# Spectral Theorem: Multiplication Operator Form

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.

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

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

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.

• 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. – Student May 31 '18 at 17:13
• @Student : I'm not convinced that the multiplication version of the spectral theorem holds in the non-separable case, are you? – DisintegratingByParts May 31 '18 at 22:16