Witold Bednorz
Institute of Mathematics, Warsaw University 02-097 Warszawa, Poland
E-mail: wbednorz@mimuw.edu.pl
Abstract
The Talagrand’s proof of the sufficiency of a majorizing measure existence for the sample boundedness of processes with bounded in- crements used contraction from a certain ultrametric space. We give a short proof of such an ultrametric construction existence using the admissible sequences of nets approach.
1 Introduction
Majorizing measures provide a tool to prove sample boundedness of stochas- tic processes. Consider a compact metric space (T, d), and a Young function ϕ, i.e. ϕ : R+→ R+ such that ϕ(0) = 0, convex and increasing. We say that a process X(t), t ∈ T is of bounded increments, if
(1.1) Eϕ(|X(s) − X(t)|
d(s, t) ) 6 1, for s, t ∈ T.
Note that under (1.1) the process X(t), t ∈ T has a separable modification, which we refer to from now on. We say that a separable process X(t), t ∈ T is sample bounded whenever supt∈T X(t) < ∞, a.s. (which is a well defined r.v. due to the separability). The main idea is that the geometric structure of (T, d) can be used to decide whether or not the process X(t), t ∈ T is sample bounded.
The simplest approach is to consider the entropy of T , i.e. let N (T, d, ε) be the smallest number of closed balls B(t, ε) = {s ∈ T : d(s, t) 6 ε} that
2010 Mathematics Subject Classification: Primary 60G17; Secondary 28A99.
Key words and phrases: sample boundedness, majorizing measure, ultrametric.
Research partially supported by MNiSW Grant N N201 397437
1
cover T . Let also D(T ) = sups,t∈Td(s, t). It turns out [4, 6] that for each process X(t), t ∈ T that satisfies (1.1) the following inequality holds
E sup
t∈T X(t) 6 CE
Z D(T ) 0
ϕ−1(N (T, d, ε))dε, where C < ∞ is a universal constant. Therefore
Z D(T ) 0
ϕ−1(N (T, d, ε))dε < ∞
is the sufficient condition for the sample boundedness of all processes that satisfy (1.1). The problem is that the structure of (T, d) may be to compli- cated for the entropy to be sufficient and therefore more advanced tools are required. We say that a probability measure µ is majorizing if
sup
t∈T
Z D(T ) 0
ϕ−1( 1
µ(B(t, ε)))dε < ∞.
The existence of a majorizing measure is always sufficient for the sample boundedness. There were several results e.g [3, 7] that in many cases (at least for ϕ(x) ≡ xp, p > 1) showed that
Theorem 1.1. There exists a universal constant K < ∞ such that for each X(t), t ∈ T that satisfies (1.1) the following inequality holds
E sup
s,t∈T|X(s) − X(t)| 6 K sup
t∈T
Z D(T ) 0
ϕ−1( 1
µ(B(t, ε)))dε < ∞, where µ is a Borel probability measure on (T, d).
Finally the above theorem was proved in [1] to be a general rule. On the other hand in [8] that was invented that some finite set approximation works at least in the Gaussian like setting (where ϕ is of exponential growth). We say that a sequence of subsets (Tk)∞k=0 of T is admissible if |T0| = 1 and
|Tk| 6 ϕ(Rk), k > 1, where R > 2. The admissible nets were studied in [2], in particular there was proved that admissible sequnces of nets are at least as good as majorizing measures, i.e.
Theorem 1.2. There are universal constants A, B < ∞ such that for each admissible sequence of nets (Tk)∞k=0 and X(t), t ∈ T that satisfies (1.1) the following inequality holds
E sup
s,t∈T|X(t) − X(s)| 6 A sup
t∈T
∞
X
k=0
d(t, Tk)Rk+ B
∞
X
k=1
X
t∈Tk−1
d(t, Tk)Rk ϕ(Rk)
and
Theorem 1.3. There are universal constants A, B < ∞ such that for each probability measure µ on (T, d) there exists an admissible sequence of nets (Tk)∞k=0 that satisfies
sup
t∈T
∞
X
k=0
d(t, Tk)Rk 6 A
Z D(T ) 0
ϕ−1( 1
µ(B(t, ε)))dε,
∞
X
k=1
X
t∈Tk−1
d(t, Tk)Rk ϕ(Rk) 6 B
Z
T
Z D(T ) 0
ϕ−1( 1
µ(B(t, ε)))dεµ(dt).
