Say, we have a Hilbert space $H$ with a semibounded self-adjoint operator $A:D(A)\to H$ generating a strongly continuous semigroup $T(t):H\to H$. Is it possible to restrict $T(t)$ to a form domain of $A$ in such a way that the restriction is again a strongly continuous semigroup, say $T_1(t)$?

How does the generator of $T(t)$ relate to the generator of $T_1(t)$? Is the latter self-adjoint w.r.t. a product $(x,y)_1:=(x,Ay)+(x,y)$?

Is there any relation of $Q(A)$ to the fractional Favard space $F_{1/2}$?

  • $\begingroup$ Thanks to the spectral theorem, it should suffice to consider the case where $A$ is multiplication by a measurable function $h$ which is bounded below. In that case $T(t)$ is multiplication by $e^{-th}$. It seems like the question may be easy to answer in that case. Slogan: "The spectral theorem: reducing functional analysis to measure theory since 1929." $\endgroup$ Aug 24, 2015 at 18:07
  • $\begingroup$ I should have said, of course, that this is the case where $H = L^2(X, \mu)$ for an arbitrary measure space $(X,\mu)$. $\endgroup$ Aug 24, 2015 at 18:48
  • $\begingroup$ Thanks, I was able to show most of that using spectral theory but is there any abstract argument for these results? $\endgroup$
    – pwl
    Aug 25, 2015 at 10:17
  • 1
    $\begingroup$ @pwl: Well, it is difficult to decide whether an argument is measure theoretical or abstrat, IMHO. Anyway, a more explicit approach to just your questions can be found in §7.1 of a beautiful online lecture note by Arendt: uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.020/arendt/… $\endgroup$ Oct 12, 2015 at 11:34

1 Answer 1


In this case you know that the form domain equals the domain of the square root of $(-A)$.

You can read about the semigroup restricted to this space in an abstract way in Section II.5 of Engel-Nagel.


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