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# Bird’s-eye view of Gaussian harmonic analysis

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A presentation on the background of Gaussian harmonic analysis, the several ways to define a Gaussian Hardy space and the Gaussian maximal functions

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### Bird’s-eye view of Gaussian harmonic analysis

1. 1. Bird’s-eye view of Gaussian harmonic analysis Jonas Teuwen June 16, 2014
2. 2. Table of Contents 1 Motivation Brownian motion with drift What is diﬀusion? Ornstein-Uhlenbeck stochastic process Feynman-Kac to Fokker-Planck 2 What is known? The goal Lp theory Hardy space? Atomic Gaussian Hardy spaces Maximal and quadratic Gaussian Hardy spaces 3 New and future work 4 References
3. 3. N-dimensional Brownian motion with drift We start with the vector-valued Itˆo SDE in N dimensions: dYt = µ(Yt, t) dt + σ(Yt, t) dWt
4. 4. N-dimensional Brownian motion with drift We start with the vector-valued Itˆo SDE in N dimensions: dYt = µ(Yt, t) drift dt + σ(Yt, t) diﬀusion dWt where Wi are independent Brownian motions and Y =    Y1 ... YN   
5. 5. Diﬀusion?? Fick’s ﬁrst law postulates that the diﬀusive ﬂux goes from regions of high concentration to regions of low concentration, with a magnitude that is proportional to the concentration gradient. J = −D u, where D is the diﬀusion tensor and u the concentration.
6. 6. Diﬀusion?? Fick’s ﬁrst law postulates that the diﬀusive ﬂux goes from regions of high concentration to regions of low concentration, with a magnitude that is proportional to the concentration gradient. J = −D u, where D is the diﬀusion tensor and u the concentration. But. . . this deﬁnition is macroscopic. Where is our particle?
7. 7. Molecular diﬀusion Luckily there was a really smart guy
8. 8. Molecular diﬀusion Luckily there was a really smart guy
9. 9. Molecular diﬀusion Luckily there was a really smart guy That has shown there is a relationship between the diﬀusion coeﬃcient and the molecular level! Einstein (1905) says: D = µkBT here µ is the mobility of the particle, kB is Boltzmann’s constant and T is the absolute temperature.
10. 10. 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 .
11. 11. Feynman-Kac The Feynman-Kac formula relates the drift-diﬀusion Itˆo processes with a PDE. That is, u(x, t) = E ψ(Yt) | Y0 = x , t ∈ [0, T] u(x, 0) = ψ(x). When applied to the Itˆo process this gives the Fokker-Planck equation: ∂tu = 1 2 ∆u − x · u := Lu with stationary distribution: γ(x) = Ce−|x|2 .
12. 12. Some properties of OU 1 u · v dγ = u(Lv) dγ.
13. 13. Some properties of OU 1 u · v dγ = u(Lv) dγ. 2 Positive and symmetric with respect to γ (Arendt and ter Elst)
14. 14. Some properties of OU 1 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)
15. 15. Some properties of OU 1 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
16. 16. The OU semigroup Recall the Ornstein-Uhlenbeck operator L L := 1 2 ∆ − x · .
17. 17. The OU semigroup Recall the Ornstein-Uhlenbeck operator L L := 1 2 ∆ − x · . L has the associated Ornstein-Uhlenbeck semigroup etL which on its turn has an associated Schwartz kernel: etL u(x) = Rd Kt(x, ξ)u(ξ) dξ, u ∈ C∞ c (Rd )
18. 18. The OU semigroup Recall the Ornstein-Uhlenbeck operator L L := 1 2 ∆ − x · . L has the associated Ornstein-Uhlenbeck semigroup etL which on its turn has an associated Schwartz kernel: etL u(x) = Rd Kt(x, ξ)u(ξ) dξ, u ∈ C∞ c (Rd ) where the Mehler kernel Kt is given by Kt(x, ξ) = 1 π d 2 1 (1 − e−2t) d 2 exp − |e−tx − ξ|2 1 − e−2t .
