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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 �P
N
n=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 �P
N
n=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 �P
N
n=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 C
f�
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 C
f�
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 C
f�
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
�1
f(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
�1
f(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 matrix
2
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.)
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<1
Z
1
0
�
�f(re2⇡i✓)�
�
2
d✓ < 1o
(21)
One might observe an isomorphism from H2(D) onto `2, and anisometry from H2(D) into L2(T) from the above definitions.
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<1
Z
1
0
�
�f(re2⇡i✓)�
�
2
d✓ < 1o
(21)
One might observe an isomorphism from H2(D) onto `2, and anisometry from H2(D) into L2(T) from the above definitions.
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<1
Z
1
0
f(re2⇡i✓)g(re2⇡i✓)d✓
(22)
for f(z) =P1
n=0
�n
zn, g(z) =P1
n=0
µn
zn.
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 Szego 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<1
Z
1
0
f(re2⇡i✓)kw
(re2⇡i✓)d✓
= sup0r<1
Z
1
0
f(re2⇡i✓)
1� w · re�2⇡i✓
d✓ = f(w)
(24)
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� zf⇣
i1 + z
1� z
⌘
. (27)
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� zf⇣
i1 + z
1� z
⌘
. (27)
Reproducing Kernel Hilbert Spaces and Hardy Spaces
Questions?
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