# Reproducing Kernel Hilbert Spaces and Hardy Spaces€¦ · Reproducing Kernel Hilbert Spaces and Hardy Spaces Reproducing Kernel Hilbert Spaces and Hardy Spaces Evan Camrud Iowa State

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• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Spaces and HardySpaces

Evan Camrud

Iowa State University

June 2, 2018

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs):

1 R:⇥

1⇤

2 R3:

2

4

100

3

5,

2

4

010

3

5,

2

4

001

3

5

3 Rn :

2

6

6

6

6

6

4

100...0

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

010...0

3

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

4

00...10

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

00...01

3

7

7

7

7

7

5

Question: How could we construct an ONB for aninfinite-dimensional case?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs):

1 R:⇥

1⇤

2 R3:

2

4

100

3

5,

2

4

010

3

5,

2

4

001

3

5

3 Rn :

2

6

6

6

6

6

4

100...0

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

010...0

3

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

4

00...10

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

00...01

3

7

7

7

7

7

5

Question: How could we construct an ONB for aninfinite-dimensional case?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs):

1 R:⇥

1⇤

2 R3:

2

4

100

3

5,

2

4

010

3

5,

2

4

001

3

5

3 Rn :

2

6

6

6

6

6

4

100...0

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

010...0

3

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

4

00...10

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

00...01

3

7

7

7

7

7

5

Question: How could we construct an ONB for aninfinite-dimensional case?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs):

1 R:⇥

1⇤

2 R3:

2

4

100

3

5,

2

4

010

3

5,

2

4

001

3

5

3 Rn :

2

6

6

6

6

6

4

100...0

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

010...0

3

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

4

00...10

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

00...01

3

7

7

7

7

7

5

Question: How could we construct an ONB for aninfinite-dimensional case?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs):

1 R:⇥

1⇤

2 R3:

2

4

100

3

5,

2

4

010

3

5,

2

4

001

3

5

3 Rn :

2

6

6

6

6

6

4

100...0

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

010...0

3

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

4

00...10

3

7

7

7

7

7

5

,

2

6

6

6

6

6

4

00...01

3

7

7

7

7

7

5

Question: How could we construct an ONB for aninfinite-dimensional case?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Answer: Exactly like you would expect:

2

6

6

6

6

6

6

6

6

4

100...0...

3

7

7

7

7

7

7

7

7

5

,

2

6

6

6

6

6

6

6

6

4

010...0...

3

7

7

7

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

6

6

6

4

00...10...

3

7

7

7

7

7

7

7

7

5

,

2

6

6

6

6

6

6

6

6

4

00...01...

3

7

7

7

7

7

7

7

7

5

,...

For ease of notation, we refer to these as {en

}1n=1

where thesubscript corresponds to the only nonzero entry to the vector.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Answer: Exactly like you would expect:

2

6

6

6

6

6

6

6

6

4

100...0...

3

7

7

7

7

7

7

7

7

5

,

2

6

6

6

6

6

6

6

6

4

010...0...

3

7

7

7

7

7

7

7

7

5

, ... ,

2

6

6

6

6

6

6

6

6

4

00...10...

3

7

7

7

7

7

7

7

7

5

,

2

6

6

6

6

6

6

6

6

4

00...01...

3

7

7

7

7

7

7

7

7

5

,...

For ease of notation, we refer to these as {en

}1n=1

where thesubscript corresponds to the only nonzero entry to the vector.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Since this is an orthonormal basis, we may construct elementsof this infinite-dimensional Hilbert space by means of (infinite)linear combinations:

v =1X

n=1

�n

en

for �j

2 C (1)

and we can define an inner product as a direct extension of thedot product of vectors, such that

hu, vi =1X

n=1

µn

�n

(2)

for u =P1

n=1

µn

en

and v =P1

n=1

�n

en

.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Since this is an orthonormal basis, we may construct elementsof this infinite-dimensional Hilbert space by means of (infinite)linear combinations:

v =1X

n=1

�n

en

for �j

2 C (1)

and we can define an inner product as a direct extension of thedot product of vectors, such that

hu, vi =1X

n=1

µn

�n

(2)

for u =P1

n=1

µn

en

and v =P1

n=1

�n

en

.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: What do we require about {µn

}1n=1

, {�n

}1n=1

inorder to have well-defined inner products?

