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8/3/2019 Ft Appl Talk
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The Fourier Transform and applications
Mihalis Kolountzakis
University of Crete
January 2006
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 1 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 2/196
Groups and Haar measure
Locally compact abelian groups:
Haar measure on G = translation invariant on G : µ(A) = µ(A + t ).Unique up to scalar multiple.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 2 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 3/196
Groups and Haar measure
Locally compact abelian groups:
Integers Z = {. . . , −2, −1, 0, 1, 2, . . .}
Haar measure on G = translation invariant on G : µ(A) = µ(A + t ).Unique up to scalar multiple.
Counting measure on Z
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 2 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 4/196
Groups and Haar measure
Locally compact abelian groups:
Integers Z = {. . . , −2, −1, 0, 1, 2, . . .}Finite cyclic group Zm = {0, 1, . . . , m − 1}: addition modm
Haar measure on G = translation invariant on G : µ(A) = µ(A + t ).Unique up to scalar multiple.
Counting measure on Z
Counting measure on Zm, normalized to total measure 1 (usually)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 2 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 5/196
Groups and Haar measure
Locally compact abelian groups:
Integers Z = {. . . , −2, −1, 0, 1, 2, . . .}Finite cyclic group Zm = {0, 1, . . . , m − 1}: addition modm
Reals R
Haar measure on G = translation invariant on G : µ(A) = µ(A + t ).Unique up to scalar multiple.
Counting measure on Z
Counting measure on Zm, normalized to total measure 1 (usually)
Lebesgue measure on R
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 2 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 6/196
Groups and Haar measure
Locally compact abelian groups:
Integers Z = {. . . , −2, −1, 0, 1, 2, . . .}Finite cyclic group Zm = {0, 1, . . . , m − 1}: addition modm
Reals R
Torus T = R/Z: addition of reals mod1
Haar measure on G = translation invariant on G : µ(A) = µ(A + t ).Unique up to scalar multiple.
Counting measure on Z
Counting measure on Zm, normalized to total measure 1 (usually)Lebesgue measure on R
Lebesgue masure on T viewed as a circle
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 2 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 7/196
Groups and Haar measure
Locally compact abelian groups:
Integers Z = {. . . , −2, −1, 0, 1, 2, . . .}Finite cyclic group Zm = {0, 1, . . . , m − 1}: addition modm
Reals R
Torus T = R/Z: addition of reals mod1
Products: Zd , Rd , T× R, etc
Haar measure on G = translation invariant on G : µ(A) = µ(A + t ).Unique up to scalar multiple.
Counting measure on Z
Counting measure on Zm, normalized to total measure 1 (usually)Lebesgue measure on R
Lebesgue masure on T viewed as a circle
Product of Haar measures on the components
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 2 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 8/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to the
multiplicative group U = {z ∈C
: |z | = 1}.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
8/3/2019 Ft Appl Talk
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Characters and the dual group
Character is a (continuous) group homomorphism from G to the
multiplicative group U = {z ∈C
: |z | = 1}.χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 10/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative
groupU
= {z
∈C
: |z
| = 1}.χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 11/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative
groupU
= {z
∈C
: |z
| = 1}.χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 12/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative
groupU
= {z
∈C
: |z
| = 1}.χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ T
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 13/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ Z
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
C
8/3/2019 Ft Appl Talk
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Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ ZG = R =⇒ G = R: the functions χt (x ) = exp(2πitx ), t ∈ R
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
Ch d h d l
8/3/2019 Ft Appl Talk
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Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ ZG = R =⇒ G = R: the functions χt (x ) = exp(2πitx ), t ∈ RG = Zm =⇒
G = Zm: the functions χk (n) = exp(2πikn/m), k ∈ Zm
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
Ch d h d l
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 16/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ ZG = R =⇒ G = R: the functions χt (x ) = exp(2πitx ), t ∈ RG = Zm =⇒
G = Zm: the functions χk (n) = exp(2πikn/m), k ∈ Zm
G = A × B =⇒ G = A × B
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
Ch t d th d l
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 17/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ ZG = R =⇒ G = R: the functions χt (x ) = exp(2πitx ), t ∈ RG = Zm =⇒
G = Zm: the functions χk (n) = exp(2πikn/m), k ∈ Zm
G = A × B =⇒ G = A × B Example: G = T× R =⇒ G = Z×R. The characters areχn,t (x , y ) = exp(2πi (nx + ty )).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
Ch t d th d l
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 18/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ ZG = R =⇒ G = R: the functions χt (x ) = exp(2πitx ), t ∈ RG = Zm =⇒
G = Zm: the functions χk (n) = exp(2πikn/m), k ∈ Zm
G =
A×
B =⇒ G
= A × B
Example: G = T× R =⇒ G = Z×R. The characters areχn,t (x , y ) = exp(2πi (nx + ty )).
G is compact ⇐⇒ G is discrete
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
Ch t s d th d l
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 19/196
Characters and the dual group
Character is a (continuous) group homomorphism from G to themultiplicative group U = {z ∈ C : |z | = 1}.
χ : G → U satsifies χ(h + g ) = χ(h)χ(g )
If χ, ψ are characters then so is χψ (pointwise product). Write χ + ψfrom now on instead of χψ.
Group of characters (written additively )
G is the dual group of G
G = Z =⇒ G = T: the functions χx (n) = exp(2πixn), x ∈ TG = T =⇒ G = Z: the functions χn(x ) = exp(2πinx ), n ∈ ZG = R =⇒ G = R: the functions χt (x ) = exp(2πitx ), t ∈ RG = Zm =⇒
G = Zm: the functions χk (n) = exp(2πikn/m), k ∈ Zm
G =
A×
B =⇒ G
= A × B
Example: G = T× R =⇒ G = Z×R. The characters areχn,t (x , y ) = exp(2πi (nx + ty )).
G is compact ⇐⇒ G is discrete
Pontryagin duality: G = G .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 3 / 36
The Fourier Transform of integrable functions
8/3/2019 Ft Appl Talk
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The Fourier Transform of integrable functions
f ∈ L1(G ). That is f 1 := G
|f (x )| d µ(x ) < ∞
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 4 / 36
The Fourier Transform of integrable functions
8/3/2019 Ft Appl Talk
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The Fourier Transform of integrable functions
f ∈ L1(G ). That is f 1 := G
|f (x )| d µ(x ) < ∞If G is finite then L1(G ) is all functions G → C
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 4 / 36
The Fourier Transform of integrable functions
8/3/2019 Ft Appl Talk
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The Fourier Transform of integrable functions
f ∈ L1(G ). That is f 1 := G
|f (x )| d µ(x ) < ∞If G is finite then L1(G ) is all functions G → CThe FT of f is f : G → C defined by
f (χ) =
G
f (x )χ(x ) d µ(x ), χ ∈ G
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 4 / 36
The Fourier Transform of integrable functions
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 23/196
The Fourier Transform of integrable functions
f ∈ L1(G ). That is f 1 := G
|f (x )| d µ(x ) < ∞If G is finite then L1(G ) is all functions G → CThe FT of f is f : G → C defined by
f (χ) =
G
f (x )χ(x ) d µ(x ), χ ∈ G
Example: G = T (“Fourier coefficients”):
f (n) = T
f (x )e −2πinx dx , n ∈ Z
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 4 / 36
The Fourier Transform of integrable functions
8/3/2019 Ft Appl Talk
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The Fourier Transform of integrable functions
f ∈ L1(G ). That is f 1 := G
|f (x )| d µ(x ) < ∞If G is finite then L1(G ) is all functions G → CThe FT of f is f : G → C defined by
f (χ) =
G
f (x )χ(x ) d µ(x ), χ ∈ G
Example: G = T (“Fourier coefficients”):
f (n) = T
f (x )e −2πinx dx , n ∈ Z
Example: G = R (“Fourier transform”):
f (ξ) = T f (x )e
−2πi ξx
dx , ξ ∈ R
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 4 / 36
The Fourier Transform of integrable functions
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 25/196
The Fourier Transform of integrable functions
f ∈ L1(G ). That is f 1 := G
|f (x )| d µ(x ) < ∞If G is finite then L1(G ) is all functions G → CThe FT of f is f : G → C defined by
f (χ) =
G
f (x )χ(x ) d µ(x ), χ ∈ G
Example: G = T (“Fourier coefficients”):
f (n) = T
f (x )e −2πinx dx , n ∈ Z
Example: G = R (“Fourier transform”):
f (ξ) = T f (x )e
−2πi ξx
dx , ξ ∈ RExample: G = Zm (“Discrete Fourier transform or DFT”):
f (k ) =
1
m
m−1
j =0
f ( j )e −2πikj /m, k ∈ Zm
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 4 / 36
Elementary properties of the Fourier Transform
8/3/2019 Ft Appl Talk
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Elementary properties of the Fourier Transform
Linearity: λf + µg = λ
f + µ
g .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 5 / 36
Elementary properties of the Fourier Transform
8/3/2019 Ft Appl Talk
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Elementary properties of the Fourier Transform
Linearity: λf + µg = λ
f + µ
g .
Symmetry: f (−x ) = f (x ), f (x ) = f (−x )
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 5 / 36
Elementary properties of the Fourier Transform
8/3/2019 Ft Appl Talk
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Elementary properties of the Fourier Transform
Linearity: λf + µg = λ
f + µ
g .
