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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”?

    Answer:f({�

    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”?

    Answer:f({�

    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”?

    Answer:f({�

    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

    Questions?