bird’s-eye view of gaussian harmonic analysis

Bird’s-eye view of Gaussian harmonic analysis

Jonas Teuwen

June 16, 2014

Table of Contents

1 MotivationBrownian motion with driftWhat is diffusion?Ornstein-Uhlenbeck stochastic processFeynman-Kac to Fokker-Planck

2 What is known?The goalLp theoryHardy space?Atomic Gaussian Hardy spacesMaximal and quadratic Gaussian Hardy spaces

3 New and future work

4 References

N-dimensional Brownian motion with drift

We start with the vector-valued Ito SDE in N dimensions:

dYt = µ(Yt , t)

︸︷︷︸drift

dt + σ(Yt , t)

︸︷︷︸diffusion

where Wi are independent Brownian motions and

N-dimensional Brownian motion with drift

We start with the vector-valued Ito SDE in N dimensions:

dYt = µ(Yt , t)︸︷︷︸drift

dt + σ(Yt , t)︸︷︷︸diffusion

where Wi are independent Brownian motions and

Diffusion??

Fick’s first law postulates that the diffusive flux goesfrom regions of high concentration to regions of lowconcentration, with a magnitude that is proportional tothe concentration gradient.

J = −D∇u,

where D is the diffusion tensor and u the concentration.

But. . . this definition is macroscopic. Where is our particle?

Diffusion??

Fick’s first law postulates that the diffusive flux goesfrom regions of high concentration to regions of lowconcentration, with a magnitude that is proportional tothe concentration gradient.

J = −D∇u,

where D is the diffusion tensor and u the concentration.

But. . . this definition is macroscopic. Where is our particle?

Molecular diffusion

Luckily there was a really smart guy

That has shown there is a relationship between the diffusioncoefficient and the molecular level!Einstein (1905) says:

D = µkBT

here µ is the mobility of the particle, kB is Boltzmann’s constantand T is the absolute temperature.

Molecular diffusion

D = µkBT

Molecular diffusion

D = µkBT

The general Ornstein-Uhlenbeck process

We assume:

1 Isotropic diffusion (D = σ is a scalar)

2 µ(x , t) = θ(µv − x) for some θ > 0 and a unit vector v.

So, the Ornstein-Uhlenbeck process is:

dYt = θ(µ− Yt)dt + σdWt

We will pick µ = 0, θ = 1 and σ = 1√2

Feynman-Kac

The Feynman-Kac formula relates the drift-diffusion Ito processeswith a PDE. That is,{

u(x , t) = E[ψ(Yt) | Y0 = x

], t ∈ [0,T ]

u(x , 0) = ψ(x).

When applied to the Ito process this gives the Fokker-Planckequation:

∂tu =1

2∆u − x · ∇u := Lu

with stationary distribution:

γ(x) = Ce−|x |2.

Some properties of OU

∫∇u · ∇v dγ =

∫u(Lv)dγ.

2 Positive and symmetric with respect to γ (Arendt and ter Elst)

3 Generates an analytic semigroup etL (Arendt and ter Elst)

4 The spectrum only depends on the drift part of the operator(Metafune and Pallara)

• For p = 1: closed left half-plane• For 1 < p <∞: negative integers

5 Has an explicit integral kernel representation

∫∇u · ∇v dγ =

∫u(Lv)dγ.

∫∇u · ∇v dγ =

∫u(Lv)dγ.

∫∇u · ∇v dγ =

∫u(Lv)dγ.

The OU semigroup

Recall the Ornstein-Uhlenbeck operator L

2∆− x · ∇.

L has the associated Ornstein-Uhlenbeck semigroup etL which onits turn has an associated Schwartz kernel:

etLu(x) =

Kt(x , ξ)u(ξ)dξ, u ∈ C∞c (Rd)

where the Mehler kernel Kt is given by

Kt(x , ξ) =1

(1− e−2t)d2

(−|e

−tx − ξ|2

1− e−2t

Not nearly as convenient to work with as the heat kernel!

The OU semigroup

2∆− x · ∇.

etLu(x) =

Kt(x , ξ)u(ξ) dξ, u ∈ C∞c (Rd)

Kt(x , ξ) =1

(1− e−2t)d2

(−|e

−tx − ξ|2

1− e−2t

The OU semigroup

2∆− x · ∇.

etLu(x) =

Kt(x , ξ) =1

(1− e−2t)d2

(−|e

−tx − ξ|2

1− e−2t

The OU semigroup

2∆− x · ∇.

etLu(x) =

Kt(x , ξ) =1

(1− e−2t)d2

(−|e

−tx − ξ|2

1− e−2t

Introduction

Goal Build a satisfactory Hardy space theory for theGaussian measure

dγ(x) :=e−|x |

and the Ornstein-Uhlenbeck operator

2∆− x · ∇

Just do it? 1 Mimick Euclidean proofs?2 Use the Euclidean representation of γ?