Symmetry: f (−x ) = f (x ), f (x ) = f (−x )
Real f : then f (x ) = f (−x )
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 5 / 36
Elementary properties of the Fourier Transform
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 29/196
y p p
Linearity: λf + µg = λ
f + µ
g .
Symmetry: f (−x ) = f (x ), f (x ) = f (−x )
Real f : then f (x ) = f (−x )
Translation: if τ ∈ G , ξ ∈ G , f τ (x ) = f (x − τ ) then
f τ (ξ) = ξ(τ ) ·
f (ξ).
Example: G = T: f (x − θ)(n) = e −2πinθ f (n), for θ ∈ T, n ∈ Z.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 5 / 36
Elementary properties of the Fourier Transform
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 30/196
y p p
Linearity: λf + µg = λ
f + µ
g .
Symmetry: f (−x ) = f (x ), f (x ) = f (−x )
Real f : then f (x ) = f (−x )
Translation: if τ ∈ G , ξ ∈ G , f τ (x ) = f (x − τ ) then
f τ (ξ) = ξ(τ ) ·
f (ξ).
Example: G = T: f (x − θ)(n) = e −2πinθ f (n), for θ ∈ T, n ∈ Z.
Modulation: If χ, ξ ∈ G then χ(x )f (x )(ξ) = f (ξ − χ).
Example: G = R: e 2πitx f (x )(ξ) = f (ξ − t ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 5 / 36
Elementary properties of the Fourier Transform
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 31/196
y p p
Linearity: λf + µg = λ
f + µ
g .
Symmetry: f (−x ) = f (x ), f (x ) = f (−x )
Real f : then f (x ) = f (−x )
Translation: if τ ∈ G , ξ ∈ G , f τ (x ) = f (x − τ ) then
f τ (ξ) = ξ(τ ) ·
f (ξ).
Example: G = T: f (x − θ)(n) = e −2πinθ f (n), for θ ∈ T, n ∈ Z.
Modulation: If χ, ξ ∈ G then χ(x )f (x )(ξ) = f (ξ − χ).
Example: G = R: e 2πitx f (x )(ξ) = f (ξ − t ).
f , g ∈ L1(G ): their convolution is f ∗ g (x ) = G f (t )g (x − t ) d µ(t ).
Then f ∗ g 1 ≤ f 1g 1 and
f ∗ g (ξ) = f (ξ) · g (ξ), ξ ∈ G
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 5 / 36
Orthogonality of characters on compact groups
8/3/2019 Ft Appl Talk
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g y p g p
If G is compact (=⇒ total Haar measure = 1) then characters are inL1(G ), being bounded.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 6 / 36
Orthogonality of characters on compact groups
8/3/2019 Ft Appl Talk
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g y g
If G is compact (=⇒ total Haar measure = 1) then characters are inL1(G ), being bounded.
If χ ∈ G then G
χ(x ) dx =
G
χ(x + g ) dx = χ(g )
G
χ(x ) dx ,
so G χ = 0 if χ nontrivial, 1 if χ is trivial (= 1).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 6 / 36
Orthogonality of characters on compact groups
8/3/2019 Ft Appl Talk
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If G is compact (=⇒ total Haar measure = 1) then characters are inL1(G ), being bounded.
If χ ∈ G then G
χ(x ) dx =
G
χ(x + g ) dx = χ(g )
G
χ(x ) dx ,
so G χ = 0 if χ nontrivial, 1 if χ is trivial (= 1).
If χ, ψ ∈ G then χ(x )ψ(−x ) is also a character. Hence
χ, ψ =
G
χ(x )ψ(x ) dx =
G
χ(x )ψ(−x ) dx =
1 χ = ψ0 χ = ψ
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 6 / 36
Orthogonality of characters on compact groups
8/3/2019 Ft Appl Talk
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If G is compact (=⇒ total Haar measure = 1) then characters are inL1(G ), being bounded.
If χ ∈ G then G
χ(x ) dx =
G
χ(x + g ) dx = χ(g )
G
χ(x ) dx ,
so G χ = 0 if χ nontrivial, 1 if χ is trivial (= 1).
If χ, ψ ∈ G then χ(x )ψ(−x ) is also a character. Hence
χ, ψ =
G
χ(x )ψ(x ) dx =
G
χ(x )ψ(−x ) dx =
1 χ = ψ0 χ = ψ
Fourier representation (inversion) in Zm: G = Zm =⇒ the m
characters form a complete orthonormal set in L2(G ):
f (x ) =m−1k =0
f (·), e 2πik ·e 2πikx =m−1k =0
f (k )e 2πikx
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 6 / 36
L2 of compact G
8/3/2019 Ft Appl Talk
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Trigonometric polynomials = finite linear combinations of characters
onG
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 7 / 36
L2 of compact G
8/3/2019 Ft Appl Talk
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Trigonometric polynomials = finite linear combinations of characters
onG
Example: G = T. Trig. polynomials are of the type N k =−N c k e 2πikx .
The least such N is called the degree of the polynomial.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 7 / 36
L2 of compact G
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 38/196
Trigonometric polynomials = finite linear combinations of characters
onG
Example: G = T. Trig. polynomials are of the type N k =−N c k e 2πikx .
The least such N is called the degree of the polynomial.
Example: G = R. Trig. polynomials are of the typeK
k =1 c k e 2πi λk x ,where λ j ∈ R.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 7 / 36
L2 of compact G
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 39/196
Trigonometric polynomials = finite linear combinations of characterson G
Example: G = T. Trig. polynomials are of the type N k =−N c k e 2πikx .
The least such N is called the degree of the polynomial.
Example: G = R. Trig. polynomials are of the typeK
k =1 c k e 2πi λk x ,where λ j ∈ R.
Compact G : Stone - Weierstrass Theorem =⇒ trig. polynomialsdense in C (G ) (in ·∞).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 7 / 36
L2 of compact G
8/3/2019 Ft Appl Talk
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Trigonometric polynomials = finite linear combinations of characterson G
Example: G = T. Trig. polynomials are of the type N k =−N c k e 2πikx .
The least such N is called the degree of the polynomial.
Example: G = R. Trig. polynomials are of the typeK
k =1 c k e 2πi λk x ,where λ j ∈ R.
Compact G : Stone - Weierstrass Theorem =⇒ trig. polynomialsdense in C (G ) (in ·∞).
Fourier representation in L2(G ): Compact G : The characters form a
complete ONS. Since C (G ) is dense in L2(G ):
f = χ∈ b G
f (χ)χ d χ all f ∈ L2(G ), convergence in L2(G )
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 7 / 36
L2 of compact G
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Trigonometric polynomials = finite linear combinations of characterson G
Example: G = T. Trig. polynomials are of the type N k =−N c k e 2πikx .
The least such N is called the degree of the polynomial.
Example: G = R. Trig. polynomials are of the typeK
k =1 c k e 2πi λk x ,where λ j ∈ R.
Compact G : Stone - Weierstrass Theorem =⇒ trig. polynomialsdense in C (G ) (in ·∞).
Fourier representation in L2(G ): Compact G : The characters form a
complete ONS. Since C (G ) is dense in L2(G ):
f = χ∈ b G
f (χ)χ d χ all f ∈ L2(G ), convergence in L2(G )
G necessarily discrete in this case
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 7 / 36
L2 of compact G , continued
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Compact G : Parseval formula:
G
f (x )g (x ) dx = b G
f (χ) g (χ) d χ.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 8 / 36
L2 of compact G , continued
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Compact G : Parseval formula:
G
f (x )g (x ) dx = b G
f (χ) g (χ) d χ.
Compact G : f →
f is an isometry from L2(G ) onto L2(
G ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 8 / 36
L2 of compact G , continued
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Compact G : Parseval formula:
G
f (x )g (x ) dx = b G
f (χ) g (χ) d χ.
Compact G : f →
f is an isometry from L2(G ) onto L2(
G ).
Example: G = T T
f (x )g (x ) dx =k ∈Z
f (k ) g (k ), f , g ∈ L2(T).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 8 / 36
L2 of compact G , continued
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Compact G : Parseval formula:
G
f (x )g (x ) dx = b G
f (χ) g (χ) d χ.
Compact G : f →
f is an isometry from L2(G ) onto L2(
G ).
Example: G = T T
f (x )g (x ) dx =k ∈Z
f (k ) g (k ), f , g ∈ L2(T).
