LBIT Proofs 3: Propositions 19-22

post by Diffractor · 2020-12-16T03:40:44.852Z · LW · GW · 0 comments

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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.

Time to move onto the second one, compact-shared compact almost-support.

For this one, we're trying to prove:

$\forall C^{X_0},ϵ\exists C_{ϵ}^{{\prod }_{i=1}^{i=n+2}{X}_i}\subseteq {\prod }_{i=1}^{i=n+2}{X}_i\forall x_0\in C^{X_0},f,f^{\prime }:$

$f_{\downarrow C_{ϵ}^{{\prod }_{i=1}^{i=n+2}{X}_i}}=f_{\downarrow C_{ϵ}^{{\prod }_{i=1}^{i=n+2}{X}_i}}^{\prime }\to |K_{:n+1}\left(x_0\right)\left(f\right)-K_{:n+1}\left(x_0\right)\left(f^{\prime }\right)|\le ϵd(f,f^{\prime })$

This is the sentence that says that $K_{:n+1}$ has compact-shared compact almost-support. $f$ and $f^{\prime }$ have type signature ${\prod }_{i=1}^{i=n+2}{X}_i\to R$.

Now, this is going to be quite complicated, so pay close attention. Fix an arbitrary compact $C^{X_0}\subseteq X_0$, and an arbitrary $ϵ$. Let ${\lambda }_{:n}^{\odot }$ be the Lipschitz constant for the infrakernel $K_{:n}$, and ${\lambda }_{n+1}^{\odot }$ be the Lipschitz constant for the infrakernel $K_{n+1}$.

Due to compact-shared compact-almost-support for $K_{:n}$ which exists by our induction assumption, your set $C^{X_0}$ induces a compact $\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}$-almost-support for the family of infradistributions $K_{:n}\left(x_0\right)$ where $x_0\in C^{X_0}$. Call said almost-support $C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$.

Further, due to compact-shared compact-almost-support for $K_{n+1}$ , the set

$C^{X_0}\times C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$

induces a compact $\frac{ϵ}{2{\lambda }_{:n}^{\odot }}$-almost-support for the family of infradistributions $K_{n+1}(x_0,x_{1:n+1})$ where $(x_0,x_{1:n+1})\in C^{X_0}\times C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$

Call said almost-support $C_{\frac{ϵ}{2{\lambda }_{:n}^{\odot }}}^{X_{n+2}}$

And now let your shared $ϵ$-almost-support for $K_{:n+1}\left(x_0\right)$ where $x_0\in C^{X_0}$ be:

$C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}\times C_{\frac{ϵ}{2{\lambda }_{:n}^{\odot }}}^{X_{n+2}}$

We must show that said set is indeed a shared $ϵ$-almost-support for $K_{:n+1}\left(x_0\right)$ where $x_0\in C^{X_0}$. So, let $f$ and $f^{\prime }$ agree on said set. Then, we have:

$|K_{:n+1}\left(x_0\right)\left(f\right)-K_{:n+1}\left(x_0\right)\left(f^{\prime }\right)|$

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

$=|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2})\left)\right)|$

This is just unpacking the definition of the iterated semidirect product, no issues here. Now, we use Lemma 2 and the fact that $C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$ is a $\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}$-almost-support for $K_{:n}\left(x_0\right)$ when $x_0\in C^{X_0}$, to get:

$\le {\lambda }_{:n}^{\odot }{sup}_{x_{1:n+1}\in C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}}|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|$

$+\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}{sup}_{x_{1:n+1}}|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|$

Ok, this is a mess. Let's try to unpack

${sup}_{x_{1:n+1}\in C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}}|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|$

first. What we can do is use that, regardless of what is picked in the supremum, we have:

$(x_0,x_{1:n+1})\in C^{X_0}\times C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$

So this means that

$C_{\frac{ϵ}{2{\lambda }_{:n}^{\odot }}}^{X_{n+2}}$

is a $\frac{ϵ}{2{\lambda }_{:n}^{\odot }}$-almost-support for $K_{n+1}(x_0,x_{1:n+1})$. Further, because $f$ and $f^{\prime }$ are identical on

$C_{\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}}^{{\prod }_{i=1}^{i=n+1}{X}_i}\times C_{\frac{ϵ}{2{\lambda }_{:n}^{\odot }}}^{X_{n+2}}$

and $x_{1:n+1}$ was being selected from the former of those, then the functions $\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})$ (and the same for $f^{\prime }$) agree on $C_{\frac{ϵ}{2{\lambda }_{:n}^{\odot }}}^{X_{n+2}}$, the almost-support. So, the supremum is upper-bounded by

$\le \frac{ϵ}{2{\lambda }_{:n}^{\odot }}d(f,f^{\prime })$

Substituting this back in, we get:

$\le {\lambda }_{:n}^{\odot }\frac{ϵ}{2{\lambda }_{:n}^{\odot }}d(f,f^{\prime })+\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}{sup}_{x_{1:n+1}}|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|$

Now let's try to unpack 

${sup}_{x_{1:n+1}}|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|$

We don't have much leverage on it, besides using the basic Lipschitz constant upper bound, so let's try that.

