spectrum of unbounded self-adjoint operators

Question

I'm self-studying Lax's functional analysis, and I'm stuck in the chapter introducing spectral theory for unbounded self-adjoint operators.

In his book, Lax proved the spectral theorem of this version in p.378 using Nevanlinna's lemma, or Tao called Herglotz representation theorem :

Let $A$ be a self-adjoint operator in a Hilbert space $H$; denote the domain of $A$ by $D$. There is a spectral resolution for $A$, that is, orthogonal projection-valued measure $E$ defined for all Borel measurable subsets of $R$, with the following properties:

• $E(\emptyset)=0, E(\mathbb{R})=1$
• $E(S\cap T)=E(S)E(T)$
• $E^*=E$
• $E$ commutes with $A$
• $D=\left\{u\in H \mid \int t^2dE(t)u<\infty \right\}$

I'm fairly well following the proof given in this book, but I found out several points I cannot make clear about. My question is:

1. It seems to me that Lax does not completely prove the theorem; at the final step he just showed $D\subseteq \left\{u\in H \mid \int t^2dE(t)u<\infty \right\}$. But some cross validation, like Tao's post, says inverse direction is also true, and I cannot fill in the missing proof for myself.

2. In p.390 in Lax, he give an exercise showing unbounded self-adjoint operator has closed unbounded subset of $\mathbb{R}$ as the spectrum. I succeed in showing that the spectrum is a closed subset of the real line(it seems to be exactly the same as in bounded case), but failed to show the unboundedness.

Any hint or reference would be really appreciated.

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I've came up with an idea about 2nd question that if $\sigma(A)$ is bounded, say $|\sigma(A)|<M$, then $A=\int_{\sigma(A)} tdE(t)$ has bounded norm $|A|<\int_{-M}^M |t|dt=M^2$, which is contradiction. Is my argument make sense?

Answer 1

You know that $E(S)x\in\mathcal{D}(A)$ for all $x$ if $S$ is a bounded measurable subset of $\mathbb{R}$. And $$ AE(-R,R]x = \int_{-R}^{R}\lambda dE(\lambda)x. $$ Furthermore, $$ \|AE(-R,R]x\|^2 = \int_{-R}^{R}\lambda^2d\|E(\lambda)x\|^2. $$

If $x\in\mathcal{D}(A)$, then $AE(-R,R]x=E(-R,R]Ax$ converges to $Ax$ as $R\rightarrow\infty$ and, therefore, the above gives $$ x \in\mathcal{D}(A)\implies \int_{-\infty}^{\infty}\lambda^2d\|E(\lambda)x\|^2 < \infty. \tag{*} $$ Conversely, if $\int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty$, then $AE(-R,R]x=\int_{-R}^{R}\lambda dE(\lambda)x$ converges as $R\rightarrow\infty$ to some $y$. Because $E(-R,R]x\rightarrow x$ as $R\rightarrow\infty$, and $AE(-R,R]x\rightarrow y$ as $R\rightarrow\infty$, and because $A$ is closed, then $x\in\mathcal{D}(A)$ and $Ax=\int_{-\infty}^{\infty}\lambda dE(\lambda)x$. Hence, the reverse implication is proved for $(*)$, which gives the full characterization of $\mathcal{D}(A)$.

Answer 2

I just answered the question spectrum of unbounded self-adjoint operators. You can find a proof for your second point in

Weidmann, Joachim. Linear operators in Hilbert spaces. Vol. 68. Springer Science & Business Media, 2012.

It seems, however, that this book uses a somewhat different concept than Lax - that of a spectral family instead of a spectral resolution. But perhaps you can translate the relevant ideas into Lax's framework.