Example: G = Zm
m−1 j =0
f ( j )g ( j ) =m−1k =0
f (k ) g (k ), all f , g : Zm → C
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 8 / 36
Triple correlations in Zp : an application
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Problem of significance in (a) crystallography, (b) astrophysics:determine a subset E ⊆ Zn from its triple correlation:
N E (a, b ) = #{x ∈ Zn : x , x + a, x + b ∈ E }, a, b ∈ Zn
=x ∈Zn
1E (x )1E (x + a)1E (x + b )
Counts number of occurences of translated 3-point patterns {0, a, b }.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 9 / 36
Triple correlations in Zp : an application
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Problem of significance in (a) crystallography, (b) astrophysics:determine a subset E ⊆ Zn from its triple correlation:
N E (a, b ) = #{x ∈ Zn : x , x + a, x + b ∈ E }, a, b ∈ Zn
=x ∈Zn
1E (x )1E (x + a)1E (x + b )
Counts number of occurences of translated 3-point patterns {0, a, b }.E can only be determined up to translation: E and E + t have thesame N (·, ·).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 9 / 36
Triple correlations in Zp : an application
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Problem of significance in (a) crystallography, (b) astrophysics:determine a subset E ⊆ Zn from its triple correlation:
N E (a, b ) = #{x ∈ Zn : x , x + a, x + b ∈ E }, a, b ∈ Zn
=x ∈Zn
1E (x )1E (x + a)1E (x + b )
Counts number of occurences of translated 3-point patterns {0, a, b }.E can only be determined up to translation: E and E + t have thesame N (·, ·).For general n it has been proved that N (·, ·) cannot determine E evenup to translation (non-trivial).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 9 / 36
Triple correlations in Zp : an application
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Problem of significance in (a) crystallography, (b) astrophysics:determine a subset E ⊆ Zn from its triple correlation:
N E (a, b ) = #{x ∈ Zn : x , x + a, x + b ∈ E }, a, b ∈ Zn
=x ∈Zn
1E (x )1E (x + a)1E (x + b )
Counts number of occurences of translated 3-point patterns {0, a, b }.E can only be determined up to translation: E and E + t have thesame N (·, ·).For general n it has been proved that N (·, ·) cannot determine E evenup to translation (non-trivial).
Special case: E can be determined up to translation from N (·, ·) if n = p is a prime.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 9 / 36
Triple correlations in Zp : an application
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Problem of significance in (a) crystallography, (b) astrophysics:determine a subset E ⊆ Zn from its triple correlation:
N E (a, b ) = #{x ∈ Zn : x , x + a, x + b ∈ E }, a, b ∈ Zn
=x ∈Zn
1E (x )1E (x + a)1E (x + b )
Counts number of occurences of translated 3-point patterns {0, a, b }.E can only be determined up to translation: E and E + t have thesame N (·, ·).For general n it has been proved that N (·, ·) cannot determine E evenup to translation (non-trivial).
Special case: E can be determined up to translation from N (·, ·) if n = p is a prime.Fourier transform of N E : Zn × Zn → R is easily computed:
N E (ξ, η) =
1E (ξ)
1E (η)
1E (−(ξ + η)), ξ, η ∈ Zn.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 9 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Setting η = 0, and using
f (−x ) =
f (x ) for real f , we get
1E
≡
1F
.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Setting η = 0, and using
f (−x ) =
f (x ) for real f , we get
1E
≡
1F
.
If 1F is never 0 we divide (1) by its RHS to get
φ(ξ)φ(η) = φ(ξ + η), where φ = 1E 1F (2)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Setting η = 0, and using
f (−x ) =
f (x ) for real f , we get
1E
≡
1F
.
If 1F is never 0 we divide (1) by its RHS to get
φ(ξ)φ(η) = φ(ξ + η), where φ = 1E 1F (2)
Hence φ : Zn → C is a character and 1E ≡ φ 1F .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Setting η = 0, and using
f (−x ) =
f (x ) for real f , we get
1E
≡
1F
.
If 1F is never 0 we divide (1) by its RHS to get
φ(ξ)φ(η) = φ(ξ + η), where φ = 1E 1F (2)
Hence φ : Zn → C is a character and 1E ≡ φ 1F .Since Zn = Zn we have φ(ξ) = e 2πit ξ/n for some t ∈ Zn
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Setting η = 0, and using
f (−x ) =
f (x ) for real f , we get
1E
≡
1F
.
If 1F is never 0 we divide (1) by its RHS to get
φ(ξ)φ(η) = φ(ξ + η), where φ = 1E 1F (2)
Hence φ : Zn → C is a character and 1E ≡ φ 1F .Since Zn = Zn we have φ(ξ) = e 2πit ξ/n for some t ∈ Zn
Hence E = F + t
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (continued)
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If N E ≡ N F for E , F ⊆ Zn then
1E (ξ) 1E (η) 1E (−(ξ + η)) = 1F (ξ) 1F (η) 1F (−(ξ + η)), ξ , η ∈ Zn (1)
Setting ξ = η = 0 we deduce #E = #F .
Setting η = 0, and using
f (−x ) =
f (x ) for real f , we get
1E
≡
1F
.
If 1F is never 0 we divide (1) by its RHS to get
φ(ξ)φ(η) = φ(ξ + η), where φ = 1E 1F (2)
Hence φ : Zn → C is a character and 1E ≡ φ 1F .Since Zn = Zn we have φ(ξ) = e 2πit ξ/n for some t ∈ Zn
Hence E = F + t
So N E determines E up to translation if
1E is never 0
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 10 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Each ζ ξ, ξ = 0, is a primitive p -th root of unity itself.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Each ζ ξ, ξ = 0, is a primitive p -th root of unity itself.
All powers (ζ ξ)s are distinct, so 1E (ξ) is a subset sum of all primitivep -th roots of unity (ξ = 0).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Each ζ ξ, ξ = 0, is a primitive p -th root of unity itself.
All powers (ζ ξ)s are distinct, so 1E (ξ) is a subset sum of all primitivep -th roots of unity (ξ = 0).
The polynomial 1 + x + x 2 + · · · + x p −1 is the minimal polynomial
over Q of each primitive root of unity (there are p − 1 of them).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Each ζ ξ, ξ = 0, is a primitive p -th root of unity itself.
All powers (ζ ξ)s are distinct, so 1E (ξ) is a subset sum of all primitivep -th roots of unity (ξ = 0).
The polynomial 1 + x + x 2 + · · · + x p −1 is the minimal polynomial
over Q of each primitive root of unity (there are p − 1 of them).
It divides any polynomial in Q[x ] which vanishes on some primitive
p -th root of unity
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Each ζ ξ, ξ = 0, is a primitive p -th root of unity itself.
All powers (ζ ξ)s are distinct, so 1E (ξ) is a subset sum of all primitivep -th roots of unity (ξ = 0).
The polynomial 1 + x + x 2 + · · · + x p −1 is the minimal polynomial
over Q of each primitive root of unity (there are p − 1 of them).
It divides any polynomial in Q[x ] which vanishes on some primitive
p -th root of unityThe only subset sums of all roots of unity which vanish are the emptyand the full sum (E = or E = Zp ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
Triple correlations in Zp : an application (conclusion)
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Suppose n = p is a prime, E ⊆ Zp . Then
1E (ξ) = 1p s ∈E
(ζ ξ)s , ζ = e −2πi /p is a p -root of unity. (3)
Each ζ ξ, ξ = 0, is a primitive p -th root of unity itself.
All powers (ζ ξ)s are distinct, so 1E (ξ) is a subset sum of all primitivep -th roots of unity (ξ = 0).
The polynomial 1 + x + x 2 + · · · + x p −1 is the minimal polynomial
over Q of each primitive root of unity (there are p − 1 of them).
It divides any polynomial in Q[x ] which vanishes on some primitive
p -th root of unityThe only subset sums of all roots of unity which vanish are the emptyand the full sum (E = or E = Zp ).
So in Zp the triple correlation N E (·, ·) determines E up to translation.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 11 / 36
The basics of the FT on the torus (circle) T = R/2πZ
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1 ≤ p ≤ q ⇐⇒ Lq (T) ⊆ Lp (T): nested Lp spaces. True on compactgroups.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 12 / 36
The basics of the FT on the torus (circle) T = R/2πZ
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1 ≤ p ≤ q ⇐⇒ Lq (T) ⊆ Lp (T): nested Lp spaces. True on compactgroups.
f ∈ L1(T): we write f (x ) ∼ ∞k =−∞ f (k )e 2πikx to denote the
Fourier series of f . No claim of convergence is made.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 12 / 36
The basics of the FT on the torus (circle) T = R/2πZ
( ) ( )
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1 ≤ p ≤ q ⇐⇒ Lq (T) ⊆ Lp (T): nested Lp spaces. True on compactgroups.
f ∈ L1(T): we write f (x ) ∼ ∞k =−∞ f (k )e 2πikx to denote the
Fourier series of f . No claim of convergence is made.
The Fourier coefficients of f (x ) = e 2πikx is the sequence f (n) = δk ,n.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 12 / 36
The basics of the FT on the torus (circle) T = R/2πZ
1 Lq(T) Lp(T) d Lp T
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1 ≤ p ≤ q ⇐⇒ Lq (T) ⊆ Lp (T): nested Lp spaces. True on compactgroups.
f ∈ L1(T): we write f (x ) ∼ ∞k =−∞ f (k )e 2πikx to denote the
Fourier series of f . No claim of convergence is made.
The Fourier coefficients of f (x ) = e 2πikx is the sequence f (n) = δk ,n.
The Fourier series of a trig. poly. f (x ) = N k =−N ak e 2πikx is the
sequence . . . , 0, 0, a−N , a−N +1, . . . , a0, . . . , aN , 0, 0, . . ..
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 12 / 36
The basics of the FT on the torus (circle) T = R/2πZ
1 ≤ ≤ Lq(T) ⊆ Lp(T) d Lp T
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1 ≤ p ≤ q ⇐⇒ Lq (T) ⊆ Lp (T): nested Lp spaces. True on compactgroups.
f ∈ L1(T): we write f (x ) ∼ ∞k =−∞ f (k )e 2πikx to denote the
Fourier series of f . No claim of convergence is made.
The Fourier coefficients of f (x ) = e 2πikx is the sequence f (n) = δk ,n.