Introduction

dγ(x) :=e−|x |

2∆− x · ∇

Introduction

dγ(x) :=e−|x |

2∆− x · ∇

The Lp theory is well-known

1 The Riesz transforms are bounded from Lp(γ) to Lp(γ) with1 < p <∞ (Muckenhoupt)

2 The Riesz transforms are of weak-type (1, 1) (Muckenhoupt)

The Lp theory is well-known

1 The Riesz transforms are bounded from Lp(γ) to Lp(γ) with1 < p <∞ (Muckenhoupt)

2 The Riesz transforms are of weak-type (1, 1) (Muckenhoupt)

How is it defined? – Classical HardySpaces

There are several equivalent ways to define Hardy spaces:

1 Atomic decomposition: a atom is a function a such that∫Qa(x)dx = 0 and ‖a‖∞ 6

to define

H1at(Rd) :=

λjaj : aj atoms, λj ∈ C, ‖λ‖`1 <∞

2 Via Riesz transforms

H1(Rd) := {u ∈ L1(Rd) : Rju ∈ L1(Rd), 1 6 j 6 d}

3 Maximal and square functions

4 . . .

to define

H1at(Rd) :=

4 . . .

to define

H1at(Rd) :=

4 . . .

to define

H1at(Rd) :=

4 . . .

to define

H1at(Rd) :=

4 . . .

Atomic Gaussian Hardy spaces

Mauceri and Meda take the atomic route. The replace theLebesgue measure by the Gaussian measure in the atomicdefinition. That is:∫

Qa(x)dγ(x) = 0 and ‖a‖∞ 6

with as before:

H1at(Rd , γ) :=

with as before:

H1at(Rd , γ) :=

with as before:

H1at(Rd , γ) :=

Mauceri and Meda’s space has nice properties, as we keep manyclassical properties:

1 Its dual is BMO(γ).

2 Their BMO(γ) has a John-Nirenberg inequality.

It also has less nice properties:

• Some Riesz transforms are unbounded from L1 to H1 indimensions higher than one.

But they do introduce useful tools such as a “locally”doubling property for the Gaussian measure!

Pierre Portal’s Hardy spaces

Defined through maximal and quadratic functions.

T ∗a u(x) := sup(y ,t)∈Γa

x (γ)|et2Lu(y)|

Sau(x) :=

(∫Γax (γ)

γ(Bt(y))|t∇et2Lu(y)|2 dγ(y)

and norms,

‖u‖h1max

:= ‖T ∗a u‖L1(γ)

‖u‖h1quad

:= ‖Sau‖L1(γ) + ‖u‖L1(γ).

Gaussian cones

Gaussian harmonic analysis is local in the way that we use a cut-offcone for our maximal and quadratic functions.

Γ(A,a)x (γ) := {(x , y) ∈ R2d : |x − y | < At and t 6 am(x)}.

Where m(x) = min{1, |x |−1}.

For Pierre Portal’s space it is yet unknown what:

1 BMO(γ) should mean.

2 If that BMO(γ) has a John-Nirenberg inequality.

But it is has the nice properties that:

• The Riesz transforms are bounded from L1 to H1.

• They interpolate as you might expect (unpublished).

But what is BMO?

Gaussian cones and the doubling property

Mimicking Euclidean proofs will usually not work.

Most theoryrelies on the doubling property of the measure µ:

µ(B2r (x)) 6 Cµ(Br (x))

for some C > 0 uniformly in x and r .

Gaussian cones and the doubling property

Mimicking Euclidean proofs will usually not work. Most theoryrelies on the doubling property of the measure µ:

µ(B2r (x)) 6 Cµ(Br (x))

for some C > 0 uniformly in x and r .

As you might guess. . .

. . . the Gaussian measure is non-doubling

(but. . . maybe local?)

Indeed, there is a kind of local doubling property due to Mauceriand Meda! For this, we define the admissible balls

Ba := {B(x , r) : r 6 m(x)},

where,

m(x) := min

For our admissible balls we then get the following lemma:

LemmaFor all Br (x) ∈ Ba we have that

γ(B2r (x)) 6 Cγ(Br (x)).

