Contracts and domination in incomplete markets: what is a true core?

Abstract

The goal of the paper is to propose and study a concept of domination by coalitions for incomplete markets. Previously elaborated in the context of contractual approach, now it is presented in more or less standard terms and style. This concept is described as a set of allocations implemented by the net trades (webs of contracts) that are characterized by a special kind of stability in future markets: (i) for every state of the world inducted allocation has to be Pareto optimal and individually rational and (ii) there is no coalition which is able to dominate the allocation via financially feasible trades in future spot markets using real assets and relative to prices specified by (i) (partial Pareto prices). This core converts into a classical one when the market turns complete. Under perfect competition conditions core allocations are GEI-equilibria. These properties prove the validity of suggested core.

Appendix: Proofs

Proof of Lemma 2

Let us write in matrix form the conditions, which define the allocations from $H=H(p^\mathbf{1})$. In fact for $x\in X$ being $x\in H\cap \mathcal{A}({X})$ $\iff$ there are $z_i\in \mathbb {R}^k,\;i\in \mathcal{I}$ such that

$$\begin{aligned} \left\{ \begin{array}{l} \sum \limits _{i=1}^n x_i=\sum \limits _{i=1}^n \mathbf{e}_i;\\ \sum \limits _{i=1}^n z_i=0;\\ p_\sigma x_i^\sigma -p_\sigma A_\sigma z_i=p_\sigma \mathbf{e}_i^\sigma ,\; \sigma =1,\ldots ,s,\; i\in \mathcal{I}\end{array}\right. \end{aligned}$$

holds. Notice that if balance relations and budget constraints $p_\sigma x_i^\sigma -p_\sigma A_\sigma z_i=p_\sigma \mathbf{e}_i^\sigma$ are satisfied for some fixed $\sigma \ge 1$ and all $i\in \mathcal{I}\setminus \{ i_0\}$, then the last budget constraint is also true automatically. This is why all agent $i_0$’s budget constraints, being linear dependent, may be removed from the system of linear equations defining the space H. One may think without loss of generality that $i_0=n$. Denote by $\mathrm{B}$ the following $(l+sl+k+n-1)\times n(l+sl+k)$ matrix

$$\begin{aligned} {\normalsize B=}{\left( \begin{array}{ccccclcccccccccccc} E_l &{} 0 &{}\ldots &{} 0 &{}.&{} E_l&{} 0 &{}\ldots &{} 0 &{}E_l&{} 0 &{}\ldots &{} 0 &{} 0 &{}\ldots &{} 0 &{}0 \\ 0 &{}E_l &{}.&{} 0 &{}\cdots &{} 0 &{} E_l &{}.&{} 0 &{} 0 &{}E_l&{}.&{} 0 &{} 0 &{}\ldots &{} 0 &{}0 \\ \vdots &{} . &{}.&{} . &{} . &{}\vdots &{} . &{}.&{} &{} . &{} . &{}.&{} . &{}\vdots &{}\vdots &{}\vdots &{}\vdots \\ 0 &{} 0 &{}.&{}E_l&{} . &{} 0 &{} 0 &{}.&{} E_l &{} 0 &{} 0 &{}.&{} E_l &{} 0 &{}\ldots &{} 0 &{} 0 \\ 0 &{} 0 &{}\ldots &{} 0 &{}\ldots &{} 0 &{} 0 &{}\ldots &{} 0 &{} 0 &{} 0&{}\ldots &{} 0 &{} E_k &{}\ldots &{} E_k &{} E_k \\ 0 &{}p_1 &{}.&{} 0 &{}\ldots &{} 0 &{} 0&{}\ldots &{} 0 &{} 0 &{} 0&{}\ldots &{} 0 &{}-p_1A_1&{} . &{} 0 &{} 0 \\ \vdots &{}. &{}.&{} . &{} . &{}\vdots &{}\vdots &{}.&{}\vdots &{}\vdots &{}\vdots &{}.&{}\vdots &{}\vdots &{} . &{}\vdots &{}\vdots \\ 0 &{} 0 &{}.&{}p_s&{} . &{} 0 &{} 0 &{}\ldots &{} 0 &{} 0&{} 0&{}\ldots &{} 0 &{}-p_sA_s&{} . &{}0 &{}0 \\ \vdots &{}\vdots &{}.&{} . &{} . &{}\vdots &{} . &{}.&{}\vdots &{}\vdots &{}\vdots &{}.&{}\vdots &{} . &{} . &{} . &{}\vdots \\ 0 &{} 0 &{}\ldots &{} 0 &{}\ldots &{} 0 &{} p_1 &{}.&{} 0 &{} 0 &{} 0 &{}\ldots &{} 0 &{} 0 &{} . &{}-p_1A_1&{} 0 \\ \vdots &{}\vdots &{}.&{}\vdots &{} . &{}\vdots &{} . &{}.&{} .&{}\vdots &{}\vdots &{}.&{}\vdots &{}\vdots &{} . &{}\vdots &{}\vdots \\ 0 &{} 0 &{}\ldots &{} 0 &{}\ldots &{} 0 &{} 0 &{}.&{}p_s&{} 0 &{} 0 &{}\ldots &{} 0 &{} 0 &{} . &{}-p_sA_s&{} 0 \end{array}\right) . } \end{aligned}$$