The Fourier series of a trig. poly. f (x ) = N k =−N ak e 2πikx is the
sequence . . . , 0, 0, a−N , a−N +1, . . . , a0, . . . , aN , 0, 0, . . ..Symmetric partial sums of the Fourier series of f :S N (f ; x ) =
N k =−N
f (k )e 2πikx
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 12 / 36
The basics of the FT on the torus (circle) T = R/2πZ
1 ≤ ≤ ⇐⇒ Lq(T) ⊆ Lp(T) t d Lp T t
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1 ≤ p ≤ q ⇐⇒ Lq (T) ⊆ Lp (T): nested Lp spaces. True on compactgroups.
f ∈ L1(T): we write f (x ) ∼ ∞k =−∞ f (k )e 2πikx to denote the
Fourier series of f . No claim of convergence is made.
The Fourier coefficients of f (x ) = e 2πikx is the sequence f (n) = δk ,n.
The Fourier series of a trig. poly. f (x ) = N k =−N ak e 2πikx is the
sequence . . . , 0, 0, a−N , a−N +1, . . . , a0, . . . , aN , 0, 0, . . ..Symmetric partial sums of the Fourier series of f :S N (f ; x ) =
N k =−N
f (k )e 2πikx
From f ∗ g =
f ·
g we get easily S N (f ; x ) = f (x ) ∗ D N (x ), where
D N (x ) =N
k =−N
e 2πikx =sin2π(N + 1
2 )x
sin πx (Dirichlet kernel of order N
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 12 / 36
The Dirichlet kernel
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-5
0
5
10
15
20
25
-0.4 -0.2 0 0.2 0.4
The Dirichlet kernel D N (x ) for N = 10
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 13 / 36
Pointwise convergence
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Important: D N 1 ≥ C log N , as N → ∞
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 14 / 36
Pointwise convergence
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Important: D N 1 ≥ C log N , as N → ∞
T N : f → S N (f ; x ) = D N ∗ f (x ) is a (continuous) linear functionalC (T) → C. From the inequality D N ∗ f ∞ ≤ D N 1f ∞
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 14 / 36
Pointwise convergence
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Important: D N 1 ≥ C log N , as N → ∞
T N : f → S N (f ; x ) = D N ∗ f (x ) is a (continuous) linear functionalC (T) → C. From the inequality D N ∗ f ∞ ≤ D N 1f ∞T N = D N 1 is unbounded
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 14 / 36
Pointwise convergence
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Important: D N 1 ≥ C log N , as N → ∞
T N : f → S N (f ; x ) = D N ∗ f (x ) is a (continuous) linear functionalC (T) → C. From the inequality D N ∗ f ∞ ≤ D N 1f ∞T N = D N 1 is unbounded
Banach-Steinhaus (uniform boundedness principle) =⇒Given x there are many continuous functions f such that T N (f ) isunbounded
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 14 / 36
Pointwise convergence
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Important: D N 1 ≥ C log N , as N → ∞
T N : f → S N (f ; x ) = D N ∗ f (x ) is a (continuous) linear functionalC (T) → C. From the inequality D N ∗ f ∞ ≤ D N 1f ∞T N = D N 1 is unbounded
Banach-Steinhaus (uniform boundedness principle) =⇒Given x there are many continuous functions f such that T N (f ) isunbounded
Consequence: In general S N (f ; x ) does not converge pointwise to
f (x ), even for continuous f
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 14 / 36
Summability
Look at the arithmetical means of S (f ; x)
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Look at the arithmetical means of S N (f ; x )
σN (f ; x ) =1
N + 1
N n=0
S n(f ; x ) = K N ∗ f (x )
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 15 / 36
Summability
Look at the arithmetical means of S (f ; x)
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Look at the arithmetical means of S N (f ; x )
σN (f ; x ) =1
N + 1
N n=0
S n(f ; x ) = K N ∗ f (x )
The Fejer kernel K N (x ) is the mean of the Dirichlet kernels
K N (x ) =N
n=−N
1 −
|n|
N + 1
e 2πinx =
1
N + 1
sin π(N + 1)x
sin πx
2
≥ 0.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 15 / 36
Summability
Look at the arithmetical means of SN(f ; x)
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Look at the arithmetical means of S N (f ; x )
σN (f ; x ) =1
N + 1
N n=0
S n(f ; x ) = K N ∗ f (x )
The Fejer kernel K N (x ) is the mean of the Dirichlet kernels
K N (x ) =N
n=−N
1 −
|n|
N + 1
e 2πinx =
1
N + 1
sin π(N + 1)x
sin πx
2
≥ 0.
K N (x ) is an approximate identity:
(a) T
K N (x ) dx = K N (0) = 1,(b) K N 1 is bounded (K N 1 = 1, from nonnegativity and (a)),(c) for any > 0 we have
|x |> |K N (x )| dx → 0, as N → ∞
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 15 / 36
The Fejer kernel
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0
2
4
6
8
10
12
-0.4 -0.2 0 0.2 0.4
The Fej’er kernel D N (x ) for N = 10
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 16 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some Banach
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spaces. These can be:
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 17 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some Banach
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spaces. These can be:
C (T) normed with ·∞: If f ∈ C (T) then σN (f ; x ) → f (x )uniformly in T.
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 17 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some Banach
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spaces. These can be:
C (T) normed with ·∞: If f ∈ C (T) then σN (f ; x ) → f (x )uniformly in T.
Lp (T), 1 ≤ p < ∞: If f ∈ Lp (T) then σN (f ; x ) − f (x )p → 0
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 17 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some BanachTh b
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spaces. These can be:
C (T) normed with ·∞: If f ∈ C (T) then σN (f ; x ) → f (x )uniformly in T.
Lp (T), 1 ≤ p < ∞: If f ∈ Lp (T) then σN (f ; x ) − f (x )p → 0
C n(T), all n-times C -differentiable functions, normed withf C n =
nk =0 f (k )∞
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 17 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some BanachTh b
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spaces. These can be:
C (T) normed with ·∞: If f ∈ C (T) then σN (f ; x ) → f (x )uniformly in T.
Lp (T), 1 ≤ p < ∞: If f ∈ Lp (T) then σN (f ; x ) − f (x )p → 0
C n(T), all n-times C -differentiable functions, normed withf C n =
nk =0 f (k )∞Summability implies uniqueness: the Fourier series of f ∈ L1(T)
determines the function.
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 17 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some BanachTh b
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spaces. These can be:
C (T) normed with ·∞: If f ∈ C (T) then σN (f ; x ) → f (x )uniformly in T.
Lp (T), 1 ≤ p < ∞: If f ∈ Lp (T) then σN (f ; x ) − f (x )p → 0
C n(T), all n-times C -differentiable functions, normed withf C n =
nk =0 f (k )∞Summability implies uniqueness: the Fourier series of f ∈ L1(T)
determines the function.
Another consequence: trig. polynomials are dense inLp (T), C (T), C n(T)
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 17 / 36
Summability (continued)
K N approximate identity =⇒ K N ∗ f (x ) → f (x ), in some Banachs aces These ca be
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spaces. These can be:
C (T) normed with ·∞: If f ∈ C (T) then σN (f ; x ) → f (x )uniformly in T.
Lp (T), 1 ≤ p < ∞: If f ∈ Lp (T) then σN (f ; x ) − f (x )p → 0
C n(T), all n-times C -differentiable functions, normed withf C n =
nk =0 f (k )∞Summability implies uniqueness: the Fourier series of f ∈ L1(T)
determines the function.
Another consequence: trig. polynomials are dense inLp (T), C (T), C n(T)
Another important summability kernel: thePoisson
kernelP (r , x ) =
k ∈Z
r k e 2πikx , 0 < r < 1: absolute convergence obvious
Significant for the theory of analytic functions.
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 17 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
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Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).Obviously true for trig. polynomials and they are dense in L1(T).
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
f (x ) = x
0 g (t ) dt , where
g = 0:
f (n) = 1
2πin
g (n) (Fubini)
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
f (x ) = x
0 g (t ) dt , where
g = 0:
f (n) = 1
2πin
g (n) (Fubini)
Previous implies: f (|n|) = − f (−|n|) ≥ 0 =⇒ n=0 f (n)n < ∞.
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
f (x ) = x
0 g (t ) dt , where
g = 0:
f (n) = 1
2πin
g (n) (Fubini)
Previous implies: f (|n|) = − f (−|n|) ≥ 0 =⇒ n=0 f (n)n < ∞.n>0
sin nx log n is not a Fourier series.
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
f (x ) = x
0 g (t ) dt , where
g = 0:
f (n) = 1
2πin
g (n) (Fubini)
Previous implies: f (|n|) = − f (−|n|) ≥ 0 =⇒ n=0 f (n)n < ∞.n>0
sin nx log n is not a Fourier series.
f is an integral =⇒ f (n) = o (1/n): the “smoother” f is the betterdecay for the FT of f
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
1
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).
Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
f (x ) = x
0 g (t ) dt , where
g = 0:
f (n) = 1
2πin
g (n) (Fubini)
Previous implies: f (|n|) = − f (−|n|) ≥ 0 =⇒ n=0 f (n)n < ∞.n>0
sin nx log n is not a Fourier series.
f is an integral =⇒ f (n) = o (1/n): the “smoother” f is the betterdecay for the FT of f
f ∈ C 2
(T) =⇒ absolute convergence for the Fourier Series of f .
