Let $T: X \rightarrow X$ be a continuous, linear operator on some Banach space $X$.
We defined the approximate point spectrum $AP\sigma(T)$ as the set
$$
\{ \lambda \in \mathbb{C} : \lambda – T \;\text{is not injective or}\; \text{Im}(\lambda – T) \;\text{is not closed in X} \}.
$$
I want to show equivalence with the definition
$$
\lambda \in AP\sigma(T) :\Leftrightarrow \exists (x_n) \subset X, \Vert x_n \Vert =1 \;\text{with}\; \Vert \lambda x_n – Tx_n\Vert \rightarrow 0.
$$
What I have done: "$\Leftarrow$".
What I need: Show that if $\lambda -T$ is injective and $\text{Im}(\lambda – T)$ is not closed in $X$, then there is a sequence such as above.
Best Answer
Suppose that $\lambda I-T$ is injective and that the range $\mathcal{R}(\lambda I -T)$ is not closed. Then $(\lambda I-T)^{-1}$ cannot be bounded $(*)$. This implies the existence of a sequence of unit vectors $\{ e_n \}\subset \mathcal{R}(\lambda I-T)$ such that $\|(\lambda I-T)^{-1}e_n\|\rightarrow\infty$. Then $$ f_n= \frac{1}{\|(\lambda I-T)^{-1}e_n\|}(\lambda I-T)^{-1}e_n $$ is a sequence of unit vectors such that $$ \Vert(\lambda I-T)f_n\Vert =\frac{1}{\|(\lambda I-T)^{-1}e_n\|} \Vert e_n\Vert \rightarrow 0. $$ Hence $\lambda$ is in the approximate point spectrum of $T$.
$(*)$ Assume that $(\lambda I-T)^{-1}\colon\mathcal{R}(\lambda I-T)\to X$ is bounded. Then, it can be extended to a bounded linear operator acting on the closure of its domain, i.e. $R_{\lambda}\colon\overline{\mathcal{R}(\lambda I-T)}\to X$. Now, by continuity $$ (\lambda I-T)(\lambda I-T)^{-1}x=x ,\;\;\; \forall x\in\mathcal{R}(\lambda I-T) $$ extends to $$ (\lambda I-T)R_{\lambda}x=x,\;\;\; \forall x\in\overline{\mathcal{R}(\lambda I-T)}. $$
This means that $\overline{\mathcal{R}(\lambda I - T)}\subseteq \mathcal{R}(\lambda I - T)$ implying that $\lambda I-T$ has closed range.