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Smooth Interpolation Arie Israel Courant Institute June 18, 2012 - - PowerPoint PPT Presentation

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Smooth Interpolation Arie Israel Courant Institute June 18, 2012 Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 1 / 45 Contributions from Whitney (1930s) Glaeser (1950s) Brudnyi-Shvartsman (1980s-present)

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Smooth Interpolation

Arie Israel

Courant Institute

June 18, 2012

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 1 / 45

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Contributions from Whitney (1930’s) Glaeser (1950’s) Brudnyi-Shvartsman (1980’s-present) Bierstone-Milman-Pawlucki (2000’s-present) Fefferman/Fefferman-Klartag (2003-present) Fefferman-I-Luli (2010-present)

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Notation

Let F : Rn → R be sufficiently smooth. For any multi-index α = (α1, . . . , αn), ∂αF(x) := ∂α1

1 · · · ∂αn n F(x);

|α| := α1 + · · · + αn. For k ≥ 1, ∇kF(x) :=

  • ∂αF(x)
  • |α|=k.

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Notation

Let F : Rn → R be sufficiently smooth. For m ≥ 1, FC m := sup

x∈Rn |∇mF(x)|.

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The Problem

Given: Finite subset E ⊂ Rn with cardinality N; Function f : E → R.

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The Problem

Given: Finite subset E ⊂ Rn with cardinality N; Function f : E → R. Compute a C-optimal interpolant: F : Rn → R with (a) F = f on E; (b) FC m ≤ C · GC m whenever G = f on E.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 5 / 45

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The Problem

Given: Finite subset E ⊂ Rn with cardinality N; Function f : E → R. Compute a C-optimal interpolant: F : Rn → R with (a) F = f on E; (b) FC m ≤ C · GC m whenever G = f on E. Side Questions: Estimate the nearly minimal norm FC m. How long do these computations take?

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Theorem (Fefferman-Klartag (’09))

Can construct C1-optimal interpolants in time C2N log(N).

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A Variant Problem

For m ≥ 1 and p ≥ 1, let FLm,p :=

  • x∈Rn |∇mF(x)|pdx

1/p . Compute a C-optimal Sobolev interpolant: F : Rn → R with F = f on E; FLm,p ≤ C · GLm,p whenever G = f on E.

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Theorem (Fefferman-I-Luli (’11))

Can construct C-optimal Sobolev interpolants.

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Theorem (Fefferman-I-Luli (’11))

Can construct C-optimal Sobolev interpolants. Plausible running-time bound is Om,n,p(N log(∆)r), where ∆ := max{|x − y| : x, y ∈ E} min{|x − y| : x, y ∈ E}

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Theorem (Fefferman-I-Luli (’11))

Can construct C-optimal Sobolev interpolants. Plausible running-time bound is Om,n,p(N log(∆)r), where ∆ := max{|x − y| : x, y ∈ E} min{|x − y| : x, y ∈ E} Can we prove this? Can we achieve O(N log(N))?

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 8 / 45

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Example I

Given: t1, . . . , tN ∈ R p1, . . . , pN ∈ R Construct p : R → R with (a) p(t1) = p1, · · · , p(tN) = pN; (b) supt∈R |p′(t)| ≤ supt∈R |q′(t)|, for any other interpolant q. Estimate: M = sup

t∈R

|p′(t)|.

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t

p

(t1,p1) (t2,p2) (t3,p3) (t4−6,p4−6) (t7,p7) (t8,p8)

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SLIDE 15

t

p

(t1,p1) (t2,p2) (t3,p3) (t4−6,p4−6) (t7,p7) (t8,p8)

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(1) sup |p′(t)| =

  • p2 − p3

t2 − t3

  • .

