Reproducing Kernel Hilbert Spaces and Hardy Spaces

Evan Camrud

Iowa State University

June 2, 2018 1 R: 1 ⇥ ⇤1 0 0 2 R3: 0 , 1 , 0 2 3 2 3 2 3 0 0 1 4 15 4 05 4 5 0 0 0 1 0 0 2 3 2 3 2 3 2 3 n 0 0 . . 3 R : , ,..., . , . 6.7 6.7 6 7 6 7 6.7 6.7 617 607 6 7 6 7 6 7 6 7 607 607 607 617 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 Question: How could we construct an ONB for an inﬁnite-dimensional case?

Inﬁnite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs): 1 0 0 2 R3: 0 , 1 , 0 2 3 2 3 2 3 0 0 1 4 15 4 05 4 5 0 0 0 1 0 0 2 3 2 3 2 3 2 3 n 0 0 . . 3 R : , ,..., . , . 6.7 6.7 6 7 6 7 6.7 6.7 617 607 6 7 6 7 6 7 6 7 607 607 607 617 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 Question: How could we construct an ONB for an inﬁnite-dimensional case?

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs): 1 R: 1 ⇥ ⇤ 1 0 0 0 0 1 0 0 2 3 2 3 2 3 2 3 n 0 0 . . 3 R : , ,..., . , . 6.7 6.7 6 7 6 7 6.7 6.7 617 607 6 7 6 7 6 7 6 7 607 607 607 617 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 Question: How could we construct an ONB for an inﬁnite-dimensional case?

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs): 1 R: 1 ⇥ ⇤1 0 0 2 R3: 0 , 1 , 0 2 3 2 3 2 3 0 0 1 4 5 4 5 4 5 Question: How could we construct an ONB for an inﬁnite-dimensional case?

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Let’s look at some orthonormal bases (ONBs): 1 R: 1 ⇥ ⇤1 0 0 2 R3: 0 , 1 , 0 2 3 2 3 2 3 0 0 1 4 15 4 05 4 5 0 0 0 1 0 0 2 3 2 3 2 3 2 3 n 0 0 . . 3 R : , ,..., . , . 6.7 6.7 6 7 6 7 6.7 6.7 617 607 6 7 6 7 6 7 6 7 607 607 607 617 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 Question: How could we construct an ONB for an inﬁnite-dimensional case? For ease of notation, we refer to these as e where the { n}n1=1 subscript 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:

1 0 0 0 0 1 0 0 2 3 2 3 2 3 2 3 0 0 . . . . 6.7, 6.7,...,6 7, 6 7,... 6.7 6.7 617 607 6 7 6 7 6 7 6 7 607 607 607 617 6 7 6 7 6 7 6 7 6.7 6.7 6.7 6.7 6.7 6.7 6.7 6.7 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Answer: Exactly like you would expect:

1 0 0 0 0 1 0 0 2 3 2 3 2 3 2 3 0 0 . . . . 6.7, 6.7,...,6 7, 6 7,... 6.7 6.7 617 607 6 7 6 7 6 7 6 7 607 607 607 617 6 7 6 7 6 7 6 7 6.7 6.7 6.7 6.7 6.7 6.7 6.7 6.7 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 For ease of notation, we refer to these as e where the { n}n1=1 subscript corresponds to the only nonzero entry to the vector. and we can deﬁne an inner product as a direct extension of the dot product of vectors, such that

1 u, v = µ (2) h i n n n=1 X

for u = n1=1 µnen and v = n1=1 nen. P P

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

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

1 v = nen for j C (1) 2 n=1 X Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

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

1 v = nen for j C (1) 2 n=1 X and we can deﬁne an inner product as a direct extension of the dot product of vectors, such that

1 u, v = µ (2) h i n n n=1 X

for u = n1=1 µnen and v = n1=1 nen. P P Answer: It must be that

1 1 µ 2 < and 2 < (3) | n| 1 | n| 1 n=1 n=1 X X Note that these conditions are both necessary and sucient for ﬁnite inner products as well as norms, a fact arising from the Cauchy-Schwartz inequality. This very special inﬁnite-dimensional Hilbert space is known as `2(N), the space of square-summable sequences. (Often this may just be written as `2.)

