Compact convex sets free of inner points in infinite‐dimensional topological vector spaces

An inner point of a non‐singleton convex set M$M$ is a point x∈M$x\in M$ satisfying that for all m∈M∖{x}$m\in M\setminus \lbrace x\rbrace$ there exists n∈M∖{m,x}$n\in M\setminus \lbrace m,x\rbrace$ such that x∈(m,n)$x\in (m,n)$ . We prove the existence of convex compact subsets free of inner points in the infinite‐dimensional setting. Following our pathway to this result, we come up with other several geometric hits, such as the existence of a non‐convex subset that coincides with its starlike envelope.

The geometrical structure of compact sets in general topological vector spaces has also been deeply studied, ever since the famous Krein-Milman theorem irrupted into the literature.Moreover, this geometrical study has been mostly carried out from the perspective of extremal theory.However, the inner geometric structure of compact sets has not been profoundly developed.By this inner geometric structure, we mean the existence of inner points, internal points, or even interior points.
In [11], the inner structure of a (not necessarily convex) set is formerly introduced in the literature.This inner structure (inner points) has the particularity that it is independent of the ambient space.It is not an easy task, as it can be seen in [6,9,11], to find non-trivial convex sets free of inner points.It is even harder to find non-trivial closed convex sets lacking inner points.Such an example can be found in [11,Theorem 5.4].This brings the following problem up.
Problem 1.3.Let  be a Hausdorff locally convex topological vector space.Does there exist a non-singleton compact and convex subset  ⊆  free of inner points?
The main purpose of this paper is to provide a full solution to Problem 1.3 in the positive for normed spaces with the weak topology.In the meantime, by trying to solve Problem 1.3, we had to investigate different geometric aspects of convex sets which led to the development of the sections of this paper.

A NON-CONVEX STARLIKE GENERATED SET
In [2, Definition 17], the following concepts were introduced in the literature of the Banach space geometry: let  be a normed space.Let  ⊆   .The starlike envelope of  is defined as where st(,   ) ∶= { ∈   ∶ ‖ + ‖ = 2} = { ∈   ∶ [, ] ⊆   } is known as the starlike set of .Furthermore, we will say that: •  is almost flat if [, ] ⊆   for all ,  ∈ .
In [2, Lemma 6], the following essential properties were proved for a normed space  and a subset  ⊆   : (1) If  is convex, then  is flat and starlike compatible. ( If  is convex and starlike generated, then  is a convex component of   .
A novel three-dimensional unit ball is constructed in [2, Example 3] which serves to show the existence of an almost flat set which is not flat.
This unit ball is a convex polyhedron whose facets (faces with non-empty interior relative the unit sphere) are six regular diamonds.An easy way to construct this unit ball is by taking a regular octahedron and placing a regular tetrahedron (with the same triangles) on top of one facet of the octahedron and the opposite tetrahedron on the bottom of the opposite facet.If  denotes the set of all four vertices of the upper regular tetrahedron, then  is clearly almost flat but not flat, since co() is the whole regular tetrahedron, which is clearly not contained in the boundary of the previous unit ball.We will make use of this original unit ball to construct a non-convex starlike generated set.Proof.First off,  is clearly not convex.However,  is trivially almost flat, thus  is starlike compatible by [2, Lemma 6(2)], hence  ⊆ st().It only remains to show that st() ⊆ .Observe that st() ⊆ st(,   ) = co({, , , −}) ∪ co({, −, , }) ∪ co({, , −, }).Note that co({, , , −}), co({, −, , }) and co({, , −, }) are the three upper diamonds.Assume on the contrary that there exists  ∈ st() ⧵ .We may assume without any loss of generality that  lies is in the diamond co({, , , −}).Since  ∉ , we have that  ∉ [, ] ∪ [, ].Then,  lies either in the interior of the diamond co({, , , −}) or in (, −] ∪ (, −].Either way, we conclude that (, ) ⊆   , which contradicts the fact that  ∈ st() ⊆ st(,   ).□

INNER POINTS COINCIDE WITH NON-SUPPORT POINTS
Let  be a vector space.Let  be a convex subset of  with at least two points.We define the set of inner points of  as inn() ∶= { ∈  ∶ ∀ ∈  ⧵ {} ∃ ∈  ⧵ {, } such that ∈ (, )}.
The inner structure was formerly introduced for the first time in [11,Definition 1.2] for non-convex sets, although it appears implicitly in [1,5,12] for convex sets.In this paper, we will only make use of the inner structure of convex sets.We refer the reader to [6,8,9,11] for a wider perspective on these concepts.In [11,Theorem 5.1], it is proved that every non-singleton convex subset of any finite-dimensional vector space has inner points.However, in [11,Corollary 5.3] it was shown that every infinite-dimensional vector space possesses a non-singleton convex subset free of inner points.Remark 3.1.Let  be a vector space.Let  be a convex subset of .We convey that if  is a singleton, then inn() = ∅.In any case, it is trivial that inn() ⊆ .In view of [9, Remark 1.1], if  ∈ inn(), then [, ) ⊆ inn() for all  ∈ .As a consequence, inn() is convex and cl(inn()) = cl().On the other hand, in [9, Lemma 2.1], it was proved that if  is an extremal subset of , then  ∩ inn() = ∅.
Recall that, given a subset  of a real vector space , a subset  ⊆  is said to be extremal in  if it satisfies the extremal property: if ,  ∈  and there exists  ∈ (0, 1) with  + (1 − ) ∈ , then ,  ∈ .The following remark can be found in [9, Lemma 2.1].Remark 3.2.In a topological vector space , if  is extremal in a convex subset  ⊆  and  ∩ int() ≠ ∅, then  = .

