It is advised to contact me if you wish to read this, the proofs aren't very edited.
Lemma 1: If is a Polish space and is a compact subset and , and , then there is a third bounded continuous function which fulfills the following three properties: First, . Second, . Third, .
To prove this, we will use the Michael selection theorem to craft a continuous function with these properties. Accordingly, let the set-valued function be defined as: if , , and if , then:
Assuming there was a continous function where , it'd get us our desired results. This is because:
And then, because is a selection from , we can make these quantities bigger by selecting from in the supremum as well, to get:
And then we can use the definition of in the two cases (the latter case is because implies ) to get:
And then we use the fact that we abbreviated the supremum of the difference between and as to get:
So, we'd have one of our three results, that . Our second desired result that mimicks on is trivial by the definition of . And finally,
And then, we do the ususual transition from f'' to the \psi function that it's a selection of,
And then, because is always when , and is always bounded in for the latter part, this turns into:
So we have our desired result of . As we have established that all three results follow from finding a continuous selection of , we just have to do that now. For this, we will be using the Michael selection theorem. In order to invoke it, we need to check that is paracompact (all Polish spaces are paracompact), that is a Banach space (it is), and verify some conditions on the sets produced.
We need that the function is nonempty for all . This is trivial by the definition of for , but for the other case, we need to verify that
Is a nonempty set. This is easy because witnesses nonemptiness. We need that the function is convex for all . This is easy because it's the intersection of two convex sets when , which is the trickier case to check. We also need that the function is closed for all , which is true because it's either a point or it's an intersection of two closed sets.
All that we're missing in order to invoke the Michael Selection theorem to get a continuous function that works is verifying lower hemicontinuity for the function .
Lower hemicontinuity is: If limits to , and , there's some subsequence and where , and limits to .
In order to show this, we can take three different cases.
The first possible case is where infinitely many of the lie in , so the limit point must lie in as well. Then, can be the subsequence which lies in , and can be the subsequence , which is the only possible choice of value that lies in . Due to continuity of , this obviously converges to , which is the only possible choice of value for because .
The second possible case is where only finitely many of the lie in , but yet the limit point lies in as well, like limiting to the border of a closed set from outside the closed set. Let be the subsequence where you get rid of all the x's in the approximating sequence that lie in . Notice that \, instead of being written as
Can be written as the single interval
And now, let be defined as:
Due to continuity of all these functions, and limiting to , the limiting value is:
Now, because
and
Therefore,
Therefore,
So our limiting y value reduces to:
Also, by similar arguments, we have:
and
Therefore
So, our limiting value reduces to merely , ie, our point selected from , and we've shown lower hemicontinuity in this case. That just leaves one last case left over, the case where .
Again, as before, in this case, is the interval
Let be how close the point is to the bottom ( for the bottom, or the top, when ). For your sequence limiting to , it's made by picking out all the which lie in , only finitely many, and the are given by:
Because of the continuity of the functions (supremum of two continuous functions) and (inf of two continuous functions), the limit to:
Which is just . So, we have lower hemicontinuity in this last case.
And therefore, we have lower hemicontinuity overall for , and so has a continuous selection function by the Michael Selection Theorem, and said selection function fulfills the requisite properties.
Lemma 2: For all and , if a functional has a set as an -almost-support, and has , then
where is the Lipschitz constant of .
Proof: Via Lemma 1, there's a function with the properties that:
Now, we can go:
And we're done. The critical steps in the second inequality were due to Lipschitzness of , and the fact that and agree on which is an -almost-support for , respectively.
Proposition 1: For a continuous function , the metric completely metrizes the set equipped with the subspace topology.
