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Let $H$ be a Hilbert space and suppose that $T : H \rightarrow H$ is a self-adjoint compact operator. It has discrete spectrum of eigenvalues of finite multiplicity: $$ \lambda_1 \geq \lambda_2 \geq \cdots $$

We assume that $T_n : H \rightarrow H$ is a sequence of self-adjoint compact operators. All these operators have got a discrete spectrum of eigenvalues with finite multiplicity: $$ \lambda_{n,1} \geq \lambda_{n,2} \geq \cdots $$

I am looking for a sense of convergence $T_n \rightarrow T$ based on the eigenvalues.

  • What are conditions under which the sequences of eigenvalues of the operators $T_n$ converge to the ones of $T$?
  • In which sense do the eigenvectors converge?
  • How does that relate to other notions of convergence?

I have not found such a notion of convergence in the literature. Obviously, it is a very weak notion, as the case of finite-dimensional Hilbert spaces demonstrates: self-adjoint compact operators with the same eigenvalues (and multiplicities) are at best conjugate to each other.

shuhalo
  • 8,084
  • Perhaps it would be more reasonable to say that $T_n \to T$ if the eigenvalues of $T_n - T$ converge to $0$ in some sense. I believe that this is equivalent to norm convergence. – Smiley1000 Oct 26 '24 at 10:32
  • You can probably consider Hausdorff convergence of the spectra. Now, of course, if you just consider $\sigma(T_n) \to \sigma(T)$ this would be an extremely weak notion of convergence and there’s nothing to be said about eigenvectors. If, instead, you consider $\sigma(T_n - T) \to {0}$, then this is equivalent to norm convergence, as @Smiley1000 mentioned in their comment. – David Gao Oct 26 '24 at 11:39
  • Possibly related: link – Frederik vom Ende Oct 26 '24 at 13:59

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