# Integration by parts for a general negative-definite self-adjoint operator.

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.

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Am I right that you want to study this problem in the abstract (i.e. $\pi$ is a measure on some unspecified measurable space etc.)? Then your assumption that the operators $D_i$ are only skew-Hermitian seems too weak to me. A more natural assumption would be that the $D_i$ are skew-adjoint, that is, the domain of the adjoint $D_i^*$ equals that of $D_i$ (in addition to the skew-symmetry). –  Florian Jun 30 '11 at 15:11
Yes, actually I should have mentioned that assumption also. Thanks for pointing this out. –  RadonNikodym Jun 30 '11 at 15:14

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