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.

  • 1
    $\begingroup$ 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

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 $(T-\lambda I)z_n \rightarrow 0$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.