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Mar 7, 2023 at 15:02 comment added Liviu Nicolaescu See Thm. 10.4.20 of this book www3.nd.edu/~lnicolae/Lectures.pdf
Mar 7, 2023 at 12:43 comment added terceira The statement of 1, holds for the Laplace operator on a compact Riemannian manfold. The crucial fact is that the eigenvalues go to infinity like a power ($>1$) of $n$ and so it is valid in many situations (Schrödinger operators, Laplacian operator on a bounded manifold with boundary, given suitable boundary conditions--Dirichlet or Neumann). The growth conditions on the eigenvalues can be obtained by classical results in the interesting special cases or by variants of the Weyl inequality in the more abstract situations.
Mar 7, 2023 at 11:48 comment added Hao Yin Thank you. I think, following your advice, I can give a proof to both the above claims.
Mar 7, 2023 at 11:09 comment added Leo Moos Does the answer to this question answer your first question? (The convergence considered there is in $L^\infty$, but I think the argument ought to work in $C^k$ also.)
S Mar 7, 2023 at 10:12 review First questions
Mar 7, 2023 at 12:45
S Mar 7, 2023 at 10:12 history asked Hao Yin CC BY-SA 4.0