Theorem 1:Any sequence of infrakernels Kn:∏i=ni=0Xiik→Xn+1 which fulfill the niceness conditions have the infinite semidirect product K:∞ being well-defined and fulfilling the niceness conditions.
As a recap, the niceness conditions are:
1: Lower-semicontinuity for inputs: For all continuous bounded f, x0:n↦Kn(x0:n)(f) is a lower-semicontinuous function.
2: 1-Lipschitzness for functions: For all x0:n, f↦Kn(x0:n)(f) is 1-Lipschitz.
3: Compact-shared compact-almost-support: For all compact sets C⊆∏i=ni=0Xi and ϵ, there is a compact set Cn+1⊆Xn+1 which is an ϵ-almost-support for all the Kn(x0:n) inframeasures where x0:n∈C.
4: Constants increase: For all x0:n and constant functions c, Kn(x0:n)(c)≥c.
5: 1-normalization (only for the [0,1] type signature): For all x0:n, Kn(x0:n)(1)=1
Now, K:∞:X0ik→∏∞i=1Xi is defined as: Fixing an arbitrary sequence of compact sets Ci⊆Xi,
Our task is to show that K:∞ is a well-defined infrakernel which fulfills these five properties.
We will be reusing the proofs from "Less Basic Inframeasure Theory" (henceforth abbreviated as LBIT) when possible, and commenting on the differences when applicable.
The proof sketch is:
Part 1: Show that the functions you're feeding into those K:n infrakernels are guaranteed to be continuous.
Part 2: Show that all the K:n are infrakernels which fulfill the niceness conditions, via induction.
Part 3: Show that if a function only depends on the first n coordinates of the input, then the K:n+m(x0)(f) values monotonically increase as m does.
Part 4: Give a general procedure for taking a compact subset of the space X0 and making a compact subset of the space ∏∞i=1Xi with nice properties related to compact almost-support, that preserves its nice properties when projected down to any finite stage.
Almost-Monotone Lemma: This is a sufficiently useful utility tool in other theorems to be broken out into its own result. It says you can take any f and ϵ and compact set C0⊆X0 and pick a sequence of compact sets Ci for the defining sequence of the infinite semidirect product such that the defining sequence for all the K:∞(x0)(f) (with x0∈C0) is 4ϵ||f||-almost-monotone.
Part 5: Use parts 2, 3, and 4 to show that for any two different Ci sequences, the defining sequences for the infinite infrakernel will limit to each each other (so if one converges, all do, if one diverges, all do, and if one wiggles around forever, all do), and then apply the Almost-Monotone Lemma to rule out the "wiggle around forever" case. Thus, the limit to define K:∞(x0)(f) exists, and doesn't depend on the choice of the sequence of compact sets. This solves well-definedness.
Part 6: Using parts 2 and 5, clean up the basic inframeasure conditions for K:∞(x0) (monotonicity and concavity), and show 1-Lipschitzness, increase of constants, and 1-normalization for K:∞ while we're at it. That's 3/5 niceness conditions down, and all inframeasure conditions except for compact almost-support.
Part 7: Use our trick from Part 4 and our freedom of picking our compact set sequence from Part 5 to show the compact-shared compact-almost-support property for K:∞. This also means that individual outputs of the kernel fulfill CAS, which verifies the last condition we need to conclude that K:∞(x0) is always an inframeasure. Only lower-semicontinuity of the kernel remains to be verified.
Part 8: We use the Almost-Monotone Lemma to wrap up the last condition, the lower-semicontinuity of the infinite infrakernel.
We often use the usual "start with thing we're trying to prove, and keep reducing it to a simpler problem" technique from the last time this proof path was used.
T1.1 Our desired result is whether the function λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞) is continuous. This was already shown in "Less Basic Inframeasure Theory" (LBIT).
T1.2 The desired result is "if all the Kn have the five niceness properties, then all the K:n are infrakernels with the five niceness properties as well".
The proof sketch for this is that this splits into 6 properties we need to show induct up. Monotonicity (phase 1), concavity (phase 2), 1-Lipschitzness (phase 3), increasing constants (phase 4), shared compact-almost-support (phase 5), and lower-semicontinuity (phase 6), which splits into two parts, one where we show a particular function is continuous, and then show lower-semicontinuity in general. Then, we finish up by arguing that this gets everything we need.
For the base case, because K:0:=K0 and we're assuming all the Kn have the niceness properties and are infrakernels, K:0 trivially fulfills them.
Time for the induction step. Remember that K:n+1(x0):=K:n(x0)⋉(λx1:n+1.Kn+1(x0,x1:n+1))
T1.2.4 Proof of increasing constants (and 1-normalization): We have
K:n+1(x0)(c)=K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(c))
And then by increasing constants for Kn+1, and monotonicity for K:n (induction assumption), we have
≥K:n(x0)(λx1:n+1.c)
Now just apply increasing constants for K:n (induction assumption) to get ≥c
For 1-normalization in the [0,1] type signature, we can have c=1, and then instead of an inequality, we have an equality, and the same proof path works, just using 1-normalization instead of increasing constants. They both induct up.
T1.2.5 Proof of the compact-shared CAS property: Our task is to take an arbitrary compact set CX0⊆X0, and find a compact set C∏i=n+2i=1Xiϵ⊆∏i=n+2i=1Xi where two functions f,f′ identical on the set only differ in expectation by ϵd(f,f′) according to all K:n+1(x0) with x0∈CX0.
Let x0∈CX0, and define the set C∏i=n+2i=1Xiϵ as C∏i=n+1i=1Xiϵ2×CXn+2ϵ2
where the first compact set is a shared ϵ2-almost-support for the K:n(x0) where x0∈CX0 (exists by compact-shared CAS for K:n, an induction assumption) and the second compact set is a shared ϵ2-almost-support for the Kn+1(x0,x1:n+1) where x0,x1:n+1 is in CX0×C∏i=n+1i=1Xiϵ2 (exists by compact-shared CAS for Kn+1, an assumed niceness condition). Let f,f′ be continuous and identical on C∏i=n+2i=1Xiϵ. Then, we have
Now, we apply Lemma 2 from LBIT, where we can upper-bound this quantity by "Lipschitz constant of the inframeasure times how much the functions differ on the almost-support + level of almost-support times how much the functions differ in general". We use C∏i=n+1i=1Xiϵ2 as our compact set for the inner function. It was defined as a ϵ2-almost support for all K:n(x0) with x0∈CX0, and K:n is 1-Lipschitz (induction assumption). So our upper bound is:
For the first chunk, we can apply Lemma 2 from LBIT again, with CXn+2ϵ2 as our compact set, which is a ϵ2-almost support for all Kn+1(x0,x1:n+1) with x0,x1:n+1∈CX0×C∏i=n+1i=1Xiϵ2, and Kn+1 is 1-Lipschitz. So our upper bound is:
And we're done, the compact-shared CAS property inducts up.