To do this, we'll need four parts. First, we need to show that is even a metric. Second, we'll need to show that for all open balls in the metric, you can fit an open ball in the original metric within it (so all the open sets induced by the metric were present in the original subspace topology). Second, we'll need to show that for all open balls in the original metric that lie within the support of , we can fit an open ball induced by the metric within it (so all the open sets in the original subspace topology can be induced by the metric), and parts 2 and 3 let us show that the metric induces the subspace topology on the support of . Finally, part 4 is showing that any Cauchy sequence in the metric is Cauchy in the original complete metric, so we can't have any cases of missing limits points, and showing that all Cauchy sequences in the support of , according to the metric, must have their limit point lying in the support of , so limits never lead outside of the support of . This then shows that is a complete metrization of the support of (it induces the same topology as the subspace topology), and all limit points lie in the same space,
So, to begin with, if is your original Polish space, pick a complete metrization of the space. Then, ensure that the maximum distance is 1 (this can always be done while preserving the exact same Cauchy sequences, and not affecting the topology in any way, and it's still a metric), and call that .
Also, there's some continuous function that you'll be updating. The set is defined as , and is clearly an open subset of because the preimage of (an open set) must be open due to the continuity of .
Now, define
To show it's a metric, we first need symmetry (which is obvious, because and and absolute value of difference are all symmetric). For identity of indiscernibles, the forward direction, the fact that is bounded below 1 proves that , always, so for to be 0, then , so . For the reverse direction, must be 0, because the equation for distance would reduce to .
That just leaves the triangle inequality, and here, we'll critically use the fact that the original metric was clipped so it's never above 1. Our goal is to show that
Without loss of generality, we can assume that (otherwise flip and ), so we can split into three cases, which are:
For the first case, we have:
For the second case, we can do the same exact argument, just replace that in the second line with . The third case is the tricky one, we'll work backwards. Remember that our inequalities are:
Now, let's proceed.
Then we do some regrouping, to yield:
At point, we remember that because we bounded our initial distance, and because of the problem case we're in, to get:
And then, we just remember that in this problem case, to get:
And we're done with the triangle inequality, so d|L is indeed a metric.
Well, is it a complete metric for ? Well, by looking at the definition of , and remembering that is always 1 or more because the likelihood function is bounded in , we have that:
When and are in the support of , so any Cauchy sequence in must also be Cauchy in . Now, either a Cauchy sequence has its limit point also lying in the support of , in which case we're good, or it has its limit point lying on the edge of the support of (and outside the set), where . However, in that case, the sequence in cannot be Cauchy, because would diverge to infinity due to the continuity of and the limit point being on the edge where , so it's impossible that you could have some finite point that'd be close to all the rest, the absolute value term in the distance would forbid Cauchyness. So, it's a complete metric for the support of .
All that remains is to show that it induces the same topology as the subspace topology that the open set should have. Because the support of is itself open, and the space is metrized by the distance metric , any open set in the support of can be written as the intersection of an open set in and the open set , and so it's open in , and can be written as the union of tiny open balls in the original distance metric.
So, we'll proceed by showing that any open ball (w.r.t. the metric) within the support of , can fit some open ball (w.r.t. the metric) in it that engulfs the center point, and any open ball (w.r.t. the metric) within the support of , can fit some open ball (w.r.t. the metric) in it that engulfs the center point.
If we can do it, than because any open set in the subspace topology can be written as the union of a bunch of open balls centered at each point in the open set according to the metric, and each of those open balls according to the metric has their center point engulfed by a smaller open ball according to the metric, we'd be able to build the open set in the subspace topology out of a union of open sets in the induced topology.
And also, any open set in the -induced topology can be written as a union of infinitely many open balls centered at each relevant point, and because each of those open balls has their center point engulfed by a smaller open ball according to the metric, so we can build any open set in the -inducd topology out of a union of open sets in the original topology.
The net result is that the induced topology and the subspace topology have the same open sets. So, let's get working on showing that we can fit the two sorts of balls inside each other.
In one direction, if you have a ball of size according to the metric , then because always, a ball of size according to the metric centered at the same point will fit entirely within the original ball, so we have one half of our topology argument done.
