Some solutions for generalized games with restricted cooperation

Abstract

We consider generalizations of Transferable Utility games with restricted cooperation in partition function form and propose their interpretation as allocation problems with several public resources. Either all resources are goods or all resources are bads. Each resource is distributed between points of its set and permissible coalitions are subsets of the union these sets. Each permissible coalition estimates each allocation of resources by its gain/loss function, that depends on the restriction of the allocation on that coalition. Moreover, we define objections at an allocation between permissible coalitions and their feasibility is described by a directed graph $\varGamma$, where permissible coalitions are its vertices. We define new solution concepts (positive envy stable solution w.r.t. $\varGamma$ for gain functions and negative envy stable solution w.r.t. $\varGamma$ for loss functions). These solutions are simplifications of the generalized kernel of cooperative games and generalize the equal sacrifice solution for claim problems. An allocation belongs to these solutions if there do not exist objections at this allocation between permissible coalitions. We describe completely conditions on $\varGamma$ that ensure the existence of these envy stable solutions and conditions that ensure the inclusion of the generalized nucleolus, the generalized anti-nucleolus, and the Wardrop equilibria in these envy stable solutions.

Appendix

Proof of Theorem 1

Denote ${\bar{c}}=\sum _{B\in \tau } c(B)$.

Let $\varGamma$, $\tau$ satisfy Condition $C^+(\varGamma , \tau )$ and $( G_S)_{S\in \mathcal A}$ be a collection of continuous strictly increasing in each variable functions.

For each $x\in X({\bar{c}})$ we define a binary relation $\succ _x$ on $\mathcal{A}\cup \tau$ as follows.

$K\succ _x L$ iff either $K,L\in \mathcal A$, (K, L) is an edge of $\varGamma$, $G_K(x)< G_L(x)$ and $x(L)>0$ or $K,L\in \tau$ and $x(L)/c(L)>x(K)/c(K)$.

Then $\succ _x$ is an acyclic binary relation on $\mathcal{A}\cup \tau$ and $x^0\in X({\bar{c}})$ is an equilibrium vector on $\mathcal{A}\cup \tau$ w.r.t. $(\succ _x )_{x\in X({\bar{c}})}$ iff $x^0\in X(\tau ,C)$ and $x^0$ belongs to the positive $( G_S)_{S\in \mathcal A}$ - envy stable solution w.r.t. $\varGamma$. Hence, it is sufficient to check the conditions of Lemma 1 for $(\succ _x )_{x\in X({\bar{c}})}$.

For $L\in \mathcal{A}\cup \tau$, denote

$$\begin{aligned} F^L({\bar{c}})=\{ x\in X({\bar{c}}):\, K\not \succ _x L\quad \text{ for } \text{ all }\quad K\in \mathcal{A}\cup \tau \}. \end{aligned}$$

Then $F^L({\bar{c}})$ is a closed set because the functions $G_S$ are continuous ones, i.e., Condition 1 is fulfilled. Condition 2 is evidently fulfilled.

Let us check Condition 3. For each $x\in X({\bar{c}})$, there exists $B^0\in \tau$ such that $x(B^0)/c(B^0)\le x(B)/c(B)$ for each $B\in \tau$, i.e., $x\in F^{B^0}({\bar{c}})$ and since the relation $\succ _x$ is acyclic, there exists $\mathcal{L} \subset \mathcal{A}$ such that $\mathcal L$ is a vertex base of $\varGamma$ and $x\in F^L({\bar{c}})$ for each $L\in \mathcal{L}$.

If we take out from $\mathcal{A}\cup \tau$ all $L\in \mathcal{L}$ and $B^0$, then the remaning elements of $\mathcal{A}\cup \tau$ do not cover N in view of Condition $C^+(\varGamma , \tau )$. Hence, Condition 3 of Lemma 1 is also fulfilled, so there exists $x^0\in X(\tau ,C)$ that belongs to the positive $( G_S)_{S\in \mathcal A}$ - envy stable solution w.r.t. $\varGamma$.

