Let $A$ be the generator of a Semigroup of linear contractions on $L$. Then, $(R_{\lambda}, \lambda \in \rho(A))$ is called the resolvent of the semigroup $(P_t)_{t \geq 0}$.
($L$ is in the following our considered Banach space)
And for $g \in L_0 = \{ f \in L: \text{s-lim}_{h \rightarrow 0} (P_hf-f)=0\}$ we can write
$R_{\lambda}g= \int_0^\infty e^{-\lambda t}P_tg ~dt$
It follows that $R_{\lambda}g \in L_0$. But in a paper I read that $R_{\lambda}g \in D(A)=\{f \in L_0: \text{s-lim}_{h \rightarrow 0} \frac{1}{h}(P_hf-f) \text{ exists in } L\} $.
But $D(A) \subset L_0$. Because of that I can't understand the implication.
- Is the implication $R_{\lambda}g \in D(A)$ satisfied?
- If yes, why is $R_{\lambda}g \in D(A)$?
\begin{align} \lVert P_h (R_{\lambda}g) - R_{\lambda}g \rVert & = \lVert \int_0^{\infty} e^{-\lambda t} (P_{t+h}g - P_t g) \,d t \rVert \\ & \le \int_0^{\infty} e^{-\lambda t} \lVert P_t (P_h g - g) \rVert \,d t\\ & \le \int_0^{\infty} e^{-\lambda t} \lVert P_h g - g \rVert \,d t \\ & \le C \lVert P_h g - g\rVert \to 0 \end{align} as $h \to 0$, using the fact that $g \in L_0$. So $R_{\lambda}g \in L_0$.
I think you need $g \in D(A)$ to conclude $R_{\lambda}g \in D(A)$. When $g \in D(A)$ write $$ G = \lim_{h \to \infty} \frac{P_h g - g}{h} \in L$$ and you can proceed similarly as above and get \begin{align} \lVert \frac{1}{h}[P_h( R_{\lambda}g) - R_{\lambda}g] - R_{\lambda G}\rVert & \le \int_0^{\infty} e^{-\lambda t} \lVert P_t( \frac{1}{h}[P_h g - g] - G)\rVert \,d t \\ & \le C \lVert \frac{1}{h}[P_h g - g]- G\rVert \to 0 \end{align} as $h \to 0$.