19. 19. The OU semigroup Recall the Ornstein-Uhlenbeck operator L L := 1 2 ∆ − x · . L has the associated Ornstein-Uhlenbeck semigroup etL which on its turn has an associated Schwartz kernel: etL u(x) = Rd Kt(x, ξ)u(ξ) dξ, u ∈ C∞ c (Rd ) where the Mehler kernel Kt is given by Kt(x, ξ) = 1 π d 2 1 (1 − e−2t) d 2 exp − |e−tx − ξ|2 1 − e−2t . Not nearly as convenient to work with as the heat kernel!
20. 20. Introduction Goal Build a satisfactory Hardy space theory for the Gaussian measure dγ(x) := e−|x|2 π d 2 dx.
21. 21. Introduction Goal Build a satisfactory Hardy space theory for the Gaussian measure dγ(x) := e−|x|2 π d 2 dx. and the Ornstein-Uhlenbeck operator L := 1 2 ∆ − x ·
22. 22. Introduction Goal Build a satisfactory Hardy space theory for the Gaussian measure dγ(x) := e−|x|2 π d 2 dx. and the Ornstein-Uhlenbeck operator L := 1 2 ∆ − x · Just do it? 1 Mimick Euclidean proofs? 2 Use the Euclidean representation of γ?
23. 23. The Lp theory is well-known 1 The Riesz transforms are bounded from Lp(γ) to Lp(γ) with 1 < p < ∞ (Muckenhoupt)
24. 24. The Lp theory is well-known 1 The Riesz transforms are bounded from Lp(γ) to Lp(γ) with 1 < p < ∞ (Muckenhoupt) 2 The Riesz transforms are of weak-type (1, 1) (Muckenhoupt)
25. 25. How is it deﬁned? – Classical Hardy Spaces There are several equivalent ways to deﬁne Hardy spaces: 1 Atomic decomposition: a atom is a function a such that Q a(x)dx = 0 and a ∞ 1 |Q|
26. 26. How is it deﬁned? – Classical Hardy Spaces There are several equivalent ways to deﬁne Hardy spaces: 1 Atomic decomposition: a atom is a function a such that Q a(x)dx = 0 and a ∞ 1 |Q| to deﬁne H1 at(Rd ) := j λj aj : aj atoms, λj ∈ C, λ 1 < ∞ .
27. 27. How is it deﬁned? – Classical Hardy Spaces There are several equivalent ways to deﬁne Hardy spaces: 1 Atomic decomposition: a atom is a function a such that Q a(x)dx = 0 and a ∞ 1 |Q| to deﬁne H1 at(Rd ) := j λj aj : aj atoms, λj ∈ C, λ 1 < ∞ . 2 Via Riesz transforms
28. 28. How is it deﬁned? – Classical Hardy Spaces There are several equivalent ways to deﬁne Hardy spaces: 1 Atomic decomposition: a atom is a function a such that Q a(x)dx = 0 and a ∞ 1 |Q| to deﬁne H1 at(Rd ) := j λj aj : aj atoms, λj ∈ C, λ 1 < ∞ . 2 Via Riesz transforms H1 (Rd ) := {u ∈ L1 (Rd ) : Rj u ∈ L1 (Rd ), 1 j d} 3 Maximal and square functions
29. 29. How is it deﬁned? – Classical Hardy Spaces There are several equivalent ways to deﬁne Hardy spaces: 1 Atomic decomposition: a atom is a function a such that Q a(x)dx = 0 and a ∞ 1 |Q| to deﬁne H1 at(Rd ) := j λj aj : aj atoms, λj ∈ C, λ 1 < ∞ . 2 Via Riesz transforms H1 (Rd ) := {u ∈ L1 (Rd ) : Rj u ∈ L1 (Rd ), 1 j d} 3 Maximal and square functions 4 . . .
30. 30. Atomic Gaussian Hardy spaces Mauceri and Meda take the atomic route. The replace the Lebesgue measure by the Gaussian measure in the atomic deﬁnition. That is: Q a(x)dγ(x) = 0 and a ∞ 1 γ(Q)
31. 31. Atomic Gaussian Hardy spaces Mauceri and Meda take the atomic route. The replace the Lebesgue measure by the Gaussian measure in the atomic deﬁnition. That is: Q a(x)dγ(x) = 0 and a ∞ 1 γ(Q) with as before: H1 at(Rd , γ) := j λj aj : aj atoms, λj ∈ C, λ 1 < ∞ .
