Theorem 3.1.

For , , if there is such that , then is an essential weakly efficient solution of (VMP). Hence, the component that contains the point is an essential component.

Proof.

Suppose that , , and there exists such that . For any open neighborhood of in , there is an open neighborhood of in such that , where denotes the closure of

Since , and , then

Take such that

Then for any with , one has

Then

by (3.2) and (3.4), one has

therefore

Then, , and hence , which implies . By Definition 2.6, is an essential weakly efficient solution to (VMP). Hence, the component that contains the point is an essential component.

Corollary 3.2 (see [2]).

When , for , if is a singleton, then is an essential optimum solution.

Lemma 3.3.

Let be a nonempty compact convex subset of a Banach space, the function continuous and strongly quasiconvex, then is a singleton.

Proof.

Suppose , and . By Definition 2.11,

which is a contradiction, then is a singleton.

By Theorem 3.1 and Lemma 3.3, we have the following Theorem 3.4.

Theorem 3.4.

Let be a nonempty compact convex subset of a Banach space, for , if there exists such that is strongly quasiconvex, then has an essential weakly efficient solution, consequently, has an essential component.

Theorem 3.5.

For , if has only one component , then is an essential component.

Proof.

By Lemma 2.4, the mapping is upper semicontinuous at , hence, for any open set with , there exists such that for all with , one has , and hence, . By Definition 2.5, is an essential component.

Remark 3.6.

When , by [3], the optimum solution set (where ) has an essential component if and only if is connected. But, when , it is not true.

Example 3.7.

Let , define

is disconnected; however, is an essential weakly efficient solution, is an essential component, and is an essential component.

Example 3.8.

Let , define

Then , . By Theorem 3.1, is an essential weakly efficient solution. But, and are not essential weakly efficient solutions. In fact, for , take , , for all , take as the following:

Then , and

But . In fact, for all , if , , take , we have

If , take , we have

Thus ; therefore, is not an essential weakly efficient solution. Similarly, is not an essential weakly efficient solution.