Lemma 3: If $F$ is a continuous function of type $X\to K\left(Y\right)$, where $K\left(Y\right)$ is the space of nonempty compact subsets of the space $Y$, then given any compact set $C^X\subseteq X$, ${\bigcup }_{x\in C^X}F\left(x\right)$ will be compact in $Y$.

Fix some compact set $C^X\subseteq X$, and continuous function $F:X\to K\left(Y\right)$. We will operate by taking an arbitrary open cover of ${\bigcup }_{x\in C^X}F\left(x\right)$ and finding a finite subcover.

Let $\{O_i{\}}_{i\in I}$ be an open cover of ${\bigcup }_{x\in C^X}F\left(x\right)$. The $O_i$ are subsets of $Y$. The topology compatible with Hausdorff distance on $K\left(Y\right)$ (space of compact subsets of $Y$) is the Vietoris topology, where the basis opens are given by finite collections of open sets in $Y$. You take the set of all compact subsets of $Y$ which are subsets of the union of your finite collection of open sets, and intersect every open set in your finite collection

Accordingly, let $J=P_{fin}\left(I\right)$ (the set of all finite subsets of $I$, the index set for our open cover), and fix a collection of open sets in $K\left(Y\right)$, $\{O_j{\}}_{j\in J}$. The sets $O_j$ are defined as:

$O_j:=\{C^Y\in K(Y)|C^Y\subseteq {\bigcup }_{i\in j}{O}_i\wedge \forall i\in j:C^Y\cap O_i\ne \varnothing \}$

Now, all the $F\left(x\right)$ with $x\in C^X$ are compact ($F$ produces compact sets as output), and they are all subsets of ${\bigcup }_{x\in C^X}F\left(x\right)$, so $\{O_i{\}}_{i\in I}$ is a cover of $F\left(x\right)$, and due to its compactness we can identify a finite subcover, and prune away every open st which doesn't intersect $F\left(x\right)$. $F\left(x\right)$ is a subset of the union of those finitely many open sets, and intersects all of them, so the point $F\left(x\right)\in K\left(Y\right)$ lies in the open set $O_j$ induced by that finite cover of open sets.

This argument works for arbitrary $F\left(x\right)$ with $x\in C^X$, so the collection $\{O_j{\}}_{j\in J}$ is an open cover of $F\left(C^X\right)$. Also, because $F$ is continuous and $C^X$ is compact, $F\left(C^X\right)$ is compact, so we can identify a finite subcover from $\{O_j{\}}_{j\in J}$.

Then, consider the collection of open sets $O_i$ where $i\in j$ for some $O_j$ which is part of the finite cover of $F\left(C^X\right)$. This is finitely many opens, we're unioning together finitely many (finitely many $O_j$ selected) finite sets of open sets (each $O_j$ is associated with finitely many $O_i$ that it was built from).

Now we just have to show that this collection covers ${\bigcup }_{x\in C^X}F\left(x\right)$, and we'll have made our finite subcover and shown that said set is compact. Assume our finite collection of opens doesn't cover the set. Then there's some $F\left(x\right)$ which wasn't covered completely. However, the point corresponding to $F\left(x\right)$ in $K\left(Y\right)$ lies in some $O_j$, and from its definition, the corresponding $O_i$ manage to cover $F\left(x\right)$, and we have a contradiction. We're done.

Proposition 19: $h⋉K$ is an infradistribution, and preserves all properties indicated in the diagram at the start of this section if $h$ and all the $K\left(x\right)$ have said property.

To show this, we'll verify that it's well-defined at all, normalization, monotonicity, concavity, Lipschitzness, compact almost-support, and preservation of the properties.

$(h⋉K)\left(f\right):=h(\lambda x.K(x\left)\right(\lambda y.f(x,y)\left)\right)$

Our first order of business is verifying that

$\lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right)$

is even a continuous function to be able to show that $h$ can accept it as input.

For continuity, let $x_n$ limit to $x$, and we'll try to show that $K\left(x_n\right)(\lambda y.f(x_n,y\left)\right)$ limits to $K\left(x\right)(\lambda y.f(x,y\left)\right)$. Let ${\lambda }_K^{\odot }$ be the Lipschitz constant upper bound of $K$.

