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UNIFORM SPACES AND THE NEWTONIAN STRUCTURE OF (BIG)DATA AFFINITY KERNELS HUGOAIMARANDIVANAGO´MEZ Abstract. Let X be a (data) set. Let K(x,y) > 0 be a measure of the 7 affinitybetweenthedatapointsxandy.WeprovethatKhasthestructure 1 0 ofaNewtonianpotentialK(x,y) = ϕ(d(x,y))withϕdecreasingandd a 2 quasi-metric on X under two mild conditions on K. The first is that n the affinity of each x to itself is infinite and that for x , y the affinity a is positive and finite. The second is a quantitative transitivity; if the J affinitybetween x andy islargerthanλ > 0 andthe affinityof y andz 3 isalsolargerthanλ,thentheaffinitybetweenxandzislargerthanν(λ). 1 Thefunctionνisconcave,increasing,continuousfromR+ ontoR+ with ν(λ)<λforeveryλ>0. ] N G . h 1. Introduction t a m Isaac Newton in the seventeenth century started the endless quantitative approach to the understanding of nature. The quantitative character of the [ formulation of the Law of Universal Gravitation, should not hide its deep 1 v qualitativeaspects. Now, more than 300 years later, we are able to explain 6 the central fields as gradients of radial potentials centered at the “object” 4 generatingthefield. UsuallythepotentialtaketheformV(x) = ϕ(|x|)where 7 3 ϕ is a decreasing profile and |x| is the distance of the point x, where the 0 field is to be measured, to the origin of coordinates supporting the mass or . 1 the charge that generates the field. In the Euclidean n-dimensional space 0 the profile ϕ(r) = r−n+2 gives the fundamental solution of the Laplacian. 7 1 And the kernel K(x,y) = ϕ(|x−y|) provides the basic information in order : v to produce the continuous models by convolution with the mass or charge Xi densities that determinethe system. These facts are also the starting points forharmonicanalysis. r a Our aim in this paper is to use arguments and results strongly related to the theory of uniform spaces, in order to give sufficient conditions on a kernel function K(x,y) defined on an abstract setting, in such a way that K(x,y) = ϕ(d(x,y))withϕadecreasing profileand d a(quasi)metricon X. In other words, we aim to obtain an abstract form of Newtonian potentials for general kernels. Our approach in the search of conditions on the kernel K(x,y) will be based in the heuristic associated to the interpretation of K(x,y) as an affinity matrix for (big) data. Amazingly enough the abstract ofthe paper [2] of the Coifman Group leading the harmonicanalysis 2010MathematicsSubjectClassification. Primary54E15,54E35. Keywordsandphrases. uniformspaces;metricspaces;affinitykernel. 1 2 approach to determine the structure of big data sets, reads “The process of iteratingordiffusingtheMarkovmatrixisseenasageneralizationofsome aspectsof theNewtonianparadigm,in whichlocalinfinitesimaltransitions of a system lead to global macroscopic by integration.” In some sense the result of this paper shows that also the potential theoretic Newtonian view of nature is still close to these problems. By the way, our approach could beagoodexampleofhowqualitativeaspectsofasystemleadstostructural