Note that it is important that (Tk)∞k=0 may be any admissible sequence of nets with no additional regularity structure. For example we do not know whether Tk ⊂ Tk+1, k > 0.
The Talagrand’s approach (see Theorem 4.3 in [7]) to Theorem 1.1 was based on the following idea: first prove (Proposition 3.4 and Theorem 4.3 in [7]) that for a given ϕ the sample boundedness for all processes of bounded increments is equivalent to the the existence of a majorizing measure, then prove (Theorem 4.3 [7]) that there exists an ultrametric ρ on T such that d(s, t) 6 ρ(s, t) and still there is a majorizing measure on (T, ρ). In other words Talagrand showed that there exists an ultrametric structure ρ on T such that identity mapping I : (T, ρ) → (T, d) is 1-Lipschitz and all processes on (T, ρ) with bounded increments are sample bounded. However the proof given in [7] is quite complicated and requires some regularity condition on ϕ (namely the 42 condition for ϕ) which necessity is not obvious from the paper.
In this article we show that whenever there exists a universal C < ∞ then (1.2)
∞
X
j=k
Rj
ϕ(Rj) 6 C Rk
ϕ(Rk), for all k > 0,
one can find admissible ( ¯Tk) such that ¯Tk ⊂ ¯Tk+1. It turns out that this regularity suffices for the existence of an ultrametric ρ such that d(s, t) 6 ρ(s, t) and
A sup
t∈T
∞
X
k=0
d(t, Tk)Rk+ B
∞
X
k=1
X
t∈Tk−1
d(t, Tk)Rk ϕ(Rk) < ∞ implies that
A sup
t∈T
∞
X
k=0
ρ(t, ¯Tk)Rk+ B
∞
X
k=1
X
t∈ ¯Tk−1
ρ(t, ¯Tk)Rk ϕ(Rk) < ∞.
Consequently Theorems 1.3 and 1.4 yield
Theorem 1.4. Let (T,d) be a compact metric space, and ϕ that satisfies (1.2). If there exists a majorizing measure on (T, d) then there exists an ultrametric δ on T such that d(s, t) 6 δ(s, t) and all process of bounded increments on (T, ρ) are sample bounded.
We give also a short proof of a fact that the sample boundedness of all processes that have bounded increments on (T, ρ), where ρ is an ultrametric, implies the existence of a majorizing measure on the space. The measure can be obtained in a effective way.
2 The increasing sequence of nets
The consequence Theorems 1.2 and 1.3 is that there exists an admissible sequence of nets (Tk)∞k=0 with properties:
1. |Tk| 6 ϕ(Rk), |T0| = 1, T0 = {t0};
2. supt∈T P∞
k=0d(t, Tk)Rk < ∞;
3. P∞ k=1
P
u∈Tk
d(u,Tk−1)Rk ϕ(Rk) < ∞.
We show that (1.2) allows to improve the admissible sequence upon (Tk)∞k=0 in a way that it is also increasing. Let
T¯0 = T0, and ¯Tk =
k−1
[
l=0
Tl for k > 1.
We show that under condition (1.2) the sequence ( ¯Tk)∞k=0 has the properties 1-3.
Lemma 2.1. The sequence ( ¯Tk)∞k=0 is admissible. Moreover
(2.1) sup
t∈T
∞
X
k=0
d(t, ¯Tk)Rk< ∞
and if (1.2) holds then also (2.2)
∞
X
k=1
X
u∈ ¯Tk
d(u, ¯Tk−1)Rk ϕ(Rk) < ∞.
