On Linear Transformations of Intersections

Abstract

For any linear transformation and two convex closed sets, we provide necessary and sufficient conditions for the transformation of the intersection of the sets to coincide with the intersection of their images. We also identify conditions for non-convex closed sets, continuous transformations, and multiple sets. We demonstrate the usefulness of our results via an application to the economics literature of mechanism design.

Appendix

Theorem 1

Consider two convex closed sets$A, B\subset \mathbb {R}^{n}$. If for all linear transformations$T:\mathbb {R}^{n}\rightarrow \mathbb {R}^{n-1}$we have T(A ∩ B) = TA ∩ TB, then A ∪ B is convex.

Proof

Take some a ∈ A and b ∈ B such that a≠b, and consider any linear transformation $T: \mathbb {R}^{n}\rightarrow \mathbb {R}^{n-1}$ with $\ker (T)=\{v\in \mathbb {R}^{n}: v=k\cdot (a-b), k\in \mathbb {R}\}$. By construction, the kernel of T amounts to the lines spanned by a − b, and we have $a-b\in \ker (T)$ and Ta = Tb = t for some $t\in \mathbb {R}^{n-1}$. As t ∈ TA ∩ TB = T(A ∩ B), there exists c ∈ A ∩ B such that Tc = t. As $\dim (\ker (T))=1$, points a, b, c must lie on one straight line. As A and B are convex and c ∈ A ∩ B, we have [a, c] ⊂ A and [b, c] ⊂ B, which implies that [a, b] ⊂ A ∪ B. Hence, A ∪ B is convex. □

Proof of Theorem 8

We want to prove that if $\forall d\in \ker (T)^{\perp }~E(A\cap B, d)\subseteq E(A, d)\cup E(B, d)$, then T(A ∩ B) = TA ∩ TB. As T(A ∩ B) ⊂ TA ∩ TB and both T(A ∩ B) and TA ∩ TB are convex compact sets, it suffices to show that E(T(A ∩ B), d) ⊂ E(TA ∩ TB, d) for all directions $d\in \mathbb {R}^{m}$, that is, every support point of T(A ∩ B) is also a support point of TA ∩ TB in the same direction. For this purpose, consider any direction $d\in \mathbb {R}^{m}$. Let $T^{*}: \mathbb {R}^{m}\rightarrow \mathbb {R}^{n}$ be the adjoint operator (or the transpose) of T. As $\text {image}(T^{*})=\ker (T)^{\perp }$ (see p. 120 in [2]) we have $T^{*}(d)\in \ker (T)^{\perp }$. The condition of the theorem then ensures that $E(A\cap B, T^{*}(d))\subseteq E(A, T^{*}(d))\cup E(B, T^{*}(d))$. By the definition of adjoint operators, we further have $E(T(A\cap B), d)\subseteq E(TA, d)\cup E(TB, d).$

Now consider any $t\in E(T(A\cap B), d)\subseteq E(TA, d)\cup E(TB, d)$. Without loss, we suppose that t ∈ E(TA, d). T(A ∩ B) ⊂ TA ∩ TB implies that t ∈ TA ∩ TB, and since TA ∩ TB is a subset of TA, from the relation t ∈ E(TA, d) we can conclude that t ∈ E(TA ∩ TB, d). □

Proof of Theorem 9

We want to prove that T(A ∩ B) = TA ∩ TB if and only if $\forall d\in \ker (T)^{\perp }$ and ∀u ∈ E(A ∩ B, d), $\exists d^{\prime }, d^{\prime \prime }\in \ker (T)^{\perp }$ such that $d^{\prime }+d^{\prime \prime }=d$ and $u\in E(A, d^{\prime })\cap E(B, d^{\prime \prime })$.

(If statement) For arbitrary direction $z\in \mathbb {R}^{m}$ the support function of the image of the intersection equals

$$ S^{T(A\cap B)}(z)=\sup_{t\in T(A\cap B)} t\cdot z=\sup_{x\in A\cap B} x\cdot T^{*}(z)=S^{A\cap B}(T^{*}(z)), $$

where T^∗ is the adjoint operator. As $\text {image}(T^{*})=\ker (T)^{\perp }$ we have $T^{*}(z)\in \ker (T)^{\perp }$. The support function for the intersection of convex sets having non-empty intersection of their relative interiors (riA ∩riB≠∅) can be conveniently characterized by (see Corollary 16.4.1 in [17])

$$ S^{A\cap B}(T^{*}(z))=\underset{d^{\prime},d^{\prime\prime} \in \mathbb{R}^{n} }{\underset{d^{\prime}+d^{\prime\prime}=T^{*}(z)}{\inf}} (S^{A}(d^{\prime})+S^{B}(d^{\prime\prime})). $$

(1)