Answer: It must be that

1X

n=1

|µn

|2 < 1 and1X

n=1

|�n

|2 < 1 (3)

Note that these conditions are both necessary and su�cient forfinite inner products as well as norms, a fact arising from theCauchy-Schwartz inequality.

This very special infinite-dimensional Hilbert space is known as`2(N), the space of square-summable sequences. (Often thismay just be written as `2.)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: What do we require about {µn

}1n=1

, {�n

}1n=1

inorder to have well-defined inner products?

Answer: It must be that

1X

n=1

|µn

|2 < 1 and1X

n=1

|�n

|2 < 1 (3)

Note that these conditions are both necessary and su�cient forfinite inner products as well as norms, a fact arising from theCauchy-Schwartz inequality.

This very special infinite-dimensional Hilbert space is known as`2(N), the space of square-summable sequences. (Often thismay just be written as `2.)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: What do we require about {µn

}1n=1

, {�n

}1n=1

inorder to have well-defined inner products?

Answer: It must be that

1X

n=1

|µn

|2 < 1 and1X

n=1

|�n

|2 < 1 (3)

Note that these conditions are both necessary and su�cient forfinite inner products as well as norms, a fact arising from theCauchy-Schwartz inequality.

This very special infinite-dimensional Hilbert space is known as`2(N), the space of square-summable sequences. (Often thismay just be written as `2.)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: What do we require about {µn

}1n=1

, {�n

}1n=1

inorder to have well-defined inner products?

Answer: It must be that

1X

n=1

|µn

|2 < 1 and1X

n=1

|�n

|2 < 1 (3)

Note that these conditions are both necessary and su�cient forfinite inner products as well as norms, a fact arising from theCauchy-Schwartz inequality.

This very special infinite-dimensional Hilbert space is known as`2(N), the space of square-summable sequences. (Often thismay just be written as `2.)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Definition

A normed linear space is called separable if it contains acountable subset of vectors whose span is dense in the vectorspace. (i.e. There exists {v

n

}1n=1

such that for v 2 V , ✏ > 0,�

�v �PNn=1

�n

vn

� < ✏.)

Question: Is R a separable vector space?

Answer: Yes. Consider Q.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Definition

A normed linear space is called separable if it contains acountable subset of vectors whose span is dense in the vectorspace. (i.e. There exists {v

n

}1n=1

such that for v 2 V , ✏ > 0,�

�v �PNn=1

�n

vn

� < ✏.)

Question: Is R a separable vector space?

Answer: Yes. Consider Q.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Definition

A normed linear space is called separable if it contains acountable subset of vectors whose span is dense in the vectorspace. (i.e. There exists {v

n

}1n=1

such that for v 2 V , ✏ > 0,�

�v �PNn=1

�n

vn

� < ✏.)

Question: Is R a separable vector space?

Answer: Yes. Consider Q.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Theorem

A Hilbert space is separable if and only if it has a countableorthonormal basis.

Proof.

(=)) Existence of such a basis is guaranteed by Zorn’s Lemma.(A maximal orthonormal set is known to be a basis.)((=) The ONB is the countable dense subset.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Theorem

A Hilbert space is separable if and only if it has a countableorthonormal basis.

Proof.

(=)) Existence of such a basis is guaranteed by Zorn’s Lemma.(A maximal orthonormal set is known to be a basis.)((=) The ONB is the countable dense subset.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Introduction

Theorem

All separable infinite-dimensional Hilbert spaces areisomorphic to `2.

Proof.

Consider a separable Hilbert space H, and let {"n

}1n=1

be anONB of H.Let ' : H ! `2 be defined to be linear, and also

'("n

) = en

(4)

where {en

}1n=1

is the ONB of `2. It is trivial to check that ' isan isomorphism.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Introduction

Theorem

All separable infinite-dimensional Hilbert spaces areisomorphic to `2.