T1.2.6.1 Proof of lower-semicontinuity: We'll start by taking an extended detour to show that the function
λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
is lower-semicontinuous. Fix a convergent sequence of inputs, xm0,xm1:n+1 which limit to x∞0,x∞1:n+1. First up, the set
{xm0,xm1:n+1}m∈N⊔{∞}
is compact in ∏i=n+1i=0Xi since it converges and we have the limit point added in. By the compact-shared CAS property for Kn+1, we can take any ϵ and make a compact set CXn+2ϵ⊆Xn+2 which is a shared ϵ-almost-support for all the Kn+1(xm0,xm1:n+1) inframeasures. For any m, we can then apply the Lemma 2 (from LBIT) decomposition with the CXn+2ϵϵ-almost-support for all the Kn+1(xm0,xm1:n+1) inframeasures (and 1-Lipschitzness of Kn+1) to get the inequality
Now, the set {xm0,xm1:n+1}m∈N⊔{∞}×CXn+2ϵ is a compact set, so any continuous bounded function f is uniformly continuous on it. For all ϵ, there is some δ where two points only δ apart within this set have f only differing by ϵ on them. Since xm1:n+1 limits to x∞1:n+1, for sufficiently large m, those two points will be within δ of each either, so eventually
supxn+2∈CXn+2ϵ|f(xm1:n+1,xn+2)−f(x∞1:n+1,xn+2)|
will be ϵ or less for late enough m. So, for late enough m, we have
We now run through the same proof path as earlier again. The set {xm0}m∈N⊔{∞} is compact in X0 since it converges and we have the limit point added in. By the compact-shared CAS property for K:n (induction assumption), we can take any ϵ and make a compact set C∏i=n+1i=1Xiϵ which is a shared ϵ-almost-support for all the K:n(xm0) inframeasures. For any m, we can then apply the Lemma 2 (from LBIT) decomposition with the C∏i=n+1i=1Xiϵϵ-almost-support for all the K:n(xm0) inframeasures (and 1-Lipschitzness of K:n by induction assumption) to get the inequality
Now, the set {xm0}m∈N⊔{∞}×C∏i=n+1i=1Xiϵ is a compact set, so any continuous bounded function f is uniformly continuous on it. For all ϵ, there is some δ where two points only δ apart within this set have f only differing by ϵ on them. Since xm0 limits to x∞0, for sufficiently large m, those two points will be within δ of each either, so eventually
and we're done, induction applies for lower-semicontinuity.
All our induction steps have been shown. First, for showing that all the K:n(x0) are inframeasures, we have monotonicity and concavity. Lipschitzness is taken care of by our 1-Lipschitzness induction. Compact almost support is taken care of by our compact-shared CAS induction for the K:n, because a single point is compact. And weak normalization (0 must map to 0 or more) is taken care of by our Constants Increase induction. So all the K:n(x0) are inframeasures.
For the 5 niceness conditions, we proved all of lower-semicontinuity, 1-Lipschitzness, compact-shared CAS, increasing constants, and 1-normalization by induction. So we can now assume that all K:n are nice.
T1.3 We must show that if we go far enough out in the K:n, the value assigned to functions which only depend on finitely many inputs starts monotonically increasing. The result that we'd like to show at this point is:
Since m≥0, that function at the end is a constant (doesn't depend on its input), so by Increasing Constants for Kn+m+1 and Monotonicity for K:n+m, we have
This says that if you pick any compact subset of X0, you can make a compact subset of ∏∞i=1Xi where the projection of it to coordinates 1 through n+1 acts as a compact ϵ(1−12n+1)-almost-support for all the K:n(x0) infradistributions when x0 lies in your compact set CX0.
The proof of this is word-for-word identical to the proof of this from LBIT, nothing changes. To recap the construction, fix some CX0 and ϵ. Now, we recursively build up compact subsets of all the Xn via the following inductive process.
Ci+1[CX0,ϵ]⊆Xi+1 is an arbitrary compact shared ϵ2i+1-almost-support for Ki(x0:i) where x0:i is in CX0×∏j=ij=1Cj[CX0,ϵ] , these sets exist by compact-shared CAS for all the Ki.
We define the abbreviations
C1:n[CX0,ϵ]:=∏i=ni=1Ci[CX0,ϵ]
and
C1:∞[CX0,ϵ]:=∏∞i=1Ci[CX0,ϵ]
This is the product of all the compact sets, and is compact. Notice the dependence of these on the starting compact set and ϵ. Notice that the projection of C1:∞[CX0,ϵ] to coordinates 1 through n is exactly C1:n[CX0,ϵ].
Almost-Monotone Lemma:If {Kn}n∈N is a sequence of infrakernels fulfilling the niceness conditions, then
This says that for any compact subset of X0 and ϵ you name, I can find a "good" sequence of compact sets where the defining sequence for all the K:∞(x0)(f) quantities is "almost monotone". Every time the sequence hits a certain value (at time n), it will never again be able to dip below 4ϵ||f|| of that value as the sequence progresses. This proof progresses in 5 phases. The first phase is just doing some rewrites of our proof goal, the second, third, and fourth are showing setup results on functions undershooting other functions, and certain pairs of quantities being similar. The last phase is wrapping up the actual result.
L1.1 So, for the rewrites, let CX0 be an arbitrary compact subset of X0, and ϵ be arbitrary. Our proof goal is now
If m=0, this result is trivial, we have equality. So now we'll assume m>0. Observe that it is now possible to factorize Cn+2:∞[CX0,ϵ] as Cn+2:n+m+1[CX0,ϵ]×Cn+m+2:∞[CX0,ϵ], therefore,
For this, since f♢n+m+1≥f♢n+1 on the set C1:n+m+1[CX0,ϵ], as we showed in our first desired result, sup(f♢n+m+1,f♢n+1) equals f♢n+m+1 on the set C1:n+m+1[CX0,ϵ].
Further, C1:n+m+1[CX0,ϵ] is an ϵ(1−12n+m+1)-almost-support for all the K:n+m(x0) with x0∈CX0, as our x0 is. So, in particular, it's an ϵ-almost-support. Accordingly, we have
Now, for any input, either f♢n+m+1 is higher (and so the distance between the functions at that point is 0), or f♢n+1 is higher, and so the distance between the functions at that point is upper-bounded by d(f♢n+m+1,f♢n+1). So, we have
≤ϵd(f♢n+1,f♢n+m+1)
At this point, we recall that both f♢n+m+1 and f♢n+1 were just f, but taking worst-case inputs past a certain point, so a trivial bound on this quantity is 2||f||. So, our net result is
For this, since f♢n+m+1≥f♢n+1 on the set C1:n+m+1[CX0,ϵ], as we showed in our first desired result, inf(f♢n+m+1,f♢n+1) equals f♢n+1 on the set C1:n+m+1[CX0,ϵ].
Further, C1:n+m+1[CX0,ϵ] is an ϵ(1−12n+m+1)-almost-support for all the K:n+m(x0) with x0∈CX0, as our x0 is. So, in particular, it's an ϵ-almost-support. Accordingly, we have
|K:n+m(x0)(f♢n+1)−K:n+m(x0)(inf(f♢n+m+1,f♢n+1))|
≤ϵd(inf(f♢n+m+1,f♢n+1),f♢n+1)
Now, for any input, either f♢n+1 is lower (and so the distance between the functions at that point is 0), or f♢n+m+1 is lower, and so the distance between the functions at that point is upper-bounded by d(f♢n+m+1,f♢n+1). So, we have
≤ϵd(f♢n+1,f♢n+m+1)
At this point, we recall that both f♢n+m+1 and f♢n+1 were just f, but taking worst-case inputs past a certain point, so a trivial bound on this quantity is 2||f||. So, our net result is
T1.5 Our aim here is to show that no matter what sequence of compact sets you have, they all limit to the same result. In order to do this, we'll have to show the result (letting Ci,1 being your first sequence of nonempty compact sets to attempt to define the limit and Ci,2 being your second sequence of nonempty compact sets, and abbreviating Cn+m+2:∞,1 as the product of the Ci,1 from n+m+2 to ∞) that,
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:∞(x0)(f) associated with using Ci,1 as your compact sets, and the sequence associated with using Ci,2 as your compact sets (as long as they're both past some shared threshold), they'll be close. So all choices of compact set must limit to each other, because ϵ can be arbitrary.
