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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.

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  • $\begingroup$ Did you mean to include the point spectrum in the approximate point spectrum? $\endgroup$ – DisintegratingByParts Jan 16 '19 at 4:24
  • $\begingroup$ The point spectrum is included as $\{ \lambda : \lambda - T \;\text{not injective} \}$ is a subset of $AP\sigma(T)$. $\endgroup$ – fpmoo Jan 16 '19 at 10:25
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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; otherwise this bounded operator would extend continuously to a bounded linear operator $R_{\lambda}$ on the closure $\mathcal{R}(\lambda I-T)^c$ of the range, and $$ (\lambda I-T)(\lambda I-T)^{-1}=I $$ would automatically extend (by continuity) to

$$ (\lambda I-T)R_{\lambda}x=x,\;\;\; x\in\mathcal{R}(\lambda I-T)^c. $$

But that would contradict the fact that the range of $\lambda I-T$ is not closed. So $(\lambda I-T)^{-1}$ cannot be bounded, which 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 $$ (\lambda I-T)f_n =\frac{1}{\|(\lambda I-T)^{-1}e_n\|}e_n \rightarrow 0. $$ Hence $\lambda$ is in the approximate point spectrum of $T$.

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