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: What you've written is kind of confused: if you define $Q(x,y)$ by $\langle Ax, y \rangle$ then of course $Q(x,y)$ only makes sense when $x \in D(A)$.  And it's obvious that if $x \in D(A)$ we have $Q(x,x) \in \mathbb{R}$ (indeed it is nonnegative, by the positivity of $A$).  But $D(A^{1/2})$ is typically strictly larger than $D(A)$.
As Christian Remling says, the usual way to define the unbounded operator $A^{1/2}$, along with its domain $D(A^{1/2})$ is via the spectral theorem.  The result you are probably thinking of is the following: define, as you did, the quadratic form $Q$ by $Q(x,y) = \langle Ax, y \rangle$, with domain $D(Q) = D(A)$.  This form is closable, and if we let $\bar{Q}$ be its closure (with its domain $D(\bar{Q})$) then $D(\bar{Q}) = D(A^{1/2})$.
So we could express this as follows:

We have $f \in D(A^{1/2})$ if and only if there exists a sequence $f_n \in D(A)$ such that $f_n \to f$ in $H$ and $f_n$ is Cauchy in $Q$-norm (i.e. $\lim_{m,n \to \infty} Q(f_n - f_m, f_n - f_m) = 0$.)

The proof is a rather straightforward exercise.  For use in a paper, I think most people would just state it as common knowledge without a proof or reference.
I guess another way of saying this is just that $D(A)$ is a core for $A^{1/2}$.
