Georgia Institute of Technology June 14, 2004
Outline Background Main Results 1 The Background of the Main Theorem Besicovitch Set Zygmund Conjecture 2 Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series
Besicovitch Set Background Main Results Besicovitch Set Zygmund Conjecture Besicovitch set is a compact set of zero measure, that contains a line in every direction of the plane.
Besicovitch Set Background Main Results Besicovitch Set Zygmund Conjecture Besicovitch set is a compact set of zero measure, that contains a line in every direction of the plane. This set necessarily has Hausdorff dimension two, a fundamental fact in this subject.
Besicovitch Set Background Main Results Besicovitch Set Zygmund Conjecture Besicovitch set is a compact set of zero measure, that contains a line in every direction of the plane. This set necessarily has Hausdorff dimension two, a fundamental fact in this subject. One constructs highly eccentric rectangles which have small union, but the translates along their long direction, their reach, are essentially disjoint.
Besicovitch Set Background Main Results Besicovitch Set Zygmund Conjecture Besicovitch set is a compact set of zero measure, that contains a line in every direction of the plane. This set necessarily has Hausdorff dimension two, a fundamental fact in this subject. One constructs highly eccentric rectangles which have small union, but the translates along their long direction, their reach, are essentially disjoint. We briefly outline the construction of this set.
Background Main Results Besicovitch Set Construction Besicovitch Set Zygmund Conjecture The triangle contains unit length line segments in a full angle of directions.
Background Main Results Besicovitch Set Construction Besicovitch Set Zygmund Conjecture The thirds of the triangle are moved so that they share a common base.
Background Main Results Besicovitch Set Construction Besicovitch Set Zygmund Conjecture Reflect the triangles about their vertexes.
Background Main Results Besicovitch Set Construction Besicovitch Set Zygmund Conjecture The red triangles are essentially disjoint, and are called the reach of the set.
Background Main Results Besicovitch Set Construction Besicovitch Set Zygmund Conjecture The red triangles are essentially disjoint, and are called the reach of the set. A vector field, which points into the set, defined in the reach can be Hölder continuous, of any index strictly less than one.
Background Main Results Besicovitch Set Construction Besicovitch Set Zygmund Conjecture The red triangles are essentially disjoint, and are called the reach of the set. A vector field, which points into the set, defined in the reach can be Hölder continuous, of any index strictly less than one. Conversely, if v is Lipschitz, then the Besicovitch set is has can t have zero Lebesgue measure.
Background Main Results Besicovitch Set Zygmund Conjecture Zygmund Conjecture If v is Lipschitz, then for all f L 2 (R 2 ), f (x) = lim ɛ 0 (2ɛ) 1 ɛ ɛ f (x yv(x)) dy a.e.(x)
Background Main Results Besicovitch Set Zygmund Conjecture Zygmund Conjecture If v is Lipschitz, then for all f L 2 (R 2 ), f (x) = lim ɛ 0 (2ɛ) 1 ɛ ɛ f (x yv(x)) dy a.e.(x) E.M. Stein s Conjecture E.M. Stein s Conjecture: For all Lipschitz v the operator below maps L 2 into itself. H v f (x) = 1 1 f (x yv(x)) dy y
Background Main Results Besicovitch Set Zygmund Conjecture Zygmund Conjecture If v is Lipschitz, then for all f L 2 (R 2 ), f (x) = lim ɛ 0 (2ɛ) 1 ɛ ɛ f (x yv(x)) dy a.e.(x) E.M. Stein s Conjecture E.M. Stein s Conjecture: For all Lipschitz v the operator below maps L 2 into itself. H v f (x) = 1 1 f (x yv(x)) dy y Both conjectures are open. They represent very subtle statements about the nature of the Besicovitch set.
Background Main Results Main Results of Xiaochun Li and L. Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Theorem (L. & Li) For all ɛ > 0, if v has 1 + ɛ derivatives, then H v 2 (1 + log v C 1+ɛ) 2.
Background Main Results Main Results of Xiaochun Li and L. Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Theorem (L. & Li) For all ɛ > 0, if v has 1 + ɛ derivatives, then H v 2 (1 + log v C 1+ɛ) 2. Let λ be a smooth Schwartz function on the plane supported in frequency in 1 ξ 2,and λ k (y) := 2 2k λ(y2 k ), so that we state things in dilation invariant setting.
