Why are inner products in Reproducing Kernel Hilbert Spaces Dirac evaluation functionals? [closed]

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 ChoiFeb 20 '11 at 1:06

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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