This is a very basic question about the definition of a reproducing kernel Hilbert space (RKHS).

It seems the standard definition of a RKHS is as a Hilbert space $H$ of functions on some set $X$ where the evaluation functionals are continuous. It is then noted that the Riesz representation theorem yields a map $k$ from $X$ to $H$ such that evaluation at $x$ is the same as taking the inner product with $k(x)$. My question is, why not simply identify $x$ with $k(x)$? Then we would have the following perhaps more abstract definition

(*) A RKHS is a Hilbert space $H$ together with a distinguished subset $K$ with dense linear span.

The density condition is to ensure that elements of $H$ can indeed be identified with the function you get by taking inner products with elements of $K$. The main difference I can see is that (*) ignores the fact the functions in the original definition may not separate the points of $X$. But this would be the degenerate case no? In any case, if really needed this could be dealt with by an extra function from $K$ to cardinals to indicate their 'multiplicity'.

There's nothing deep about what I'm saying here, I'm just wondering why this alternative definition is never mentioned? Is it too obvious to even worth noting or am I missing something fundamental about the original definition? I guess most RKHS's are already given as spaces of functions and maybe this is usually the best way to think of them. But I imagine (*) would be good for some things, e.g. Hilbert space frames could then be considered as special kinds of RKHS's.

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    $\begingroup$ Your question seems to be "Why not use this alternate definition, which does not handle some special cases, even though those special cases could be handled by extra machinery?" I think this nearly answers itself, especially when the original definition is quite simple. I also think of the "Why Whittaker models?" question; sometimes having a preference among isomorphic options is meaningful. $\endgroup$ – LSpice Feb 23 '17 at 18:23
  • $\begingroup$ It seems you're saying "Your definition is valid, I just don't like it." That's fine, I'm not claiming (*) is superior, just that it provides a more abstract perspective. The fact (*) is never even mentioned made me wonder if I was missing some key aspect of the original definition. $\endgroup$ – Tristan Bice Feb 24 '17 at 12:25
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    $\begingroup$ I apologise if my tone was poor. I did not mean it to be. I have no opinion on the definition, either its validity or its desireability. I have never studied RKHS's, and so went only by your own description. To me, it seems that a proposed alternate definition that, as you say, does not describe exactly the same objects as the original cannot really be considered an alternate definition at all, and must instead be considered as a proposal for an alternate collection of objects to study—in which capacity it may well be an excellent proposal! $\endgroup$ – LSpice Feb 24 '17 at 20:09
  • $\begingroup$ You can view a RKHS not as a specified set of continuous linear functionals on a Hilbert space, but rather a map from your given set into the dual of a Hilbert space (which has a total image). Then you can have equal point evaluations. I am pretty sure, this is the alternative definition that you are looking for. $\endgroup$ – erz Jun 17 '17 at 21:28

As an interested observer of the "reproducing kernel Hilbert space" world, it seems to me that, for purposes of "outsiders", the typical substance of a discussion about such spaces is an assertion that some naturally occurring space of functions (on a physical space with various further attributes) is a RKHS... because one can prove a Sobolev-like inequality, namely, that local sup-norms are dominated by whatever given $L^2$ norm. The latter implies that everything in that $L^2$ space is continuous, and that evaluation-at-a-point is continuous... so (Riesz-Frechet) point-evaluations are given by vectors in the Hilbert space itself, via the pairing.

As you speculate, some aspects of this are easily abstracted, ... and then it becomes unclear why this might be so interesting?

But, it is somewhat akin to observing that infinite-dimensional but separable Hilbert spaces are all isomorphic to $\ell^2$. Boring. But/and if we merely add an operator $T:\ell^2\to \ell^2$, suddenly things are vastly more complicated (and interesting, for suitable $T$).

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