The competitor q interpolates the data, so MVT = ⇒ (2) ∃t∗ ∈ [t2, t3] with q′(t∗) = p2 − p3 t2 − t3 . Finally, (1) and (2) = ⇒ (3) sup |p′(t)| ≤ C sup |q′(t)|.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 12 / 45

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Example II

Given: t1, . . . , tN ∈ R p1, . . . , pN ∈ R Construct p : R → R with (a) p(t1) = p1, · · · , p(tN) = pN; (b) supt∈R |p′′(t)| ≤ supt∈R |q′′(t)|, for any other interpolant q. Estimate: M = sup

t∈R

|p′′(t)|.

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SLIDE 18

t

p

(t1,p1) (t2,p2) (t3,p3) (t4−6,p4−6) (t7,p7) (t8,p8)

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SLIDE 19

t

p

(t1,p1) (t2,p2) (t3,p3) (t4−6,p4−6) (t7,p7) (t8,p8)

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SLIDE 20

t

p

(t1,p1) (t3,p3) (t4−6,p4−6) (t7,p7) (t8,p8) (t2,p2)

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Higher Dimensions

Given: Finite subset E ⊂ [0, 1]2; Function f : E → R

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Higher Dimensions

Given: Finite subset E ⊂ [0, 1]2; Function f : E → R There’s a Competitor: G : R2 → R with G = f on E; |∇2G| ≤ 1 on R2.

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Higher Dimensions

Given: Finite subset E ⊂ [0, 1]2; Function f : E → R There’s a Competitor: G : R2 → R with G = f on E; |∇2G| ≤ 1 on R2. Goal: Construct F : [0, 1]2 → R with F = f on E; |∇2F| ≤ C on [0, 1]2.

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Two Examples

(a) E contained in a line. (b) E contained in a smooth curve.

(a) (b) Figure: Sets with 1D structure

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The Straight Line

Suppose that E = {(0, y1), . . . , (0, yN)}; f : E → R.

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The Straight Line

Suppose that E = {(0, y1), . . . , (0, yN)}; f : E → R. Step 1: Let g : R → R be the cubic spline with g(yk) = f (0, yk) for k = 1, . . . , N, and |g′′(y)| ≤ C.

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The Straight Line

Suppose that E = {(0, y1), . . . , (0, yN)}; f : E → R. Step 1: Let g : R → R be the cubic spline with g(yk) = f (0, yk) for k = 1, . . . , N, and |g′′(y)| ≤ C. Step 2: Define F(x, y) := g(y). Then |∇2F(x, y)| = |g′′(y)| ≤ C for all (x, y).

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 19 / 45

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The Smooth Curve

Suppose that E ⊂ {(φ(y), x)}, where |φ′′| ≤ 1.

(a) (b) Figure: Sets with 1D structure

Consider the diffeomorphism Φ : R2 → R2: Φ(x, y) = (x − φ(y), y).

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The Smooth Curve

Suppose that E ⊂ {(φ(y), x)}, where |φ′′| ≤ 1.

(c) (d) Figure: Sets with 1D structure

Consider the diffeomorphism Φ : R2 → R2: Φ(x, y) = (x − φ(y), y). Note that Φ maps E onto a line segment.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 20 / 45

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The Smooth Curve

Suppose that E ⊂ {(φ(y), x)}, where |φ′′| ≤ 1.

(e) (f) Figure: Sets with 1D structure

Consider the diffeomorphism Φ : R2 → R2: Φ(x, y) = (x − φ(y), y). Note that Φ maps E onto a line segment. There is a 1 − 1 correspondence between interpolation problems on E and on Φ(E).

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 20 / 45

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Some Notation

S(x, δ) := square with center x and sidelength δ. δ(S) := sidelength of the square S. A · S := A-dilate of S about its center.

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Definition (Neat Squares)

A square S is neat if 3S ∩ E lies on the graph of a function h with |h′′| ≤ δ(S)−1 uniformly.