resultsofthemodel. Let X be a set (data set). Each element x of X is understood as a data point. Let K(x,y) be a nonnegative number measuring the affinity between the two data points x and y. We shall consider some basic properties of affinity which will be sufficient to obtain the Newtonian potential type structure for K. Symmetry; affinity is a symmetric relation on X × X (K(x,y) = K(y,x) for every x, y ∈ X). Positivity; there is positive affinity between any couple of data points x and y (K(x,y) ≥ 0 for every x, y ∈ X). Diagonalsingularity;theselfaffinityisunimprovable. Precisely,theaffin- ityofeach datapoint xwithitselfis+∞ (K(x,x) = +∞ forevery x)butfor x , y the affinity is finite (K(x,y) < ∞ for x , y). Quantitative Transiti- vity; if the affinity between the data points x and y is larger than λ > 0 and the affinity between y and z is larger than λ then the affinity between the points x and z is larger than ν(λ) < λ. Here ν is a nonnegative, concave, increasing and continuous function defined on R+ onto R+ such that ν(λ) < λ. A quasi-metric in X is a nonnegative symmetric function d defined on X ×X whichvanishesonlyonthediagonal∆of X ×X and satisfiesaweak form of the triangle inequality, there exists a constant τ (≥ 1) such that the inequalityd(x,z) ≤ τ(d(x,y)+d(y,z))holdsforevery x, yand zin X. Themain resultofthispapercan bestatedas follows. Theorem 1. Let X be a set. Let K : X × X → R+ be a symmetric function satisfying the singularity and the quantitative transitivityconditions. Then there exist a decreasing and continuous function ϕ defined in R+ and a quasi-metricd on X suchthat K(x,y) = ϕ(d(x,y)). Moreover, d(x,y) = h(x,y)ρ(x,y) with ρ a metric on X and h a symmetric function such that for some constants 0 < c < c < ∞ satisfies c ≤ 1 2 1 h(x,y) ≤ c forevery xand yin X. 2 The deepest results on the structure of quasi-metrics are due to Mac´ıas andSegoviaandarecontainedin[5]. See also[1]. Themostsignificantfor our purposes is the fact that each quasi-metricis equivalentto a power ofa metric. In other words, given a quasi-metric d on X with constant τ, there existβ ≥ 1andametricρon X suchthatforsomepositiveconstantsa and 1 a theinequalities 2 a d(x,y) ≤ ρβ(x,y) ≤ a d(x,y) 1 2 3 holdforevery xandyin X. ActuallytheproofisbasedinFrink’slemmaof metrizationofuniformstructures withacountablebasis([3], [4]). The rest of the paper is organized in the following way. Section 2 gives a characterization of quasi-metrics on a set in terms of properties of the family of stripes in X × X induced by the quasi-metric. Section 3 contains the construction of the monotonic profile. In Section 4 we prove the main result. 2. Quasi-metrics and families ofstripes aroundthe diagonal Let X be a set. The composition of two subsets U and V of X × X is given by V ◦ U = {(x,z) ∈ X × X : thereexistsy ∈ X such that(x,y) ∈ U and(y,z) ∈ V}. A subset U in X × X is said to be symmetric if (x,y) ∈ U if and only if (y,x) ∈ U. Set ∆ to denote the diagonal in X × X, i.e. ∆ = {(x,x) : x ∈ X}. When a quasi-metric δ with constant τ is given in X, it is easy to check that the one parameter family V(r) = {(x,y) ∈ X × X : δ(x,y) < r};r > 0,ofstripesaround thediagonalof X ×X, satisfies (S1) each V(r) issymmetric; (S2) ∆ ⊆ V(r), forevery r > 0; (S3) V(r ) ⊆ V(r ), for0 < r ≤ r ; 1 2 1 2 (S4) ∪ V(r) = X ×X; r>0 (S5) ∩ V(r) ⊆ ∆; r>0 (S6) thereexistsT ≥ 1 such thatV(r)◦V(r) ⊆ V(Tr), foreveryr > 0. Actually,theconstantT in(S6)canbetakentobethetriangleconstantτof δ. Set P(X ×X) to denotethesetofsubsetsof X ×X. Theorem 2. Let V : R+ → P(X × X) be a one parameter family of the subsetsofX×X thatsatisfies(S1)to(S6)above. Thenthefunctionδdefined on X ×X by δ(x,y) = inf{r > 0 : (x,y) ∈ V(r)} is a quasi-metricon X with τ ≤ T. Moreover, foreachγ > 0, wehave V (r) ⊆ V(r) ⊆ V ((1+γ)r) (2.1) δ δ holdforeveryr > 0, whereV (s) = {(x,y) ∈ X ×X : δ(x,y) < s}. δ Proof. From (S4) for the family V we see that δ(x,y) is well defined as a nonnegative real number. The symmetry of δ follows from (S1). The fact that δ vanishes on the diagonal ∆ follows from ∆ ⊆ ∩ V(r) which is r>0 containedin(S2). Now,if(x,y) ∈ X×X andδ(x,y) = 0,then,from(S3)for eachr > 0,(x,y) ∈ V(r). Now,from(S5)wenecessarilyhavethat(x,y) ∈ ∆ or, in other words x = y. Let us check that δ satisfies a triangle inequality. Let x, y and z be three points in X. Let ε > 0. Take r > 0 and r > 0 such 1 2 thatr < δ(x,y)+ε,r < δ(y,z)+ε,(x,y) ∈ V(r )and(y,z) ∈ V(r ). From 1 2 1 2 (S6)withr∗ = sup{r ,r },wehave(x,z) ∈ V(r )◦V(r ) ⊆ V(r∗)◦V(r∗) ⊆ 1 2 2 1 V(Tr∗). Henceδ(x,z) ≤ Tr∗ ≤ T(r +r ) ≤ T(δ(x,y)+δ(y,z))+2εT andwe 1 2 get the triangle inequality with τ = T. Notice first that from (S3), V (r) ⊆ δ V(r) for every r > 0. Take now (x,y) ∈ V(r), then δ(x,y) ≤ r < (1 +γ)r, sothat V(r) ⊆ V ((1+γ)r)foreveryγ > 0 andevery r > 0. (cid:3) δ 4 The next result follows from the above and the metrization theorem of quasi-metric spaces proved in [5] as an application of Frink’s Lemma on metrizabilityofuniformspaces withcountablebases. Theorem 3. Let V and δ be as in Theorem 2. Then, there exist a constant β ≥ 1 anda metricρ on X suchthat (i) 4−βδ ≤ ρβ ≤ 2βδ; (ii) V r ⊆ V(r) ⊆ V (2β+1r) where V (r) = {(x,y) ∈ X × X : ρβ 4β ρβ ρβ ρβ(x,(cid:16)y)(cid:17)< r}. Proof. Following the proof of Theorem 2 on page 261 in [5] take α < 1 such that (3T2)α = 2 and β = 1 > 1. With ρ the metric provided by the α metrizationtheorem for uniform spaces with countablebases, we havethat 4ρ(x,y) ≥ δ(x,y)α ≥ 1ρ(x,y). So that 2 1 δ(x,y) ≤ ρ(x,y)β ≤ 2βδ(x,y), 4β and (ii)followfromtheseinequalitiesand (2.1)withγ = 1. (cid:3) 3. Building thebasic profileshape Theclassicalinverseproportionalitytothesquareofthedistancebetween thetwobodiesforthegravitationfield, translatesintoinverseproportional- ity to the distance for the potential. That is ϕ(r) = 1 for the gravitational r potential. Thissectionisdevotedtotheconstructionofthebasicshapesofthepro- files that allow theNewtonian representation of the kernels. This construction requires to solve a functional inequality involving the function ν that controlsthequantitativetransitivepropertyof K. Proposition4. Letνbeaconcave,continuous,nonnegativeandincreasing function defined on R+ onto R+ such that ν(λ) < λ for every λ > 0. Then, given M > 