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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:

  1. $A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{1},0)$ on $X$.
  2. $A$ 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:

  1. $A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{1},\omega)$ on $X$.
  2. $A$ 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:

  1. $A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{M},\omega)$ on $X$.
  2. $A$ 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:

  1. $A$ is the infinitesimal generator of a $C_0$-semigroup of type $(\color{red}{M},\omega)$ on $X$.
  2. $A$ 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?
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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 $(1)$?

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  • $\begingroup$ My question was motivated by a silly calculation mistake of mine. So, there is no reason to believe that $5.$ is equivalent to $6.$ and thus such operator probably exists. Do you have an example? $\endgroup$ – Pedro Oct 13 '16 at 5:12
  • $\begingroup$ @Pedro : I believe if you use a normal operator and you put the spectrum in the complement of a cone given by $|y| \ge x$, $x > 0$ I believe you have such an operator. The resolvent for the normal operator is bounded by the reciprocal distance by the real axis to the cone. I think that does it, but check me on that. $\endgroup$ – DisintegratingByParts Oct 13 '16 at 13:52
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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.

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