In the other direction, if you have a ball of size according to the metric , around some point , then there's some distance around where the function only varies by .
At this point, we can then fit a ball of size around the point (according to the original metric), and it will lie within the ball of size around the point according to the metric. The reason for this is:
And then, because we selected our distance between and so that and only differ by at most (because a ball of size suffices to accomplish this), we have:
And then, because and is very small, we have
So, this small of a ball in the original distance metric suffices to slot entirely inside a ball of size centered at in the metric.
And that's all we need!
Theorem 1: The set of infradistributions (set form) is isomorphic to the set of infradistributions (functional form). The part of the isomorphism is given by , and the part of the isomorphism is given by , where and is the convex conjugate of .
Our first order of business is establishing the isomorphism. Our first direction is to and back is exactly. By upper completion, and reproved analogues of Proposition 2 and Theorem 1 from "Basic inframeasure theory", which an interested party can reprove if they want to see it, we can characterize as
And then, our can further be reexpressed as
Also, , so we can rewrite this as:
and, by the definition of the convex conjugate, and the space of finite signed measures being the dual space of , and being a functional applied to an element, this is...
So, our original set is identical to the convex-conjugate set, when we go from to back to a set of sa-measures.
Proof Phase 2: In the reverse direction for isomorphism, assume that fulfills the conditions (we'll really only need continuity and concavity)
We want to show that
\mathbb{E}_{\{(m,b)|b\ge(h')^{*}(m)\}}(f)=h(f)
Let's begin.
Given an , we have a natural candidate for minimizing the , just set it equal to . So then we get
And this is just... , and, because is continuous over , and concave, then is continuous over , and convex, so . From that, we get
and we're done with isomorphism.
So, in our first direction, we're going to derive the conditions on the functional from the condition on the set, so we can assume nonemptiness, closure, convexity, upper completion, projected-compactness, and normalization, and derive monotonicity, concavity, normalization, Lipschitzness, and compact almost-support (CAS) from that.
For monotonicity, remember that all points in the infradistribution set are a-measures, so if , then
We could do that because all the measure components are actual measures.
For concavity,
And we're done with that. For normalization,
And
So we have normalization.
For Lipschitzness, we first observe that compact-projection (the minimal points, when projected down to their measure components, make a set with compact closure) enforces that there's an upper bound on the value of a minimal point , because otherwise you could pick a sequence with unbounded , and it'd have no convergent subsequence of measures, which contradicts precompactness of the minimal points projected down to their measure components.
Then, we observe that points in correspond perfectly to hyperplanes that lie above the graph of , and a minimal point is "you shift your hyperplane down as much as you can as much as you can until you can't shift it down any more without starting to cut into the function ". Further, for every function , you can make a hyperplane tangent to the function at that point by the Hahn-Banach theorem, which must correspond to a minimal point.
Putting it together, the hypograph of is exactly the region below all its tangent hyperplanes. And we know all the tangent hyperplanes correspond to minimal points, and their Lipschitz constants correspond to the value of the minimal points. Which are bounded. So, Compact-Projection in implies is Lipschitz.
Finally, we'll want compact almost-support. A set of measures is compact iff the amount of measure is upper-bounded, and, for all , there is a compact set where all the measures have measure outside of .
So, given that the set of measures corresponding to is compact by the compact-projection property, we want to show that the functional has compact almost-support. To do this, we'll observe that if is the inf of a bunch of functions, and all functions think two different points are only a little ways apart in value, then must think they're only a little distance apart in value. Keeping that in mind, we have:
And then, we can think of a minimal point as corresponding to a hyperplane and , and is the inf of all of them, so to bound the distance between these two values, we just need to assess the maximum size of the gap between those values over all minimal points/tangent hyperplanes. Thus, we can get:
And then, we can do some canceling and get:
And then, because f' was selected to be 0 on , which makes up all but of the measure for all measures present in , we can upper-bound by , so we have that
And so, is a compact -almost-support for , and this argument works for all , so is CAS, and that's the last condition we need. Thus, if is an infradistribution (set form), the expectation functional h is an infradistribution (expectation form).
