Lp theory for outer measures: Application to time-frequency analzysis Christoph Thiele Santander,...
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Transcript of Lp theory for outer measures: Application to time-frequency analzysis Christoph Thiele Santander,...
Lp theory for outer measures:Application to time-frequency analzysis
Christoph ThieleSantander, September 2014
Recall Tents (or Carleson boxes)
X is the open upper half plane, generating sets are tents T(x,s) :
Define outer measure on X by
€
σ(T(x,s)) := s
Sizes
Sp average of function on a tent
Also Sinfty
€
S p ( f )(T(x,s)) = (1
sf (y, t)p dy
dt
tT (x,s)
∫∫ )1/ p
€
S∞( f )(T(x,s)) = sup(y,t )∈T (x,s) f (y, t)
Outer essential supremum
Space of functions with finite out.ess.supremum
Outer essential supremum on a subset F:€
outess( f ) = sup{S( f )(E) : E ∈Σ}
€
L∞(X,μ,S)
€
outessF ( f ) = sup{S( f 1F )(E) : E ∈Σ}
Outer Lp spaces
Define super level measures
Define Lp norms
Also weak Lp (Lorentz space) €
μ( f > λ ) := inf{μ(F) : outessF c f ≤ λ}
€
fLp (X ,μ ,S )
:= ( pλ p
0
∞
∫ μ( f > λ )dλ /λ )1/ p
€
fLp ,∞ (X ,μ ,S )
:= supλ λ pμ( f > λ )
Embedding theorems
Thm: Define for fixed Schwartz function ϕ
Then we have:
If φ has integral zero:
€
Fφ ( f )(y, t) := f (z)t−1φ(t−1(z − y))dz∫
€
Fφ ( f )Lp (X ,μ ,S ∞ )
≤ Cφ fp
€
Fφ ( f )Lp (X ,μ ,S 2 )
≤ Cφ fp
Paraproduct estimates
For two phi-I with mean 0 and one with mean 1
Hoelder with two qi equal 2 and one equal infty
€
Λ( f1, f2, f3) := ( Fφ i( f i)(y, t))dy
dt
ti=1
3
∏∫∫
€
Λ( f1, f2, f3) ≤ Fφ i( f i)
i=1
3
∏L1 (X ,μ ,S1 )
≤ C Fφ i( f i)
i=1
3
∏Lpi (X ,μ ,Sqi
)
≤ C f ii=1
3
∏Lpi
Lerner’s sparse operator estimates
For two phi-I with mean 1 and some G
Hoelder with two qi equal infty and one equal 1
€
ΛG ( f1, f2) := G(y, t)( Fφ i( f i)(y, t))dy
dt
ti=1
2
∏∫∫
€
ΛG ( f1, f2) ≤ G Fφ i( f i)
i=1
2
∏L1 (X ,μ ,S1 )
≤ C GL∞ ( X , μ ,S1 )
Fφ i( f i)
Lpi ( X , μ ,S∞ )i=1
2
∏ ≤ C f ii=1
2
∏Lpi
Bilinear Hilbert transform
Consider vectorDefine the trilinear form
The bilinear Hilbert transform is any of the pre-dual bilinear operators.
May assume beta nonzero. By scaling may assume it is unit length. By change of variables may assume it is perpendicular to (1,1,1).
€
(β1,β2,β3)∈R3
€
Λ( f1, f2, f3) := p.v. ( f i(x − βit)i=1
3
∏ )dxdt
t∫∫
Degenerate case
If two beta-I are equal, degenerate case.
Reduce to pointwise product and Hilbert transf.We consider only one fixed non-degenerate case
€
Λ( f1, f2, f3) := p.v. f1(x − βt) f2(x − βt) f3(x + 2βt)dxdt
t∫∫
€
Λ( f1, f2, f3) := p.v. ( f1 f2)(x − 3βt)dt
t∫∫ f3(x)dx
Boundedness of BHT
Theorem: Fix nondegenerate direction beta. Let
Then there is an a priori estimate
Extensions exist for p-I less than 2, but that is for some other time. €
Λ( f1, f2, f3) ≤ C f i p ii=1
3
∏€
2 < p1, p2, p3 < ∞
€
1
p1
+1
p2
+1
p3
=1
Symmetries
Like CZ operators, BHT is invariant under simultaneous translation of f-i and has proper scaling under simultaneous dilation of f-i.
It has additional one parameter modulation symm.
This equals original BHT if alpha perpendicular to beta and (1,1,1). BHT shares symmetries with
€
Mη f (x) = e ixη f (x)
€
Λ(Mα 1η f1,Mα 1η f2,Mα 1η f3) := p.v. ( f i(x − βit)ei(x −β i t )α iη
i=1
3
∏ )dxdt
t∫∫
€
˜ Λ ( f1, f2, f3) = f1(x)R
∫ f2(x) f3(x)dx
New embedding map
Additional symmetry makes the upper half plane embedding inappropriate and useless.
Need embedding map reflecting the orbit of entire symmetry group :
Phi assumed to have FT nonnegative and with small support near 0. The testing wavelets become arbitrarily high oscillating wavepacket. €
Fφ (y,η, t) = f (z)e izη∫ t−1φ(t−1(z − y))dz
BHT in terms of embedded fcts
(Remark: The RHS is the bilinear Cauchy projection.)Two ways of proving this algebraic identity:1) Exercise in applying Plancherel, akin to the proof of
the Calderon reproducing formula.2) Checking RHS satisfies the symmetries, and noting
that any form with these symmetries is a linear combination as on LHS. Prove RHS not multiple of Lambda tilde to show a not 0.
€
aΛ( f1, f2, f3) + b ˜ Λ ( f1, f2, f3) = F j (y,α jη + β j t−1, t)dη dydt
j =1
3
∏R
∫R
∫0
∞
∫
Proof of symmetries
By change of variables.
Translation symmetry: y to y+y’Modulation symmetry: eta to eta+eta’Dilation symmetry: t to ts, y to ys, eta to eta/s
€
F j (y,α jη + β j t−1, t)dηdydt
j =1
3
∏R
∫R
∫0
∞
∫
€
F j (y,η, t) = f j (z)e izη∫ t−1φ(t−1(z − y))dz
€
Not multiple of Lambda tilde
For fixed eta,y,t we have a pairing of with a certain nonnegative bump.
This bump is supported on the half space If is supported outside this half space,
then form vanishes. If supported and positive in half space, then form does not vanish.
So form not invariant under beta-modulation.
€
F j (y,α jη + β j t−1, t)dηdydt
j =1
3
∏R
∫R
∫0
∞
∫
€
F(y,η, t) = f (z)e izη∫ t−1φ(t−1(z − y))dz
€
€
ς⋅β > 0
€
ˆ f 1(ς1)ˆ f 2(ς 2) ˆ f 3(ς 3)
€
ˆ f 1(ς1)ˆ f 2(ς 2) ˆ f 3(ς 3)
Define outer measure on upper 3space
€
ξ
€
T(x,ξ,s) = {(y,η, t) : y − x < s − t,|η − ξ |< t−1}
€
η
€
σ(T(x,ξ,s)) = s
€
S(F)(T(x,ξ,s)) =1
sF(y,η, t)
T (x,ξ ,s)
∫∫∫ dydη dt
Outer triangle inequality
where
€
F j (y,α jη + β j t−1, t)dηdydt
j =1
3
∏R
∫R
∫0
∞
∫
€
= G j (y,η, t)dη dydtj =1
3
∏R
∫R
∫0
∞
∫
€
≤ G1G2G3 L1 (X ,μ ,S )
€
G j (y,η, t) = F j (y,α jη + β j t−1, t)
Outer Hoelder inequality
Where we need to identify appropriate sizesS1,S2,S3 which satisfy size-Hoelder€
G1G2G3 L1 (X ,μ ,S )≤ C G1 Lp1 (X ,μ ,S1 )
G2 Lp2 (X ,μ ,S2 )G3 Lp3 (X ,μ ,S3 )
€
S(G1G2G3)(T) ≤ S1(G1)(T)S2(G2)(T)S3(G3)(T)
Idea of sizes
The j-size is a sum of L2 and Linfty average. The L2 is taken outside an exceptional region A-j. The three exceptional regions are disjoint.
€
S j (G)(T(x,ξ,s)) =1
sG(y,η, t)
2
T (x,ξ ,s)\ A j
∫∫∫ dydηdt ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
1/ 2
+ sup(y,η ,s)∈T (x,ξ ,s) G(y,η, t)
Size-Hoelder
To proveWe split the integral over tent on LHS into four
regions: three orange regions and the rest.On each exceptional region we do the
appropriate 2-2-infty Hoelder, since we control two functions in L2. On the rest, either of the Hoelders is OK.
€
S(G1G2G3)(T) ≤ S1(G1)(T)S2(G2)(T)S3(G3)(T)
Modified outer measure space
To formulate sizes precisely, express everything for F rather than G (change variables). Fix j.
Tents:
Size (pick small parameter b): Note: regions AWill be disjoint for different j after change vrbls.€
T(x,ξ,s) = {(y,η, t) : y − x < s − t,| α (η − ξ) + βt−1 |< t−1}
€
σ(T(x,ξ,s)) = s
€
S(F)(T(x,ξ,s)) =1
sF(y,η, t)
2
T (x,ξ ,s)\ A
∫∫∫ dydηdt ⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2
+ sup(y,η ,s)∈T (x,ξ ,s) F(y,η, t)
€
A = {(y,η, t) : y − x < s − t,|η − ξ |< bt−1}
Main embedding theorem
After this preparation, bounds for BHT are reduced to embedding theorem
For 2<p<infty
€
Fφ Lp (X ,μ ,S )≤ C f
p
Few comments on the proof
Story told by pictures…
ξ
x