Mihalis Kolountzakis (U of Crete) FT and applications January 2006 18 / 36
The decay of the Fourier coefficients at ∞
Obvious: f (n) ≤ f 1
( ) 1( )
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Riemann-Lebesgue Lemma: lim|n|→∞ f (n) = 0 if f ∈ L1(T).
Obviously true for trig. polynomials and they are dense in L1(T).
Can go to 0 arbitrarily slowly if we only assume f ∈ L1.
f (x ) = x
0 g (t ) dt , where
g = 0:
f (n) = 1
2πin
g (n) (Fubini)
Previous implies: f (|n|) = − f (−|n|) ≥ 0 =⇒ n=0 f (n)n < ∞.n>0
sin nx log n is not a Fourier series.
f is an integral =⇒ f (n) = o (1/n): the “smoother” f is the betterdecay for the FT of f
f ∈ C 2
(T) =⇒ absolute convergence for the Fourier Series of f .Another condition that imposes “decay”:
f ∈ L2(T) =⇒
n
f (n)2 < ∞.
Mih lis K l t kis (U f C t ) FT d li ti s J 2006 18 / 36
Interpolation of operators
T is bounded linear operator on dense subsets of Lp 1 and Lp 2 :
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T is bounded linear operator on dense subsets of Lp and Lp :
Tf q 1 ≤ C 1f p 1 , Tf q 2 ≤ C 2f p 2
Mih li K l t ki (U f C t ) FT d li ti J 2006 19 / 36
Interpolation of operators
T is bounded linear operator on dense subsets of Lp 1 and Lp 2 :
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T is bounded linear operator on dense subsets of L and L :
Tf q 1 ≤ C 1f p 1 , Tf q 2 ≤ C 2f p 2
Riesz-Thorin interpolation theorem: T : Lp → Lq for any p
between p 1, p 2 (all p ’s and q ’s ≥ 1).
Mih li K l t ki (U f C t ) FT d li ti J 2006 19 / 36
Interpolation of operators
T is bounded linear operator on dense subsets of Lp 1 and Lp 2 :
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T is bounded linear operator on dense subsets of L and L :
Tf q 1 ≤ C 1f p 1 , Tf q 2 ≤ C 2f p 2
Riesz-Thorin interpolation theorem: T : Lp → Lq for any p
between p 1, p 2 (all p ’s and q ’s ≥ 1).
p and q are related by:
1
p = t
1
p 1+ (1 − t )
1
p 2=⇒
1
q = t
1
q 1+ (1 − t )
1
q 2
Mih li K l t ki (U f C t ) FT d li ti J 2006 19 / 36
Interpolation of operators
T is bounded linear operator on dense subsets of Lp 1 and Lp 2 :
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T is bounded linear operator on dense subsets of L and L :
Tf q 1 ≤ C 1f p 1 , Tf q 2 ≤ C 2f p 2
Riesz-Thorin interpolation theorem: T : Lp → Lq for any p
between p 1, p 2 (all p ’s and q ’s ≥ 1).
p and q are related by:
1
p = t
1
p 1+ (1 − t )
1
p 2=⇒
1
q = t
1
q 1+ (1 − t )
1
q 2
T Lp →Lq ≤ C t 1 C (1−t )2
Mih li K l ki (U f C ) FT d li i J 2006 19 / 36
Interpolation of operators
T is bounded linear operator on dense subsets of Lp 1 and Lp 2 :
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p
Tf q 1 ≤ C 1f p 1 , Tf q 2 ≤ C 2f p 2
Riesz-Thorin interpolation theorem: T : Lp → Lq for any p
between p 1, p 2 (all p ’s and q ’s ≥ 1).
p and q are related by:
1
p = t
1
p 1+ (1 − t )
1
p 2=⇒
1
q = t
1
q 1+ (1 − t )
1
q 2
T Lp →Lq ≤ C t 1 C (1−t )2
The exponents p , q , . . . are allowed to be ∞.
Mih li K l ki (U f C ) FT d li i J 2006 19 / 36
Interpolation of operators: the 1p , 1q plane
1
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1
p
s
s
S S S S S S
S S S S S
0 1
0
1
q
(1/p 1, 1/q 1)
(1/p 2, 1/q 2)
(1/p , 1/q )
s
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 20 / 36
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The Hausdorff-Young inequality
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Hausdorff-Young: Suppose 1 ≤ p ≤ 2, 1p + 1q = 1, andf ∈ Lp (T). It follows that
f Lq (Z)
≤ C p f Lp (T)
False if p > 2.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 21 / 36
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The Hausdorff-Young inequality
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Hausdorff-Young: Suppose 1 ≤ p ≤ 2, 1p + 1q = 1, andf ∈ Lp (T). It follows that
f Lq (Z)
≤ C p f Lp (T)
False if p > 2.
Clearly true if p = 1 (trivial) or p = 2 (Parseval).
Use Riesz-Thorin interpolation for 1 < p < 2 for the operatorf → f from Lp (T) → Lq (Z).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 21 / 36
An application: the isoperimetric inequality
Suppose Γ is a simple closed curve in the plane with perimeter L
enclosing area A.
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A ≤ 14π
L2 (isoperimetric inequality)
Equality holds only when Γ is a circle.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 22 / 36
An application: the isoperimetric inequality
Suppose Γ is a simple closed curve in the plane with perimeter L
enclosing area A.
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A ≤ 14π
L2 (isoperimetric inequality)
Equality holds only when Γ is a circle.
Wirtinger’s inequality: if f ∈ C ∞(T) then
1
0f (x ) − f (0)2 dx ≤ 1
4π2 1
0f (x )2 dx . (4)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 22 / 36
An application: the isoperimetric inequality
Suppose Γ is a simple closed curve in the plane with perimeter L
enclosing area A.
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A ≤ 14π
L2 (isoperimetric inequality)
Equality holds only when Γ is a circle.
Wirtinger’s inequality: if f ∈ C ∞(T) then
1
0f (x ) − f (0)2 dx ≤ 1
4π2 1
0f (x )2 dx . (4)
By smoothness f (x ) equals its Fourier series and so doesf (x ) = 2πi
n n
f (n)e 2πinx
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 22 / 36
An application: the isoperimetric inequality
Suppose Γ is a simple closed curve in the plane with perimeter L
enclosing area A.
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A ≤ 14πL2 (isoperimetric inequality)
Equality holds only when Γ is a circle.
Wirtinger’s inequality: if f ∈ C ∞(T) then
1
0f (x ) − f (0)2 dx ≤ 1
4π2 1
0f (x )2 dx . (4)
By smoothness f (x ) equals its Fourier series and so doesf (x ) = 2πi
n n
f (n)e 2πinx
FT is an isometry (Parseval) so LHS of (4) is n=0 f (n)2 while the
RHS is
n=0 n f (n)
2 so (4) holds.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 22 / 36
An application: the isoperimetric inequality
Suppose Γ is a simple closed curve in the plane with perimeter L
enclosing area A.
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A ≤ 14πL2 (isoperimetric inequality)
Equality holds only when Γ is a circle.
Wirtinger’s inequality: if f ∈ C ∞(T) then
1
0f (x ) − f (0)2 dx ≤ 1
4π2 1
0f (x )2 dx . (4)
By smoothness f (x ) equals its Fourier series and so doesf (x ) = 2πi
n n
f (n)e 2πinx
FT is an isometry (Parseval) so LHS of (4) is n=0 f (n)2 while the
RHS is
n=0 n f (n)
2 so (4) holds.
Equality in (4) precisely when f (x ) = f (−1)e −2πix + f (0) + f (1)e 2πix .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 22 / 36
An application: the isoperimetric inequality (continued)
Hurwitz’ proof. First assume Γ is smooth, has L = 1.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 23 / 36
An application: the isoperimetric inequality (continued)
Hurwitz’ proof. First assume Γ is smooth, has L = 1.Parametrization of Γ: (x (s ), y (s )), 0 ≤ s ≤ 1 w.r.t. arc length s
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 23 / 36
An application: the isoperimetric inequality (continued)
Hurwitz’ proof. First assume Γ is smooth, has L = 1.Parametrization of Γ: (x (s ), y (s )), 0 ≤ s ≤ 1 w.r.t. arc length s
x , y ∈ C ∞(T), (x (s ))2 + (y (s ))2 = 1.
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, y ( ), ( ( )) + (y ( ))
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 23 / 36
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An application: the isoperimetric inequality (continued)
Hurwitz’ proof. First assume Γ is smooth, has L = 1.Parametrization of Γ: (x (s ), y (s )), 0 ≤ s ≤ 1 w.r.t. arc length s
x , y ∈ C ∞(T), (x (s ))2 + (y (s ))2 = 1.
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, y ( ), ( ( )) (y ( ))
Green’s Theorem =⇒ area A = 10 x (s )y (s ) ds :
A =
(x (s ) − x (0))y (s ) =
=
1
4π (2π(x (s ) − x (0)))
2
+ y
(s )
2
− (2π(x (s ) − x (0)) − y
(s ))
2
≤ 1
4π
4π2(x (s ) − x (0))2 + y (s )2 (drop last term)
≤ 14π x (s )2 + y (s )2 (Wirtinger’s ineq)
= 1
4π
For equality must have x (s ) = a cos2πs + b sin2πs + c ,y (s ) = 2π(x (s ) − x (0)). So x (s )2 + y (s )2 constant if c = 0.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 23 / 36
Fourier transform on Rn
Initially defined only for f ∈ L1(Rn). f (ξ) = Rn f (x )e −2πi ξ·x dx .