Fix our x0,ϵ,¯¯¯¯¯¯C1,¯¯¯¯¯¯C2,f (input value, closeness parameter, two sequences of compact sets, function), and now we'll find our n to make
true. Do it in the following way. Use {x0} as your compact seed set to invoke the technique in part 4 to construct your sequence Ci[x0,ϵ], and then consider the set:
∏∞i=1(Ci,1∪Ci,2∪Ci[x0,ϵ])
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(x1:∞,x′1:∞)=∑∞i=12−iinf(1,dXi(xi,x′i))
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 δ difference in inputs that only leads to an ϵ difference in values. Now we can define our n as log2(δ).
Now that we've picked our n, let m be arbitrary. Our goal is to now prove:
We can upper-bound the second term with just 2||f||, as the difference in values can't be any higher than the gap between the maximum value of f and the minimum value of f. As for the first term, note that if x1:n+m+1∈C1:n+m+1[x0,ϵ] and xn+m+2:∞∈∏∞i=n+m+2Ci,1 and x′n+m+2:∞∈∏∞i=n+m+2Ci,2(the latter two are the inf terms for a particular x1:n+m+1) then we have
x1:n+m+1,xn+m+2:∞∈∏∞i=1(Ci,1∪Ci,2∪Ci[x0,ϵ])
And same for x1:n+m+1,x′n+m+2:∞. And these two points agree on x1:n+m+1, and n is large enough that the function itself varies by only ϵ for any two points in that set which share the same prefix up to length n (and x1:n+m+1 fulfills that) So, we can break down the upper bound as
So all the defining sequences for K:∞(x0)(f) limit to each other. We've got three possible cases. One is divergence, that liminf of the defining sequences goes to infinity. This is fine, the limit exists. One is convergence, that liminf and limsup are finite and equal, ie, the limit exists. This is fine. And the third case, that liminf and limsup aren't equal, must be ruled out. We'll invoke the Almost-Monotone Lemma, that
Specializing to the compact set consisting of a single arbitrary point x0, and recalling that the defining sequence for that was the Ci[x0,ϵ] sequence of almost-supports, and an arbitrary function f, the lemma turns into
Since the limsup is always equal or greater than the liminf, this implies that the limsup and liminf can be at most 4ϵ||f|| apart for this sequence. And since all the defining sequences limit to each other, they all have the same limsup and liminf. Since ϵ was arbitrary, we get that all the defining sequences limit to each other, and either converge to a fixed number, or diverge to infinity. The infinite semidirect product is now well-defined, and we can use whichever sequence of compact sets is the most convenient.
T1.6 Showing the following infradistribution properties for the infinite semidirect product: monotonicity and concavity. Also, the 1-Lipschitz property and 1-normalization property and Constants Increase property, for 4/5 inframeasure properties, and 3/5 niceness properties.
For monotonicity, concavity, and 1-Lipschitzness, the exact proof from LBIT works.
For constants increasing, observe that regardless of n,
The same proof applies to 1-normalization, but with equality signs, if we plug in 1, and the limit of the all-1 sequence is always 1.
T1.7 Showing compact-shared compact-almost-support for K:∞. Again, the exact proof from LBIT works here. We're up to 5/5 inframeasure properties, and 4/5 niceness properties for the infinite semidirect product.
T1.8 Our task is to show lower-semicontinuity for K:∞. This will take some cunning. At the start, fix an ϵ, we'll see why later. Our task is to show that
liminfm→∞K:∞(xm0)(f)≥K:∞(x∞0)(f)
For some arbitrary sequence of xm0 which limits to x∞0. What we can do is first unpack.
Now, the sequence of sets we picked was important. It means we can now apply the Almost-Monotone Lemma. Regardless of m, the defining sequence for K:∞(xm0)(f) is almost monotone now. We can fix a particular n∗ (arbitrary) to get a lower bound of
Theorem 1: Any sequence of infrakernels Kn:∏i=ni=0Xiik→Xn+1 which fulfill the niceness conditions have the infinite semidirect product K:∞ being well-defined and fulfilling the niceness conditions.
As a recap, the niceness conditions are:
1: Lower-semicontinuity for inputs: For all continuous bounded f, x0:n↦Kn(x0:n)(f) is a lower-semicontinuous function.
2: 1-Lipschitzness for functions: For all x0:n, f↦Kn(x0:n)(f) is 1-Lipschitz.
3: Compact-shared compact-almost-support: For all compact sets C⊆∏i=ni=0Xi and ϵ, there is a compact set Cn+1⊆Xn+1 which is an ϵ-almost-support for all the Kn(x0:n) inframeasures where x0:n∈C.
4: Constants increase: For all x0:n and constant functions c, Kn(x0:n)(c)≥c.
5: 1-normalization (only for the [0,1] type signature): For all x0:n, Kn(x0:n)(1)=1
Now, K:∞:X0ik→∏∞i=1Xi is defined as: Fixing an arbitrary sequence of compact sets Ci⊆Xi,
K:∞(x0)(f):=limn→∞K:n(x0)(λx1:n+1infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))
Our task is to show that K:∞ is a well-defined infrakernel which fulfills these five properties.
We will be reusing the proofs from "Less Basic Inframeasure Theory" (henceforth abbreviated as LBIT) when possible, and commenting on the differences when applicable.
The proof sketch is:
Part 1: Show that the functions you're feeding into those K:n infrakernels are guaranteed to be continuous.
Part 2: Show that all the K:n are infrakernels which fulfill the niceness conditions, via induction.
Part 3: Show that if a function only depends on the first n coordinates of the input, then the K:n+m(x0)(f) values monotonically increase as m does.
Part 4: Give a general procedure for taking a compact subset of the space X0 and making a compact subset of the space ∏∞i=1Xi with nice properties related to compact almost-support, that preserves its nice properties when projected down to any finite stage.
Almost-Monotone Lemma: This is a sufficiently useful utility tool in other theorems to be broken out into its own result. It says you can take any f and ϵ and compact set C0⊆X0 and pick a sequence of compact sets Ci for the defining sequence of the infinite semidirect product such that the defining sequence for all the K:∞(x0)(f) (with x0∈C0) is 4ϵ||f||-almost-monotone.
Part 5: Use parts 2, 3, and 4 to show that for any two different Ci sequences, the defining sequences for the infinite infrakernel will limit to each each other (so if one converges, all do, if one diverges, all do, and if one wiggles around forever, all do), and then apply the Almost-Monotone Lemma to rule out the "wiggle around forever" case. Thus, the limit to define K:∞(x0)(f) exists, and doesn't depend on the choice of the sequence of compact sets. This solves well-definedness.