Here in the standard manner $E_l$ and $E_k$ denote the unit matrices of an appropriate size and $p_1$, $p_s$, $p_1A_1$ and $p_sA_s$ are row-vectors. Clearly for $x\in X$ we have the following equivalence: $x\in H$, $\sum _\mathcal{I}x_i=\sum _\mathcal{I}\mathbf{e}_i$ $\Longleftrightarrow$ there exists $z=(z_1,\ldots ,z_n)$ such that

$$\begin{aligned} \mathrm{B}\quad *\quad \left( \begin{array}{c} x_1^0\\ \vdots \\ x_1^s \\ \vdots \\ x_n^0 \\ \vdots \\ x_n^s\\ z_1 \\ \vdots \\ z_n \end{array}\right) \;=\; \left( \begin{array}{c} \bar{\mathbf{e}}^0 \\ \vdots \\ \bar{\mathbf{e}}^s \\ \mathbf{0}_k \\ p_1\mathbf{e}_1^1 \\ \vdots \\ p_s\mathbf{e}_1^s \\ \vdots \\ p_1\mathbf{e}_{n-1}^1 \\ \vdots \\ p_s\mathbf{e}_{n-1}^s \end{array}\right) . \end{aligned}$$

Consider the subspace

$$\begin{aligned} H^Z=\{((x_1,z_1),.,(x_n,z_n))\in (\mathbb {R}^{l(s+1)}\times \mathbb {R}^k)^\mathcal{I}\mid {B}(x_1,\ldots ,x_n,z_1,\ldots ,z_n)^t=0\}; \end{aligned}$$

this is the kernel of operator $\mathrm{B}(\cdot )$, in which the order of components is changed for the convenience of the below considerations.

Now let an allocation $\bar{x}\in H$ and be Pareto-optimal relative to H. Therefore

$$\begin{aligned} \prod \limits _{\mathcal{I}} { \mathcal{P}}_i(\bar{x}_i)\cap H(p^\mathbf{1})=\emptyset \iff \prod \limits _{\mathcal{I}} [({\mathcal{P}}_i(\bar{x}_i)-\bar{x}_i)\times \mathbb {R}^k]\cap H^Z=\emptyset \end{aligned}$$

takes place (note that via $\mathbf (S)$ each of these sets is nonempty). Now by the separation theorem we may find a linear functional $f=(f_1,\ldots ,f_n)\ne 0$, $f_i=(f_i^x,f_i^z)$, $f_i^x\in \mathbb {R}^{l(s+1)}$ and $f_i^z\in \mathbb {R}^{k}$, $i\in \mathcal{I}$ such that

$$\begin{aligned} \left\langle f,\prod \limits _{ \mathcal{I}} [({ \mathcal{P}}_i(\bar{x}_i)-\bar{x}_i)\times \mathbb {R}^k] \right\rangle \ge \langle f, H^Z\rangle \end{aligned}$$

holds. Notice the functional f is constant (and hence is equal to zero) onto subspace $H^Z$, since the right-hand side of the last inequality is bounded. Therefore

$$\begin{aligned} \left\langle f,\prod \limits _{ \mathcal{I}} [({ \mathcal{P}}_i(\bar{x}_i)-\bar{x}_i)\times \mathbb {R}^k] \right\rangle \ge 0 \end{aligned}$$

