# Finiteness of the Measure Space appearing in Spectral Theorem of Bounded Operators

Consider the multiplication operator version of the spectral theorem for bounded operators in a Hilbert space. Quoted from Wikipedia:

Let $$A$$ be a bounded self-adjoint operator on a Hilbert space $$H$$. Then there is a measure space $$(X, \Sigma, \mu)$$ and a real-valued essentially bounded measurable function $$f$$ on $$X$$ and a unitary operator $$U:H \to L^2(X, \mu)$$ such that $$U^\dagger T_f U = A$$ where $$T_f$$ is the multiplication operator: $$T_f [\varphi](x) = f(x) \varphi(x)$$ and $$\|T_f\| = \|f\|_\infty$$.

It is not much work to find an example where this measure space is not unique. For example, one can look at the sum of the left shift and right shift operators in $$L^2(\mathbb{Z})$$ and use Fourier series to see that the space $$X$$ can be any real interval.

My question is, using the boundedness of $$A$$, can we always find a measure space with finite measure satisfying this theorem? In particular, can we always find one such that $$\mu(X)=1$$?

• I think the answer is yes if $H$ is separable Oct 31, 2023 at 22:02
• @EvangelopoulosPhoevos In his book Spectral Theory and Differential Operators, Davies says that $H$ doesn't need to be separable as you can apparently still decompose $H$ as a sum of cyclic subspaces for $A$ with the help of Zorn's lemma (but doesn't elaborate and leaves it to the reader...). Nov 1, 2023 at 0:13
• I'll correct myself, he doesn't say that outright and only that you can modify the statement appropriately in the non separable case. Just in case you do lose the finiteness of the measure, I don't really like saying possibly wrong things with confidence. Nov 1, 2023 at 0:32
• The boundedness of $A$ implies boundedness of $f$.
– daw
Nov 1, 2023 at 9:33

The theorem you quote as the spectral theorem is viewed as a quick corollary of the "real" spectral theorem, Reed-Simon Theorem VII.3 where $$A$$ is conjugated to a direct sum: $$U:H\to \oplus_{n=1}^N L^2(\mathbb R,d\mu_n)\\ (UAU^*\psi)_n(\lambda)=\lambda\psi_n$$ where $$N$$ may be infinite. I.e. U conjugates $$A$$ to a direct sum of operators which are just multiplication by the independent variable on each of the (direct) summands. What you quote as the spectral theorem is then a quick corollary of the above (immediately below in the book), with the added property that the measure $$\mu$$ may be taken to be finite (and in their construction, it in fact satisfies $$\mu(X)=1$$).
Given a self-adjoint operator $$A$$ (it can be unbounded) on $$H$$, you can decompose $$H$$ as a countable sum of cyclic subspaces $$L_n$$ for $$A$$, on each of which you can apply the Riesz-Markov theorem on a well-chosen linear functional to get a finite measure $$\mu_n$$ such that on $$L_n$$ the theorem is true for $$X = \sigma(A)$$ and $$f$$ is just the identity function. Then, you "gather" all these copies of $$\sigma(A)$$ and those $$\mu_n$$. Hence at the end $$X$$ can be chosen as $$\sigma(A) \times \mathbb{N}$$ and $$f$$ as the function $$f : (s,n) \mapsto s$$ (still the identity function in terms of $$s$$), with the (finite!) measure being $$\mu_n$$ on each $$\sigma(A) \times \{n\}$$. I don't know what happens in the general non-separable case.