Representation Theorems of $\mathbb{R}$-trees

0.42 MB·

by  Asuman G. Aksoy

R
Representation Theorems of -trees
6
1
Asuman Gu¨ven AKSOY, Monairah AL-ANSARI and Qidi PENG
0
2
Abstract. In this paper, we provide a new representation of an R-tree by using

n a set of graphs. We have captured the four-point condition from these graphs and
a
identified the radial metric and river metric by some particular graphical representa-
J
tions. In stochastic analysis, these representation theorems are of particular interest
3 in identifying Brownian motions indexed by R-trees.
1

] 1 Introduction
G
M
OneofthecentralobjectofprobabilityisBrownianmotion(Bm),whichisthemicro-
scopic picture emerging from a particle moving in n-dimensional space and of course
.
h the nature of Brownian paths is of special interest. In this paper, we study the fea-
at tures of Brownian motion indexed by an R-tree. An R-tree is a 0-hyperbolic metric
m space with desirable properties. Note that, J. Istas in [8] proved that the fractional
Brownian motion can be well defined on a hyperbolic space when 0 < H ≤ 1. Fur-
[ 2
thermore,in[2]theauthorsuseDirichletformmethodstoconstructBrownianmotion

1 on any given locally compact R-tree, additionally in [7], representation of a Gaussian
v field via a set of independent increments were discussed. In this paper we focus on
8 “radial”and“river”metricandclarifytherelationshipbetweenmetrictreesgenerated
0
by these metrics and a particular metric ray denoted by C (A,B). Our investigation
1 d
is motivated by the questions: under what conditions on the set {C (A,B)}
3 d A,B∈M
does (M,d) become an R-tree? and when can an R-tree be identified by the sets
0
{C (A,B)} ? Itisourhopethatthisworkcouldleadtotheinterestofapplying
. d A,B∈M
1 those results to random fields indexed by metric spaces.
0 The study of injective envelopes of metric spaces, also known as R-trees (metric
6
treesorT-theory)beganwithJ.Titsin[12]in1977andsincethen,applicationshave
1
been found within many fields of mathematics. For a complete discussion of these
:
v spaces and their relation to global metric spaces of nonpositive curvature we refer to
i [4]. Applications of metric trees in biology and medicine stems from the construction
X
of phylogenetic trees [11]. Concepts of “string matching” in computer science are
ar closelyrelatedwiththestructureofmetrictrees[3]. R-treesareageneralizationofan
ordinarytreewhichallowsfordifferentweightsonedges. InordertodefineanR-tree,
wefirstintroducethenotionofmetricsegment. Let(M,d)beametricspace. Forany
A,B ∈M, the metric segment [A,B] is defined by
[A,B]={X ∈M : d(A,X)+d(X,B)=d(A,B)<+∞}.
Notice that by this definition, [A,B](cid:54)=∅ if and only if A,B are connected in (M,d).
Mathematics Subject Classification (2010): 05C05,05C62,60J65,54E35.
Key words: R-tree,BrownianFields,IndependentIncrements
1
Definition 1.1 (see [10]) An R-tree is a nonempty metric space (M,d) satisfying:
(a) Any two points of A,B ∈M are joined by a unique metric segment [A,B].
(b) If A,B,C ∈M, then
[A,B]∩[A,C]=[A,O] for some O∈M.
(c) If A,B,C ∈M and [A,B]∩[B,C]={B}, then
[A,B]∪[B,C]=[A,C].
Through out this paper we only consider the class of metric spaces satisfying (a) in
Definition1.1above. Wecallthismetricspaceuniquelygeodesicmetricspace. Inthe
following we characterize an R-tree by the theorem below (given in [5]):
Theorem 1.1 A uniquely geodesic metric space (M,d) is an R-tree if and only if it
is connected, contains no triangles and satisfies the four-point condition (4PC).
Note that, we say a metric d(·,·) satisfies the four-point condition (4PC) if, for any
A,B,C,D in M the following inequality holds:
d(A,B)+d(C,D)≤max{d(A,C)+d(B,D), d(A,D)+d(B,C)}.
