1
$\begingroup$

Let $H$ be a complex Hilbert space (not necessary separable).

Spectral Theorem: Let $A_1$ and $A_2$ be two commuting normal operators, then there exists a measure space $(X,\mathcal{E},\mu)$, two functions $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:H\longrightarrow L^2(\mu)$, such that each $A_k$ is unitarily equivalent to multiplication by $\varphi_k$, $k=1,2$. i.e. $$UA_kU^*f=\varphi_kf,\;\forall f\in H,\,k=1,2.$$

I look for a reference which contains the proof of the above theorem. More precisely I hope to find a proof which show that $\mu$ can be taken semifinite: i.e. every set of infinite measure contains a subset of positive finite measure.

$\endgroup$
  • $\begingroup$ Do you know about spectral measures, and the spectral measure version of the spectral theorem for normal operators? $\endgroup$ – DisintegratingByParts May 31 '18 at 4:31
  • $\begingroup$ @DisintegratingByParts you mean the integral version? $\endgroup$ – Student May 31 '18 at 6:44
  • $\begingroup$ Yes, the integral version $N=\int_{\sigma}\lambda dE(\lambda)$. $\endgroup$ – DisintegratingByParts May 31 '18 at 15:11
  • $\begingroup$ @DisintegratingByParts Yes i know it $\endgroup$ – Student May 31 '18 at 15:28
3
$\begingroup$

In the book "A course in abstract harmonic analysis" by "Gerald B. Folland"., we have the follwoing theorem

enter image description here


enter image description here

$\endgroup$
1
$\begingroup$

Start with the normal operator $A_1$ and its spectral resolution $A_1 = \int \lambda dE_1(\lambda)$. Let $x$ be a non-zero vector, and consider the closed subspace $\mathscr{A}_1(x)$ generated by $\{ x,A_1 x,A_1^*,A_1^2 x,A_1^{*2}x\cdots\}$. Every element of $\mathscr{A}_1(x)$ can be written as a limit of vectors of the form $p_n(A_1,A_1^*)x$ as $n\rightarrow\infty$, where $p_n(\lambda,\overline{\lambda})$ is a polynomial in the two variables. Knowing the spectral resolution of $A_1$ allows you to trade this limit for a function limit in $L^2(d\|E_1(\lambda)x\|^2)$ because $$ \|p_n(A_1,A_1^*)x\|^2 = \int |p_n(\lambda,\overline{\lambda})|^2d\|E_1(\lambda)x\|^2. $$ So there is a map $U_x : \mathscr{A}_1(x) \rightarrow L^2_{\mu_x}$ where $\mu_x(S)=\|E_1(S)x\|^2$, and $$ U_x A_1 x = \lambda U_x x \\ U_x A_1^* x = \overline{\lambda}U_x x. $$ Because $A_2$ commutes with $A_1$, then $U_x A_2 U_x^*$ commutes with multiplication by all polynomials in $\lambda,\overline{\lambda}$ on $L^2_{\mu_x}$, which forces $U_xA_2U_x^*f=a_2(\lambda)f$ for all $f\in L^2_{\mu_x}$ because the operator space of bounded multiplication operators on $L^2_{\mu_x}$ is maximally Abelian. Finally, there is a trick that I do not recall at the moment for piecing together mutually orthogonal spaces $U_x\mathscr{A}_1(x)=L^2_{\mu_x}$ together in order to write $A_1$ as a multiplication operator by $a_1\in L^{\infty}_{\mu}$ on the composite $L^2_\mu$ space; the corresponding representation for $A_2$ as a multiplication follows as well. So this is a partial answer, but it gives you the basic building block coming from the spectral theorem.

$\endgroup$
  • $\begingroup$ Thank you. I hope that you give me a reference which contains the proof of spectral theorem when H is not separable because i would like to cite it. $\endgroup$ – Student May 31 '18 at 17:13
  • $\begingroup$ @Student : I'm not convinced that the multiplication version of the spectral theorem holds in the non-separable case, are you? $\endgroup$ – DisintegratingByParts May 31 '18 at 22:16

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.