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how can I show the following theorem?

Let $H$ a Hilbert space and $T:H \to H$ a linear, continuos 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 basicly $\sigma(T) \subseteq \sigma_{ap}(T)$. Thanks for your help.

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    $\begingroup$ if $\lambda \in \sigma(T)$ then there exists $x$ such that $(T-\lambda I)x = 0$, or $(T-\lambda I)^{-1}$ is unbounded. in the second case, you have a sequence $\|y_n\| = 1$ such that $\|(T-\lambda I)^{-1} y_n\| \to \infty$ hence $\|(T-\lambda I) x_n \| \to 0$ with $x_n = \displaystyle\frac{(T-\lambda I)^{-1} y_n}{\|(T-\lambda I)^{-1} y_n\|}$ $\endgroup$ – reuns May 4 '16 at 12:28
  • $\begingroup$ Where exactly do I need the normality then? $\endgroup$ – Yaddle May 4 '16 at 14:36
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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 $(T-\lambda I)z_n \rightarrow 0$.

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  • $\begingroup$ I am not quite sure, how $\overline{\mathcal{R}(T-\lambda I)}=\mathcal{N}(T^*-\overline{\lambda}I)^{\perp}$ follows. Can you elaborate that? $\endgroup$ – Yaddle May 5 '16 at 11:49
  • $\begingroup$ @Yaddle : $M^{TT}=\overline{M}$ for a subspace. $y\in\mathcal{R}(T-\lambda I)^T$ iff $((T-\lambda I)x,y)$ for all $x$; equivalently, $(x,(T^*-\overline{\lambda}I)y)=0$ for all $x$ or $(T^*-\overline{\lambda}I)y=0$. $\endgroup$ – DisintegratingByParts May 5 '16 at 12:37
  • $\begingroup$ Thanks! Sorry, that was quite obvious. $\endgroup$ – Yaddle May 5 '16 at 12:48
  • $\begingroup$ @Yaddle : Glad to help. $\endgroup$ – DisintegratingByParts May 5 '16 at 13:29
  • $\begingroup$ @Student : I suspect that every book on Functional Analysis would give such a proof. $\endgroup$ – DisintegratingByParts Feb 7 at 14:13

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