Proof of fungibility theorem

Appendix to: A fungibility theorem

Suppose that $P$ is a set and we have functions $v_{1}, \dots, v_{n} : P \to R$ . Recall that for $p, q \in P$ , we say that $p$ is a Pareto improvement over $q$ if for all $i$ , we have $v_{i} (p) \geq v_{i} (q)$ . And we say that it is a strong Pareto improvement if in addition there is some $i$ for which $v_{i} (p) > v_{i} (q)$ . We call $p$ a Pareto optimum if there is no strong Pareto improvement over it.

Theorem. Let $P$ be a set and suppose $v_{i} : P \to R$ for $i = 1, \dots, n$ are functions satisfying the following property: For any $p, q \in P$ and any $α \in [0, 1]$ , there exists an $r \in P$ such that for all $i$ , we have $v_{i} (r) = α v_{i} (p) (1 - α) v_{i} (q)$ .

Then if an element $p$ of $P$ is a Pareto optimum, then there exist nonnegative constants $c_{1}, \dots, c_{n}$ such that the function $\sum c_{i} v_{i}$ achieves a maximum at $p$ .

Proof. Let $V = v_{1} \times \dots \times v_{n} : P \to R^{n}$ . By hypothesis, the image $V (P)$ is convex.

For $p \in P$ , let the Pareto volume of $p$ be the set

$p v (p) = {(x_{1}, \dots, x_{n}) \in R^{n} ∣ x_{i} \geq v_{i} (p) for all i}$

This is a closed convex set. Note that $p \in P$ is a Pareto optimum precisely when $p v (p) \cap V (P) = {V (p)}$ . Let’s assume that this is the case; we just have to prove that $p$ maximizes $\sum c_{i} v_{i}$ for some choice of $c_{i} \geq 0$ .

It suffices to find a hyperplane $H \subset R^{n}$ that contains $V (p)$ and that supports $V (P)$ . Then the desired function $\sum c_{i} v_{i}$ can be constructed by ensuring that $H$ is a level set.

If $V (P)$ lies in a proper affine subspace of $R^{n}$ , let $Z$ be the smallest such subspace. Let $A$ be the interior of $V (P)$ in $Z$ and let $B$ be the interior of $p v (p)$ . The case where $V (P)$ is a point is trivial; suppose it’s not, so $A$ is nonempty. By convexity, $V (P)$ is the closure of $A$ and $p v (p)$ is the closure of $B$ .

Since $V (P)$ is convex, $A$ is convex, and we can exhaust $A$ with a nested sequence of nonempty compact convex sets $A_{j}$ . And $p v (p)$ is convex, so we can exhaust $B$ with a nested sequence of nonempty compact convex sets $B_{j}$ . By the hyperplane separation theorem, for each $j$ , there is a hyperplane $H_{j}$ separating $A_{j}$ and $B_{j}$ . I claim that $H_{j}$ has a convergent subsequence. Indeed, each $H_{j}$ must intersect the convex hull of $A_{1} \cup B_{1}$ , and the space of hyperplanes intersecting that convex hull is compact. So $H_{j}$ has a subsequence that converges to a hyperplane $H$ .

It’s easy to see that $H$ separates $A_{j}$ from $B_{j}$ for each $j$ , and so $H$ separates $A$ from $B$ . So $H$ must contain $V (p)$ and support $V (P)$ .

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Note that the theorem does not guarantee the existence of a Pareto optimum. But if $V (P)$ is closed and bounded, then a Pareto optimum exists.

A limitation of the theorem is that it assumes a finite list of values $v_{1}, \dots, v_{n}$ , not an infinite one.