Kernels, Sequences and Approximations P . Zinterhof Department of - - PowerPoint PPT Presentation

Kernels, Sequences and Approximations

P . Zinterhof

Department of Computer Sciences University of Salzburg

Smolenice, 07.09 - 09.09.16

Overview

1

Introduction

2

Hausdorff-Distances and Dispersion

3

Quality of Interpolation

4

Computational complexity

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Outline

1

Introduction

2

Hausdorff-Distances and Dispersion

3

Quality of Interpolation

4

Computational complexity

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Standard Problem

Given a domain E and a function f(x), find x1, ..., xN such that |

• E

f(x)dµ(x) − 1 N

N

• n=1

f(xn)| is small! The answer depends on the domain E and the class H of functions f ∈ H. The problem of approximation and interpolation of functions is similar. Because of the generality of the questions the literature is knowingly infinite.

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I consider the following problem: Given a domain E and a class of functions H and points x1, x2, ...xN ∈ E. What can one say about the quality of the ”mesh” x1, ..., xN in view of integration, interpolation and approximation?

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A possible starting point

let E = [0, 1)s, s = 1, 2, 3... Let Hα

s = {f(x1, x2, ...xs) = ∞

• m1,...ms=−∞

ˆ f(m1, ...ms)exp(2πi(m1x1 + ... + msxs))}, such that for ¯ m = max(1, |m|), m ∈ Z, holds |ˆ f(m1, ...ms)| ≤ c (m1, ...ms)α = O

• 1

(m1, ...ms)α

• , α > 1,

The exponent α is connected with differentiability properties of the function f. ”Korobow-classes”

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Discrepancy

A mesh x1, ..., xN ∈ E will be good, if it is ”uniform distributed” in E. A classical measure for the quality of uniform distribution is essentially due to H. Weyl, who introduced the so called discrepancy, which we will not consider in this frame.

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Diaphony

Some years ago I introduced the so called Diaphony FN of the mesh

• x1, ...,

xN ∈ E[0, 1)s : xn = (x1n, ..., xsn), n = 1, ..., N : FN := 1 N2

N

• k,l=1

∞

• m=−∞
• m=

1 (m1, m2...ms)2 e(2πi(m1(x1k−x1l)+...+ms(xsk−xsl)))

1 2

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It could be made much simpler: Let B2(x) = 1 − π2 6 + π2 2 (1 − 2{x})2 =

• m

1 m2 exp(2πixm) We get now FN = ( 1 N2

N

• k,l=1

s

• j=1

B2(xkj − xlj) − 1)

1 2

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curse of dimension

Theorem: FN

N−>∞

− − − − → 0 ⇔ ( xn)∞

n=1 is u.d. mod 1

Theorem: | 1 N

N

• n=1

f( xn) −

• E

f( x)d x| ≤ CF α

N

Typically: FN = O

• 1

N1−ǫ

• ,

FN = O lns N N

• Very Bad: cassical cartesian product rules:

FN = O 1 N

1 s

• , it is sharp!
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The frequent occurrences of 2 and 1

2 call for Hilbert space approach!!

Leitmotiv: Put ϕn(x) = 1 nexp(2πinx), n ∈ Z, and put < ϕn, ϕm >= δnm, H := {f(x) =

• n∈R

˜ fnϕn(x)},

• n

|˜ fn|2 < ∞, and you get a Hilbert space of continuous functions.

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More general

Let E = ∅, K(x, y) : ExE → C strictly pos. def. matrix. Let < K(x, y1), K(x, y2) >:= K(y2, y1), then span{a1K(x, y1) + a2K(x, y2) + ... + anK(x, yn)} = H defines a Hilbertspace with rep. Kernel K(x,y): < f(x), K(x, y) >= f(y), f ∈ H f ⇒ f(x) being a cont. functional on H Bergman, Aronszajn, Moore, Zaremba, ...

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We define a metric on E := d(x, y) : ||K(t, x) − K(t, y)|| and assume compactness of the metric space (E, d). Let now (xn)∞

n=1 be a sequence of points in E, xn = xm, n = m. We

apply Gram-Schmidt to K(x, x1), K(x, x2), ..., K(x, xn), ... and get an

• rthonormal sequence τn(x), n ∈ N.

(τn(x))∞

n=1 has the important property τm(xn) = 0 for m > n.

Let f(x) = ∞

n=1 ˜

fnτn(x) ∈ H. We compute ˜ fn, n = 1, 2, ... recursively: ˜ f1 = f(x1)/τ1(x1),˜ f2 = (f(x2) − ˜ f1τ1(x2))/τ2(x2), ...˜ fn = (f(xn) − ˜ f1τ1(xn) − ... − ˜ fn−1τn−1(xn))/τn(xn) The τn(xn) = 0, n ∈ N, by Cholesky’s theorem. e.g.

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Main question

In which cases is the ONS (τn(x))∞

n=1 an ONB:

f(x) =

∞

• n=1

˜ fnτn(x) for all f(x) ∈ H?? Theorem: (τn(x))n is an ONB ⇔ (K(t, xn))n is total in H ⇔ (f(xn) = 0 => f = 0 ∀f ∈ H) Let now KN(x, y) := N

n=1 τn(x)τn(y), K ⊥ N (x, y) = K(x, y) − KN(x, y).