Follows: f ∞
≤ f 1. f is continuous.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 24 / 36
Fourier transform on Rn
Initially defined only for f ∈ L1(Rn). f (ξ) = Rn f (x )e −2πi ξ·x dx .
Follows: f ∞
≤ f 1. f is continuous.
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Trig. polynomials are not dense anymore in the usual spaces.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 24 / 36
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8/3/2019 Ft Appl Talk
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Fourier transform on Rn
Initially defined only for f ∈ L1(Rn). f (ξ) = Rn f (x )e −2πi ξ·x dx .
Follows: f ∞
≤ f 1. f is continuous.
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Trig. polynomials are not dense anymore in the usual spaces.
But Riemann-Lebesgue is true. First for indicator function of aninterval
[a1, b 1] × · · · × [an, b n].
Then approximate an L1 function by finite linear combinations of such.
Multi-index notation α = (α1, . . . , αn) ∈ Nn:(a) |α| = α1 + · · · + αn.
(b) x α
= x α1
1 x α2
2 · · · x αn
n(c) ∂ α = (∂
∂ 1)α1 · · · (∂
∂ n)αn
Diff operators D j φ := 12πi (∂
∂ x j ), D αφ = (1
2πi )|α|∂ α.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 24 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
8/3/2019 Ft Appl Talk
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
Schwartz class S : those φ ∈ C ∞(Rn) s.t. for all multiindices α, γ
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φα,γ := sup
x ∈Rn
|x γ ∂ αφ(x )| < ∞.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
Schwartz class S : those φ ∈ C ∞(Rn) s.t. for all multiindices α, γ
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φα,γ := sup
x ∈Rn
|x γ ∂ αφ(x )| < ∞.
The φα,γ are seminorms . They determine the topology of S .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
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Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
Schwartz class S : those φ ∈ C ∞(Rn) s.t. for all multiindices α, γ
8/3/2019 Ft Appl Talk
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φα,γ := supx ∈Rn
|x γ ∂ αφ(x )| < ∞.
The φα,γ are seminorms . They determine the topology of S .
C ∞0 (Rn) ⊆ S Easy to see that D j (φ)(ξ) = ξ j φ(ξ) and x j φ(x )(ξ) = −D j φ(ξ).
More generally ξαD γ φ(ξ) = D α(−x )γ φ(x )(ξ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
Schwartz class S : those φ ∈ C ∞(Rn) s.t. for all multiindices α, γ
8/3/2019 Ft Appl Talk
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φα,γ := supx ∈Rn
|x γ ∂ αφ(x )| < ∞.
The φα,γ are seminorms . They determine the topology of S .
C ∞0 (Rn) ⊆ S Easy to see that D j (φ)(ξ) = ξ j φ(ξ) and x j φ(x )(ξ) = −D j φ(ξ).
More generally ξαD γ φ(ξ) = D α(−x )γ φ(x )(ξ).
φ ∈ S =⇒
φ ∈ S (smoothness =⇒ decay, decay =⇒ smoothness)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
Schwartz class S : those φ ∈ C ∞(Rn) s.t. for all multiindices α, γ
8/3/2019 Ft Appl Talk
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φα,γ := supx ∈Rn
|x γ ∂ αφ(x )| < ∞.
The φα,γ are seminorms . They determine the topology of S .
C ∞0 (Rn
) ⊆ S Easy to see that D j (φ)(ξ) = ξ j φ(ξ) and x j φ(x )(ξ) = −D j φ(ξ).
More generally ξαD γ φ(ξ) = D α(−x )γ φ(x )(ξ).
φ ∈ S =⇒
φ ∈ S (smoothness =⇒ decay, decay =⇒ smoothness)
Fourier inversion formula: φ(x ) = φ(ξ)e 2πi ξx d ξ.
Can also write as φ(−x ) = φ(x ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Schwartzfunctions on R
n
Lp (Rn) spaces are not nested.
Not clear how to define
f for f ∈ L2.
Schwartz class S : those φ ∈ C ∞(Rn) s.t. for all multiindices α, γ
8/3/2019 Ft Appl Talk
http://slidepdf.com/reader/full/ft-appl-talk 131/196
φα,γ := supx ∈Rn
|x γ ∂ αφ(x )| < ∞.
The φα,γ are seminorms . They determine the topology of S .
C ∞0 (Rn
) ⊆ S Easy to see that D j (φ)(ξ) = ξ j φ(ξ) and x j φ(x )(ξ) = −D j φ(ξ).
More generally ξαD γ φ(ξ) = D α(−x )γ φ(x )(ξ).
φ ∈ S =⇒
φ ∈ S (smoothness =⇒ decay, decay =⇒ smoothness)
Fourier inversion formula: φ(x ) = φ(ξ)e 2πi ξx d ξ.
Can also write as φ(−x ) = φ(x ).
We first show its validity for φ ∈ S .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 25 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
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Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x) = |x |2
g (x) = 1
8/3/2019 Ft Appl Talk
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This normalization gives g (x ) = |x | g (x ) = 1.Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x) = |x |2
g (x) = 1
8/3/2019 Ft Appl Talk
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This normalization gives g (x ) = |x | g (x ) = 1.Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x) = |x |2
g (x) = 1
8/3/2019 Ft Appl Talk
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This normalization gives g (x ) |x | g (x ) 1.Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
We have g (ξ) = (2π)n/2
g (2πξ), lim→0 g (ξ) = 1.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x) = |x |2
g (x) = 1.
8/3/2019 Ft Appl Talk
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This normalization gives g (x ) |x | g (x ) 1.Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
We have g (ξ) = (2π)n/2
g (2πξ), lim→0 g (ξ) = 1. φ(−x ) =
e 2πi ξx φ(ξ) d ξ = lim→0
e 2πi ξx φ(ξ) g (ξ) d ξ (dom. conv.)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x ) = |x |2
g (x ) = 1.
8/3/2019 Ft Appl Talk
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This normalization gives g ( ) | | g ( ) 1.Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
We have g (ξ) = (2π)n/2
g (2πξ), lim→0 g (ξ) = 1. φ(−x ) =
e 2πi ξx φ(ξ) d ξ = lim→0
e 2πi ξx φ(ξ) g (ξ) d ξ (dom. conv.)
= lim→0
φ(· + x )(ξ)
g (ξ) d ξ (FT of translation)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula onS
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x ) = |x |2
g (x ) = 1.
8/3/2019 Ft Appl Talk
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g g ( ) | | g ( )Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
We have g (ξ) = (2π)n/2
g (2πξ), lim→0 g (ξ) = 1. φ(−x ) =
e 2πi ξx φ(ξ) d ξ = lim→0
e 2πi ξx φ(ξ) g (ξ) d ξ (dom. conv.)
= lim→0
φ(· + x )(ξ)
g (ξ) d ξ (FT of translation)
= lim→0 φ(x + y ) g (y ) dy ( shifting)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x ) = |x |2
g (x ) = 1.
8/3/2019 Ft Appl Talk
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g g ( ) | | g ( )Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
We have g (ξ) = (2π)n/2
g (2πξ), lim→0 g (ξ) = 1. φ(−x ) =
e 2πi ξx φ(ξ) d ξ = lim→0
e 2πi ξx φ(ξ) g (ξ) d ξ (dom. conv.)
= lim→0
φ(· + x )(ξ)
g (ξ) d ξ (FT of translation)
= lim→0 φ(x + y ) g (y ) dy ( shifting)
= lim→0 φ(x + y )g (−y ) dy (FT inversion for g ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
Fourier inversion formula on S
shifting: f , g ∈ L1(Rn) =⇒
f g = f g (Fubini)
Define the Gaussian function g (x ) = (2π)−n/2e −|x |2/2, x ∈ Rn.
This normalization gives g (x ) = |x |2
g (x ) = 1.
8/3/2019 Ft Appl Talk
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g g ( ) | | g ( )Using Cauchy’s integral formula for analytic functions we prove g (ξ) = (2π)n/2g (2πξ). The Fourier inversion formula holds.
Write g (x ) = −ng (x /), an approximate identity.
We have g (ξ) = (2π)n/2
g (2πξ), lim→0 g (ξ) = 1. φ(−x ) =
e 2πi ξx φ(ξ) d ξ = lim→0
e 2πi ξx φ(ξ) g (ξ) d ξ (dom. conv.)
= lim→0
φ(· + x )(ξ)
g (ξ) d ξ (FT of translation)
= lim→0 φ(x + y ) g (y ) dy ( shifting)
= lim→0 φ(x + y )g (−y ) dy (FT inversion for g ).
= φ(x ) (g is an approximate identity)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 26 / 36
FT on L
2
(R
n
)
Preservation of inner product:
φψ =
φ
ψ, for φ, ψ ∈ S
Fourier inversion implies ψ = ψ. Use shifting.
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Fourier inversion implies ψ ψ. Use shifting.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 27 / 36
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FT on L
2
(Rn
)
Preservation of inner product:
φψ =
φ
ψ, for φ, ψ ∈ S
Fourier inversion implies ψ = ψ. Use shifting.
