I suspect I am asking a very stupid question.


Suppose you have self-adjoint negative-definite operator $L$ densely defined on a space $L^2(\pi)$, with $Lf = \nabla \cdot ( A(x)\nabla f)$, for some symm. pos. def matrix A. Here assume that differentiation $\nabla = (D\_i)\_{i=1,..,n}$ is a skew-adjoint operator densely defined on $L^2(\pi)$, that is, $\mathcal{D}(D\_i) = \mathcal{D}(D\_i^\*)$ and $D\_i^\* = -D\_i$, for $i=1,\ldots n$.


Now, I'm trying to make sense of the statement that $f \in \mathcal{D}((-L)^\frac{1}{2})$. This would imply that $((-L)f, f) < \infty$, but I'm not sure if we can ALWAYS write this as:


$$((-L)f, f) = \int\_\Omega \nabla f\cdot A(x) \nabla f \space \pi(dx)$$


I'm sorry if this is a stupid question but for some reason I can't convince myself of this fact.



Integration by parts for a general negative-definite self-adjoint operator.
It is possible to make sense of $T^{1/2}$ without some of the particulars mentioned, when $T$ is a positive self-adjoint (densely-defined) operator on a Hilbert space. Namely, Friedrichs' argument (as in Riesz-Nagy, for example) shows that the resolvent $(T-\lambda)^{-1}$ exists and is a bounded operator for $\lambda$ not positive real. In particular, $T^{-1}$ is a bounded operator. It is also positive, so by standard (bounded-operator) spectral calculus admits a positive square root, whose inverse is the desired $T^{1/2}$.


Edit: after seeing some reactions, it is worth clarifying, as follows. Again, the square root of a positive *unbounded* (densely defined) \_truly\_self-adjoint\_ operator exists without necessarily expressing the square root in terms of differential operators. The domain $D(\sqrt{T})$ is the same as the domains $D(\sqrt{T-\lambda})$ for $\lambda$ not in $[0,\infty)$.


There is the subordinate issue of whether one knows that the operator is *genuinely* self-adjoint, or only "symmetric". In the latter case, the question of self-adjoint extensions is non-trivial, depending on things abstracting "boundary conditions", tho' there *is* always the Friedrichs "minimal" extension. 


In a similar vein, if we truly know a-priori that the operators $D\_i$ are well-behaved in the sense that their domains and their adjoints' domains agree (the skew-adjoint version of self-adjoint), and assuming the domain(s) of the various expressions genuinely agree with that of the original operator, the seemingly formal computations are (by fiat) correct.


In practice, yes, there would be a non-trivial issue of specifying a common domain for the $D\_i$ so that they are "genuinely" skew-adjoint, and so that the implied domain of the symmetric second-order operator is equal to that of its adjoint (so it is truly self-adjoint). 


G. Grubb's book "Distributions and Operators" discusses many concrete examples of such things.


