Talk:Reproducing kernel Hilbert space

From Wikipedia, the free encyclopedia

You have new messages (last change).

What is this sesquilinearity convention of which you speak? Do you mean linear in the second variable and conjucate-linear in the first? If so, it seems a strange convention to me. Lupin 08:39, 6 Aug 2004 (UTC)

Contents

[edit] Norm-continuous

The article states that H is a reproducing kernel Hilbert space iff the linear map f→f(x) is norm-continuous for any element x of X. -- what does norm-continuous mean in this context? linas 15:06, 25 Jun 2005 (UTC)

H is a Hilbert space, thus carries a norm. This f → f(x) has a weell-defined topology on both source and target.--CSTAR 15:38, 25 Jun 2005 (UTC)

[edit] Examples, too

Some examples would be handy too. Reading this article naively, one has the Dirac delta function playing the role of K, so that f(x)=\int \delta(x-y) f(y) dy and δ is usually naively understood to be the "identity operator" on the Hilbert space in question (that is, its the sum over all states, e.g. the sum over all orthogonal polynomials on an interval). Is this naive reading correct? Can I correctly say that an example is

K(x,y)=\sum_{n=0}^\infty T_n(x) T_n(y)

where T_n ar the Chebyshev polynomials (and the norm/volume elt/weight is taken appropriately)? It seems to fulfill the requirements described in this article, I think ... linas 15:21, 25 Jun 2005 (UTC)

That's just it, the delta function is not in the space.--CSTAR 15:38, 25 Jun 2005 (UTC)
Would you object if I added the above as an example to the article? linas 18:31, 25 Jun 2005 (UTC)

[edit] Counter-examples, too

Assuming the example given above is correct, then, naively, every hilbert space encountered by a naive physicist is a reproducing kernel hilbert space. So it now becomes hard to imagine a Hilbert space that doesn't have such a beast associated with it. (Since K seems to be defined as "the identity operator", and these are always easily constructed as a sum over states). linas 15:27, 25 Jun 2005 (UTC)

What's relevant here is the realization of the Hilbert space as a space of functions on a set X.--CSTAR 15:38, 25 Jun 2005 (UTC)
Well, then, for the record, are there any hilbert spaces that are not realized as a space of functions on a set? Alternately, are there any hilbert spaces that are realized as a space of functions on a set, but do not have the norm-continuous property? I'm one of these people who don't feel comfortable with a topic until I have understood both an example and a counter-example. I'll try to think of something, but if you have something up your sleeve, that's certainly better than whatever I might dream up. linas 18:31, 25 Jun 2005 (UTC)
Yes: any L2μ for a non-atomic measure μ. These are realized as equivalence classes of functions.--CSTAR 18:41, 25 Jun 2005 (UTC)
late late comment: i think linas's original post is right. every Hilbert space H is isomorphic to a reproducing kernel Hilbert space, meaning there exist a set X, and a positive kernel K: X × X → C such that the resulting Hilbert space is isomorphic H, and, yes, K can be taken as, essentially, the identity operator.Mct mht 19:38, 30 March 2007 (UTC)

[edit] Edit

User:Oleg Alexandrov states that the sentence I edited was wrong, e.g [1]. Hoever, several sentences were modified. Which sentence are you referring to here?--CSTAR 17:34, 25 Jun 2005 (UTC)

Well, I modified there only one sentence, everything else is just link insertion. I was referring to the sentence:
H is a reproducing kernel Hilbert space iff the linear map
f \mapsto f(x)
is norm-continuous for any element x of X.

I thought that the part "for any element x of X" is wrong, but then I realized I was wrong, so I put it back in the next edit. Does that make more sence now? Oleg Alexandrov 17:43, 25 Jun 2005 (UTC)

Seems to. I'll give this more thought later tonight. linas 18:35, 25 Jun 2005 (UTC)
So, basically x is fixed, and then you get a linear map with input f and output f(x). I understand that this is weird, because usually the function f is fixed and input is x while output is f(x). Oleg Alexandrov 19:29, 25 Jun 2005 (UTC)
An important class of examples are Hilbert spaces of analytic functions.--CSTAR 05:59, 26 Jun 2005 (UTC)

[edit] article is not consistent

what is meant by "reproducing kernel Hilbert space" here? sections of the article disagree.

  1. one can say that a Hilbert space of functions on X is a RKHS if it admits a positive definite kernel K on X × X. this is the approach in the first section. with this definition, as CSTAR said above, any L2μ for a non-atomic measure μ would not be a RKHS. (BTW, the positivity of the kernel is really the central theme whether one adopts this formulation or the one below, but that's not pointed out at all in the intro.)
  2. one can start with a positive kernel K on X × X. then there exists a Hilbert space H for which K is a reproducing kernel. this is alluded to in the Moore-Aronszajn theorem section. but then "is a reproducing kernel for" now means something slightly different. it is no longer insisted that H consists of functions on X. after quotiening out the degenrate subspace and taking the completion, we'd end up with equivalence classes in general. in this formulation, any Hilbert space is a RKHS, because the existence of orthonormal basis.

this should be resolved. if it's ok, i am going to attach clean-up tag to the article. Mct mht 01:40, 31 March 2007 (UTC)

also, the the Moore-Aronszajn theorem section contains some unusual notation and language ("first class citizens"?), and the very brief outline of the construction doesn't look quite right. Mct mht 02:04, 31 March 2007 (UTC)

Retrieved from "http://en.wikipedia.org../../../r/e/p/Talk%7EReproducing_kernel_Hilbert_space_5b41.html"