5
$\begingroup$

I have a question regarding the spectral theorem for bounded self-adjoint operators. The book "Functional Analysis, an Introduction" by Eidelman, Milman, and Tsolomitis says that if an operator $T$ has a simple spectrum then it is unitarily equivalent to the operator $T\varphi(\lambda)=\lambda\cdot \varphi(\lambda)$ acting on $L^2$ equipped with the measure generated by the right-continuous function $\lambda\mapsto\langle E_\lambda x_0, x_0\rangle$. Here, $\{E_{\lambda}\}$ is our resolution of identity. Hence, is this unitary equivalence to the multiplication operator only possible when $T$ has a simple spectrum, or is it possible in the general self-adjoint case? Wikipedia says that it's always true, but my functional book along with the book "Treastie on the Shift Operator" say that it's only possible when $T$ has simple spectrum.

Also, the definition of simple spectrum is when there exists an $x_0\in H$ such that $$\{ (E_{\lambda_1}-E_{\lambda_2})x_0\ : \ \lambda_2<\lambda_1\}$$ is a dense subset of our Hilbert Space. Can someone give me some intuition as to why this definition is natural.

$\endgroup$
4
  • $\begingroup$ How's $x_0$ related to $T$? $\endgroup$
    – Michael
    Commented Jul 12, 2013 at 3:07
  • $\begingroup$ I might be too sleepy, but I don't understand "or is it possible in the general self-adjoint case? " $\endgroup$
    – Julien
    Commented Jul 12, 2013 at 13:21
  • $\begingroup$ My books proves that operators with simple spectra are unitarily equivalent to multiplication by t. Is it true in general though? $\endgroup$ Commented Jul 12, 2013 at 16:00
  • $\begingroup$ @Alex Lapanowski: it is true. I would recommend the book by Reed, Simon "Functional analysis" Chapters 6,7. $\endgroup$ Commented Jul 15, 2013 at 11:44

1 Answer 1

2
$\begingroup$

Just to give you some intuition, consider the case of finite-dimensional space.

Let $A$ be a self-adjoint operator on $\mathbb C^n.$ That is $Ae_k=\lambda_k e_k,\ \lambda\in\mathbb R$ for some ONB $e_k,\ k=1,\dots, n.$ Then:

$A$ has simple spectrum iff $\lambda_k\neq \lambda_m$ for all $k,m$ iff there is a cyclic vector $v\in\mathbb C^n$ for $A.$

(Cyclic means that $\{p(A)v\mid p\ \mbox{polynomial}\}$ is dense in $\mathbb C^n$. In this case you can take $v=(1,1,\dots,1)$).

Thus you can define a bounded s.-a. operator to have a simple spectrum if it has a cyclic vector. (You can easily check that multiplication operator $f(x)\mapsto xf(x)$ on $L^2(\mathbb R,\mu)$ has a cyclic vector $v\equiv 1,$ $\mu$ is Borel measure with compact support).

Another characterization: the spectrum of $A$ is simple iff there exists an $x_0\in H$ such that the linear hull of $$\{ (E_{\lambda_1}-E_{\lambda_2})x_0\ : \ \lambda_2<\lambda_1\}$$ is a dense subset of $H$. For the multiplication operator $f(x)\mapsto xf(x)$ it means that the set of step functions is dense in $L^2(\mathbb R,\mu).$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .