## 3 Properties of the Set of Normalized Immobile Indices  
In this section, we consider a copositive problem in the form (1) with the corresponding sets \(X\), \(\mathcal{Z}\), \(T_{im}(X)\), and \(T_{im}(\mathcal{Z})\) defined in (5) and (8)-(10), respectively. The propositions below describe the properties of these sets, show the relations between the sets, and allow one to recognize the infeasibility of the problem (1). These results are then used in the next section to construct the algorithms for regularization of the CoP problem without a feasibility assumption. Proposition 1: _If \(T_{im}(\mathcal{Z})=\emptyset\), then \(X\
eq\emptyset\) and \(T_{im}(X)=\emptyset\)._  
Proof: Note that the set \(\mathcal{Z}\) defined in (8) is not empty. Since \(T_{im}(\mathcal{Z})=\emptyset\), there exist a vector \((\check{y},\check{y}_{0})\in\mathcal{Z}\) and a number \(\mu\) such that  
\[t^{\top}\mathcal{B}(\check{y},\check{y}_{0})t\geq\mu>0\ \forall t\in T. \tag{16}\]If \(\tilde{y}_{0}>0\), consider vector \(\bar{x}:=\tilde{y}/\tilde{y}_{0}\). From (16), it follows that \(t^{\top}\mathcal{A}(\bar{x})t\geq\mu/\tilde{y}_{0}>0\) for all \(t\in T\). Hence, \(X\
eq\emptyset\) and \(T_{im}(X)=\emptyset\). Suppose now that \(\tilde{y}_{0}=0\). Then relations (16) take the form  
\[t^{\top}\Big{(}\sum_{j=1}^{n}A_{j}\tilde{y}_{j}\Big{)}t\geq\mu>0\ \forall t\in T.\]  
Set: \(\alpha_{0}:=\min\limits_{t\in T}t^{\top}A_{0}t\) and \(\theta:=\left\{ \begin{array}{ll}1&\text{if }\alpha_{0}\geq 0,\\ -2\alpha_{0}/\mu&\text{if }\alpha_{0}<0,\end{array} \right.\) and consider the vector \(\bar{x}:=\theta\tilde{y}\). Calculate  
\[\min\limits_{t\in T}t^{\top}\mathcal{A}(\bar{x})t\geq\theta\min\limits_{t\in T }t^{\top}\Big{(}\sum_{j=1}^{n}A_{j}\tilde{y}_{j}\Big{)}t+\alpha_{0}\geq\theta \mu+\alpha_{0}=\left\{ \begin{array}{ll}\mu+\alpha_{0}>0,&\text{if }\alpha_{0}\geq 0,\\ -\alpha_{0}>0,&\text{if }\alpha_{0}<0.\end{array} \right.\]  
This implies that \(\bar{x}\in X\
eq\emptyset\) and \(T_{im}(X)=\emptyset\). The proposition is proved. **Proposition 2**. : _If \(X\
eq\emptyset\), then \(T_{im}(X)=T_{im}(\mathcal{Z})\)._  
Proof. : Let us, first, show that  
\[T_{im}(X)\subset T_{im}(\mathcal{Z}). \tag{17}\]  
Suppose the contrary: there exists \(\bar{t}\in T_{im}(X)\) such that \(\bar{t}\
ot\in T_{im}(\mathcal{Z})\). The latter relation implies the following condition:  
\[\exists\ (\tilde{y},\tilde{y}_{0})\in\mathcal{Z}:\ \bar{t}^{\top}\mathcal{B}( \tilde{y},\tilde{y}_{0})\bar{t}>0,\ \tilde{y}_{0}\geq 0.\]  
In the case \(\tilde{y}_{0}>0\), let us set \(\bar{x}:=\tilde{y}/\tilde{y}_{0}\) and in the case \(\tilde{y}_{0}=0\), set \(\bar{x}:=x^{*}+\tilde{y}\), where \(x^{*}\) is an arbitrary vector from \(X\). It is easy to check that by construction, we have \(\bar{x}\in X\) and \(\bar{t}^{\top}\mathcal{A}(\bar{x})\bar{t}>0\), but this contradicts the assumption \(\bar{t}\in T_{im}(X)\). Hence, inclusion (17) is proved. Now let us show that  
\[T_{im}(\mathcal{Z})\subset T_{im}(X). \tag{18}\]  
Suppose the contrary: there exists \(\bar{t}\in T_{im}(\mathcal{Z})\) such that \(\bar{t}\
ot\in T_{im}(X)\). Then for some \(\bar{x}\in X\), it holds that \(\bar{t}^{\top}\mathcal{A}(\bar{x})\bar{t}>0\). Consider vector \(\tilde{z}:=(\tilde{y}=\bar{x},\ \tilde{y}_{0}=1)\). By construction, \(\tilde{z}\in\mathcal{Z}\) and \(\bar{t}^{\top}\mathcal{B}(\tilde{z})\bar{t}>0\), but this contradicts the assumption \(\bar{t}\in T_{im}(\mathcal{Z})\). Hence, inclusion (18) is valid and the proposition is proved. **Proposition 3**. : _Given \(\tau(i)\in T_{im}(\mathcal{Z}),i\in I\), suppose that the set_  
\[\Delta X:=\left\{x\in\mathbb{R}^{n}:\ \mathcal{A}(x)\tau(i)\geq 0\ \forall i\in I\right\} \tag{19}\]  
_is empty. Then the set \(X\) is empty too._  
Proof. : Suppose the contrary: \(\Delta X=\emptyset\) but \(X\
eq\emptyset\). Since \(X\
eq\emptyset\), then by Proposition 2 we have \(\tau(i)\in T_{im}(X)\), \(i\in I\).