$\le {sup}_{x_{1:n+1}}{\lambda }_{n+1}^{\odot }d(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}),\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)$

$={\lambda }_{n+1}^{\odot }d(f,f^{\prime })$

And substituting this back in, we get:

$\le {\lambda }_{:n}^{\odot }\frac{ϵ}{2{\lambda }_{:n}^{\odot }}d(f,f^{\prime })+\frac{ϵ}{2{\lambda }_{n+1}^{\odot }}{\lambda }_{n+1}^{\odot }d(f,f^{\prime })$

$=ϵd(f,f^{\prime })$

And so we've shown that the functions are only $ϵ$ times their distance apart, so the compact set we cooked up is indeed an $ϵ$-almost-support for $K_{:n+1}\left(x_0\right)$ whenever $x_0\in C^{X_0}$, and because $ϵ$ and $C^{X_0}$ was arbitrary, we have compact-shared compact-almost-support for $K_{:n+1}$.

Time to move onto the third one, pointwise convergence. If $x_{0,m}$ limits to $x_{0,\infty }$, we want $K_{:n+1}\left(x_{0,m}\right)\left(f\right)$ to limit to $K_{:n+1}\left(x_{0,\infty }\right)\left(f\right)$. As usual, we use ${\lambda }_{n+1}^{\odot }$ for the Lipschitz constant of $K_{n+1}$ and ${\lambda }_{:n}^{\odot }$ for the Lipschitz constant of $K_{:n}$.

To begin with, fix an arbitrary $ϵ$ and bounded continuous function $f$, and note that $\{x_{0,m}{\}}_{m\in N\cup \left\{\infty \right\}}$ is a compact subset of $X_0$. Because $K_{:n}:X_0\stackrel{ik}{\to }{\prod }_{i=1}^{i=n+1}{X}_i$ is assumed to be an infrakernel by induction, $\{x_{0,m}{\}}_{m\in N\cup \left\{\infty \right\}}$ acts as a compact set for it. So, by compact-shared compact-almost-support for $K_{:n}$, we can find a compact set $C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$ which is a $\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}$-almost-support for $K_{:n}$.

Also, it is important to note that 

$\lambda x_{:n+1}.K_{n+1}\left(x_{:n+1}\right)(\lambda x_{n+2}.f(x_{:n+1},x_{n+2}\left)\right)$

Is a continuous function (as it must be for semidirect products with $K_{n+1}$ to have the functions on the inside be continuous). Accordingly, this means that the function:

$\lambda x_0,x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

must be uniformly continuous when restricted to the set $\{x_{0,m}{\}}_{m\in N\cup \left\{\infty \right\}}\times C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$
And so, by uniform continuity, given any $ϵ$, there is some $\delta$ difference in inputs which gives rise to a $\frac{ϵ}{2{\lambda }_{:n}^{\odot }}$ difference in output.

Now, here's what we'll be doing. We'll attempt to show the result that

$\forall ϵ\exists m^{\ast }\forall m\ge m^{\ast }:|K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,m},x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)|\le ϵ$

Straight off the bat, we can apply Lemma 2 to decompose this difference into "starting Lipschitz constant times the difference of the inner functions on the compact set of interest" and "level of almost-support times the difference of the two functions", yielding:

$|K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,m},x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)|$

$\le {\lambda }_{:n}^{\odot }{sup}_{x_{1:n+1}\in C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}}|K_{n+1}(x_{0,m},x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_{0,\infty },x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)|$

$+\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}{sup}_{x_{1:n+1}}|K_{n+1}(x_{0,m},x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_{0,\infty },x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)|$

Time to start breaking this down. First, to break down

${sup}_{x_{1:n+1}}|K_{n+1}(x_{0,m},x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_{0,\infty },x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)|$

we can realize that the maximum value of one of these would be ${\lambda }_{n+1}^{\odot }||f||$, and the minimum possible value of one of these is $-{\lambda }_{n+1}^{\odot }||f||$, from Lipschitzness of $K_{n+1}$, producing an upper bound of:

$\le 2{\lambda }_{n+1}^{\odot }||f||$

Substituting this back in, we have:

$\le {\lambda }_{:n}^{\odot }{sup}_{x_{1:n+1}\in C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}}|K_{n+1}(x_{0,m},x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_{0,\infty },x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)|+\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}2{\lambda }_{n+1}^{\odot }||f||$

And now, we can use the fact that there is always some $\delta$ difference in inputs which gives rise to a $\frac{ϵ}{2{\lambda }_{:n}^{\odot }}$ difference in output of the function

$\lambda x_0,x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

when restricted to the set $\{x_{0,m}{\}}_{m\in N\cup \left\{\infty \right\}}\times C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$
in order to find some $m^{\ast }$ where all future $m$ have $x_{0,m}$ being within $\delta$ of $x_{0,\infty }$.

