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. $\forall y\in E:(0,y)\in A\Rightarrow y=0$.


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 $$Bx=\lim_{t\to0+}\frac{T(t)x-x}t\in\overline{\mathcal D(A)}.\tag8$$


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