8/3/2019 Ft Appl Talk
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p ψ ψ g
Parseval: f ∈ S : f 2 = f
2.
S dense in L2(Rn): FT extends to L2 and f →
f is an isometry on L2.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 27 / 36
FT onL2
(Rn
)
Preservation of inner product:
φψ =
φ
ψ, for φ, ψ ∈ S
Fourier inversion implies ψ = ψ. Use shifting.
8/3/2019 Ft Appl Talk
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p ψ ψ g
Parseval: f ∈ S : f 2 = f
2.
S dense in L2(Rn): FT extends to L2 and f →
f is an isometry on L2.
By interpolation FT is defined on Lp , 1 ≤ p ≤ 2, and satisfies theHausdorff-Young inequality:
f q
≤ C p f p ,1
p +
1
q = 1.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 27 / 36
Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
FT defined on S : for u ∈ S we define u (φ) = u ( φ), for φ ∈ S .
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
FT defined on S : for u ∈ S we define u (φ) = u ( φ), for φ ∈ S .
Fourier inversion for S : u (φ(x )) = u (φ(−x )).
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(φ( )) (φ( ))
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
FT defined on S : for u ∈ S we define u (φ) = u ( φ), for φ ∈ S .
Fourier inversion for S : u (φ(x )) = u (φ(−x )).
8/3/2019 Ft Appl Talk
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(φ( )) (φ( ))
u → u is an isomorphism on S
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
FT defined on S : for u ∈ S we define u (φ) = u ( φ), for φ ∈ S .
Fourier inversion for S : u (φ(x )) = u (φ(−x )).
8/3/2019 Ft Appl Talk
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( ( )) ( ( ))
u → u is an isomorphism on S
1 ≤ p ≤ ∞ : Lp ⊆ S . If f ∈ Lp this mapping is in S :
φ → f φ
Also S ⊆ S .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
FT defined on S : for u ∈ S we define u (φ) = u ( φ), for φ ∈ S .
Fourier inversion for S : u (φ(x )) = u (φ(−x )).
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u → u is an isomorphism on S
1 ≤ p ≤ ∞ : Lp ⊆ S . If f ∈ Lp this mapping is in S :
φ → f φ
Also S ⊆ S .
Tempered measures:
(1 + |x |)−k d |µ|(x ) < ∞, for some k ∈ N.
These are in S .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
8/3/2019 Ft Appl Talk
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Tempered distributions
Tempered distributions: S is the space of continuous linearfunctionals on S .
FT defined on S : for u ∈ S we define u (φ) = u ( φ), for φ ∈ S .
Fourier inversion for S : u (φ(x )) = u (φ(−x )).
8/3/2019 Ft Appl Talk
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u → u is an isomorphism on S
1 ≤ p ≤ ∞ : Lp ⊆ S . If f ∈ Lp this mapping is in S :
φ → f φ
Also S ⊆ S .
Tempered measures:
(1 + |x |)−k d |µ|(x ) < ∞, for some k ∈ N.
These are in S .Differentiation defined as: (∂ αu )(φ) = (−1)|α|u (∂ αφ).
FT of Lp functions or tempered measures defined in S
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 28 / 36
Examples of tempered distributions and their FT
u = δ0,
u = 1
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 29 / 36
Examples of tempered distributions and their FT
u = δ0,
u = 1
u = D α
δ0. To find its FT
( ) ( )() ( )| | ( ) ( )| | ( ( ) ( ))
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D αδ0(φ) = (D αδ0)( φ) = (−1)|α|δ0(D α φ) = (−1)|α|δ0((−x )αφ(x ))
= (−1)|α|
δ0((−x )αφ(x )) =
x αφ(x )
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 29 / 36
Examples of tempered distributions and their FT
u = δ0,
u = 1
u = D α
δ0. To find its FT
D δ (φ) (Dαδ )(φ) ( )|α|δ (Dαφ) ( )|α|δ ( ( ) φ( ))
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D αδ0(φ) = (D αδ0)( φ) = (−1)|α|δ0(D α φ) = (−1)|α|δ0((−x )αφ(x ))
= (−1)|α|
δ0((−x )αφ(x )) =
x αφ(x )
So D αδ0 = x α.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 29 / 36
Examples of tempered distributions and their FT
u = δ0,
u = 1
u = D α
δ0. To find its FT
D δ (φ) (Dαδ )(φ) ( 1)|α|δ (Dαφ) ( 1)|α|δ ( ( ) φ( ))
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D αδ0(φ) = (D αδ0)( φ) = (−1)|α|δ0(D α φ) = (−1)|α|δ0((−x )αφ(x ))
= (−1)|α|
δ0((−x )αφ(x )) =
x αφ(x )
So D αδ0 = x α.
u = x α, u = D αδ0
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 29 / 36
Examples of tempered distributions and their FT
u = δ0,
u = 1
u = D
α
δ0. To find its FT
Dαδ (φ) (Dαδ )(φ) ( 1)|α|δ (Dαφ) ( 1)|α|δ ( ( )αφ( ))
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D αδ0(φ) = (D αδ0)( φ) = (−1)|α|δ0(D α φ) = (−1)|α|δ0((−x )αφ(x ))
= (−1)|α|
δ0((−x )αφ(x )) =
x αφ(x )
So D αδ0 = x α.
u = x α, u = D αδ0
u =
J j =1 a j δp j ,
u (ξ) =
J j =1 a j e 2πip j ξ.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 29 / 36
Examples of tempered distributions and their FT
u = δ0,
u = 1
u = D
α
δ0. To find its FT
Dαδ (φ) (Dαδ )(φ) ( 1)|α|δ (Dαφ) ( 1)|α|δ ( ( )αφ( ))
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D αδ0(φ) = (D αδ0)( φ) = (−1)|α|δ0(D α φ) = (−1)|α|δ0((−x )αφ(x ))
= (−1)|α|
δ0((−x )αφ(x )) =
x αφ(x )
So D αδ0 = x α.
u = x α, u = D αδ0
u =
J j =1 a j δp j ,
u (ξ) =
J j =1 a j e 2πip j ξ.
Poisson Summation Formula (PSF): u = k ∈Zn δk , u = u
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 29 / 36
Proof of the Poisson Summation Formula
φ ∈ S . Define g (x ) =
k ∈Zn φ(x + k ).
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 30 / 36
Proof of the Poisson Summation Formula
φ ∈ S . Define g (x ) =
k ∈Zn φ(x + k ).
g has Zn as a period lattice: g (x + k ) = g (x ), x ∈ Rn, k ∈ Zn.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 30 / 36
Proof of the Poisson Summation Formula
φ ∈ S . Define g (x ) =
k ∈Zn φ(x + k ).
g has Zn as a period lattice: g (x + k ) = g (x ), x ∈ Rn, k ∈ Zn.
The periodization g may be viewed as g : Tn
→ C.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 30 / 36
Proof of the Poisson Summation Formula
φ ∈ S . Define g (x ) =
k ∈Zn φ(x + k ).
g has Zn as a period lattice: g (x + k ) = g (x ), x ∈ Rn, k ∈ Zn.
The periodization g may be viewed as g : Tn
→ C.FT of g lives on Tn = Zn. The Fourier coefficients are
8/3/2019 Ft Appl Talk
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g
g (k ) =
Tn m∈Zn
φ(x + m)e −2πikx dx =
φ(k ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 30 / 36
8/3/2019 Ft Appl Talk
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Proof of the Poisson Summation Formula
φ ∈ S . Define g (x ) =
k ∈Zn φ(x + k ).
g has Zn as a period lattice: g (x + k ) = g (x ), x ∈ Rn, k ∈ Zn.
The periodization g may be viewed as g : Tn
→ C.FT of g lives on Tn = Zn. The Fourier coefficients are
8/3/2019 Ft Appl Talk
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g (k ) =
Tn m∈Zn
φ(x + m)e −2πikx dx =
φ(k ).
From decay of φ follows that the FS of g (x ) converges absolutely anduniformly and
g (x ) =
k ∈Zn φ(k )e 2πikx .
x = 0 gives the PSF:
k ∈Zn φ(k ) =
k ∈Znφ(k ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 30 / 36
FT behavior under linear transformation
Suppose T : Rn → Rn is a non-singular linear operator. u ∈ S ,v = u ◦ T .
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 31 / 36
FT behavior under linear transformation
Suppose T : Rn → Rn is a non-singular linear operator. u ∈ S ,v = u ◦ T .
Change of variables formula for integration implies
1 T−
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v =|det T |
u ◦ T −.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 31 / 36
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FT behavior under linear transformation
Suppose T : Rn → Rn is a non-singular linear operator. u ∈ S ,v = u ◦ T .
Change of variables formula for integration implies
1 T−
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v =|det T |
u ◦ T .
WriteRn
=H
⊕H ⊥
,H
a linear subspace.Projection onto subspace defined by
(πH f )(h) :=
H
f (h + x ) dx , (h ∈ H ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 31 / 36
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Analyticity of the FT
Compact support: f : R→ C, f ∈ L1(R), f (x ) = 0 for |x | > R .
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 32 / 36
Analyticity of the FT
Compact support: f : R→ C, f ∈ L1(R), f (x ) = 0 for |x | > R .FT defined by
f (ξ) =
Rf (x )e −2πi ξx dx .