Proof.

Consider a separable Hilbert space H, and let {"n

}1n=1

be anONB of H.Let ' : H ! `2 be defined to be linear, and also

'("n

) = en

(4)

where {en

}1n=1

is the ONB of `2. It is trivial to check that ' isan isomorphism.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: Do non-separable infinite-dimensional Hilbertspaces exist?

Answer: Yes. Consider L2(R, n), where n is the countingmeasure.

Remark

Non-separable infinite-dimensional Hilbert spaces are hardlyever considered, so in the future, when a Hilbert space isintroduced, it is seen as either finite-dimensional (isomorphic toCn) or separable infinite-dimensional (isomorphic to `2).

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: Do non-separable infinite-dimensional Hilbertspaces exist?

Answer: Yes. Consider L2(R, n), where n is the countingmeasure.

Remark

Non-separable infinite-dimensional Hilbert spaces are hardlyever considered, so in the future, when a Hilbert space isintroduced, it is seen as either finite-dimensional (isomorphic toCn) or separable infinite-dimensional (isomorphic to `2).

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Infinite Dimensional Hilbert Spaces

Question: Do non-separable infinite-dimensional Hilbertspaces exist?

Answer: Yes. Consider L2(R, n), where n is the countingmeasure.

Remark

Non-separable infinite-dimensional Hilbert spaces are hardlyever considered, so in the future, when a Hilbert space isintroduced, it is seen as either finite-dimensional (isomorphic toCn) or separable infinite-dimensional (isomorphic to `2).

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R ! R such that for all x1

, x2

2 R, a, b 2 C

f(ax1

+ bx2

) = af(x1

) + bf(x2

). (5)

That is, suppose f is a linear functional.

Question: What does f “look like”?

Answer: f(x) = �x for some � 2 R.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R ! R such that for all x1

, x2

2 R, a, b 2 C

f(ax1

+ bx2

) = af(x1

) + bf(x2

). (5)

That is, suppose f is a linear functional.

Question: What does f “look like”?

Answer: f(x) = �x for some � 2 R.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R ! R such that for all x1

, x2

2 R, a, b 2 C

f(ax1

+ bx2

) = af(x1

) + bf(x2

). (5)

That is, suppose f is a linear functional.

Question: What does f “look like”?

Answer: f(x) = �x for some � 2 R.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R3 ! R such that for all ~x1

, ~x2

2 R3, a, b 2 C

f(a~x1

+ b~x2

) = af(~x1

) + bf(~x2

). (6)

That is, once again, suppose f is a linear functional.

Question: What does f “look like”?

Answer: For ~x =

2

4

x1

x2

x3

3

5,

f(~x) = �1

x1

+ �2

x2

+ �3

x3

= ~� · ~x = h~�, ~xi for some ~� 2 R3.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R3 ! R such that for all ~x1

, ~x2

2 R3, a, b 2 C

f(a~x1

+ b~x2

) = af(~x1

) + bf(~x2

). (6)

That is, once again, suppose f is a linear functional.

Question: What does f “look like”?

Answer: For ~x =

2

4

x1

x2

x3

3

5,

f(~x) = �1

x1

+ �2

x2

+ �3

x3

= ~� · ~x = h~�, ~xi for some ~� 2 R3.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R3 ! R such that for all ~x1

, ~x2

2 R3, a, b 2 C

f(a~x1

+ b~x2

) = af(~x1

) + bf(~x2

). (6)

That is, once again, suppose f is a linear functional.

Question: What does f “look like”?

Answer: For ~x =

2

4

x1

x2

x3

3

5,

f(~x) = �1

x1

+ �2

x2

+ �3

x3

= ~� · ~x = h~�, ~xi for some ~� 2 R3.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : Rn ! R such that for all ~x1

, ~x2

2 Rn, a, b 2 C

f(a~x1

+ b~x2

) = af(~x1

) + bf(~x2

). (7)

That is, once again, suppose f is a linear functional.