Now suppose that Condition $C^+(\varGamma , \tau )$ is not fulfilled, i.e., there exist $B^0\in \tau$ and a vertex base $\mathcal L$ of $\varGamma$ such that

$$\begin{aligned} B^0\subseteq \bigcup _{T\in \mathcal{A}\setminus \mathcal{L}} T. \end{aligned}$$

Let $0< \epsilon < c(B^0)/(|B^0| {\bar{c}})$. Take the following collection of functions $( G_S)_{S\in \mathcal A}$.

$G_S(x)=x(S)$ for $S\in \mathcal{L}$,

$G_S(x)=x(S)/{\epsilon }$ otherwise.

Let $y\in X(\tau , C)$ and y belong to the positive $( G_S)_{S\in \mathcal A}$ - envy stable solution w.r.t. $\varGamma$.

For defined collection $( G_S)_{S\in \mathcal A}$, if (Q, P) is an edge of $\varGamma$, then $G_Q(y)\ge G_P(y)$.

If (Q, P) is an edge of $\varGamma$ and $Q\in \mathcal{L}$, then $P\not \in \mathcal{L}$ and ${\bar{c}}\ge y(Q)\ge y(P)/{\epsilon }$. Thus, for all $T\in \mathcal{A}\setminus \mathcal{L}$,

$$\begin{aligned} y(T)\le \epsilon {\bar{c}}<c(B^0)/|B^0|. \end{aligned}$$

Then $y_i<c(B^0)/|B^0|$ at each $i\in T$, hence $y_i<c(B^0)/|B^0|$ at each $i\in B^0$, so $y(B^0)<c(B^0)$ and $y\not \in X(\tau , C)$. $\square$

Proof of Theorem 4

Let $\mathcal A$ be a $(\varGamma ,\tau )$–positive weakly mixed collection of coalitions and x belong to the generalized anti-nucleolus of $X(\tau , C)$ w.r.t. $(G_S)_{S\in \mathcal{A}}$. Suppose that x does not belong to the positive $( G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$, i.e., there exists a directed edge (P, Q) of $\varGamma$ such that $G_Q(x)>G_P(x)$ and $x(Q)>0$. Take $i_0\in Q$ such that $x_{i_0}>0$. Then $i_0\not \in P$ as if $P\cap Q=\emptyset$. Let $B\ni i_0$, $B\in \tau$.

Since $\mathcal A$ is a $(\varGamma ,\tau )$–positive weakly mixed collection of coalitions, there exists $j\in B$ such that $\mathcal{A}_j \supset \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$.

For $\delta >0$, let $y^{\delta }=(y_i)_{i\in N}$, where $y_{i_0}=x_{i_0}-\delta$, $y_j=x_j+\delta$, $y_t=x_t$ otherwise. Take $\delta >0$ such that $\delta < x_{i_0}$ and

$$\begin{aligned} G_Q(y^{\delta })> G_P(y^{\delta } ). \end{aligned}$$

Then $y^{\delta }\in X(\tau , C)$, $G_T(y^{\delta }) < G_T(x)$ only for $T=Q$ because $G_T\in \mathcal{G}^{\tau }_T$ and $G_P(y^{\delta })>G_P(x)$.

Since $G_P(y^{\delta } )< G_Q(y^{\delta } )$, we obtain $\theta (y^{\delta })>_{lex}\theta (x)$ and this contradicts the definition of the generalized anti-nucleolus of $X(\tau , C)$ w.r.t. $(G_S)_{S\in \mathcal{A}}$.

Now let the generalized anti-nucleolus of $X(\tau , C)$ w.r.t. $(G_S)_{S\in \mathcal{A}}$ be always contained in the positive $( G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$.

Suppose that $\mathcal A$ is not a $(\varGamma ,\tau )$–positive weakly mixed collection of coalitions. Then either there exist $i_0\in N$, an edge (P, Q) of $\varGamma$ with $i_0\in P\cap Q$ and $B^0\in \tau$ with $B^0\ni i_0$ or $K\cap L=\emptyset$ for each edge (K, L) of $\varGamma$ and there exist an edge (P, Q) of $\varGamma$, $B^0\in \tau$, $i_0\in B^0$ such that $Q\in \mathcal{A}_{i_0}$ and $\mathcal{A}_j \not \supset \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$ for all $j\in B^0$.

Let $0<\epsilon <1/(|\tau ||N|)$.