32. 32. Atomic Gaussian Hardy spaces Mauceri and Meda take the atomic route. The replace the Lebesgue measure by the Gaussian measure in the atomic deﬁnition. That is: Q a(x)dγ(x) = 0 and a ∞ 1 γ(Q) with as before: H1 at(Rd , γ) := j λj aj : aj atoms, λj ∈ C, λ 1 < ∞ .
33. 33. Atomic Gaussian Hardy spaces Mauceri and Meda’s space has nice properties, as we keep many classical properties: 1 Its dual is BMO(γ).
34. 34. Atomic Gaussian Hardy spaces Mauceri and Meda’s space has nice properties, as we keep many classical properties: 1 Its dual is BMO(γ). 2 Their BMO(γ) has a John-Nirenberg inequality.
35. 35. Atomic Gaussian Hardy spaces Mauceri and Meda’s space has nice properties, as we keep many classical 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 in dimensions higher than one.
36. 36. Atomic Gaussian Hardy spaces Mauceri and Meda’s space has nice properties, as we keep many classical 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 in dimensions higher than one. But they do introduce useful tools such as a “locally” doubling property for the Gaussian measure!
37. 37. Pierre Portal’s Hardy spaces Deﬁned through maximal and quadratic functions. T∗ a u(x) := sup (y,t)∈Γa x (γ) |et2L u(y)| Sau(x) := Γa x (γ) 1 γ(Bt(y)) |t et2L u(y)|2 dγ(y) dt t 1 2 and norms, u h1 max := T∗ a u L1(γ) u h1 quad := Sau L1(γ) + u L1(γ).
38. 38. Gaussian cones Gaussian harmonic analysis is local in the way that we use a cut-oﬀ cone for our maximal and quadratic functions. Γ (A,a) x (γ) := {(x, y) ∈ R2d : |x − y| < At and t am(x)}. −1 1 −1 1 1 x y t Where m(x) = min{1, |x|−1}.
39. 39. Atomic Gaussian Hardy spaces For Pierre Portal’s space it is yet unknown what: 1 BMO(γ) should mean.
40. 40. Atomic Gaussian Hardy spaces 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.
41. 41. Atomic Gaussian Hardy spaces 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?
42. 42. Gaussian cones and the doubling property Mimicking Euclidean proofs will usually not work.
43. 43. Gaussian cones and the doubling property Mimicking Euclidean proofs will usually not work. Most theory relies on the doubling property of the measure µ: µ(B2r (x)) Cµ(Br (x)) for some C > 0 uniformly in x and r.
44. 44. As you might guess. . . . . . the Gaussian measure is non-doubling
45. 45. As you might guess. . . . . . the Gaussian measure is non-doubling (but. . . maybe local?) Indeed, there is a kind of local doubling property due to Mauceri and Meda! For this, we deﬁne the admissible balls Ba := {B(x, r) : r m(x)}, where, m(x) := min 1, 1 |x| . For our admissible balls we then get the following lemma: Lemma For all Br (x) ∈ Ba we have that γ(B2r (x)) Cγ(Br (x)).
46. 46. Gaussian cones – Useful consequences 1 On the cone Γ (A,a) x (γ) we have t|x| aA (Maas, van Neerven, Portal),
47. 47. Gaussian cones – Useful consequences 1 On the cone Γ (A,a) x (γ) we have t|x| aA (Maas, van Neerven, Portal), 2 If |x − y| < At and t am(x) then |x| ∼ |y|.
48. 48. Gaussian cones – Useful consequences 1 On the cone Γ (A,a) x (γ) we have t|x| aA (Maas, van Neerven, Portal), 2 If |x − y| < At and t 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.
