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.(adsbygoogle = window.adsbygoogle || []).push({});

We proved in our class notes that if [itex]T:X\to X[/itex] is a compact operator defined on a Banach space [itex]X[/itex], [itex]\lambda \neq 0[/itex], and [itex]\lambda \in \sigma_p(T)[/itex] (the point spectrum; i.e., the set of eigenvalues of [itex]T[/itex]), then the range [itex]\mathcal R(T_\lambda) = \mathcal R(T-\lambda) \neq X[/itex]. 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 thevery next theorem, we show that if [itex]\lambda \neq 0[/itex] and [itex]\lambda \in \sigma(T)[/itex], then [itex]\lambda \in \sigma_p(T)[/itex]. The argument is broken down into cases: either [itex]\mathcal R(T_\lambda) = X[/itex] or [itex]\mathcal R(T_\lambda) \neq X[/itex]. The [itex]\mathcal R(T_\lambda) = X[/itex] case is presented as follows: If [itex]\lambda \neq 0[/itex] and [itex]\lambda \in \sigma(T)[/itex] but [itex]\mathcal R(T_\lambda) = X[/itex], then [itex](T-\lambda)^{-1} = T_\lambda^{-1}[/itex] cannot exist (otherwise we would have [itex]\lambda \in \rho(T)[/itex]), so we must have [itex]\ker T_\lambda \neq \{ 0 \}[/itex]; hence [itex]\lambda \in \sigma_p(T)[/itex], since we then have [itex]x \neq 0[/itex] that satisfies [itex]T_\lambda x = (T-\lambda)x = 0[/itex].

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

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# Confusing result about the spectrum of compact operators

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