Consider a set X and the normed vector space (ℝbX,∥·∥∞), where ℝbX is the space of all bounded real-valued functions on X and ∥·∥_∞ is the usual uniform norm, i.e., for any f∈ℝbX, ∥f∥∞:=supx∈X|f(x)|. On the set X, the space of transformations is given by elements of Aut(X), i.e., the group of bijections from X to itself. Then, we can consider the right group action R defined as follows (we represent composition as a juxtaposition of functions):

Remark 3.1. For every s ∈ Aut(X), the map Rs:ℝbX→ℝbX, with Rs(φ):=φs preserves the distances. In fact, for any φ1,φ2∈ℝbX, by bijectivity of s, we have that

In our model, our data sets are represented as two sets Φ and Φ′ of bounded real-valued measurements on X. In particular, X represents the space where the measurements can be made, Φ is the space of permissible measurements, and Φ′ is a space which Φ can be transformed into, without changing the interpretation of its measurements after a transformation is applied. In other words, we want to be able to apply some admissible transformations on the space X so that the resulting changes in the measurements in Φ are contained in the space Φ′. Thus, in our model, we consider operations on X in the following way:

Definition 3.2. A (Φ, Φ′)-operation is an element s of Aut(X) such that, for any measurement φ ∈ Φ, the composition φs belongs to Φ′. The set of all (Φ, Φ′) operations is denoted by AutΦ,Φ′(X).

Remark 3.3. We can observe that the identity function id_X is an element of AutΦ,Φ′(X) if Φ ⊆ Φ′.

For any s∈AutΦ,Φ′(X), the restriction to Φ×AutΦ,Φ′(X) of the map Rs takes values in Φ′ since Rs(φ):=φs∈Φ′ for any φ ∈ Φ. We can consider the restriction of the map R (for simplicity, we will continue to use the same symbol to denote this restriction):

where R(φ,s)=Rs(φ), for every s∈AutΦ,Φ′(X) and every φ ∈ Φ.

Definition 3.4. Let X be a set. A perception triple is a triple (Φ, Φ′, S) with Φ,Φ′⊆ℝbX and S⊆AutΦ,Φ′(X). The set X is called the domain of the perception triple and is denoted by dom(Φ, Φ′, S).

Example 3.5. Given X = ℝ², consider two rectangles R and R′ in X. Assume Φ: = {φ: X → [0, 1]: supp(φ) ⊆ R} and Φ′: = {φ′:X → [0, 1]: supp(φ′) ⊆ R′}. We recall that, if we consider a function f: X → ℝ, the support of f is the set of points in the domain, where the function does not vanish, i.e., supp(f) = {x ∈ X | f(x) ≠ 0}. Consider S as the set of translations that bring R into R′. The triple (Φ, Φ′, S) is a perception triple. If Φ represents a set of gray level images, S determines which translations can be applied to our pictures.

In our model, considering a generic set X, data are represented by a space Ω⊆ℝbX of bounded real-valued functions. We endow the real line ℝ with the usual Euclidean metric and the space X with an extended pseudo-metric induced by Ω:

for every x₁, x₂ ∈ X. The choice of this pseudo-metric over X means that two points can only be distinguished if they assume different values for some measurements. For example, if Φ contains only a constant function and X contains at least two points, the distance between any two points of X is always null.

The pseudo-metric space XΩ:=(X,DXΩ) can be considered as a topological space with the basis

and the induced topology is denoted by τ_Ω. The reason for considering a topological space X, rather than just a set, follows from the need of formalizing the assumption that data are stable under small perturbations.

Remark 3.6. In our case, there are two collections of functions Φ and Φ′ in ℝbX representing our data, both of which induce a topology on X. Hence, in the model, we consider two pseudo-metric spaces X_Φ and XΦ′ with the same underlying set X. If Φ⊆Φ′⊆ℝbX, the topologies τ_Φ and τΦ′ are comparable and, in particular, τΦ′ is finer than τ_Φ.

Now, given a set Ω⊆ℝbX, we will prove a result about the compactness of the pseudo-metric space X_Ω. Before proceeding, let us recall the following lemma (e.g., see Gaal, 1964):

Lemma 3.7. Let (P,d) be a pseudo-metric space. The following conditions are equivalent:

Theorem 3.8. If Ω is totally bounded, X_Ω is totally bounded.