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Definition (Neat Squares)

A square S is neat if 3S ∩ E lies on the graph of a function h with |h′′| ≤ δ(S)−1 uniformly. Equivalently, S neat when δ(S)−1 · (3S ∩ E) lies on the graph of a function h with |h′′| ≤ 1 uniformly.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 22 / 45

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Definition (Neat Squares)

A square S is neat if 3S ∩ E lies on the graph of a function h with |h′′| ≤ δ(S)−1 uniformly. Equivalently, S neat when δ(S)−1 · (3S ∩ E) lies on the graph of a function h with |h′′| ≤ 1 uniformly. Small enough squares are neat. If S is neat and S′ ⊂ S then S′ is neat.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 22 / 45

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Lemma

Suppose that S is neat. Then we can construct F : 3S → R with F = f

  • n E ∩ 3S and |∇2F| ≤ C on 3S.

(a) A Neat S... (b) Rescaled

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Definition (Messy Squares)

A square S is messy if S is not neat.

Figure: Some Messy Squares

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The CZ Decomposition

Keep bisecting S ⊂ [0, 1]2 until S is neat. Define CZ as the collection of nonbisected squares.

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SLIDE 43

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SLIDE 44

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Properties of the CZ Decomposition

Note that CZ = {Sν} partitions [0, 1]2. (a) If S ∈ CZ, then S is neat. (b) If S ∈ CZ, then 3S is messy. (c) Good Geometry: If S, S′ ∈ CZ touch, then 1 2δ(S′) ≤ δ(S) ≤ 2δ(S′).

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Properties of the CZ Decomposition

Note that CZ = {Sν} partitions [0, 1]2. (a) If S ∈ CZ, then S is neat. (b) If S ∈ CZ, then 3S is messy. (c) Good Geometry: If S, S′ ∈ CZ touch, then 1 2δ(S′) ≤ δ(S) ≤ 2δ(S′). One-Line Proofs: (a) That was our stopping rule! (b) 3S contains the dyadic parent S+. (c) If S, S′ ∈ CZ touch and δ(S) ≤ δ(S′)/4, then 3S+ ⊂ 3S′.

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The Naive Plan: Step 1

Construct local interpolants for the CZ squares: Functions Fν : 3Sν → R that satisfy: (a) Fν = f

  • n

E ∩ (1.1)Sν. (b) |∇2Fν| ≤ C

  • n

3Sν.

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The Naive Plan: Step 2

Introduce a partition of unity adapted to the CZ squares: Functions θν : [0, 1]2 → R that satisfy (a) 0 ≤ θν ≤ 1; (b) supp(θν) ⊂ (1.1)Sν; (c) |∇θν| ≤ C · δ(Sν)−1 and |∇2θν| ≤ C · δ(Sν)−2; (d)

  • ν

θν = 1

  • n

[0, 1]2.

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The Naive Plan: Step 3

Define: F =

  • ν

θνFν.

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The Naive Plan: Step 3

Define: F =

  • ν

θνFν. By Local Interpolation and support properties of the partition of unity, F = f

  • n

E.

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The Naive Plan: Step 3

Define: F =

  • ν

θνFν. By Local Interpolation and support properties of the partition of unity, F = f

  • n

E. Question: Is |∇2F| ≤ C

  • n

[0, 1]2?

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SLIDE 52

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Arranging Consistency

Lemma

Let S be any messy square. Then there exists a “non-degenerate” triplet T ⊂ E ∩ 9S.

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SLIDE 54

Arranging Consistency

Lemma

Let S be any messy square. Then there exists a “non-degenerate” triplet T ⊂ E ∩ 9S. Associate to each Sν some “non-degenerate” triplet: Tν ⊂ E ∩ 9Sν.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 38 / 45

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Arranging Consistency

Lemma

Let S be any messy square. Then there exists a “non-degenerate” triplet T ⊂ E ∩ 9S. Associate to each Sν some “non-degenerate” triplet: Tν ⊂ E ∩ 9Sν. Let Lν be affine with Lν = f on Tν.

