Minkowski deviation measures

Lemma A.1.

Let f : X → R ∪ { + ∞ } . If f is positive homogeneous, then the set

has empty interior.

Proof.

Let us proceed by contraposition by showing that if E has non-empty interior, then f is not positive homogeneous. Assume, then, that X ∈ int ⁡ E , and let V denote an open neighborhood of X with V ⊆ E . By continuity of scalar multiplication, for small enough u > 0 we have ( 1 + u ) ⁢ X ∈ V ⊆ E . But then f ⁢ ( ( 1 + u ) ⁢ X ) = 1 < ( 1 + u ) ⁢ f ⁢ ( X ) , so f is not positive homogeneous. ∎

Lemma A.2.

Let A ⊆ X be non-empty Then, we have the following:

1. If A is absorbing, then 𝒟 A is finite-valued.

2. If A is star-shaped and y ≔ 𝒟 A ⁢ ( X ) > 0 , then y - 1 ⁢ X ∈ bd ⁡ ( A ) , i.e., 𝒟 A ⁢ ( X ) = 1 implies X lies in the boundary of A.

3. If A is strongly star-shaped, X ∈ bd ⁡ ( A ) and X ≠ 0 , then 𝒟 A ⁢ ( X ) = 1 .

4. If R X ∩ A = ∅ then 𝒟 A ⁢ ( X ) = + ∞ . In particular, if 0 ∉ A , then 𝒟 A ⁢ ( 0 ) = + ∞ .

5. If A is closed, absorbing and radially bounded, then the infimum in equation ( 2.1 ) is attained for any X ∈ 𝒳 ∖ { 0 } , that is, X ∈ 𝒟 A ⁢ ( X ) ⁢ A for any X ∈ 𝒳 .

6. If A is closed, then the infimum in equation ( 2.1 ) is attained for any X such that 𝒟 A ⁢ ( X ) > 0 .

Proof.

Let X ∈ 𝒳 and write y ≔ 𝒟 A ⁢ ( X ) .

For item (i), there exists – by the absorbing property – some δ X > 0 such that the inclusion [ 0 , δ X ] ⁢ X ⊆ A holds. It is straightforward to see that in this case the set { m > 0 : m - 1 ⁢ X ∈ A } is never empty. Therefore, we have 𝒟 A ⁢ ( X ) < ∞ .

For item (ii), it suffices to consider the case 𝒟 A ⁢ ( X ) = 1 , as the general case then easily follows from positive homogeneity. In order to verify that X ∈ bd ⁡ ( A ) , we only have to exhibit sequences { Y n } ⊆ A and { Z n } ⊆ A ∁ such that lim ⁡ Y n = lim ⁡ Z n = X . Let, then, Y n be defined through Y n ≔ ( 1 + 1 2 n ) - 1 ⁢ X and, similarly, Z n ≔ ( 1 - 1 2 n ) - 1 ⁢ X . Continuity of scalar multiplication immediately yields the desired equality of limits, so it only remains to show that Y n ∈ A and Z n ∈ A ∁ for all n. For such, just notice that – due to star-shapedness – if m > 1 , then m - 1 ⁢ X ∈ A , so Y n ∈ A , and – since 𝒟 A ⁢ ( X ) = 1 and by the definition of 𝒟 A – if 0 < m < 1 , then m - 1 ⁢ X ∉ A , so Z n ∉ A .

For item (iii), as X ≠ 0 by assumption, we see that whenever the ray R X has a non-empty intersection with bd ⁡ ( A ) , it necessarily also holds that 𝒟 A ⁢ ( X ) > 0 . Therefore, as X ∈ bd ⁡ ( A ) , we also have that 𝒟 A ⁢ ( X ) - 1 ⁢ X ∈ bd ⁡ ( A ) , by item (ii). Thus, we have X ∈ R X ∩ bd ⁡ A and 𝒟 A ⁢ ( X ) - 1 ⁢ X ∈ R X ∩ bd ⁡ A , and hence strong star-shapedness of A tells us that 𝒟 A ⁢ ( X ) = 1 .

Item (iv) is clear, since if { λ ⁢ X : λ > 0 } ∩ A = ∅ , then the set { m > 0 : m - 1 ⁢ X ∈ A } is empty, and we have adopted the convention that the infimum of such set is + ∞ .

For item (v), notice that if A is radially bounded, then by Theorem 3.2, item (ii), it holds that 𝒟 A ⁢ ( X ) > 0 , for every non-zero X ∈ 𝒳 . Now, let T X : ( 0 , + ∞ ) → 𝒳 be defined by T X ⁢ ( m ) = m - 1 ⁢ X . Clearly, T X is continuous. Thus, if A is a closed subset of 𝒳 , so is T X - 1 ⁢ ( A ) a closed subset of ( 0 , + ∞ ) . Also, if A is absorbing, then T X - 1 ⁢ ( A ) is non-empty. Finally, since radial boundedness ensures 𝒟 A ⁢ ( X ) > 0 , it follows that inf ⁡ T X - 1 ⁢ ( A ) ∈ T X - 1 ⁢ ( A ) as stated.

The proof of the last item is identical to the previous one. ∎

Lemma A.3.

If D is positive homogeneous and finite valued, then A D k is absorbing for all k > 0 .

Proof.

As D is a positive homogeneous finite function and k > 0 , it follows that, for any X ∈ 𝒳 such that D ⁢ ( X ) > 0 , one has D ⁢ ( k ⁢ X D ⁢ ( X ) ) = k . Therefore, we have t ⁢ X ∈ 𝒜 D k for any 0 ⩽ t ⩽ δ X ≔ k D ⁢ ( X ) . Of course, if D ⁢ ( X ) ⩽ 0 , then there is nothing to prove, as in this case we have D ⁢ ( X ) ⩽ k , that is, it holds that X ∈ 𝒜 D k . ∎

Definition A.4.

For a dual pair 〈 𝒳 , 𝒳 ′ 〉 , the polar A ⊙ ⊆ 𝒳 ′ of a non-empty set A ⊆ 𝒳 is defined through

and the bipolar of A is the set A ⊙ ⊙ ⊆ 𝒳 given by

Lemma A.5.

Given a dual pair 〈 X , X ′ 〉 , let A ⊆ X . If A is star-shaped and translation insensitive, then 〈 1 , X ′ 〉 = 0 for all X ′ ∈ A ⊙ .

Proof.

Let X ′ ∈ A ⊙ . Then – as ℝ ⊆ A and A + ℝ ⊆ A by assumption – we have, for any X ∈ A and y ∈ ℝ ,

and, as y is arbitrary, it is necessarily true that 〈 1 , X ′ 〉 = 0 . ∎