Consider a one-parameter family $A_t$ of unbounded positive self-adjoint operators with discrete spectrum (for example, one can consider a one-parameter family of Laplacians on a compact Riemannian manifold). Kato's real-analytic perturbation theory says that the eigenvalues $\lambda_t$ of $A_t$ vary real-analytically, and we can also select a complete orthonormal basis of eigenfunctions $\varphi_k(t)$ that vary real-analytically in $t$. My question is, let's say that the perturbation is smooth instead of real-analytic. How do the eigenvalues vary then? Also, can we still find an orthonormal basis of eigenfunctions that vary at a certain scale of regularity (in particular, are the eigenfunctions differentiable with respect to $t$)? I am particularly interested in the second question, as I am curious if something nasty happens when eigenvalues have multiplicities. Thanks!
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1$\begingroup$ Something nasty does happen when eigenvalues have multiplicities. I believe these are all discussed already in Kato's book. Summary: when eigenvalues are distinct the smoothness of the eigenprojections correspond to that of the parameter. When they are not distinct the eigenvectors need not even be continuous. (You can even see this for symmetric $2\times2$ matrices.) $\endgroup$– Willie WongCommented Nov 17, 2015 at 18:09
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1$\begingroup$ See also: mathoverflow.net/questions/43124/… and math.upenn.edu/~kazdan/509S07/eigenv5b.pdf $\endgroup$– Willie WongCommented Nov 17, 2015 at 18:12
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1$\begingroup$ The papers cited in mathoverflow.net/a/108630/26935 give a quite complete answer to your question. In particular, see the beginning of mat.univie.ac.at/~michor/DC-perturb.pdf. $\endgroup$– Peter MichorCommented Nov 17, 2015 at 18:37
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$\begingroup$ @WillieWong Nice comment, that is what I was sure is correct. However, I just found this paper: arxiv.org/pdf/0710.3947.pdf. Please take a look at Remark 1.2 on page 2. If the perturbation in question is Ricci flow, how is the author guaranteeing that the eigenfunctions are $C^1$? I am a bit confused now. I don't see that the Ricci flow gives any additional handle on whether the eigenvalues are distinct. $\endgroup$– SMSCommented Nov 20, 2015 at 17:57
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$\begingroup$ The author gave 3 references to that claim. Do none of them answer you question? @HSM $\endgroup$– Willie WongCommented Nov 20, 2015 at 18:47
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