1
$\begingroup$

I am interested on some connections between the spectral theorem for normal operators and a corresponding functional calculus which can be assigned to them. I'll begin by recalling these two results, the first of which is

The Spectral Theorem for Normal Operators: Let T be a normal operator on a Hilbert space, H. Then there exists a measure space $(\Omega, \Sigma, \mu)$ and a function $\phi\in L^\infty(\Omega,\mu)$ such that $T$ is unitarily equivalent to the multiplication operator $M_\phi$ on $L^2(\Omega,\mu)$.

and the second of which is

The Continuous Functional Calculus for Normal Operators: Let $T\in\mathcal B(H)$ be a normal operator. Then the map $$C^0\left(\sigma(T)\right)\to\mathcal B(H),\,\,f\to f(T)$$ is an isometric $^*$-homomorphism.

  1. The spectral theorem for normal operators says that a normal operator is unitarily equivalent to a multiplication operator. Is this the same as saying that the normal operator has a spectral representation as a multiplication operator? In a set of notes I am considering, this identification is made. If it is not the same, then what is the relation between the two concepts? And under what conditions does one imply the other?

  2. With the above question in mind, it is interesting to note that the functional calculus mapping assigns elements in the given function class (above, this is just the continuous functions) which are, in particular, defined on the spectrum of the original normal operator. How does this relate to the above question?

For the next question we recall the definition of unitary equivalence between two operators.

Unitary Equivalence: Let $H, K$ be Hilbert spaces and let $T\in\mathcal B(H)$, $S\in\mathcal B(K)$. We say that $T$ and $S$ are unitarily equivalent, if there exists a unitary operator $U:H\to K$ such that $$T=U^*SU.$$

  1. The functional calculus mapping assigns to $f\in C^0\left(\sigma(T)\right)$ the operator $f(T)$, for the given normal operator T. How can this be applied to the unitarily equivalent multiplication operator that we obtain via the spectral theorem? Indeed, is it reasonable to wish to apply the functional calculus to this unitarily equivalent operator?

  2. In particular, how does it follow that $$f(T)=f(U^*M_\phi U)=U^*f(M_\phi)U?$$ What enables such $f$ to commute with the unitary operator $U^*$?

$\endgroup$
  • $\begingroup$ 1) I think (according to Reed and Simon) a spectral representation is the representation of the hilbert space as $\mathcal{H} = \bigoplus_{\alpha} L^2(\sigma (T), \Sigma, \mu)$, such that on each summand the operator acts as mult by identity. The relation to your mentioned reult is formally joining all these summands into a single $L^2$ space on a maybe none sigma finite measure space. See this for example math.stackexchange.com/questions/3216082/… $\endgroup$ – pitariver Jun 5 at 5:57

Your Answer

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

Browse other questions tagged or ask your own question.