Solvability of a Regular Polynomial Vector Optimization Problem without Convexity^††thanks: This work was partially supported by the National Natural Science Foundation of China (11471230).

In this subsection we shall show that the regularity of $\text{PVOP}(K,f)$ is closely related to the regularity of $\text{PSOP}(K,f_{i})$. To do so, we first give a characterization of $SOL^{w}(K_{\infty},f^{\infty})=\emptyset$.

Proof We only need to prove the sufficiency. Suppose that $0\notin SOL^{w}(K_{\infty},f^{\infty})$. Then there exists $v_{0}\in K_{\infty}\backslash\{0\}$ such that

which yields

Let $v\in K_{\infty}$. It follows that

for all sufficiently large $t>0$. Since $v\in K_{\infty}$ is arbitrary, $SOL^{w}(K_{\infty},f^{\infty})=\emptyset$. ∎

Proof Since $SOL^{w}(K_{\infty},f^{\infty})\neq\emptyset$, by Proposition 1, $0\in SOL^{w}(K_{\infty},f^{\infty})$. Suppose on the contrary that there exist $v_{0}\in SOL^{w}(K_{\infty},f^{\infty})$ and $t_{0}>0$ such that $t_{0}v_{0}\notin SOL^{w}(K_{\infty},f^{\infty})$. Then there exists $v_{1}\in K_{\infty}$ such that

Dividing the both sides of the inequality (1) by $t^{d_{i}}_{0}$, we obtain $f^{\infty}_{i}(\frac{v_{1}}{t_{0}})-f^{\infty}_{i}(v_{0})<0$ for all $i\in\{1,\dots,s\}$. This reaches a contradiction to $v_{0}\in SOL^{w}(K_{\infty},f^{\infty})$ since $\frac{v_{1}}{t_{0}}\in K_{\infty}$. Thus, $SOL^{w}(K_{\infty},f^{\infty})$ is a nonempty cone. ∎

Proof Suppose on the contrary that there exists $i_{0}\in\{1,2,\dots,s\}$ such that $f_{i_{0}}$ is bounded from below on $K$. Then $SOL(K_{\infty},f^{\infty}_{i_{0}})=\emptyset$ since $SOL(K_{\infty},f^{\infty}_{i_{0}})\subseteq SOL^{w}(K_{\infty},f^{\infty})$. By Lemma 1, there exists $v_{0}\in K_{\infty}\backslash\{0\}$ such that $f^{\infty}_{i_{0}}(v_{0})<f^{\infty}_{i_{0}}(0)=0$. Since $v_{0}\in K_{\infty}\backslash\{0\}$, there exist $t_{k}>0$ with $t_{k}\rightarrow+\infty$ and $x_{k}\in K$ such that $t^{-1}_{k}x_{k}\rightarrow v_{0}$ as $k\rightarrow+\infty$. Since $f_{i_{0}}$ is bounded from below on $K$, there exists a constant $l$ such that

Letting $k\rightarrow+\infty$, we obtain $f^{\infty}_{i_{0}}(v_{0})\geq 0$ which is a contradiction to $f^{\infty}_{i_{0}}(v_{0})<0$. ∎

When $f_{i}$ is bounded from below on $K$ for some $i\in\{1,\cdots,s\}$, by Proposition 3 and Proposition 1, we know that $0\in SOL^{w}(K_{\infty},f^{\infty})$. In the following proposition, we further show $0\in SOL^{s}(K_{\infty},f^{\infty})$ when each component $f_{i}$ is bounded from below on $K$.

Proof Suppose that $f$ is bounded from below on $K$. Then $f_{i}$ is bounded from below on $K$ for all $i$. By Proposition 3 and Proposition 1, $f^{\infty}_{i}(x)\geq f^{\infty}_{i}(0)=0$ for all $x\in K_{\infty}$ and for all $i\in\{1,\dots,s\}$. This means $0\in SOL^{s}(K_{\infty},f^{\infty})$. ∎

The following result shows that the weak (strong) regularity of $\text{ PVOP }(K,f)$ is closely related to the regularity of $\text{ PVOP }(K,f_{i})$.