Part 6: Using parts 2 and 5, clean up the basic inframeasure conditions for K:∞(x0) (monotonicity and concavity), and show 1-Lipschitzness, increase of constants, and 1-normalization for K:∞ while we're at it. That's 3/5 niceness conditions down, and all inframeasure conditions except for compact almost-support.
Part 7: Use our trick from Part 4 and our freedom of picking our compact set sequence from Part 5 to show the compact-shared compact-almost-support property for K:∞. This also means that individual outputs of the kernel fulfill CAS, which verifies the last condition we need to conclude that K:∞(x0) is always an inframeasure. Only lower-semicontinuity of the kernel remains to be verified.
Part 8: We use the Almost-Monotone Lemma to wrap up the last condition, the lower-semicontinuity of the infinite infrakernel.
We often use the usual "start with thing we're trying to prove, and keep reducing it to a simpler problem" technique from the last time this proof path was used.
T1.1 Our desired result is whether the function λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞)
is continuous. This was already shown in "Less Basic Inframeasure Theory" (LBIT).
T1.2 The desired result is "if all the Kn have the five niceness properties, then all the K:n are infrakernels with the five niceness properties as well".
The proof sketch for this is that this splits into 6 properties we need to show induct up. Monotonicity (phase 1), concavity (phase 2), 1-Lipschitzness (phase 3), increasing constants (phase 4), shared compact-almost-support (phase 5), and lower-semicontinuity (phase 6), which splits into two parts, one where we show a particular function is continuous, and then show lower-semicontinuity in general. Then, we finish up by arguing that this gets everything we need.
For the base case, because K:0:=K0 and we're assuming all the Kn have the niceness properties and are infrakernels, K:0 trivially fulfills them.
Time for the induction step. Remember that K:n+1(x0):=K:n(x0)⋉(λx1:n+1.Kn+1(x0,x1:n+1))
T1.2.1 Proof of monotonicity: Assume f′≥f. Then
K:n+1(x0)(f′)=K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2)))
And the same goes for f so if we could prove
K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2)))
≥K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2)))
we'd be successful. Now, by monotonicity for K:n (by induction assumption), we'd have that result if
∀x1:n+1:Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2))
≥Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
Letting x1:n+1 be arbitrary, by applying monotonicity for Kn+1, we'd have that result if
∀xn+2:f′(x1:n+1,xn+2)≥f(x1:n+1,xn+2)
Which is true since f′≥f. And so we're done, montonicity inducts up.
T1.2.2 Proof of concavity:
K:n+1(x0)(pf+(1−p)f′)
=K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.pf(x1:n+1,xn+2)+(1−p)f′(x1:n+1,xn+2)))
and then, by concavity for all Kn+1, and monotonicity for K:n (by induction assumption), we have
≥K:n(x0)(λx1:n+1.pKn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
+(1−p)Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2)))
and then, by concavity for K:n (by induction assumption), we have
≥pK:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2)))
+(1−p)K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2)))
and then this packs up as
=pK:n+1(x0)(f)+(1−p)K:n+1(f′)
and concavity inducts on up.
T1.2.3 Proof of 1-Lipschitzness: we have
|K:n+1(x0)(f)−K:n+1(x0)(f′)|
=|K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2)))
−K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2)))|
And by 1-Lipschitzness for K:n (by induction assumption), we have
≤supx1:n+1|Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
−Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2))|
And by 1-Lipschitzness for Kn+1, we have
≤supx1:n+1,xn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|=d(f,f′)
And 1-Lipschitzness inducts on up.
T1.2.4 Proof of increasing constants (and 1-normalization): We have
K:n+1(x0)(c)=K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(c))
And then by increasing constants for Kn+1, and monotonicity for K:n (induction assumption), we have
≥K:n(x0)(λx1:n+1.c)
Now just apply increasing constants for K:n (induction assumption) to get ≥c
For 1-normalization in the [0,1] type signature, we can have c=1, and then instead of an inequality, we have an equality, and the same proof path works, just using 1-normalization instead of increasing constants. They both induct up.
T1.2.5 Proof of the compact-shared CAS property: Our task is to take an arbitrary compact set CX0⊆X0, and find a compact set C∏i=n+2i=1Xiϵ⊆∏i=n+2i=1Xi where two functions f,f′ identical on the set only differ in expectation by ϵd(f,f′) according to all K:n+1(x0) with x0∈CX0.
Let x0∈CX0, and define the set C∏i=n+2i=1Xiϵ as C∏i=n+1i=1Xiϵ2×CXn+2ϵ2
where the first compact set is a shared ϵ2-almost-support for the K:n(x0) where x0∈CX0 (exists by compact-shared CAS for K:n, an induction assumption) and the second compact set is a shared ϵ2-almost-support for the Kn+1(x0,x1:n+1) where x0,x1:n+1 is in CX0×C∏i=n+1i=1Xiϵ2 (exists by compact-shared CAS for Kn+1, an assumed niceness condition). Let f,f′ be continuous and identical on C∏i=n+2i=1Xiϵ. Then, we have
|K:n+1(x0)(f)−K:n+1(x0)(f′)|
=|K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2)))
−K:n(x0)(λx1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2)))|
Now, we apply Lemma 2 from LBIT, where we can upper-bound this quantity by "Lipschitz constant of the inframeasure times how much the functions differ on the almost-support + level of almost-support times how much the functions differ in general". We use C∏i=n+1i=1Xiϵ2 as our compact set for the inner function. It was defined as a ϵ2-almost support for all K:n(x0) with x0∈CX0, and K:n is 1-Lipschitz (induction assumption). So our upper bound is:
≤supx1:n+1∈C∏i=n+1i=1Xiϵ2|Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
−Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2))|
+ϵ2⋅supx1:n+1|Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
−Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2))|
Focusing on that second chunk, we can apply 1-Lipschitzness of Kn+1 to yield
≤supx1:n+1∈C∏i=n+1i=1Xiϵ2|Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
−Kn+1(x0,x1:n+1)(λxn+2.f′(x1:n+1,xn+2))|
+ϵ2⋅supx1:n+1,xn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|
For the first chunk, we can apply Lemma 2 from LBIT again, with CXn+2ϵ2 as our compact set, which is a ϵ2-almost support for all Kn+1(x0,x1:n+1) with x0,x1:n+1∈CX0×C∏i=n+1i=1Xiϵ2, and Kn+1 is 1-Lipschitz. So our upper bound is:
≤supx1:n+1∈C∏i=n+1i=1Xiϵ2(supxn+2∈CXn+2ϵ2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|
+ϵ2⋅supxn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|)
+ϵ2⋅supx1:n+1,xn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|
Now, since f and f′ are identical on C∏i=n+1i=1Xiϵ2×CXn+2ϵ2 that first term just vanishes, yielding
=supx1:n+1∈C∏i=n+1i=1Xiϵ2(ϵ2⋅supxn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|)
+ϵ2⋅supx1:n+1,xn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|
We upper-bound this by
≤ϵ2supx1:n+1,xn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|
+ϵ2⋅supx1:n+1,xn+2|f(x1:n+1,xn+2)−f′(x1:n+1,xn+2)|
Which then clearly just reduces to
=ϵd(f,f′)
And we're done, the compact-shared CAS property inducts up.