AN EXAMPLE OF A NON-TRIVIAL COMPACT CONVEX SET FREE OF INNER POINTS
Keeping in mind the study realized in [11], where the authors gave several examples of convex and closed convex subsets without inner points, we bring up Problem 1.3.In the first place, note that the above question has to be tackled in an infinite-dimensional setting, because, as we have already mentioned in the previous section, in [11,Theorem 5.1], it is proved that every non-singleton convex subset of any finite-dimensional vector space has inner points.However, in [11,Corollary 5.3], it was shown that every infinite-dimensional vector space possesses a non-singleton convex subset free of inner points.
We will fully solve Problem 1.3 in the positive for normed spaces with the weak topology.For this, several technical remarks and lemmas are going to be made use of.
Remark 4.1.Let  be an infinite-dimensional Banach space.Let (  ) ∈ℕ ⊆  be a bounded basic sequence.The following operator: is well defined, one-to-one, linear, and continuous.Even more, ‖‖ = ‖(‖  ‖) ∈ℕ ‖ ∞ = sup{‖  ‖ ∶  ∈ ℕ}.In particular,  is - continuous, so it maps -compact subsets of  1 to -compact subsets of .Also,  is an isomorphism of vector spaces over its image, so  preserves inner points in virtue of [11, Section 1], hence it maps convex subsets of  1 free of inner points to convex subsets of  free of inner points.
The following technical lemma, needed to achieve our goal, is implicitly used in [10, Lemma 2.3] and justified with strong machinery such as the Banach-Steinhauss theorem.Here, we will provide a simple proof not requiring any machinery.
We are finally in the right position to provide a full solution to Problem 1.3 in the normed space setting endowed with the weak topology.Theorem 4.8.Let  be an infinite-dimensional Banach space.There exists a non-singleton weakly compact and convex subset  ⊆  free of inner points.
Observe that   ≥ 0 for all  ∈ ℕ and In  Given a vector space  and a subset  ⊆ , a face of  is a convex extremal subset of .Recall that a point  ∈   in the unit ball of a normed space  is said to be a rotund point of   provided that {} is a maximal face of   , or equivalently, whenever  ∈   and ‖ ‖ ‖ is not strongly maximal.
Observe that rotund points are strongly maximal faces of the unit ball.

F I G U R E 1
Pacheco-Campos unit ball.Theorem 2.1.Consider the upper regular tetrahedron of the unit ball given in Figure 1.Denote by  to the upper vertex and by , ,  to the base vertices of the upper regular tetrahedron.Then,  ∶= [, ] ∪ [, ] ∪ [, ] is a non-convex starlike generated set.

1 ,Definition 5 . 5 .Lemma 5 . 6 .Example 5 . 7 .
then  = .Before proving our last result in this paper, we will introduce a novel definition in extremal theory.Let  be a normed space.We say that a convex subset  ⊆   is a strongly maximal face of   provided that ⋃ ∈ st (,   ) = .Let  be a real vector space.If  ⊆   is a strongly maximal face of   , then  is a maximal face of   .Proof.By definition,  is convex.Let  ⊆   be a face of   containing .Take any  ∈  and any  ∈ .Then, [, ] ⊆  ⊆   .As a consequence,  ∈ st (,   ) ⊆ .□ Not every maximal face of the unit ball is strongly maximal.Any of the maximal faces of the unit ball of  2 ∞ ∶=
Let  be an infinite-dimensional Banach space.Let (  ) ∈ℕ ⊆  be a bounded basic sequence.Let  be the operator given by Equation (1).If  is a  * -closed bounded subset of  1 , then  () is -sequentially closed in .If, in addition,  is convex, then () is -closed.For every  ∈ ℕ, let  *  ∈  * denote a Hahn-Banach extension of the coordinate functional associated with   .Note that Now, we are in the right position to call on Lemma 4.3 or Lemma 4.5 because ( (   ) ∈ℕ ) ∈ℕ is a bounded sequence in  1 which is pointwise convergent to the sequence ( *  ()) ∈ℕ .So, we conclude that ( *  ()) ∈ℕ ∈  1 and [10,Lemma 2.3].Remark 4.4.Let  be a topological space.Let  ∶  →  be a net in , where  is a directed set.Let  ∈ .If every subnet  of  has a further subnet  with  ∈ lim(), then  ∈ lim().Next, lemma generalizes Remark 4.2 and Lemma 4.3 at once.Lemma 4.5.Let  be a normed space.If ( *  ) ∈ℕ ⊆  * is bounded and pointwise convergent, then there exists  * -convergent to  * 0 .□Thefollowinglemma is a refinement of[10, Lemma 2.3].Here, we will provide a much simpler proof by using less machinery.Lemma 4.6.)∈ℕbe a sequence in  such that))∈ℕ is -convergent to some  ∈ .∈ℕconverges to  *  () for each  ∈ ℕ. Keep in mind that  *  for all ,  ∈ ℕ.
[11,onstruction of , (1 − )   +   ≥ 0 for all  ∈ ℕ.In particular, for  ∈ ℕ, (1 − )    +    ≥ 0, meaning that Note that the set constructed in[11, Theorem 5.4]does not serve to solve Problem 1.3 since it is not weakly compact (see next proposition).However, it is closed, convex, and free of inner points[11, Theorem 5.4].Suppose  is weakly compact.Then, it is weakly sequentially compact by the Eberlein-Smulian theorem.However, sequential weak convergence and sequential norm convergence are equivalent in  1 by Schur's theorem.Then,  is sequentially compact, thus compact since it is a metric space.Finally, the sequence (  ) ∈ℕ of canonical vectors of  1 is contained in  and this sequence satisfies that ‖  −   ‖ 1 = 2 whenever  ≠ , meaning that (  ) ∈ℕ does not have convergent subsequences.This is a contradiction.□