a class of calderón-zygmund operators arising …cz martingales a class of calder on-zygmund...
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CZ Martingales
A Class of Calderon-Zygmund Operators Arisingfrom the Projections of Martingale Transforms
Michael Perlmutter
Department of MathematicsPurdue University
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CZ Martingales
Calderon-Zygmund Operators
Let T be an operator on S(Rn), which is bounded on L2(Rn) of the form
Tf (x) = p.v .
∫Rn
K (x , x)f (x)dx .
T is called Calderon-Zygmund if K is C 1 on {x 6= x} and
|K (x , x)| ≤ κ
|x − x |n
|∇xK (x , x)| ≤ κ
|x − x |n+1
|∇xK (x , x)| ≤ κ
|x − x |n+1,
for some universal constant κ whenever x 6= x .
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Examples
Hilbert Transform
Hf (x) =1
π
∫R
1
x − xf (x)dx .
Riesz Transform
Ri f (x) = Cn
∫Rn
xi − xi|x − x |n+1
f (x)dx ,
1 ≤ i ≤ n, Cn =Γ( n+1
2 )π(n+1)/2 .
Beurling-Ahlfors Transform
Bf (z) = − 1
π
∫C
1
(z − w)2f (w)dw .
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Fourier Multipliers and Pseudo-Differential Operators
Fourier Multipliers
Rj f (ξ) =iξj|ξ|
f (ξ)
Bf (ξ) =ξ2
1 − ξ22 − 2iξ1ξ2
|ξ|2f (ξ)
B = R22 − R2
1 + 2iR1R2.
Pseudo-Differential Operators
Rj f = ∂xj (−4)−1/2f
Bf =∂z∂z
f .
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Symmetric α-stable Processes
Characteristic Function
E(e iξ·Xt ) = e−t|ξ|α, 0 < α ≤ 2.
Examples:
α = 2: Xt is Brownian Motion with density ht(x) = 1(4πt)n/2 e
−|x |2/4t .
α = 1: Xt is the Cauchy process with density pt(x) = Cnt
(|x |2+t2)(n+1)/2 .
Properties
Semigroup: ψs ∗ ψt = ψs+t .Scaling: ψt(x) = 1
tn/αψ1( x
t1/α ).
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Setup
I (Xt)t>0, rotationally-invariant α-stable process on Rn, 0 < α ≤ 2.
I For y > 0, ϕy (x) = 1ynϕ( xy ), where ϕ denotes the density of X1.
I Let A(x , y) = (ai ,j(x , y)) be an (n + 1)× (n + 1) matrix-valuedfunction
‖A‖ = ‖ sup|v |≤1
(|A(x , y)v |)‖L∞(Rn×[0,∞)) <∞.
I ai ,j(x , y) = ai ,j(y) whenever i or j = n + 1.
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Results
Theorem:
Consider the kernel
KA(x , x) =
∫ ∞0
∫Rn
2yA(x , y)∇ϕy (x − x)∇ϕy (x − x)dxdy ,
where ∇ = (∂x1 , . . . , ∂xn , ∂y ). Then the operator
TAf (x) =
∫Rn
K (x , x)f (x)dx
is a CZ operator.
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Results cont’d
Remark:
If ai ,j(y) = 0 whenever i or j = n + 1, we may also write our kernel interms of the density of Xt .The scaling relation ψt(x) = 1
tn/αψ1( x
t1/α ) implies ϕt1/α = ψt .Therefore,
KA(x , x) =
∫ ∞0
∫Rn
2
αt
2α−1A(x , t1/α)∇ψt(x − x)∇ψt(x − x)dxdt.
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Boundedness of CZ Operators
Strong type
‖Tf ‖p ≤ Cp,n,κ‖f ‖p, 1 < p <∞ (1)
Weak type
|{x : |Tf (x)| > λ}| ≤ C1,n,κ
λ‖f ‖1. (2)
Questions
For which operators can the constant in (1) and (2) be taken independentof n?What are the best possible constants?
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Probabilistic Representations of the Riesz Transform
Basic Idea
Lp → Mp → Mp → Lp
Background Radiation
“Time-reversed Brownian motion” in Rn+1+ . B−∞ has Lebesgue
distribution on Rn × {∞}, B0 has Lebesgue distribution on Rn × {0}.
Embedding into Mp
For f ∈ Lp, let uf be it’s Poisson extension to Rn+1+ . Then
(Xt)t≤0 = (uf (Bt))t≤0 is a martingale and
uf (Bt) =
∫ t
−∞∇uf (Bs) · dBs .
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Probabilistic Representations of the Riesz Transform cont’d
Martingale Transforms
Let A(x , y) an (n + 1)× (n + 1) matrix-valued function, ‖A‖ <∞.
(A ∗ f )t =
∫ t
−∞A(Bs)∇uf (Bs) · dBs .
Theorem: (Banuelos, Wang, Burkholder)
‖(A ∗ f )‖p ≤ (p∗ − 1)‖A‖‖X‖p 1 < p <∞,
p∗ = max{p, pp−1}.
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Probabilistic Representations of the Riesz Transform cont’d
Projection of the Martingale Transform
TAf (x) = E((A ∗ f )0|B0 = (x , 0))
Theorem: (Banuelos and Wang) ‖TAf ‖p ≤ (p∗ − 1)‖A‖‖f ‖p for1 < p <∞.
Orthogonal Martingale Transforms
Theorem: (Banuelos and Wang) Suppose further that A(x , y)v · v = 0 forall (x , y) ∈ Rn+1
+ , v ∈ Rn+1, then
‖TAf ‖p ≤ cot
(2π
p∗
)‖A‖‖f ‖p for 1 < p <∞.
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Riesz Transforms
Choose Aj = (ajl ,m),
ajl ,m =
1 l = n + 1, m = j−1 l = j , m = n + 10 otherwise
.
Then TAj f = Rj f and consequently ‖Rj f ‖ ≤ cot(
2πp∗
)‖f ‖p.
Beurling-Ahlfors Transform
Let
B =
2 2i 02i −2 00 0 0
.
Then TB f = Bf and consequently ‖Bf ‖ ≤ 4(p∗ − 1)‖f ‖p.
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∫Rn
TAf (x)g(x)dx =
∫Rn
E
(∫ 0
−∞A(Bs)∇uf (Bs) · dBs |X0 = x
)g(x)dx
= E
(∫ 0
−∞A(Bs)∇uf (Bs) · dBsug (B0)
)= E
(∫ 0
−∞A(Bs)∇uf (Bs) · dBs
∫ 0
−∞∇ug (Bs) · dBs
)= E
(∫ 0
−∞A(Bs)∇uf (Bs) · ∇ug (Bs)ds
)=
∫ ∞0
∫Rn
2yA(x , y)∇uf (x , y) · ∇ug (x , y)dxdy .
Using the fact that ∇uf (x , y) = ((∇py ) ∗ f )(x) and applying Fubini’stheorem, we see that we have
KA(x , x) =
∫ ∞0
∫Rn
2yA(x , y)∇py (x − x)∇py (x − x)dxdy .
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Space-time Brownian Motion and Heat Martingales
Space-time Brownian Motion
Fix T > 0 and let (Xt)0≤t≤T be Brownian motion with initial distributiongiven by Lebesgue measure on Rn. Let
(Zt)0≤t≤T = (Xt ,T − t)0≤t≤T .
Heat martingales
For f ∈ Lp, let uf (x , t) = (ht ∗ f )(x). By Ito’s formula, (uf (Zt))0≤t≤T is amartingale and
uf (Zt) =
∫ t
0∇xuf (Zs)dXs .
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Projections of Heat Martingales
Projection Operator
If A is an n × n matrix valued function, ‖A‖ <∞
STA f (x) = E
(∫ T
0A(Zt)∇xuf (Zs)dXs |XT = x)
),
SAf (x) = limT→∞
STA f (x)
Theorem: (Banuelos and Mendez) ‖SAf (x)‖p ≤ (p∗ − 1)‖A‖‖f ‖p for1 < p <∞.
