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This question follows from this one and especially from Willie Wong's answer: link.

In Reed & Simon's book Methods of modern mathematical physics, vol. I, pag.277, the form domain of a self-adjoint operator $(A, D(A))$ on a Hilbert space $H$ is defined by passing to a spectral representation, that is by taking a unitary isomorphism

$$U \colon H \to \bigoplus_{j=1}^N L^2(\mathbb{R}, d\mu_j), $$

(where $N\in \{1, 2 \ldots +\infty\}$ and $\mu_j$ are finite Borel measures) such that $(UA)\varphi=(x\psi_j(x))_{j=1}^N$ [$A$ is unitarily equivalent to multiplication by $x$]. The sought domain is then said to be

$$Q(q)=\left\{ (\psi_j)_{j=1}^N \ :\ \sum_{j=1}^N \int_{-\infty}^\infty \lvert x \rvert \lvert \psi_j(x)\rvert^2\, d\mu_j <+\infty \right\}.$$

Question. Let

$$D(\lvert A\rvert^{1/2})=\left\{ \varphi \in H\ :\ \int_{-\infty}^\infty \lvert \lambda \rvert\, d\big(E_A(\lambda)\varphi, \varphi\big)<+\infty\right\},$$

where $\{E_A(\lambda)\}_{\lambda \in \mathbb{R}}$ is the spectral family of $A$ (cfr. Reed & Simon vol. I Theorem VIII.6).

Is it true that $Q(q)=D(\lvert A\rvert^{1/2})$?

I believe the answer to be affirmative. This should make for a characterization of the form domain a bit more transparent than the one based on spectral representations. For example, it is not immediately clear that the latter is independent on the particular representation chosen.

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An alternative treatment for positive operators can be found on Berezin - Shubin's The Schrödinger equation, pagg.436-437 link. –  Giuseppe Negro Apr 9 '12 at 0:15
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up vote 1 down vote accepted

Yes, it is true. First, note that $Q(q) = D(|A|^{1/2})$ holds in the case that $A$ is "multiplication by $x$". Then observe that the spectral family is preseved by unitary transformations.

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Thank you Nate! To sum things up let us make a concrete example and please correct me if I'm wrong. We take a differential operator $p(D)$ ($p$ is a polynomial s.t. $p(i\xi)$ is real) acting on $\mathscr{D}(\mathbb{R}^n)$. In Fourier space it converts to a multiplication by $p(i\xi)$. Then: $$$$ 1. The domain of self-adjointness of $p(D)$ is $\{f\in L^2(\mathbb{R}^n)\mid \int \lvert p(i\xi)\tilde{f}(\xi)\rvert^2\, d\xi < \infty\}$; $$$$ 2. The form domain of $p(D)$ is $\{f \in L^2(\mathbb{R}^n)\mid \int \lvert p(i\xi)\rvert \lvert \tilde{f}(\xi)\rvert^2\, d\xi < \infty\}$. –  Giuseppe Negro Apr 8 '12 at 20:10
    
Domain 2. is bigger than domain 1. because of Cauchy-Schwarz inequality. If $p(D)=-\Delta$ we get the spaces $H^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$ that Willie Wong mentions in the other question. –  Giuseppe Negro Apr 8 '12 at 20:15
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@GiuseppeNegro: Yes, that looks right. –  Nate Eldredge Apr 8 '12 at 20:24
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