Q:

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

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

A:

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.

