Part 6: Showing the basic four infradistribution properties: Normalization, monotonicity, concavity, and Lipschitzness (specifically 1-Lipschitzness). For normalization, observe that regardless of n,

This is because, since f′≥f, the "extend with worst-case outputs" function is bigger for f′ than f, and then by monotonicity for K:n(x0), the inequality transfers to the outside.

This was by monotonicity (minimizing over the two parts separately produces lower values, and monotonicity transfers that to the outside), and concavity respectively, because K:n(x0) is an infradistribution. And now:

Then we'd be able to pair that with the 1-Lipschitzness of K:n(x0) to get our desired result. So let this be our new proof target. Reexpressing it a bit, it's:

Now, if two functions are only d(f,f′) apart, then their minimal values over a compact set can only be d(f,f′) apart at most. Let your compact set be ∏i=n+1i=1{xi}×∏∞i=n+2Ci to see the result. Thus, we have proved our proof target and we have the result that, for all n,f,f′

Since they're always only d(f,f′) apart, the same property transfers to the limit, so we get:

|K:∞(x0)(f)−K:∞(x0)(f′)|≤d(f,f′)

And so, K:∞(x0) is 1-Lipschitz for all x0. This takes care of the Lipschitzness condition on infradistributions and the uniform Lipschitz bound condition on infrakernels. All that's left is the compact almost-support condition on infradistributions and the compact-shared compact almost-support condition on infrakernels, and the pointwise convergence condition on infrakernels.

Part 7: This will use our result from Part 4 about making a suitable compact set. Our desired thing we want to prove is that

Because this property is exactly the compact-shared compact-almost-support property for K:∞.

So, here's our plan of attack. Fix our CX0 and ϵ, so our target is

∃Cϵ⊆∏∞i=1Xi∀x0∈CX0f,f′∈CB(∏∞i=1Xi):

f↓Cϵ=f↓Cϵ→|K:∞(x0)(f)−K:∞(x0)(f′)|≤ϵd(f,f′)

Let our Cϵ be C1:∞[CX0,ϵ], as defined from Part 4, and let x0 be arbitrary in CX0 and f,f′ be arbitrary and agree with each other on C1:∞[CX0,ϵ]. Then our goal becomes

|K:∞(x0)(f)−K:∞(x0)(f′)|≤ϵd(f,f′)

Now, letting our chosen defining sequence be Ci[CX0,ϵ], our goal is to show

If we could do this, all the approximating points being only ϵ away from each other shows that the same property transfers to the limit and we get our result. Accordingly, let n be arbitrary, and our proof goal is now:

Then we could conclude the two functions were equal on the set C1:n+1[CX0,ϵ], which, from Part 4, is an ϵ-almost support for all the K:n(x0) where x0∈CX0, and it would yield our result. So our proof target is now

However, f and f′ are equal on the set C1:∞[CX0,ϵ], which breaks down as C1:n+1[CX0,ϵ]×Cn+2:∞[CX0,ϵ]. And we know that x1:n+1∈C1:n+1[CX0,ϵ], and since the second part is being selected from Cn+2:∞[CX0,ϵ], the two values are equal, and we've proved our proof target. Thus, we have compactly-shared compact almost-support, and that's the second condition for K:∞ to be an infrakernel, as well as the last condition needed for all the K:∞(x0) to be infradistributions. Just pointwise convergence left!

Part 8: Alright, for showing pointwise convergence, we'll need a lemma. We need to show:

This is roughly saying that the rate of convergence of K:n(x0)(f) to K:∞(x0)(f) is uniform over compact sets. Let's begin. Let CX0,¯¯¯¯C,ϵ,f be arbitrary, so our proof target is now:

Here's what we'll do. Because we have a CX0 and ϵ, we can craft our sequence Ci[CX0,ϵ] of compact ϵ-supports. Now, on the set:

∏∞i=1(Ci[CX0,ϵ]∪Ci)

which is compact, we know that f is uniformly continuous, so there's some δ you need to go to to ensure that function values are only ϵ away from each other, which translates into an n via n:=log2(δ). Any two inputs which only differ beyond that point will be only δ apart and can only get an ϵ-difference in values from f. Now that that's defined, let m and x0∈CX0 be arbitrary. Our proof target is now to show:

Then, because C1:n+m+1[CX0,ϵ] is an ϵ-almost-support for K:n+m(x0) as long as x0∈CX0 (which we know to be the case), we could apply Lemma 2 to get our desired proof target.