Proof (i) Assume that $SOL^{w}(K_{\infty},f^{\infty})=\{0\}$. Since

it suffices to show $SOL(K_{\infty},f^{\infty}_{i})\neq\emptyset$ for all $i\in\{1,\dots,s\}$. Suppose on the contrary that there exists $i_{0}\in\{1,\dots,s\}$ such that $SOL(K_{\infty},f^{\infty}_{i_{0}})=\emptyset$ which is equivalent to $0\notin SOL(K_{\infty},f^{\infty}_{i_{0}})$ (by Lemma 1). Then there exists $v_{0}\in K_{\infty}\backslash\{0\}$ such that $f^{\infty}_{i_{0}}(v_{0})<0$. Let $\lambda\in\mbox{ int }\mathbf{R}^{s}_{+}$. Consider the function

Then $S_{v_{0}}$ is nonempty and closed. Next we shall show that $S_{v_{0}}$ is bounded. If not, then there exists the sequence $\{x_{k}\}\subseteq S_{v_{0}}$ such that $\|x_{k}\|\to+\infty$ as $k\to+\infty$. Without loss of generality, we assume that $\|x_{k}\|\neq 0$ and $\frac{x_{k}}{\|x_{k}\|}\to x^{*}\in K_{\infty}\backslash\{0\}$. Since $x_{k}\in S_{v_{0}}$, we have

Dividing the both sides of the above inequality by $\|x_{k}\|^{d_{i}}$ and letting $k\rightarrow+\infty$, we get

which together with $0\in SOL^{w}(K_{\infty},f^{\infty})$ yields $x^{*}\in SOL^{w}(K_{\infty},f^{\infty})$, a contradiction. Thus, $S_{v_{0}}$ is bounded. By the known Weierstrass’ theorem, we have $SOL(S_{v_{0}},F_{\lambda})\neq\emptyset$. It follows from Lemma 1 that

which implies $SOL(S_{v_{0}},F_{\lambda})=\{0\}$, and so $0\in S_{v_{0}}$. By the definition of $S_{v_{0}}$, we get $f^{\infty}_{i_{0}}(v_{0})\geq 0$, a contradiction to $f^{\infty}_{i_{0}}(v_{0})<0$. Therefore, $SOL(K_{\infty},f^{\infty}_{i})\neq\emptyset$ for all $i\in\{1,\dots,s\}$.

For the converse, assume that $SOL(K_{\infty},f^{\infty}_{i})=\{0\}$ for all $i\in\{1,\dots,s\}$ and there exists $v_{0}\in SOL^{w}(K_{\infty},f^{\infty})\backslash\{0\}$. Then

which implies that $f^{\infty}_{i_{0}}(v_{0})\leq f^{\infty}_{i_{0}}(0)$ for some $i_{0}\in\{1,\dots,s\}$. This reaches a contradiction to $SOL(K_{\infty},f^{\infty}_{i_{0}})=\{0\}$.

(ii) Suppose on the contrary that there exists $i_{0}\in\{1,\dots,s\}$ such that $SOL(K_{\infty},f^{\infty}_{i_{0}})=\emptyset$. By Lemma 1, $0\notin SOL(K_{\infty},f^{\infty}_{i_{0}})$. Then there exists $v_{0}\in K_{\infty}\backslash\{0\}$ such that $f^{\infty}_{i_{0}}(v_{0})<0$. Let $\lambda\in\mbox{ int }\mathbf{R}^{s}_{+}$. By similar arguments as in the proof of (i), we have

where $S_{v_{0}}$ and $F_{\lambda}$ are defined as in (2). The rest is same as the one of (i), and so we omit it. ∎

Proof The sufficiency follows immediately from Proposition 4 and Theorem 3.1 (i).