1, there exists a continuous, decreasing and convex function ψ defined onR+ withψ(1) = 1suchthattheinequality ψ(ν(λ)) ≤ Mψ(λ) (3.1) holdsforeveryλ > 0. Proof. Set λ = 1, λ = ν(1) and λ = ν−1(1). Notice that λ < 1 and 0 1 −1 1 λ > 1. In fact, λ = ν(1) < 1 and 1 = ν−1(ν(1)) = ν−1(λ ) < ν−1(1) = λ . −1 1 1 −1 Setfork ∈ N,λ = ν(λ )andλ = ν−1(λ ). Noticethatλ decreasesas k k−1 −k −k+1 k k → ∞ and increases when k → −∞. The continuity of ν and the property ν(λ) < λ for every positive λ imply that λ → 0 as k → ∞ and λ → ∞ k k as k → −∞ monotonically. This basic sequence {λ : k ∈ Z} allows to k construct a function ψ in the following way. Set ψ(λ ) = Mk, k ∈ Z and k for λ ∈ [λ ,λ ] define ψ(λ) by linear interpolation. It is clear that ψ is k+1 k continuous, decreasing, ψ(0+) = +∞, ψ(∞) = 0, ψ(1) = ψ(λ ) = M0 = 1 0 and that ψ is convex. We only have to check that ψ solves inequality (3.1). 5 On thesequence{λ : k ∈ Z},(3.1)becomesan equality. In fact, ψ(ν(λ )) = k k ψ(λ ) = Mk+1 = MMk = Mψ(λ ). k+1 k Letus nowtakeλ ∈ (λ ,λ )for k ∈ Z. Forsuchλ, ψ(λ)satisfies k+1 k Mk+1 − Mk Mk+1 −ψ(λ) = . (3.2) λ −λ λ−λ k k+1 k+1 On the other hand, since λ < λ < λ , we have that λ = ν(λ ) < k+1 k k+2 k+1 ν(λ) < ν(λ ) = λ . Hence ψ(ν(λ)) satisfies k k+1 Mk+2 − Mk+1 Mk+2 −ψ(ν(λ)) = . (3.3) λ −λ ν(λ)−λ k+1 k+2 k+2 From (3.2) and(3.3)weget λ−λ ψ(λ) = Mk+1 − Mk(M −1) k+1 λ −λ k k+1 and ν(λ)−λ ψ(ν(λ)) = M Mk+1 − Mk(M −1) k+2 . λk+1 −λk+2! Now,sinceν isconcave, wehaveforλ < λ < λ that k+1 k ν(λ)−ν(λ ) ν(λ )−ν(λ ) k+1 ≥ k k+1 , λ−λ λ −λ k+1 k k+1 hence, ψ(ν(λ)) ≤ Mψ(λ). (cid:3) The basic shapes for the profiles ϕ in our main result will be given as compositionoftheinverseηofψ withpowerlaws. 4. Proofofthe main result Letusstartbyrewriting,formally,thepropertiesofsymmetry,positivity, singularityandtransitivityofadataaffinitykernelK(x,y)definedontheset X ×X. Let K : X ×X → Rsuchthat (K1) K(x,y) = K(y,x), forevery x and yin X; (K2) K(x,y) > 0, forevery x andyin X; (K3) K(x,y) = +∞ ifand onlyif x = y; (K4) there exists a continuous, concave, increasing and nonnegative function ν defined on R+ onto R+, with ν(λ) < λ, λ > 0, such that K(x,z) > ν(λ) whenever there exists y ∈ X with K(x,y) > λ and K(y,z) > λ,foreveryλ > 0. Withtheseproperties,Theorem 1can berestated as follows. Theorem 5. Let X be a set. Let K be a kernel on X × X satisfying properties(K1) to (K4). Then, there exist a metric ρ on X, a real number β ≥ 1, a function h(x,y) defined on X × X with 2−1/β ≤ h(x,y) ≤ 4 and a function ϕ : R+ → R+ continuous,decreasing with ϕ(0+) = +∞ and ϕ(∞) = 0, such that K(x,y) = ϕ(h(x,y)ρ(x,y)). 6 Proof. Let ν be the function provided by (K4). Let ψ be given by Proposi- tion 4 with this function ν, and M = 2. Hence ψ(ν(λ)) ≤ 2ψ(λ) for every λ > 0. Takeη = ψ−1 and V : R+ → P(X ×X) givenby V(r) = E = {(x,y) : K(x,y) > η(r)}. η(r) Let is check that V satisfies properties (S1) to (S6) in Section 2 with constant T = 2 (= M). From (K1) we see that each E is symmetric, in par- λ ticular V(r) is symmetricfor every r > 0. Since K(x,x) = +∞, from (K3), we have that ∆ ⊆ E = V(r) for r > 0. In order to check (S3) take η(r) 0 < r < r < ∞. Hence η(r ) > η(r ), so that K(x,y) > η(r ) implies 1 2 1 2 1 K(x,y) > η(r ). Or, in other words E ⊂ E . Or V(r ) ⊆ V(r ). 