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Suppose $H$ is the reproducing kernel Hilbert space on a space $X$ with reproducing kernel $K$. If, say, $K - c$ is a positive definite kernel for some $c>0$ then $H$ contains the constant functions on $X$. But this condition is not necessarily easy to verify.

Are there other known sufficient conditions that imply that a reproducing kernel Hilbert space on a space $X$ contains the constant functions? In particular I'm interested in the case in which $X$ is a compact metric space and $K(x,y) = e^{-d(x,y)}$. (This kernel is not necessarily positive definite; I'm interested in those $X$ for which it is.) Certain special cases, like subsets of Euclidean space or spheres, I can handle with Fourier analysis via the above condition, but the more general case is unclear.

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  • $\begingroup$ Your edits appeared after I posted my answer, but I don't think you want $\delta_{xy}$ in the condition you give, rather you subtract off the kernel that is identically $1$ on $X\times X$. Also, for most of the kernels I'm familiar with this condition is not so hard to check; do you have a particular kernel that you're looking at? $\endgroup$
    – Mike Jury
    Commented Jun 4, 2013 at 15:14
  • $\begingroup$ Mike: You're right, I was sloppy about which basis I was thinking about. I'll also add in the specific kernel I'm interested in. $\endgroup$
    – Mark Meckes
    Commented Jun 4, 2013 at 15:17
  • $\begingroup$ There is another condition in terms of the geometry of the embedding $X\to H$ via $e:x\to k_x$ (assuming the normalization $k(x,x)\equiv 1$), though I'm not sure it's any easier to check: The space $H$ contains the constant functions if and only if the image of $X$ under $e$ lies in an affine hyperplane of $H$. (If $w\in H$ is constant on $X$, then the $k_x$'s lie in an affine hyperplane orthogonal to $w$, and conversely.) Though maybe this is the conclusion you want rather than a condition you want to check... $\endgroup$
    – Mike Jury
    Commented Jun 4, 2013 at 16:59
  • $\begingroup$ Mike: Your guess is correct --- this is actually the conclusion I want. $\endgroup$
    – Mark Meckes
    Commented Jun 4, 2013 at 18:27

1 Answer 1

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Yes. A (fairly standard) fact about RKHS's is that a function $f:X\to \mathbb C$ belongs to the RKHS (with kernel $k$) if and only if there is a $c>0$ such that the Hermitian kernel $$ k(x,y) -c^2f(x)\overline{f(y)} $$ is positive semidefinite; the norm of $f$ is then equal to $1/c$ for the best possible $c$. Thus the constants belong to the RKHS if and only if $$ k(x,y)-c^2 $$ is positive semidefinite for some positive $c$.

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  • $\begingroup$ Thanks for that. As you noted, I added an edit while you were writing your answer to point out that I know this condition, but I wasn't positive that it was necessary as well as sufficient. $\endgroup$
    – Mark Meckes
    Commented Jun 4, 2013 at 15:25
  • $\begingroup$ Do you know if this fact applies to real valued RKHSs too? Also, do you know where I might find a reference for this? $\endgroup$
    – user27182
    Commented Apr 25, 2021 at 12:15
  • $\begingroup$ Do you have a reference for this? $\endgroup$
    – user27182
    Commented Feb 24, 2023 at 11:58
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    $\begingroup$ Theorem 4.15 here: math.uh.edu/~vern/rkhs.pdf $\endgroup$
    – user27182
    Commented Mar 8, 2023 at 22:42
  • $\begingroup$ @user27182: link doesn't work, do you have an update? $\endgroup$
    – Wicher
    Commented Jun 30 at 20:32

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