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I'd like to know why inner products in Reproducing kernel Hilbert spaces are (linear) evaluation functionals.

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.

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

$\mathcal{F}_y(x)\;=\;\langle \; x \;,\;y\;\rangle$

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.

PS: I've posted a similar question on math.stackexchange but didn't get any replies.

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closed as too localized by Bill Johnson, Jonas Meyer, Dmitri Pavlov, Willie Wong, Yemon Choi Feb 20 '11 at 1:06

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

You have it backwards--In a reproducing kernel Hilbert space, a pointwise evaluation functional is continuous and hence is given by inner product with some member of the Hilbert space. Not every continuous linear functional is given by pointwise evaluation at a point of the domain. See en.wikipedia.org/wiki/Reproducing_kernel_Hilbert_space for some basics. –  Bill Johnson Feb 19 '11 at 14:17
Olumide, I noticed this here just now, after I answered your question on math.SE. You waited only a few hours, and you should not expect your questions there to be answered immediately. I think you were correct to ask it over there, and I've cast the second vote to close this as too localized. –  Jonas Meyer Feb 19 '11 at 18:00