Proof: By Lemma 3.7, it will suffice to prove that every sequence in X admits a Cauchy subsequence with respect to the pseudo-metric DXΩ. A sequence (x_i)_i∈ℕ in X_Ω is considered and a real number ε > 0 is taken. Since Ω is totally bounded, we can find a finite subset Ω_ε = {ω₁, …, ω_n} such that for every ω ∈ Ω, there exists ω_r ∈ Ω for which ||ω−ω_r||_∞ < ε. We can consider now the real sequence (ω₁(x_i))_i∈ℕ, which is bounded since ω1∈ℝbX. From Bolzano-Weierstrass Theorem, it follows that we can extract a convergent subsequence (ω₁(x_ih))_h∈ℕ. Again, we can extract from (ω₂(x_ih))_h∈ℕ another convergent subsequence (ω₂(x_iht))_t∈ℕ. Repeating the process, we are able to extract a subsequence of (x_i)_i∈ℕ, that for simplicity of notation we can indicate as (x_ij)_j∈ℕ, such that (ω_k(x_ij))_j∈ℕ is a convergent subsequence in ℝ, and hence a Cauchy sequence in ℝ, for every k ∈ {1, …, n}. By construction, Ω_ε is finite, then we can find an index ȷ̄ such that for any k ∈ {1, …, n}

Furthermore, we have that, for any ω ∈ Ω, any ω_k ∈ Ω_ε, and any ℓ, m ∈ ℕ

We observe that the choice of ȷ̄ depends only on ε and Ω_ε not on k. Then, choosing a ω_k ∈ Ω_ε such that ||ω_k−ω||_∞ < ε, we get ||ω(x_iℓ)−ω(x_im)||_∞ < 3ε for every ω ∈ Ω and every ℓ,m≥ȷ̄. Then,

Then (x_ij)_j∈ℕ is a Cauchy sequence in X_Ω. For Lemma 3.7 the statement holds.

Corollary 3.9. If Ω is totally bounded and X_Ω is complete, X_Ω is compact.

Proof: From Theorem 3.8, we have that X_Ω is totally bounded, and since by hypothesis it is also complete, it is compact.

Now, we will prove that the choice of the pseudo-metric DXΩ on X makes the functions in Ω non-expansive.

Definition 3.10. Two pseudo-metric spaces (P, d_P) and (Q, d_Q) are considered. A non-expansive function from (P, d_P) to (Q, d_Q) is a function f: P → Q such that d_Q(f(p₁), f(p₂)) ≤ d_P(p₁, p₂) for any p₁, p₂ ∈ P.

We denote as NE(P, Q) the space of all non-expansive functions from (P, d_P) to (Q, d_Q).

Proposition 3.11. Ω ⊆ NE(X_Ω, ℝ).

Proof: For any x₁, x₂ ∈ X, we have that

Then, the topology on X induced by DXΩ naturally makes the measurements in Ω continuous. In particular, since the previous results hold for a generic Ω⊆ℝbX, they are also true for Φ and Φ′ in our model.

Remark 3.12. Assuming that (Φ, Φ′, S) is a perception triple. A function φ′ ∈ Φ′ may not be continuous from X_Φ to ℝ and a function φ ∈ Φ may not be continuous from XΦ′ to ℝ. In other words, the topology on X induced by the pseudo-metric of one of the function spaces does not make the functions in the other continuous.

Example 3.13. Assuming X = ℝ, for every a, b ∈ ℝ the functions φ_a:X → ℝ and φb′:X→ℝ are defined by setting

Suppose Φ: = {φ_a:a ≥ 0} and Φ′:={φb′:b≤0}. Consider the symmetry with respect to the y-axis, i.e., the map s(x) = −x. Surely, s∈AutΦ,Φ′(X). We can observe that the function φ₁ ∈ Φ is not continuous from XΦ′ to ℝ; indeed DXΦ′(0,2)=0, but |φ₁(0)−φ₁(2)| = 1.

However, if Φ ⊆ Φ′, we have that the functions in Φ are also continuous on XΦ′, indeed:

Corollary 3.14. If Φ ⊆ Φ′, then Φ⊆NE(XΦ′,ℝ).

Proof: By Proposition 3.11, the statement trivially holds since Φ⊆Φ′⊆NE(XΦ′,ℝ).

Proposition 3.15. Every element of AutΦ,Φ′(X) is non-expansive from XΦ′ to X_Φ.

Proof: Considering a bijection s∈AutΦ,Φ′(X) we have that

for every x₁, x₂ ∈ X, where Φs = {φs, φ ∈ Φ}. Then, s∈NE(XΦ′,XΦ) and the statement is proved.