49. 49. Only one theorem. . . – The Euclidean case Theorem Let u ∈ C∞ c (Rd ), then sup (y,t)∈Rd+1 + |x−y|<t |et2∆ u(y)| sup r>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) d 2 . is the heat kernel. (There are many theorems about such convolution-type operators)
50. 50. How to (ad hoc) prove it? Let Ck := 2k+1B 2kB as before then |et2∆ u(y)| 1 (4πt2) d 2 Rd e−|y−ξ|2/4t2 |u(ξ)| dξ
51. 51. How to (ad hoc) prove it? Let Ck := 2k+1B 2kB as before then |et2∆ u(y)| 1 (4πt2) d 2 Rd e−|y−ξ|2/4t2 |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k Ck (Bt (x)) |u(ξ)| dξ
52. 52. How to (ad hoc) prove it? Let Ck := 2k+1B 2kB as before then |et2∆ u(y)| 1 (4πt2) d 2 Rd e−|y−ξ|2/4t2 |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k Ck (Bt (x)) |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k |B2k+1t(x)|Mu(y)
53. 53. How to (ad hoc) prove it? Let Ck := 2k+1B 2kB as before then |et2∆ u(y)| 1 (4πt2) d 2 Rd e−|y−ξ|2/4t2 |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k Ck (Bt (x)) |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k |B2k+1t(x)|Mu(y) Mu(y) 1 td ∞ k=0 e−c4k td 2d(k+1)
54. 54. How to (ad hoc) prove it? Let Ck := 2k+1B 2kB as before then |et2∆ u(y)| 1 (4πt2) d 2 Rd e−|y−ξ|2/4t2 |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k Ck (Bt (x)) |u(ξ)| dξ 1 (4πt2) d 2 ∞ k=0 e−c4k |B2k+1t(x)|Mu(y) Mu(y) 1 td ∞ k=0 e−c4k td 2d(k+1) Mu(y). taking the supremum, and we are done.
55. 55. Same ad hoc proof? What goes wrong when we blundly replace ∆ by L and the Lebesgue measure by the Gaussian? On Ck we have a lower bound for |x − ξ|, so, |e−t x − 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 bit unsatisfactory. We require a Gaussian measure. So what can be done?
56. 56. Unsatisfactory? – Some observations 1 We use the Lebesgue measure again, while we want a Gaussian theory. Perspective,. . .
57. 57. Unsatisfactory? – Some observations 1 We use the Lebesgue measure again, while we want a Gaussian theory. Perspective,. . . 2 In the end we want all admissibility paramaters and apertures a and A. Proof gets very messy (e.g., Urbina and Pineda),
58. 58. Unsatisfactory? – Some observations 1 We use the Lebesgue measure again, while we want a Gaussian theory. Perspective,. . . 2 In the end we want all admissibility paramaters and apertures a and A. Proof gets very messy (e.g., Urbina and Pineda), 3 Simple observations shows that the kernel should be symmetric in its arguments against the Gaussian measure.
59. 59. Unsatisfactory? – Some observations 1 We use the Lebesgue measure again, while we want a Gaussian theory. Perspective,. . . 2 In the end we want all admissibility paramaters and apertures a and A. Proof gets very messy (e.g., Urbina and Pineda), 3 Simple observations shows that the kernel should be symmetric in its arguments against the Gaussian measure. So, honouring these observations we come to. . . etL u(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) d 2 exp −2e−t x, ξ 1 + e−t (1 + e−t) d 2 .