This tiny difference means that the values

$K_{n+1}(x_{0,m},x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

and

$K_{n+1}(x_{0,\infty },x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

will only differ by $\frac{ϵ}{2{\lambda }_{:n}^{\odot }}$ for all $x_{1:n+1}$ which lie in

$C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}$

Therefore, we have that for all $m$ past $m^{\ast }$, 

${sup}_{x_{1:n+1}\in C_{\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}}^{{\prod }_{i=1}^{i=n+1}{X}_i}}|K_{n+1}(x_{0,m},x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_{0,\infty },x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)|\le \frac{ϵ}{2{\lambda }_{:n}^{\odot }}$

And substituting this back in, we have:

$\le {\lambda }_{:n}^{\odot }\frac{ϵ}{2{\lambda }_{:n}^{\odot }}+\frac{ϵ}{4{\lambda }_{n+1}^{\odot }||f||}2{\lambda }_{n+1}^{\odot }||f||=ϵ$

And $ϵ$ was arbitrary. Therefore we have our desired result that, regardless of bounded continuous function $f$,

$\forall ϵ\exists m^{\ast }\forall m\ge m^{\ast }:|K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,m},x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)|\le ϵ$

Therefore,

${lim}_{m\to \infty }|K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,m},x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)|=0$

These two things limit increasingly close to each other. Further,

${lim}_{m\to \infty }{K}_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$=K_{:n}\left(x_{0,\infty }\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

By pointwise convergence for $K_{:n}$ which is an infrakernel by our induction assumption. Putting these two parts together, we have:

${lim}_{m\to \infty }{K}_{:n}\left(x_{0,m}\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,m},x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$=K_{:n}\left(x_{0,\infty }\right)(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$
so

${lim}_{m\to \infty }\left(K_{:n}\right(x_{0,m})⋉(\lambda x_{1:n+1}.K_{n+1}(x_{0,m},x_{1:n+1})\left)\right)\left(f\right)$

$=\left(K_{:n}\right(x_{0,\infty })⋉(\lambda x_{1:n+1}.K_{n+1}(x_{0,\infty },x_{1:n+1})\left)\right)\left(f\right)$

so

${lim}_{m\to \infty }{K}_{:n+1}\left(x_{0,m}\right)\left(f\right)=K_{:n+1}\left(x_{0,\infty }\right)\left(f\right)$

And we're done, we showed pointwise convergence for $K_{:n+1}$ which is the last condition necessary to show it's an infrakernel, and the induction proof goes through to show that all the $K_{:n}$ are infrakernels.

Now all that's left is to show that

$h⋉K_{:n+1}=(h⋉K_{:n})⋉K_{n+1}$

using induction, we have the base case set up. We can go:

$(h⋉K_{:n+1})\left(f\right)$

$=h(\lambda x_0.K_{:n+1}(x_0\left)\right(\lambda x_{1:n+2}.f(x_0,x_{1:n+2})\left)\right)$

$=h(\lambda x_0.(K_{:n}\left(x_0\right)⋉(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right)\left)\right(\lambda x_{1:n+2}.f(x_0,x_{1:n+2})\left)\right)$

$=h(\lambda x_0.K_{:n}(x_0\left)\right(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_0,x_{1:n+1},x_{n+2}\left)\right)\left)\right)$

$=(h⋉K_{:n})(\lambda x_{:n+1}.K_{n+1}(x_{:n+1}\left)\right(\lambda x_{n+2}.f(x_{:n+1},x_{n+2})\left)\right)$

$=\left(\right(h⋉K_{:n})⋉K_{n+1})\left(f\right)$

And we're done. Because 

$h⋉K_{:n+1}=(h⋉K_{:n})⋉K_{n+1}$

and we know that $h⋉K_{:0}=h⋉K_0$, this means that

$\forall m:(...((h⋉K_0)⋉K_1)...⋉K_m)=h⋉K_{:m}$

Proposition 22: $K_{:\infty }$ is an infrakernel (C-additive, specifically) if all the $K_n$ are C-additive infrakernels. It is unchanged by altering the $C_i$ sequence of compact sets. In addition, if all the $K_n$ are homogenous/cohomogenous/crisp/sharp, then $K_{:\infty }$ will be so as well.

So, $K_{:\infty }:X_0\stackrel{ik}{\to }{\prod }_{i=1}^{\infty }{X}_i$ is defined as: Fixing an arbitrary sequence of compact sets $C_i\in X_i$,
$K_{:\infty }\left(x_0\right)\left(f\right):={lim}_{n\to \infty }{K}_{:n}\left(x_0\right)\left(\lambda x_{1:n+1}{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f\right(x_{1:n+1},x_{n+2:\infty }\left)\right)$
Is it an infrakernel?

This is going to suck unbelievably much, we're gonna need a ton of results. The game plan is:

Part 1: Show that the functions you're feeding into those infrakernels are guaranteed to be continuous, to make some progress towards showing that $K_{:\infty }$ is well-defined.

Part 2: Show that all the $K_{:n}$ are 1-Lipschitz, and also preserve all nice properties we'd want if all the $K_n$ do (homogenity, cohomogenity, C-additity, crispness, sharpness).

Part 3: Show that if a function only depends on the first n coordinates of the input, then all the $K_{:n+m}$ start agreeing on the expectation value of the function.

Part 4: Give a general procedure for taking a compact subset of the space $X_0$ and making a compact subset of the space ${\prod }_{i=1}^{\infty }{X}_i$ with nice properties related to compact almost-support, that preserves its nice properties when projected down to any finite stage.

Part 5: Use parts 2, 3, 4, and a complicated chain of reasoning to get a result which implies that it doesn't matter which $C_i$ sequence you pick, the limit will exist and be same for all of them, so $K_{:\infty }$ actually exists and is well-defined.