Now, we are ready to put more structure on AutΦ,Φ′(X). Considering a set Ω⊆ℝbX of bounded real-valued functions, we can endow the set Aut(X) with a pseudo-metric inherited from Ω

for any s₁, s₂ in Aut(X).

Remark 3.16. Analogously to what happens in Remark 3.6 for X, the sets Φ and Φ′ can endow Aut(X) with two possibly different pseudo-metrics DAutΦ and DAutΦ′. In particular, we can consider AutΦ,Φ′(X) as a pseudo-metric subspace of Aut(X) with the induced pseudo-metrics.

Remark 3.17. We observe that, for any s₁, s₂ in Aut(X),

In other words, the pseudo-metric DAutΩ, which is based on the action of the elements of Aut(X) on the set Ω, is exactly the usual uniform pseudo-metric on X_Ω.

Since we are only interested in transformations of functions in Φ, it would be natural to just endow AutΦ,Φ′(X) with the pseudo-metric DAutΦ. However, it is sometimes necessary to consider the pseudo-metric DAutΦ′ in order to guarantee the continuity of the composition of elements in AutΦ,Φ′(X), whenever it is admissible. Considering two elements s, t in AutΦ,Φ′(X) such that st is still an element of AutΦ,Φ′(X), i.e., for every function φ ∈ Φ, we have that φst ∈ Φ′. Then, for any φ ∈ Φ we have that

Therefore, t is also an element of AutΦs,Φ′(X). By definition, Φs is contained in Φ′ for every s∈AutΦ,Φ′(X), and this justifies the choice of considering in AutΦ,Φ′(X) also the pseudo-metric DAutΦ′. We have shown, in particular, that if s, t are elements of AutΦ,Φ′(X) such that st is still an element of AutΦ,Φ′(X), t is an element of AutΦs,Φ′(X), which is an implication of the following proposition:

Proposition 3.18. Let s,t∈AutΦ,Φ′(X). Then, st∈AutΦ,Φ′(X) if t∈AutΦs,Φ′(X).

Proof: If the composition st belongs to AutΦ,Φ′(X), we have already proven that t∈AutΦs,Φ′(X). On the other hand, if t∈AutΦs,Φ′(X), we have that φ̄t∈Φ′ for every φ̄∈Φs. Since φ(st) = (φs)t, it follows that φ(st) ∈ Φ′ for every φ ∈ Φ. Therefore, st∈AutΦ,Φ′(X) and the statement is proved.

Remark 3.19. Let t∈AutΦ,Φ′(X). We can observe that if s ∈ Aut_Φ(X), Φs ⊆ Φ and st∈AutΦ,Φ′(X).

Lemma 3.20. Consider r, s, t ∈ Aut(X). For any Ω⊆ℝbX, it holds that

Proof: Since Rt preserves the distances, we have that:

Lemma 3.21. Consider r, s ∈ Aut(X) and t∈AutΦ,Φ′(X). It holds that

Proof: Since Φt ⊆ Φ′, we have that:

Let Π be the set of all pairs (s, t) such that s,t,st∈AutΦ,Φ′(X). We endow Π with the pseudo-metric

and the corresponding topology.

Proposition 3.22. The function ◦:Π→(AutΦ,Φ′(X),DAutΦ) that maps (s, t) to st is non-expansive and hence continuous.

Proof: Consider two elements (s₁, t₁), (s₂, t₂) of Π. By Lemma 3.20 and Lemma 3.21,

Therefore, the statement is proved.

Let Υ be the set of all s with s,s-1∈AutΦ,Φ′(X).

Proposition 3.23. The function (·)-1:(Υ,DAutΦ′)→(AutΦ,Φ′(X),DAutΦ), that maps s to s⁻¹, is non-expansive, and hence continuous.

Proof: Consider two bijections s₁, s₂ ∈ Υ. Because of Lemma 3.20 and Lemma 3.21, we obtain that

We have previously defined the map

where R(Φ,s)=Rs(Φ), for every s∈AutΦ,Φ′(X).

Proposition 3.24. The function R is continuous by choosing the pseudo-metric DAutΦ on AutΦ,Φ′(X).

Proof: We have that

for any φ,φ¯∈Φ and any t,s∈AutΦ,Φ′(X). This proves that R is continuous.

Now, we can give a result about the compactness of (AutΦ,Φ′(X),DAutΦ) under suitable assumptions.

Proposition 3.25. If Φ and Φ′ are totally bounded, (AutΦ,Φ′(X),DAutΦ) is totally bounded.