We take allocation problem with $c(B)=1/|\tau |$ for each $B\in \tau$ and the following $( G_S)_{S\in \mathcal A}$:

$G_P(x)=x(P)$,

$G_T(x)=x(T)/|N|^2$ for $T\in \mathcal{A}_{i_0}\setminus \{ Q\}\setminus \{ P\}$,

$G_T(x)=x(T)/{\epsilon }$ otherwise.

Let x belong to the generalized anti-nucleolus of $X(\tau , C)$ w.r.t. $(G_S)_{S\in \mathcal{A}}$ and to the positive $(G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$.

Since x belongs to the positive $( G_S)_{S\in \mathcal A}$–envy stable solution and (P, Q) is an edge of $\varGamma$, we have either $x(Q)=0$ or $G_Q(x)\le G_P(x)$, hence $x(Q)\le \epsilon x(P)\le \epsilon .$

There exists $j_0\in B^0$ such that $x_{j_0}\ge 1/(|\tau ||B^0|)\ge 1/(|\tau ||N|)$. Then $j_0\not \in Q$ and $j_0\ne i_0$. Hence, $|B^0|\ge 2$ and $|\tau |< |N|$.

Let $\delta >0$, $y^{\delta }=(y_i)_{i\in N}$, where $y_{i_0}=x_{i_0}+\delta$, $y_{j_0}=x_{j_0}-\delta$, $y_i=x_i$ otherwise.

We can take $\delta$ such that $\delta <1/(|\tau ||N|)$ and for each $K,L\in \mathcal A$,

$$\begin{aligned} G_K(x)<G_L(x) \quad \text{ implies }\quad G_K(y^{\delta }) < G_L(y^{\delta }). \end{aligned}$$

Then $y^{\delta }\in X(\tau , C)$, $y^{\delta }(T) > x(T)$ for each $T\in \mathcal{A}_{i_0} \setminus \mathcal{A}_{j_0}$. In particular, $y^{\delta }(Q) > x(Q)$, i.e., $G_Q(y^{\delta }) > G_Q(x)$.

It remains to prove that $\theta (y^{\delta })>_{lex} \theta (x)$, i.e., x does not belong to the generalized anti-nucleolus of $X(\tau , C)$ w.r.t. $(G_S)_{S\in \mathcal{A}}$. In order to prove that, we shall prove the following.

There exists $M\in \mathcal A$ such that $G_M(y^{\delta }) > G_M(x)$ and $G_T(y^{\delta }) < G_T(x)$ implies $G_M(y^{\delta }) < G_T(y^{\delta })$. By the choice of $\delta$, it is sufficient to check that $G_M(x) < G_T(x)$ in this case.

Consider 2 cases.

Case 1. Either $i_0\in P\cap Q$ or $K\cap L=\emptyset$ for each edge (K, L) of $\varGamma$ and $j_0\not \in P$. Then $G_P(y^{\delta })\ge G_P(x)$. Let $G_T(y^{\delta }) < G_T(x)$, then $T\ni j_0$ and $G_T(x)=x(T)/{\epsilon }$, hence $G_T(x)\ge x_{j_0}/{\epsilon }>1$. Since $G_Q(x)=x(Q)/{\epsilon } <1$ and $y^{\delta }(Q)>x(Q)$, we can take $M=Q$.

Case 2. $j_0\in P$, $K\cap L=\emptyset$ for each edge (K, L) of $\varGamma$, and there exists $M\in \mathcal{A}_{i_0}\setminus \mathcal{A}_{j_0} \setminus \{ Q\}$. Then $G_M(x)=x(M)/(|N|^2)\le 1/(|N|^2)$ and $G_M(y^{\delta })>G_M(x)$.

If $y^{\delta }(T)<x(T)$ then either $T=P$ and $G_P(x)=x(P)\ge 1/(|\tau ||N|)> 1/(|N|^2)$ or $G_T(x)=x(T)/{\epsilon }$ and $G_T(x)\ge x_{j_0}/{\epsilon }>1$. Thus, $y^{\delta }(T)<x(T)$ implies $G_T(x) > G_M(x)$. $\square$

Proof of Theorem 5

Let $\mathcal A$ be a $(\varGamma ,\tau )$–negative weakly mixed collection of coalitions and x belong to the generalized nucleolus of $X(\tau , C)$ w.r.t. $(G_S)_{S\in \mathcal{A}}$. Suppose that x does not belong to the negative $(G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$, i.e., there exists a directed edge (P, Q) of $\varGamma$ such that $G_P(x)>G_Q(x)$ and $x(P)>0$. Take $i_0\in P$ such that $x_{i_0}>0$. Let $i_0\in B^0\in \tau$.

