Reproducing Kernel Hilbert Spaces and Hardy Spaces 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? Reproducing Kernel Hilbert Spaces and Hardy Spaces 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 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 λ (2) h i n n n=1 X for u = n1=1 µnen and v = n1=1 λnen. 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 = λnen for λj 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 = λnen for λj 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 λ (2) h i n n n=1 X for u = n1=1 µnen and v = n1=1 λnen. P P Answer: It must be that 1 1 µ 2 < and λ 2 < (3) | n| 1 | n| 1 n=1 n=1 X X Note that these conditions are both necessary and sufficient 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 , λ in { n}n1=1 { n}n1=1 order to have well-defined inner products? Note that these conditions are both necessary and sufficient 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 , λ in { n}n1=1 { n}n1=1 order to have well-defined inner products? Answer: It must be that 1 1 µ 2 < and λ 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 , λ in { n}n1=1 { n}n1=1 order to have well-defined inner products? Answer: It must be that 1 1 µ 2 < and λ 2 < (3) | n| 1 | n| 1 n=1 n=1 X X Note that these conditions are both necessary and sufficient for finite inner products as well as norms, a fact arising from the Cauchy-Schwartz inequality. Reproducing Kernel Hilbert Spaces and Hardy Spaces Infinite Dimensional Hilbert Spaces Question: What do we require about µ , λ in { n}n1=1 { n}n1=1 order to have well-defined inner products? Answer: It must be that 1 1 µ 2 < and λ 2 < (3) | n| 1 | n| 1 n=1 n=1 X X Note that these conditions are both necessary and sufficient 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.) 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 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 λ v <✏.) − n=1 n n P 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 λ v <✏.) − n=1 n n Question:P Is Ra separable vector space? 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.