Appendix to: A fungibility theorem

Suppose that is a set and we have functions . Recall that for , we say that is a Pareto improvement over if for all , we have . And we say that it is a strong Pareto improvement if in addition there is some for which . We call Pareto optimum if there is no strong Pareto improvement over it.

Theorem. Let be a set and suppose for are functions satisfying the following property: For any and any , there exists an such that for all , we have .

Then if an element of is a Pareto optimum, then there exist nonnegative constants such that the function achieves a maximum at .

Proof. Let . By hypothesis, the image is convex.

For , let the Pareto volume of be the set

This is a closed convex set. Note that is a Pareto optimum precisely when . Let's assume that this is the case; we just have to prove that maximizes for some choice of .

It suffices to find a hyperplane that contains and that supports . Then the desired function can be constructed by ensuring that is a level set.

If lies in a proper affine subspace of , let be the smallest such subspace. Let be the interior of in and let be the interior of . The case where is a point is trivial; suppose it's not, so is nonempty. By convexity, is the closure of and is the closure of .

Since is convex, is convex, and we can exhaust with a nested sequence of nonempty compact convex sets . And is convex, so we can exhaust with a nested sequence of nonempty compact convex sets . By the hyperplane separation theorem, for each , there is a hyperplane separating and . I claim that has a convergent subsequence. Indeed, each must intersect the convex hull of , and the space of hyperplanes intersecting that convex hull is compact. So has a subsequence that converges to a hyperplane .

It's easy to see that separates from for each , and so separates from . So must contain and support .

 

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

A limitation of the theorem is that it assumes a finite list of values , not an infinite one.

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10 comments, sorted by Click to highlight new comments since: Today at 3:39 PM

Couldn't this have been in a comment to the original post or on the wiki?

Would you prefer it that way?

Yeah, just to keep the list of posts tidy.

I'll do it that way in the future.

It suffices to find a hyperplane that contains ) and that supports ).

I assume you also want it to support ?

I love this proof, by the way.

Ah, you're right. The important thing is that it supports V(P). Thanks.

The definition of pv is the next line. Does that not show up for you?

Nope. In fact, the one I wrote no longer shows up for me in my comment either. How odd.

Does it show up for you in the html source?

Maybe an unescaped character is causing the trouble.

Problem appears to have been on my end. I can see both now.