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It is known that the spectrum of a compact operator $T \in B(X)$, where $X$ is an infinite-dimensional Banach space, is given by

$\sigma(T)=\sigma_p(T) \cup \{0\}$

and $0$ is the only accumulation point of $\sigma(T)$.

Answers to similar questions suggest that the converse is true if $X$ is a Hilbert space and $T$ is self-adjoint.

Does the converse also hold in this more general setting or does it fail for Banach spaces?

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  • $\begingroup$ Actually, more is true. For $\lambda\in\sigma(T)\setminus\{0\}$, the null space $\ker (T-\lambda)$ should be of finite dimensional if $T$ is compact. $\endgroup$ – Song Jan 16 at 9:07
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The converse does not hold, even in Hilbert spaces! Let us take $X = \ell^2$. We consider the operator $T_a$ induced by multiplication with a real sequence $a \in \ell^\infty$, i.e., $$T_a \, x = (a_1 \, x_1, a_2 \, x_2, \ldots ).$$ Properties of this operator:

  • $T_a$ is always self-adjoint.
  • The point spectrum is $\{a_1, a_2, \ldots\}$.
  • $T_a$ is compact iff $a_n \to 0$.

To give a precise example, we consider $$a_n = \begin{cases} 0 & \text{if $n = 1$}\\1 & \text{if $n>1$ is odd}\\ 2/n & \text{if $n$ is even}.\end{cases}$$ Then, $T_a$ is not compact, but $\sigma(T_a) = \sigma_p(T_a) = \{1/n \mid n \in \mathbb N\} \cup \{0\}$.

I think that the converse holds (in Hilbert spaces) if you assume that the eigenspaces associated with $\lambda \in \sigma_p \setminus\{0\}$ are finite-dimensional.

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