An Example of
Ultracontractive Lévy
semigroup
Patrick Maheux (Mapmo.Orléans)
Joint work with Alexander Bendikov
(Wroclaw.Pl).
( Toulouse. November 1st, 2008)
Contents:
I. Motivation
II. Symmetric Lévy generators on R
III. General results
IV. Very oscillating symbol
V. Simulations
1
I. Motivation
Let (Tt) be a symmetric submarkovian
semigroup L2(X, µ) with µ
σ-finite measure on X
0 ≤ f ≤ 1 =⇒ 0 ≤ Ttf ≤ 1 for all t > 0.
Tt : contraction semigroup on Lp,
1 ≤ p ≤ ∞.
2
(−A, D) be its generator.
Ttf − f
t→0
t
−Af = lim
(in L2)
Tt = e−tA
For n > 2, the following conditions are
equivalent
3
(S)
(N )
(U )
|| f ||2
2n/n−2 ≤ c(Af, f ),
2+4/n
|| f ||2
∀f ∈ D
4/n
≤ c1(Af, f ) || f ||1
|| Ttf ||∞≤ c2 t−n/2 || f ||1,
,
∀f ∈ D
∀t > 0.
(Varopoulos, Carlen-Kusuoka-Stroock )
4
Let M be a non-increasing non-negative
function
defined on (0, +∞) with M (0+) = +∞
N (x) = sup (xt − tM (1/t)) .
t>0
When M is C 1, we set
Q(x) = −M 0 o M −1(x)
if
x > M (∞)
and Q(x) = 0 otherwise.
5
Thm: (Coulhon.T) Let −A be the
infinitesimal generator of a sub-Markovian
semigroup on L2(X, µ). Let M, N, Q as
above.
(1)
(GU )
If ||Ttf ||∞ ≤ eM (t)||f ||2, t > 0
Then
(N T I1)
2
2
||f ||2 N log ||f ||2 ≤ (Af, f ), ||f ||1 ≤ 1,
6
(2)
If
2
2
(N T I2)||f ||2 Q log ||f ||2 ≤ (Af, f ),
||f ||1 ≤ 1,
Then,
||Ttf ||∞ ≤ eM (t)||f ||2, t > 0
7
Examples:
(1) Assume that M (t) = c log+( 1t ). Then
−1
N (x) = ce exp xc = c0 Q(x), x >> 1
(2) Assume that M (t) = tcα with α > 0.
N (x) = c3 x1+1/α = c4 Q(x), x > 0,
8
(3) Assume that M (t) = exp( t1α ) with
α > 0. Then
N (x) ∼ x (log x)1/α ,
Q(x) = α x [log(x)]
as x → +∞.
1
1+ α
,
∀x > 1.
Observation: We see that in the first two
examples that N and Q are comparable,
whereas in Example 3. the ratio Q/N goes
to infinity at infinity.
9
Thm: (Bendikov,Coulhon,Saloff-Coste):
M convex non-increasing function s.t.
−tM 0(t) ≤ bM (t), t → 0
Then the following properties are ∼
(1) ∃ c1, c2 > 0 s.t.
∀t > 0,
||Tt||2,∞ ≤ ec1M (c2t)
(2) ∃ c3, c4 > 0 s.t. ∀f ∈ D,
||f ||1 ≤ 1,
2
2
c3||f ||2 N c4 log ||f ||2 ≤ (Af, f )
10
Remark: Function t → M (t) as at most
polynomial explosion so the heat kernel has
at most exponential explosion at t = 0.
In this case, Nash-type inequality gives an
exact description of the behavior of the
function t → ||Tt||2,∞.
The situation changes in the double
exponential case that is
log log ||Ttf ||2,∞ ∼ t1α as t → 0.
11
By Coulhon’s Thm. : (NTI): for all f s.t.
||f ||1 ≤ 1,
||f ||2
2
log ||f ||2
2
h
i1/α
2
≤ (Af, f )
log(log ||f ||2)
+
Question:
Does this Nash-type inequality imply
estimate of the function t → ||e−tA||2,∞ for
small t?