Since set A ∩ B is compact, the set of support points E(A ∩ B, T^∗(z)) is non-empty. The condition of the theorem then implies that for any u ∈ E(A ∩ B, T^∗(z)), there exist $d^{\prime }, d^{\prime \prime }\in \ker (T)^{\perp }$, such that $d^{\prime }+d^{\prime \prime }=T^{*}(z)$ and $u\in E(A, d^{\prime })\cap E(B, d^{\prime \prime })$. Therefore, we have

$$ S^{A\cap B}\left( T^{*}(z)\right)=u\cdot T^{*}(z)=u\cdot d^{\prime}+u\cdot d^{\prime\prime}=S^{A}(d^{\prime})+S^{B}(d^{\prime\prime}). $$

(2)

Equations (1) and (2) then imply that

$$ S^{A\cap B}(T^{*}(z))= \underset{d^{\prime},d^{\prime\prime} \in \ker(T)^{\perp}}{\underset{d^{\prime}+d^{\prime\prime}=T^{*}(z)}{\inf}} (S^{A}(d^{\prime})+S^{B}(d^{\prime\prime})). $$

(3)

As $d^{\prime },d^{\prime \prime }\in \ker (T)^{\perp }=\text {image}(T^{*})$ there exist $z^{\prime }, z^{\prime \prime }\in \mathbb {R}^{m}$ such that $d^{\prime }=T^{*}(z^{\prime })$, $d^{\prime \prime }=T^{*}(z^{\prime \prime })$, and $z^{\prime }+z^{\prime \prime }=z$.

Hence,

$$ \begin{array}{@{}rcl@{}} S^{A\cap B}(T^{*}(z))&=&\inf_{{z^{\prime}+z^{\prime\prime}=z} \atop {z^{\prime},z^{\prime\prime} \in\mathbb{R}^{m} }} (S^{A}(T^{*}(z^{\prime}))+S^{B}(T^{*}(z^{\prime\prime})))\\ &=& \underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} (S^{TA}(z^{\prime})+S^{TB}(z^{\prime\prime}))=S^{TA\cap TB}(z). \end{array} $$

Overall, S^T(A∩B)(z) = S^TA∩TB(z) for all directions $z\in \mathbb {R}^{m}$. Hence, T(A ∩ B) = TA ∩ TB.

(Only-if statement) To establish the necessity part, we assume that T(A ∩ B) = TA ∩ TB and consider any direction $d\in \ker (T)^{\perp }$ and support point u ∈ E(A ∩ B, d). As $\ker (T)^{\perp }=\text {image}(T^{*})$ there must exist $z\in \mathbb {R}^{m}$ such that d = T^∗(z). For this direction, we have

$$ S^{TA\cap TB}(z)=\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m}}{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} \left( S^{TA}(z^{\prime})+S^{TB}(z^{\prime\prime})\right)=\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} \left( S^{A}(T^{*}(z^{\prime}))+S^{B}(T^{*}(z^{\prime\prime}))\right). $$

(4)

As T(A ∩ B) = TA ∩ TB there must exist $z^{\prime }, z^{\prime \prime }\in \mathbb {R}^{m}$ such that $z^{\prime }+z^{\prime \prime }=z$, and

$$ u\cdot d=S^{A\cap B}(T^{*}(z))=S^{T(A\cap B)}(z)=S^{TA\cap TB}(z)=S^{A}(d^{\prime})+S^{B}(d^{\prime\prime}), $$

where $d^{\prime }=T^{*}(z^{\prime })$, $d^{\prime \prime }=T^{*}(z^{\prime \prime })$, and the last equality follows from Corollary 16.4.1 in [17], which asserts that the infimum of (4) is achieved when the relative interiors of the two sets have a point in common. Also, as u ∈ A ∩ B, we have $u\cdot d^{\prime }\leq S^{A}(d^{\prime })$ and $u\cdot d^{\prime \prime }\leq S^{B}(d^{\prime \prime })$. As a result, we must have $u\cdot d^{\prime }=S^{A}(d^{\prime }), ~\text {and}~u\cdot d^{\prime \prime }=S^{B}(d^{\prime \prime }).$ In other words, $u\in E(A, d^{\prime })\cap E(B, d^{\prime \prime })$, where $d^{\prime }, d^{\prime \prime }\in \ker (T)^{\perp }$ and $d^{\prime }+d^{\prime \prime }=d=T^{*}(z)$. As the choice of z is arbitrary and $\text {image}(T^{*})=\ker (T)^{\perp }$ the only-if statement follows. □

Proof of Theorem 10

We want to prove that under the assumptions of the theorem, $\overline {T(A\cap B)}=\overline {TA\cap TB}$ if and only if for any d ∈ ker(T)^⊥,

$$ \underset{d^{\prime},d^{\prime\prime} \in \mathbb{R}^{n} }{\underset{d^{\prime}+d^{\prime\prime}=d}{\inf}} (S^{A}(d^{\prime})+S^{B}(d^{\prime\prime})) = \underset{d^{\prime},d^{\prime\prime} \in \ker(T)^{\perp}}{\underset{d^{\prime}+d^{\prime\prime}=d}{\inf}} (S^{A}(d^{\prime})+S^{B}(d^{\prime\prime})). $$