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Question: What do we require about µ , in { n}n1=1 { n}n1=1 order to have well-defined inner products? Note that these conditions are both necessary and sucient for finite inner products as well as norms, a fact arising from the Cauchy-Schwartz inequality. This very special infinite-dimensional Hilbert space is known as `2(N), the space of square-summable sequences. (Often this may just be written as `2.)

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Question: What do we require about µ , in { n}n1=1 { n}n1=1 order to have well-deﬁned inner products?

Answer: It must be that

1 1 µ 2 < and 2 < (3) | n| 1 | n| 1 n=1 n=1 X X This very special inﬁnite-dimensional Hilbert space is known as `2(N), the space of square-summable sequences. (Often this may just be written as `2.)

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Question: What do we require about µ , in { n}n1=1 { n}n1=1 order to have well-deﬁned inner products?

Answer: It must be that

Inﬁnite Dimensional Hilbert Spaces

Question: What do we require about µ , in { n}n1=1 { n}n1=1 order to have well-deﬁned inner products?

Answer: It must be that

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 a countable subset of vectors whose span is dense in the vector space. (i.e. There exists v such that for v V , ✏>0, { n}n1=1 2 v N v <✏.) n=1 n n P Answer: Yes. Consider Q.

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Answer: Yes. Consider Q. 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 countable orthonormal basis. Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Theorem A Hilbert space is separable if and only if it has a countable orthonormal 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. ( Proof. Consider a separable Hilbert space , and let " be an H { n}n1=1 ONB of . H Let ' : `2 be deﬁned to be linear, and also H!

'("n)=en (4)

where e is the ONB of `2. It is trivial to check that ' is { n}n1=1 an isomorphism.

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Introduction

Theorem All separable inﬁnite-dimensional Hilbert spaces are isomorphic to `2. Reproducing Kernel Hilbert Spaces and Hardy Spaces

Introduction

Theorem All separable inﬁnite-dimensional Hilbert spaces are isomorphic to `2.

Proof. Consider a separable Hilbert space , and let " be an H { n}n1=1 ONB of . H Let ' : `2 be deﬁned to be linear, and also H!

'("n)=en (4)

where e is the ONB of `2. It is trivial to check that ' is { n}n1=1 an isomorphism. Answer: Yes. Consider L2(R,n), where n is the counting measure.

Remark Non-separable infinite-dimensional Hilbert spaces are hardly ever considered, so in the future, when a Hilbert space is introduced, it is seen as either finite-dimensional (isomorphic to Cn) or separable infinite-dimensional (isomorphic to `2).

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Question: Do non-separable infinite-dimensional Hilbert spaces exist? Remark Non-separable infinite-dimensional Hilbert spaces are hardly ever considered, so in the future, when a Hilbert space is introduced, it is seen as either finite-dimensional (isomorphic to Cn) or separable infinite-dimensional (isomorphic to `2).

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Question: Do non-separable inﬁnite-dimensional Hilbert spaces exist?

Answer: Yes. Consider L2(R,n), where n is the counting measure. Reproducing Kernel Hilbert Spaces and Hardy Spaces

Inﬁnite Dimensional Hilbert Spaces

Question: Do non-separable inﬁnite-dimensional Hilbert spaces exist?

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

Answer: f(x)=x for some R. 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R R such that for all x1,x2 R, a, b C ! 2 2

f(ax1 + bx2)=af(x1)+bf(x2). (5) That is, suppose f is a linear functional. Answer: f(x)=x for some R. 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R R such that for all x1,x2 R, a, b C ! 2 2

f(ax1 + bx2)=af(x1)+bf(x2). (5) That is, suppose f is a linear functional.