T1.2.6.1 Proof of lower-semicontinuity: We'll start by taking an extended detour to show that the function
λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
is lower-semicontinuous. Fix a convergent sequence of inputs, xm0,xm1:n+1 which limit to x∞0,x∞1:n+1. First up, the set
{xm0,xm1:n+1}m∈N⊔{∞}
is compact in ∏i=n+1i=0Xi since it converges and we have the limit point added in. By the compact-shared CAS property for Kn+1, we can take any ϵ and make a compact set CXn+2ϵ⊆Xn+2 which is a shared ϵ-almost-support for all the Kn+1(xm0,xm1:n+1) inframeasures. For any m, we can then apply the Lemma 2 (from LBIT) decomposition with the CXn+2ϵ ϵ-almost-support for all the Kn+1(xm0,xm1:n+1) inframeasures (and 1-Lipschitzness of Kn+1) to get the inequality
|Kn+1(xm0,xm1:n+1)(λxn+2.f(xm1:n+1,xn+2))−Kn+1(xm0,xm1:n+1)(λxn+2.f(x∞1:n+1,xn+2))|
≤supxn+2∈CXn+2ϵ|f(xm1:n+1,xn+2)−f(x∞1:n+1,xn+2)|
+ϵsupxn+2|f(xm1:n+1,xn+2)−f(x∞1:n+1,xn+2)|
That second term can be upper-bounded by 2||f||, so doing that, we have
≤supxn+2∈CXn+2ϵ|f(xm1:n+1,xn+2)−f(x∞1:n+1,xn+2)|+2ϵ||f||
Now, the set {xm0,xm1:n+1}m∈N⊔{∞}×CXn+2ϵ is a compact set, so any continuous bounded function f is uniformly continuous on it. For all ϵ, there is some δ where two points only δ apart within this set have f only differing by ϵ on them. Since xm1:n+1 limits to x∞1:n+1, for sufficiently large m, those two points will be within δ of each either, so eventually
supxn+2∈CXn+2ϵ|f(xm1:n+1,xn+2)−f(x∞1:n+1,xn+2)|
will be ϵ or less for late enough m. So, for late enough m, we have
|Kn+1(xm0,xm1:n+1)(λxn+2.f(xm1:n+1,xn+2))−Kn+1(xm0,xm1:n+1)(λxn+2.f(x∞1:n+1,xn+2))|
≤ϵ+2ϵ||f||
Since ϵ can be arbitrary, we have that the sequences
Kn+1(xm0,xm1:n+1)(λxn+2.f(xm1:n+1,xn+2))
and
Kn+1(xm0,xm1:n+1)(λxn+2.f(x∞1:n+1,xn+2))
limit to each other. Therefore,
liminfm→∞Kn+1(xm0,xm1:n+1)(λxn+2.f(xm1:n+1,xn+2))
=liminfm→∞Kn+1(xm0,xm:n+1)(λxn+2.f(x∞1:n+1,xn+2))
And then, by lower-semicontinuity for Kn+1, we have
≥Kn+1(x∞0,x∞1:n+1)(λxn+2.f(x∞1:n+1,xn+2))
and so, lower-semicontinuity of the function
λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
is shown. Time to return to the usual induction proof.
T1.2.6.2 For lower-semicontinuity in inputs, we first unpack the semidirect product. Fix a sequence xm0 which limits to x∞0.
liminfm→∞K:n+1(xm0)(f)
=liminfm→∞K:n(xm0)(λx1:n+1.Kn+1(xm0,x1:n+1)(λxn+2.f(x1:n+1,xn+2)))
Since we showed that the function
λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
is lower-semicontinuous, we can apply the monotone convergence theorem for inframeasures and rewrite as
=liminfm→∞supf′≤λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))K:n(xm0)(λx1:n+1.f′(xm0,x1:n+1))
For any individual f′, call it f∗, we have
≥liminfm→∞K:n(xm0)(λx1:n+1.f∗(xm0,x1:n+1))
We now run through the same proof path as earlier again. The set {xm0}m∈N⊔{∞} is compact in X0 since it converges and we have the limit point added in. By the compact-shared CAS property for K:n (induction assumption), we can take any ϵ and make a compact set C∏i=n+1i=1Xiϵ which is a shared ϵ-almost-support for all the K:n(xm0) inframeasures. For any m, we can then apply the Lemma 2 (from LBIT) decomposition with the C∏i=n+1i=1Xiϵ ϵ-almost-support for all the K:n(xm0) inframeasures (and 1-Lipschitzness of K:n by induction assumption) to get the inequality
|K:n(xm0)(λx1:n+2.f∗(xm0,x1:n+1))−K:n(xm0)(λx1:n+2.f∗(x∞0,x1:n+1))|
≤supx1:n+1∈C∏i=n+1i=1Xiϵ|f∗(xm0,x1:n+1)−f(x∞0,x1:n+1)|
+ϵsupx1:n+1|f∗(xm0,x1:n+1)−f∗(x∞0,x1:n+1)|
That second term can be upper-bounded by 2||f||, so doing that, we have
≤supx1:n+1∈C∏i=n+1i=1Xiϵ|f∗(xm0,x1:n+1)−f∗(x∞0,x1:n+1)|+2ϵ||f||
Now, the set {xm0}m∈N⊔{∞}×C∏i=n+1i=1Xiϵ is a compact set, so any continuous bounded function f is uniformly continuous on it. For all ϵ, there is some δ where two points only δ apart within this set have f only differing by ϵ on them. Since xm0 limits to x∞0, for sufficiently large m, those two points will be within δ of each either, so eventually
supx1:n+1∈C∏i=n+1i=1Xiϵ|f∗(xm0,x1:n+1)−f∗(x∞0,x1:n+1)|
will be ϵ or less for late enough m. So, for late enough m, we have
|K:n(xm0)(λx1:n+2.f∗(xm0,x1:n+1))−K:n(xm0)(λx1:n+2.f∗(x∞0,x1:n+1))|≤ϵ+2ϵ||f||
Since ϵ can be arbitrary, we have that the sequences
K:n(xm0)(λx1:n+2.f∗(xm0,x1:n+1))
and
K:n(xm0)(λx1:n+2.f∗(x∞0,x1:n+1))
limit to each other. Taking a break to recap, we had
liminfm→∞K:n+1(xm0)(f)
=liminfm→∞supf′≤λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))K:n(xm0)(λx1:n+1.f′(xm0,x1:n+1))
and for any f∗ fulfilling those bounding conditions for the sup, we have
≥liminfm→∞K:n(xm0)(λx1:n+1.f∗(xm0,x1:n+1))
and from our fresh new result, we then have
=liminfm→∞K:n(xm0)(λx1:n+1.f∗(x∞0,x1:n+1))
and, by lower-semicontinuity for K:n (induction assumption), we have
≥K:n(x∞0)(λx1:n+1.f∗(x∞0,x1:n+1))
Since f∗ was arbitrary below
λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))
this means that
liminfm→∞K:n+1(xm0)(f)
≥supf′≤λx0,x1:n+1.Kn+1(x0,x1:n+1)(λxn+2.f(x1:n+1,xn+2))K:n(x∞0)(λx1:n+1.f′(x∞0,x1:n+1))
Now, we can put our lower-semicontinuous upper bound back in to get
=K:n(x∞0)(λx1:n+1.Kn+1(x∞0,x1:n+1)(λxn+2.f(x1:n+1,xn+2)))
=K:n+1(x∞0)(f)
and we're done, induction applies for lower-semicontinuity.