(19)

is true. Let us show further that $f_i^z=0,\;i\in {\mathcal{I}}$, i.e.,

$$\begin{aligned} f_i=(f_i^0,\ldots ,f_i^s,\underbrace{0,\ldots ,0}_k), \quad {\forall }i\in {\mathcal{I}}. \end{aligned}$$

In fact, consider fixed $\hat{x}=(\hat{x}_1,\ldots ,\hat{x}_n)$, $\hat{x}_i\in ({\mathcal{P}}_i(\bar{x}_i)-\bar{x}_i)$, $i\in {\mathcal{I}}$, $y_j\in \mathbb {R}^k$, $j\ne i_0$ for some $i_0\in \mathcal{I}$. In view of (19), for any $y\in \mathbb {R}^k$ we have

$$\begin{aligned} \sum \limits _{j\ne i_0}\langle f_j,(\hat{x}_j,y_j)\rangle + \langle f_{i_0},(\hat{x}_{i_0},y)\rangle \ge 0, \end{aligned}$$

which is possible only if $f_{i_0}^z=0$ and in view of the arbitrariness of $i_0$, for all $i_0 \in { \mathcal{I}}$.

For the convenience of the below notations I will identify the functional $f_i$ with $f_i^x$, i.e., by convention let us put $f_i=(f_i^x,0)=(f_i,0)$. Let us show now that

$$\begin{aligned} \langle f_i,({\mathcal{P}}_i(\bar{x}_i)-\bar{x}_i)\rangle \ge 0,\quad i\in {\mathcal{I}}, \end{aligned}$$

(20)

and moreover, if $f_i\not = 0$ and $y_i\in {\mathcal{P}}_i(\bar{x}_i)\cap \mathop {\text {int}}X_i$, then the inequality is strict: $\langle f_i,(y_i-\bar{x}_i)\rangle >0$. Indeed due to the preferences are locally non-satiated, we have $\bar{x}_j\in \mathop {\text {cl}}\nolimits ({{ \mathcal{P}}_j(\bar{x}_j)})$; now the substitution of $\bar{x}_j$ instead of ${{ \mathcal{P}}_j(\bar{x}_j)}$ in (19) for all $j\not =i$ and due to $f_j^z=0,\;j\in {\mathcal{I}}$ immediately gives us the result. So, (20) is true in the non-strict form of inequalities. Now assumption $y_i\in {\mathcal{P}}_i(\bar{x}_i)\cap \mathop {\text {int}}X_i$ and the fact that each ${ \mathcal{P}}_i(\bar{x})$ is an open in $X_i$ standardly implies the strict inequalities for $f_i\not =0$.

Further, the fact that functional $f=(f_1,\ldots , f_n)$ is constant onto subspace $H^Z$ implies that this functional presenting vector can be represented as a linear combination of the vector-rows of matrix $\mathrm B$. Now using the structure of matrix $\mathrm B$, one can conclude the existence of such real $\lambda _i^\sigma$, $\sigma \ge 1$, $i\in \mathcal{I},~ i\not = n$ and such vectors $q\in { \mathbb {R}}^k,\; \bar{p}=(\bar{p}_0,\ldots ,\bar{p}_s)\in E'$, that for all $i\not =n$ the following system of linear equations is true:

$$\begin{aligned} \left\{ \begin{array}{l} f_i^0=\bar{p}_0,\\ f_i^\sigma =\bar{p}_\sigma +\lambda _i^\sigma p_\sigma , \; ~~~~~\sigma \ge 1,\\ f_i^z=0=-q+\sum \limits _{\sigma =1}^s\lambda _i^\sigma p_\sigma A_\sigma , \end{array}\right. \end{aligned}$$

(21)

and for $i=n$ we have $f_n^z=0=-q+0$ and $f_n^x=\bar{p}$. Putting $\lambda _n^\sigma =0$ for all $\sigma \ge 1$, one may think (21) is true for all $i\in \mathcal{I}$. Moreover, system (21) implies that $q= \sum \nolimits _{\sigma =1}^s\lambda _i^\sigma p_\sigma A_\sigma =0$ for all i. Note also that assumption $\bar{p}_0=0$ contradicts $\mathbf (S)$ (we have the local non-satiation in each spot market⁵). Therefore $f_i\not = 0$ and we have strict inequality in (20) for $\mathop {\text {int}}\mathcal{P}_i(\bar{x})$ and all i. Moreover, as soon as $f_n=\bar{p}\not =0$, due to the similar arguments (from the local-nonsatiation) we conclude that $\bar{p}_\sigma \not =0$ for all $\sigma$. It is also clear that due to (20) and $\mathbf (S)$, for $x_i\in \mathop {\text {int}}X_i$ we have $f_i^\sigma \not =0$ for all i and $\sigma$.

