Version of Hille-Yosida Theorem for non contractive semigroups We say that a semigroup $\{T(t)\}_{t\geq 0}$ of bounded linear operators on a Banach space $X$ is of type $(M,\omega)$ if there are constants $\omega\geq0$ and $M\geq 1$ such that
$$\|T(t)\|_{\mathcal{L}}\leq M\mathrm{e}^{\omega t},\qquad\forall\ t\geq 0.$$
Let $A:D(A)\subset X\to X$ be a linear operator. The Hille-Yosida Theorem states:

The following conditions are equivalent:
  
  
*
  
*$A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{1},0)$ on $X$.  
  
*$A$ is closed, $D(A)$ is dense in $X$, $(0,\infty)\subseteq\rho(A)$ and
  $$ \|(\lambda-A)^{-1}\|_{\mathcal{L}}\leq\frac{\color{red}{1}}{\lambda},\quad\forall\ \lambda>0. $$
  

By considering the rescaled semigroup $S(t)=\mathrm{e}^{-\omega t}T(t)$, we get the version below.

The following conditions are equivalent:

  
*$A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{1},\omega)$ on $X$.  
  
*$A$ is closed, $D(A)$ is dense in $X$, $(\omega,\infty)\subseteq\rho(A)$ and
  $$ \|(\lambda-A)^{-1}\|_{\mathcal{L}}\leq\frac{\color{red}{1}}{\lambda-\omega},\quad\forall\ \lambda>\omega. $$
  

Concerning to Pazy's proof, it seems to me that the argument also works with $\color{red}{1}$ replaced by $\color{red}{M}$. So, I'd like to confirm if the following conditions are equivalent:


  
*$A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{M},\omega)$ on $X$.  
  
*$A$ is closed, $D(A)$ is dense in $X$, $(\omega,\infty)\subseteq\rho(A)$ and
  $$ \|(\lambda-A)^{-1}\|_{\mathcal{L}}\leq\frac{\color{red}{M}}{\lambda-\omega},\quad\forall\ \lambda>\omega.\tag{A}$$
  

I've never seen this version in any book. The usual generalization states:

The following conditions are equivalent:

  
*$A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{M},\omega)$ on $X$.  
  
*$A$ is closed, $D(A)$ is dense in $X$, $(\omega,\infty)\subseteq\rho(A)$ and
  $$ \|(\lambda-A)^{-\color{red}{n}}\|_{\mathcal{L}}\leq\frac{\color{red}{M}}{(\lambda-\omega)^{\color{red}{n}}},\quad\forall\ \lambda>\omega,\;\color{red}{\forall\ n\in\mathbb{N}}. \tag{B}$$
  

The Wikipedia says that this version "is mainly of theoretical importance since the estimates on the powers of the resolvent operator that appear in $(B)$ can usually not be checked in concrete examples".
On the other hand, it seems to me that if "$5.\Leftrightarrow 6.$" were true, then it would be of practical importance (because $(A)$ can be checked). So,


*

*if it is true, why it is not in the books? 

*if it is not true, where the Pazy's argument fails?
 A: If $T(t)$ is a bounded $C_0$ semigroup with bound $M$, then the resolvent exists for $\Re\lambda > 0$ and
$$
          (\lambda I-A)^{-1}=\int_{0}^{\infty}T(t)e^{-\lambda t}dt 
$$
That gives you a uniform bound on the resolvent
$$
             \|(\lambda I - A)^{-1}\| \le \frac{M}{\Re\lambda},\;\;\Im\lambda > 0. \tag{1}
$$
Your question becomes the following: Do you believe it is possible to find an operator $A$ with resolvent set that includes the right half plane such that
$$
               \|(\lambda I-A)^{-1}\| \le \frac{M}{\lambda},\;\;\lambda > 0,
$$
but which does not satisfy the estimate $(1)$?
A: My question was motivated by a silly calculation mistake of mine. Here is the conclusion:
The Pazy's proof with $\color{red}{1}$ replaced by $\color{red}{M}$ fails in Lemma 3.4.
With the hypothesis $(A)$, estimate (3.11) becomes
$$\|e^{t A_\lambda}x-e^{t A_\mu}x\|\leq t\mathrm{e}^{t\lambda (M+1)}\|A_\lambda x-A_\mu x\|,$$
which is not good enough to yield the convergence of $e^{t A_\lambda}x$ as $\lambda\to\infty$ (in fact we need a uniform convergence w.r.t. $t$ on bounded intervals).
Note: With the usual hypothesis $(B)$, estimate (3.11) becomes
$$\|e^{t A_\lambda}x-e^{t A_\mu}x\|\leq tM\|A_\lambda x-A_\mu x\|,$$
which is good enough.