Background Main Results Main Results of Xiaochun Li and L. Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Theorem (L. & Li) For all ɛ > 0, if v has 1 + ɛ derivatives, then H v 2 (1 + log v C 1+ɛ) 2. Let λ be a smooth Schwartz function on the plane supported in frequency in 1 ξ 2,and λ k (y) := 2 2k λ(y2 k ), so that we state things in dilation invariant setting. Key Proposition (Scale Invariant Formulation) If v is Lipschitz, then H v λ k 2 1 + log 2 k v Lip.
Prior Results Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Previously, these results where known if v were analytic, a result of Nagel, Stein and Wainger [8], or real analytic a result of Bourgain [1].
Prior Results Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Previously, these results where known if v were analytic, a result of Nagel, Stein and Wainger [8], or real analytic a result of Bourgain [1]. There is a rich and beautiful theory of Radon Transforms, as developed by Christ, Nagel, Stein, and Wainger.
Prior Results Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Previously, these results where known if v were analytic, a result of Nagel, Stein and Wainger [8], or real analytic a result of Bourgain [1]. There is a rich and beautiful theory of Radon Transforms, as developed by Christ, Nagel, Stein, and Wainger. The main point is that these results are true in absence of (a) geometric conditions on v (b) minimal smoothness conditions.
Prior Results Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Previously, these results where known if v were analytic, a result of Nagel, Stein and Wainger [8], or real analytic a result of Bourgain [1]. There is a rich and beautiful theory of Radon Transforms, as developed by Christ, Nagel, Stein, and Wainger. The main point is that these results are true in absence of (a) geometric conditions on v (b) minimal smoothness conditions. Genuinely two dimensional time frequency analysis.
Background Main Results Key Proposition Implies Main Theorem Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series With 1 + ɛ smoothness, one can show this: H v λ k f λ k (H v λ k f ) 2 2 ɛ k k f 2 And this proves the Main Theorem from the Key Proposition. Important Obstacle in Extensions 3 This decouples R 2 scales in th crudest possible way. We need a far more sophisticated decoupling of 2-dim l scales to significantly improve the Theorem.
Background Main Results Key Proposition Implies Main Theorem Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series With 1 + ɛ smoothness, one can show this: H v λ k f λ k (H v λ k f ) 2 2 ɛ k k f 2 And this proves the Main Theorem from the Key Proposition. This orthogonality takes some care to formalize correctly. Important Obstacle in Extensions 3 This decouples R 2 scales in th crudest possible way. We need a far more sophisticated decoupling of 2-dim l scales to significantly improve the Theorem.
Background Main Results Role of Carleson s Theorem Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Our Main Theorem has an implication: Pointwise Convergence of Fourier Series in L 2 (R).
Background Main Results Role of Carleson s Theorem Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Our Main Theorem has an implication: Pointwise Convergence of Fourier Series in L 2 (R). Carleson s Theorem For all measurable functions N(x), the operator below is bounded from L 2 into itself. C N f (x) := p.v. 1 1 in(x)y dy f (x y)e y
Background Main Results Role of Carleson s Theorem Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Our Main Theorem has an implication: Pointwise Convergence of Fourier Series in L 2 (R). Carleson s Theorem For all measurable functions N(x), the operator below is bounded from L 2 into itself. C N f (x) := p.v. 1 1 in(x)y dy f (x y)e y Assuming our Main Theorem, we can show that for smooth N(x), the operator above is bounded, with norm independent of N C 2.
Construction of v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Calculate the symbols of both operators where σ(ξ) = 1 1 eixξ dy/y. H v f (x) = σ(ξ v(x)) f (ξ)e ix ξ dξ R 2 C N g(x) = σ(θ N(x))ĝ(θ) dx. R
Construction of v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Calculate the symbols of both operators where σ(ξ) = 1 1 eixξ dy/y. H v f (x) = σ(ξ v(x)) f (ξ)e ix ξ dξ R 2 C N g(x) = σ(θ N(x))ĝ(θ) dx. R View g L 2 (R) as being on the frequency line ξ 2 = J on the plane, where J is large constant.