Pick an $ϵ$, we'll show that there's some $m_0$ where

$\forall n\forall m\ge m_0:|K\left(x_n\right)(\lambda y.f(x_m,y\left)\right)-K\left(x_n\right)(\lambda y.f(x,y\left)\right)|\le ϵ(2||f||+{\lambda }_K^{\odot })$

First, note that $\{x_n{\}}_{n\in N}\cup \{x\}$ is a compact set because $x_n$ limits to $x$. Thus, by the compact-shared compact almost-support condition on an infrakernel, there must be some compact set $C_{ϵ}\subseteq Y$ where all the $K\left(x_n\right)$ agree that functions $f,f^{\prime }$ agreeing on $C_{ϵ}$ have values only $ϵd(f,f^{\prime })$ apart from each other.

Now, because f is a continuous bounded function $X\times Y\to R$, it's uniformly continuous when restricted to 

$\left(\right\{x_n{\}}_{n\in N}\cup \left\{x\right\})\times C_{ϵ}$

as this is the product of two compact sets and is compact. Due to the uniform continuity of $f$ restricted to that set, there is some number $\delta$ where points only $\delta$ apart in that set have their values only differing by $ϵ$. Further, there is some number $m_0$ where, for all $m\ge m_0$, $d(x_m,x)<\delta$.

Additionally, the maximum difference between $\lambda y.f(x,y)$ and $\lambda y.f(x^{\prime },y)$ is $2||f||$.

Now that we know our number $m_0$ we can pick an arbitrary $m$ above it, and go:

$\forall m\ge m_0\forall y\in C_{ϵ}:d\left(\right(x_m,y),(x,y\left)\right)=d(x_m,x)\le \delta$

$\forall m\ge m_0\forall y\in C_{ϵ}:|f(x_m,y)-f(x,y)|\le ϵ$

$\forall m\ge m_0:d\left(\right(\lambda y.f(x_m,y){)}_{\downarrow C_{ϵ}},(\lambda y.f(x,y\left){)}_{\downarrow C_{ϵ}}\right)\le \delta$

And now, because these two functions restricted to $C_{ϵ}$ are only $ϵ$ apart, we can apply Lemma 2 to conclude that (since $C_{ϵ}$ and ${\lambda }_K^{\odot }$ work for all the $K\left(x_n\right)$)

$\forall n:|K\left(x_n\right)(\lambda y.f(x_m,y\left)\right)-K\left(x_n\right)(\lambda y.f(x,y\left)\right)|\le ϵ\cdot 2||f||+ϵ{\lambda }_K^{\odot }$

This argument works for any $m\ge m_0$, so we have:

$\exists m_0\forall n\forall m\ge m_0:|K\left(x_n\right)(\lambda y.f(x_m,y\left)\right)-K\left(x_n\right)(\lambda y.f(x,y\left)\right)|\le ϵ(2||f||+{\lambda }_K^{\odot })$

Letting $n=m$ in particular,

$\exists m_0\forall n\ge m_0:|K\left(x_n\right)(\lambda y.f(x_n,y\left)\right)-K\left(x_n\right)(\lambda y.f(x,y\left)\right)|\le ϵ(2||f||+{\lambda }_K^{\odot })$

And for each \eps we can construct a m_0 in this way, concluding that

${lim}_{n\to \infty }|K\left(x_n\right)(\lambda y.f(x_x,y\left)\right)-K\left(x_n\right)(\lambda y.f(x,y\left)\right)|=0$

Also, from our pointwise convergence condition on infradistributions,

${lim}_{n\to \infty }K\left(x_n\right)(\lambda y.f(x,y\left)\right)=K\left(x\right)(\lambda y.f(x,y\left)\right)$

Therefore,

${lim}_{n\to \infty }K\left(x_n\right)(\lambda y.f(x_n,y\left)\right)=K\left(x\right)(\lambda y.f(x,y\left)\right)$

and so, we now know that

$\lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right)$

is a continuous function $X\to R$. For boundedness, upper and lower bounds on ${\lambda }^y.f(x,y)$ are $||f||$ (and the negative version of it). Due to the shared Lipschitz constant on the $K\left(x\right)$, an upper and lower-bound on $\lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right)$ is ${\lambda }_K^{\odot }||f||$ (and the negative version.) Thus, we can safely feed said function into the infradistribution $h$, so the semidirect product is well-defined. We must still show that it makes an infradistribution.

