I'd like to know whether a certain unbounded operator on a Hilbert space is the generator of a strongly continuous contraction semigroup. (Such operators are known as maximally dissipative operators.)

•    Let $A$ be an unbounded skew-adjoint operator on $H$.

•    Let $P$ be an unbounded positive self-adjoint operator on $H$.

Suppose that $\mathcal D:=\mathcal D(A)\cap \mathcal D(P)$ is a common dense core for $A$ and $P$, and let $A-P$ be the strong sum of $A$ and $-P$ (the closure of $(A-P)|_{\mathcal D}$).

Is $A-P$ the generator of a strongly continuous contraction semigroup on $H$ (i.e., is $A-P$ a maximally dissipative operator)? Are there reasonable extra assumptions that I could add which would imply that $A-P$ is the generator of a strongly continuous contraction semigroup?

  • $\begingroup$ Isn't that obvious, or am I missing something here? For $x\in D$, we have $\textrm{Re}\langle x, (A-P)x\rangle = -\langle x, Px\rangle\le 0$, as desired, and then we also get the same inequality for the operator closure. $\endgroup$ – Christian Remling Jul 5 '17 at 17:44
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    $\begingroup$ @ Christian: Yes, the fact that it's dissipative is obvious. The difficulty is in showing that it is maximally dissipative (see Wikipedia page for definitions). Sorry--the first formulation of the question was confusing. $\endgroup$ – André Henriques Jul 5 '17 at 20:54
  • $\begingroup$ Let us take a look at the evolution equation, with $A=-iQ$, $Q$ self-adjoint, $$ \frac{du}{dt}+(P+iQ)u=0, \quad u(0)=u_0. $$ With $u=e^{-it Q}v$, we have $e^{it Q}$ unitary and $$ e^{-it Q}\dot v- iQ e^{-it Q} v+iQe^{-it Q} v+Pe^{-it Q}v=0,\quad\text{ i.e.,} $$ $$ \dot v+e^{it Q}Pe^{-it Q}v=0. $$ In the simple case where $P$ and $Q$ commute, then $v= e^{-t P}v(0)$. With $P(t)=e^{it Q}Pe^{-it Q},$ we have $$ \Vert v(t)\Vert^2+ 2\int_0^t\Vert P^{1/2} e^{-is Q} v(s)\Vert^2 ds=\Vert v(0)\Vert^2. $$It seems that whenever $[P,iQ]\ge 0$ the situation should not be much different. $\endgroup$ – Bazin Jul 17 '17 at 20:03

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