Construction of v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series Calculate the symbols of both operators where σ(ξ) = 1 1 eixξ dy/y. H v f (x) = σ(ξ v(x)) f (ξ)e ix ξ dξ R 2 C N g(x) = σ(θ N(x))ĝ(θ) dx. R View g L 2 (R) as being on the frequency line ξ 2 = J on the plane, where J is large constant. Then you choose v(x 1, x 2 ) (1, N(x 1 )/J), so that ξ v(x) = x 1 ξ 1 N(x 1 ) on the line ξ 2 = J.
The Picture for v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series ξ 2 = J N(x 1 ) The Blue line is the function σ(ξ 1 N(x 1 )). v(x 1 )
The Picture for v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series ξ 2 = J N(x 1 ) The Blue line is the function σ(ξ 1 N(x 1 )). Then v(x 1 ) (1, N(x 1 )/J), so take J >> v C 2 v(x 1 )
The Picture for v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series ξ 2 = J N(x 1 ) The Blue line is the function σ(ξ 1 N(x 1 )). Then v(x 1 ) (1, N(x 1 )/J), so take J >> v C 2 The symbol of H v and C N agree on the line ξ 2 = J. v(x 1 )
The Picture for v Background Main Results Main Theorem and Key Proposition Key Proposition Implies Main Theorem Corollary: Carleson s Theorem on Fourier Series ξ 2 = J v(x 1 ) N(x 1 ) The Blue line is the function σ(ξ 1 N(x 1 )). Then v(x 1 ) (1, N(x 1 )/J), so take J >> v C 2 The symbol of H v and C N agree on the line ξ 2 = J. Note that small oscillations give oscillations in frequency, that increase as frequency increases.
Lemma Related To Carleson s Theorem Annular Tiles The Proof of Main Theorem 3 Lemma Related To Carleson s Theorem The Weak L 2 estimate is Sharp 4 Annular Tiles The Functions associated to a Tile
Lemma Related To Carleson s Theorem Annular Tiles The Weak L 2 estimate is Sharp Lemma Related to Carleson s Theorem Using the methods of Carleson s Theorem, as proved in [7], one can show that Lemma for Measurable Vector Fields If v is measurable, one has H v λ 0 2 2, <, H v λ 0 p <, 2 < p <.
Lemma Related To Carleson s Theorem Annular Tiles The Weak L 2 estimate is Sharp Lemma Related to Carleson s Theorem Using the methods of Carleson s Theorem, as proved in [7], one can show that Lemma for Measurable Vector Fields If v is measurable, one has H v λ 0 2 2, <, H v λ 0 p <, 2 < p <. The L 2 to weak L 2 estimate is optimal.
Lemma Related To Carleson s Theorem Annular Tiles The Weak L 2 estimate is Sharp Lemma Related to Carleson s Theorem Using the methods of Carleson s Theorem, as proved in [7], one can show that Lemma for Measurable Vector Fields If v is measurable, one has H v λ 0 2 2, <, H v λ 0 p <, 2 < p <. The L 2 to weak L 2 estimate is optimal. And these estimate are critical to an interpolation argument that we use to prove the Key Proposition.
Lemma Related To Carleson s Theorem Annular Tiles Weak L 2 estimate Is Sharp The Weak L 2 estimate is Sharp Consider the radial vector field v, and a smooth bump function ϕ.
Lemma Related To Carleson s Theorem Annular Tiles Weak L 2 estimate Is Sharp The Weak L 2 estimate is Sharp Consider the radial vector field v, and a smooth bump function ϕ. Modulate ϕ so that it is supported in frequency in the annulus 1 ξ 2.
Lemma Related To Carleson s Theorem Annular Tiles Weak L 2 estimate Is Sharp The Weak L 2 estimate is Sharp H v ϕ(x) 1 x Consider the radial vector field v, and a smooth bump function ϕ. Modulate ϕ so that it is supported in frequency in the annulus 1 ξ 2. Calculate H v at any point inside the lines, and there will be no cancellation.
Lemma Related To Carleson s Theorem Annular Tiles Measurable Vector Fields Below L 2 The Weak L 2 estimate is Sharp The proof of the positive result for measurable vector fields follows the Lacey Thiele approach.
Lemma Related To Carleson s Theorem Annular Tiles Measurable Vector Fields Below L 2 The Weak L 2 estimate is Sharp The proof of the positive result for measurable vector fields follows the Lacey Thiele approach. The proof breaks down completely below L 2, as it requires essentially the boundedness of the Kakeya maximal function.
Lemma Related To Carleson s Theorem Annular Tiles Measurable Vector Fields Below L 2 The Weak L 2 estimate is Sharp The proof of the positive result for measurable vector fields follows the Lacey Thiele approach. The proof breaks down completely below L 2, as it requires essentially the boundedness of the Kakeya maximal function. A key innovation is to replace the Kakeya maximal function by a variant associated to the vector field v.
Annular Tiles Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile A Tile is a product of dual rectangles. R s
Annular Tiles Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile A Tile is a product of dual rectangles. The Frequency Rectangle ω s is tangent to a circle of radius r, and spans the annulus r < ξ < 2r. ω s R s
Annular Tiles Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile A Tile is a product of dual rectangles. The Frequency Rectangle ω s is tangent to a circle of radius r, and spans the annulus r < ξ < 2r. But otherwise vary arbitrarily. ω s R s
Annular Tiles Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile A Tile is a product of dual rectangles. The Frequency Rectangle ω s is tangent to a circle of radius r, and spans the annulus r < ξ < 2r. But otherwise vary arbitrarily. A tile is ω s R s. ω s R s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles More Pictures of Tiles The Functions associated to a Tile R s ω s
Lemma Related To Carleson s Theorem Annular Tiles The Uncertainty Intervals The Functions associated to a Tile µ(r) R Interval of uncertainty associated with rectangle R is a sub arc of the unit circle. Its center is the long direction of the rectangle. Its length is the width of R divided by length of R.
Lemma Related To Carleson s Theorem Annular Tiles The Uncertainty Intervals The Functions associated to a Tile µ(r) R Interval of uncertainty associated with rectangle R is a sub arc of the unit circle. Its center is the long direction of the rectangle. Its length is the width of R divided by length of R. Basis changes are permitted, up to a tolerance level dictated by the angle of uncertainty.
Lemma Related To Carleson s Theorem Annular Tiles The Functions Associated to a Tile s The Functions associated to a Tile ϕ s = Modulate c(ωs) Dilate 2 R s ϕ
Lemma Related To Carleson s Theorem Annular Tiles The Functions Associated to a Tile s The Functions associated to a Tile c(ω s ) = center of ω s, ϕ s = Modulate c(ωs) Dilate 2 R s ϕ Dilate 2 R s = L 2 norm one dilation adapted to scale and location of R s.
Lemma Related To Carleson s Theorem Annular Tiles The Functions Associated to a Tile s The Functions associated to a Tile c(ω s ) = center of ω s, ϕ s = Modulate c(ωs) Dilate 2 R s ϕ Dilate 2 R s = L 2 norm one dilation adapted to scale and location of R s. φ s = 1 µ(rs)(v(x)) ϕ s (x yv(x)) scl(s)ψ(scl(s)y) dy R
Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile The Functions Associated to a Tile s c(ω s ) = center of ω s, ϕ s = Modulate c(ωs) Dilate 2 R s ϕ Dilate 2 R s = L 2 norm one dilation adapted to scale and location of R s. φ s = 1 µ(rs)(v(x)) ϕ s (x yv(x)) scl(s)ψ(scl(s)y) dy R scl(s) = the short side of ω s. ψ is a Schwartz function on R, with Fourier support in a small neighborhood of one.
Lemma Related To Carleson s Theorem Annular Tiles The Functions Associated to a Tile s The Functions associated to a Tile c(ω s ) = center of ω s, ϕ s = Modulate c(ωs) Dilate 2 R s ϕ Dilate 2 R s = L 2 norm one dilation adapted to scale and location of R s. φ s = 1 µ(rs)(v(x)) ϕ s (x yv(x)) scl(s)ψ(scl(s)y) dy R scl(s) = the short side of ω s. ψ is a Schwartz function on R, with Fourier support in a small neighborhood of one. Let AT be all annular tiles with ω s contained in 1 ξ 2.
Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile The Main Lemma for Sums Over Tiles For a measurable vector field, we have the estimate 2, f, ϕ s φ s f 2 s AT We also have the L p inequality for p > 2.
Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile The Main Lemma for Sums Over Tiles For a measurable vector field, we have the estimate 2, f, ϕ s φ s f 2 s AT We also have the L p inequality for p > 2. If in addition v Lip <, then we have the L 2 inequality f, ϕ s φ s s AT scl(s)>100 v Lip 2 f 2
The 1 Trees Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile A 1 tree has frequency intervals that progress, counterclockwise, around the circle, at a regular rate. ω R
The 1 Trees Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile A 1 tree has frequency intervals that progress, counterclockwise, around the circle, at a regular rate. The spatial intervals are all contained in R, up to an angle of uncertainty. ω R
The 1 Trees Lemma Related To Carleson s Theorem Annular Tiles The Functions associated to a Tile ω R A 1 tree has frequency intervals that progress, counterclockwise, around the circle, at a regular rate. The spatial intervals are all contained in R, up to an angle of uncertainty. This looks like a singular integral computed in the direction v.
Outline Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement 5 Uncertainty and Density of Rectangles 6 Maximal Function Lemma Two Lemmas on Lipschitz vector fields 7 The Covering Lemma Statement Selection of R
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement The Uncertainty Intervals µ(r) R Interval of uncertainty associated with rectangle R is a sub arc of the unit circle. Its center is the long direction of the rectangle. Its length is the width of R divided by length of R.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement The Uncertainty Intervals µ(r) R Interval of uncertainty associated with rectangle R is a sub arc of the unit circle. Its center is the long direction of the rectangle. Its length is the width of R divided by length of R. Basis changes are permitted, up to a tolerance level dictated by the angle of uncertainty.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Density for Tiles dense(r) = R v 1 (µ(r)) R This measures the percentage of time v(x) point in the long direction of R.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Density for Tiles dense(r) = R v 1 (µ(r)) R This measures the percentage of time v(x) point in the long direction of R. A rectangle and three vectors in the angle of uncertainty
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Definition of the Maximal Function Let R be a collection of rectangles with dense(r) > δ for all R R. and they all have the same width, and lengths at most v Lip /100. Let 1 R M R f = sup f (y) dy R R R R
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Definition of the Maximal Function Let R be a collection of rectangles with dense(r) > δ for all R R. and they all have the same width, and lengths at most v Lip /100. Let 1 R M R f = sup f (y) dy R R R R Previously, one considered maximal function over all rectangles of a given eccentricity. Then the maximal function has norm on L 2 that blows up like the log of the eccentricity.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Definition of the Maximal Function Let R be a collection of rectangles with dense(r) > δ for all R R. and they all have the same width, and lengths at most v Lip /100. Let 1 R M R f = sup f (y) dy R R R R Previously, one considered maximal function over all rectangles of a given eccentricity. Then the maximal function has norm on L 2 that blows up like the log of the eccentricity. Here, it is essential that the estimate be independent of eccentricity.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Maximal Function Lemma For all 0 < δ < 1, the maximal function satisfies M R p p, δ 3, 1 < p <
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Maximal Function Lemma For all 0 < δ < 1, the maximal function satisfies M R p p, δ 3, 1 < p < We need the lemma for some 1 < p < 2.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Maximal Function Lemma For all 0 < δ < 1, the maximal function satisfies M R p p, δ 3, 1 < p < We need the lemma for some 1 < p < 2. And a norm estimate of δ N for any N <.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Two Lemmas on Lipschitz vector fields Maximal Function Lemma For all 0 < δ < 1, the maximal function satisfies M R p p, δ 3, 1 < p < We need the lemma for some 1 < p < 2. And a norm estimate of δ N for any N <. The method of proof is a careful analysis, in the style of Fefferman and Cordoba.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement First Lemma on Lipschitz vector fields Two Lemmas on Lipschitz vector fields Lemma 1: Suppose there are R 1,..., R n R, all containing a common point. Then, n < δ 1.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement First Lemma on Lipschitz vector fields Two Lemmas on Lipschitz vector fields... Lemma 1: Suppose there are R 1,..., R n R, all containing a common point. Then, n < δ 1. Suppose not. Then we can find two rectangles, and points in the rectangles, where the vector field points in the long direction of the rectangle.