For normalization,

$(h⋉K)\left(1\right)=h(\lambda x.K(x\left)\right(\lambda y.1\left)\right)=h\left(\lambda x.1\right)=1$

$(h⋉K)\left(0\right)=h(\lambda x.K(x\left)\right(\lambda y.0\left)\right)=h\left(\lambda x.0\right)=0$

For monotonicity, if $f^{\prime }\ge f$,

$\forall x:\lambda y.f^{\prime }(x,y)\ge \lambda y.f(x,y)$

$\forall x:K\left(x\right)(\lambda y.f^{\prime }(x,y\left)\right)\ge K\left(x\right)(\lambda y.f(x,y\left)\right)$

$\lambda x.K\left(x\right)(\lambda y.f^{\prime }(x,y\left)\right)\ge \lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right)$

$(h⋉K)\left(f^{\prime }\right)=h(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)\ge h(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)=(h⋉K)\left(f\right)$

For concavity,

$(h⋉K)(pf+(1-p\left)f^{\prime }\right)=h(\lambda x.K(x\left)\right(\lambda y.pf(x,y)+(1-p)f^{\prime }(x,y)\left)\right)$

$\ge h(\lambda x.pK(x\left)\right(\lambda y.f(x,y))+(1-p\left)K\right(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)$

$\ge ph(\lambda x.K(x\left)\right(\lambda y.f(x,y)\left)\right)+(1-p)h(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)$

$=p(h⋉K)\left(f\right)+(1-p)(h⋉K)\left(f^{\prime }\right)$

In order, this was the definition of the semidirect product, all the $K\left(x\right)$ being concave so splitting them up produces a lower value (and then monotonicity for $h$), then $h$ being concave.

This leaves Lipschitzness and CAS. For Lipschitzness, given some $f$ and $f^{\prime }$, and letting ${\lambda }_h^{\odot }$ be the Lipschitz constant of $h$, we have:

$|(h⋉K)\left(f\right)-(h⋉K)\left(f^{\prime }\right)|=|h(\lambda x.K(x\left)\right(\lambda y.f(x,y)\left)\right)-h(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)|$

$\le {\lambda }_h^{\odot }d(\lambda x.K(x\left)\right(\lambda y.f(x,y)),\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)$

$={\lambda }_h^{\odot }{sup}_x|K\left(x\right)(\lambda y.f(x,y\left)\right)-K\left(x\right)(\lambda y.f^{\prime }(x,y\left)\right)|$

$\le {\lambda }_h^{\odot }{sup}_x{\lambda }_K^{\odot }d(\lambda y.f(x,y),\lambda y.f^{\prime }(x,y\left)\right)={\lambda }_h^{\odot }{\lambda }_K^{\odot }{sup}_{x}d(\lambda y.f(x,y),\lambda y.f^{\prime }(x,y\left)\right)$

$\le {\lambda }_h^{\odot }{\lambda }_K^{\odot }d(f,f^{\prime })$

Thus, that final thing shows that there's a finite Lipschitz constant for $h⋉K$.

This leaves compact almost-support. Pick any $ϵ$. This induces a compact set $C_{ϵ}^X$ which is an $ϵ$-almost-support for $h$, and then this compact set induces a compact set $C_{ϵ}^Y$ which an $ϵ$-almost-support for all the $K\left(x\right)$ where $x\in C_{ϵ}^X$. Now, we can apply Lemma 2 to go:

$|h(\lambda x.K(x\left)\right(\lambda y.f(x,y)\left)\right)-h(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)|$

$\le ϵd(\lambda x.K(x\left)\right(\lambda y.f(x,y)),\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)$

$+{\lambda }_h^{\odot }d\left(\right(\lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right){)}_{\downarrow C_{ϵ}^X},(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left){)}_{\downarrow C_{ϵ}^X}\right)$

Pretty much, that first part is the "$C_{ϵ}^X$ is an $ϵ$-almost-support for $h$" piece, and the second piece is the "hey, these two functions may be a bit different on said compact set, we've gotta multiply that by the Lipschitz constant" piece. So, let's work on unpacking these two distances. For the first one, we can go:

$d(\lambda x.K(x\left)\right(\lambda y.f(x,y)),\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left)\right)$

$={sup}_x|K\left(x\right)(\lambda y.f(x,y\left)\right)-K\left(x\right)(\lambda y.f^{\prime }(x,y\left)\right)|$

$\le {sup}_x{\lambda }_K^{\odot }d(\lambda y.f(x,y),\lambda y.f^{\prime }(x,y\left)\right)$

$={\lambda }_K^{\odot }{sup}_{x}d(\lambda y.f(x,y),\lambda y.f^{\prime }(x,y\left)\right)$

$={\lambda }_K^{\odot }{sup}_x{sup}_y|f(x,y)-f^{\prime }(x,y)|$

$={\lambda }_K^{\odot }{sup}_{x,y}|f(x,y)-f^{\prime }(x,y)|={\lambda }_K^{\odot }d(f,f^{\prime })$

Substituting this back in produces:

$\le ϵ{\lambda }_K^{\odot }d(f,f^{\prime })+{\lambda }_h^{\odot }d\left(\right(\lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right){)}_{\downarrow C_{ϵ}^X},(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left){)}_{\downarrow C_{ϵ}^X}\right)$

Time to go after the second distance piece. We have:

$d\left(\right(\lambda x.K\left(x\right)(\lambda y.f(x,y\left)\right){)}_{\downarrow C_{ϵ}^X},(\lambda x.K(x\left)\right(\lambda y.f^{\prime }(x,y)\left){)}_{\downarrow C_{ϵ}^X}\right)$

$={sup}_{x\in C_{ϵ}^X}|K\left(x\right)(\lambda y.f(x,y\left)\right)-K\left(x\right)(\lambda y.f^{\prime }(x,y\left)\right)|$

And, because $f$ and $f^{\prime }$ agree on $C_{ϵ}^X\times C_{ϵ}^Y$, we have $\lambda y.f(x,y)$ and $\lambda y.f^{\prime }(x,y)$ agreeing on $C_{ϵ}^Y$, which is an $ϵ$-almost-support for all the $K\left(x\right)$ where $x\in C_{ϵ}^X$, so we have:

$\le {sup}_{x\in C_{ϵ}^X}ϵd(\lambda y.f(x,y),\lambda y.f^{\prime }(x,y\left)\right)$

$=ϵ{sup}_{x\in C_{ϵ}^X}{sup}_y|f(x,y)-f^{\prime }(x,y)|$

$\le ϵ{sup}_{x,y}|f(x,y)-f^{\prime }(x,y)|=ϵd(f,f^{\prime })$

Substituting this back in produces:

$\le ϵ{\lambda }_K^{\odot }d(f,f^{\prime })+ϵ{\lambda }_h^{\odot }d(f,f^{\prime })$

And regrouping this and recapping means that we have:

$|(h⋉K)\left(f\right)-(h⋉K)\left(f^{\prime }\right)|\le ϵ({\lambda }_K^{\odot }+{\lambda }_h^{\odot })d(f,f^{\prime })$

So we have crafted a compact $ϵ({\lambda }_K^{\odot }+{\lambda }_h^{\odot })$-support for $h⋉K$, and we can make $ϵ$ arbitrarily small, so the semidirect product has compact almost-support, which is the last condition we needed.

Time for property verification.

Homogenity:

$(h⋉K)\left(af\right)=h(\lambda x.K(x\left)\right(\lambda y.af(x,y)\left)\right)=h(\lambda x.aK(x\left)\right(\lambda y.f(x,y)\left)\right)$

$=ah(\lambda x.K(x\left)\right(\lambda y.f(x,y)\left)\right)=a(h⋉K)\left(f\right)$

1-Lipschitz: We showed in the Lipschitz section that an upper bound on the Lipschitz constant of $h⋉K$ is the product of the Lipschitz constants of the kernel and the original infradistribution, so $1\cdot 1=1$ and 1-Lipschitzness is preserved.

