The Gaussian Hardy-Littlewood Maximal Function

The Gaussian Hardy-Littlewood MaximalFunction

YFAW ’14 Lancaster, UK

Jonas Teuwen

Delft Institute for Applied MathematicsDelft University of Technology, The Netherlands

24 April 2014

Introduction

Goal Build a satisfactory harmonic analytic theory for theGaussian measure

dγ(x) :=e−|x |

2

πd2

dx .

and the Ornstein-Uhlenbeck operator

L :=1

2∆− x · ∇

Just do it? Most theory relies on the doubling property of themeasure µ:

µ(B2r (x)) 6 Cµ(Br (x))

uniformly in r and x .

Introduction

Goal Build a satisfactory harmonic analytic theory for theGaussian measure

dγ(x) :=e−|x |

2

πd2

dx .

and the Ornstein-Uhlenbeck operator

L :=1

2∆− x · ∇

Just do it? Most theory relies on the doubling property of themeasure µ:

µ(B2r (x)) 6 Cµ(Br (x))

uniformly in r and x .

As you might guess. . .

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

However, there is a kind of local doubling property! First this, wedefine the admissible balls

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

where,

m(x) := min

{1,

1

|x |

}.

For our admissible balls we then get the following lemma (Mauceri& Meda):

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?)

However, there is a kind of local doubling property! First this, wedefine the admissible balls

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

where,

m(x) := min

{1,

1

|x |

}.

For our admissible balls we then get the following lemma (Mauceri& Meda):

LemmaFor all Br (x) ∈ Ba we have that

γ(B2r (x)) 6 Cγ(Br (x)).

Some modifications – Gaussian conesGaussian harmonic analysis is local in the way that we use a cut-offcone for our non-tangential maximal function

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

−1

1−1

1

1

x

y

t

Figure: This is a cut-off cone!

Gaussian cones – Useful consequences

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

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

Using these two we have

1. If we have this then additionally t|y | . 1

2. e−|x |2 ∼ e−|y |

2.

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

Ck(x) :=

{2Bt(x) if k = 0,

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

Gaussian cones – Useful consequences

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

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

Using these two we have

1. If we have this then additionally t|y | . 1

2. e−|x |2 ∼ e−|y |

2.

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

Ck(x) :=

{2Bt(x) if k = 0,

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

Gaussian cones – Useful consequences

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

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

Using these two we have

1. If we have this then additionally t|y | . 1

2. e−|x |2 ∼ e−|y |

2.

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

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

1

|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|

2/4t

(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

∫Rd

e−|y−ξ|2/4t2 |u(ξ)| dξ

61

(4πt2)d2

∞∑k=0

e−c4k∫Ck (Bt(x))

|u(ξ)| dξ

61

(4πt2)d2

∞∑k=0

e−c4k |B2k+1t(x)|Mu(y)

. Mu(y)1

td

∞∑k=0

e−c4k td2kd .

. Mu(y).

taking the supremum, and we are done.

Intermezzo: The Ornstein-Uhlenbeck semigroup

Remember the Ornstein-Uhlenbeck operator L?

L :=1

2∆− x · ∇.

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

etLu(x) =

∫Rd

Mt(x , ξ)u(ξ)dξ, u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =1

πd2

1

(1− e−2t)d2

exp

(−|e

−tx − ξ|2

1− e−2t

).

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

Intermezzo: The Ornstein-Uhlenbeck semigroup

Remember the Ornstein-Uhlenbeck operator L?

L :=1

2∆− x · ∇.

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

etLu(x) =

∫Rd

Mt(x , ξ)u(ξ)dξ, u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =1

πd2

1

(1− e−2t)d2

exp

(−|e

−tx − ξ|2

1− e−2t

).

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

Same ad hoc proof?

What goes wrong when we bluntly 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?

Same ad hoc proof?

What goes wrong when we bluntly 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) =

∫Rd

Mt(x , ξ)u(ξ) γ(dξ), u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =

exp

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

d2

exp

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

d2

.

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) =

∫Rd

Mt(x , ξ)u(ξ) γ(dξ), u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =

exp

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

d2

exp

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

d2

.

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) =

∫Rd

Mt(x , ξ)u(ξ) γ(dξ), u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =

exp

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

d2

exp

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

d2

.

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) =

∫Rd

Mt(x , ξ)u(ξ) γ(dξ), u ∈ C∞c (Rd)

where the Mehler kernel Mt is given by

Mt(x , ξ) =

exp

(−e−2t |x − ξ|2

1− e−2t

)(1− e−t)

d2

exp

(−2e−t

〈x , ξ〉1 + e−t

)(1 + e−t)

d2

.

Estimating the Mehler kernel

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

Mt2(x , ξ) 6e−c4k

(1− e−2t2)d2

exp

(−2e−t

2 〈x , ξ〉1 + e−t2

)

6e−c4k

(1− e−2t2)d2

exp(−|〈x , ξ〉|)

6e−c4k

(1− e−2t2)d2

exp(|〈x , ξ − x〉|)e|x |2

6e−c4k

(1− e−2t2)d2

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

Was that enough? – Putting the things together

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

|et2Lu(y)| 6 1

(1− e−2t2)d2

∞∑k=0

e−c4k e2k+1t|y |e|y |2∫Ck (Bt(x))

|u| dγ

6 Mγu(y)e|y |

2

(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.

Was that enough? – Putting the things together

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

|et2Lu(y)| 6 1

(1− e−2t2)d2

∞∑k=0

e−c4k e2k+1t|y |e|y |2∫Ck (Bt(x))

|u| dγ

6 Mγu(y)e|y |

2

(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.

We do need to use the locality

As before, locally:

1. |x | ∼ |y |2. t|x | . 1 and t|y | . 1.

Combining we neatly get

|et2Lu(y)| . Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

Which is bounded for bounded t.(Thanks for your attention!)

We do need to use the locality

As before, locally:

1. |x | ∼ |y |2. t|x | . 1 and t|y | . 1.

Combining we neatly get

|et2Lu(y)| . Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

Which is bounded for bounded t.

(Thanks for your attention!)

We do need to use the locality

As before, locally:

1. |x | ∼ |y |2. t|x | . 1 and t|y | . 1.

Combining we neatly get

|et2Lu(y)| . Mγu(y)td

(1− e−2t2)d2

∞∑k=0

e−c4k eC2k 2dk

Which is bounded for bounded t.(Thanks for your attention!)