Proof. Observe that | ¯T0| = 1 and for k > 1
| ¯Tk| 6
k−1
X
l=0
ϕ(Rl) 6
k
X
l=1
R−lϕ(Rk) 6 ϕ(Rk),
due to the convexity of ϕ. The inequality (2.1) is trivial since d(t, ¯Tk) 6 d(t, Tk−1). Now suppose that (1.2) holds, the condition implies
∞
X
k=1
X
u∈ ¯Tk
d(u, ¯Tk−1)Rk ϕ(Rk) 6
∞
X
k=1 k
X
l=1
X
u∈Tl
d(u, Tl−1)Rk ϕ(Rk) 6 6
∞
X
l=1
X
u∈Tl
d(u, Tl−1)
∞
X
k=l
Rk
ϕ(Rk) 6 C
∞
X
l=1
X
u∈Tl
d(u, Tl−1)Rl ϕ(Rl) < ∞.
This completes the proof.
3 The ultrametric construction
Proposition 3.1. There exists an ultrametric ρ on T that dominates d and
(3.1) sup
t∈T
∞
X
k=0
ρ(t, ¯Tk)Rk < ∞.
Moreover if (1.2) holds then (3.2)
∞
X
k=1
X
u∈ ¯Tk−1
ρ(u, ¯Tk)Rk ϕ(Rk) < ∞.
Proof. Let πl : T → ¯Tl, l > 0 satisfies d(t, πl(t)) = d(t, ¯Tl). Consider ¯Tl⊂ T , l > 0. For any t ∈ ¯Tl, l > 0 denote tll = t and tlk = πk(tlk+1) for 0 6 k 6 l − 1. We start the ultrametric construction. First define ρ on ¯T × ¯T , where T =¯ S∞
k=0T¯k. Let ρ(t, t) = 0 for t ∈ ¯T and if s 6= t there exists l, m > 0 such that t ∈ ¯Tl\ ¯Tl−1 and s ∈ ¯Tm\ ¯Tm−1, (for simplicity let ¯T−1 = ∅). We define ρ(s, t) = 2 max{d(smm, smm−1) + ... + d(smn+1, smn), d(tll, tll−1) + ... + d(tln+1, tln)}, where n satisfies tln= smn and tln+1 6= smn+1. Clearly
ρ(u, v) 6 max(ρ(u, w), ρ(v, w)), u, v, w ∈ ¯T ,
so ρ is an ultrametric on ¯T × ¯T . It is also clear by the construction that (3.3) d(s, t) 6 ρ(s, t), s, t ∈ ¯T .
Observe that if t ∈ ¯Tk, k > 1 then 2d(t, πk−1(t)) = ρ(t, πk−1(t)) (in par- ticular ρ(t, πk−1(t)) = 0 if t ∈ ¯Tk−1). By the construction we have also ρ(t, πk−1(t)) = ρ(t, ¯Tk−1) and thus
(3.4)
∞
X
k=1
X
u∈ ¯Tk−1
ρ(u, ¯Tk−1)Rk ϕ(Rk) =
∞
X
k=1
X
u∈ ¯Tk−1
2d(u, ¯Tk−1)Rk ϕ(Rk) .
Combining (3.4) with (2.2) in Lemma 2.1 we deduce (3.2).
Now fix t ∈ Tl, l > 0 and observe that
l
X
k=0
ρ(t, ¯Tk)Rk6
l−1
X
k=0 l
X
j=k+1
ρ(tlj, tlj−1)Rk =
=
l
X
j=1
ρ(tlj, tlj−1)
j−1
X
k=0
Rk 6
l
X
j=1
ρ(tlj, tlj−1)Rj,
sinceP∞
k=1R−k 6 1. Consequently using that ρ(tlj, πj−1(tjl)) = 2d(tlj, πj−1(tlj)) we have
l
X
k=0
ρ(t, ¯Tk)Rk 6 2
l
X
j=1
d(tlj, πj−1(tlj))Rj. By the definition d(tlj, πj−1(tlj)) 6 d(tlj, πj−1(t)), so
(3.5)
l
X
k=0
ρ(t, ¯Tk)Rk6 2
l
X
j=1
d(tlj, πj−1(t))Rj.