Cohomogenity:

$(h⋉K)(1+af)=h(\lambda x.K(x\left)\right(\lambda y.1+af(x,y)\left)\right)=h(\lambda x.1-a+aK(x\left)\right(\lambda y.1+f(x,y)\left)\right)$

$=h(\lambda x.1+a(-1+K\left(x\right)(\lambda y.1+f(x,y\left)\right)\left)\right)=1-a+ah(\lambda x.1-1+K(x\left)\right(\lambda y.1+f(x,y)\left)\right))$

$=1-a+ah(\lambda x.K(x\left)\right(\lambda y.1+f(x,y)\left)\right))=1-a+a(h⋉K\left)\right(1+f)$

C-additivity:

$(h⋉K)\left(c\right)=h(\lambda x.K(x\left)\right(\lambda y.c\left)\right)=h(\lambda x.c)=c$

Crispness: Both homogenity and C-additivity are preserved, so crispness is too.

Sharpness:

$(h⋉K)\left(f\right)=h(\lambda x.K(x\left)\right(\lambda y.f(x,y)\left)\right)=h(\lambda x.{inf}_{y\in C_{K\left(x\right)}}f(x,y\left)\right)$

$={inf}_{x\in C_h}\left({inf}_{y\in C_{K\left(x\right)}}f\right(x,y\left)\right)={inf}_{(x,y)\in {\bigcup }_{x\in C_h}\left(\right\{x\}\times C_{K\left(x\right)})}f(x,y)$

Our task now is to show that ${\bigcup }_{x\in C_h}\left(\right\{x\}\times C_{K\left(x\right)})$ is compact, which will take a fair amount of topology work. Our first piece that we'll need is that if $x_n$ limits to $x$, then $C_{K\left(x_n\right)}$ limits to $C_{K\left(x\right)}$ in Hausdorff-distance.

To show this, we'll split it into two parts. First, we'll assume that there is an $ϵ$ where, infinitely often, there is a point in $C_{K\left(x_n\right)}$ that is $ϵ$ away from $C_{K\left(x\right)}$ and disprove that. Second, we'll assume there is an $ϵ$ where, infinitely often, there is a point in $C_{K\left(x\right)}$ that is $ϵ$ away from $C_{K\left(x_n\right)}$, and disprove that.

For the first part, assume that there is an $ϵ$ where, infinitely often, there is a point in $C_{K\left(x_n\right)}$ that is $ϵ$ away from $C_{K\left(x\right)}$. Craft the continuous function

$f_1:=\lambda y.sup(1-\frac{1}{ϵ}{inf}_{y^{\prime }\in C_{K\left(x\right)}}d(y,y^{\prime }),0)$

What this does is it's 1 on the set $C_{K\left(x\right)}$, and 0 on anything more than $ϵ$ away from it. One of our conditions on an infrakernel was that ${lim}_{n\to \infty }K\left(x_n\right)\left(f\right)=K\left(x\right)\left(f\right)$, so:

${lim}_{n\to \infty }{inf}_{y\in C_{K\left(x_n\right)}}{f}_1\left(y\right)={inf}_{y\in C_{K\left(x\right)}}{f}_1\left(y\right)$

The latter term is 1 because $f_1$ is 1 over $C_{K\left(x\right)}$. However, because we're assuming that infinitely often, there's a point in $C_{K\left(x_n\right)}$ that is $ϵ$ away from $C_{K\left(x\right)}$, the sequence on the left-hand side is infinitely often 0, so it doesn't converge and we have a contradiction.

For the second part, assume there is an $ϵ$ where, infinitely often, there is a point in $C_{K\left(x\right)}$ that is $ϵ$ away from $C_{K\left(x_n\right)}$. By compactness of $C_{K\left(x\right)}$, we can find finitely many points $y_i$ in it s.t. every point in $C_{K\left(x\right)}$ is only $\frac{ϵ}{2}$ away from one of the $y_i$ (cover $C_{K\left(x\right)}$ with $\frac{ϵ}{2}$-size open balls centered on points in it and take a finite subcover). Now, for each of these, we can craft a function

$f_i:=\lambda y.inf(1,\frac{2}{ϵ}d(y,y_i\left)\right)$

So, this is 0 at the point $y_i$, and 1 at any distance $\frac{ϵ}{2}$ or more away from it.