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement First Lemma on Lipschitz vector fields Two Lemmas on Lipschitz vector fields... Lemma 1: Suppose there are R 1,..., R n R, all containing a common point. Then, n < δ 1. Suppose not. Then we can find two rectangles, and points in the rectangles, where the vector field points in the long direction of the rectangle. At these two points, vector field is nearly radial. But these two points are very close together. Less than angle length. That is a contradiction.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Second Lemma on Lipschitz Two Lemmas on Lipschitz vector fields R R R 0 Lemma 2 Consider three rectangle, R 0, R, and R as pictured. Key assumption is that there is a point x R with v(x) µ(r), and x µ(r ), which share the same projection onto R 0. Then the uncertainty intervals µ(r) and µ(r ) are very close.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Covering Lemma Given the collection R, of rectangles of density δ, it suffices to show that there is an R R for which
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Covering Lemma Given the collection R, of rectangles of density δ, it suffices to show that there is an R R for which R δ 1 R R R R R,
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Covering Lemma Given the collection R, of rectangles of density δ, it suffices to show that there is an R R for which R δ 1 R R R R lngth(r)<lngth(r 0 ) R R R, 1 R R0 n δ 2 R 0 1/n R 0 R.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Covering Lemma Given the collection R, of rectangles of density δ, it suffices to show that there is an R R for which R δ 1 R R R R lngth(r)<lngth(r 0 ) R R R, 1 R R0 n δ 2 R 0 1/n R 0 R. Standard Arguments then prove the Maximal Function Estimate.
How To Select R Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Let M 100 be a maximal function computed in 100 uniformly distributed directions of the plane. This operator maps L 1 (R 2 ) to weak L 1.
How To Select R Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Let M 100 be a maximal function computed in 100 uniformly distributed directions of the plane. This operator maps L 1 (R 2 ) to weak L 1. Initialize R stock := R, R =.
How To Select R Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Let M 100 be a maximal function computed in 100 uniformly distributed directions of the plane. This operator maps L 1 (R 2 ) to weak L 1. Initialize R stock := R, R =. Inductive stage: Select R R stock with maximal length.
How To Select R Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Let M 100 be a maximal function computed in 100 uniformly distributed directions of the plane. This operator maps L 1 (R 2 ) to weak L 1. Initialize R stock := R, R =. Inductive stage: Select R R stock with maximal length. Update, R := R {R }. Remove from R stock any rectangle R such that } R {M 100. R R 1 R δ 1
How To Select R Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Let M 100 be a maximal function computed in 100 uniformly distributed directions of the plane. This operator maps L 1 (R 2 ) to weak L 1. Initialize R stock := R, R =. Inductive stage: Select R R stock with maximal length. Update, R := R {R }. Remove from R stock any rectangle R such that } R {M 100. Repeat until R stock is exhausted. R R 1 R δ 1
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Focusing on the Principal Inequality The main point to prove is n 1 R R0 δ 2 R 0 1/n R 0 R. R R lngth(r)<lngth(r 0 )
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Focusing on the Principal Inequality The main point to prove is n 1 R R0 δ 2 R 0 1/n R 0 R. R R lngth(r)<lngth(r 0 ) Fix R 0 as in this inequality. R 0 = I J, in standard coordinates. Length of R is in the first coordinate.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Focusing on the Principal Inequality The main point to prove is n 1 R R0 δ 2 R 0 1/n R 0 R. R R lngth(r)<lngth(r 0 ) Fix R 0 as in this inequality. R 0 = I J, in standard coordinates. Length of R is in the first coordinate. We only consider those R R that intersect R 0 and have a smaller length. Call this collection R 0.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Focusing on the Principal Inequality The main point to prove is n 1 R R0 δ 2 R 0 1/n R 0 R. R R lngth(r)<lngth(r 0 ) Fix R 0 as in this inequality. R 0 = I J, in standard coordinates. Length of R is in the first coordinate. We only consider those R R that intersect R 0 and have a smaller length. Call this collection R 0. We will need to select a distinguished subset of R 0.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 1 R R 0
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 1 R R 0 I (R) is the projection of all of R onto the top side of R 0. I 0 is the top side of R 0.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 1 R R 0 The rectangle R has density at least δ.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 1 R F (R) R 0 The rectangle R has density at least δ. Project those x R with v(x) in the angle of uncertainty of R, onto the top side of R 0. Call that set F (R).
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 1 R F (R) R 0 The rectangle R has density at least δ. Project those x R with v(x) in the angle of uncertainty of R, onto the top side of R 0. Call that set F (R). This set is at least as big as δlngth(r).
