# Spectral theorem and multiplication operator

Let $$\mathscr{H}$$ be a complex Hilbert space and $$A$$ be a densely-defined self-adjoint operator on its domain $$D(A) \subset \mathscr{H}$$. There is one version of the spectral theorem which says that there exists a measure space $$(X, \mathcal{F},\mu)$$, a unitary operator $$U: \mathscr{H} \to L^{2}(X,d\mu)$$ and a real-valued measurable function $$f$$ which is finite $$\mu$$-almost everywhere and satisfies: $$UAU^{-1} = M_{f}$$ where $$M_{f}$$ denotes the multiplication operator $$L^{2}(M,d\mu) \ni g \mapsto (M_{f}g)(x) := f(x)g(x) \in L^{2}(M,d\mu)$$.

There is no uniqueness on this theorem. Does it mean that any particular measure space $$(X, \mathcal{F},\mu)$$, unitary operator $$U$$ and measurable function $$f$$ will do the job?

I am interested in the case where $$\mathscr{H} = L^{2}(\mathbb{R}^{d},dx)$$ and $$A$$ is the multiplication by the identity operator $$M_{\operatorname{Id}}: D(M_{\operatorname{Id}})\subset L^{2}(\mathbb{R}^{d},dx) \to L^{2}(\mathbb{R}^{d},dx)$$ given by $$(M_{\operatorname{Id}}g)(x) := xg(x)$$. In the above: $$D(M_{\operatorname{Id}}) := \{g \in L^{2}(\mathbb{R}^{d},dx): \int_{\mathbb{R}^{d}}|x|^{2}|g(x)|^{2}dx < +\infty\}$$ is a dense domain for $$M_{\operatorname{Id}}$$.

In this case, it seems that the conditions of the theorem follow trivially. For instance, I could take $$U = \operatorname{Id}$$, $$f = \operatorname{Id}$$ and $$X = \mathbb{R}^{d}$$ with the Borel $$\sigma$$-algebra and Borel measure.

I now come back to my question: can I simply take these as the objects in the spectral theorem? Do all consequences of the Borel calculus still follow in this case?

The spectral theorem specifically says that these objects exist, not that you can take just any measure space, unitary operator und measurable function.

But yes, in your case you can trivially use $$U = \text{Id}$$, $$f=\text{id}$$ and $$\mathbb{R}^d$$ with the Borel $$\sigma$$-algebra and the Lebesgue measure $$\lambda$$. Although I don't quite understand why you would like to use this version of the spectral theorem as your operator is already a multiplication operator.