No for α ≥ 1:
12
Counter- example of Davies-Simon:
There exists A s.t. ||f ||1 ≤ 1
||f ||2
2
log ||f ||2
2
h
i
2
log(log ||f ||2)
≤ (Af, f )
+
(α = 1)
but ||e−tA||2,∞ = ∞ at least for small t > 0
13
Yes for 0 < α < 1 but with some loss.
1
||e−tA||2,∞ ≤ exp exp α0
t
α0 =
,
t > 0.
α
>α
1−α
14
Aim of the paper:
In the setting of Lévy semigroup (on R),
we are interested in relationships between
log log ||Ttf ||2,∞ ∼ t1α as t → 0.
and
||f ||2
2
log ||f ||2
2
h
i1/α
2
log(log ||f ||2)
≤ (Af, f )
with ||f ||1 ≤ 1.
15
II. Symmetric Lévy generators on R
Af (x) =
1 ∞
−
[f (x + y) − 2f (x) + f (x − y)] dΠ(y),
2 −∞
Z
where the Lévy measure Π is a symmetric
measure such that
Z
R
∗
(1 ∧ y 2) dΠ(y) < ∞.
16
A is the generator of a
translation invariant Markov semigroup
Tt = e−tA.
Let µt be the measure such that
Ttf = µt ∗ f.
Fourier transform
(Ttf )ˆ(y) = e−tψ(y)fˆ(y)
17
where
ψ(y) = 2
Z ∞
0
[1 − cos(xy)] dΠ(x)
Lévy symbol (or exponent)
(Af, f ) =
Z
ψ(x) |fˆ(x)|2 dx
R
18
• The function ψ is a continuous negative
definite function
µ̂t(x) = e−tψ(x),
x∈R
• The asymptotic behavior of Π(x), x > 0,
as x → 0 reflects into the growth of ψ at
infinity, and ultimately in the behavior of
µt(0) = 2
Z ∞
0
e−tψ(x) dx
t small
19
Examples:
(1)
α ∈ (0, 1)
dΠ(x) = cα|x|−1−2αdx, .
ψ(y) = |y|2α
A = (∆)α
Tt is the symmetric α-stable semigroup.
20
(2)
dΠ(x) = |x|−1e−|x|dx.
ψ(y) = 2 log(1 + |y|2)
∀u ∈ C0∞,
Au = 2 log (1 + ∆)u
Tt is the symmetric Γ-semigroup.
21
( Berg-Forst )
µt is absolutely continuous and
has a continuous density
(with respect to Lebesgue measure)
iff
its Fourier transform µ̂t is in L1.
22
if µ̂t ∈ L1,
||e−(t/2)A||2
2,∞ = µt(0) = 2
=2
Z ∞
0
Z ∞
0
e−tψ(x) dx
e−ts dV (s)
where (Volume function)
V (s) = |{t > 0 : ψ(t) ≤ s}|,
(t → µt(x) denotes the density of µt.)
23
III. General results
ψ a be a real (even) negative definite
function on Γ = Rd.
ψ is radial, we define ψ̃ by ψ̃(r) = ψ(θ)
for any r = |θ| > 0, θ ∈ Rd
ψ is said radially non-decreasing
if ψ is radial and if r −→ ψ̃(r) is
non-increasing.
24
Thm:[In a forthcoming paper B-M]
On Rd, let ψ be a Lévy symbol
radially non-decreasing.
Then ∀ρ > 0, we have
(1 − wd ρd) ||f ||2
2 ψ̃
2/d
ρ||f ||2
≤ (Af, f ) (1)
with ||f ||1 ≤ 1.
25
Thm:
(On LCA groups) [In a forthcoming
paper B-M]
Let ψ is real and non- negative definite
function on Γ (dual group of G).
We set V (t) = dθ{θ ∈ Γ : ψ(θ) ≤ t}, t > 0.
Assume V (t) < ∞, ∀t > 0.
