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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.