Confusing result about the spectrum of compact operators

[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?

 

[micromass]

(\mathcal R(T_\lambda) = X) \Rightarrow (\lambda \notin \sigma_p(T))[/itex] if \lambda \neq 0? Is there a subtle point in the logic I'm missing, or is the argument given somehow unsound?

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.