Now for the other direction, where we assume monotonicity, concavity, normalization, Lipschitzness, and CAS on an infradistribution (expectation form) and show that the induced form fulfills nonemptiness, convexity, closure, upper completion, projection-compactness, normalization, and being a set of a-measures.
Remember, our specification of the corresponding set was:
Where is the function given by , and is the convex conjugate of .
First, being a nonempty set of a-measures. Because there's an isomorphism linking points of the set and hyperplanes above the graph of , we just need to establish that no hyperplanes above the graph of can slope down in the direction of a nonnegative function (as this certifies that the measure component must be an actual measure), and no hyperplanes above the graph of can assign 0 a value below 0 (as this corresponds to the term, and can be immediately shown by normalization).
What we do is go "assume there's a where the linear functional corresponding to isn't a measure, ie, there's some nonnegative function where ". Well, because of monotonicity for (one of the assumed properties), we have .... And, because all affine functionals are made by taking a linear functional and displacing it, ..., decreases at a linear rate, so eventually the hyperplane and cross over, but was assumed to be above always, so we have a contradiction.
Therefore, all hyperplanes above must have their linear functional component corresponding to an actual measure, ie, being an a-measure. And we get nonemptiness from the concavity of , so we can pick any function and use the Hahn-Banach theorem to make a tangent hyperplane to h that touches at that point, certifying nonemptiness.
By the way, the convex conjugate, , can be reexpressed as .
For closure and convexity: By monotonicity of and normalization, , and is continuous (Lipschitz) on , and concave, so is proper, continuous on , and convex, so, by the Wikipedia page on "Closed Convex Function", is a closed convex function, and then by the Wikipedia page on "Convex Conjugate" in the Properties section, is convex and closed. From the Wikipedia page on "Closed Convex Function", this means that the epigraph of is closed, and also the epigraph of a convex function is convex. This takes care of closure and convexity for our .
Time for upper-completeness. Assume that lies in the epigraph. Our task now is to show that lies in the epigraph. This is equivalent to showing that . Let's begin.
And we're done.
Normalization of the resulting set is easy. Going from to a (maybe)-inframeasure back to is identity as established earlier, so all we have to do is show that a failure of normalization in a (maybe)-inframeasure makes the resulting not normalized. Thus, if our is normalized, and it makes an that isn't normalized, then going back makes a non-normalized , which contradicts isomorphism. So, assume there's a failure of normalization in . Then , or , so either or and we get a failure of normalization for which is impossible. So must be normalized.
That just leaves compact-projection. We know that a set of measures is precompact iff there's a bound on their values, and for all , there's a compact set where all the measure components have measure outside of that set.
First, we can observe that no hyperplane above can have a Lipschitz constant above the maximal Lipschitz constant for the function , because if it increased more steeply in some direction, you could go in the other direction to decrease as steeply as possible, and would be constrained to decrease strictly less steeply in that direction, so if you went far enough in that direction, your hyperplane and would intersect, which is impossible. Thus, Lipschitzness of enforces that there can be no point in the set with too much measure, which gives us one half of compact-projection for .
For the other half, CAS for ensures that for all , there is a compact set where
What we'll do is establish that no hyperplane lying above can have a slope more than in the direction of a function that's in and is 0 on . Let be such a function fulfilling those properties that makes some hyperplane above h slope down too hard. Then
Because going in the direction of a negative function decreases your value, so the Gateaux derivative would be negative, and is in , and we have CAS on .
Now, we can realize that as we travel from 0 to to to , our vector of travel is always in the direction of , which can't be too negative. Each additional added drops the value of by at most . However, each additional added drops the value of (our assumed functional that's sloping down too hard in the direction) by more than that quantity, so eventually will cross over to be lower than , so can't correspond to an a-measure in , and we have a contradiction.