Part 6: Using parts 2 and 5, clean up the normalization, monotonicity, concavity, and C-additivity properties of $K_{:\infty }$. Showing that all the $K_{:\infty }\left(x_0\right)$ are C-additive trivially nets the bounded Lipschitz constant property to show that $K_{:\infty }$ is an infrakernel and $K_{:\infty }\left(x_0\right)$ is an infradistribution.

Part 7: Use our trick from Part 4 and our freedom of picking our compact set sequence from Part 5 to show compact-shared compact almost-support for $K_{:\infty }$, netting us the second infrakernel property, and the compact almost-support property for all the individual components of kernel, verifying the last condition we need to conclude that $K_{:\infty }\left(x_0\right)$ is an infradistribution.

Part 8: We recap one of the arguments for part 5, and it lets us get uniform convergence for a certain limit on any compact set, which is a critical lemma for Part 9.

Part 9: We use our result from Part 8 to invoke the Moore-Osgood theorem in order to show pointwise convergence for $K_{:\infty }$, wrapping up the last condition for it to be an infrakernel.

Part 10: Show that if all the $K_{:n}$ have some nice property, then the limit $K_{:\infty }$ inherits it too.

The proofs will proceed in a strange way to keep track of all the moving parts in places. We'll first present the thing we're trying to prove, and repeatedly go "we could prove it if we could prove this other thing", and keep chaining back until we get something that's easy to show.

Proof Part 1: Our desired result is whether the function $\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty })$
is continuous. So, letting $x_{1:n+1}^m$ limit to $x_{1:n+1}^{\infty }$, our task is to show that:

${lim}_{m\to \infty }{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1}^m,x_{n+2:\infty })={inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1}^{\infty },x_{n+2:\infty })$

Now, what we can do is consider the compact subset of ${\prod }_{i=1}^{i=\infty }{X}_i$ to be
$\{x_{1:n+1}^m{\}}_{m\in N\cup \left\{\infty \right\}}\times {\prod }_{i=n+2}^{\infty }{C}_i$

And then $f$ must be uniformly continuous on it, so given any $ϵ$, there is some $\delta$ where points only $\delta$ away lead to only an $ϵ$ differ in value. You can consider $m$ to be big enough to guarantee that all future values of $x_{1:n+1}^m$ are within $\delta$ of $x_{1:n+1}^{\infty }$, and then this gets that the function values can only differ by $ϵ$ between $(x_{1:n+1}^m,x_{n+2:\infty })$ and $(x_{1:n+1}^{\infty },x_{n+2:\infty })$ if $x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i$, which it is. This ensures that the worst-case function values are only $ϵ$ apart. This works for all $ϵ$, showing

${lim}_{m\to \infty }{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1}^m,x_{n+2:\infty })={inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1}^{\infty },x_{n+2:\infty })$

And so, all the functions we're feeding into the $K_{:n}\left(x_0\right)$ are continuous.

Proof Part 2: Desired result is "if all the $K_n$ have a nice property, then all the $K_{:n}$ have it too".

This can be simply addressed by noting that, for the base case, because $K_{:0}=K_0$ and we're assuming all the $K_n$ have (C-additivity/cohomogenity/homogenity/crispness/sharpness), $K_{:0}$ trivially fulfills it.

And for the induction step, if we assume that $K_{:n}$ is 1-Lipschitz, note that:

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

And, by our results on semidirect products preserving nice properties, if $K_{:n}\left(x_0\right)$ has the nice property (by induction assumption) and $K_{n+1}$ does, then we get that $K_{:n+1}\left(x_0\right)$ preserves the same property, and it holds all the way up the $K_{:n}$.
And we can move on to Part 3.

Part 3: Showing that, if we go far enough out in the $K_{:n}$, the value assigned to functions which only depend on finitely many inputs stabilizes. The result that we'd like to show at this point is:

$\forall x_0\in X_0,n,m\in N,f\in C_B\left({\prod }_{i=1}^{i=n+1}{X}_i\right):K_{:n+m}\left(x_0\right)\left(f\right)=K_{:n}\left(x_0\right)\left(f\right)$

Admittedly, f is not of the proper type signature to be evaluated by $K_{:n+m}\left(x_0\right)$, but we're abusing notation so that we can feed it in anyways and it just ignores all the coordinates it doesn't need. Accordingly, fix an arbitrary $x_0,n,f,$ and our proof target will now be:

$\forall m\in N:K_{:n+m}\left(x_0\right)\left(f\right)=K_{:n+m+1}\left(x_0\right)\left(f\right)$

Proving this would let you apply induction, because we have a base case where $K_{:n+0}\left(x_0\right)\left(f\right)=K_{:n}\left(x_0\right)\left(f\right)$. Let $m$ be arbitrary. Then, we can go:

$K_{:n+m+1}\left(x_0\right)\left(f\right)=K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.K_{n+m+1}(x_0,x_{1:n+m+1}\left)\right(\lambda x_{n+m+2}.f\left(x_{1:n+1}\right)\left)\right)$

And then, since the function doesn't depend on the choice of $x_{n+m+2}$, it's a constant and C-additivity of $K_{n+m+1}$ lets us pull it out, yielding

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

And we're done.

