Porosity of Convex Nowhere Dense Subsets of Normed Linear Spaces

and Applied Analysis 3 Definition 2.4. M is called c-porous if for any x ∈ X and every r > 0, there are y ∈ B x, r and φ ∈ X∗ \ {0} such that { z ∈ X : φ z > φy ∩M ∅. 2.2 C-porosity turns out to be the suitable notion to describe the smallness of convex nowhere dense sets see Proposition 2.5 and is a stronger form of 0-angle porosity x ∈ X instead x ∈ M . Indeed, consider the unit sphere S of any nontrivial normed space. S is not c-porous simply take x 0 and r 1/2 and is 0-angle porous—to see it, use the HahnBanach separation theorem cf. 8 for sets B 0, 1 the closure of B 0, 1 and { 1 r/2 y}, where y ∈ S and r > 0. If a set M is a countable union of c-porous sets, then we say that M is σ-c-porous. In the same way we define σ0-angle porosity. If M ⋃ n∈N Mn and each Mn is Rn-ball porous for some Rn > 0, then we say that M is ball small. The next result shows that c-porosity is the best approximation of smallness in the sense of porosity of convex nowhere dense sets in the proof we extend an argument suggested by Zajı́ček 4, page 518 . Proposition 2.5. A subset M of a normed space X is c-porous if and only if conv M is nowhere dense. Proof. “⇒” It is obvious that for any φ ∈ X∗ and y ∈ X, we have { z : φ z > φ ( y )} ∩M ∅ ⇐⇒ z : φ z > φy ∩ convM ∅. 2.3 Hence ifM is c-porous, then convM is also c-porous and, in particular, nowhere dense. “⇐” Fix any x ∈ X and r > 0. Since convM the closure of convM is nowhere dense, there exists y ∈ B x, r \ convM. Sets convM and {y} satisfy the assumptions of the HahnBanach separation theorem, so there exist φ ∈ X∗ and c ∈ R such that φ y > c and for any z ∈ convM,φ z < c. ThenM ∩ {z : φ z > φ y } ∅. Corollary 2.6. Let X be any normed space and let ∗ be any condition such that if A ⊂ X satisfies ∗ and B ⊂ A, then B satisfies ∗ . 2.4 If every convex and nowhere dense subset ofX satisfies ∗ , then any c-porous subset ofX satisfies ∗ . The notions of 0-angle porosity and c-porosity involve the space X∗; however, in its origin the porosity was defined in metric spaces. In the next part of this section we will show what kind of porosity without using X∗ is implied by them see Proposition 2.8 . Note that we will use this result in Sections 3 and 4. We omit the proof of the following result since it is technical and can be easily deduced from the proof of 5, Lemma 1 . 4 Abstract and Applied Analysis Lemma 2.7. Let R > 0, r ∈ 0, 1/2 , x0, y0 ∈ X, φ ∈ X∗ \ {0}. If ‖y0 − x0‖ < rR, then there exists y ∈ X such that ‖y − x0‖ R and B ( y, 1 − 2r R ⊂ x ∈ X : φ x > φy0 )} . 2.5 Proposition 2.8. The following statements hold. i IfM is 0-angle porous, thenM is R-ball porous for every R > 0, that is, for every R > 0, x ∈ M and every α ∈ 0, 1 , there exists y ∈ X with ∥y − x∥ R and By, αR ∩M ∅. 2.6 ii IfM is c-porous, then for every R > 0, x ∈ X and every α ∈ 0, 1 , there exists y ∈ X with ∥y − x∥ R and By, αR ∩M ∅. 2.7 Proof. We will prove only i , since the proof of ii is very similar. Fix R > 0, x0 ∈ M and α ∈ 0, 1 . Let r > 0 be such that 1 − 2r > α, and let y0 ∈ X and φ ∈ X∗ \ {0} be such that ‖y0 −x0‖ < rR andM∩{x : φ x > φ y0 } ∅. By Lemma 2.7, we have that there exists y ∈ X such that ‖y−x0‖ R andB y, 1−2r R ⊂ {x : φ x > φ y0 }. SinceB y, αR ⊂ B y, 1−2r R , the result follows. Note that 2.7 is stronger than 2.6 . Indeed, the unit sphere in any normed space satisfies 2.6 and does not satisfy 2.7 . In the sequel, we will extend this observation see Theorem 3.2 . The next example shows, in particular, that the converse of the Proposition 2.8 is not true. Example 2.9. Let X, ‖ · ‖ be one of the following real Banach spaces: c0 or lp, p ∈ 1,∞ . Let us define the set M : ⋃ n∈N{±ne1,±ne2, . . . ,±nen}, where en r : ⎧ ⎨ ⎩ 1, r n, 0, r / n. 2.8 Now we will show that M satisfies the following condition, which is stronger than 2.7 and, in particular, than 2.6 : for every R > 0, x ∈ X, there exists y ∈ X s.t. ∥y − x∥ R and By,R ∩M ∅. 2.9 Abstract and Applied Analysis 5 To see it, take any x ∈ X and R > 0. Since x n → 0, there exists n0 > 3R such that |x n0 | < R. Assume, without loss of generality, that x n0 ≥ 0. Now let y ∈ X be such thatand Applied Analysis 5 To see it, take any x ∈ X and R > 0. Since x n → 0, there exists n0 > 3R such that |x n0 | < R. Assume, without loss of generality, that x n0 ≥ 0. Now let y ∈ X be such that y k : ⎧ ⎨ ⎩ x k , k / n0,