The "Lipschitz constant times deviation over compact set" part would make one ϵ, and the "degree of almost-support times difference in the two functions" part would make 2ϵ||f|| in the worst-case due to being an ϵ-almost-support and the worst possible case where the two functions differ by 2||f|| somewhere.

Thus, let x1:n+m+1 be arbitrary within the appropriate set, so our proof target is now:

|infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)

−infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞)|≤ϵ

Because we're minimizing over a compact set in both instances, we can consider the infs to be attained by a xn+2:∞ and a xn+m+2:∞, so our proof target is now

|f(x1:n+1,xn+2:∞)−f(x1:n+m+1,xn+m+2:∞)|≤ϵ

Now, because x1:n+1∈C1:n+1[CX0,ϵ], and xn+2:∞∈∏∞i=n+2Ci, we have:

(x1:n+1,xn+2:∞)∈C1:n+1[CX0,ϵ]×∏∞i=n+2Ci

=∏i=n+1i=1Ci[CX0,ϵ]×∏∞i=n+2Ci⊆∏∞i=1(Ci[CX0,ϵ]∪Ci)

And also, because x1:n+m+1∈C1:n+m+1[CX0,ϵ], and xn+m+2:∞∈∏∞i=n+m+2Ci,

So, both inputs lie in the relevant compact set, and they agree on x1:n+1, so the inputs are only δ apart, so they only have value differing by ϵ, and we're done, our desired result of

Part 9: Now, for showing our pointwise continuity property, fix a sequence x0,m which limits to x0,∞, and an arbitrary f. Our task is now to show that:

limm→∞K:∞(x0,m)(f)=K:∞(x0,∞)(f)

Fixing some arbitrary Ci sequence, we could do this if we could show that

uniformly inm. However, back in Part 8 we established uniform convergence on any compact set. And {x0,m}m∈N∪{∞} is a compact set! So we have uniform convergence and can invoke Moore-Osgood to swap the limits and get our proof target, showing our result, that

limm→∞K:∞(x0,m)(f)=K:∞(x0,∞)(f)

So, K:∞ fulfills pointwise convergence and we've verified all properties needed to show it's an infrakernel.

Part 10: Showing that if all the K:n have some nice property, then K:∞ inherits it too. Homogenity, Cohomogenity, C-additivity, crispness, and sharpness.

Proof of crispness preservation: Crispness is equivalent to the conjunction of C-additivity and homogenity and both of those are preserved.

Proof of sharpness preservation: So, this will take a bit of work because we do have to get a fairly explicit form for the set you're minimizing over. Remember that when h and K are crisp, with associated compact sets Ch and CK(x), then the compact minimizing set for h⋉K is ⋃x∈ChCK(x), from the proof of sharpness preservation for semidirect product. Another way of writing this minimizing set of the semidirect product is as the set

{(x,y)|x∈Ch,y∈CK(x)}

Now, we'll give a form for the associated compact set of K:n(x0). It is:

Part 6: Showing the basic four infradistribution properties: Normalization, monotonicity, concavity, and Lipschitzness (specifically 1-Lipschitzness). For normalization, observe that regardless of n,

K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Ci1)=K:n(x0)(1)=1

K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Ci0)=K:n(x0)(0)=0

The limit of the all-1 sequence is 1 and the limit of the all-0 sequence is 0, so:

K:∞(x0)(1)=1

K:∞(x0)(0)=0

For monotonicity, let f′≥f. Then, regardless of n,

K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))

≥K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

This is because, since f′≥f, the "extend with worst-case outputs" function is bigger for f′ than f, and then by monotonicity for K:n(x0), the inequality transfers to the outside.

