Let $\Omega$ a bounded open set of $R^n$, $\omega$ a non-empty open subset of $\Omega$ and 


$\chi\_{\omega} : L^2(\Omega) \longrightarrow L^2(\omega)$
 be the restriction operator to $\omega$, while
 $\chi\_{\omega}^\*$ denotes the adjoint operator of $\chi\_{\omega}$ and given by
 \begin{equation}\label{v1}
 (\chi\_{\omega}^\*y)(x)=\left\{
 \begin{array}{lll}
 y(x),\quad x\in \omega\\
 0, \quad otherwise 
 \end{array}
 \right.
 \end{equation}
 and denote $i\_{\omega} = \chi\_{\omega}^\*\chi\_{\omega}$ ($i\_{\omega}$ is a positive operator).


On the other hand, let $A :\mathcal{D}(A)\subset L^2(\Omega) \longrightarrow L^2(\Omega)$ be a linear
 unbounded operator such that $\langle Ay,y\rangle\leq 0, \forall y\in\mathcal{D}(A)$:


My question is : can we prove that $\langle i\_{\omega}Ay,y\rangle\leq 0$ ?



Composition of a negative operator and a positive one
No, of course not. Even if $A$ is bounded, this would imply that $i\_\omega A$ is self-adjoint. So you only have a chance if $A$ and $i\_\omega$ commute.


