# Equivalence of definitions for the approximate point spectrum

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.

• Did you mean to include the point spectrum in the approximate point spectrum? – DisintegratingByParts Jan 16 '19 at 4:24
• The point spectrum is included as $\{ \lambda : \lambda - T \;\text{not injective} \}$ is a subset of $AP\sigma(T)$. – fpmoo Jan 16 '19 at 10:25

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 $$T$$.