Timeline for Reproducing kernel Hilbert spaces and the isomorphism theorem

9 events

when toggle format	what		by	license	comment
May 23, 2011 at 13:19	vote	accept	Simon
May 22, 2011 at 9:47	history	edited	t.b.	CC BY-SA 3.0	edited title
May 22, 2011 at 8:01	answer	added	Tim van Beek		timeline score: 14
May 22, 2011 at 6:42	history	edited	Phira	CC BY-SA 3.0	edited body
May 22, 2011 at 1:33	comment	added	Jonas Meyer		To make part of Theo's first comment more explicit: Elements of $L^2[0,1]$ are not functions. The problem isn't that evaluation at $x\in[0,1]$ isn't continuous, but that the evaluation doesn't exist. Also, a minor nitpick: 2 Hilbert spaces are isomorphic if they have orthonormal bases with the same cardinality (as is the case for $\ell_2$ and $L^2[0,1]$), not if the spaces themselves have the same cardinality.
May 21, 2011 at 23:27	comment	added	t.b.		More to the point, of course the functions can be interpreted on the respective sets, but the evaluation functionals won't necessarily be continuous anymore.
May 21, 2011 at 23:12	comment	added	Simon		Beautiful, thank you.
May 21, 2011 at 23:11	comment	added	t.b.		It is of fundamental importance that the elements of a reproducing kernel Hilbert space are functions on a set* $X$*. The reproducing kernel itself is a function on $X \times X$. The isomorphism between $\ell^2$ and $L^2[0,1]$ won't respect this interpretation of an element as a function on $\mathbb{N}$ or $[0,1]$. Jonas Meyer's answer to this question might clarify things a little.
May 21, 2011 at 23:03	history	asked	Simon	CC BY-SA 3.0