One of our conditions on an infrakernel was that ${lim}_{n\to \infty }K\left(x_n\right)\left(f\right)=K\left(x\right)\left(f\right)$, and there are finitely many $f_i$, so there's some time where all of them nearly converge, ie:

${lim}_{n\to \infty }{sup}_i|K\left(x_n\right)\left(f_i\right)-K\left(x\right)\left(f_i\right)|=0$

However, infinitely often there's a point $y_n\in C_{K\left(x\right)}$ that is $ϵ$ away from $C_{K\left(x_n\right)}$. $y_n$ is $\frac{ϵ}{2}$ away from some $y_i$, so that $y_i$ can't be closer than $\frac{ϵ}{2}$ to $C_{K\left(x_n\right)}$. (if it was closer, then we could pick some point in $C_{K\left(x_n\right)}$ that's closer than $\frac{ϵ}{2}$ to $y_i$, and then since it's only $\frac{ϵ}{2}$ away from $y_n$, we'd have that the distance from $y_n$ to $C_{K\left(x_n\right)}$ is below $\frac{ϵ}{2}$, an impossibility).

Because the distance from $y_i$ to any point in $C_{K\left(x_n\right)}$ is above $\frac{ϵ}{2}$, then

$|K\left(x_n\right)\left(f_i\right)-K\left(x\right)\left(f_i\right)|=|{inf}_{y\in C_{K\left(x_n\right)}}{f}_i\left(y\right)-{inf}_{y\in C_{K\left(x\right)}}{f}_i\left(y\right)|=|1-0|=1$

This is because $y_i\in C_{K\left(x\right)}$ and attains a value of 0 according to $f_i$, while $C_{K\left(x_n\right)}$ stays away from $y_i$ and all its points must have a value of 1. This situation happens infinitely often, which leads to a contradiction with

${lim}_{n\to \infty }{sup}_i|K\left(x_n\right)\left(f_i\right)-K\left(x\right)\left(f_i\right)|=0$

Because infinitely often, one of these $f_i$ has very different values, so the sequence is 1 infinitely often and can't limit to 0.

So, we've ruled out that there is an $ϵ$ where, infinitely often, there is a point in $C_{K\left(x_n\right)}$ that is $ϵ$ away from $C_{K\left(x\right)}$. And we've ruled out that there is an $ϵ$ where, infinitely often, there is a point in $C_{K\left(x\right)}$ that is $ϵ$ away from $C_{K\left(x_n\right)}$. Fixing any $ϵ$, in the tail of the sequence, $C_{K\left(x\right)}$ and $C_{K\left(x_n\right)}$ are $ϵ$ distance or closer in Hausdorff distance because you can't find points in either set which are far away from the other set. So, $C_{K\left(x_n\right)}$ limits to $C_{K\left(x\right)}$ in Hausdorff-distance when $x_n$ limits to $x$, and we know that $x\mapsto C_{K\left(x\right)}$ is a continuous function $X\to K\left(Y\right)$.

This lets us show that the set

${\bigcup }_{x\in C_h}\left(\right\{x\}\times C_{K\left(x\right)})$

is closed, because if $x_n$ limits to $x$ and $y_n\in C_{K\left(x_n\right)}$ and $y_n$ limits to $y$, we have that $y\in C_{K\left(x\right)}$ because $C_{K\left(x_n\right)}$ limits to $C_{K\left(x\right)}$ in Hausdorff distance, so we've got closed graph.

Also, by invoking Lemma 3, we know that

${\bigcup }_{x\in C_h}{C}_{K\left(x\right)}$

is compact.

Time to wrap this all up. We know that ${\bigcup }_{x\in C_h}\left\{x\right\}\times C_{K\left(x\right)}$ is closed in $X\times Y$ from our Hausdorff limit argument. This set is also a subset of:

$C_h\times {\bigcup }_{x\in C_h}{C}_{K\left(x\right)}$

Which is a product of two sets known to be compact, and is compact. It's a closed subset of a compact set, so it's compact. Therefore, 

${\bigcup }_{x\in C_h}\left\{x\right\}\times C_{K\left(x\right)}$

is a compact set, and from way back,

$(h⋉K)\left(f\right)={inf}_{(x,y)\in {\bigcup }_{x\in C_h}\left(\right\{x\}\times C_{K\left(x\right)})}f(x,y)$

And we've shown that set is compact, so $h⋉K$ where $h$ and all the $K\left(x\right)$ are sharp can be written as minimizing over a compact set, so $h⋉K$ is sharp. Thus, semidirect product preserves all the nice properties, and we're finally done with this proof.

