## 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.

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## Acknowledgments

We are very grateful to Itai Ashlagi, Egon Balas, Heinz Bauschke, Jérôme Bolte, Boris Bukh, Keenan Crane, Patrick Combettes, Gerard Cornuejols, Federico Echenique, Alfred Galichon, Ben Golub, Sergiu Hart, Fatma Kılınç-Karzan, Michael McCoy, Javier Pena, Marek Pycia, R Ravi, Stephen Spear, Rakesh Vohra, Josephine Yu, Weijie Zhong, two anonymous referees, and seminar participants at Columbia University, Carnegie Mellon University, Higher School of Economics, New Economic School, and the University of Pittsburgh for discussions and useful suggestions. Shuo Liu acknowledges the financial support by the Forschungskredit of the University of Zurich, grant no. FK-17-018.

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## Appendix

### 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*) = *T**A* ∩ *T**B*, *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 *T**a* = *T**b* = *t* for some \(t\in \mathbb {R}^{n-1}\). As *t* ∈ *T**A* ∩ *T**B* = *T*(*A* ∩ *B*), there exists *c* ∈ *A* ∩ *B* such that *T**c* = *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*) = *T**A* ∩ *T**B*. As *T*(*A* ∩ *B*) ⊂ *T**A* ∩ *T**B* and both *T*(*A* ∩ *B*) and *T**A* ∩ *T**B* are convex compact sets, it suffices to show that *E*(*T*(*A* ∩ *B*), *d*) ⊂ *E*(*T**A* ∩ *T**B*, *d*) for all directions \(d\in \mathbb {R}^{m}\), that is, every support point of *T*(*A* ∩ *B*) is also a support point of *T**A* ∩ *T**B* 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*(*T**A*, *d*). *T*(*A* ∩ *B*) ⊂ *T**A* ∩ *T**B* implies that *t* ∈ *T**A* ∩ *T**B*, and since *T**A* ∩ *T**B* is a subset of *TA*, from the relation *t* ∈ *E*(*T**A*, *d*) we can conclude that *t* ∈ *E*(*T**A* ∩ *T**B*, *d*). □

###
*Proof of Theorem 9*

We want to prove that *T*(*A* ∩ *B*) = *T**A* ∩ *T**B* 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

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 (ri*A* ∩ri*B*≠*∅*) can be conveniently characterized by (see Corollary 16.4.1 in [17])

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

Equations (1) and (2) then imply that

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,

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

*(Only-if statement)* To establish the necessity part, we assume that *T*(*A* ∩ *B*) = *T**A* ∩ *T**B* 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

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

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* ∈ *k**e**r*(*T*)^{⊥},

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

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

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,

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 *T**A* ∩ *T**B* 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

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,

At the same time, we have

Finally, as image(*T*^{∗}) = *k**e**r*(*T*)^{⊥}, we can conclude that for any *d* ∈ *k**e**r*(*T*)^{⊥},

□

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Kushnir, A., Liu, S. On Linear Transformations of Intersections.
*Set-Valued Var. Anal* **28, **475–489 (2020). https://doi.org/10.1007/s11228-019-00525-0

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### Keywords

- Linear transformation
- Convex closed set
- Intersection
- Directional convexity
- Mechanism design
- Dominant-strategy implementation
- Bayesian implementation

### Mathematics Subject Classification (2010)

- 52A20
- 52A35
- 91A99