Consider the following theorem taken from Reed & Simon's book.

Theorem [Spectral Theorem - Multiplication Operator Form] Let $A$ be a (unbounded) self-adjoint operator on a separable Hilbert space $\mathcal{H}$ with domain $D(A)$. Then there is a measure space $(M,d\mu)$ with $\mu$ finite measure, a unitary operator $U: \mathcal{H}\to L^{2}(M,d\mu)$ and a real-valued function $f$ on $M$ which is finite a.e. so that:

(a) $\psi \in D(A)$ iff $f(\cdot)(U\psi)(\cdot) \in L^{2}(M,d\mu)$

(b) If $\varphi \in U[D(A)]$, then $(UAU^{-1}\varphi)(m) = f(m)\varphi(m)$

Now, before posting my question, let me introduce some notation. With the above notation, suppose that $f$ is a real-valed measurable function on $M$ which is finite a.e. Let $T_{f}:\mathcal{D}(T_{f}) \to L^{2}(M,d\mu)$ be the multiplication operator $T_{f}(\varphi) := f\varphi$. Here, $D(T_{f})$ is the subset of $L^{2}(M,d\mu)$ of all functions $\varphi$ for which $f\varphi$ is in $L^{2}(M,d\mu)$.

On Reed & Simon's book, it is proved that $T_{f}$ is self-adjoint. Then, they state the following:

Unless $f \in L^{\infty}(M,d\mu)$ the operator $T_{f}$ will be unbounded. Thus, we have found a large class of unbounded self-adjoint operators. In fact, we have found all of them.

The above is stated right before the mentioned theorem, so that is seems that the theorem should justify the claim.

Question 1: I understand the content of the above theorem but I don't understand how to extract from it the information that the operators $T_{f}$ represent 'all self-adjoint operators'. How does the theorem justifies Reed & Simon's claim?

Question 2: What is the natural interpretation of the above Theorem? Why is it called a spectral theorem? It doesn't seem to have anything about the spectrum of $A$.

  • $\begingroup$ An important detail worth mentioning: the function $f$ doesn't have to be a function in the usual sense – if the spectrum of $A$ is discrete, then $M = \mathbb{N}$ and $\mu$ is the counting measure. Hence, $f$ will be a sequence and $T_f$ will be a diagonal matrix on $\ell^2$. $\endgroup$
    – csha
    Commented Aug 29, 2021 at 20:01

2 Answers 2

  1. The spectral theorem proves that for each self-adjoint $A$, $UAU^{-1} = T_f$ for some $f$. Thus $A$ is unitarily equivalent to a multiplication operator, so in this sense multiplication operators represent all self-adjoint operators on a Hilbert space.

  2. Suppose $A$ is a $1 \times 1$ matrix (or more precisely, a linear operator on a $1$-dimensional Hilbert space). Then $Av = \lambda v$ for a scalar $\lambda$. So we can view $f$ as a function on one-point measure space, taking the single value $\lambda$. Bumping it up, let $A$ instead be an $n \times n$ matrix. The traditional spectral theorem says $UAU^{-1} = \Lambda$, where $\Lambda$ is the diagonal matrix of eigenvalues. Take $M$ to be a discrete measure space with $n$ atoms. Then diagonal matrix $\Lambda$ can be viewed as a function on $M$, taking the values $\lambda_1, \ldots, \lambda_n$ (the eigenvalues), composed with projections onto the $n$ orthogonal eigenspaces of $A$. In this way, the traditional spectral theorem can be seen as a discrete-measure special case of the general unbounded spectral theorem.


Second question first. The theorem is a continuous version of the diagonalization theorem from linear algebra. If we imagine that $M$ is finite, and take functions equal to $1$ at some point $m$ and zero elsewhere as the basis then it says that any self-adjoint operator is unitarily equivalent to multiplication by a diagonal matrix, with $f(m)$ on the diagonal, i.e. $f(m)$ are its eigenvalues (spectrum). In general, $f(m)$ may represent continuous spectrum if the measure $\mu$ does not assign a positive weight to $m$, but $m$ is still in its support.

Now the first question. The theorem says that any self-adjoint operator is unitarily equivalent to some $T_f$, in other words, it is $T_f$ in some unitarily equivalent representation. In finite dimensional case the change of representation amounts to simply choosing a different basis. In general, one needs rigged Hilbert spaces and generalized eigenfunctions, such as $\delta$-functions that do not belong to the original space itself, to express it in this form. The multiplicative formulation avoids such complications. Hence the claim of Reed and Simon that $T_f$, when $f$ is essentially unbounded, represent all unbounded self-adjoint operators.

  • $\begingroup$ Great! Very helpful indeed! Thank you! $\endgroup$
    – Idontgetit
    Commented Sep 6, 2020 at 20:17

You must log in to answer this question.

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