Question: What does f “look like”? Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Consider f : R R such that for all x1,x2 R, a, b C ! 2 2

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 R. 2 Question: What does f “look like”?

x1 Answer: For ~x = x , 2 23 x3 ~ ~ ~ 3 f(~x )=1x1 + 2x24+ 53x3 = ~x = ,~x for some R . · h i 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

3 3 Consider f : R R such that for all ~x 1,~x2 R , a, b C ! 2 2

f(a~x1 + b~x2)=af(~x 1)+bf(~x 2). (6)

That is, once again, suppose f is a linear functional. x1 Answer: For ~x = x , 2 23 x3 ~ ~ ~ 3 f(~x )=1x1 + 2x24+ 53x3 = ~x = ,~x for some R . · h i 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

3 3 Consider f : R R such that for all ~x 1,~x2 R , a, b C ! 2 2

f(a~x1 + b~x2)=af(~x 1)+bf(~x 2). (6)

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

Question: What does f “look like”? Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

3 3 Consider f : R R such that for all ~x 1,~x2 R , a, b C ! 2 2

f(a~x1 + b~x2)=af(~x 1)+bf(~x 2). (6)

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

Question: What does f “look like”?

x1 Answer: For ~x = x , 2 23 x3 ~ ~ ~ 3 f(~x )=1x1 + 2x24+ 53x3 = ~x = ,~x for some R . · h i 2 Question: What does f “look like”?

x1 x2 Answer: For ~x = 2 . 3, . 6 7 6x 7 6 n7 ~ ~ ~ n f(~x )=1x1 + 2x24+ ...5+ nxn = ~x = ,~x for some R . · h i 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

n n Consider f : R R such that for all ~x 1,~x2 R , a, b C ! 2 2

f(a~x1 + b~x2)=af(~x 1)+bf(~x 2). (7)

That is, once again, suppose f is a linear functional. x1 x2 Answer: For ~x = 2 . 3, . 6 7 6x 7 6 n7 ~ ~ ~ n f(~x )=1x1 + 2x24+ ...5+ nxn = ~x = ,~x for some R . · h i 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

n n Consider f : R R such that for all ~x 1,~x2 R , a, b C ! 2 2

f(a~x1 + b~x2)=af(~x 1)+bf(~x 2). (7)

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

Question: What does f “look like”? Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

n n Consider f : R R such that for all ~x 1,~x2 R , a, b C ! 2 2

f(a~x1 + b~x2)=af(~x 1)+bf(~x 2). (7)

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

Question: What does f “look like”?

x1 x2 Answer: For ~x = 2 . 3, . 6 7 6x 7 6 n7 ~ ~ ~ n f(~x )=1x1 + 2x24+ ...5+ nxn = ~x = ,~x for some R . · h i 2 Question: What does f “look like”?

Answer: f( )=⌫ + ⌫ + ... + ⌫ + ... = ⌫ , { n}n1=1 1 1 2 2 n n h{ n}n1=1 { n}n1=1i for some ⌫ `2. { }n1=1 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

2 2 Consider f : ` R, such that for all n 1 , µn 1 ` , ! { }n=1 { }n=1 2 a, b C 2 f a 1 + b µ 1 = af 1 + bf( µ 1 . (8) { n}n=1 { n}n=1 { n}n=1 { n}n=1 as well as f( 1 ) M 1 (9) { n}n=1  · { n}n=1 for some M R. That is, once again, suppose f is a linear 2 functional, but also that it is bounded . This boundedness is equivalent to continuity of the functional. Answer: f( )=⌫ + ⌫ + ... + ⌫ + ... = ⌫ , { n}n1=1 1 1 2 2 n n h{ n}n1=1 { n}n1=1i for some ⌫ `2. { }n1=1 2

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Question: What does f “look like”? Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Question: What does f “look like”?

