I'm trying to show that given an operator $T \in B(X)$ with $X$ Banach we have $$\sigma(T) = \sigma_{ap}(T) \cup \sigma_p(T') $$
Where $T' \in B(X')$ is the dual operator.
I know that $\sigma(T) = \sigma(T')$ and certainly $\sigma_{ap}(T) \subset \sigma(T)$ so we have $\sigma_{ap}(T) \cup \sigma_p(T') \subset \sigma(T)$ and I'm struggling with the other inclusion.
If we write $R_{\lambda} = {\lambda}I - T$ and now we want to show if $\lambda$ is outside of $ \sigma_{ap}(T) \cup \sigma_p(T')$ then $R_{\lambda}$ is invertible. Here is what I have so far:
- $R_{\lambda}$ has closed image. ($\lambda$ is not in the approx. point spectrum of $T$)
- $R_{\lambda}$ is injective - the point spectrum of $T$ is contained in the approximate point spectrum of $T$.
- I'd like to show that $R_{\lambda}$ has dense image. We know that $\ker(R_{\lambda}')$ is empty - so if $g \in Y'$ is such that $g(R_{\lambda})(x) = 0$ then $g$ must be zero. I'd like to maybe use the Hahn-Banach theorem on an element in $X$ not in the closure of $\operatorname{im} R_{\lambda}$ or perhaps by extending a functional on $\operatorname{im}R_{\lambda}$ but unfortunately I can't seem to get anywhere with this!
Thanks for any help!
CASE 1: If $\mathcal{R}(T-\lambda I)$ is not dense, then $\lambda\in\sigma_{p}(T')$.
CASE 2: $\mathcal{R}(T-\lambda I)=X$.
2a. $T-\lambda I$ is injective, which forces $(T-\lambda I)^{-1}$ to be continuous and, hence, $\lambda\in\rho(T)$; or
2b. $T-\lambda I$ is not injective, which gives $\lambda \in\sigma_{p}(T)$.
CASE 3: $\mathcal{R}(T-\lambda I)$ is dense but $\mathcal{R}(T-\lambda I)\ne X$.
3a. $T-\lambda I$ is not injective and, hence, $\lambda\in\sigma_{p}(T)$.
3b. $T-\lambda I$ is injective. In this case $T-\lambda I$ is not continuously invertible because $\mathcal{R}(T-\lambda I)$ is not dense, which means $\lambda\in\sigma_{ap}(T)$.
That covers all cases: $\sigma(T)\subseteq\sigma_{p}(T')\cup\sigma_{p}(T)\cup\sigma_{ap}(T)$.