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I'm trying to prove the following about the closure of the point spectrum: $\overline \sigma_p \subseteq \sigma_c \cup \sigma_p$ where $\sigma_p$ is the point spectrum and $\sigma_c$ is the continuous spectrum. What I noticed so far is that the union of the closure of the point spectrum and the eigenvalues of infinity multiplicity forms the continuous spectra. But I think I'm missing something since I can't conclude.

Best Regards,

Lievet.

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    Do you define $\lambda\in\sigma_c(A)$ to require that $A-\lambda I$ has a dense range? – 2017-02-04
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    Yes, I do. Why? – 2017-02-04

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I don't believe that your assertion is true. For example, let $H = \ell^2(\mathbb{N}) \times \ell^2(\mathbb{N})$, and define $A : H\rightarrow H$ by $$ A(e_n,e_m)=(\frac{1}{n}e_n,e_{m+1}),\;\;\; n,m=1,2,3,\cdots. $$ Then $A(e_n,0)=\frac{1}{n}(e_n,0)$, which means that $0\in\overline{\sigma_p(A)}$. However, the range of $A$ is not dense because $(0,e_1)\perp \mathcal{R}(A)$. Therefore $0\notin\sigma_c(A)$. And it is easy to check that $0\notin\sigma_p(A)$, meaning that $\mathcal{N}(A)=\{0\}$.

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    Actually you are right, but I need to think a little bit about this. Thanks for your answer. – 2017-02-08