Question: What does f “look like”?

Answer: For ~x =

2

6

6

6

4

x1

x2

...xn

3

7

7

7

5

,

f(~x) = �1

x1

+�2

x2

+ ...+�n

xn

= ~� · ~x = h~�, ~xi for some ~� 2 Rn.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : Rn ! R such that for all ~x1

, ~x2

2 Rn, a, b 2 C

f(a~x1

+ b~x2

) = af(~x1

) + bf(~x2

). (7)

That is, once again, suppose f is a linear functional.

Question: What does f “look like”?

Answer: For ~x =

2

6

6

6

4

x1

x2

...xn

3

7

7

7

5

,

f(~x) = �1

x1

+�2

x2

+ ...+�n

xn

= ~� · ~x = h~�, ~xi for some ~� 2 Rn.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : Rn ! R such that for all ~x1

, ~x2

2 Rn, a, b 2 C

f(a~x1

+ b~x2

) = af(~x1

) + bf(~x2

). (7)

That is, once again, suppose f is a linear functional.

Question: What does f “look like”?

Answer: For ~x =

2

6

6

6

4

x1

x2

...xn

3

7

7

7

5

,

f(~x) = �1

x1

+�2

x2

+ ...+�n

xn

= ~� · ~x = h~�, ~xi for some ~� 2 Rn.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : `2 ! R, such that for all {�n

}1n=1

, {µn

}1n=1

2 `2,a, b 2 Cf�

a{�n

}1n=1

+ b{µn

}1n=1

= af�{�

n

}1n=1

+ bf({µn

}1n=1

. (8)

as well asf({�

n

}1n=1

) M · ��{�n

}1n=1

� (9)

for some M 2 R. That is, once again, suppose f is a linearfunctional, but also that it is bounded. This boundedness isequivalent to continuity of the functional.

Question: What does f “look like”?

n

}1n=1

) = ⌫1

�1

+ ⌫2

�2

+ ...+ ⌫n

�n

+ ... = h{⌫n

}1n=1

, {�n

}1n=1

ifor some {⌫}1

n=1

2 `2.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : `2 ! R, such that for all {�n

}1n=1

, {µn

}1n=1

2 `2,a, b 2 Cf�

a{�n

}1n=1

+ b{µn

}1n=1

= af�{�

n

}1n=1

+ bf({µn

}1n=1

. (8)

as well asf({�

n

}1n=1

) M · ��{�n

}1n=1

� (9)

for some M 2 R. That is, once again, suppose f is a linearfunctional, but also that it is bounded. This boundedness isequivalent to continuity of the functional.

Question: What does f “look like”?

n

}1n=1

) = ⌫1

�1

+ ⌫2

�2

+ ...+ ⌫n

�n

+ ... = h{⌫n

}1n=1

, {�n

}1n=1

ifor some {⌫}1

n=1

2 `2.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : `2 ! R, such that for all {�n

}1n=1

, {µn

}1n=1

2 `2,a, b 2 Cf�

a{�n

}1n=1

+ b{µn

}1n=1

= af�{�

n

}1n=1

+ bf({µn

}1n=1

. (8)

as well asf({�

n

}1n=1

) M · ��{�n

}1n=1

� (9)

for some M 2 R. That is, once again, suppose f is a linearfunctional, but also that it is bounded. This boundedness isequivalent to continuity of the functional.

Question: What does f “look like”?

n

}1n=1

) = ⌫1

�1

+ ⌫2

�2

+ ...+ ⌫n

�n

+ ... = h{⌫n

}1n=1

, {�n

}1n=1

ifor some {⌫}1

n=1

2 `2.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Aside from the non-separable infinite dimensional case, we havejust “proved” the Riesz-Representation theorem for Hilbertspaces.

Theorem (Riesz-Represenation for Hilbert spaces)

Let H be a Hilbert space. Let f : H ! R be a bounded linearfunctional. Then there is a v 2 H such that for all u 2 H

f(u) = hu, vi. (10)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Aside from the non-separable infinite dimensional case, we havejust “proved” the Riesz-Representation theorem for Hilbertspaces.