Now let us show that $\bar{p}$ satisfies the other requirements of Lemma 2. First note that by subspaces $\mathcal{H}_i$ specification for every $x_i\in \mathcal{H}_i$ we have

$$\begin{aligned} p_\sigma (x_i^\sigma -\mathbf{e}_i^\sigma )=p_\sigma A_\sigma z_i, \;\sigma =1,\ldots ,s \end{aligned}$$

for some $z_i\in \mathbb {R}^k$. Now multiplying equalities by $\lambda _i^\sigma$ and then summing them by $\sigma =1,\ldots ,s$, one obtains

$$\begin{aligned} \sum \limits _{\sigma =1}^s \lambda _i^\sigma p_\sigma (x_i^\sigma - \mathbf{e}_i^\sigma )= \sum \limits _{\sigma =1}^s \lambda _i^\sigma p_\sigma A_\sigma z_i= \left( \sum \limits _{\sigma =1}^s\lambda _i^\sigma p_\sigma A_\sigma \right) z_i=qz_i. \end{aligned}$$

(22)

Next let us recall that due to (20) and the above considerations we also have

$$\begin{aligned} \langle f_i,({ \mathcal{P}}_i(\bar{x}_i)-\bar{x}_i)\rangle> 0 ~\Rightarrow ~ \langle f_i,(y_i-\bar{x}_i)\rangle > 0 \quad \forall y_i\in {\mathcal{P}}_i(\bar{x})\cap \mathcal{H}_i\ne \emptyset \end{aligned}$$

(23)

for all $i\in \mathcal{I}$. Now substituting the representation of $f_i$ from (21), we obtain

$$\begin{aligned} \bar{p}_0(y_i^0-\bar{x}_i^0)+ \sum \limits _{\sigma =1}^s\langle (\bar{p}_\sigma + \lambda _i^\sigma p_\sigma ), (y_i^\sigma -\bar{x}_i^\sigma )\rangle >0~~ \forall y_i\in {\mathcal{P}}_i(\bar{x})\cap \mathcal{H}_i. \end{aligned}$$

Subtracting from the left and right-hand sides of the inequality the value $\sum \nolimits _{\sigma =1}^s\lambda _i^\sigma (p_\sigma \mathbf{e}_i^\sigma )$, after transformations we obtain

$$\begin{aligned} \bar{p}_0y_i^0+\!\sum \limits _{\sigma =1}^s\bar{p}_\sigma y_i^\sigma +\!\sum \limits _{\sigma =1}^s\lambda _i^\sigma p_\sigma (y_i^\sigma - \mathbf{e}_i^\sigma )> \bar{p}_0\bar{x}_i^0+\!\sum \limits _{\sigma =1}^s\bar{p}_\sigma \bar{x}_i^\sigma +\!\sum \limits _{\sigma =1}^s\lambda _i^\sigma p_\sigma (\bar{x}_i^\sigma - \mathbf{e}_i^\sigma ). \end{aligned}$$

Since $\bar{x}_i,y_i\in \mathcal{H}_i$, there are $\bar{z}_i=\bar{z}_i(\bar{x}_i)$, $z_i={z}_i(y_i)$ such that relations (22) are true. Now due to the previous inequality we get

$$\begin{aligned} \langle \bar{p},y_i\rangle +q z_i>\langle \bar{p},\bar{x}_i\rangle +q\bar{z}_i. \end{aligned}$$

However, from the last equation of (21) we have $q=f_n^z=0$, which gives

$$\begin{aligned} \langle \bar{p},y_i\rangle>\langle \bar{p},\bar{x}_i\rangle ~~\forall y_i\in { \mathcal{P}}_i(\bar{x}_i)\cap \mathcal{H}_i \iff \langle \bar{p},(({\mathcal{P}}_i(\bar{x}_i)\cap \mathcal{H}_i)- \bar{x}_i)\rangle > 0. \end{aligned}$$

