# Domain of square root of a self-adjoint positive operator [closed]

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.

## closed as off-topic by Christian Remling, Nate Eldredge, András Bátkai, Dima Pasechnik, Alex DegtyarevApr 3 '15 at 20:58

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• $D(A^{1/2})$ is most naturally defined via the spectral theorem, but this question is really better suited for math.stackexchange.com – Christian Remling Apr 3 '15 at 19:19

## 1 Answer

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 $A^{1/2}$.

• Can the following actually happen: start with a densely defined positive self-adjoint operator $A$, then the domain of $A^{1/2}$ is defined as $\mathcal{D}(A^{1/2}) = \{ x \in H : \langle A'x, x\rangle < \infty\}$, where $A'$ is now the Friedrichs extension of $A$? – anonymous Apr 3 '15 at 20:59
• @anonymous: I believe you're still confused. An expression like $A'x$ only makes sense when $x$ is in the domain of $A'$. So $\langle A' x, x \rangle$ is clearly finite when $x \in D(A')$ and is undefined otherwise, and the set on the right side of what you wrote is just a fancy way of writing $D(A')$. Moreover, it doesn't make sense to ask whether $D(A^{1/2})$ "is defined as" something: the spectral theorem defines what $D(A^{1/2})$ is. Finally, if $A$ is self-adjoint then it is its own Friedrichs extension. – Nate Eldredge Apr 3 '15 at 21:27
• @anonymous: Normally the Friedrichs extension is used for operators which are symmetric but not self-adjoint. So taken literally, your comment asks "For positive self-adjoint $A$, can $D(A^{1/2}) = D(A)$?" The answer is, this happens iff $A$ is bounded (in which case both sides equal $H$). – Nate Eldredge Apr 3 '15 at 21:28