The four-point condition is stronger than the triangle inequality (take C = D), but it
should not be confused with the ultrametric definition. An ultrametric satisfies the
condition d(A,B) ≤ max{d(A,C),d(B,C)}, and this is stronger than the four-point
condition. Moreover, we say A,B,C form a triangle if all the triangle inequalities
involving A,B,C are strict and [X,Y]∩[Y,Z]={Y} for any {X,Y,Z}={A,B,C}.
d(.,.) is said to be a tree metric if it satisfies the (4PC). Given a metric space (M,d),
wewouldcapturethetreemetricpropertiesof(M,d)byintroducingthefollowingsets
{C (A,B)} .
d A,B∈M
Definition 1.2 For any A,B ∈M, we define
C (A,B)={X ∈M : d(X,A)=d(X,B)+d(A,B)<+∞}.
d
Note that two points A,B ∈M are connected if and only if C (A,B)(cid:54)=∅.
d
WeremarkthataBrownianmotionisuniquelydeterminedbyindependentincre-
ments and furthermore, since the set C (P ,P ) is also defined as:
d 1 2
C (P ,P )={X ∈M : B(X)−B(P ) is independent of B(P )−B(P )},
d 1 2 2 2 1
It is of interest to ask the following questions:
Question1: Underwhatconditionsontheset{C (A,B)} does(M,d)become
d A,B∈M
an R-tree?
Question 2: When can an R-tree be identified by the set {C (A,B)} ?
d A,B∈M
In this paper we give complete solution to Question 1 (see Section 2 below), namely,
weprovideasufficientandnecessaryconditionon{C (A,B)} suchthat(M,d)
d A,B∈M
is an R-tree. In Section 3.1, we study Question 2 by considering radial metric and
river metric. We show that the answer to Question 2 is positive when M =Rn and
d(A,B)=g (|A−B|) for A,B ∈Π ,
k k
where|·|istheEuclideannorm,(Π ) issomepartitionofRn andg : R →
k k=1,...,N k +
R is a continuous function.
+
2
2 Results : An Equivalence of R-tree Properties
We start by introducing the following conditions that will be used in the proof of the
Theorem 2.1:
Condition (A): For any 3 distinct points A,B,C ∈M, there exists unique O∈M
such that
{X,Y}⊂C (Z,O) for any distinct X,Y,Z ∈{A,B,C}.
d
Condition (B): For any distinct A,B,C ∈M, there exists O∈M such that
[A,B]∩[B,C]∩[A,C]={O}.
Notethatifthecardinality|M|=1or2,then(M,d)isobviouslyanR-tree,since
any2pointsarejoinedbyauniquegeodesic. When|M|≥3,Condition(A)guarantees
that(M,d)containsnocircuit. ThefollowingwillbeusedintheproofoftheTheorem
2.1:
Lemma 2.1 Condition (A) is equivalent to Condition (B).
Proof. We only consider the case where M contains at least 3 distinct points. Let’s
pick 3 distinct points A,B,C ∈ M. Then by observing that for any distinct X,Y ∈
{A,B,C},
X ∈C (Y,O) is equivalent to O∈[X,Y].
d
Thus Lemma 2.1 holds. (cid:3)
Theorem 2.1 A uniquely geodesic metric space (M,d) is an R-tree if and only if
Condition (A) holds.
Proof By Lemma 2.1, it is sufficient to prove Theorem 2.1 holds under Condition
(B). First we show that if (M,d) is an R-tree, then Condition (B) is satisfied, and
thenshowthatifCondition (B)holds,then(M,d)isanR-tree. Suppose(M,d)isan
R-tree, since (M,d) is connected, then [A,B](cid:54)=∅ for all A,B ∈M. For any 3 points
A,B,C ∈M we have:
• if [A,B]∩[B,C]={B}, then by (c) in Definition 1.1,
{B}=[A,B]∩[B,C]⊂[A,B]∪[B,C]=[A,C].
This yields
[A,B]∩[B,C]∩[A,C]={B}∩[A,C]={B}.
• If there exists O ∈ M, O (cid:54)= B such that [A,B]∩[B,C] = [B,O], then O ∈
[A,B]∩[B,C]∩[A,C]. Thus, condition (B) is verified.
