Let $H$ be a Reproducing Kernel Hilbert Space with elements $f:X\rightarrow \mathbb{C}$, with kernel $K(x, y)$. My question is whether, for some choice of $x_i\in X$, it is the case that $u_i:=K(x_i, \cdot)$ is a basis for $H$. What additional conditions should be imposed on the $x_i$?
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2$\begingroup$ It will depend on what sense of "basis" is meant...orthonormal basis? Riesz basis? something else? (In most reproducing kernel applications, the kernel functions are not orthogonal.) $\endgroup$ – Mike Jury Apr 2 '13 at 18:18

$\begingroup$ In the weakest sense, i.e., not necessarily an orthobasis nor a Riesz sequence. The span of $K(x, \cdot)$ is dense in $H$. I think that if one adds the requirement that $X$ be a separable space, then the result follows. $\endgroup$ – gappy3000 Apr 12 '13 at 2:39

$\begingroup$ In case you are still reading: what precisely do you mean by "weakest sense"? It sounds like you mean something different from "Schauder basis", which is the weakest notion that I've usually seen of "basis" $\endgroup$ – Yemon Choi Aug 13 '13 at 3:13
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If {z_n} is not a zero set for the Hilbert space (separable) then K(z_n, .) spans, but is not necessarily a Schauder basis. I believe the sequence needs to be an interpolating sequence for the Kernels to form a Shauder basis.