As you might guess. . .

. . . the Gaussian measure is non-doubling(but. . . maybe local?)

Indeed, there is a kind of local doubling property due to Mauceriand Meda! For this, we define the admissible balls

Ba := {B(x , r) : r 6 m(x)},

where,

m(x) := min

For our admissible balls we then get the following lemma:

LemmaFor all Br (x) ∈ Ba we have that

γ(B2r (x)) 6 Cγ(Br (x)).

Gaussian cones – Useful consequences

1 On the cone Γ(A,a)x (γ) we have t|x | 6 aA (Maas, van Neerven,

Portal),

2 If |x − y | < At and t 6 am(x) then |x | ∼ |y |.

Additionally we define the annuli (Ck)k>0 through

Ck(x) :=

{2Bt(x) if k = 0,

2k+1Bt(x) \ 2kBt(x) if k > 1.

Portal),

Ck(x) :=

{2Bt(x) if k = 0,

2k+1Bt(x) \ 2kBt(x) if k > 1.

Portal),

Ck(x) :=

{2Bt(x) if k = 0,

2k+1Bt(x) \ 2kBt(x) if k > 1.

Only one theorem. . . – The Euclidean case

TheoremLet u ∈ C∞c (Rd), then

sup(y ,t)∈Rd+1

|x−y |<t

|et2∆u(y)| . supr>0

|Br (x)|

∫Br (x)

|u|dλ︸︷︷︸Mu(x)

where λ is the Lebesgue measure.

The proof is straightforward, as et∆ is a convolution-type operator.So et∆ = ρt ∗ u where

ρt(ξ) :=e−|ξ|

(4πt)d2

is the heat kernel.(There are many theorems about such convolution-type operators)

How to (ad hoc) prove it?

Let Ck := 2k+1B \ 2kB as before then

|et2∆u(y)| 6 1

(4πt2)d2

e−|y−ξ|2/4t2 |u(ξ)| dξ

(4πt2)d2

∞∑k=0

e−c4k∫Ck (Bt(x))

|u(ξ)| dξ

(4πt2)d2

∞∑k=0

e−c4k |B2k+1t(x)|Mu(y)

. Mu(y)1

∞∑k=0

e−c4k td2d(k+1)

. Mu(y).

taking the supremum, and we are done.

|et2∆u(y)| 6 1

(4πt2)d2

e−|y−ξ|2/4t2 |u(ξ)| dξ

(4πt2)d2

∞∑k=0

|u(ξ)| dξ

(4πt2)d2

∞∑k=0

. Mu(y)1

∞∑k=0

e−c4k td2d(k+1)

. Mu(y).

|et2∆u(y)| 6 1

(4πt2)d2

e−|y−ξ|2/4t2 |u(ξ)| dξ

(4πt2)d2

∞∑k=0

|u(ξ)| dξ

(4πt2)d2

∞∑k=0

. Mu(y)1

∞∑k=0

e−c4k td2d(k+1)

. Mu(y).

|et2∆u(y)| 6 1

(4πt2)d2

e−|y−ξ|2/4t2 |u(ξ)| dξ

(4πt2)d2

∞∑k=0

|u(ξ)| dξ

(4πt2)d2

∞∑k=0

. Mu(y)1

∞∑k=0

e−c4k td2d(k+1)

. Mu(y).

|et2∆u(y)| 6 1

(4πt2)d2

e−|y−ξ|2/4t2 |u(ξ)| dξ

(4πt2)d2

∞∑k=0

|u(ξ)| dξ

(4πt2)d2

∞∑k=0

. Mu(y)1

∞∑k=0

e−c4k td2d(k+1)

. Mu(y).

Same ad hoc proof?

What goes wrong when we blundly replace ∆ by L and theLebesgue measure by the Gaussian?On Ck we have a lower bound for |x − ξ|, so,

|e−tx − ξ| > |x − ξ| − (1− e−t)|x |> |x − ξ| − t|x |.

Here the cone condition t|x | . 1 comes into play. Still, this is a bitunsatisfactory. We require a Gaussian measure. So what can bedone?