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 2 1 Select R 1 R 0 by initializing R stock := R 0, R 1 :=.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 2 1 Select R 1 R 0 by initializing R stock := R 0, R 1 :=. 2 In the inductive stage, take the longest R R stock for which F (R) is disjoint from {F (R) : R R 1 }.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 2 1 Select R 1 R 0 by initializing R stock := R 0, R 1 :=. 2 In the inductive stage, take the longest R R stock for which F (R) is disjoint from {F (R) : R R 1 }. 3 Then, set R(R) to be those R R stock with I (R ) 2I (R) and F (R ) F (R) not empty.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R Selecting a Distinguished Subset of R 0, Part 2 1 Select R 1 R 0 by initializing R stock := R 0, R 1 :=. 2 In the inductive stage, take the longest R R stock for which F (R) is disjoint from {F (R) : R R 1 }. 3 Then, set R(R) to be those R R stock with I (R ) 2I (R) and F (R ) F (R) not empty. 4 Remove this collection from R stock, and repeat until R stock is exhausted.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Principal Lemma of the Maximal Function Estimate For any subinterval I I 0, we have the two estimates
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Principal Lemma of the Maximal Function Estimate For any subinterval I I 0, we have the two estimates I R1 J δ 1 I J, R 1 R 1 I R1 I
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Principal Lemma of the Maximal Function Estimate For any subinterval I I 0, we have the two estimates I R1 J δ 1 I J, R 1 R 1 I R1 I This is essentially immediate from the disjointness of the sets F (R) for R R 1.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Principal Lemma of the Maximal Function Estimate For any subinterval I I 0, we have the two estimates I R1 J δ 1 I J, R 1 R 1 I R1 I R R 1 (R 1 ) R R 0 I J R R 0 δ 1 I J, R 1 R 1.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Principal Lemma of the Maximal Function Estimate For any subinterval I I 0, we have the two estimates I R1 J δ 1 I J, R 1 R 1 I R1 I R R 1 (R 1 ) R R 0 I J R R 0 δ 1 I J, R 1 R 1. This is essentially a BMO estimate, so it implies the higher moments condition we need.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Essential Geometric Observation Suppose that there is an interval I I 0 and a choice of R 1 R 1 such that R I J 10 3 δ 1 I J. R R(R 1 ) lngth(r) 4 I
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Essential Geometric Observation Suppose that there is an interval I I 0 and a choice of R 1 R 1 such that R I J 10 3 δ 1 I J. R R(R 1 ) lngth(r) 4 I Then, for either ε = +1 or ε = 1, there can be no R R(R 1 ) with 2lngth(R ) < I and R intersects 1 2 (I + ε I ) J.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Essential Geometric Observation Suppose that there is an interval I I 0 and a choice of R 1 R 1 such that R I J 10 3 δ 1 I J. R R(R 1 ) lngth(r) 4 I Then, for either ε = +1 or ε = 1, there can be no R R(R 1 ) with 2lngth(R ) < I and R intersects 1 2 (I + ε I ) J. A final inductive/recursive scheme will then prove the Lemma.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Proof of the Essential Geometric Observation I J I 0 J The proof is by contradiction to the construction of R, and in particular, the use of the maximal function in 100 different directions.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Proof of the Essential Geometric Observation I J I 0 J Select I so that I J is covered more and 10 3 δ 1 times by longer intervals.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Proof of the Essential Geometric Observation I J I 0 J Remember that all the rectangles that cover I J have to have angles of uncertainty that are very close.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Proof of the Essential Geometric Observation I J R I 0 J Consider a rectangle that is rotated by 90 o, has width the same as all other rectangles, and is translated by about I.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Proof of the Essential Geometric Observation I J R I 0 J Consider a rectangle that is rotated by 90 o, has width the same as all other rectangles, and is translated by about I. And height approximately angle I.
Uncertainty and Density of Rectangles Maximal Function Lemma The Covering Lemma Statement Selection of R The Proof of the Essential Geometric Observation I J R I 0 J Consider a rectangle that is rotated by 90 o, has width the same as all other rectangles, and is translated by about I. And height approximately angle I. This rectangle is contained in the removed set. A contradiction. This proves the Essential Geometric Observation.
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