Theorem (Riesz-Represenation for Hilbert spaces)

Let H be a Hilbert space. Let f : H ! R be a bounded linearfunctional. Then there is a v 2 H such that for all u 2 H

f(u) = hu, vi. (10)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Example

Consider a Hilbert space H, whose elements are continuousfunctions defined on a compact set K, then point-evaluation offunctions is a bounded linear functional.

That is, if �x

: H ! R (for some fixed x 2 K) is defined by

�x

(f) = f(x) (11)

then �x

is both bounded and linear. Hence there must be somegx

2 H such that

�x

(f) = hf, gx

i = f(x). (12)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Example

Consider a Hilbert space H, whose elements are continuousfunctions defined on a compact set K, then point-evaluation offunctions is a bounded linear functional.

That is, if �x

: H ! R (for some fixed x 2 K) is defined by

�x

(f) = f(x) (11)

then �x

is both bounded and linear. Hence there must be somegx

2 H such that

�x

(f) = hf, gx

i = f(x). (12)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Definition

Let H be a Hilbert space of functions. Let �x

: H ! R be apoint-evaluation functional such that for all f 2 H

�x

(f) = f(x). (13)

If there exists gx

2 H such that

�x

(f) = hf, gx

i = f(x) (14)

then H is called a reproducing kernel Hilbert space.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Exercise: Consider the Hilbert space of functions spanned by{ 1p

2

, x, x2} on the domain [�1, 1]. The inner product on thisspace is

hf, gi =Z

1

�1f(x)g(x)dx. (15)

What is the reproducing kernel of this Hilbert space?

Hint: First construct an ONB by means of the Gram-Schmidtmethod:

1 e1

= 1p2

2 e2

= x� hx, e1

ie1

then normalize.

3 e3

= x2 � hx2, e2

ie2

� hx2, e1

ie1

then normalize.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Exercise: Consider the Hilbert space of functions spanned by{ 1p

2

, x, x2} on the domain [�1, 1]. The inner product on thisspace is

hf, gi =Z

1

�1f(x)g(x)dx. (15)

What is the reproducing kernel of this Hilbert space?

Hint: First construct an ONB by means of the Gram-Schmidtmethod:

1 e1

= 1p2

2 e2

= x� hx, e1

ie1

then normalize.

3 e3

= x2 � hx2, e2

ie2

� hx2, e1

ie1

then normalize.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: Since we have an orthonormal basis, f =P

3

n=1

hf, en

ien

.Hence for a fixed x 2 [�1, 1]

f(x) =3

X

n=1

hf, en

ien

(x). (16)

Hint: But en

(x) 2 R, so we can pull it inside.

f(x) =3

X

n=1

hf, en

(x) · en

i (17)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: Since we have an orthonormal basis, f =P

3

n=1

hf, en

ien

.Hence for a fixed x 2 [�1, 1]

f(x) =3

X

n=1

hf, en

ien

(x). (16)

Hint: But en

(x) 2 R, so we can pull it inside.

f(x) =3

X

n=1

hf, en

(x) · en

i (17)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: The sum is finite, so we can certainly pull it inside of theinner product.

f(x) =

f,3

X

n=1

en

(x) · en

(18)

Thus for gx

(y) =P

3

n=1

en

(x) · en

(y) we have a reproducingkernel.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Just so, if we can justify an exchange-of-limit process, werecover the following.

Theorem

Let H be a separable infinite dimensional reproducing kernelHilbert space. Let {e

n

}1n=1

be an ONB of H. Then thereproducing kernel of H is given by

gx

(y) =1X

n=1

en

(x)en

(y). (19)

Notice that we may consider gx

(y) = K(x, y) as a function oftwo variables.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Definition

A positive definite function is a functionK(x, y) : D ⇥D ! C such that for all finite sequencest1

, t2

, ..., tn

2 D, the matrix2

6

6

6

6

4

K(t1

, t1

) K(t1

, t2

) ... K(t1

, tn

)

K(t2

, t1

). . . K(t

2

, tn

)...

