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Suppose that I have a sequence of compact, injective operators $\{T_\delta\}_{\delta>0}$ on a Hilbert space $H$ such that each operator $T_\delta$ has infinitely many eigenvalues. My question is the following.

Question: If $T$ is a compact, injective operator on $H$ and $T_\delta\to T$ in operator norm as $\delta\to0^+$, then must $T$ have infinitely many eigenvalues?

I know that each eigenvalue of $T$ must be a limit of eigenvalues of $T_\delta$, but I am not sure how to establish the converse in the above sense. According to Convergence of spectrum with multiplicity under norm convergence, any $\lambda\in\mathbb{C}$ which is the limit point of eigenvalues of $T_\delta$ (with multiplicity taken into account) must be an eigenvalue of $T$, but I am not sure how to guarantee that infinitely many such points exist. I have mainly used Kato's book Perturbation Theory for Linear Operators as a reference, but I have been unable to find the right result.

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An example of a compact injective operator on $\ell^2(\mathbb Z)$ (i.e. square-summable two-sided sequences) with no eigenvalues is $T$ defined by $$ (Tx)_i = x_{i+1}/(|i|+1)$$ This is the norm limit of compact injective operators $T_N$ where $$ (T_N x)_i = \cases{x_{i+1}/(|i|+1) & if $|i| \le N$\cr x_{-N}/N & if $i = N+1$\cr x_{i}/(|i|+1) & otherwise\cr} $$ which have infinitely many eigenvalues.

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  • $\begingroup$ Thank you! I originally tried to construct a counterexample using this operator, but I couldn't see how to obtain the right sequence that converged in norm. This was really helpful. $\endgroup$ – gar May 10 at 15:31

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