Definition 2.6.1 | Brownian Motion and Stochastic Calculus

Definition 2.6.1.label Let $E$ be a Banach space and $\bracs{T_t|t \ge 0}$ be a strongly continuous contraction semigroup on $E$. Let

\[D = \bracs{x \in E \bigg | \lim_{t \to 0}\frac{T_{t}x - x}{t} \text{ exists}}\]

then:

(1)
For each $t \ge 0$, let
\[A_{t}: E \to E \quad x \mapsto \int_{0}^{t} T_{s}x ds\]

then for any $x \in E$, $A_{t}x \in D$ with
\[A(A_{t}x) = T_{t}x - x\]
(2)
$D$ is dense in $E$.
(3)
The operator
\[A: D(A) \to E \quad x \mapsto \lim_{t \to 0}\frac{T_{t}x - x}{t}\]

is closed.

and operator $A$ in $(2)$ is the generator of $\bracs{T_t|t \ge 0}$.

Proof. (1): For any $r \in (0, t)$,

\begin{align*}\frac{T_{r}A_{t}x - A_{t}x}{r}&= \frac{1}{r}\braks{T_r\int_0^tT_sx ds - \int_0^tT_sxds}\\&= \frac{1}{r}\braks{\int_r^{t+r}T_sx ds - \int_0^tT_sxds}\\&= \frac{1}{r}\braks{\int_t^{t+r}T_sx ds - \int_0^rT_sxds}\end{align*}

By the Fundamental Theorem of Calculus,

\[\lim_{r \downto 0}\frac{T_{r}A_{t}x - A_{t}x}{r}= T_{t}x - x\]

so for any $x \in E$ and $t > 0$, $A_{t}x \in D$.

(2): By continuity of $s \mapsto T_{s}x$ and the Fundamental Theorem of Calculus, $A_{t}x/t \to x$ strongly as $t \downto 0$. Therefore $D$ is dense in $E$.

(3): Let $\seq{x_n}\subset D$, $x \in E$, and $y \in E$ such that $x_{n} \to x$ and $Ax_{n} \to y$ as $n \to \infty$. By the Fundamental Theorem of Calculus,

\[T_{t}x_{n} - x_{n} = \int_{0}^{t} T_{s}Ax_{n} ds\]

By the Dominated Convergence Theorem,

\[T_{t}x - x = \int_{0}^{t}T_{s}yds\]

Using the Fundamental Theorem of Calculus again,

\[\lim_{t \downto 0}\frac{T_{t}x - x}{t}= y\]

Hence $y \in D$ with $Ax = y$.$\square$