(j_{0})\). However, this contradicts the assumption that \(k_{0}\in M(j_{0})\). The resulting contradiction proves that equalities (A9) hold true.  
Now we are ready to prove Lemma 2. In fact, under the assumption \(X\
eq\emptyset\), from Propositions 2 and 6 it follows that  
\[T_{im}(X)=T_{im}(\mathcal{Z})\ \text{and}\ \{\tau^{*}(j),\ j\in J\}\subset\{ \tau(i),\ i\in I_{m_{*}}\}\subset T_{im}(\mathcal{Z}).\]  
Hence, \(\{\tau(i),\ i\in I_{m_{*}}\}\subset\text{conv}\{\tau^{*}(j),\ j\in J\}\). Using this inclusion, it is easy to show that \(Z_{*}=Z(m_{*})\). From the latest equality, the inclusion \(\{\tau^{*}(j),\ j\in J\}\subset\{\tau(i),\ i\in I_{m_{*}}\}\), and the definitions of the sets \(\mathcal{F}_{m_{*}}^{3}\) and \(F_{*}\) (see (39) and (A7)), it follows that \(\mathcal{F}_{m_{*}}^{3}\subset F_{*}\) and both faces, \(F_{*}\) and \(\mathcal{F}_{m_{*}}^{3}\), contain the set \(\mathcal{D}\).  
On the other hand, Proposition A1 states that \(F_{*}\) is the minimal face containing the set \(\mathcal{D}\). This implies that \(\mathcal{F}_{m_{*}}^{3}=F_{*}=\mathbf{F_{min}}\). Lemma 2 is proved.