Next we prove the sufficiency. By Theorem 3.1(i), $\text{ PVOP }(K,f)$ is strongly regular. Suppose on the contrary that there exists $i_{0}\in\{1,\dots,s\}$ such that $f_{i_{0}}$ is not bounded from below on $K$. Let $x\in K$. Then there exists $\{x_{k}\}^{\infty}_{k=1}\subseteq K$ such that

for all sufficiently large $k$. We claim that $\{x_{k}\}^{\infty}_{k=1}\subseteq K$ is unbounded. Indeed, if not, then we may assume that $\|x_{k}\|\to+\infty$ and $\frac{x_{k}}{\|x_{k}\|}\to v\in K_{\infty}\backslash\{0\}$ as $k\to+\infty$. Dividing the both sides of (3) by $\|x_{k}\|^{d_{i}}$ and letting $k\rightarrow+\infty$, we get $f^{\infty}_{i}(v)\leq 0$, a contradiction to $SOL(K_{\infty},f^{\infty}_{i_{0}})=\{0\}$. Hence, $\{x_{k}\}^{\infty}_{k=1}$ is bounded. Without loss of generality, we may assume that $x_{k}\to x^{*}\in K$ as $k\to+\infty$. It follows from (3) that

a contradiction. Thus, $f$ is bounded from below on $K$. ∎

It has been shown in Oett ; FYP ; FangH ; Gowda ; Lop that $R_{0}$-property plays an important role in studying the compactness of the solution sets of complementarity problems as well as the upper continuity of the solution mappings. In this subsection, we shall show that the weak (strong) regularity of $\mbox{ PVOP }(K,f)$ is closely related to the $R_{0}$-property of the following weak vector complementarity problem CGY ; Yangxq

where $\nabla f^{\infty}=(\nabla f^{\infty}_{1},\nabla f^{\infty}_{2},\dots,\nabla f^{\infty}_{s})$ is the gradient of $f^{\infty}$,

is the weak dual cone of $K_{\infty}$ with respect to $\mathbf{R}^{s}_{+}$, and $L(\mathbf{R}^{n},\mathbf{R}^{s})$ is the space of all bounded linear operators from $\mathbf{R}^{n}$ to $\mathbf{R}^{s}$. We denote the solution set of $\mbox{WVCP}(K_{\infty},\nabla f^{\infty})$ by $SOL_{WVCP}(K_{\infty},\nabla f^{\infty})$. Recall that $\mbox{WVCP}(K_{\infty},\nabla f^{\infty})$ is of type $R_{0}$ FYP if $SOL_{WVCP}(K_{\infty},\nabla f^{\infty})=\{0\}$.

Proof It is clear that $0\in SOL_{WVCP}(K_{\infty},\nabla f^{\infty})$.

Suppose that $SOL^{w}(K_{\infty},f^{\infty})=\{0\}$. By Theorem 3.1(i), we have

This implies $f^{\infty}_{i}(v)>0$ for all $v\in K_{\infty}\backslash\{0\},i=1,\dots,s$. This together with the Euler’s Homogeneous Function Theorem yields

for all $v\in K_{\infty}\backslash\{0\}$, where $d_{i}$ is the degree of $f_{i},i=1,\cdots,s$. As a consequence, $SOL_{WVCP}(K_{\infty},\nabla f^{\infty})=\{0\}$ and so $\text{WVCP}(K_{\infty},\nabla f^{\infty})$ is of type $R_{0}$.

For the converse, suppose $\mbox{WVCP}(K_{\infty},\nabla f^{\infty})$ is of type $R_{0}$. To complete the proof, it suffices to show $SOL^{w}(K_{\infty},f^{\infty})=\{0\}$ (by Remark 3). Suppose on the contrary that there exists $v^{*}\in SOL^{w}(K_{\infty},f^{\infty})\backslash\{0\}$. By (LKY, , Theorem 4), we get

This implies $v\in SOL_{WVCP}(K_{\infty},\nabla f^{\infty})$ since $K_{\infty}$ is a closed convex cone. This reaches a contradiction. ∎

The following example illustrates the conclusion of Theorem 3.2.