. . ....

K(tn

, t1

) K(tn

, t2

) ... K(tn

, tn

)

3

7

7

7

7

5

(20)

is a positive, semidefinite matrix. (That is, the above matrix Mis such that ~xTM~x � 0 for all ~x with non-negative elements.)

Weber, Eric [MATH]

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Theorem (Moore-Aronszajn)

A function is positive definite if and only if it is thereproducing kernel of a Hilbert space.

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Definition

The Hardy space of the unit disk D ⇢ C is

H2(D) =n

f : f(z) =1X

n=0

�n

zn for {�n

}1n=0

2 `2o

=n

f analytic : sup0r

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Definition

The Hardy space of the unit disk D ⇢ C is

H2(D) =n

f : f(z) =1X

n=0

�n

zn for {�n

}1n=0

2 `2o

=n

f analytic : sup0r

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy Space

The inner product on H2(D) is given by

hf, gi =1X

n=0

�n

µn

= sup0r

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f 2 H2(D) implies f is analytic (hencecontinuous), H2(D) must be a reproducing kernel Hilbert space.

Question: What is the ONB for H2(D)?

Answer: By observation we can see that {zn}1n=0

is a basis.Fourier analysis gives us orthonormality of this set, as on theboundary of D we have zn 7! e2⇡inx.

Question: What is the reproducing kernel for H2(D)? Can itbe simplified?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f 2 H2(D) implies f is analytic (hencecontinuous), H2(D) must be a reproducing kernel Hilbert space.

Question: What is the ONB for H2(D)?

Answer: By observation we can see that {zn}1n=0

is a basis.Fourier analysis gives us orthonormality of this set, as on theboundary of D we have zn 7! e2⇡inx.

Question: What is the reproducing kernel for H2(D)? Can itbe simplified?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f 2 H2(D) implies f is analytic (hencecontinuous), H2(D) must be a reproducing kernel Hilbert space.

Question: What is the ONB for H2(D)?

Answer: By observation we can see that {zn}1n=0

is a basis.Fourier analysis gives us orthonormality of this set, as on theboundary of D we have zn 7! e2⇡inx.

Question: What is the reproducing kernel for H2(D)? Can itbe simplified?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f 2 H2(D) implies f is analytic (hencecontinuous), H2(D) must be a reproducing kernel Hilbert space.

Question: What is the ONB for H2(D)?

Answer: By observation we can see that {zn}1n=0

is a basis.Fourier analysis gives us orthonormality of this set, as on theboundary of D we have zn 7! e2⇡inx.

Question: What is the reproducing kernel for H2(D)? Can itbe simplified?

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Definition

The Szegö kernel is the reproducing kernel of H2(D), given by

kw

(z) =1X

n=0

wnzn =1

1� wz (23)

since |w|, |z| < 1.Thus we have

hf(z), kw

(z)i = sup0r

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Definition

The Hardy space of the upper half-plane H ⇢ C is

H2(H) =n

f analytic : supy>0

Z 1

�1

�f(x+ iy)�

2

dx < 1o

=n

f : f(z) =

Z 1

0

g(y)e�iyzdy, g 2 L2(R+)o

(25)

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

The inner product of H2(H) is given by

hf, gi = supy>0

Z 1

�1f(x+ iy)g(x+ iy)dx. (26)

With some e↵ort, one can produce an isomorphism' : H2(H) ! H2(D) given by

['f ](z) =

p⇡

1� z f⇣

i1 + z

1� z⌘

. (27)

Weber, Eric [MATH]

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

The inner product of H2(H) is given by

hf, gi = supy>0

Z 1

�1f(x+ iy)g(x+ iy)dx. (26)

With some e↵ort, one can produce an isomorphism' : H2(H) ! H2(D) given by

['f ](z) =

p⇡

1� z f⇣

i1 + z

1� z⌘

. (27)

Weber, Eric [MATH]

• Reproducing Kernel Hilbert Spaces and Hardy Spaces

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