Part 4: Our desired result here is:

$\forall C^{X_0}\subseteq X_0,ϵ>0:\exists C_{1:\infty }[C^{X_0},ϵ]\subseteq {\prod }_{i=1}^{\infty }{X}_i:\forall n\in N,x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+1}{X}_i\right):$

$f_{\downarrow p{r}_{1:n+1}\left(C_{1:\infty }\right[C^{X_0},ϵ\left]\right)}=f_{\downarrow p{r}_{1:n+1}\left(C_{1:\infty }\right[C^{X_0},ϵ\left]\right)}^{\prime }$

$\to |K_{:n}\left(x_0\right)\left(f\right)-K_{:n}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })$

This is a bit complicated. It's saying that if you pick any compact subset of $X_0$, you can make a compact subset of ${\prod }_{i=1}^{\infty }{X}_i$ where the projection of it to coordinates 1 through $n+1$ acts as a compact $ϵ(1-\frac{1}{2^{n+1}})$-almost-support for all the $K_{:n}\left(x_0\right)$ infradistributions when $x_0$ lies in your compact subset of $X_0$. Regardless of what $n$ is.

Accordingly, fix some $C^{X_0}$ and $ϵ$. Now, we can recursively build up compact subsets of all the $X_n$ in the following way.

$C_{\frac{ϵ}{2^{n+1}}}^{X_{n+1}}\subseteq X_{n+1}$ is a $\frac{ϵ}{2^{n+1}}$-almost-support for all the $K_n\left(x_{1:n}\right)$ where $x_{1:n}\in C^{X_0}\times {\prod }_{i=1}^{i=n}{C}_{\frac{ϵ}{2^i}}^{X_i}$. So, basically, we're recursively building up compact subsets of ${\prod }_{i=0}^{i=n}{X}_i$ by taking products of earlier compact subsets (with your base case being $C^{X_0}$), and then going "that's a compact subset of the input to $K_n$, we must be able to find a compact subset of $X_{n+1}$ that's a $\frac{ϵ}{2^{n+1}}$-almost-support for all the $K_n\left(x_{:n}\right)$ where $x_{:n}$ lies in our compact subset of input, because of the compact-shared almost-support condition for all the $K_n$" to go to the next compact set.

To establish some notation to make this a bit easier, let

$C_i[C^{X_0},ϵ]:=C_{\frac{ϵ}{2^{i+1}}}^{X_i}$

(the i'th compact set in the sequence, defined with $C^{X_0}$ to start building your sequence), and let

$C_{1:n}[C^{X_0},ϵ]:={\prod }_{i=1}^{i=n}{C}_i\left[C^{X_0}\right]$

(the product of compact sets 1 through $n$, which is compact)

And let

$C_{1:\infty }[C^{X_0},ϵ]:={\prod }_{i=1}^{\infty }{C}_i\left[C^{X_0}\right]$

This is the product of all the compact sets, and is compact.

Note the dependence of these on the starting compact sets. Notice that the projection of $C_{1:\infty }[C^{X_0},ϵ]$ to coordinates 1 through $n$ is exactly $C_{1:n}[C^{X_0},ϵ]$.

Now that this is established, our proof target is (using our new notation):

$\forall n\in N,x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+1}{X}_i\right):$

$f_{\downarrow C_{1:n+1}[C^{X_0},ϵ])}=f_{\downarrow C_{1:n+1}[C^{X_0},ϵ])}^{\prime }\to |K_{:n}\left(x_0\right)\left(f\right)-K_{:n}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })$

This structure naturally suggests an induction proof, so for the base case, let our number be 0. Our proof target then turns into:

$\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left(X_1\right):$

$f_{\downarrow C_1[C^{X_0},ϵ])}=f_{\downarrow C_1[C^{X_0},ϵ])}^{\prime }\to |K_{:0}\left(x_0\right)\left(f\right)-K_{:0}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2})d(f,f^{\prime })$

Using that $K_{:0}=K_0$ and that $C_1[C^{X_0},ϵ]=C_{\frac{ϵ}{2}}^{X_1}$ and $ϵ(1-\frac{1}{2})=\frac{ϵ}{2}$ our proof target is now:

$\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left(X_1\right):$

$f_{\downarrow C_{\frac{ϵ}{2}}^{X_1}}=f_{\downarrow C_{\frac{ϵ}{2}}^{X_1}}^{\prime }\to |K_0\left(x_0\right)\left(f\right)-K_0\left(x_0\right)\left(f^{\prime }\right)|\le \frac{ϵ}{2}d(f,f^{\prime })$

However, we constructed $C_{\frac{ϵ}{2}}^{X_1}$ to be a $\frac{ϵ}{2}$-almost-support for all the $K_0\left(x_0\right)$ where $x_0\in C^{X_0}$, so this statement is just true, and we're done with our base case.