60. 60. Estimating the Mehler kernel On Ck this is now easier, for t 1 and t|x| 1: Mt2 (y, ξ) e−c4k (1 − e−2t2 ) d 2 exp −2e−t2 y, ξ 1 + e−t2
61. 61. Estimating the Mehler kernel On Ck this is now easier, for t 1 and t|x| 1: Mt2 (y, ξ) e−c4k (1 − e−2t2 ) d 2 exp −2e−t2 y, ξ 1 + e−t2 e−c4k (1 − e−2t2 ) d 2 exp(−| y, ξ |)
62. 62. Estimating the Mehler kernel On Ck this is now easier, for t 1 and t|x| 1: Mt2 (y, ξ) e−c4k (1 − e−2t2 ) d 2 exp −2e−t2 y, ξ 1 + e−t2 e−c4k (1 − e−2t2 ) d 2 exp(−| y, ξ |) e−c4k (1 − e−2t2 ) d 2 exp(| y, ξ − y |)e|y|2
63. 63. Estimating the Mehler kernel On Ck this is now easier, for t 1 and t|x| 1: Mt2 (y, ξ) e−c4k (1 − e−2t2 ) d 2 exp −2e−t2 y, ξ 1 + e−t2 e−c4k (1 − e−2t2 ) d 2 exp(−| y, ξ |) e−c4k (1 − e−2t2 ) d 2 exp(| y, ξ − y |)e|y|2 e−c4k (1 − e−2t2 ) d 2 exp(2k+1 t|y|)e|y|2
64. 64. Was that enough? – Putting the things together Let Ck := 2k+1B 2kB as before then |et2L u(y)| ∞ k=0 Ck (Bt (x)) Mt2 (y, ξ)|u(ξ)| dγ(ξ)
65. 65. Was that enough? – Putting the things together Let Ck := 2k+1B 2kB as before then |et2L u(y)| ∞ k=0 Ck (Bt (x)) Mt2 (y, ξ)|u(ξ)| dγ(ξ) 1 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| e|y|2 Ck (Bt (x)) |u| dγ
66. 66. Was that enough? – Putting the things together Let Ck := 2k+1B 2kB as before then |et2L u(y)| ∞ k=0 Ck (Bt (x)) Mt2 (y, ξ)|u(ξ)| dγ(ξ) 1 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| e|y|2 Ck (Bt (x)) |u| dγ Mγu(y) e|y|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| γ(B2k+1t(x)).
67. 67. Was that enough? – Putting the things together Let Ck := 2k+1B 2kB as before then |et2L u(y)| ∞ k=0 Ck (Bt (x)) Mt2 (y, ξ)|u(ξ)| dγ(ξ) 1 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| e|y|2 Ck (Bt (x)) |u| dγ Mγu(y) e|y|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| γ(B2k+1t(x)). Estimating Gaussian balls: γ(B2k+1t(x)) d 2d(k+1) td e2k+2t|x| e−|x|2 .
68. 68. We do need to use the locality As before, locally we have |x| ∼ |y| which gives t|x| 1 and t|y| 1. Combining we neatly get |et2L u(y)| Mγu(y) e|y|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| γ(B2k+1t(x))
69. 69. We do need to use the locality As before, locally we have |x| ∼ |y| which gives t|x| 1 and t|y| 1. Combining we neatly get |et2L u(y)| Mγu(y) e|y|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| γ(B2k+1t(x)) Mγu(y) td e|y|2 e−|x|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k eC2k 2d(k+1)
70. 70. We do need to use the locality As before, locally we have |x| ∼ |y| which gives t|x| 1 and t|y| 1. Combining we neatly get |et2L u(y)| Mγu(y) e|y|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| γ(B2k+1t(x)) Mγu(y) td e|y|2 e−|x|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k eC2k 2d(k+1) Mγu(y) td (1 − e−2t2 ) d 2 ∞ k=0 e−c4k eC2k 2dk
71. 71. We do need to use the locality As before, locally we have |x| ∼ |y| which gives t|x| 1 and t|y| 1. Combining we neatly get |et2L u(y)| Mγu(y) e|y|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k e2k+1t|y| γ(B2k+1t(x)) Mγu(y) td e|y|2 e−|x|2 (1 − e−2t2 ) d 2 ∞ k=0 e−c4k eC2k 2d(k+1) Mγu(y) td (1 − e−2t2 ) d 2 ∞ k=0 e−c4k eC2k 2dk Which is bounded for bounded t.
72. 72. Future work 1 BMO 2 Singular integrals 3 Atomic decompositions
73. 73. Literature From Forms to Semigroups Wolfgang Arendt and A.F.M. ter Elst Spectrum of Ornstein-Uhlenbeck operators in Lp spaces with respect to invariant measures G. Metafune and D. Pallara Hermite conjugate expansions B. Muckenhoupt Maximal and quadratic Gaussian Hardy Spaces Pierre Portal A note on the Gaussian maximal functions Jonas Teuwen