Unsatisfactory? – Some observations

1 We use the Lebesgue measure again, while we want aGaussian theory. Perspective,. . .

2 In the end we want all admissibility paramaters and aperturesa and A. Proof gets very messy (e.g., Urbina and Pineda),

3 Simple observations shows that the kernel should besymmetric in its arguments against the Gaussian measure.

So, honouring these observations we come to. . .

etLu(x) =

Mt(x , ξ)u(ξ) γ(dξ), u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

etLu(x) =

Mt(x , ξ) =

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

etLu(x) =

Mt(x , ξ) =

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

etLu(x) =

Mt(x , ξ) =

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

Estimating the Mehler kernel

On Ck this is now easier, for t . 1 and t|x | . 1:

Mt2(y , ξ) 6e−c4k

(1− e−2t2)d2

(−2e−t

2 〈y , ξ〉1 + e−t2

6e−c4k

(1− e−2t2)d2

exp(−|〈y , ξ〉|)

6e−c4k

(1− e−2t2)d2

exp(|〈y , ξ − y〉|)e|y |2

6e−c4k

(1− e−2t2)d2

exp(2k+1t|y |)e|y |2

(1− e−2t2)d2

(−2e−t

2 〈y , ξ〉1 + e−t2

6e−c4k

(1− e−2t2)d2

exp(−|〈y , ξ〉|)

6e−c4k

(1− e−2t2)d2

exp(|〈y , ξ − y〉|)e|y |2

6e−c4k

(1− e−2t2)d2

(−2e−t

2 〈y , ξ〉1 + e−t2

6e−c4k

(1− e−2t2)d2

exp(−|〈y , ξ〉|)

6e−c4k

(1− e−2t2)d2

exp(|〈y , ξ − y〉|)e|y |2

6e−c4k

(1− e−2t2)d2

(−2e−t

2 〈y , ξ〉1 + e−t2

6e−c4k

(1− e−2t2)d2

exp(−|〈y , ξ〉|)

6e−c4k

(1− e−2t2)d2

exp(|〈y , ξ − y〉|)e|y |2

6e−c4k

(1− e−2t2)d2

Was that enough? – Putting the thingstogether

|et2Lu(y)| 6∞∑k=0

∫Ck (Bt(x))

Mt2(y , ξ)|u(ξ)|dγ(ξ)

(1− e−2t2)d2

∞∑k=0

e−c4k e2k+1t|y |e|y |2∫Ck (Bt(x))

|u|dγ

6 Mγu(y)e|y |

(1− e−2t2)d2

∞∑k=0

e−c4k e2k+1t|y |γ(B2k+1t(x)).

Estimating Gaussian balls:

γ(B2k+1t(x)) .d 2d(k+1)tde2k+2t|x |e−|x |2.

|et2Lu(y)| 6∞∑k=0

∫Ck (Bt(x))

(1− e−2t2)d2

∞∑k=0

|u| dγ

6 Mγu(y)e|y |

(1− e−2t2)d2

∞∑k=0

|et2Lu(y)| 6∞∑k=0

∫Ck (Bt(x))

(1− e−2t2)d2

∞∑k=0

|u| dγ

6 Mγu(y)e|y |

(1− e−2t2)d2

∞∑k=0

|et2Lu(y)| 6∞∑k=0

∫Ck (Bt(x))

(1− e−2t2)d2

∞∑k=0

|u| dγ

6 Mγu(y)e|y |

(1− e−2t2)d2

∞∑k=0

We do need to use the locality

As before, locally we have |x | ∼ |y | which gives t|x | . 1 andt|y | . 1. Combining we neatly get

|et2Lu(y)| 6 Mγu(y)e|y |

(1− e−2t2)d2

∞∑k=0

e−c4k e2k+1t|y |γ(B2k+1t(x))

6 Mγu(y)tde|y |

2e−|x |

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2d(k+1)

. Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

Which is bounded for bounded t.

(1− e−2t2)d2

∞∑k=0

6 Mγu(y)tde|y |

2e−|x |

(1− e−2t2)d2

∞∑k=0

. Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

(1− e−2t2)d2

∞∑k=0

6 Mγu(y)tde|y |

2e−|x |

(1− e−2t2)d2

∞∑k=0

. Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

(1− e−2t2)d2

∞∑k=0

6 Mγu(y)tde|y |

2e−|x |

(1− e−2t2)d2

∞∑k=0

. Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

Future work

2 Singular integrals

3 Atomic decompositions

Literature

From Forms to SemigroupsWolfgang Arendt and A.F.M. ter Elst

Spectrum of Ornstein-Uhlenbeck operators in Lp spaces withrespect to invariant measuresG. Metafune and D. Pallara

Hermite conjugate expansionsB. Muckenhoupt

Maximal and quadratic Gaussian Hardy SpacesPierre Portal

A note on the Gaussian maximal functionsJonas Teuwen

bird’s-eye view of gaussian harmonic analysis

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