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 32 / 36
Analyticity of the FT
Compact support: f : R→ C, f ∈ L1(R), f (x ) = 0 for |x | > R .FT defined by
f (ξ) =
R
f (x )e −2πi ξx dx .
Allow ξ ∈ C, ξ = s + it in the formula.
f ( )
f ( ) 2πtx 2πisx d
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f (s + it ) =
f (x )e 2πtx e −2πisx dx
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 32 / 36
Analyticity of the FT
Compact support: f : R→ C, f ∈ L1(R), f (x ) = 0 for |x | > R .FT defined by
f (ξ) =
R
f (x )e −2πi ξx dx .
Allow ξ ∈ C, ξ = s + it in the formula.
f ( i )
f ( ) 2πtx 2πisx d
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f (s + it ) =
f (x )e 2πtx e −2πisx dx
Compact support implies f (x )e 2πtx ∈ L1(R), so integral is defined.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 32 / 36
Analyticity of the FT
Compact support: f : R→ C, f ∈ L1(R), f (x ) = 0 for |x | > R .FT defined by
f (ξ) =
R
f (x )e −2πi ξx dx .
Allow ξ ∈ C, ξ = s + it in the formula.
f ( + it)
f ( ) 2πtx −2πisx d
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f (s + it ) =
f (x )e 2πtx e 2πisx dx
Compact support implies f (x )e 2πtx ∈ L1(R), so integral is defined.Since e −2πix ξ is analytic for all ξ ∈ C, so is f (ξ).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 32 / 36
Analyticity of the FT
Compact support: f : R→ C, f ∈ L1(R), f (x ) = 0 for |x | > R .FT defined by
f (ξ) =
R
f (x )e −2πi ξx dx .
Allow ξ ∈ C, ξ = s + it in the formula.
f ( + it)
f ( ) 2πtx −2πisx d
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f (s + it ) =
f (x )e 2πtx e 2πisx dx
Compact support implies f (x )e 2πtx ∈ L1(R), so integral is defined.Since e −2πix ξ is analytic for all ξ ∈ C, so is f (ξ).Paley–Wiener: f ∈ L2(R). The following are equivalent:(a) f is the restriction on R of a function F holomorphic in the strip{z : |z | < a} which satisfies
|F (x + iy )|2 dx ≤ C , (|y | < a)
(b) e a|ξ| f (ξ) ∈ L2(R).
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 32 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions: E is required to be measurable or not
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions:E
is required to be measurable or notNon-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions:E
is required to be measurable or notNon-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
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Measurable version remains open.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions:E
is required to be measurable or notNon-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
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Measurable version remains open.
Equivalent form (R θ is rotation by θ):
k ∈Z2
1R θE (t + k ) = 1, (0 ≤ θ < 2π, t ∈ R2). (5)
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions: E is required to be measurable or not
Non-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
M bl i i
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Measurable version remains open.
Equivalent form (R θ is rotation by θ):
k ∈Z2
1R θE (t + k ) = 1, (0 ≤ θ < 2π, t ∈ R2). (5)
We prove: there is no bounded measurable Steinhaus set.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions: E is required to be measurable or not
Non-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
M bl i i
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Measurable version remains open.
Equivalent form (R θ is rotation by θ):
k ∈Z2
1R θE (t + k ) = 1, (0 ≤ θ < 2π, t ∈ R2). (5)
We prove: there is no bounded measurable Steinhaus set.
Integrating (5) for t ∈ [0, 1]2 we obtain |E | = 1.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions: E is required to be measurable or not
Non-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
M bl i i
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Measurable version remains open.
Equivalent form (R θ is rotation by θ):
k ∈Z2
1R θE (t + k ) = 1, (0 ≤ θ < 2π, t ∈ R2). (5)
We prove: there is no bounded measurable Steinhaus set.
Integrating (5) for t ∈ [0, 1]2 we obtain |E | = 1.
LHS of (5) is the Z2-periodization of 1R θE . Hence 1R θE (k ) = 0,k ∈ Z2 \ {0}.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
Application: the Steinhaus tiling problem
Question of Steinhaus: Is there E ⊆ R2 such that no matter howtranslated and rotated it always contains exactly one point withinteger coordinates?
Two versions: E is required to be measurable or not
Non-measurable version was answered in the affirmative by Jackson
and Mauldin a few years ago.
M bl i i
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Measurable version remains open.
Equivalent form (R θ is rotation by θ):
k ∈Z2
1R θE (t + k ) = 1, (0 ≤ θ < 2π, t ∈ R2). (5)
We prove: there is no bounded measurable Steinhaus set.
Integrating (5) for t ∈ [0, 1]2 we obtain |E | = 1.
LHS of (5) is the Z2-periodization of 1R θE . Hence 1R θE (k ) = 0,k ∈ Z2 \ {0}. 1E (ξ) = 0, whenever ξ on a circle through a lattice point
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 33 / 36
The circles on which 1E must vanish
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 34 / 36
Application: the Steinhaus tiling problem: conclusion
Consider the projection f of 1E on R.E bounded =⇒ f has compact support, say in [−B , B ].
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 35 / 36
Application: the Steinhaus tiling problem: conclusion
Consider the projection f of 1E on R.E bounded =⇒ f has compact support, say in [−B , B ].
For ξ ∈ R we have f (ξ) = 1E (ξ, 0), hence
f m2 + n2
= 0 (m n) ∈ Z2 \ {0}
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f
m + n
= 0, (m, n) ∈ Z \ {0}.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 35 / 36
Application: the Steinhaus tiling problem: conclusion
Consider the projection f of 1E on R.E bounded =⇒ f has compact support, say in [−B , B ].
For ξ ∈ R we have f (ξ) = 1E (ξ, 0), hence
f m2 + n2
= 0, (m, n) ∈ Z2 \ {0}.
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f
m + n
0, (m, n) ∈ Z \ {0}.
Landau: The number of integers up to x which are sums of twosquares is ∼ Cx / log1/2 x .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 35 / 36
Application: the Steinhaus tiling problem: conclusion
Consider the projection f of 1E on R.E bounded =⇒ f has compact support, say in [−B , B ].
For ξ ∈ R we have f (ξ) = 1E (ξ, 0), hence
f m2 + n2
= 0, (m, n) ∈ Z2 \ {0}.
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f
m + n
0, (m, n) ∈ Z \ {0}.
Landau: The number of integers up to x which are sums of twosquares is ∼ Cx / log1/2 x .
Hence f has almost R 2 zeros from 0 to R .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 35 / 36
Application: the Steinhaus tiling problem: conclusion
Consider the projection f of 1E on R.E bounded =⇒ f has compact support, say in [−B , B ].
For ξ ∈ R we have f (ξ) = 1E (ξ, 0), hence
f
m2 + n2
= 0, (m, n) ∈ Z2 \ {0}.
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+
, ( , ) ∈ \ { }
Landau: The number of integers up to x which are sums of twosquares is ∼ Cx / log1/2 x .
Hence f has almost R 2 zeros from 0 to R .
supp f ⊆ [−B , B ] implies f (z )
≤ f 1e 2πB |z |, z ∈ C
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 35 / 36
Application: the Steinhaus tiling problem: conclusion
Consider the projection f of 1E on R.E bounded =⇒ f has compact support, say in [−B , B ].
For ξ ∈ R we have f (ξ) = 1E (ξ, 0), hence
f
m2 + n2
= 0, (m, n) ∈ Z2 \ {0}.
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, ( , ) \ { }
Landau: The number of integers up to x which are sums of twosquares is ∼ Cx / log1/2 x .
Hence f has almost R 2 zeros from 0 to R .
supp f ⊆ [−B , B ] implies f (z )
≤ f 1e 2πB |z |, z ∈ C
But such a function can only have O (R ) zeros from 0 to R .
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 35 / 36
Zeros of entire functions of exponential type
Jensen’s formula: F analytic in the disk {|z | ≤ R }, z k are the zerosof F in that disk. Then
k
log R |z k |
= 1
0log F (Re 2πi θ) d θ.
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Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 36 / 36
Zeros of entire functions of exponential type
Jensen’s formula: F analytic in the disk {|z | ≤ R }, z k are the zerosof F in that disk. Then
k
log R |z k |
= 1
0log F (Re 2πi θ) d θ.
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It follows
#k : |z k | ≤ R e ≤ 1
0log F (Re 2πi θ) d θ
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 36 / 36
Zeros of entire functions of exponential type
Jensen’s formula: F analytic in the disk {|z | ≤ R }, z k are the zerosof F in that disk. Then
k
log R |z k |
= 1
0log F (Re 2πi θ) d θ.
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It follows
#k : |z k | ≤ R e ≤ 1
0log F (Re 2πi θ) d θ
Suppose |F (z )| ≤ Ae B |z |. Then RHS above is ≤ BR + log A.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 36 / 36
Zeros of entire functions of exponential type
Jensen’s formula: F analytic in the disk {|z | ≤ R }, z k are the zerosof F in that disk. Then
k
log R |z k |
= 1
0log F (Re 2πi θ) d θ.
I f ll
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It follows
#k : |z k | ≤ R e ≤ 1
0log F (Re 2πi θ) d θ
Suppose |F (z )| ≤ Ae B |z |. Then RHS above is ≤ BR + log A.
Such a function F can therefore have only O (R ) zeros in the disk{|z | ≤ R }.
Mihalis Kolountzakis (U. of Crete) FT and applications January 2006 36 / 36