Therefore, regardless of the point in , its measure component must assign any function that's 0 on and bounded in a value of at most. We can then realize that this can only happen to a measure that assigns or less measure to the set (otherwise you can back off from your continuous function to a discontinuous indicator function).
Thus, given our , we've found a compact set where the measure component of all points in assign or less value to the outside of that set, and this can be done for all , certifying the last missing piece for compact-projection of (because the projection is precompact iff the set of measures is bounded above in amount of measure present, and for all , there's a compact set where all the measures assign measure outside of that set.
And that's the last condition we need to conclude that the set form of an infradistribution (functional form) is an infradistribution (set form), and we're done.
Proposition 2: For any infradistribution there is a unique closed set which is the intersection of all supports for , and is a support itself.
Proof sketch: The proof splits into three parts. First, we show that if is a support and is an -almost support, then is a -almost support, by picking two functions which agree on and using them to construct a third function which has similar expectations to both of them, due to agreeing with the first function on and agreeing with the second function on , so our arbitrary functions which agree on the intersection have similar expectation values. A support is precisely an -almost support for all , so actual supports are preserved under finite intersection.
Second, we show that supports are preserved under countable intersections, which critically relies on compact -almost supports for all . Roughly, that part of the proof proceeds by using our compact almost-support in order to make a sequence of compact almost-supports nested in each other, and showing they converge in Hausdorff distance, so two functions which agree on the intersection of all the compact sets are very close to agreeing on some finite almost-support and have similar expectation values, so the intersection of compact sets is an almost-support. The property of being an almost-support is hereditary upwards, and we can show our countable intersection of compact sets is a subset of the countable intersection we're interested in, so it's an almost-suppport. However, we can shrink to 0 so it's a support.
Finally, we show that supports are preserved under arbitrary intersections by using strong Lindelofness of (because it's Polish) to reduce the arbitrary intersection case to the countable intersection case which has already been solved.
For the finite intersection case, let be a support and be an -almost support. Our task is to show that is an -almost support, by showing that two functions which agree on it can't be more than apart in expectation value.
Pick any two functions and where . Let's define a on as follows. If , then . If , and , then . Our first task is to show that is continuous.
If limits to , we can split into three cases. Our first case is where , and . Then past a certain n, will always not be in (complement of a closed set is open), and so its values are dictated by , which is continuous, and we have continuity of at that point. The second case where and is symmetric, but with the values being dictated by instead. Our third case where is the only one which takes some care. As limits to , then both and (not the restrictions, the original functions!) limit to (because and agree on ) so no matter how our toggles back and forth between and , we have continuity.
Now that we know is continuous on , we can use the Tietze Extension Theorem to extend to a continuous function on the entire space , because the union of two closed sets is closed, and we can ensure that the extension stays in as well. (because is copying either or ), call this extension .
Now, because
and is a support, then . However, since
(because and agree on ), then
The last inequality is because was selected to be bounded in the relevant range, and the first inequality is because and are identical on which is a -almost-support.
Thus,
That last inequality was because and agree on which is a support, and for the latter, we already derived that inequality. Because and were arbitrary functions which agreed on , and was an arbitrary support and was an arbitrary -almost-support, we have that the intersection of any support and -almost-support is an -almost support. As a special case of this, the intersection of two supports is a support, so being a support is preserved under finite intersection.
Now for supports being preserved under countable intersection. Fix a countable family of supports, . We will show that is a support.
First, by our "compact almost-support" property, for all , there's a compact set where any two functions agreeing on will only differ in expectation by . Fix an arbitrary to use through the rest of the argument, which induces a compact set .
Let be defined as , and . Polish spaces are Hausdorff, so the intersection of compact sets with closed sets is compact. Therefore, the are compact and all nested in each other. All the are -supports, because they're the intersection of finitely many supports (which by the earlier result is a support) intersected with an -support.
