My book says that a self-adjoint bounded linear operator $A:H\to H$ on a complex Hilbert (not sure if separability is needed) space has a real spectrum.

I guess that the key is in the fact that any $f\in H^{\ast}$ can be represented by a functional of the form $\langle-,x_0\rangle$ for some $x_0\in H$, but I am not able to use that fact. I also see that $\forall x,y\in H\quad\langle(A-\lambda I)x,y\rangle=\langle x,(A-\bar{\lambda} I)y\rangle$ but I'm not sure that can be useful to prove the statement...

Thank you very much for any help!!!

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    $\begingroup$ I think you forgot to write self-adjoint in your first sentence, even though you write self-adjoint in the title. $\endgroup$ – DisintegratingByParts Sep 9 '14 at 18:43
  • $\begingroup$ Thank you so much: yes, I did. Edited. $\endgroup$ – Self-teaching worker Sep 9 '14 at 19:23

If $\Im\lambda \ne 0$, and $x \in X$, then $$ \Im\lambda \|x\|^{2} = \Im((A-\lambda I)x,x),\\ |\Im\lambda|\|x\|^{2} \le |((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\|,\\ |\Im\lambda|\|x\| \le \|(A-\lambda I)x\|. $$ So $A-\lambda I$ is injective for all $\lambda\notin\mathbb{R}$. The above inequality can be used to show that the range $\mathcal{R}(A-\lambda I)$ is closed for $\lambda\notin\mathbb{R}$. So $A-\lambda I$ is surjective for $\lambda\notin\mathbb{R}$ because $$ \begin{align} \mathcal{R}(A-\lambda I)& =\overline{\mathcal{R}(A-\lambda I)} \\ & =\mathcal{N}(A^{\star}-\overline{\lambda}I)^{\perp} \\ & = \mathcal{N}(A-\overline{\lambda}I)^{\perp}=\{0\}^{\perp}=H. \end{align} $$ Therefore, $A-\lambda I$ is injective and surjective for $\lambda\notin\mathbb{R}$, which leaves $\sigma(A)\subseteq\mathbb{R}$.

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    $\begingroup$ Thank you so much! My text proposes lemmas to be proven as exercises of a level often much higher than the preparation given in theory part, so I have found material on line to understand why $\overline{\mathcal{R}(A-\lambda I)}=\mathcal{N}(A^{\ast}-\bar{\lambda}I)^{\perp}$, but I still don't understand how to realise that such a range is closed. Moreover why $\Im\lambda\|x\|^{2}=\Im((A-\lambda I)x,x)$ (I only see that $((A-\lambda I)x,x)=(Ax,x)-\lambda\|x\|^2$) and why $\mathcal{N}(A^{\ast}-\bar{\lambda}I)^{\perp}=\mathcal{N}(A-\bar{\lambda}I)^{\perp}$? $\infty$ thanks!!! $\endgroup$ – Self-teaching worker Sep 9 '14 at 23:52
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    $\begingroup$ @DavideZena : $A^{\star}=A$ which is why $\mathcal{N}(A^{\star}-\overline{\lambda}I)=\mathcal{N}(A-\overline{\lambda}I)$. Because $A=A^{\star}$, you also know that $(Ax,x)$ is real because $(Ax,x)^{\star}=(x,Ax)=(A^{\star}x,x)=(Ax,x)$. Finally, the inequality shows the range is closed because $(A-\lambda I)x_{n}\rightarrow y$ implies $\Im\lambda\|x_{n}-x_{m}\| \le \|(A-\lambda I)(x_{n}-x_{m})\|\rightarrow 0$ as $n,m\rightarrow \infty$; that then gives $\lim_{n}x_{n}=x$ for some $x$ and, so, $(A-\lambda I)x_{n}\rightarrow (A-\lambda)x$ which gives $y=(A-\lambda I)x$ is in the range. $\endgroup$ – DisintegratingByParts Sep 10 '14 at 0:18
  • $\begingroup$ As to not seeing that $A^{\ast}=A$, I'm quite embarassed :-s. $\aleph_1$ thanks! $\endgroup$ – Self-teaching worker Sep 10 '14 at 8:27
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    $\begingroup$ @DavideZena : You're welcome. Glad I could help. $\endgroup$ – DisintegratingByParts Sep 10 '14 at 13:03
  • $\begingroup$ That $(A-\lambda I)^{-1}$ is continuous for $\lambda\notin\mathbb{R}$ seems to be considered obvious. Is this because of the open mapping theorem or for more elementary reasons? $\endgroup$ – Stefan Feb 3 at 11:50

The representation of the elements of $H^*$ is not needed.

Let $\lambda$ be in the point spectrum of $A$. Then there is $0\neq x\in H=(H,(\cdot,\cdot))$ such that $Ax=\lambda x$ and by the self-adjointness of $A$: $\lambda(x,x)=(\lambda x,x)=(Ax,x)=(x,Ax)=(x,\lambda x)=\overline{\lambda}(x,x)$. Hence $\lambda=\overline{\lambda}$, thus $\lambda\in\mathbb{R}$.

See e.g. http://en.wikipedia.org/wiki/Spectral_theorem for related results in greater generality.

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    $\begingroup$ Forgive me if I don't understand: why $A-\lambda I=0$? I only know that, if $\lambda$ is in the (continuous or punctual) spectrum, $A-\lambda I$ isn't bijective... Thank you so much!!! $\endgroup$ – Self-teaching worker Sep 9 '14 at 18:44
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    $\begingroup$ Sorry, that was a mistake. I've edited the answer, but the case where $\lambda\in\sigma\setminus\sigma_p$ still needs to be done. $\endgroup$ – Bati Sep 9 '14 at 19:02
  • $\begingroup$ No problem: thank you again! $\endgroup$ – Self-teaching worker Sep 9 '14 at 23:41

the matters are not o straightforward. Please consult

R. K. GOODRICH, The spectral theorem for real Hilbert space, Acta Sci. Math. (Szeged), 33 (1972), 123–127.

(open access)


  • $\begingroup$ Thank you for the resource! $\endgroup$ – Self-teaching worker Jun 25 '15 at 8:22

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