Answer: f( )=⌫ + ⌫ + ... + ⌫ + ... = ⌫ , { n}n1=1 1 1 2 2 n n h{ n}n1=1 { n}n1=1i for some ⌫ `2. { }n1=1 2 Theorem (Riesz-Represenation for Hilbert spaces) Let be a Hilbert space. Let f : R be a bounded linear H H! functional. Then there is a v such that for all u 2H 2H f(u)= u, v . (10) h i

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Aside from the non-separable inﬁnite dimensional case, we have just “proved” the Riesz-Representation theorem for Hilbert spaces. Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Aside from the non-separable inﬁnite dimensional case, we have just “proved” the Riesz-Representation theorem for Hilbert spaces.

Theorem (Riesz-Represenation for Hilbert spaces) Let be a Hilbert space. Let f : R be a bounded linear H H! functional. Then there is a v such that for all u 2H 2H f(u)= u, v . (10) h i That is, if x : R (for some ﬁxed x K)isdeﬁnedby H! 2

x(f)=f(x) (11)

then x is both bounded and linear. Hence there must be some g such that x 2H (f)= f,g = f(x). (12) x h xi

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Example Consider a Hilbert space , whose elements are continuous H functions deﬁned on a compact set K, then point-evaluation of functions is a bounded linear functional. Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Example Consider a Hilbert space , whose elements are continuous H functions deﬁned on a compact set K, then point-evaluation of functions is a bounded linear functional.

That is, if x : R (for some ﬁxed x K)isdeﬁnedby H! 2

x(f)=f(x) (11)

then x is both bounded and linear. Hence there must be some g such that x 2H (f)= f,g = f(x). (12) x h xi Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Deﬁnition

Let be a Hilbert space of functions. Let x : R be a H H! point-evaluation functional such that for all f 2H

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

If there exists g such that x 2H (f)= f,g = f(x) (14) x h xi then is called a reproducing kernel Hilbert space. H Hint: First construct an ONB by means of the Gram-Schmidt method: 1 1 e = 1 p2 2 e = x x, e e then normalize. 2 h 1i 1 3 e = x2 x2,e e x2,e e then normalize. 3 h 2i 2 h 1i 1

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Exercise: Consider the Hilbert space of functions spanned by 1 ,x,x2 on the domain [ 1, 1]. The inner product on this { p2 } space is 1 f,g = f(x)g(x)dx. (15) h i 1 Z What is the reproducing kernel of this Hilbert space? Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: First construct an ONB by means of the Gram-Schmidt method: 1 1 e = 1 p2 2 e = x x, e e then normalize. 2 h 1i 1 3 e = x2 x2,e e x2,e e then normalize. 3 h 2i 2 h 1i 1 Hint: But en(x) R, so we can pull it inside. 2 3 f(x)= f,e (x) e (17) h n · ni n=1 X

Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: Since we have an orthonormal basis, f = 3 f,e e . n=1h ni n Hence for a ﬁxed x [ 1, 1] 2 P 3 f(x)= f,e e (x). (16) h ni n n=1 X Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: Since we have an orthonormal basis, f = 3 f,e e . n=1h ni n Hence for a ﬁxed x [ 1, 1] 2 P 3 f(x)= f,e e (x). (16) h ni n n=1 X

Hint: But en(x) R, so we can pull it inside. 2 3 f(x)= f,e (x) e (17) h n · ni n=1 X Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Hint: The sum is ﬁnite, so we can certainly pull it inside of the inner product. 3 f(x)= f, e (x) e (18) n · n n=1 ⌧ X Thus for g (y)= 3 e (x) e (y)wehaveareproducing x n=1 n · n kernel. P Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Just so, if we can justify an exchange-of-limit process,we recover the following. Theorem Let be a separable inﬁnite dimensional reproducing kernel H Hilbert space. Let e be an ONB of . Then the { n}n1=1 H reproducing kernel of is given by H 1 gx(y)= en(x)en(y). (19) n=1 X

Notice that we may consider gx(y)=K(x, y) as a function of two variables. Reproducing Kernel Hilbert Spaces and Hardy Spaces

Reproducing Kernel Hilbert Space

Definition A positive definite function is a function K(x, y):D D C such that for all finite sequences ⇥ ! t ,t ,...,t D, the matrix 1 2 n 2

