Talk:Reproducing kernel Hilbert space

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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)

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[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)

[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)
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