Next assume that Condition (B) holds. By taking any A(cid:54)=B =C, one easily shows
that [A,B] (cid:54)= ∅, thus (M,d) is connected. The fact that [A,B]∩[B,C]∩[A,C] (cid:54)= ∅
leads to the fact that there is no triangles in (M,d). Then it is sufficient to prove
thatd(·,·)satisfiesthe(4PC).Letuspick4distinctpointsA,B,C,DfromM. Under
Condition (B), we have two possibilities to the positions of A,B,C,D, namely:
3
1. A,B,C,D are in the same metric segment.
2. Case 1 above does not hold.
Case 1 is equivalent to
(cid:26)
d(W,X)=d(W,Z)+d(X,Z);
d(W,Y)=d(W,Z)+d(Y,Z).
Thiseasilyleadstothe(4PC).Forthesecondcase,forany{X,Y,Z,W}={A,B,C,D},
if C (X,Z) = C (Y,Z), then one necessarily has W ∈/ C (X,Z) = C (Y,Z). This is
d d d d
equivalent to
(cid:26)
d(W,X)<d(W,Z)+d(X,Z);
d(W,Y)<d(W,Z)+d(Y,Z).
This easily leads to the (4PC). It is easy to check that in each case, the (4PC) is
satisfied. (cid:3)
2.1 Characterization of C (P ,P ) via Radial Metric
d 1 2
We define an R-tree (Rn,d ) (n≥1) with root 0 and the metric
1
(cid:26) |A−B| if A=aB for some a∈R;
d (A,B)=
1 |A|+|B| otherwise.
We explicitly represent the set C (P ,P ) for all P ,P ∈Rn in the following propo-
d1 1 2 1 2
sition.
Proposition 2.1 For any P ,P ∈(Rn,d ),
1 2 1
Cd1(P1,P2)= R[RPnn2\,(+P∞2,)+−0−P∞→2)−0−P→1 iiifff PPP22 ∈∈=/ [[P00,,,PP11])−0−0−−PP→→11;; (2.1)
1 2
where for any A,B ∈ Rn, [A,B)−−→ denotes the segment {(1−a)A+aB;a ∈ [0,1)}
AB
and (A,+∞)−→ denotes {aA+bB;a > 1,b > 0} under Euclidean distance. These
0B
notations shouldn’t be confused with the metric segments [A,B] of a metric space.
Proof. Since it is always true that C (P ,P ) = Rn for P = P , then we only
d1 1 2 1 2
consider the case when P (cid:54)= P . There are 3 different situations to the positions of
1 2
P , P : (1) P , P are on the same ray (which means, P =aP for some a∈R) and
1 2 1 2 2 1
0≤|P |<|P |; (2) P , P are on the same ray and 0≤|P |<|P |; (3) P ,P are on
1 2 1 2 2 1 1 2
different rays.
Case (1): P , P are on the same ray and 0≤|P |<|P |.
1 2 1 2
In this case one necessarily has
d (A,P )=d (A,P )+|P −P |. (2.2)
1 1 1 2 1 2
Case (1.1): If A is on a different ray as P , then (2.2) becomes
2
|A|+|P |=|A|+|P |+|P −P |.
1 2 1 2
This together with the fact that P (cid:54)=P implies
1 2
|P |=|P |+|P −P |>|P |.
1 2 1 2 2
4
This is impossible, because it is assumed that |P |<|P |.
1 2
Case (1.2): Suppose A is on the same ray as P . Now (2.2) is equivalent to
2
|A−P |=|A−P |+|P −P |.
1 2 1 2
The solution space for A is then the segment [P2,+∞)−0−P→2 under Euclidean distance.
One concludes that in Case (1),
Cd1(P1,P2)=[P2,+∞)−0−P→2. (2.3)
Case (2): P ,P are on the same ray and |P |>|P |≥0. Note that (2.2) still holds.
1 2 1 2
Case (2.1): Suppose that A is on a different ray as P . (2.2) is then equivalent to
1
|P |=|P |+|P −P |.
1 2 1 2
Theaboveequationalwaysholdstrue. ThereforeanyAonadifferentrayasP belongs
1
to C (P ,P ). Case (2.2): A is on the same ray as P . Equation (2.2) then becomes
d1 1 2 1
|A−P |=|A−P |+|P −P |,
1 2 1 2
and its solution space is segment [0,P2]−0−P→2 under Euclidean distance.
