Let $T$ be a (possibly unbounded) self-adjoint operator on a Hilbert space. Assume that we for some reason know that the point spectrum of $T$ consists of a finite number of eigenvalues $\lambda _1, \lambda _2, \ldots ,\lambda _N$ and that $T$ in addition to this has some continuous spectrum. Let $f(T)$ be defined for instance as in


(or in some other familiar way, e.g. the Cauchy-Green formula or using the Fourier transform) for appropriate functions $f$.

If $\Pi _p$ denotes the projection onto the point spectrum is it then true that $\Pi _p f(T)$ is a trace class operator? In that case, does it hold that $$ \operatorname{Tr} (\Pi _p f(T))= \sum _{j=1}^N f(\lambda _j) \quad ? $$

  • 1
    $\begingroup$ No, for instance take $T$ to be the identity operator on an infinite-dimensional Hilbert space and $f(x) = x$. You'd have to at least assume that the eigenspaces corresponding to $\lambda_1, \ldots, \lambda_N$ are finite-dimensional. $\endgroup$
    – Nik Weaver
    Sep 5 '13 at 14:28
  • 1
    $\begingroup$ With eigenvalues, you mean finite dimensional eigenspaces? Then the answer is yes. $\endgroup$ Sep 5 '13 at 14:28
  • $\begingroup$ @András: no, it's still false because $f$ could collapse some continuous spectrum into point spectrum. E.g., $f(x) = 1$. $\endgroup$
    – Nik Weaver
    Sep 5 '13 at 14:30
  • $\begingroup$ @NikWeaver: I confess I do not understand. Then $\Pi_p f(T)$ is an operator acting on a finite dimensional space (if I understand the definitions correctly). $\endgroup$ Sep 5 '13 at 14:34
  • 2
    $\begingroup$ Well, to summarize the comments of @NikWeaver and myself, if the eigenspaces are finite dimensional, then it is true and your formula follows in a straightforward way from the spectral theory of selfadjoint operators. Otherwise in general it is not true, you need assumptions on your function. $\endgroup$ Sep 6 '13 at 10:45

You should see the paper on Berezin inequality:



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