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 a 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.
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
a 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.