Now, for the induction step, our proof target is:

$\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+1}{X}_i\right):f_{\downarrow C_{1:n+1}[C^{X_0},ϵ]}=f_{\downarrow C_{1:n+1}[C^{X_0},ϵ]}^{\prime }$

$\to |K_{:n}\left(x_0\right)\left(f\right)-K_{:n}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })$

implies

$\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+2}{X}_i\right):f_{\downarrow C_{1:n+2}[C^{X_0},ϵ]}=f_{\downarrow C_{1:n+2}[C^{X_0},ϵ]}^{\prime }$

$\to |K_{:n+1}\left(x_0\right)\left(f\right)-K_{:n+1}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+2}})d(f,f^{\prime })$

Accordingly, assume

$\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+1}{X}_i\right):f_{\downarrow C_{1:n+1}[C^{X_0},ϵ]}=f_{\downarrow C_{1:n+1}[C^{X_0},ϵ]}^{\prime }$

$\to |K_{:n}\left(x_0\right)\left(f\right)-K_{:n}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })$

And our task is to prove

$\forall x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+2}{X}_i\right):f_{\downarrow C_{1:n+2}[C^{X_0},ϵ]}=f_{\downarrow C_{1:n+2}[C^{X_0},ϵ]}^{\prime }$

$\to |K_{:n+1}\left(x_0\right)\left(f\right)-K_{:n+1}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+2}})d(f,f^{\prime })$

Therefore let $x_0,f,f^{\prime }$ be arbitrary, and remember that they have the indicated properties, and that $f,f^{\prime }$ agree with each other on the indicated set $C_{1:n+2}[C^{X_0},ϵ]$. Our proof target is now:

$|K_{:n+1}\left(x_0\right)\left(f\right)-K_{:n+1}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+2}})d(f,f^{\prime })$

Unpacking the definition of $K_{:n+1}$ and rewriting the thing on the end, this is equivalent to (we now take this as the proof target)

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2})\left)\right)|$

$\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })+\frac{ϵ}{2^{n+2}}d(f,f^{\prime })$

We can apply the Lemma 2 decomposition, to split this into "level of support of compact set x distance of functions + distance of functions on compact set x lipschitz constant of infradistribution". So, theoretically, if we had the following two results:

$\forall x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]:|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|\le \frac{ϵ}{2^{n+2}}d(f,f^{\prime })$

and

$\forall x_{1:n+1}:|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|\le d(f,f^{\prime })$

then applying Lemma 2 would get us

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})\left)\right)$

$-K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.K_{n+1}(x_0,x_{1:n+1}\left)\right(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2})\left)\right)|$

$\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })+\frac{ϵ}{2^{n+2}}d(f,f^{\prime })\cdot 1$

Which is exactly what we need. This works because $C_{1:n+1}[C^{X_0},ϵ]$ is an $ϵ(1-\frac{1}{2^{n+1}})$-almost-support for $K_{:n}\left(x_0\right)$ by our induction assumption, our 2 assumptions, $K_{n+1}$ be a 1-Lipschitz infrakernel, and Lemma 2. Accordingly, let's try to show our two proof targets we need to wrap up this result. We'll start with

$\forall x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]:|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|\le \frac{ϵ}{2^{n+2}}d(f,f^{\prime })$

Now, we know that $f$ and $f^{\prime }$ agree on $C_{1:n+2}[C^{X_0},ϵ]$ by assumption, which is a set that factorizes as $C_{1:n+1}[C^{X_0},ϵ]\times C_{n+2}[C^{X_0},ϵ]$, and we have a promise that $x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]$, so $\lambda x_{n+2}.f(x_{1:n+1},x_{n+2})$ equals $\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2})$ on the set $C_{n+2}[C^{X_0},ϵ]$, which is a $\frac{ϵ}{2^{n+2}}$-almost-support for all the $K_{n+1}(x_0,x_{1:n+1})$ where $x_0\in C^{X_0}$ (which is the case by assumption) and $x_{1:n+1}\in C_{1:n+1}[C^{X_0},ϵ]$ (also the case), so we have our result.

Now for the other branch,

$\forall x_{1:n+1}:|K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}\left)\right)$

$-K_{n+1}(x_0,x_{1:n+1})(\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)|\le d(f,f^{\prime })$

Due to 1-Lipschitzness of $K_{n+1}(x_0,x_{1:n+1})$ regardless of $x_0$ and $x_{1:n+1}$, we could prove it if, regardless of $x_{1:n+1}$,

$d(\lambda x_{n+2}.f(x_{1:n+1},x_{n+2}),\lambda x_{n+2}.f^{\prime }(x_{1:n+1},x_{n+2}\left)\right)\le d(f,f^{\prime })$

which is the case, so we have our result.

That's everything wrapped up, so our induction proof goes through, yielding our result of:

$\forall C^{X_0}\subseteq X_0,ϵ>0:\exists C_{1:\infty }[C^{X_0},ϵ]\subseteq {\prod }_{i=1}^{\infty }{X}_i:\forall n\in N,x_0\in C^{X_0},f,f^{\prime }\in C_B\left({\prod }_{i=1}^{n+1}{X}_i\right):$

$f_{\downarrow C_{1:n+1}[C^{X_0},ϵ]}=f_{\downarrow C_{1:n+1}[C^{X_0},ϵ]}^{\prime }\to |K_{:n}\left(x_0\right)\left(f\right)-K_{:n}\left(x_0\right)\left(f^{\prime }\right)|\le ϵ(1-\frac{1}{2^{n+1}})d(f,f^{\prime })$

Now to begin our fifth part.