Analytic Representation
SAf (x) =∫Rn K (x , x)dx where
K (x , x) =
∫ ∞0
∫Rn
A(x , t)∇xht(x − x) · ∇xht(x − x)dxdt.
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Projections of Heat Martingales cont’d
Beurling-Ahlfors Transform
Let
B =
(1 ii −1
).
Then SB f = Bf , consequently, ‖Bf ‖p ≤ 2(p∗ − 1)‖f ‖p for 1 < p <∞.
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Proof of Theorem 1
K (i ,j)(x , x) =
∫ ∞0
∫Rn
2yai ,j(x , y)∂xiϕy (x − x)∂xjϕy (x − x)dxdy
Case 1: i or j = n + 1:
K (i ,j)(x , x) =
∫ ∞0
∫Rn
2yai ,j(x , y)∂xiϕy (x − x)∂xjϕy (x − x)dxdy
=
∫ ∞0
2yai ,j(y)∂xi∂xjϕ21/αy (x − x)dy
≤ ‖ai ,j‖∞K (x − x)
K (x) =
∫ ∞0
2y |∂xi∂xjϕ21/αy (x)|dy
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The Proof of Theorem 1, cont’d
|K (x)| =
∫ ∞0
2y |∂xi∂xjϕ21/αy (x)|dy
=
∫ ∞0
2y |∂xi∂xjϕ21/α|x | y|x|(|x |x ′)|dy
=1
|x |n
∫ ∞0
2y |∂xi∂xjϕ21/αt(x′)|dt.
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Proof of Theorem 1, cont’d
Case 2: 1 ≤ i , j ,≤ n
|K (i ,j)(x , x)| ≤ ‖ai ,j‖∞∫ ∞
0
∫Rn
2y |∂xiϕy (w)||∂xjϕy (w − (x − x))|dwdy .
K (x) =
∫Rn
∫ ∞0
2y |∂xiϕy (w)||∂xjϕy (w − x)|dwdy
=1
|x |n
∫Rn
∫ ∞0
2y |∂xiϕt(z)||∂xjϕt(z − x ′)|dzdt.
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Estimates on the Derivatives
Lemma
There exists a constant Cn,α, depending only on n and α, such that for allx ∈ Rn, 1 ≤ i , j ≤ n,
|ϕ(x)| ≤ Cn,α
(1 + |x |2)(n+α)/2
|∂xiϕ(x)| ≤ Cn,α|x |(1 + |x |2)(n+2+α)/2
≤ Cn,α
(1 + |x |2)(n+1+α)/2
and
|∂xi∂xjϕ(x)| ≤ Cn,α
(1 + |x |2)(n+2+α)/2.
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Subordinated Brownian Motion
Stable Subordinator
Xt = BTt ,
Bt is a standard Brownian motion and Tt is the α/2 stable subordinatorwith density ηα/2(t, ·) .
Estimates
ψt(x) =
∫ ∞0
1
(4πs)n/2e−|x |
2/4sηα/2(t, s)ds,
∂xiϕ(x) =
∫ ∞0
1
(4πs)n/2
xise−|x |
2/4sηα/2 (1, s) ds
ηα/2(t, s) ≤ Cαts−1−α/2,
|∂xiϕ(x)| ≤ Cα|x |n+1+α
∫ ∞0
u(n+α)/2e−udu.
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Estimate Near 0
Fourier Inversion
ϕ(x) =
∫Rn
e−ix ·ξe−|ξ|α.
|∂xiϕ(x)| =
∣∣∣∣∫Rn
ξie−ix ·ξe−|ξ|
αdξ
∣∣∣∣=
∣∣∣∣∫Rn
ξi (e−ix ·ξ − 1)e−|ξ|
αdξ
∣∣∣∣≤∫Rn
|ξ| |e−ix ·ξ − 1|e−|ξ|αdξ
≤ 2
∫Rn
|ξ|2|x |e−|ξ|αdξ ≤ Cn,α|x |,
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THANK YOU!
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