K(t1,t1) K(t1,t2) ... K(t1,tn) .. 2K(t2,t1) . K(t2,tn)3 . . . (20) 6 . .. . 7 6 7 6K(tn,t1) K(tn,t2) ... K(tn,tn)7 6 7 4 5 is a positive, semideﬁnite matrix. (That is, the above matrix M is such that ~x T M~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 the reproducing kernel of a Hilbert space. One might observe an isomorphism from H2(D) onto `2, and an isometry 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 ⇢

2 1 n 2 H (D)= f : f(z)= nz for n 1 ` { }n=0 2 n n=0 o (21) X 1 = f analytic : sup f(re2⇡i✓) 2d✓< 0 r<1 0 1  Z n o Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Deﬁnition The Hardy space of the unit disk D C is ⇢

2 1 n 2 H (D)= f : f(z)= nz for n 1 ` { }n=0 2 n n=0 o (21) X 1 = f analytic : sup f(re2⇡i✓) 2d✓< 0 r<1 0 1  Z n o

One might observe an isomorphism from H2(D) onto `2, and an isometry 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

1 f,g = µ h i n n n=0 (22) X 1 =sup f(re2⇡i✓)g(re2⇡i✓)d✓ 0 r<1 0  Z n n for f(z)= n1=0 nz , g(z)= n1=0 µnz . P P Question: What is the ONB for H2(D)?

Answer: By observation we can see that zn is a basis. { }n1=0 Fourier analysis gives us orthonormality of this set, as on the boundary of D we have zn e2⇡inx. 7!

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

Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f H2(D)impliesf is analytic (hence 2 continuous), H2(D) must be a reproducing kernel Hilbert space. Answer: By observation we can see that zn is a basis. { }n1=0 Fourier analysis gives us orthonormality of this set, as on the boundary of D we have zn e2⇡inx. 7!

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

Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

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

Question: What is the ONB for H2(D)? Question: What is the reproducing kernel for H2(D)? Can it be simpliﬁed?

Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f H2(D)impliesf is analytic (hence 2 continuous), 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 is a basis. { }n1=0 Fourier analysis gives us orthonormality of this set, as on the boundary of D we have zn e2⇡inx. 7! Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Since D is compact, and f H2(D)impliesf is analytic (hence 2 continuous), 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 is a basis. { }n1=0 Fourier analysis gives us orthonormality of this set, as on the boundary of D we have zn e2⇡inx. 7!

Question: What is the reproducing kernel for H2(D)? Can it be simpliﬁed? Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

Deﬁnition The Szeg¨okernel is the reproducing kernel of H2(D), given by

1 1 k (z)= wnzn = (23) w 1 wz n=0 X since w , z < 1. | | | | Thus we have

1 2⇡i✓ 2⇡i✓ f(z),kw(z) =sup f(re )kw(re )d✓ h i 0 r<1 0  Z (24) 1 f(re2⇡i✓) =sup 2⇡i✓ d✓ = f(w) 0 r<1 0 1 w re  Z · Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

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

2 1 2 H (H)= f analytic : sup f(x + iy) dx < 1 y>0 Z n 1 o (25) 1 iyz 2 + = f : f(z)= g(y)e dy, g L (R ) 0 2 n Z o With some e↵ort, one can produce an isomorphism ' : H2(H) H2(D) given by ! p⇡ 1+z ['f](z)= f i . (27) 1 z 1 z ⇣ ⌘

Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

The inner product of H2(H) is given by

f,g =sup 1 f(x + iy)g(x + iy)dx. (26) h i y>0 Z1 Reproducing Kernel Hilbert Spaces and Hardy Spaces

The Hardy space

The inner product of H2(H) is given by

f,g =sup 1 f(x + iy)g(x + iy)dx. (26) h i y>0 Z1

With some e↵ort, one can produce an isomorphism ' : H2(H) H2(D) given by ! p⇡ 1+z ['f](z)= f i . (27) 1 z 1 z ⇣ ⌘ Reproducing Kernel Hilbert Spaces and Hardy Spaces

Questions?