Part 5: Our aim here is to show that no matter what sequence of compact sets you have, they all limit to the same result, so our limit is going to be well defined. In order to do this, we'll have to show the result (letting $C_{i,1}$ being your first sequence of nonempty compact sets to attempt to define the limit and $C_{i,2}$ being your second sequence of nonempty compact sets, and abbreviating $C_{n:\infty ,1}$ as the product of the $C_{i,1}$ from $n$ to $\infty$) that,

$\forall x_0\in X_0,ϵ>0,\overline{C_1},\overline{C_2}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),f\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right)\exists n\in N\forall m_1,m_2\in N:$

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le 2ϵ(2||f||+1)$

In words, this is saying that for any two sequences of compact sets, there exists some threshold where if you pick any value of the defining sequence for $K_{:\infty }\left(x_0\right)\left(f\right)$ associated with using $C_{i,1}$ as your compact sets, and the sequence associated with using $C_{i,2}$ as your compact sets (as long as they're both past some shared threshold), they'll be close. If you have both sequences being identical, then this result is basically saying that the sequence used to define $K_{:\infty }\left(x_0\right)\left(f\right)$ is Cauchy (never varies too far from itself after a finite time), and thus the limit exists. And if you have the sequences being different, then this can be used to show that they limit to each other, so $K_{:\infty }$ is well-defined and you always get the same result no matter which particular sequence of compact sets you fix.

This is going to be rough. Fix our $x_0,ϵ,\overline{C_1},\overline{C_2},f$ (input value, closeness parameter, two sequences of compact sets, function), and now we'll find our $n$ to make

$\forall m_1,m_2\in N:$

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le 2ϵ(2||f||+1)$

true. Do it in the following way. Use $\left\{x_0\right\}$ as your compact seed set to invoke the technique in part 4 to construct your sequence $C_i[x_0,ϵ]$, and then consider the set:

${\prod }_{i=1}^{\infty }(C_{i,1}\cup C_{i,2}\cup C_i[x_0,ϵ\left]\right)$

A finite union of compact sets is compact, and the product of compact sets is compact. If we restrict f to this set, it's uniformly continuous. In particular, using the standard product metric (which metrizes the product topology), defined as:

$d(x_{1:\infty },x_{1:\infty }^{\prime })={\sum }_{i=1}^{\infty }{2}^{-i}inf(1,d_{X_i}(x_i,x_i^{\prime }\left)\right)$

We can conclude that two sequences which agree up till time $n$ must be, at most, distance $2^{-n}$ apart. Since $f$ restricted to our compact set of interest is uniformly continuous, there is some $\delta$ difference inputs that only leads to an $ϵ$ difference in values. Now we can define our $n$ as ${\log }_2\left(\delta \right)$.

Now that we've picked our $n$, let $m_1$ and $m_2$ be arbitrary. Our goal is to now prove:

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le 2ϵ(2||f||+1)$

We can break up the distance between these two quantities as:

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le |K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)|$

$+|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)|$

$+|K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

The distances group into three "chunks". What we'll do is show that chunks 1 and 3 have value upper-bounded by $ϵ(2||f||+1)$, and chunk 2 has a value of 0, producing our net $2ϵ(2||f||+1)$ upper-bound, and we'd be done. So, let's try to show the first one, that: 

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)|\le ϵ(2||f||+1)$

This one is perfectly symmetric to the third distance chunk, so disposing of this will also deal with showing

$|K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le ϵ(2||f||+1)$

The way we'll deal with this one is by using our good old Lemma 2, where we split up the difference of the two quantities into "how much of a support is this set times how different are the two functions" and "how close are the two functions on this set times the Lipschitz constant of the infradistributions". We'll be picking the set $C_{1:n+m_1+1}[x_0,ϵ]$, which is an $ϵ$-almost-support for $K_{1:n+m_1}\left(x_0\right)$ by our discussions in Section 4. Because the infrakernels are always 1-Lipschitz because of C-additivity, the maximum/minimum expectation value the functions

$\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty })$

and

$\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty })$

can have is $||f||$ (or $-||f||$) respectively. This produces a $2ϵ||f||$ bound on the value of that piece produced via Lemma 2. All that remains is to prove that

${sup}_{x_{1:n+m_1+1}\in C_{1:n+m_1+1}[x_0,ϵ]}$

$|{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty })$

$-{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty })|\le ϵ$

(because of 1-Lipschitzness of the infrakernel) And we'll be done. We can reformulate this proof objective as:

$\forall x_{1:n+m_1+1}\in C_{1:n+m_1+1}[x_0,ϵ]:$

$|{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty })$

$-{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty })|\le ϵ$
Accordingly, let $x_{1:n+m_1+1}$ be arbitrary in said set, so our objective is now:

$|{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty })$

$-{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty })|\le ϵ$

We'd be able to prove this if we could show:

$\forall x_{n+m+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1},x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]:$

$|f(x_{1:n+m_1+1},x_{n+m_1+2:\infty })-f(x_{1:n+1},x_{n+2:\infty })|\le ϵ$

Accordingly, let $x_{n+m_1+2:\infty }$ and $x_{n+2:\infty }$ be selected from the appropriate sets, and our goal is now:

$|f(x_{1:n+m_1+1},x_{n+m_1+2:\infty })-f(x_{1:n+1},x_{n+2:\infty })|\le ϵ$

At this point, we should remember that if you have a promise that your input to $f$ will be within the set ${\prod }_{i=1}^{\infty }(C_{i,1}\cup C_{i,2}\cup C_i[x_0,ϵ\left]\right)$

then knowing the first n coordinates fully pins down the value of your function $f$ to within $ϵ$, by how we picked our $n$. And then we can notice something interesting. By how they were selected,

$(x_{1:n+m_1+1},x_{n+m_1+2:\infty })\in {\prod }_{i=1}^{i=n+m_1+1}{C}_i[x_0,ϵ]\times {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}$

And also, $(x_{1:n+1},x_{n+2:\infty })\in {\prod }_{i=1}^{\infty }{C}_i[x_0,ϵ]$

So, the two inputs are both in the relevant compact set, and agree on the first $n$ coordinates, so they agree to within $ϵ$ value, and our desired result is the case. We've showed

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)|\le ϵ(2||f||+1)$

Which symmetrically establishes

$|K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le ϵ(2||f||+1)$

By pretty much exactly the same line of argument. That leaves one last distance equality to establish. All we need now is to show that

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)|=0$

Which is the same as the proof target:

$K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$=K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

However, this inner function only depends on the inputs from 1 to $n+1$, so by our Part 3, both of these equal the value

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i[x_0,ϵ]}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

And so, we're done.

So, our entire result goes through, as we've shown all our proof targets, and we have:

$\forall x_0\in X_0,ϵ>0,\overline{C_1},\overline{C_2}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),f\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right)\exists n\in N\forall m_1,m_2\in N:$

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|$

$\le 2ϵ(2||f||+1)$

As a result. Let's clean this up a little bit. It can be cleaned up into the equivalent form:

$\forall x_0\in X_0,\overline{C_1},\overline{C_2}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),f\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right),ϵ>0,\exists n\in N\forall m_1,m_2\in N:$

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|\le ϵ$

By shifting the $ϵ$ to the front and using that $||f||$ is finite so we can just let the old $ϵ$ be small enough.

And now this gets interesting. Let $x_0$ and $f$ be arbitrary, specialize to $\overline{C_1}=\overline{C_2}$, and $m_1=0$, and abbreviate $m_2$ as $m$. Then this turns into:

$\forall \overline{C_i}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),ϵ>0,\exists n\in N\forall m\in N:$

$|K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

$-K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_i}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)|\le ϵ$

Ie, this is pretty much saying that, regardless of your series of compact sets, the sequence in n given by:

$K_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

is Cauchy (and therefore must converge to a particular value, regardless of which choice of compact sets is made). So, when we defined $K_{:\infty }\left(x_0\right)\left(f\right)$ as

${lim}_{n\to \infty }{K}_{:n}\left(x_0\right)(\lambda x_{1:n+1}.{inf}_{x_{n+2:\infty }\in {\prod }_{i=n+2}^{\infty }{C}_i}f(x_{1:n+1},x_{n+2:\infty }\left)\right)$

The limit does indeed exist. But, we can actually get something even stronger. All these limits must be the same. Given the thing we showed, 

$\forall x_0\in X_0,\overline{C_1},\overline{C_2}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),f\in C_B\left({\prod }_{i=1}^{\infty }{X}_i\right),ϵ>0,\exists n\in N\forall m_1,m_2\in N:$

$|K_{:n+m_1}\left(x_0\right)(\lambda x_{1:n+m_1+1}.{inf}_{x_{n+m_1+2:\infty }\in {\prod }_{i=n+m_1+2}^{\infty }{C}_{i,1}}f(x_{1:n+m_1+1},x_{n+m_1+2:\infty }\left)\right)$

$-K_{:n+m_2}\left(x_0\right)(\lambda x_{1:n+m_2+1}.{inf}_{x_{n+m_2+2:\infty }\in {\prod }_{i=n+m_2+2}^{\infty }{C}_{i,2}}f(x_{1:n+m_2+1},x_{n+m_2+2:\infty }\left)\right)|\le ϵ$

We can let $x_0$and $f$ be arbitrary, and $m_1=m_2$ to get:

$\forall \overline{C_1},\overline{C_2}\in {\prod }_{i=1}^{\infty }K\left(X_i\right),ϵ>0,\exists n\in N\forall m\in N:$

$|K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_{i,1}}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)$

$-K_{:n+m}\left(x_0\right)(\lambda x_{1:n+m+1}.{inf}_{x_{n+m+2:\infty }\in {\prod }_{i=n+m+2}^{\infty }{C}_{i,2}}f(x_{1:n+m+1},x_{n+m+2:\infty }\left)\right)|\le ϵ$

Ie, no matter which sequence of compact sets is selected, the two convergent sequences get arbitrarily close to each other, so our definition of $K_{:\infty }\left(x_0\right)\left(f\right)$ doesn't just have the limit being well-defined, it has it being the same regardless of which sequence of compact sets $C_i$ was selected.

With this, now we can let the compact sequence be whatever is most convenient for arguments, as it always produces the same limit no matter what. (continued in next post)

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