I asked this question on Mathematics Stackexchange, but got no answer.
In Pavel's book: Nonlinear Evolution Operators and Semigroups - Applications to Partial Differential Equations, we have the following definitions and result:
Consider the following semilinear equation:
$$\tag{5.59}\begin{cases}u_t(t)=Au(t)+F(u(t)), ~0 \leq t \leq T,\\u(0)=u_0 \in H. \end{cases} $$ where $A:D(A)\subset H \rightarrow H$ be the infinitesimal generator of $C_0$-semigroup $S(t)$ with $\|S(t)\|\leq 1$.
Definition 5.1. (1)A function $u:[0,T]\rightarrow H$ is said to be a classical solution of $(5.59)$ if $u(0)=u_0$ and $$\tag{5.68}u \in C([0,T];D(A))\cap C^1([0,T];H)$$ and $(5.59)$ is satisfied. (2) The function $u$ is a classical solution on $[0,\infty[$ of $(5.59)$ if $(5.68)$ holds for every $T>0$ and $(5.59)$ is satisfied for every $t>0$.
By a mild solution of $(5.59)$ we mean a continuous function $u:[0,T] \rightarrow H$ satisfying $$\tag{5.69} u(t)=S(t)u_0+\int_{0}^{t} S(t-s)F(u(s))\, ds, ~0 \leq t \leq T.$$
Theorem 5.8 If $F:H\rightarrow H$ is locally Lipschitz then for every $u \in H$, there is a unique mild solution $u:[0,T_{\max}[\rightarrow H$ of $(5.59)$ with either $T_{\max}=+\infty$, or $T_{\max}<\infty$ and $\lim_{t \uparrow T_{\max}}\|u(t)\|=+\infty$. If $u_0 \in D(A)$, the mild solution $u$ is classical. (2) If $F$ is Lispschitz continuous, then $T_{max}=+\infty.$
My question:
I would like to know if Theorem 5.8 is valid for $ A $ being just the infinitesimal generator of a $C_0$-semigroup, that is, without the hypothesis that $\|S(t)\|\leq 1$.
I imagine that the answer is positive in the case that $S(t)$ is uniformly bounded, that is, there is $M>0$ such that $\|S(t)\|\leq M$ for $t \geq 0$.
Assuming I have proven that $A-\lambda I$ is the infinitesimal generator of a $C_0$-semigroup of contractions, for some $\lambda>0$. Then, $A$ is the infinitesimal generator of a $C_0$-semigroup $T(t)$ satisfyng $\|T(t)\|\leq e^{\lambda t}$, for all $t \geq 0$. I don't know if the result is valid in this case.
Can anyone help me? Please.