The argument is similar to the proof of Theorem 9. Let us first consider the sufficiency part. For arbitrary direction $z\in \mathbb {R}^{m}$, we have

$$ S^{T(A\cap B)}(z)=\sup_{t\in T(A\cap B)} t\cdot z=\sup_{x\in A\cap B} x\cdot T^{*}(z)=S^{A\cap B}(T^{*}(z)), $$

Given that $T^{*}(z)\in \ker (T)^{\perp }$, the condition of the theorem implies

$$ S^{A\cap B}(T^{*}(z))=\underset{d^{\prime},d^{\prime\prime} \in \mathbb{R}^{n}}{\underset{d^{\prime}+d^{\prime\prime}=T^{*}(z)}{\inf}} (S^{A}(d^{\prime})+S^{B}(d^{\prime\prime}))=\underset{d^{\prime},d^{\prime\prime} \in \ker(T)^{\perp} }{\underset{d^{\prime}+d^{\prime\prime}=T^{*}(z)}{\inf}} (S^{A}(d^{\prime})+S^{B}(d^{\prime\prime})). $$

Similar to the proof of Theorem 9, as $d^{\prime },d^{\prime \prime }\in \ker (T)^{\perp }=\text {image}(T^{*})$ there exist $z^{\prime }, z^{\prime \prime }\in \mathbb {R}^{m}$ such that $d^{\prime }=T^{*}(z^{\prime })$, $d^{\prime \prime }=T^{*}(z^{\prime \prime })$, and $z^{\prime }+z^{\prime \prime }=z$. Hence,

$$ \begin{array}{@{}rcl@{}} S^{A\cap B}(T^{*}(z))&=&\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} (S^{A}(T^{*}(z^{\prime}))+S^{B}(T^{*}(z^{\prime\prime})))\\ &=&\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} (S^{TA}(z^{\prime})+S^{TB}(z^{\prime\prime}))=S^{TA\cap TB}(z). \end{array} $$

Overall, S^T(A∩B)(z) = S^TA∩TB(z) for all directions $z\in \mathbb {R}^{m}$. In contrast to Theorem 9, since T(A ∩ B) and TA ∩ TB are not necessary closed, this only implies that $\overline {T(A\cap B)}=\overline {TA\cap TB}$.

Let us now establish the necessity part. Assume that $\overline {T(A\cap B)}=\overline {TA \cap TB}$. This implies that for any $z\in \mathbb {R}^{m}$ we have S^T(A∩B)(z) = S^TA∩TB(z). We know that

$$ \begin{array}{@{}rcl@{}} S^{TA\cap TB}(z)&=&\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} \left( S^{TA}(z^{\prime})+S^{TB}(z^{\prime\prime})\right)\\ &=&\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{m} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} \left( S^{A}(T^{*}(z^{\prime}))+S^{B}(T^{*}(z^{\prime\prime}))\right). \end{array} $$

As $\text {image}(T^{*})=\ker (T)^{\perp }$, there must exist $d^{\prime }, d^{\prime \prime }\in \ker (T)^{\perp }$ such that $d^{\prime }=T^{*}(z^{\prime }), d^{\prime \prime }=T^{*}(z^{\prime \prime })$. Since T^∗ is linear and bijective we also have that sets $\{d^{\prime },d^{\prime \prime }\in ker(T)^{\perp } | d^{\prime }+d^{\prime \prime }=T^{*}(z)\}$ and $\{d^{\prime },d^{\prime \prime }\in ker(T)^{\perp } | d^{\prime }=T^{*}(z^{\prime }), d^{\prime \prime }=T^{*}(z^{\prime \prime }), z^{\prime }+z^{\prime \prime }=z\}$ coincide. Therefore,

$$ S^{TA\cap TB}(z)=\underset{z^{\prime},z^{\prime\prime} \in \ker(T)^{\perp} }{\underset{z^{\prime}+z^{\prime\prime}=T^{*}(z)}{\inf}}\left( S^{A}(z^{\prime})+S^{B}(z^{\prime\prime})\right). $$

At the same time, we have

$$ S^{T(A\cap B)}(z)=S^{A\cap B}(T^{*}(z))=\underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{n} }{\underset{z^{\prime}+z^{\prime\prime}=T^{*}(z)}{\inf}} \left( S^{A}(z^{\prime})+S^{B}(z^{\prime\prime})\right). $$

Finally, as image(T^∗) = ker(T)^⊥, we can conclude that for any d ∈ ker(T)^⊥,

$$ \underset{z^{\prime},z^{\prime\prime} \in \mathbb{R}^{n} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} (S^{A}(z^{\prime})+S^{B}(z^{\prime\prime})) = \underset{z^{\prime},z^{\prime\prime} \in \ker(T)^{\perp} }{\underset{z^{\prime}+z^{\prime\prime}=z}{\inf}} (S^{A}(z^{\prime})+S^{B}(z^{\prime\prime})). $$

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