# Confusing result about the spectrum of compact operators

1. Aug 11, 2011

### AxiomOfChoice

I have been posting on here pretty frequently; please forgive me. I have an exam coming up in functional analysis in a little over a week, and my professor is (conveniently) out of town.

We proved in our class notes that if $T:X\to X$ is a compact operator defined on a Banach space $X$, $\lambda \neq 0$, and $\lambda \in \sigma_p(T)$ (the point spectrum; i.e., the set of eigenvalues of $T$), then the range $\mathcal R(T_\lambda) = \mathcal R(T-\lambda) \neq X$. The argument given to support this conclusion is complicated and relies (among other things) on the Riesz lemma, so I won't reproduce it, unless I'm asked to do so.

However, in the very next theorem, we show that if $\lambda \neq 0$ and $\lambda \in \sigma(T)$, then $\lambda \in \sigma_p(T)$. The argument is broken down into cases: either $\mathcal R(T_\lambda) = X$ or $\mathcal R(T_\lambda) \neq X$. The $\mathcal R(T_\lambda) = X$ case is presented as follows: If $\lambda \neq 0$ and $\lambda \in \sigma(T)$ but $\mathcal R(T_\lambda) = X$, then $(T-\lambda)^{-1} = T_\lambda^{-1}$ cannot exist (otherwise we would have $\lambda \in \rho(T)$), so we must have $\ker T_\lambda \neq \{ 0 \}$; hence $\lambda \in \sigma_p(T)$, since we then have $x \neq 0$ that satisfies $T_\lambda x = (T-\lambda)x = 0$.

Here is my question: This argument makes sense, but doesn't the contrapositive of the first theorem I mentioned give $(\mathcal R(T_\lambda) = X) \Rightarrow (\lambda \notin \sigma_p(T))$ if $\lambda \neq 0$? Is there a subtle point in the logic I'm missing, or is the argument given somehow unsound?

Last edited: Aug 11, 2011
2. Aug 11, 2011

### micromass

This is true. So together with that "very next theorem" implies that there are no $\lambda\neq 0$ such that $\mathcal{R}(T_\lambda)=X$. But you do not know that a priori.

So you have here a very curious phenomenon. The theorem implies that all $\lambda\neq 0$ such that $\mathcal{R}(T_\lambda)=X$ will belong to the point spectrum. But this will means that there are no such $\lambda$.