2 η(r1) η(r2) 1 2 Thepositivity(K2)of K(x,y) shows(S4). Property(S5)ofV followsfrom (S3). Toprove(S6)forV, taker > 0. If(x,z) ∈ V(r)◦V(r) = E ◦E , η(r) η(r) then there exists y ∈ X such that K(x,y) > η(r) and K(y,z) > η(r). From (K4), K(x,z) > ν(η(r)). Now applying (3.1) with λ = ψ−1(r) we get K(x,z) > ν(ψ−1(r)) ≥ η(2r), or (x,z) ∈ E = V(2r). Hence (S6) η(2r) for V holds with T = 2. We can, then apply the results of Section 2. First to produce a quasi-metric δ as in Theorem 2 and then to provide the metric ρ and the exponent β given in Theorem 3. Thus, for every r > 0, V r ⊆ V(r) = E ⊆ V (2β+1r), where ρ is a metric in X and, since ρβ 4β η(r) ρβ T ca(cid:16)n b(cid:17)e taken to be equal to 2, β = log 12 works. The aboveinclusiones, 2 taking s = r1/β, areequivalentto ρ < s ⊆ (ψ◦K)1/β > s ⊆ ρ < 21+1/βs (4.1) 4 n o n o n o foreverys > 0. LetxandybetwodifferentpointsinX. Since0 < K(x,y) < ∞sois(ψ◦K)1/β(x,y). Thereexists,then,aunique j ∈ Z(j = j(x,y))such that 2j ≤ (ψ ◦ K)1/β(x,y) < 2j+1. By the second inclusion in (4.1) we see that ρ(x,y) < 21/β2j ≤ 21/β(ψ ◦ K)1/β(x,y). On the other hand, since (ψ◦K)1/β(x,y) < 2j+1, from the first inclusionin (4.1) we necessarily have thatρ(x,y) ≥ 2j+1 > 1(ψ◦K)1/β(x,y). Hencefor x , ywehave 4 4 1 (ψ◦K)1/β ≤ ρ ≤ 21/β(ψ◦K)1/β. 4 Set h(x,y) = (ψ◦K)1/β for x , y and h(x,x) = 1. Then 1 ≤ h ≤ 4 and ρ(x,y) 21/β K(x,y) = ψ−1((h(x,y)ρ(x,y))β) = ϕ(h(x,y)ρ(x,y)) with ϕ(r) = ψ−1(rβ) = η(rβ). (cid:3) Notice that since h is symmetric and bounded above and below by positive constants the function d(x,y) = h(x,y)ρ(x,y) is a quasi-metric. But actuallydisbetterthanageneralquasi-metricsinceitstriangularinequality constant can be taken to be independent of the length of chains. Precisely, d(x ,x ) ≤ 22+1/β m−1d(x ,x ). 1 m j=1 j j+1 Let us observe also that Newtonian type power laws are obtained when P ν(λ) = aλ for a < 1. In fact, with m = 1 < 0, ψ(r) = rm solves the log a equation(aλ)m = 2λm. Hence ϕbecomes also2apowerlaw. 7 References [1] HugoAimar,BibianaIaffei, andLilianaNitti. OntheMac´ıas-Segoviametrizationof quasi-metricspaces.Rev.Un.Mat.Argentina,41(2):67–75,1998. [2] R. R. Coifman, S. Lafon, A. B. Lee, M. Maggioni, B. Nadler, F. Warner, and S. W. Zucker.Geometricdiffusionsasatoolforharmonicanalysisandstructuredefinitionof data: Diffusionmaps.ProceedingsoftheNationalAcademyofSciencesoftheUnited StatesofAmerica,102(21):7426–7431,2005. [3] A.H.Frink.Distancefunctionsandthemetrizationproblem.Bull.Amer.Math.Soc., 43(2):133–142,1937. [4] JohnL.Kelley.Generaltopology.Springer-Verlag,NewYork-Berlin,1975.Reprintof the 1955edition[VanNostrand,Toronto,Ont.],GraduateTextsinMathematics, No. 27. [5] RobertoA.Mac´ıasandCarlosSegovia.Lipschitzfunctionsonspacesofhomogeneous type.Adv.inMath.,33(3):257–270,1979. InstitutodeMatema´ticaAplicadadelLitoral(IMAL),UNL,CONICET,FIQ. CCT-CONICET-SantaFe,Predio“Dr. AlbertoCassano”,ColectoraRutaNac.168km0, ParajeElPozo,3000SantaFe,Argentina. E-mailaddress:[email protected] E-mailaddress:[email protected]