We will show that converge to in Hausdorff-distance. Assume that they don't converge in Hausdorff-distance. Then there's some where all have a point more than distance away from the set . We can pick a sequence of those points from the , and they're all in which is compact, so we can isolate a convergent subsequence. Then, for all i, the subsequence enters the closed set and doesn't leave, so the limit point must be in all the and thus in , contradicting that the sequence of points is always away from . This shows that the converge to in Hausdorff-distance.
At this point fix a and which agree with each other on . and restricted to any (which is compact) must be uniformly continuous, and as we restrict to smaller and smaller , the function saying how close two points need to be for and to change little gets more tolerant.
Pick any you want. By uniform continuity for the restrictions of and to a compact set we can find some where points in (a superset of all later ) must be that close for and to only vary by between the two points. And for any there's some i where is within Hausdorff-distance of because they limit to each other in Hausdorff-distance. Any point in is only away from , so and are only or less apart from each other on suitably late .
By Lemma 2, since and are only away from each other on which is an -almost support, we have:
And, because was arbitrary, we can take the limit as it approaches 0 to get:
Also, and were arbitrary functions which agreed on , so is an -almost-support.
Being an almost-support is hereditary upwards, because two functions which agree on a superset agree on a subset and are therefore close. So, is an -almost-support. But was arbitrary, so it's a support, and we have that countable intersections of supports are supports.
Finally, we'll show that any uncountable intersection of supports is identical to a countable intersection of supports. Let's say you intersect closed sets which are supports. Biject that with the ordinal by the well-ordering principle, so now all supports are indexed like with . Our task is now to write as a countable intersection of closed supports.
Separable metric spaces are strongly Lindelof, which implies that every open set can be written as a countable union of open sets from a countable basis. By taking the complement, we can get the converse. Every closed set can be written as a countable intersection of closed sets from a countable basis. Call the countable basis closed sets .
Now, define . It is the family of basis closed sets which is a subset of. Because every closed set is the intersection of closed sets from the basis, .
Now, let's consider
Ie, the set made by intersecting all basis closed sets which contain some within them. We will show that this equals , our arbitrary intersection of choice.
In one direction,
Now, assume . From this, we derive . The i was arbitrary, so we've derived:
Which is the same as:
which is the same as:
which is the same as:
which is the same as:
So, there's one direction done, we just showed that:
For the other direction, observe that
Now, because ,
which is equivalent to:
Swapping the quantifiers, we get:
Which can be reexpressed as:
And by our work in the first direction of trying to establish equality between the two sets (specifically, all our equivalences towards the end), this is the same as:
So, we have our bidirectional implication and equality,
We're trying to show the former set is a support. It can be written as the intersection of a bunch of by our equality above, but how does that help us? Well, if an is present in , then it's present in some . Which means that it's a closed superset of the corresponding . Which, by assumption, is a support. So, all our that we're intersecting are supports, since the property of being a support is closed upwards. There are countably many , so we have rewritten our uncountable intersection of supports as a countable intersection of supports, which we know from earlier is a support.
Therefore, the intersection of all supports is a support.
Proposition 3: for all iff all minimal points in have .
Proof sketch: From LF-duality, we recall our usual technique to prove things about minimal points. If you have a proposed minimal point violating a property, which corresponds to a hyperplane , and we can find a hyperplane where , and isn't itself, then isn't an actual minimal point. We'll take a where , and undershoot it. The reverse direction of the iff proof is much easier.
Proof: Assume that there's a minimal point in with . This corresponds to a minimal hyperplane where and . Now, consider the sets and which are defined as follows:
the set is the hypograph of , and the set is like the convex hull of the region above , and the point , but not the actual point itself.
is the hypograph of a concave continuous function, so it's obviously convex and closed. As for , it is routine but rather tedious to verify that it is convex, this is left for the reader. Further, is open. Given any point in it, we can make a tiny little open ball around where everything in that ball has , and scale down the ball by multiplying by . Because p isn't 1, the ball doesn't collapse to a point when scaled, so this gets you a tiny little open ball around your arbitrary point in , showing it's open.