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Arranging Consistency

Lemma

Let S be any messy square. Then there exists a “non-degenerate” triplet T ⊂ E ∩ 9S. Associate to each Sν some “non-degenerate” triplet: Tν ⊂ E ∩ 9Sν. Let Lν be affine with Lν = f on Tν. This gives our rough guess for the affine structure of our interpolant.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 38 / 45

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Let’s check consistency:

Lemma

Suppose that Sν and Sν′ are neighboring squares. Then |∇Lν − ∇Lν′| ≤ Cδ(Sν) and |Lν − Lν′| ≤ Cδ(Sν)2

  • n

100Sν.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 39 / 45

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Need this version of Rolle’s Theorem:

Lemma

Suppose that H vanishes on a “non-degenerate” triplet T ⊂ S and HC 2 ≤ 1. Then, |∇H| ≤ Cδ(S) and |H| ≤ Cδ(S)2

  • n

S.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 40 / 45

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SLIDE 59

Need this version of Rolle’s Theorem:

Lemma

Suppose that H vanishes on a “non-degenerate” triplet T ⊂ S and HC 2 ≤ 1. Then, |∇H| ≤ Cδ(S) and |H| ≤ Cδ(S)2

  • n

S. Recall that G = f on E and GC 2 ≤ 1. Lν = f on Tν. Lν′ = f on Tν′.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 40 / 45

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SLIDE 60

Need this version of Rolle’s Theorem:

Lemma

Suppose that H vanishes on a “non-degenerate” triplet T ⊂ S and HC 2 ≤ 1. Then, |∇H| ≤ Cδ(S) and |H| ≤ Cδ(S)2

  • n

S. Recall that G = f on E and GC 2 ≤ 1. Lν = f on Tν. Lν′ = f on Tν′. For any x ∈ 100Sν, |∇Lν − ∇Lν′| ≤ |∇Lν − ∇G(x)| + |∇Lν′ − ∇G(x)| ≤ Cδ(S),

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Lemma

Suppose that Sν and Sν′ are neighboring squares. Then |∇Lν − ∇Lν′| ≤ Cδ(Sν) and |(Lν − Lν′)(x)| ≤ Cδ(Sν)2

  • n

100Sν.

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Lemma

Suppose that Sν and Sν′ are neighboring squares. Then |∇Lν − ∇Lν′| ≤ Cδ(Sν) and |(Lν − Lν′)(x)| ≤ Cδ(Sν)2

  • n

100Sν. Define Fν := Lν whenever E ∩ (1.1)Sν = ∅.

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SLIDE 63

Lemma

Suppose that Sν and Sν′ are neighboring squares. Then |∇Lν − ∇Lν′| ≤ Cδ(Sν) and |(Lν − Lν′)(x)| ≤ Cδ(Sν)2

  • n

100Sν. Define Fν := Lν whenever E ∩ (1.1)Sν = ∅. Do something similar for all other CZ squares.

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SLIDE 64

Lemma

Suppose that Sν and Sν′ are neighboring squares. Then |∇Lν − ∇Lν′| ≤ Cδ(Sν) and |(Lν − Lν′)(x)| ≤ Cδ(Sν)2

  • n

100Sν. Define Fν := Lν whenever E ∩ (1.1)Sν = ∅. Do something similar for all other CZ squares. Set F =

  • ν

Fνθν. We obtain FC 2 ≤ C ′.

Arie Israel (Courant Institute) Smooth Interpolation June 18, 2012 41 / 45

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SLIDE 65

Keystone Squares

Definition

Sµ ∈ CZ is keystone if every CZ square that intersects 9Sµ has sidelength larger than Sµ.

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SLIDE 66

Definition

Sµ ∈ CZ is keystone if every CZ square that intersects 9Sµ has sidelength larger than Sµ.

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SLIDE 67

Definition

Sµ ∈ CZ is keystone if every CZ square that intersects 9Sµ has sidelength larger than Sµ.

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SLIDE 68

Diverging Paths

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