if T is normal, then $‎\sigma(T)=‎\sigma‎‎_{‎ap‎}‎(T)‎‎‎‎‎$‎

Question

I want to show that if the operator T is normal, then $‎\sigma(T)=‎\sigma‎‎_{‎ap‎}‎(T)‎‎‎‎‎$‎

Its proof is obvious from one hand.But i cant prove that $‎\sigma(T)\subseteq‎‎‎‎‎\sigma‎‎_{‎ap‎}‎(T)‎‎‎‎‎$‎

Recall that $\sigma‎‎_{‎ap‎}‎(T)‎‎‎‎‎=\{‎\lambda \in\mathbb{C}~|~T-\lambda~is~not~bounded~below‎‎\}‎$

Answer 1

We show that $\sigma_{ap}(T)^c \subset \sigma(T)^c$ :

Suppose $\lambda \notin \sigma_{ap}(T)$, then we want to show that $\lambda \notin \sigma(T)$. Note that $\exists c > 0$ such that $$ \|(T-\lambda I)x\| \geq c\|x\| \quad \forall x\in H $$ and hence $(T-\lambda I)$ is injective, so it suffices to prove that $R(T-\lambda I) = H$.

Also, $R(T-\lambda I)$ is complete (why?), and so it is closed in $H$. It suffices to show that $R(T-\lambda I)^{\perp} = \{0\}$. So choose $x \in R(T-\lambda I)^{\perp}$, then $$ 0 = \langle x,(T-\lambda I)(T^{\ast} - \overline{\lambda} I)x \rangle $$ $$ = \langle x, (T^{\ast} - \overline{\lambda}I)(T-\lambda I)x\rangle \quad\text{ (since $T$ is normal)} $$ $$ = \langle (T-\lambda I)x, (T-\lambda I)x \rangle \geq c^2\|x\|^2 $$ and hence $x=0$. Thus $R(T-\lambda I) = H$, and so $T-\lambda I$ is invertible.

Answer 2

a normal operator is invertible if and only if it is bounded below(you can use the fact that the operator is one to one if and only if its adjoint is one to one)