# spectrum of unbounded self-adjoint operators

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.

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?

• (by definition ?) an unbounded operator has a sequence of singular values which tend $\to \infty$. and the singular values of a self-adjoint operators are the same as its eigenvalues – reuns Apr 9 '16 at 16:17
• @user1952009 : Sorry I'm newbie in this stuff, why unbounded operator has divergent seq of singular values? – cjackal Apr 9 '16 at 16:24
• I'm not sure of the proof, but every bounded operator has a SVD, hence intuitively any unbounded operator too, and its singular values are then unbounded (otherwise the operator would be bounded). – reuns Apr 9 '16 at 16:33
• Now for your first question you need a trick: Prove that $f(A)^*=f^*(A)$ and $(fg)(A)=\overline{f(A)g(A)}$. So they are all normal! But normal operators are maximal in the sense if $N\subseteq N'$ then $N=N'$. – C-Star-W-Star Apr 10 '16 at 10:29
• You might have a look into Spectral Measures and Normal Operators. – C-Star-W-Star Apr 10 '16 at 10:35

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 $\mathcal{D}(A)$.