In a Banach space $E$, we consider the Cauchy problem $$u'(t)=Au(t),\quad u(0)=u_0 \quad (t\geq0)$$ where $A:D(A)\subseteq E \to E$ denote a linear operator with domain dense.
If the problem is uniformly well posed we define the solution operator $\{S(t)\}_{t\geq0}$ as $S(t)u_0=u(t)$ where $u(t)$ is the solution of the problem. We can prove that $\{S(t)\}_{t\geq0}$ is a strongly continuous semigroup.
Let $(\tilde{A},D(\tilde{A}))$ be the infinitesimal generator of $\{S(t)\}_{t\geq0}$ i.e $$D(\tilde{A})=\{u \in E \colon \lim_{h\to0} \frac{1}{h} (S(h)u-u) \text{ exists}\},\quad \tilde{A}u=\lim_{h\to0} \frac{1}{h} (S(h)u-u)$$ Can we prove that $A=\tilde{A}$?
It is easy to see that $$\tilde{A}u=\lim_{h\to0} \frac{1}{h} (S(h)u-u)=\lim_{h\to0} \frac{1}{h} (u(h)-u)=u'(0)=Au,$$ but do we have $D(A)=D(\tilde{A})$?