Now, we must show that and are disjoint to invoke Hahn-Banach. Assume there's some point that lies in both of them at once. Then, by the defining condition of , and and . And also, since the hyperplane corresponding to lies above the graph of , . Putting these together, we get:
The first strict inequality is because isn't 1 or higher, and . The is because , the critical equality is because we're assuming that . And the is because of the defining condition of . This is impossible, we just showed a number is above itself. So and are disjoint.
and are disjoint, both convex, and is open, so we can invoke Hahn-Banach and separate them with a hyperplane . At the start we assumed that was a minimal point, and , so if we can show that over , and , then isn't minimal and we have a contradiction, so all minimal points must map 0 to 0 (it can't be less because any hyperplane corresponding to a minimal point must lie above h and by normalization)
Now, assuming that there is a where , then lies in because and . Thus, cuts into the set , but it doesn't because it's a separating hyperplane, so we have a contradiction and over . Also, since separates and , it doesn't cut into the set , which is the hypograph of , so .
Now that we have , all we need is to show that to wrap things up. First, and , so . Second, there are points arbitrarily close to that lie within the set , by letting be very very close to 1. So, if , then would be strictly below , which is incompatible with it having points arbitrarily close to it that lie within . Thus, .
Hang on, is a minimal and , so . Also, . Thus, isn't minimal and we have a contradiction. The flaw must be with our very first assumption, that there's a minimal point with . So all minimal points must have , and we're done with the first half, implies all minimals have .
In the other direction, assume all minimal points have . Then, for
And we've shown the equivalence.
Proposition 4: iff all minimal points in have .
We know from past proofs in Basic Inframeasure Theory that the maximum Lipschitz constant of a function is the maximum value of one of the minimal points, so this is immediate.
Proposition 5: iff all minimal points in have .
Proof sketch: This is very similar to the proof of proposition 3 of homogenity, the same proof path occurs, except we assume there's a minimal point with and derive a contradiction from that, and then it's easy to clean up the other direction from there.
Proof: Assume that there's a minimal point in with . This corresponds to a minimal hyperplane where and . Now, consider the sets and which are defined as follows:
The set is the hypograph of , and the set is like the convex hull of the region above , and the point , but not the actual point itself.
is the hypograph of a concave continuous function, so it's obviously convex and closed. As for , it is routine but rather tedious to verify that it is convex, this is left for the reader. Further, is open. Given any point in it, we can make a tiny little open ball around where everything in that ball has , and scale down the ball by multiplying by and adding which shifts it. Because isn't 1, the ball doesn't collapse to a point when scaled, so this gets you a tiny little open ball around your arbitrary point in , showing it's open.
Now, we must show that and are disjoint to invoke Hahn-Banach. Assume there's some point that lies in both of them at once. Then, by the defining condition of , and and . And also, since the hyperplane corresponding to lies above the graph of , . Putting these together, we get:
The first strict inequality is because isn't 1 or higher, and . The is because . The equality in the first line is just expanding accordingly. The critical equality that starts off the second line is because our infradistribution is cohomogenous, so , and we specialize this accordingly. Then the next equality is just expanding, canceling out like terms, and reexpressing, and the final is because of the defining condition of . As a whole, this is impossible, we just showed a number is above itself. So and are disjoint.
A and B are disjoint, both convex, and is open, so we can invoke Hahn-Banach and separate them with a hyperplane . At the start we assumed that was a minimal point, and , so if we can show that and , then isn't minimal and we have a contradiction, so all minimal points must map 1 to 1 (it can't be less because any hyperplane corresponding to a minimal point must lie above and )
Now, assuming there is a where so lies in because and . Thus, cuts into the set , but it doesn't because it's a separating hyperplane, so we have a contradiction and . Also, since separates and , it doesn't cut into the set , which is the hypograph of , so .