All our induction steps have been shown. First, for showing that all the K:n(x0) are inframeasures, we have monotonicity and concavity. Lipschitzness is taken care of by our 1-Lipschitzness induction. Compact almost support is taken care of by our compact-shared CAS induction for the K:n, because a single point is compact. And weak normalization (0 must map to 0 or more) is taken care of by our Constants Increase induction. So all the K:n(x0) are inframeasures.
For the 5 niceness conditions, we proved all of lower-semicontinuity, 1-Lipschitzness, compact-shared CAS, increasing constants, and 1-normalization by induction. So we can now assume that all K:n are nice.
T1.3 We must show that if we go far enough out in the K:n, the value assigned to functions which only depend on finitely many inputs starts monotonically increasing. The result that we'd like to show at this point is:
∀x0∈X0,n,m∈N,f∈CB(∏i=n+1i=1Xi):K:n+m+1(x0)(f)≥K:n+m(x0)(f)
Fix an arbitrary x0,n,m, and f. Then, we can just go:
K:n+m+1(x0)(f)
=K:n+m(x0)(λx1:n+m+1.Kn+m+1(x0,x1:n+m+1)(λxn+m+2.f(x1:n+1)))
Since m≥0, that function at the end is a constant (doesn't depend on its input), so by Increasing Constants for Kn+m+1 and Monotonicity for K:n+m, we have
≥K:n+m(x0)(λx1:n+m+1.f(x1:n+1))=K:n+m(x0)(f)
and we're done.
T1.4 Our desired result here is:
∀CX0∈K(X0),ϵ>0:∃C1:∞[CX0,ϵ]∈K(∏∞i=1Xi):∀n∈N,x0∈CX0,f,f′∈CB(∏n+1i=1Xi):
f↓pr1:n+1(C1:∞[CX0,ϵ])=f′↓pr1:n+1(C1:∞[CX0,ϵ])
→|K:n(x0)(f)−K:n(x0)(f′)|≤ϵ(1−12n+1)d(f,f′)
This says that if you pick any compact subset of X0, you can make a compact subset of ∏∞i=1Xi where the projection of it to coordinates 1 through n+1 acts as a compact ϵ(1−12n+1)-almost-support for all the K:n(x0) infradistributions when x0 lies in your compact set CX0.
The proof of this is word-for-word identical to the proof of this from LBIT, nothing changes. To recap the construction, fix some CX0 and ϵ. Now, we recursively build up compact subsets of all the Xn via the following inductive process.
Ci+1[CX0,ϵ]⊆Xi+1 is an arbitrary compact shared ϵ2i+1-almost-support for Ki(x0:i) where x0:i is in CX0×∏j=ij=1Cj[CX0,ϵ] , these sets exist by compact-shared CAS for all the Ki.
We define the abbreviations
C1:n[CX0,ϵ]:=∏i=ni=1Ci[CX0,ϵ]
and
C1:∞[CX0,ϵ]:=∏∞i=1Ci[CX0,ϵ]
This is the product of all the compact sets, and is compact. Notice the dependence of these on the starting compact set and ϵ. Notice that the projection of C1:∞[CX0,ϵ] to coordinates 1 through n is exactly C1:n[CX0,ϵ].
Almost-Monotone Lemma: If {Kn}n∈N is a sequence of infrakernels fulfilling the niceness conditions, then
∀CX0∈K(X0),ϵ>0:∃¯¯¯¯C∈∏∞i=1K(Xi):∀n,m∈N,x0∈CX0,f∈CB(∏∞i=1Xi):
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))−4ϵ||f||
This says that for any compact subset of X0 and ϵ you name, I can find a "good" sequence of compact sets where the defining sequence for all the K:∞(x0)(f) quantities is "almost monotone". Every time the sequence hits a certain value (at time n), it will never again be able to dip below 4ϵ||f|| of that value as the sequence progresses. This proof progresses in 5 phases. The first phase is just doing some rewrites of our proof goal, the second, third, and fourth are showing setup results on functions undershooting other functions, and certain pairs of quantities being similar. The last phase is wrapping up the actual result.
L1.1 So, for the rewrites, let CX0 be an arbitrary compact subset of X0, and ϵ be arbitrary. Our proof goal is now
∃¯¯¯¯C∈∏∞i=1K(Xi):∀n,m∈N,x0∈CX0,f∈CB(∏∞i=1Xi):
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))−4ϵ||f||
Let your sequence of compact sets be the Ci[CX0,ϵ] sequence recursively defined from CX0 and ϵ. Then our proof goal becomes
∀n,m∈N,x0∈CX0,f∈CB(∏∞i=1Xi):
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))−4ϵ||f||
Let n,m,x0, and f be arbitrary. Remember x0 is in CX0. Then our proof goal becomes
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))−4ϵ||f||
First, let's define some abbreviations for later. The function
λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞)
will be abbreviated as f♢n+m+1, and the function
λx1:n+m+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞)
will be abbreviated as f♢n+1. Both of these functions are continuous, by section 1. We'll need three key results before beginning.
L1.2 Our first desired result is:
∀x1:n+m+1∈C1:n+m+1[CX0,ϵ]:f♢n+m+1(x1:n+m+1)≥f♢n+1(x1:n+m+1)
Unpacking this proof target, it is
∀x1:n+m+1∈C1:n+m+1[CX0,ϵ]:
infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞)≥infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞)
If m=0, this result is trivial, we have equality. So now we'll assume m>0. Observe that it is now possible to factorize Cn+2:∞[CX0,ϵ] as Cn+2:n+m+1[CX0,ϵ]×Cn+m+2:∞[CX0,ϵ], therefore,
infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞)
turns into
infxn+2:n+m+1∈Cn+2:n+m+1[CX0,ϵ]infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+1,xn+2:n+m+1,xn+m+2:∞)
which is just
infxn+2:n+m+1∈Cn+2:n+m+1[CX0,ϵ]f♢n+m+1(xn+1,xn+2:n+m+1)
Packing up f♢n+m+1 on the left side of our proof target, our proof target is now
∀x1:n+m+1∈C1:n+m+1[CX0,ϵ]:
f♢n+m+1(x1:n+m+1)≥infxn+2:n+m+1∈Cn+2:n+m+1[CX0,ϵ]f♢n+m+1(xn+1,xn+2:n+m+1)
And then, since we can factorize C1:n+m+1[CX0,ϵ] as C1:n+1[CX0,ϵ]×Cn+2:n+m+1[CX0,ϵ], our proof target is now
∀x1:n+1,xn+2:n+m+1∈C1:n+1[CX0,ϵ]×Cn+2:n+m+1[CX0,ϵ]:
f♢n+m+1(x1:n+1,xn+2:n+m+1)≥infx′n+2:n+m+1∈Cn+2:n+m+1[CX0,ϵ]f♢n+m+1(xn+1,x′n+2:n+m+1)
Which is obviously true. So, we have our first result of
∀x1:n+m+1∈C1:n+m+1[CX0,ϵ]:f♢n+m+1(x1:n+m+1)≥f♢n+1(x1:n+m+1)
L1.3 Time for our second desired result. Our second desired result is that
|K:n+m(x0)(sup(f♢n+m+1,f♢n+1))−K:n+m(x0)(f♢n+m+1)|≤2ϵ||f||
For this, since f♢n+m+1≥f♢n+1 on the set C1:n+m+1[CX0,ϵ], as we showed in our first desired result, sup(f♢n+m+1,f♢n+1) equals f♢n+m+1 on the set C1:n+m+1[CX0,ϵ].
