0
$\begingroup$

Let

  • $E$ be a $\mathbb R$-Banach space
  • $(T_n(t))_{t\ge0}$ and $(T(t))_{t\ge0}$ be $C^0$-semigroups on $E$ with generator $(\mathcal D(A_n),A_n)$ and $(\mathcal D(A),A)$, respectively, for $n\in\mathbb N$

Suppose $\mathcal D(A_n)\subseteq\mathcal D(A)$ for all $n\in\mathbb N$ and $$\left\|A_nx-Ax\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }x\in\mathcal D(A)\tag1.$$ Are we able to conclude $$\left\|T_n(t)x-T(t)x\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }x\in E\tag2$$ for all $t>0$? If not, are there mild conditions which allow the conclusion?

$\endgroup$
  • $\begingroup$ an $\mathbb{R}$-Banach $\endgroup$ – mathworker21 Nov 24 '18 at 13:34
  • $\begingroup$ @mathworker21 Thank you very much, that solved my problem. $\endgroup$ – 0xbadf00d Nov 24 '18 at 13:39
  • $\begingroup$ if I knew what generator meant, I would try to help. $\endgroup$ – mathworker21 Nov 24 '18 at 13:41
  • $\begingroup$ In (1) you are implicitly assuming that $\mathcal{D}(A) \subseteq \mathcal{D}(A_n)$ for all $n$...? Otherwise $A_n x$ doesn't make sense for $x \in \mathcal{D}(A)$. $\endgroup$ – saz Nov 24 '18 at 15:43
  • $\begingroup$ @saz Yes, I forgot to add that. $\endgroup$ – 0xbadf00d Nov 24 '18 at 17:18
1
$\begingroup$

As mentioned in the comments, you have to be careful with the domains of the generators, but morally it is true that strong convergence of the generators implies strong convergence of the semigroups. Here are sufficient conditions to make it rigorous (you can probably weaken them):

Assume that

  • there are constants $M,\beta>0$ such that $\lVert T_n(t)\rVert\leq Me^{\beta t}$ for $t\geq 0$, $n\in\mathbb{N}$,
  • there is a core $D_0$ of $A$ contained in $D(A_n)$ for all $n\in\mathbb{N}$ such that $A_n x\to A x$ for $x\in D_0$.

Then $T_n(t)\to T(t)$ strongly for all $t\geq 0$ (in fact uniformly on compact intervals).

For reference see Kato, Perturbation Theory, Theorems VIII.1.5 and IX.2.16.

$\endgroup$
  • $\begingroup$ Ah, I see. It's even sufficent that there are $x_n\in\mathcal D(A_n)$ with $x_n\to x$ and $Ax_n\to Ax$, right? $\endgroup$ – 0xbadf00d Nov 24 '18 at 17:28
  • $\begingroup$ Oh, and is there any issue if the $(T_n)$ are considered in discrete time, i.e. they are semigroups $(T_n(k))_{k\in\mathbb N_0}$? In that case the generator would simply be defined to be $A_nx:=T_n(1)x-x$. $\endgroup$ – 0xbadf00d Nov 24 '18 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.