The reproducing kernel for harmonics on compact manifolds Page 39, proposition 1.1.3 here, http://www.cis.upenn.edu/~cis610/sharmonics.pdf clearly explains how for every ``level" (the parameter $k$ in the proposition) one can construct a function ("kernel") such that the function space of all level $k$ harmonics on $S^n$ is the "Reproducing Kernel Hilbert Space" (RKHS) for that kernel. Because of the orthogonality property of the harmonics it probably wasn't surprising that the kernel itself also turned out to be completely described by the harmonics themselves at the same level. 


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*Firstly I want to confirm if my above interpretation is correct at all! I am inquiring if this proposition 1.1.3 in my cited reference can at all be see as constructing a RKHS. Because as far as I understand for their $F_k$ function in the reference to be considered as a "kernel" it has to converge pointwise and also maybe uniformly on compact sets. Is that true here?

*Is this something special about $S^n$ or are there other (compact?) manifolds too on which the Hilbert space of harmonic functions similarly decomposes so that the eigenspace of each eigenvalue acts as a RKHS for the kernel being constructed similarly? (and this kernel will again have the "right" (pointwise? uniformly on compact sets?) convergent properties?) 
 A: First, the issue is not at all about "harmonic" functions on a compact manifold: the fact that eigenfunctions for the Laplace-Beltrami operator on the (round) sphere are restrictions of harmonic functions on an ambient (flat) Euclidean space is rather special, and certainly does not mean that the spherical harmonics are "harmonic" on the sphere itself. Depending on normalization, their eigenvalues for the actual Laplace-Beltrami operator are something like $-k(k+n-2)$ where $k$ is the degree and $n$ is the dimension of the ambient Euclidean space (not of the sphere). This is discussed in some form in Stein-Weiss, and many other places.
For fixed $n$, yes, the space of degree-$k$ spherical harmonics is finite-dimensional, and consists of smooth functions, so $K_k(x,y)=\sum_j f_j(x)\overline{f_j(y)}$ is a finite sum, so converges very well, and gives a reproducing kernel for that finite-dimensional eigenspace, yes.
On a compact Riemannian manifold, the Laplace-Beltrami operator has compact inverse/resolvent, so has finite-dimensional eigenspaces consisting of smooth functions. Thus, as with spherical harmonics, the functional "evaluate at point $z_o$" is continuous, and is given by inner product against an element of the eigenspace.
