Sobolev regularity via Laplace spectrum

Question

Fix a positive integer $n$ and let $\mu$ be the uniform measure on the sphere $\mathbb{S}^n$, with respect to its usual Riemannian metric $g$. Let $\nabla$ be the Laplacian on $(\mathbb{S}^n,g)$ and suppose that $f\in L^2_{\mu}(\mathbb{S}^n)$ has eigendecomposition $(\lambda_n)_{n=1}^{\infty}$ satisfying $$ \lambda_n \asymp n^{-r} $$ for some $r>0$, where $\lambda_n:=\langle f,\psi_n\rangle$ where $(\psi_n)_{n=1}^{\infty}$ are the eigenfunctions of $\nabla$ therein.

Under what conditions, if any, on the "rate" $r$ can we infer that $f\in H^{k}(\mathbb{S}^n)$ for some $k$ and bound its Sobolev norm $\|f\|_{H^k}$ as a function of $(\lambda_n)_n$?

Furthermore, (when) can we get a bound on its Lipschitz constant using this data?

Answer 1

This issue can be nearly-completely resolved. Namely, as in Stein-Weiss' book on Fourier analysis on Euclidean spaces, there is a sharp bound on the ratio of sup-norm to $L^2$-norm of spherical harmonics of a given degree. In fact, the result admits considerable abstraction...! That ratio is (up to a constant depending on measure, which obviously affects $L^2$ norm but not sup-norm) something like the square root of the dimension of the corresponding irreducible repn of the orthogonal group.

(This replaces the all-too-happy estimate $|e^{i\xi x}|\le 1$ in the tiniest case.)

Some simple algebra (!) shows that the dimensions grow polynomially (in an explicit way...).

So, yes, ignoring details to give a qualitative answer, there is indeed a very explicit good Sobolev theory on spheres. My notes https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2013-14/09_spheres.pdf do lots of background, and write out the case of sufficiently high Sobolev index implies continuity, as an example.

Addition after OP's comment/question: here, as in other cases where we have a Laplacian eigenfunction expansion, with sufficient info about the eigenfunctions and growth rate of their eigenvalues... an "expression" $f=\sum_n \langle f,u_n\rangle\cdot u_n$ for a "function" $f$ in terms of an orthonormal basis $\{u_n\}$ for $L^2$ consisting of eigenfunctions for the Laplacian immediately gives all the $H^k$ norms, even for general $k\in\mathbb R$.

Namely, first, for the definition $|f|^2_{H^k}=\langle (1-\Delta)^kf,f\rangle$ with $0\le k\in\mathbb Z$, $$ |f|^2_{H^k} \;=\; \sum_n |\langle f,u_n\rangle|^2\cdot \langle (1-\Delta)^k u_n,u_n\rangle \;=\; \sum_n |\langle f,u_n\rangle|^2\cdot (1-\lambda_n)^k $$ where I've changed notation so that $\Delta u_n=\lambda_n$.

Second, in fact the right-hand side of the latter display makes perfect sense for $k$ an arbitrary real number.