Proposition 20: If all the $K\left(x\right)$ are C-additive, then $p{r}_{X\ast }(h⋉K)=h$.

$p{r}_{X\ast }(h⋉K)\left(f\right)=(h⋉K)(f\circ p{r}_X)=h(\lambda x.K(x\left)\right(\lambda y.f\left(p{r}_X\right(x,y\left)\right)\left)\right)$

$=h(\lambda x.K(x\left)\right(\lambda y.f\left(x\right)\left)\right)=h(\lambda x.f(x\left)\right)=h\left(f\right)$

This is because, since $f\left(x\right)$ doesn't depend on $y$, it acts as a constant inside $K\left(x\right)$ and C-additivity lets us pull it out.

Proposition 21: If $K_0,K_1,K_2...$ are a sequence of infrakernels of type  $K_n:{\prod }_{i=0}^{i=n}{X}_i\stackrel{ik}{\to }{X}_{n+1}$, and $h$ is an infradistribution over $X_0$, then $(...((h⋉K_0)⋉K_1)...⋉K_m)$ can be rewritten as $h⋉K_{:m}$ where $K_{:n}$ is an infrakernel of type $X_0\stackrel{ik}{\to }{\prod }_{i=1}^{i=n+1}{X}_i$, recursively defined as $K_{:0}:=K_0$ and $K_{:n+1}\left(x_0\right):=K_{:n}\left(x_0\right)⋉(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right)$

So, for our inductive definition,

$K_{:0}\left(x_0\right):=K_0\left(x_0\right)$

$K_{:n+1}\left(x_0\right):=K_{:n}\left(x_0\right)⋉(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1})$

Our task is to show that these are all infrakernels, by induction, and that for any infradistribution $h$,

$(...((h⋉K_0)⋉K_1)...⋉K_n)=h⋉K_{:n}$

For the base case, we observe that $K_{:0}$ is an infrakernel because it equals $K_0$, which is an infrakernel, and that $h⋉K_0=h⋉K_{:0}$

Time for the induction step. We'll assume that $K_{:n}$ is an infrakernel, and show that $K_{:n+1}$ is. Further, we need to show that $h⋉K_{:n+1}=(h⋉K_{:n})⋉K_{n+1}$. This will show the result.

Our first requirement is showing that for all $x_0$, $K_{:n+1}\left(x_0\right)$ is an infradistribution. 

$K_{:n+1}\left(x_0\right):=K_{:n}\left(x_0\right)⋉(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right)$

By our induction assumption, $K_{:n}\left(x_0\right)$ is an infradistribution as $K_{:n}$ is an infrakernel. Further, $\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1})$ is an infrakernel because $K_{n+1}$ is and we're just restricting it to a subset of its domain, so it keeps being an infrakernel. And we know from earlier that the semidirect product of an infradistribution and an infrakernel is an infradistribution. So that's taken care of.

Now, we must show a common Lipschitz constant, pointwise function convergence, and compact-shared compact almost-support for $K_{:n+1}$ to certify that it's an infrakernel.

Starting with common Lipschitz constant, we can just note that, in our proof of Proposition 19, we saw that the Lipschitz constant of the semidirect product was upper-bounded by the product of the Lipschitz constants of the starting infradistributions and the kernel. Assuming that $K_{:n}$ is an infradistribution, we have that the Lipschitz constant of any $K_{:n}\left(x_0\right)$ is upper-bounded by some ${\lambda }_{:n}^{\odot }$ Lipschitz constant. Also, the Lipschitz constant of $K_{n+1}(x_0,x_{1:n+1})$ is upper-bounded by some ${\lambda }_{n+1}^{\odot }$ Lipschitz constant. Thus, ${\lambda }_{:n}^{\odot }{\lambda }_{n+1}^{\odot }$ is an upper-bound on the Lipschitz constant of any

$K_{:n}\left(x_0\right)⋉(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right)$

infradistribution, which is exactly $K_{:n+1}\left(x_0\right)$, witnessing that $K_{:n+1}$ has a uniform upper bound on its Lipschitz constants.