Further, C1:n+m+1[CX0,ϵ] is an ϵ(1−12n+m+1)-almost-support for all the K:n+m(x0) with x0∈CX0, as our x0 is. So, in particular, it's an ϵ-almost-support. Accordingly, we have
|K:n+m(x0)(sup(f♢n+m+1,f♢n+1))−K:n+m(x0)(f♢n+m+1)|
≤ϵd(sup(f♢n+m+1,f♢n+1),f♢n+m+1)
Now, for any input, either f♢n+m+1 is higher (and so the distance between the functions at that point is 0), or f♢n+1 is higher, and so the distance between the functions at that point is upper-bounded by d(f♢n+m+1,f♢n+1). So, we have
≤ϵd(f♢n+1,f♢n+m+1)
At this point, we recall that both f♢n+m+1 and f♢n+1 were just f, but taking worst-case inputs past a certain point, so a trivial bound on this quantity is 2||f||. So, our net result is
|K:n+m(x0)(sup(f♢n+m+1,f♢n+1))−K:n+m(x0)(f♢n+m+1)|≤2ϵ||f||
L1.4 And it's time to move onto our third setup result. Our desired third setup result is that
|K:n+m(x0)(f♢n+1)−K:n+m(x0)(inf(f♢n+m+1,f♢n+1))|≤2ϵ||f||
For this, since f♢n+m+1≥f♢n+1 on the set C1:n+m+1[CX0,ϵ], as we showed in our first desired result, inf(f♢n+m+1,f♢n+1) equals f♢n+1 on the set C1:n+m+1[CX0,ϵ].
Further, C1:n+m+1[CX0,ϵ] is an ϵ(1−12n+m+1)-almost-support for all the K:n+m(x0) with x0∈CX0, as our x0 is. So, in particular, it's an ϵ-almost-support. Accordingly, we have
|K:n+m(x0)(f♢n+1)−K:n+m(x0)(inf(f♢n+m+1,f♢n+1))|
≤ϵd(inf(f♢n+m+1,f♢n+1),f♢n+1)
Now, for any input, either f♢n+1 is lower (and so the distance between the functions at that point is 0), or f♢n+m+1 is lower, and so the distance between the functions at that point is upper-bounded by d(f♢n+m+1,f♢n+1). So, we have
≤ϵd(f♢n+1,f♢n+m+1)
At this point, we recall that both f♢n+m+1 and f♢n+1 were just f, but taking worst-case inputs past a certain point, so a trivial bound on this quantity is 2||f||. So, our net result is
|K:n+m(x0)(f♢n+1)−K:n+m(x0)(inf(f♢n+m+1,f♢n+1))|≤2ϵ||f||
Now that everything's in place, we can start working on our actual result.
L1.5 Remember, our overall proof target at this stage is
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))−4ϵ||f||
We can reshuffle this proof target a little to get the equivalent statement
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞))+4ϵ||f||
≥K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))
So let's get cracking on that inequality.
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[CX0,ϵ]f(x1:n+m+1,xn+m+2:∞))+4ϵ||f||
First, we abbreviate it as
=K:n+m(x0)(f♢n+m+1)+4ϵ||f||
Then, we apply our second sublemma we proved to get
≥K:n+m(x0)(f♢n+m+1)+|K:n+m(x0)(sup(f♢n+m+1,f♢n+1))−K:n+m(x0)(f♢n+m+1)|+2ϵ||f||
And undo the absolute value to get
≥K:n+m(x0)(f♢n+m+1)+K:n+m(x0)(sup(f♢n+m+1,f♢n+1))−K:n+m(x0)(f♢n+m+1)+2ϵ||f||
and cancel out to get
=K:n+m(x0)(sup(f♢n+m+1,f♢n+1))+2ϵ||f||
and then, by monotonicity we have
≥K:n+m(x0)(inf(f♢n+m+1,f♢n+1))+2ϵ||f||
and apply our third sublemma we proved to get
≥K:n+m(x0)(inf(f♢n+m+1,f♢n+1))+|K:n+m(x0)(f♢n+1)−K:n+m(x0)(inf(f♢n+m+1,f♢n+1))|
And undo the absolute value
≥K:n+m(x0)(inf(f♢n+m+1,f♢n+1))+K:n+m(x0)(f♢n+1)−K:n+m(x0)(inf(f♢n+m+1,f♢n+1))
And cancel
=K:n+m(x0)(f♢n+1)
and at this point, we can unpack the inner function to get
=K:n+m(x0)(λx1:n+m+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))
And then, we apply our Part 3 result that for functions which only depend on finitely many inputs, K:n+m(x0)(f) is monotonically increasing, to get
≥K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[CX0,ϵ]f(x1:n+1,xn+2:∞))
and we're done with the Almost-Monotone Lemma!
T1.5 Our aim here is to show that no matter what sequence of compact sets you have, they all limit to the same result. In order to do this, we'll have to show the result (letting Ci,1 being your first sequence of nonempty compact sets to attempt to define the limit and Ci,2 being your second sequence of nonempty compact sets, and abbreviating Cn+m+2:∞,1 as the product of the Ci,1 from n+m+2 to ∞) that,
∀x0∈X0¯¯¯¯¯¯C1,¯¯¯¯¯¯C2∈∏∞i=1K(Xi),f∈CB(∏∞i=1Xi),ϵ>0∃n∈N∀m∈N:
|K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞))
−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞))|≤ϵ(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:∞(x0)(f) associated with using Ci,1 as your compact sets, and the sequence associated with using Ci,2 as your compact sets (as long as they're both past some shared threshold), they'll be close. So all choices of compact set must limit to each other, because ϵ can be arbitrary.
Fix our x0,ϵ,¯¯¯¯¯¯C1,¯¯¯¯¯¯C2,f (input value, closeness parameter, two sequences of compact sets, function), and now we'll find our n to make
∀m∈N:|K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞))
−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)
true. Do it in the following way. Use {x0} as your compact seed set to invoke the technique in part 4 to construct your sequence Ci[x0,ϵ], and then consider the set:
∏∞i=1(Ci,1∪Ci,2∪Ci[x0,ϵ])
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(x1:∞,x′1:∞)=∑∞i=12−iinf(1,dXi(xi,x′i))
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 δ difference in inputs that only leads to an ϵ difference in values. Now we can define our n as log2(δ).
