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Among the so-called "classical" Hilbert spaces ($L^2$, Sobolev, Hardy, Bergman, etc.), which are very well-studied, which are RKHSs?

It is easy to construct example of RKHSs by applying the Moore–Aronszajn theorem and defining appropriate kernels, but I would like to firm up my intuition about these spaces by comparing them to "what I know". Surprisingly, I have not found much about this in the literature, although I am certain I am just looking in the wrong places.

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    $\begingroup$ Honestly pretty much any Hilbert space of functions you'll come across is a RKHS (Lest we forget, $L^2$ spaces are not actually spaces of functions). $\endgroup$
    – Yonah Borns-Weil
    Commented Aug 7, 2023 at 4:16
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    $\begingroup$ Morrey's inequality tells you which Sobolev spaces embed into the space of continuous functions, and thus are RKHSs. $L^2(X,\mu)$ is an RKHS if and only if $(X,\mu)$ is purely atomic. $\endgroup$
    – Nate Eldredge
    Commented Aug 7, 2023 at 4:41
  • $\begingroup$ For the uninitiated, presumably RKHS stands for Reproducing Kernel Hilbert Space? $\endgroup$
    – Igor Khavkine
    Commented Aug 7, 2023 at 9:34
  • $\begingroup$ A non-rigorous POV that I think is still useful: if you have some space of functions on a set $X$, how do you know if it is complete w.r.t. a given inner product? You'll need to show that given a Cauchy sequence $(f_n)$ there is some genuine function $f$ on $X$ that is the norm limit of the $f_n$, and in many cases the only way to define $f(x)$ is to know that the sequence $f_n(x)$ is Cauchy. But how do you get pointwise Cauchy estimates from a Cauchy norm condition? Well, you probably want "evaluation at x" to be continuous w.r.t. your norm -- and that is the RKHS condition. $\endgroup$
    – Yemon Choi
    Commented Aug 8, 2023 at 1:45
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    $\begingroup$ @YemonChoi + YonahBorns-Weil: Thank you for the suggestions! I guess what I was looking for is a rigorous statement of these claims. In fact, I finally found some details here: math.uh.edu/~vern/rkhs.pdf If either of you would like to your own details as an answer, I will happily accept it. Otherwise, I might self-answer with this link. $\endgroup$
    – lost_analyst
    Commented Aug 8, 2023 at 6:13

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