Combining Case (2.1) and Case (2.2), one obtains, in Case (2),
Cd1(P1,P2)=Rn\(P2,+∞)−0−P→1. (2.4)
Case (3): P ,P are on different rays with P ,P (cid:54)=0.
1 2 1 2
Case (3.1): A is on the same ray as P .
1
In this case we have
|A−P |=|A|+|P |+|P |+|P |.
1 2 1 2
By the triangle inequality,
|A|+|P |+2|P |=|A−P |≤|A|+|P |.
1 2 1 1
This yields the absurd statement P =0!
2
Case (3.2): A is on the same ray as P .
2
We have
|A|+|P |=|A−P |+|P |+|P |.
1 2 1 2
This leads to A∈[P2,+∞)−0−P→2.
Case (3.3): A is on a different ray as P , P .
1 2
We have
|A|+|P |=|A|+|P |+|P |+|P |,
1 2 1 2
which again implies P =0. Contradiction!
2
We conclude that in Case (3),
Cd1(P1,P2)=[P2,+∞)−0−P→2. (2.5)
Finally,bycombining(2.3),(2.4)and(2.5),oneprovesProposition2.1(seeFigure
1, Figure 2). (cid:3)
5
Figure 1: The thick line represents Figure 2: The shaded region repre-
thesetofC (P ,P )whenP isnot sents C (P ,P ) when P is in the
d1 1 2 2 d1 1 2 2
in the segment [0,P ]. segment [0,P ).
1 1
NowwewouldshowtheinverseofProposition2.1,namelywhetherornotwecan
recognize (Rn,d) as an R-tree using C (P ,P ). For that purpose, we first show the
d 1 2
following statement.
Proposition 2.2 Let(Rn,d)beametricspace. If(2.1)holdsforanyP ,P ∈(Rn,d),
1 2
then (Rn,d) is an R-tree.
Proof. ByTheorem2.1,weonlyneedtoshowCondition(B)holds. Letusarbitrarily
pick 3 different points A,B,C ∈ Rn. If A,B,C are in the same segment, saying,
A ∈ C (B,C), then C ∈ [A,B]∩[B,C]∩[A,C] and Condition (B) is satisfied. If
d
A,B,C arenotinthesamesegment,i.e.,X ∈/ C (Y,Z)forany{X,Y,Z}={A,B,C},
d
then we see from the definition of C (P ,P ) that
d 1 2
X ∈C (Y,0) for any distinct X,Y ∈{A,B,C},
d
whichisequivalentto0∈[A,B]∩[B,C]∩[A,C]. Hence Condition (B)issatisfied. (cid:3)
2.2 Characterization of C (P ,P ) via River Metric
d 1 2
For A∈R2, we denote by A=(A(1),A(2)). We define the river metric space (R2,d )
2
by taking
(cid:26) |A(2)−B(2)| for A(1) =B(1);
d (A,B)=
2 |A(2)|+|A(1)−B(1)|+|B(2)| for A(1) (cid:54)=B(1).
From now on we say that A,B are on the same ray in (R2,d ) if and only if A,B are
2
on a vertical Euclidean line: A(1) =B(1).
6
Proposition 2.3 Let (R2,d ) be a river metric space. For P ∈R2, denote by P∗ =
2
(P(1),0)theprojectionofP tothehorizontalaxis. ThenforanyP ,P ∈R2, wehave
1 2
Cd2(P1,P2)= RR[[PP2222(\,1()∞P,∞2),−P)∞−2∗−P−P−→1()−21−P−)−1∗P−−P−2→(→11) ×R iiiiffff PPPP221(1∈∈=/) (cid:54)=[[PPPP11∗∗.2,,(1PP)11,))P;a2(n2d) =P2(02;) (cid:54)=0; (2.6)
1 2
Proof. It is obvious that C (P ,P ) = R2 when P = P . For P (cid:54)= P , we mainly
d2 1 2 1 2 1 2
consider 2 cases: (1) P , P are on the same ray (P(1) = P(1)); (2) P ,P are on
1 2 1 2 1 2
different rays (P(1) (cid:54)=P(1)).
1 2
Case (1): P , P are on the same ray.
