# Converse of the Spectral Theorem

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?

• 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. – 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.