Now that we've picked our n, let m be arbitrary. Our goal is to now prove:
|K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞))
−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)
First-up, we use that C1:n+m+1[x0,ϵ] is an ϵ-almost support for K:n+m(x0), and 1-Lipschitzness, to apply Lemma 2 from LBIT to break up
|K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞))
−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞))|
as
≤supx1:n+m+1∈C1:n+m+1[x0,ϵ]|infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞)
−infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞)|
+ϵ⋅supx1:n+m+1|infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞)
−infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞)|
We can upper-bound the second term with just 2||f||, as the difference in values can't be any higher than the gap between the maximum value of f and the minimum value of f. As for the first term, note that if x1:n+m+1∈C1:n+m+1[x0,ϵ] and xn+m+2:∞∈∏∞i=n+m+2Ci,1 and x′n+m+2:∞∈∏∞i=n+m+2Ci,2(the latter two are the inf terms for a particular x1:n+m+1) then we have
x1:n+m+1,xn+m+2:∞∈∏∞i=1(Ci,1∪Ci,2∪Ci[x0,ϵ])
And same for x1:n+m+1,x′n+m+2:∞. And these two points agree on x1:n+m+1, and n is large enough that the function itself varies by only ϵ for any two points in that set which share the same prefix up to length n (and x1:n+m+1 fulfills that) So, we can break down the upper bound as
≤ϵ+ϵ⋅2||f||=ϵ(2||f||+1)
And so we're done here and have our result.
Moving on, the statement
∀x0∈X0¯¯¯¯¯¯C1,¯¯¯¯¯¯C2∈∏∞i=1K(Xi),f∈CB(∏∞i=1Xi),ϵ>0∃n∈N∀m∈N:
|K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,1f(x1:n+m+1,xn+m+2:∞))
−K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞,2f(x1:n+m+1,xn+m+2:∞))|≤ϵ(2||f||+1)
is the same as
∀x0∈X0¯¯¯¯¯¯C1,¯¯¯¯¯¯C2∈∏∞i=1K(Xi),f∈CB(∏∞i=1Xi):
limn→∞|K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞,1f(x1:n+1,xn+2:∞))
−K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞,2f(x1:n+1,xn+2:∞))|=0
So all the defining sequences for K:∞(x0)(f) limit to each other. We've got three possible cases. One is divergence, that liminf of the defining sequences goes to infinity. This is fine, the limit exists. One is convergence, that liminf and limsup are finite and equal, ie, the limit exists. This is fine. And the third case, that liminf and limsup aren't equal, must be ruled out. We'll invoke the Almost-Monotone Lemma, that
∀CX0∈K(X0),ϵ>0:∃¯¯¯¯C∈∏∞i=1K(Xi):∀n,m∈N,x0∈CX0,f∈CB(∏∞i=1Xi):
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈∏∞i=n+m+2Cif(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cif(x1:n+1,xn+2:∞))−4ϵ||f||
Specializing to the compact set consisting of a single arbitrary point x0, and recalling that the defining sequence for that was the Ci[x0,ϵ] sequence of almost-supports, and an arbitrary function f, the lemma turns into
∀ϵ>0,n,m∈N:
K:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[x0,ϵ]f(x1:n+m+1,xn+m+2:∞))
≥K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[x0,ϵ]f(x1:n+1,xn+2:∞))−4ϵ||f||
Fix an arbitrary ϵ and n∗ now. Just by definitions, we have
liminfn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[x0,ϵ]f(x1:n+1,xn+2:∞))
=limn→∞infmK:n+m(x0)(λx1:n+m+1.infxn+m+2:∞∈Cn+m+2:∞[x0,ϵ]f(x1:n+m+1,xn+m+2:∞))
This sequence is monotonically increasing, so a lower bound on this is
≥infmK:n∗+m(x0)(λx1:n∗+m+1.infxn∗+m+2:∞∈Cn∗+m+2:∞[x0,ϵ]f(x1:n∗+m+1,xn∗+m+2:∞))
and then, using ϵ and n∗ in the Almost-Monotone Lemma, we have
≥K:n∗(x0)(λx1:n∗+1.infxn∗+2:∞∈Cn∗+2:∞[x0,ϵ]f(x1:n∗+1,xn∗+2:∞))−4ϵ||f||
Since n∗ was an arbitrary number, we have
liminfn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[x0,ϵ]f(x1:n+1,xn+2:∞))
≥supn∗∈NK:n∗(x0)(λx1:n∗+1.infxn∗+2:∞∈Cn∗+2:∞[x0,ϵ]f(x1:n∗+1,xn∗+2:∞))−4ϵ||f||
and then we can lower-bound this by the limsup, yielding
≥limsupn→∞K:n(x0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[x0,ϵ]f(x1:n+1,xn+2:∞))−4ϵ||f||
Since the limsup is always equal or greater than the liminf, this implies that the limsup and liminf can be at most 4ϵ||f|| apart for this sequence. And since all the defining sequences limit to each other, they all have the same limsup and liminf. Since ϵ was arbitrary, we get that all the defining sequences limit to each other, and either converge to a fixed number, or diverge to infinity. The infinite semidirect product is now well-defined, and we can use whichever sequence of compact sets is the most convenient.
T1.6 Showing the following infradistribution properties for the infinite semidirect product: monotonicity and concavity. Also, the 1-Lipschitz property and 1-normalization property and Constants Increase property, for 4/5 inframeasure properties, and 3/5 niceness properties.
For monotonicity, concavity, and 1-Lipschitzness, the exact proof from LBIT works.
For constants increasing, observe that regardless of n,
K:n(x0)(λx1:n+1.infxn+2:∞∈∏∞i=n+2Cic)=K:n(x0)(c)≥c
Then, this bound transfers to the limit, so
K:∞(x0)(c)≥c
The same proof applies to 1-normalization, but with equality signs, if we plug in 1, and the limit of the all-1 sequence is always 1.
T1.7 Showing compact-shared compact-almost-support for K:∞. Again, the exact proof from LBIT works here. We're up to 5/5 inframeasure properties, and 4/5 niceness properties for the infinite semidirect product.
T1.8 Our task is to show lower-semicontinuity for K:∞. This will take some cunning. At the start, fix an ϵ, we'll see why later. Our task is to show that
liminfm→∞K:∞(xm0)(f)≥K:∞(x∞0)(f)
For some arbitrary sequence of xm0 which limits to x∞0. What we can do is first unpack.
liminfm→∞K:∞(xm0)(f)
=liminfm→∞limn→∞K:n(xm0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[{xm0}m∈N⊔{∞},ϵ]f(x1:n+1,xn+2:∞))
Now, the sequence of sets we picked was important. It means we can now apply the Almost-Monotone Lemma. Regardless of m, the defining sequence for K:∞(xm0)(f) is almost monotone now. We can fix a particular n∗ (arbitrary) to get a lower bound of
≥liminfm→∞K:n∗(xm0)(λx1:n∗+1.infxn∗+2:∞∈Cn∗+2:∞[{xm0}m∈N⊔{∞},ϵ]f(x1:n∗+1,xn∗+2:∞))−4ϵ||f||
And then, by lower-semicontinuity for K:n∗, which was already established by induction, we get
≥K:n∗(x∞0)(λx1:n∗+1.infxn∗+2:∞∈Cn∗+2:∞[{xm0}m∈N⊔{∞},ϵ]f(x1:n∗+1,xn∗+2:∞))−4ϵ||f||
But, since this holds for all n∗, we can take a limit and get
liminfm→∞K:∞(xm0)(f)
≥limn→∞K:n(x∞0)(λx1:n+1.infxn+2:∞∈Cn+2:∞[{xm0}m∈N⊔{∞},ϵ]f(x1:n+1,xn+2:∞))−4ϵ||f||
=K:∞(x∞0)(f)−4ϵ||f||
But since ϵ can be arbitrarily small, we have
liminfm→∞K:∞(xm0)(f)≥K:∞(x∞0)(f)
and thus, lower-semicontinuity. And we're finally done!!