0
$\begingroup$

Let $E$ be a $\mathbb R$-Banach space and $(T(t))_{t\ge0}$ be a contraction$^1$ semigroup on $E$. Assume $$[0,\infty)\to E\;,\;\;\;t\mapsto T(t)x\tag1$$ is Borel measurable for all $x\in E$ and hence $$\mathcal A:=\left\{(x,y)\in E\times E\mid\forall t\ge0:T(t)x-x=\int_0^tT(s)y\:{\rm d}s\right\}$$ is well-defined.

Are we able to show that$^2$ $$A:=\left\{(x,y)\in\mathcal A:y\in\overline{\mathcal D(\mathcal A)}\right\}$$ is the generator of $(T(t))_{t\ge0}$?

I was able to show the following:

  1. $A$ is single-valued$^3$ and hence the graph of a linear operator on $E$
  2. If $t\ge0$ and $(x,y)\in\mathcal A$, then $(T(t)x,T(t)y)\in\mathcal A$. In particular, $T(t)\mathcal D(\mathcal A)\subseteq\mathcal D(\mathcal A)$ for all $t\ge0$.
  3. $T(t)\overline{\mathcal D(\mathcal A)}\subseteq\overline{\mathcal D(\mathcal A)}$ for all $t\ge0$.
  4. $(T(t))_{t\ge0}$ is strongly continuous on $\overline{\mathcal D(\mathcal A)}$
  5. $\mathcal A$ is closed (with respect to the product topology on $E\times E$)

Let $(\mathcal D(B),B)$ denote the actual generator of $(T(t))_{t\ge0}$. It should be easy to show that $$A\subseteq\left\{(x,Bx):x\in\mathcal D(B)\right\}\tag2.$$ To do so, let $(x,y)\in A$. We need to show that $$\left\|\frac{T(t)x-x}t-y\right\|_E\xrightarrow{t\to0+}0.\tag3$$ Since $(x,y)\in A$, $$\left\|\frac{T(t)x-x}t-y\right\|_E=\frac1t\left\|\int_0^tT(s)y-y\:{\rm d}s\right\|_E\le\frac1t\int_0^t\left\|T(s)y-y\right\|_E\:{\rm d}s.\tag4$$ Let $\varepsilon>0$. Since $y\in\overline{\mathcal D(\mathcal A)}$, 4. yields that there is a $\delta>0$ with $$\left\|T(s)y-y\right\|<\varepsilon\;\;\;\text{for all }s\in[0,\delta)\tag5$$ and hence $$\frac1t\int_0^t\left\|T(s)y-y\right\|_E<\varepsilon\:{\rm d}s.\tag5$$ Thus, we should obtain $(3)$ and hence $(2)$.

How can we show the other inclusion?

Since $$\left\{(x,Bx):x\in\mathcal D(B)\right\}\subseteq\mathcal A\tag6$$, we only need to show that $B\mathcal D(B)\subseteq\overline{\mathcal D(\mathcal A)}$.

EDIT: Maybe an argument of the following kind is possible: Even when $(T(t))_{t\ge0}$ is not strongly continuous (on all of $E$) it is always strongly continuous on $\overline{\mathcal D(B)}$ (since $(T(t))_{t\ge0}$ is locally bounded (even contractive)). This should be enough to show that $(\mathcal D(B),B)$ is a closed operator (in the same way as it is usually done for strongly continuous semigroups). This means that $\overline{\left\{(x,Bx):x\in\mathcal D(B)\right\}}$ is closed and (since $\mathcal A$ is closed) still contained in $\mathcal A$.


$^1$ i.e. $\left\|T(t)\right\|_{\mathfrak L(E)}\le1$ for all $t\ge0$.

$^2$ As usual, if $B\subseteq E\times E$, then $$\mathcal D(B):=\left\{x\in E\mid\exists y\in E:(x,y)\in B\right\}.$$

$^3$ i.e. $\forall y\in E:(0,y)\in A\Rightarrow y=0$.

$\endgroup$
0
$\begingroup$

Okay, if I'm not missing anything, this is quite trivial: Let $x\in\mathcal D(B)$. By $(6)$ and 2., $$\frac{T(t)x-x}t\in\mathcal D(A)\tag7\;\;\;\text{for all }t>0$$ and hence $$Bx=\lim_{t\to0+}\frac{T(t)x-x}t\in\overline{\mathcal D(A)}.\tag8$$

$\endgroup$

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.