Proof: Consider a sequence (s_i)_i∈ℕ in AutΦ,Φ′(X) and a real number ε > 0. Since Φ is totally bounded, we can find a finite subset Φ_ε = {φ₁, …, φ_n} such that for every φ ∈ Φ, there exists φ_r ∈ Φ for which ||φ−φ_r||_∞ < ε. Now, consider the sequence (φ₁s_i)_i∈ℕ in Φ′. Since also Φ′ is totally bounded, from Lemma 3.7, it follows that we can extract a Cauchy subsequence (φ₁s_ih)_h∈ℕ. Again, we can extract another Cauchy subsequence (φ₂s_iht)_t∈ℕ. Repeating the process for every k ∈ {1, …, n}, we are able to extract a subsequence of (s_i)_i∈ℕ, that for simplicity of notation we can indicate as (s_ij)_j∈ℕ, such that (φ_ks_ij)_j∈ℕ is a Cauchy sequence in Φ′ for every k ∈ {1, …, n}.

Since Φ_ε is finite, we can find an index ȷ̄ such that for any k ∈ {1, …, n}

Furthermore, we have that for any φ ∈ Φ, any φ_k ∈ Φ_ε, and any ℓ, m ∈ ℕ

We observe that the choice of ȷ̄ in (3.4.1) depends only on ε and Φ_ε not on φ. Then, choosing a φ_k ∈ Φ_ε such that ||φ_k−φ||_∞ < ε, we get ||φs_iℓ−φs_im||_∞ < 3ε for every φ ∈ Φ and every ℓ,m≥ȷ̄. Hence, for every ℓ, m ∈ ℕ

Therefore, (s_ij)_j∈ℕ is a Cauchy sequence in AutΦ,Φ′(X). For Lemma 3.7, the statement holds.

Corollary 4.26. Assume that S⊆AutΦ,Φ′(X). If Φ and Φ′ are totally bounded and (S,DAutΦ) is complete, it is also compact.

Proof: From Proposition 3.25, we have that S is totally bounded, and since by hypothesis it is also complete, the statement holds.

Starting from a finite number of P-GENEOs, we will illustrate some methods to construct new P-GENEOs. First of all, the composition of two P-GENEOs is still a P-GENEO.

Proposition 4.7. Given two composable P-GENEOs, (F1,F1′,T1):(Φ,Φ′,S)→(Ψ,Ψ′,Q) and (F2,F2′,T2):(Ψ,Ψ′,Q)→(Ω,Ω′,K), their composition defined as

is a P-GENEO.

Proof: First, one could easily check that the map T = T₂ ◦ T₁ respects the second and the third property of Definition 4.1. Therefore, it remains to verify that F(Φ) ⊆ Ω, F′(Φ′) ⊆ Ω′ and the properties of equivariance and non-expansiveness are maintained.

Given a finite number of P-GENEOs with respect to the same map T, we illustrate a general method to construct a new operator as a combination of them. Given two sets X and Y, consider a finite set {H₁, …, H_n} of functions from Ω⊆ℝbX to ℝbY and a map L:ℝn→ℝ, where ℝⁿ is endowed with the norm ||(x1,…,xn)||∞:=max1≤i≤n|xi|. We define L*(H1,…,Hn):Ω→ℝbY as

for any ω ∈ Ω, where [L(H1(ω),…,Hn(ω))]:Y→ℝ is defined by setting

for any y ∈ Y. Now, consider two perception triples (Φ, Φ′, S) and (Ψ, Ψ′, Q) with domains X and Y, respectively, and a finite set of P-GENEOs (F1,F1′),…(Fn,Fn′) between them associated with the map T: S → Q. We can consider the functions L*(F1,…,Fn):Φ→ℝbY and L*(F1′,…,Fn′):Φ′→ℝbY, defined as before and state the following result.

Proposition 4.8. Assume that L:ℝn→ℝ is non-expansive. If L*(F1,…,Fn)(Φ)⊆Ψ and L*(F1′,…,Fn′)(Φ′)⊆Ψ′, (L*(F1,…,Fn),L*(F1′,…,Fn′)) is a P-GENEO from (Φ, Φ′, S) to (Ψ, Ψ′, Q) with respect to T.

Proof: By hypothesis, L*(F1,…,Fn)(Φ)⊆Ψ and L*(F1′,…,Fn′)(Φ′)⊆Ψ′, so we just need to verify the properties of equivariance and non-expansiveness.