KN(x, y) reproduces HN = span{K(x, x1), ..., K(x, xN)}, K ⊥

N (x, y)

reproduces H - HN = H⊥

N .

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Def: The N-th totality of x1, ...xN, ... is TN = max

y∈E ||K ⊥ N (x, y)|| = max y

||KN(x, y)|| − ||KN(x, y)|| = max

y (K(y, y)

1 2 − KN(y, y) 1 2 ) = max

y

TN(y) Theorem: (xn)∞

n=1 is total in E iff lim N→∞TN = 0.

The proof uses Arzela-Ascoli’s theorem.

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Outline

1

Introduction

2

Hausdorff-Distances and Dispersion

3

Quality of Interpolation

4

Computational complexity

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Hausdorff-Distances and Dispersion

The Hausdorff-Distance between {x1, ...xN} ⊆ E and E itself is defiend by δ(x1, ...xN) = max

x

min

n=1,...N d(x, xn)

Mostly δ(x1, ...xN) is called the dispersion of x1, ..., xN in E. Well known: The sequence (xn)∞

n=1 is dense in E iff lim N→∞δN = 0

The following theorem holds: 0 ≤ TN ≤ δN

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Corollary

If (xn)∞

n=1 is dense in E, it is total as well. I have a Hilbert space and

total sequences with arbitrary bad density properties. Furthermore I have a Hilbert space (E,K,H), where TN → 0 ⇔ δN → 0. It means, a sequence (xn)∞

n=1 is total iff it is dense in that special space

E.

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Outline

1

Introduction

2

Hausdorff-Distances and Dispersion

3

Quality of Interpolation

4

Computational complexity

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Problem

We want to find a function fN(x), such that fN(xn) = f(xn), n = 1, ..., N, and fN(x) ∈ span{K(x, xn), n = 1, ..., N} = HN. Let l1(x), l2(x), ..., lH(x) ∈ HN, such that lm(xn) = δnm. The l1(x), ..., lH(x) are the dual base to K(x, x1), ..., K(x, xN). So, fN(x) = N

n=1 f(xn)ln(x) ∈ Hn fulfills the requirements. On the

• ther hand f(xn) = N

k=1 ˜

fkτk(xn), because τk(xn) = 0 for k > N. So we get fN(x) = N

n=1 ˜

fnτn(x), because fN(x) and f(x) coincide at x1, ..., xN.

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It follows: |f(x) − fN(x)| = |

∞

• n=N+1

˜ fnτn(x)| ≤

• ∞
• n=N+1

|˜ fn|2 1

2

∞

• n=N+1

|τn(x)|2 1

2

max

x∈E = |f(x) − fN(x)| ≤ ( ∞

• n=N+1

|˜ fn|2)

1 2

TN = ||f ⊥

N (x)|| · TN ≤ ||f||TN

and ||f − fN|| = (

∞

• n=N+1

|˜ fn|2)

1 2

means that the ”Lagrange”-Interpolation function fN(x) is the best approximation in Hilbert space sense!

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Outline

1

Introduction

2

Hausdorff-Distances and Dispersion

3

Quality of Interpolation

4

Computational complexity

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Computational complexity

The computation of ˜ fn, τn(x), n = 1, ..., N, KN(x, y) needs O(N2)

• perations.

Important special case: E = G, compact abelian group with Haar measure λ and K(x, y) = k(x − y), k(x) = f(x) ⋆ f(−x), f(x) ∈ L2(G), and xn = x1 · n, n = 0, ..., N − 1 is a cyclic subgroup of G. Then one can apply FFT with N log(N) operations.

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I have similar results for numercical Integration of the type Ig(f) =< f, g > using f(x1), ..., f(xH). ”But that’s another story” would Rudyard Kipling say....

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The Kernel KN(x, y) = N

n=1 τn(x)τn(y) plays the main role in

interpolation and approximation of functions in H: fN(x) =< f(x), KN(x, y) > KN(x, y) reproduces the space Hn = span{K(x, x1), ..., K(x, xN)}. Let Gram(K(x, x1), ..., K(x, xN)) = GramN, then we can show, that KN(x, y) = (K(y, x1), ..., K(y, xN)) · Gram−1

N

    K(x, x1) K(x, x2) ... K(x, xN)    

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GramN is not singular and pos. def. So we have with a matrix U, U∗ = U−1 and λ2

1 ≥ λ2 2 ≥ ... ≥ λ2 N > 0

GramN = U∗   λ2

1

... λ2

N

  U and KN(x, y) = (K(y, x1), ..., K(y, xN))U∗   

1 λ2

1

...

1 λ2

N

   U   K(x, x1) ... K(x, xN)   This means explicitly, that we have KN(x, y) =

N

• n=1

Ln(x)Ln(y) λ2

n

with Ln⊥Lm, ||Ln||2 = λ2

n

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We know already: the ”better” the x1, ...xN are distributed in E the better KN(x, y) will be suitable for interpolation and for integration. So the behavior of the Eigenfunctions and Eigenvalues of GramN play an important role. In many important cases (E = [0, 1)s, E = c.a.g.) we have λ2

n ˜ N. It means, that in this cases the arithmetic mean is

asymptotically the best choice for numerical integration.

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