4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How's $x_0$ related to $T$? $\endgroup$ – Michael Jul 12 '13 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 Jul 12 '13 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? $\endgroup$ – Alex Lapanowski Jul 12 '13 at 16:00
  • $\begingroup$ @Alex Lapanowski: it is true. I would recommend the book by Reed, Simon "Functional analysis" Chapters 6,7. $\endgroup$ – Yurii Savchuk Jul 15 '13 at 11:44
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 $L^2(\mathbb R,\mu).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.