Accordingly,

K:∞(x0)(f′)=limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))

≥limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))=K:∞(x0)(f)

Now for concavity. For all n,

K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Ci(pf+(1−p)f′)(x1:n+1,xn+2:∞))

≥K:n(x0)(λx1:n+1.pinfxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)

+(1−p)infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))

≥pK:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

+(1−p)K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))

This was by monotonicity (minimizing over the two parts separately produces lower values, and monotonicity transfers that to the outside), and concavity respectively, because K:n(x0) is an infradistribution. And now:

K:∞(x0)(pf+(1−p)f′)

=limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Ci(pf+(1−p)f′)(x1:n+1,xn+2:∞))

≥limn→∞pK:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

+(1−p)K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))

=plimn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

+(1−p)limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))

=pK:∞(x0)(f)+(1−p)K:∞(x0)(f′)

Concavity is shown. Now for 1-Lipschitzness. Fix an arbitrary n. Our proof target is:

|K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))|≤d(f,f′)

Now, if we knew:

d(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞),λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))≤d(f,f′)

Then we'd be able to pair that with the 1-Lipschitzness of K:n(x0) to get our desired result. So let this be our new proof target. Reexpressing it a bit, it's:

∀x1:n+1∈∏i=n+1i=1Xi:|infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)

−infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)|≤d(f,f′)

Now, if two functions are only d(f,f′) apart, then their minimal values over a compact set can only be d(f,f′) apart at most. Let your compact set be ∏i=n+1i=1{xi}×∏∞i=n+2Ci to see the result. Thus, we have proved our proof target and we have the result that, for all n,f,f′

|K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif′(x1:n+1,xn+2:∞))|≤d(f,f′)

Since they're always only d(f,f′) apart, the same property transfers to the limit, so we get:

|K:∞(x0)(f)−K:∞(x0)(f′)|≤d(f,f′)

And so, K:∞(x0) is 1-Lipschitz for all x0. This takes care of the Lipschitzness condition on infradistributions and the uniform Lipschitz bound condition on infrakernels. All that's left is the compact almost-support condition on infradistributions and the compact-shared compact almost-support condition on infrakernels, and the pointwise convergence condition on infrakernels.

Part 7: This will use our result from Part 4 about making a suitable compact set. Our desired thing we want to prove is that

∀CX0∈K(X0),ϵ>0∃Cϵ⊆∏∞i=1Xi∀x0∈CX0,f,f′∈CB(∏∞i=1Xi):

f↓Cϵ=f′↓Cϵ→|K:∞(x0)(f)−K:∞(x0)(f′)|≤ϵd(f,f′)

Because this property is exactly the compact-shared compact-almost-support property for K:∞.

So, here's our plan of attack. Fix our CX0 and ϵ, so our target is

∃Cϵ⊆∏∞i=1Xi∀x0∈CX0f,f′∈CB(∏∞i=1Xi):

f↓Cϵ=f↓Cϵ→|K:∞(x0)(f)−K:∞(x0)(f′)|≤ϵd(f,f′)

Let our Cϵ be C1:∞[CX0,ϵ], as defined from Part 4, and let x0 be arbitrary in CX0 and f,f′ be arbitrary and agree with each other on C1:∞[CX0,ϵ]. Then our goal becomes

|K:∞(x0)(f)−K:∞(x0)(f′)|≤ϵd(f,f′)

Now, letting our chosen defining sequence be Ci[CX0,ϵ], our goal is to show

∀n:|K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))

−K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f′(x1:n+1,xn+2:∞))|≤ϵd(f,f′)

If we could do this, all the approximating points being only ϵ away from each other shows that the same property transfers to the limit and we get our result. Accordingly, let n be arbitrary, and our proof goal is now:

|K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))

−K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f′(x1:n+1,xn+2:∞))|≤ϵd(f,f′)

Hypothetically, if we were able to show:

∀x1:n+1∈C1:n+1[CX0,ϵ]:infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞)

=infxn+2:∞∈Cn+2:∞[CX0,ϵ]f′(x1:n+1,xn+2:∞)

Then we could conclude the two functions were equal on the set C1:n+1[CX0,ϵ], which, from Part 4, is an ϵ-almost support for all the K:n(x0) where x0∈CX0, and it would yield our result. So our proof target is now

∀x1:n+1∈C1:n+1[CX0,ϵ]:infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞)

=infxn+2:∞∈Cn+2:∞[CX0,ϵ]f′(x1:n+1,xn+2:∞)

Accordingly, let x1:n+1 be arbitrary within said set, turning our proof target into:

infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞)=infxn+2:∞∈Cn+2:∞[CX0,ϵ]f′(x1:n+1,xn+2:∞)

However, f and f′ are equal on the set C1:∞[CX0,ϵ], which breaks down as C1:n+1[CX0,ϵ]×Cn+2:∞[CX0,ϵ]. And we know that x1:n+1∈C1:n+1[CX0,ϵ], and since the second part is being selected from Cn+2:∞[CX0,ϵ], the two values are equal, and we've proved our proof target. Thus, we have compactly-shared compact almost-support, and that's the second condition for K:∞ to be an infrakernel, as well as the last condition needed for all the K:∞(x0) to be infradistributions. Just pointwise convergence left!

Part 8: Alright, for showing pointwise convergence, we'll need a lemma. We need to show:

∀CX0⊆X0,¯¯¯¯C∈∏∞i=1K(Xi),ϵ>0,f∈CB(∏∞i=1Xi)∃n∈N∀m∈N,x0∈CX0:

|K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)

This is roughly saying that the rate of convergence of K:n(x0)(f) to K:∞(x0)(f) is uniform over compact sets. Let's begin. Let CX0,¯¯¯¯C,ϵ,f be arbitrary, so our proof target is now:

∃n∈N∀m∈N,x0∈CX0:|K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)

Here's what we'll do. Because we have a CX0 and ϵ, we can craft our sequence Ci[CX0,ϵ] of compact ϵ-supports. Now, on the set:

∏∞i=1(Ci[CX0,ϵ]∪Ci)

which is compact, we know that f is uniformly continuous, so there's some δ you need to go to to ensure that function values are only ϵ away from each other, which translates into an n via n:=log2(δ). Any two inputs which only differ beyond that point will be only δ apart and can only get an ϵ-difference in values from f. Now that that's defined, let m and x0∈CX0 be arbitrary. Our proof target is now to show:

|K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)

We know from Part 3 that

K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

=K:n+m(x0)(λx1:n+m+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

So let's make that substitution, turning our proof target into

|K:n+m(x0)(λx1:n+m+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)

Assuming hypothetically we were able to show that:

∀x1:n+m+1∈C1:n+m+1[CX0,ϵ]:|infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)

−infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞)|≤ϵ

Then, because C1:n+m+1[CX0,ϵ] is an ϵ-almost-support for K:n+m(x0) as long as x0∈CX0 (which we know to be the case), we could apply Lemma 2 to get our desired proof target.

The "Lipschitz constant times deviation over compact set" part would make one ϵ, and the "degree of almost-support times difference in the two functions" part would make 2ϵ||f|| in the worst-case due to being an ϵ-almost-support and the worst possible case where the two functions differ by 2||f|| somewhere.

So now our proof target is:

∀x1:n+m+1∈C1:n+m+1[CX0,ϵ]:|infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)

−infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞)|≤ϵ

Thus, let x1:n+m+1 be arbitrary within the appropriate set, so our proof target is now:

|infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)

−infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞)|≤ϵ

Because we're minimizing over a compact set in both instances, we can consider the infs to be attained by a xn+2:∞ and a xn+m+2:∞, so our proof target is now

|f(x1:n+1,xn+2:∞)−f(x1:n+m+1,xn+m+2:∞)|≤ϵ

Now, because x1:n+1∈C1:n+1[CX0,ϵ], and xn+2:∞∈∏∞i=n+2Ci, we have:

(x1:n+1,xn+2:∞)∈C1:n+1[CX0,ϵ]×∏∞i=n+2Ci

=∏i=n+1i=1Ci[CX0,ϵ]×∏∞i=n+2Ci⊆∏∞i=1(Ci[CX0,ϵ]∪Ci)

