Why are inner products in Reproducing Kernel Hilbert Spaces Dirac evaluation functionals? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T14:34:17Z http://mathoverflow.net/feeds/question/55976 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55976/why-are-inner-products-in-reproducing-kernel-hilbert-spaces-dirac-evaluation-func Why are inner products in Reproducing Kernel Hilbert Spaces Dirac evaluation functionals? Olumide 2011-02-19T12:42:59Z 2011-02-19T13:27:06Z <p>I'd like to know why inner products in Reproducing kernel Hilbert spaces are (linear) evaluation functionals.</p> <p>I understand the Riesz representation theorem (in chapter 2 of "Principles of Functional Analysis" by Martin Schechter), and that inner products are linear functionals, and I know what an evaluation functional is; I just can't explain why an inner product (in a RKHS) is evaluation functional, and vise-versa.</p> <p>Edit: to make things clearer, I'm aware that given a Hilbert space $\mathcal{H}$ having the inner-product $\langle x \;,\; y \rangle$, where $x , y \in \mathcal{H}$; if $y$ is fixed $\langle x \;,\; y \rangle$ assigns to each $x$ a number. What I don't understand is why the the relation</p> <p>$\mathcal{F}_y(x)\;=\;\langle \; x \;,\;y\;\rangle$</p> <p>appears to be sufficient to make the inner product a Dirac evaluation functional. Ad I said, I know that the inner product is a functional but I can't explain why it's a (Dirac) evaluation functional.</p> <p>PS: I've posted a similar question on math.stackexchange but didn't get any replies.</p>