(24)

The sufficiency of relations (5) and (6) for an allocation $\bar{x}\in \mathcal{A}({X})\cap H(p^\mathbf{1})$ to be Pareto $H(p^\mathbf{1})$-optimal is stated quite standardly. In fact let $y=(y_i)_{\mathcal{I}} \in \mathcal{A}({X})\cap H(p^\mathbf{1})$ be such that $y_i\succ _i \bar{x}_i$ is true for all i. Then as soon as the right-hand side in (5) is fulfilled for $y=(y_i)_{\mathcal{I}} \in H(p^\mathbf{1})$ due to $H(p^\mathbf{1})$ determination, via the left-hand part of (5), we can conclude $\bar{p} y_i> \bar{p} \bar{x}_i$ for all i. Now, summing inequalities over i one finds $\bar{p} \sum _{i\in \mathcal{I}}y_i> \bar{p} \sum _{i\in \mathcal{I}}\bar{x}_i$. Since $\sum _{i\in \mathcal{I}}y_i=\sum _{i\in \mathcal{I}}\bar{x}_i=\sum _{i\in \mathcal{I}}\mathbf{e}_i$, we are coming to a contradiction. Lemma 2 is proved. $\square$

Proof of Corollary 1

We have to consider the smooth case in the context of Lemma 2. On the necessary side, for the existence of values $\alpha _i>0$ and $\lambda _i^\sigma$ ${\forall }\sigma \ge 1$, one can state it directly from relations (5) and (6), applying separation theorem (or via the necessary conditions of a convex programming problem). However, the easiest way to see it may be found from condition $\bar{x}\in \mathrm{int}X$ and relations (20), stated in the proof of Lemma 2. From this we conclude in a standard manner the existence of $\alpha _i>0$ such that $\mathrm{grad}\,u_i(\bar{x}_i)=\alpha _i f_i$ ($\alpha _i\not =0$ due to $\text {grad}\, u_i(\bar{x}_i)\not =0$). Finally, one needs to apply (21).

To state the sufficiency, let us show that relations (5) and (6) are true. For some i and $y_i\in \mathcal{P}_i(\bar{x}_i)$, assume $\exists z_i \in \mathbb {R}^k\,:~ p_\sigma (y_i^\sigma -\mathbf{e}_i^\sigma )= p_\sigma A_\sigma z_i~ \forall \sigma \ge 1$. Due to gradient’s properties for interior points we have

$$\begin{aligned} \langle \text {grad}\, u_i(\bar{x}_i),y_i\rangle > \langle \text {grad}\, u_i(\bar{x}_i),\bar{x}_i\rangle \quad {\forall }i\in \mathcal{I}. \end{aligned}$$

Now substituting the gradient presentation given in the corollary conditions, one can conclude

$$\begin{aligned} \alpha _i \bar{p} y_i+\sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma y_i^{\sigma }> \alpha _i \bar{p} \bar{x}_i+ \sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma \bar{x}_i^{\sigma }. \end{aligned}$$

However, there are $z_i, \bar{z}_i \in \mathbb {R}^k$, such that $p_\sigma y_i^\sigma =p_\sigma \mathbf{e}_i^\sigma + p_\sigma A_\sigma z_i$ & $p_\sigma \bar{x}_i^\sigma =p_\sigma \mathbf{e}_i^\sigma + p_\sigma A_\sigma \bar{z}_i$ $\forall \sigma \ge 1$. Substituting these expressions under summation in the formula, one can find

$$\begin{aligned} \alpha _i \bar{p} y_i+ \sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma \mathbf{e}_i^\sigma + \left( \sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma A_\sigma \right) z_i> \alpha _i \bar{p} \bar{x}_i+ \sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma \mathbf{e}_i^\sigma + \left( \sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma A_\sigma \right) \bar{z}_i, \end{aligned}$$

that due to $\sum _{\sigma =1}^{s}\lambda _i^\sigma p_\sigma A_\sigma =0$ and $\alpha _i>0$ gives the result. $\square$

Proof of Lemma 3

The proof of lemma is reduced to the application of the separation theorem to a convex set properly constructed; this set corresponds to the ability of fuzzy coalitions to dominate an allocation.