Now that we have , all we need is to show that to wrap things up. First, and , so . Second, there are points arbitrarily close to that lie within the set , by letting be very very close to 1. So, if , then would be strictly below , which is incompatible with it having points arbitrarily close to it that lie within . Thus, .
Hang on, is a minimal and , so . Also, . Thus, isn't minimal and we have a contradiction. The flaw must be with our very first assumption, that there's a minimal point with . So all minimal points must have , and we're done with the first half, implies all minimals have .
In the other direction, assume all minimal points have . Then
And we've shown the equivalence.
Proposition 6: iff all minimal points in H have iff .
Proof sketch: To go from to "all minimal points have " requires a contrapositive proof where we assume that there are minimal points with , and show it's incompatible with . Showing that for all minimal points implies the full form of C-additivity only takes some equality shuffling, and then it's trivial that implies , just let be 0.
Proof: Assume there's a minimal point with . Then the corresponding hyperplane slopes down at a rate of starting at 0 in the direction while h only slopes down at a rate of -1 in that same direction due to . So, eventually, the hyperplane crosses the graph of , witnessing that it can't actually be a hyperplane above the graph of and we have a contradiction. Similar arguments dispatch the existence of a minimal point with , because the hyperplane slopes up at a rate of in the direction starting at 0, while slopes up at a rate of 1 in that same direction due to , so eventually the hyperplane crosses the graph of and we have another contradiction. So, implies all minimal points have .
In the other direction,
Finally, it's trivial that implies .
Proposition 7: iff all minimal points in have and .
This proof will proceed by showing that iff both homogenity and C-additivity hold. We know from earlier that homogenity and C-additivity are equivalent to and respectively, so if we can show "crispness iff homogenity and C-additivity", we're done.
In one direction, implies both homogenity and crispness by taking and in the two respective cases. In the other direction, by homogenity and C-additivity, and we've derived crispness, so we're done.
Proposition 8: iff the set of minimal points of corresponds to the set of probability distributions supported on .
In one direction, if the set of minimal points entirely consists of probability distributions supported on and only those, then
In the other direction, we can take any infradistribution that maps a function to its minimum value over and use the above equalities to conclude that the infradistribution generated by all probability distributions supported over perfectly duplicates the expectation values of the infradistribution, and so the two sets must be equivalent modulo closure, convex hull, and upper completion. The space of distributions over is closed and convex, and so the same applies to the upper completion of it, and it equals the infradistribution set corresponding to .
Proposition 9: All sharp infradistributions are extreme points in the space of infradistributions.
Assume we can make a sharp infradistribution corresponding to a compact set as a probabilistic mix of crisp infradistributions. If one of the crisp infradistributions has a minimal point (probability distribution) which isn't supported on , then you could pick that minimal point, and mix it with other minimal points/probability distributions to make a minimal point/probability distribution in the set which has some support outside of , which is impossible, because is just .
Therefor, all crisp infradistributions which mix to make the sharp infradistribution induced by must have all their probability distribution minimal points supported entirely on .
Now, pick any point . Consider the dirac-delta distribution on , . It lies in , so . This set is a mix of the other sets corresponding to crisp infradistributions, so there must be a probability distribution from each of the crisp infradistributions that mix to make . This can only happen if every component in the mix is itself.
Thus, given any , all the crisp infradistributions that mix to make the sharp infradistribution induced by must include in them. Thus, all crisp infradistributions that mix to make a sharp infradistribution must contain all the dirac-deltas of points in within them, and by closure and convexity, all the crisp infradistributions contain all of within their minimal points.
So, all these crisp infradistributions that mix to make our sharp lack every probability distribution that isn't supported on , and contain every probability distribution that's supported on , so they're all equal to our sharp infradistribution itself. Thus, is extreme in the space of crisp infradistributions.
This argument works for any sharp infradistribution, so they're all extreme in the space of crisp infradistributions.