And also, because x1:n+m+1∈C1:n+m+1[CX0,ϵ], and xn+m+2:∞∈∏∞i=n+m+2Ci,

(x1:n+m+1,xn+m+2:∞)∈C1:n+m+1[CX0,ϵ]×∏∞i=n+m+2Ci

=∏i=n+m+1i=1Ci[CX0,ϵ]×∏∞i=n+m+2Ci⊆∏∞i=1(Ci[CX0,ϵ]∪Ci)

So, both inputs lie in the relevant compact set, and they agree on x1:n+1, so the inputs are only δ apart, so they only have value differing by ϵ, and we're done, our desired result of

∀CX0⊆X0,¯¯¯¯C∈∏∞i=1K(Xi),ϵ>0,f∈CB(∏∞i=1Xi)∃n∈N∀m∈N,x0∈CX0:

|K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)

goes through.

Part 9: Now, for showing our pointwise continuity property, fix a sequence x0,m which limits to x0,∞, and an arbitrary f. Our task is now to show that:

limm→∞K:∞(x0,m)(f)=K:∞(x0,∞)(f)

Fixing some arbitrary Ci sequence, we could do this if we could show that

limm→∞limn→∞K:n(x0,m)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

=limn→∞K:n(x0,∞)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

Now, because

K:n(x0,∞)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

=limm→∞K:n(x0,m)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

we can rewrite our desired proof target as:

limm→∞limn→∞K:n(x0,m)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

=limn→∞limm→∞K:n(x0,m)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

Huh, we just need to swap our limits! And the Moore-Osgood theorem says you can swap limits while preserving equality if, for fixed n,

limm→∞K:n(x0,m)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

=K:n(x0,∞)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

and (this part is harder)

K:n(x0,m)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Xif(x1:n+1,xn+2:∞))

limits to

K:∞(x0,m)(f)

uniformly inm. However, back in Part 8 we established uniform convergence on any compact set. And {x0,m}m∈N∪{∞} is a compact set! So we have uniform convergence and can invoke Moore-Osgood to swap the limits and get our proof target, showing our result, thatlimm→∞K:∞(x0,m)(f)=K:∞(x0,∞)(f)

So, K:∞ fulfills pointwise convergence and we've verified all properties needed to show it's an infrakernel.

Part 10: Showing that if all the K:n have some nice property, then K:∞ inherits it too. Homogenity, Cohomogenity, C-additivity, crispness, and sharpness.

Proof of homogenity preservation:

K:∞(x0)(af)=limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Ciaf(x1:n+1,xn+2:∞))

=limn→∞K:n(x0)(λx1:n+1.ainfxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

=limn→∞aK:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))

=alimn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))=aK:∞(x0)(f)

Proof of cohomogenity preservation: Let your Ci be a mere sequence of points xi, that's compact.

K:∞(x0)(1+af)=limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2{xi}1+af(x1:n+1,xn+2:∞))

=limn→∞K:n(x0)(λx1:n+1.1+af(x1:n+1,xn+2:∞))

=limn→∞1−a+aK:n(x0)(λx1:n+1.1+f(x1:n+1,xn+2:∞))

=1−a+alimn→∞K:n(x0)(λx1:n+1.1+f(x1:n+1,xn+2:∞))

=1−a+aK:∞(x0)(1+f)

Proof of C-additivity preservation:

K:∞(x0)(c)=limn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cic)

=limn→∞K:n(x0)(c)=limn→∞c=c

Proof of crispness preservation: Crispness is equivalent to the conjunction of C-additivity and homogenity and both of those are preserved.

Proof of sharpness preservation: So, this will take a bit of work because we do have to get a fairly explicit form for the set you're minimizing over. Remember that when h and K are crisp, with associated compact sets Ch and CK(x), then the compact minimizing set for h⋉K is ⋃x∈ChCK(x), from the proof of sharpness preservation for semidirect product. Another way of writing this minimizing set of the semidirect product is as the set

{(x,y)|x∈Ch,y∈CK(x)}

Now, we'll give a form for the associated compact set of K:n(x0). It is:

{x1:n+1|∀i≤n:xi+1∈C