Analogously to formula (2), let us determine the subspaces

$$\begin{aligned} \mathcal{H}_i=\mathcal{H}+\mathbf{e}_i, ~~\mathcal{H}= \{y\in {\mathbb {R}}^{l(s+1)}\mid \exists z\in {\mathbb {R}}^{k} \;:\; p_\sigma y^\sigma =p_\sigma A_\sigma z,~ \forall \sigma \ge 1\}. \end{aligned}$$

Next let us take any consumer $i_0\in \mathcal{I}$, let $i_0=1$, and determine the following set

$$\begin{aligned} G=G(x)= \mathop {\text {co}}\nolimits \left[ (\mathcal{P}_1(x_1)-\{\mathbf{e}_1\})\bigcup \left( \mathop {\cup }\limits _{i=2}^{n}[(\mathcal{P}_i(x_i)-\{\mathbf{e}_i\})\cap \mathcal{H}]\right) \right] . \end{aligned}$$

Now we show that if $0\in G$, then there is a fuzzy coalition p-dominating given allocation x in the incomplete market. In fact, $0\in G$ implies the existence of $t=(t_1,\ldots ,t_n)\ge 0$, $\sum t_i=1$ such that for some $y_i\in \mathcal{P}_i(x_i)$

$$\begin{aligned} \sum _{i\in \mathcal{I}} t_i(y_i-\mathbf{e}_i)=0~\iff ~ \sum _{i\in \mathcal{I}} t_iy_i= \sum _{i\in \mathcal{I}}t_i \mathbf{e}_i \end{aligned}$$

(25)

is true and, moreover, for $i=2,\ldots ,n$

$$\begin{aligned} \exists z_i\in \mathbb {R}^k\,:~~\mathrm{P}_{\! 1}y_i=\mathrm{P}_{\! 1}\mathbf{e}_i+\mathrm{P}_{\! 1}\mathrm{A}z_i \end{aligned}$$

takes place. To check the fuzzy domination definition in part (10), it is sufficient to state the latter relation for $i=1\in \text {supp}(t)$. To realize this, multiply (25) on matrix $\mathrm{P}_{\! 1}$, which after transformations due to $t_1\not =0$ yields

$$\begin{aligned} \textstyle \mathrm{P}_{\! 1}y_1=\mathrm{P}_{\! 1}\mathbf{e}_1+\mathrm{P}_{\! 1}\mathrm{A}\left( -\sum \limits _{i=2}^n\frac{t_i}{t_1}z_i \right) . \end{aligned}$$

Thus one can take $z_1=-\sum _{i=2}^n\frac{t_i}{t_1}z_i$ as the necessary solution (portfolio) for agent 1. As a result we conclude that coalition t can p-dominate the allocation.

Therefore, for every $x\in {C}^{ f}_\mathbf{p}(\mathcal{E}^\mathrm{in})$, it has to be that $0\notin G$ and since $\mathop {\text {int}}G\not = \emptyset$ (due to $\mathbf (S)$ we have $\mathop {\text {int}}\mathcal{P}_1(x_1)\not =\emptyset$), then one can apply the separation theorem and find nonzero $\bar{p}$ such that

$$\begin{aligned} \langle \bar{p},G\rangle \ge 0. \end{aligned}$$

Since $\mathcal{P}_1(x_1)-\{\mathbf{e}_1\}$ and $(\mathcal{P}_i(x_i)-\{\mathbf{e}_i\})\cap \mathcal{H}$, $i=2,\ldots ,n$ are the subsets of G, we conclude

$$ \begin{aligned} \langle \bar{p},\mathcal{P}_1(x_1)\rangle \ge \bar{p}\mathbf{e}_1~~ \& ~~ \langle \bar{p},\mathcal{P}_i(x_i)\cap (\{\mathbf{e}_i\}+\mathcal{H})\rangle \ge \bar{p}\mathbf{e}_i, \quad {\forall }i=2,\ldots ,n. \end{aligned}$$

Moreover, the first of these inequalities due to $x_1\in \mathop {\text {int}}X_1$ and $\bar{p}\not =0$ is realized in strict form that via $\mathbf (S)$ implies $\bar{p}_\sigma \not =0$, ${\forall }\sigma$ in a standard way. The lemma is proved. $\square$