1 2
In this case one necessarily has
d (A,P )=d (A,P )+|P(2)−P(2)|. (2.7)
2 1 2 2 1 2
Case (1.1): Suppose A is on a different ray as P , then it follows from (2.7) that
2
|A(2)|+|P(1)−A(1)|+|P(2)|=|A(2)|+|P(1)−A(1)|+|P(2)|+|P(2)−P(2)|.
1 1 2 2 1 2
Since P(1) =P(1), the above equation is simplified to
1 2
|P(2)−P(2)|+|P(2)|−|P(2)|=0.
1 2 2 1
This equation holds for all A with A(1) (cid:54)= P(1) provided that P(2) ∈ [0,P(2))−−−→.
2 2 1 0P(2)
1
When P(2) ∈/ [0,P(2))−−−→, it has no solution.
2 1 0P(2)
1
Case (1.2): A is on the same ray as P . Now one has
2
|A(2)−P(2)|=|A(2)−P(2)|+|P(2)−P(2)|.
1 2 1 2
The above equation holds only when
A∈{P(1)}×[P(2),∞)−−−→.
2 2 0P(2)
2
Therefore one concludes that when P and P are on the same ray,
1 2
Cd2(P1,P2)= R[P22\,(∞P2),−P∞−2∗−P→)2−P−1∗−P→1 iiff PP22((22)) ∈∈/ [[00,,PP11((22))))−0−0−−PP−−11((→→22));. (2.8)
Case (2): P ,P are on different rays.
1 2
Case (2.1): A is on the same ray as P . In this case we have
1
|A(2)−P(2)| = |A(2)|+|A(1)−P(1)|+|P(2)|+|P(2)|+|P(1)−P(1)|+|P(2)|
1 2 2 1 1 2 2
> |A(2)|+|P(2)|.
1
This contradicts the triangle inequality, therefore there is no solution for A in this
case.
7
Case (2.2): A is on the same ray as P . We have
2
|A(2)|+|A(1)−P(1)|+|P(2)|=|A(2)−P(2)|+|P(2)|+|P(1)−P(1)|+|P(2)|.
1 1 2 1 1 2 2
By using the fact that A(1) =P(1), the above equation becomes
2
|A(2)|=|A(2)−P(2)|+|P(2)|. (2.9)
2 2
This provides:
• if P(2) =0, then the solution space of (2.9) is {P(1)}×R;
2 2
• if P(2) (cid:54)=0, then the solution space of (2.9) is {P(1)}×[P(2),∞)−−−→.
2 2 2 0P(2)
2
Case (2.3): A is on a different ray as P , P .
1 2
We have
|A(2)|+|A(1)−P(1)|+|P(2)|
1 1
=|A(2)|+|A(1)−P(1)|+|P(2)|+|P(2)|+|P(1)−P(1)|+|P(2)|.
2 2 1 1 2 2
It is equivalent to
|A(1)−P(1)|=|A(1)−P(1)|+|P(1)−P(1)|+2|P(2)|.
1 2 1 2 2
ThisequationhassolutiononlywhenP(2) =0. Thentheaboveequationiswrittenas
2
|A(1)−P(1)|=|A(1)−P(1)|+|P(1)−P(1)|.
1 2 1 2
This implies
A(1) ∈(P(1),∞)−−−−−−→.
2 P(1)P(1)
1 2
By combining the solutions for Cases (2.1), (2.2), we obtain, in Case (2),
C (P ,P )= [P2,∞)−P−2∗−P→2 if P1(1) (cid:54)=P2(1), P2(2) (cid:54)=0; (2.10)
d2 1 2  [P2(1),∞)−P−(−1−)P−−(→1) ×R if P1(1) (cid:54)=P2(1), P2(2) =0.
1 2
Finally, putting together Cases (1),(2), one proves Proposition 2.3. (cid:3)
Figure 4: The thick line represents
Figure 3: The shaded region repre-
bseenlotsngthsetosetthoefsCedg2m(Pen1,tP[P2)∗w,Phe)n. P2 C|Pd2(2()P|1>,P|2P)(2w)|h.en P1(1) = P2(1) and
1 1 2 1
8
Figure 5: The thick line represents Figure 6: The shaded region repre-
the set of C (P ,P ) when P(1) (cid:54)= sents C (P ,P ) when P(1) (cid:54)= P(1)
d2 1 2 1 d2 1 2 1 2
P(1) and P(2) (cid:54)=0. and P(2) =0.
