Domain of square root of a self-adjoint positive operator Let $A \geq 0$ be a densely defined self-adjoint positive operator on a Hilbert space $H$ obtained by Friedrichs extension, and let $Q$ be the densely defined quadratic form associated to $A$, that is, $Q(x, y) = \langle Ax, y\rangle$. Now it is clear that $x \in D(A) \Rightarrow Q(x, x) \in \mathbb{R}$. I remember reading somewhere that $D(A^{1/2})$ is defined as $\{x \in H : Q(x, x) \in \mathbb{R}\}$. If this is true, can I please have a reference? Thanks.
 A: More generally, for any self-adjoint operator $A$ in a Hilbert space $H$ and any Borel function $f$ on $\sigma(A)$, we define $f(A)$ by the functional calculus associated to the Spectral theorem.  If $E$ is the resolution of the identity associated to $A$ by $A = \int_{\mathbb R} \lambda \; dE(\lambda)$, then $f(A) = \int_{\mathbb R} f(\lambda)\; dE(\lambda)$, with
domain
$$\mathcal D(f(A)) = \{x \in H: \int_{\mathbb R} |f(\lambda)|^2\; dE_{x,x}(\lambda) < \infty \}$$ 
In the case of $f(z) = z^{1/2}$, with $A \ge 0$ so $E$ is concentrated on $[0,\infty)$. 
$$ \eqalign{\mathcal D(A^{1/2}) &= \{x \in H: \int_{[0,\infty)} \lambda dE_{x,x}(\lambda) < \infty\}\cr
&= \{x \in H: \langle A x, x \rangle < \infty\}\cr}$$ 
See any of the standard functional analysis texts, e.g. Rudin "Functional Analysis" chap. 13.
EDIT: This isn't quite right: $\int_{[0,\infty)} \lambda\; dE_{x,x}(\lambda)  = Q(x,x)$ where $Q$ is the quadratic form associated to $A$: in fact 
$Q(x,y)$ is defined precisely when $x$ and $y$ are in the set
$\{x \in H: \int_{[0,\infty)} \lambda\; dE_{x,x}(\lambda) < \infty$.
But we can't write $Q(x,x) = \langle Ax, x\rangle$ unless $x \in \mathcal D(A)$, which as anonymous points out isn't always true.
