Approximate point spectrum of a normal operator 
Let $H$ a Hilbert space and $T:H \to H$ a linear, continuous and normal operator. Then for every $\lambda \in \sigma(T)$ there exists a sequence $(x_n)_{n \in \mathbb N}$ with $\Vert x_n \Vert = 1$ for all $n \in \mathbb N$ such that $$\lim_{n \to \infty} \Vert Tx_n - \lambda x_n \Vert = 0,$$
  what means basically $\sigma(T) \subseteq \sigma_{ap}(T)$.

Thanks for your help.
 A: The case where $\mathcal{N}(T-\lambda I)\ne \{0\}$ is covered. So assume $\mathcal{N}(T-\lambda I)=\{0\}$ and $\lambda\in\sigma(T)$. Because $T-\lambda I$ is normal, then
$$
           \|(T-\lambda I)x\|=\|(T^*-\overline{\lambda}I)x\|,\;\;x\in H,
$$
which also implies that $\mathcal{N}(T^*-\overline{\lambda}I)=\{0\}$. Therefore,
$$
      \overline{\mathcal{R}(T-\lambda I)}=\mathcal{N}(T^*-\overline{\lambda}I)^{\perp}=\{0\}^{\perp}=H.
$$
If $\mathcal{R}(T-\lambda I)$ is closed, then $T-\lambda I$ is bijective and closed, which forces $\lambda\in\rho(T)$ and gives a contradiction. Therefore, $T-\lambda I$ is injective with a dense, non-closed range and an unbounded
inverse on that range. Hence, there exists a sequence of unit vectors $\{ x_n \}$ in the range of $T-\lambda I$ such that $y_n=(T-\lambda I)^{-1}x_n$ tends to $\infty$ in norm. Renormalizing the tail of this sequence gives a sequence of unit vectors $\{ z_n = \frac{1}{\|y_n\|}y_n \}$ in $H$ such that $(T-\lambda I)z_n \rightarrow 0$.