Since $\mathcal A$ is a $(\varGamma ,\tau )$–negative weakly mixed collection of coalitions, there exists $j\in B^0$ such that $\mathcal{A}_j \subset \mathcal{A}_{i_0}\cup \{ Q\}\setminus \{ P\}$. Then $j\not \in P$.

For $\delta >0$, let $y^{\delta }=\{ y_i\}_{i\in N}$, where $y_{i_0}=x_{i_0}-\delta$, $y_j=x_j+\delta$, $y_t=x_t$ otherwise. Take $\delta >0$ such that $\delta < x_{i_0}$ and $G_Q(y^{\delta } )< G_P(y^{\delta })$.

Then $y^{\delta }\in X(\tau , C)$, $G_P(y^{\delta })< G_P(x)$ because $j\not \in P$ and $G_T(y^{\delta }) >G_T(x)$ implies $T=Q$ because $G_S\in \mathcal{G}^{\tau }_S$. Since $G_Q(y^{\delta })<G_P(y^{\delta })$, we obtain $\bar{\theta }(y^{\delta })<_{lex}\bar{\theta }(x)$ and this contradicts the definition of the generalized nucleolus of $X(\tau , C)$ with respect to $(G_S)_{S\in \mathcal{A}}$.

Now let the generalized nucleolus of $X(\tau , C)$ with respect to $(G_S)_{S\in \mathcal{A}}$ be always contained in the negative $(G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$.

Suppose that $\mathcal A$ is not a $(\varGamma ,\tau )$–negative weakly mixed collection of coalitions. Then there exist $B^0\in \tau$, $i_0\in B^0$, $Q\in \mathcal{A}_{i_0}$, and a directed edge (Q, P) of $\varGamma$ such that $\mathcal{A}_j \not \subset \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$ for all $j\in B^0$.

Let $0<\epsilon <1/(|\tau ||N|)$, $\lambda >1$.

We take the allocation problem with $c(B)=1/|\tau |$ at each $B\in \tau$ and the following $(G_S)_{S\in \mathcal A}$:

$G_P(x)=x(P)$,

$G_T(x)=x(T)/\lambda$ for $T\in \mathcal{A}_{i_0}\setminus \{ Q\}\setminus \{ P\}$,

$G_T(x)=x(T)/{\epsilon }$ otherwise.

Let x belong to the generalized nucleolus of $X(\tau , C)$ with respect to $(G_S)_{S\in \mathcal{A}}$ and to the negative $(G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$.

If $x(Q)>\epsilon$ then the case $G_Q(x)>G_P(x)$ is impossible because x belongs to the negative $(G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$, then $x(Q)/{\epsilon } \le x(P)\le 1$, hence $x(Q)\le \epsilon$.

There exists $j_0\in B^0$ such that $x_{j_0}\ge 1/(|\tau ||B|)\ge 1/(|\tau ||N|) > \epsilon$. Then $j_0\not \in Q$ and $j_0\ne i_0$.

Let $\delta >0$, $y^{\delta }=(y_i)_{i\in N}$, where $y_{i_0}=x_{i_0}+\delta$, $y_{j_0}=x_{j_0}-\delta$, $y_i=x_i$ otherwise. We can take $\delta$ such that $\delta <1/(|\tau ||N|)$ and for each $K,L\in \mathcal A$,

$$\begin{aligned} G_K(x)<G_L(x) \quad \text{ implies }\quad G_K(y^{\delta }) < G_L(y^{\delta }). \end{aligned}$$

Then $y^{\delta }\in X(\tau ,C)$.

Since $\mathcal{A}_j \not \subset \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$ for all $j\in B^0$, there exists $M\in \mathcal A$ such that $j_0\in M$ and $M\not \in \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$. Then $M\ne Q$ as if $j_0\not \in Q$, hence $i_0\not \in M$. This implies $y^{\delta }(M)<x(M)$.