Due to the triangle inequality
2
l
X
j=1
d(tlj, πj−1(t))Rj 6 2
l
X
j=1
(d(tlj, πj(t)) + d(πj(t), πj−1(t)))Rj 6
6 2
l
X
j=1
(d(tlj+1, tlj) + d(tlj+1, πj(t)) + d(πj(t), πj−1(t)))Rj 6
6 4 R
l
X
j=1
d(tlj+1, πj(t))Rj+1+ 2
l
X
j=1
d(πj(t), πj−1(t))Rj, (3.6)
where for simplicity tll+1 = tll= t. Clearly
2
l
X
j=1
d(tlj+1, πj(t))Rj+16 2
l
X
j=1
d(tlj, πj−1(t))Rj.
Since R > 2 we have 2/R < 1 and thus (3.6) implies that
(3.7) 2
l
X
j=1
d(tlj, πj−1(t))Rj 6 2(1 − 2 R)−1
l
X
j=1
d(πj(t), πj−1(t))Rj.
Again by the triangle inequality
(3.8) d(πj(t), πj−1(t)) 6 d(t, πj(t)) + d(t, πj−1(t)) = d(t, ¯Tj) + d(t, ¯Tj−1).
Combining (3.5), (3.7) and (3.8) we deduce that for each t ∈ Tl
(3.9)
l
X
k=0
ρ(t, ¯Tk)Rk 6 2(1 + R)(1 − 2 R)−1
l
X
j=0
d(t, ¯Tj)Rj.
Therefore by (3.9) and (2.1) in Lemma 2.1 we have that for each t ∈ ¯T the following inequality holds
(3.10) sup
l>0
sup
t∈ ¯T l
X
k=0
ρ(πl(t), ¯Tk)Rk < ∞.
In particular it implies that the sequence (πl(t))∞l=0 is Cauchy in ρ distance.
Therefore we can extend the definition of ρ on the whole T by ρ(s, t) = liml→∞ρ(πl(s), πl(t)). Consequently due to (3.10) and the Fatou theorem
sup
t∈T
∞
X
k=0
ρ(t, ¯Tk)Rk 6 sup
l>0
sup
t∈ ¯T l
X
k=0
ρ(πl(t), ¯Tk)Rk < ∞.
It completes the proof of (3.3).
Proposition 3.1 and Theorem 1.2 imply that
(3.11) sup
X
E sup
s,t∈T
|X(t) − X(s)| < ∞,
where the supremum is taken over all X(t), t ∈ T that satisfy
(3.12) Eϕ(|X(t) − X(s)|
ρ(s, t) ) 6 1, s, t ∈ T.
This completes the proof of Theorem 1.4.
We now recall the simple idea (see Proposition 3.4 in [7]) how to construct a majorizing measure, namely there exists a majorizing measure m on (T, ρ) if the following holds
(3.13) sup
µ
Z
T
Z Dρ(T ) 0
ϕ−1( 1
µ(Bρ(t, r)))drdt < ∞,
where Bρ(t, r) = {x ∈ T : ρ(t, x) 6 r}. Moreover if ϕ−1(1x) is convex then (3.14)
sup
t∈T
Z Dρ(T ) 0
ϕ−1( 1
m(Bρ(t, r)))dr 6 sup
µ
Z
T
Z Dρ(T ) 0
ϕ−1( 1
µ(Bρ(t, r)))drdt.
In the general case ϕ−1(1x) may not be convex , yet there exits convex ξ such that ξ(x) 6 ϕ−1(x1) 6 2ξ(x). Therefore the constant in (3.14) should
be changed from 1 to 2. To show (3.13) first construct on the probability space (T, B(T ), µ) the process
X(t, ω) =
Z Dρ(T ) ρ(t,ω)
ϕ−1( 1
µ(Bρ(ω, r)))dr, By the Jensen’s inequality
Eϕ(|X(s) − X(t)|
ρ(s, t) ) = Z
T
ϕ(ρ(s, t)−1|
Z ρ(t,ω) ρ(s,ω)
ϕ−1( 1
µ(Bρ(ω, r)))dr|)µ(dω) 6 6
Z
T
ρ(s, t)−1
Z ρ(t,ω) ρ(s,ω)
1
µ(Bρ(ω, r))drµ(dω) =
=
Z D(T ) 0
1
µ(Bρ(t, r)4Bρ(s, r))ρ(s, t)µ(B(t, r))dr, where 4 denotes the symmetric set difference. By the main property of ultrametric spaces B(t, r) = B(s, r) for r > ρ(s, t) and therefore
Eϕ(|X(s) − X(t)|
ρ(s, t) ) 6
Z ρ(s,t) 0
1
ρ(s, t)dr = 1.
