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You need two additional assumptions: the operator $A$ has to be a so-called cosine function generator, and your product space has to be $V\times H$ with a space $V\subset H$. Cosine function generat …

To expand the answer of Michael Renardy: your question concerns the important question when can we "insert" an operator into a function and what properties can we deduce from the properties of the fun …

I might change your question a little bit: Let $X=L^2[0,1]$, $Af:=f''$ with $D(A)=H^2[0,1]$ and let $Bf=f'(0)\cdot\mathbb{1}$, with $D(B)=H^2[0,1]$. Then $B$ is not closable (easy exercise from the de …

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 …

If $X$ has the so-called Wiener-regularity, then it generates an analytic semigroup, see the paper by Arendt and Bénilan. The paper shows this even for unbounded domains, along with the fact that if …

The semigroups you construct is in general only weak-* continuous. Are you looking for so-called implemented semigroups? See for example Alber, Jochen, On implemented semigroups, Semigroup Forum 63, N …

If you have an analytic semigroup (generated by a so-called sectorial operator), then the answer is well-known and yes, even convergence rates are possible. This and generalizations were proved by man …

If I understand your question correctly, this would mean that the function $t\mapsto T(t)x$ is Lipschitz continuous, which is equivalent for $x$ to be in the Favard space $\text{Fav}(A)$$. See for exa …

I do not know who used it first, but I claim that Halmos made it popular in his 1963 Monthly paper. There he makes the connection to the diagonalization of hermitian matrices and mentions that peopl …

Just to expand the answer by Alex, let me mention that the internet seminar of last year: http://isem17.unisa.it/w/index.php/Phase_1:_The_lectures was intended as a gentle introduction where a lot o …

As you write it, this is just a bounded perturbation of the sectorial operator $$\begin{pmatrix} 0 & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ It is quite standard that bounded perturbat …

First, if $A$ is symmetric, then $X$ should be a Hilbert space, but I remain in a Banach space. If $$\|T(t)x\| \leq C\|A^{-1} x\|$$ holds for all $x\in X$, then using the substitution $y=A^{-1}x$, …

I do not think it is good to ask three questions in one, since I will only address the first now. I would put the question back and ask you: how do you define a "solution"? Usually you have an exist …

No, this cannot hold in general. An argument is the following. If a Laplace transform representation of the resolvent holds for all $\Re\lambda>\omega$, then $\omega\geq \omega_1(A)$, see Arendt, …

The problem is discussed in a more general setting (operator ideals in Banach spaces) for the so-called analytic semigroups (parabolic problems) in Blunck, S.; Weis, L., Operator theoretic properties …