Moreover, $M\ne P$ implies $G_M(x)=x(M)/{\epsilon }$, hence

$$\begin{aligned} G_M(x)\ge x_{j_0}/{\epsilon }>1. \end{aligned}$$

Let $y^{\delta }(T)>x(T)$ then $i_0\in T$, $j_0\not \in T$. We have either $T=Q$ and $G_T(x)=x(Q)/{\epsilon } \le 1$ or $G_T(x)=x(T)/\lambda <1$, thus,

$$\begin{aligned} G_M(x)>G_T(x). \end{aligned}$$

It follows from the choice of $\delta$ that $G_M(y^{\delta })>G_T(y^{\delta })$ for each T with $y^{\delta }(T)>x(T)$.

Thus, $\bar{\theta }(y^{\delta })<_{lex}\bar{\theta }(x)$ and this contradicts the definition of the generalized nucleolus of $X(\tau , C)$ with respect to $(G_S)_{S\in \mathcal{A}}$. $\square$

Proof of Theorem 6

Let $\mathcal A$ be a $(\varGamma ,\tau )$– negative mixed collection of coalitions. Let x be a Wardrop equilibria with respect to $(g_S)_{S\in \mathcal A}$ and $G_S(x)=g_S(x(S))$ for all $S\in \mathcal{A}$.

Suppose that x does not belong to the negative $( G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$, i.e., there exists a directed edge of $\varGamma$ (Q, P) such that $x(Q)>0$, and $G_Q(x))>G_P(x)$. Then $g_Q(x(Q))>g_P(x(P))$. Take $i_0\in Q$ with $x_{i_0}>0$. Let $i_0\in B\in \tau$. Since $\mathcal A$ is a $(\varGamma ,\tau )$–mixed collection (negative), there exists $j_0\in B$ such that $\mathcal{A}_{j_0}=\mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$. Thus,

$\{ T\in \mathcal{A}:\, T\not \ni j_0, T\ni i_0\} =\{ Q\}$,

$\{ T\in \mathcal{A}:\, T\ni j_0, T\not \ni i_0\} =\{ P\}$, hence

$$\begin{aligned} \sum _{T\in \mathcal{A}:\, T\ni i_0} g_T(x(T)) - \sum _{T\in \mathcal{A}:\, T\ni j_0} g_T(x(T)) = g_Q(x(Q)) - g_P(x(P)) >0. \end{aligned}$$

This implies that x is not a Wardrop equilibria because $x_{i_0}>0$ and we get a contradiction. Thus, x belongs to the negative $( G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$.

Let the Wardrop equilibria with respect to $(g_S)_{S\in \mathcal A}$ be always contained in the $( G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$. Suppose that $\mathcal A$ is not a $(\varGamma ,\tau )$ – negative mixed collection of coalitions. Then there exist $B^0\in \tau$, $i_0\in B^0$, $Q\in \mathcal{A}_{i_0}$, and a directed edge (Q, P) of of $\varGamma$ such that for each $j\in B^0$, $\mathcal{A}_j\ne \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$.

Let $0<\epsilon <1/(|\tau ||N|)$, $M>1$. We take the following problem. $c(B)=1/|\tau |$ for each $B\in \tau$,

$g_T(t)=t-1$ for $T=P$,

$g_T(t)=t/M -1$ for $T\in \mathcal{A}_{i_0}\setminus \{ Q\}$,

$g_T(t)=t/{\epsilon } -1$ otherwise.

Let x be a Wardrop equilibria with respect to $(g_S)_{S\in \mathcal A}$ and x belong to the negative $( G_S)_{S\in \mathcal A}$–envy stable solution w.r.t. $\varGamma$.

First, we prove that $x(Q)\le \epsilon$. Suppose that $x(Q) > \epsilon$, then $g_Q(x(Q))\le g_P(x(P))$, i.e., $x(Q)\le \epsilon x(P)\le \epsilon$.

There exists $j_0\in B^0$ such that $x_{j_0}\ge 1/(|\tau ||N|)$. Then $j_0\not \in Q$ as if $x_{j_0}>\epsilon$.

Note that $g_Q(x(Q))\le g_P(x(P))$. Indeed, if $x(Q)=0$ then $g_Q(x(Q))=-1 \le g_P(x(P))$ and if $x(Q)>0$ then it follows from the definition of envy stable solution.