2 2 2
Proposition 2.4 Let (Rn,d) be a metric space. If for any P ,P ∈Rn, (2.6) holds,
1 2
then (Rn,d) is an R-tree.
Proof. WeonlyneedtoshowCondition(A)issatisfiedbytheexpressionofC (P ,P )
d 1 2
in(2.6). Observethatforany3distinctpointsA,B,C ∈Rn,withoutlossofgenerality,
there are 3 situations according to the positions:
Case 1 : A(1) =B(1) =C(1), A(2) ∈[0,B(2))−−−→, B(2) ∈[0,C(2))−−−→.
0B(2) 0B(2)
Case 2: A(1) =B(1) (cid:54)=C(1), A(2) ∈[0,B(2))−−−→.
0B(2)
Case 3: A(1), B(1) and C(1) are all distinct, B(1) ∈[A(1),C(1)]−−−−−−→.
A(1)C(1)
By (2.6), it is easy to see Condition (A) holds with O=B, O=A and O=(0,B(2))
respectively for Case 1, Case 2 and Case 3. Hence Proposition 2.4 is proven by using
Theorem 1.1. (cid:3)
3 Identification of Radial Metric and River Met-
ric via C (P ,P )
d 1 2
3.1 Identification of Radial Metric via C (P ,P )
d1 1 2
InProposition2.2andProposition2.4,wehaveshownthatthesets{C (P ,P )}
d 1 2 P1,P2
capture the tree properties of the metric spaces (Rn,d ) and (R2,d ). Now we claim
1 2
thatsubjecttosomeadditionalconditionsthesetwoR-treescanbeuniquelyidentified
by the sets {C (P ,P )} .
d 1 2 P1,P2
Definition 3.1 Letd˜ beametricdefinedonRn satisfyingthatthereexistsafunction
1
f :R →R such that
+ +
• f is continuous;
9
• f satisfies the following equation:
(cid:26) d˜(ax,x)=f(|ax−x|) for all x∈Rn and all a≥0;
1
f(1)=1.
Theorem 3.1 The following statements are equivalent:
(i) d˜ =d .
1 1
(ii) For any P ,P ∈(Rn,d˜), C (P ,P )=C (P ,P ) given in (2.1).
1 2 1 d˜1 1 2 d1 1 2
Before proving Theorem 3.1, we first introduce the following useful statement.
Theorem 3.2 (See Aczel [1], Theorem 1) If Cauchy’s functional equation
g(u+v)=g(u)+g(v)
is satisfied for all positive u,v, and if the function g is
• continuous at a point;
• nonnegative for small positive u−s or bounded in an interval,
then
g(u)=cu
is the general solution for all positive u.
Proof. The implication (i) =⇒ (ii) is simply Proposition 2.1. Now it remains to
prove (ii)=⇒(i).
Case (1): A,P ,0 are on the same straight line with A(cid:54)=P .
1 1
Withoutlossofgenerality,assume|A|>|P1|. ThenthereexistsP2 ∈(P1,+∞]−0−P→2
such that A∈[P2,+∞)−0−P→2. By Proposition 2.1, one has
d˜(A,P )=d˜(A,P )+d˜(P ,P ).
1 1 1 2 1 1 2
ObservethatA,P ,P ,0areonthesamestraightline,thenbythedefinitionof
1 2
d˜, the above equation is equivalent to
1
f(|A−P |)=f(|A−P |)+f(|P −P |). (3.1)
1 2 1 2
ThisisaCauchy’sequation,thenbyusingTheorem3.2,thegeneralsolutionis
f(u)=cu. Together with its initial condition f(1)=1, one finally gets
f(u)=u.
Hence,
d˜(A,P )=|A−P |, for A,P ,0 lying on the same straight line.
1 1 1 1
Case (2): A,P ,0 are not on the same straight line (in this case one necessarily has
1
A,P (cid:54)=0).
1
We take P2 =0. The fact that A∈/ (0,+∞)−0−P→1 implies
d˜(A,P )=d˜(A,0)+d˜(P ,0).
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