Consequently due to (3.11) we deduce (3.13).
The usual Fernique’s argument (see Proposition 3.2 [7]) shows by the Hanh Banach theorem that (3.13) implies that the existence of a majorizing mea- sure on (T, ρ). However there is an alternative way to show the fact, which works for ϕp(x) = xp, p > 1. First observe that
∞
X
k=1
2−kD(T )ϕ−1( 1
µ(Bρ(t, 2−k+1D(T )))) 6
Z Dρ(T ) 0
ϕ−1( 1
µ(Bρ(t, r)))dt 6 6
∞
X
k=1
2−kD(T )ϕ−1( 1
µ(Bρ(t, 2−kD(T )))).
(3.15)
For any s, t ∈ T either Bρ(t, 2−kD(T )) = Bρ(s, 2−kD(T )) or Bρ(t, 2−kD(T ))∩
Bρ(s, 2−kD(T )) = ∅. Therefore there exists an increasing sequence of par- titions Ak, k > 0 so that Ak consists of pairwise disjoint balls of radius 2−kD(T ). Let Ak(t) denotes the partition element that contains t. Conse- quently due to (3.13) we obtain that
B = sup
µ
Z
T
∞
X
k=1
2−kD(T )ϕ−1( 1
µ(Ak(t))) =
=
∞
X
k=1
2−kD(T ) X
A∈Ak
ϕ−1( 1
µ(A)) < ∞.
(3.16)
We show that whenever ϕ = ϕp for some p > 1 there exists a measure m on (T, ρ) such that
∞
X
k=0
2−k−1D(T )ϕ−1( 1
m(Ak(t))) 6 B for all t ∈ T.
Lemma 3.2. For a given l > 0 there exists measure ml such that for all t ∈ T
l
X
k=0
2−k−1D(T )ϕ−1p ( 1
ml(Ak(t))) = Bl 6 B.
Proof. The proof goes by induction on l over all spaces T with a partition structure. For l = 0 there is nothing to prove, suppose we have shown the fact for l > 0. Observe that we can consider only partitions with |A1| = N >
1, let A1 = {A1, ..., AN}. By the induction assumption, for each Ai ∈ A1, 1 6 i 6 N there exists measure µi on Ai such that for all t ∈ Ai
l
X
k=2
2−kD(T )ϕ−1p ( 1
µi(Ak(t))) = 2−1Bl−1(i).
Let αi, 1 6 i 6 N be such that αi > 0, PN
i=1αi = 1 and 2−1ϕ−1p (1
αi)(D(T ) + Bl−1(i)) = 2−1α−
1 p
i (D(T ) + Bl−1(i)) = const.
Therefore to satisfy PN
i=1αi = 1 we need that Bl = const = 2−1(
N
X
i=1
(D(T ) + Bl−1(i))p)1p.
We use the property ϕ−1p (xy) = ϕ−1p (x)ϕ−1p (y), where x, y > 0 to get for all t ∈ Ai, 1 6 i 6 N
2−1D(T )ϕ−1p ( 1 αi) +
l
X
k=2
2−kD(T )ϕ−1p ( 1
αiµi(Ak(t))) = Bl. Consequently if ml =PN
i=1αiµi then
l
X
k=1
2−kD(T )ϕ−1( 1
ml(Ak(t))) = Bl.
By (3.16) it must be that Bl 6 B. It completes the argument.
Therefore defining m as the weak limit of ml we obtain that
∞
X
k=0
2−k−1D(T )ϕ−1( 1
m(Ak(t))) 6 lim
l→∞Bl 6 B
and therefore m is the majorizing measure on T . The above idea is similar to the approach used in [5] to construct a majorizing measure.
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