We shall prove that

$$\begin{aligned} \sum _{T\in \mathcal{A}:T\ni i_0} g_T(x(T)) < \sum _{T\in \mathcal{A}:T\ni j_0}g_T(x(T)), \end{aligned}$$

(1)

and this will imply that x is not a Wardrop equilibria with respect to $(g_S)_{S\in \mathcal A}$ because $x_{j_0}>0$.

Since $\mathcal{A}_{j_0}\ne \mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}$, and $j_0\not \in Q$, the following 3 cases are possible.

• 1. $j_0\not \in P$.

• 2. $j_0\in P$, $\mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}\not \subset \mathcal{A}_{j_0}$.

• 3. $j_0\in P$, $\mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}\subset \mathcal{A}_{j_0}$, $\mathcal{A}_{i_0}\cup \{ P\}\setminus \{ Q\}\not \supset \mathcal{A}_{j_0}$.

Case 1. If $T\in \mathcal{A}_{i_0}\setminus \{ Q\}$ then $g_T(x(T))<0$ because $x(T)\le 1$, $M>1$. Moreover, $x(Q)<\epsilon$ in this case because $x(P)\le 1-x_{j_0}<1$ and $x(Q)\ge \epsilon$ implies $x(Q)\le \epsilon x(P)<\epsilon$. Thus, $g_Q(x(Q))<0$ and

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\not \ni j_0, T\ni i_0}g_T(x(T))\le g_Q(x(Q))<0. \end{aligned}$$

(2)

For all $T\in \mathcal{A}_{j_0}\setminus \mathcal{A}_{i_0}$, $x(T)>\epsilon$ since $x_{j_0}>\epsilon$ and $g_T(x(T))>0$ as if $j_0\not \in P$, therefore,

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\ni j_0, T\not \ni i_0}g_T((x(T))\ge 0, \end{aligned}$$

this and (2) imply (1).

Case 2. Since $\mathcal{A}_{i_0}\setminus \mathcal{A}_{j_0}\setminus \{ Q\}\ne \emptyset$ and $g_T(x(T))<0$ for all $T\in \mathcal{A}_{i_0}\setminus \{ Q\}\setminus \{ P\}$ we obtain

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\not \ni j_0, T\ni i_0}g_T(x(T))= g_Q(x(Q)) + \sum \limits _{T\in \mathcal{A}_{i_0}\setminus \mathcal{A}_{j_0}\setminus \{ Q\}} g_T(x(T))< g_Q(x(Q)). \end{aligned}$$

(3)

If $T\in \mathcal{A}_{j_0}\setminus \mathcal{A}_{i_0}$ then either $T=P$ or $x(T)>\epsilon$ and $g_T(x(T))>0$.

If $i_0\not \in P$, then

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\ni j_0, T\not \ni i_0}g_T(x(T))\ge g_P(x(P)), \end{aligned}$$

and we obtain (1) due to (3) and $g_Q(x(Q))\le g_P(x(P))$.

If $i_0\in P$ then

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\ni j_0, T\not \ni i_0}g_T(x(T))\ge 0, \end{aligned}$$

and we obtain (1) due to (3) and $g_Q(x(Q))\le 0$ because $x(Q)\le \epsilon$.

Case 3. Here $\{ T\in \mathcal{A}:\, T\not \ni j_0, T\ni i_0\} =\{ Q\}$, and there exists $T_0\in \mathcal{A}_{j_0}\setminus (\mathcal{A}_{i_0}\cup \{ P\})$. Thus,

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\not \ni j_0, T\ni i_0}g_T(x(T))=g_Q(x(Q)). \end{aligned}$$

If $T\in \mathcal{A}_{j_0}\setminus (\mathcal{A}_{i_0}\cup \{ P\})$ then

$$\begin{aligned} g_T(x(T))=x(T)/\epsilon -1\ge x_{j_0}/\epsilon -1> 0, \end{aligned}$$

so

$$\begin{aligned} \sum \limits _{T\in \mathcal{A}:\, T\ni j_0, T\not \ni i_0}g_T(x(T))\ge g_P(x(P))+g_{T_0}(x(T_0))> g_P(x(P)). \end{aligned}$$

Since $g_Q(x(Q))\le g_P(x(P))$, we obtain (1). $\square$