Let Λ(s) = supt>0{st − tV (t)}.
⇒ (N T I) :
Λ(||f ||2
2 ) ≤ (Af, f ),
||f ||1 ≤ 1
25
On R, we associate to x → ψ(x), x > 0
1 x
∗
ψ (x) =
ψ(τ ) dτ ,
x 0
Z
x > 0.
We extend ψ ∗(x) to be even.
Thm
(1) For any x ∈ R,
1 ψ(x) ≤ ψ ∗ (x)
3
26
(2) ψ ∗ is again a continuous
even negative definite function having
representation
ψ ∗(x) = 2
Z ∞
0
(1 − cos xξ)Π∗(ξ) dξ
with Π∗(ξ) being continuous non-increasing
function.
(3) Under one of the following two
conditions ψ ∗ is comparable to ψ:
27
a) ψ is a non-decreasing function at infinity
b) ψ has representation
ψ(x) = 2
Z ∞
0
(1 − cos xξ)Π(ξ) dξ
and Π(ξ) is a non-increasing
27
Thm: Assume that ψ ∗ is comparable to ψ
A: Lévy generator associated to ψ.
Let N and M related by
N (x) = sup (xt − tM (1/t))
t>0
The following properties are equivalent :
28
1. There exist c1, c2 ∈ (0, ∞) s.t.
∀t > 0,
||Tt||2,∞ ≤ exp (c1M (c2t)) (2)
2. There exist c3, c4 ∈ (0, ∞) : || f ||1≤ 1,
2
2
c3 || f ||2 N c4 log || f ||2 ≤ (Af, f ). (3)
29
IV. Very oscillating symbol
a) Construction of ψ
ψ = ψ1 + ψ2
For given 0 < α0 < α,
•
ψ1(x) =
X
ak
h
1 − cos(x2−k )
i
k≥1
with ak = (log k)α, k ≥ 1.
30
The Lévy measure Π is given by
dΠ1(x) =
X
ak δ2−k (x)
k≥1
where δa the Dirac mass at point a ∈ R.
•
ψ2(x) =
Z ∞
0
[1 − cos(x.ξ)] dΠ(x)
with
dΠ2(x) = x−1Φ(x−1) dx
31
Φ(τ ) = 0 for τ ≤ e and
0
Φ(τ ) = (log log τ )α for τ > e.
We have
•
0
ψ2(x) ∼ 2(log x)(log log x)α , x → +∞.
•
ψ2∗ (x) ∼ ψ2(x), x → +∞.
32
b) Properties of ψ
There exists a symmetric Lévy generator A
with symbol
ψ
(= ψ1 + ψ2) s.t.
(1) ∃ 0 < c1 < c2 < ∞ s.t. for x 1,
0
c1(log x)(log log x)α ≤ ψ(x) ≤ c2(log x)(log log x)α
33
(2) There exists a sequence (xn) s.t.
xn → +∞
ψ(xn)
0 = c3 > 0
α
n→+∞ (log xn)(log log xn)
lim
(3) There exists a sequence (yn) such that
yn → +∞
ψ(yn)
= c4 > 0
α
n→+∞ (log yn)(log log yn)
lim
34
(4) The function ψ ∗ has the following
behavior
ψ ∗(x)
lim
= c5 > 0
α
x→+∞ (log x)(log log x)
(5) In particular, ψ and ψ ∗ are
not comparable
35
Picture 1 : ψ oscillating
70
60
50
40
30
20
10
0
1000
2000
3000
4000
5000
6000
7000
36
Picture 2
• Heat kernels
(decreasing curves)
• Ratio:
log log µ1 (t)−log log µosc (t)/ log log µ2 (t)−log log µosc (t)
37
10
8
6
4
2
0
−2
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
Numerical simulations : M. Haddou/M.P.
Open problems :
• simulations of volume function and Nash
function for ψosc
•=⇒ Conjecture about the behavior of µosc
and the behavior of Nash function:
close to the upper or lower level (Or
nothing...)?
4