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Let $(T_t)_{t\geq 0}$ be a $C_0$-semigroup on a Banach space $X$ with generator $A$ such that the spectral bound $s(A)=0.$ Suppose there exists an operator $P$ on $X$ such that $$T(t) \stackrel{t\to \infty}{\to} P \text{ strongly }.$$

Then I was able to show that $$\lim_{\lambda\to 0}\lambda R(\lambda,A)f \text{ exists for each } f\in X \qquad\qquad (1).$$ I was told the converse is true if the semigroup is holomorphic.

However, I'm wondering what can be said about the limit in $(1).$

I know that if $0$ is an isolated spectral value and a pole, then the limit in $(1)$ is the spectral projection associated to $0.$

What happens when it is not an isolated spectral value? Is the limit in $(1)$ equal to $P?$

Edit: Let $\lambda>0.$ Since $T(t) \to P,$ therefore $\mathrm{Im}\, P=\ker A$ and $\overline{\mathrm{Im}\, A}\subseteq \ker P.$ The first implies that $$\lambda R(\lambda,A)P=P$$ and the second implies that $$\lim_{\lambda \to 0}\lambda R(\lambda,A)=0 \text{ on a closed subspace of } \ker P.$$

Can we now conclude by the existence of limit that, infact $$\lim_{\lambda \to 0}\lambda R(\lambda,A)=0 \text{ on } \ker P$$ and hence $$\lim_{\lambda \to 0}\lambda R(\lambda,A)=P?$$

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Part 1 of the answer.

Yes, if the semigroup converges strongly to $P$, then $\lambda R(\lambda,A)$ converges strongly to $P$, too, as $\lambda \downarrow 0$. (You do not need the assumption $s(A) = 0$ for this.)

This follows from the Laplace transform representation of the resolvent, i.e., from the formula $$ \lambda R(\lambda,A)f = \int_0^\infty \lambda e^{-t\lambda}T_t f \, dt \quad \text{for all } f \in X, $$ which is true for each complex $\lambda$ with real part $> 0$ (since convergence of the semigroups implies, by means of the uniform boundedness theorem, that its growth bound $\le 0$).

Part 2 of the answer.

I was told the converse is true if the semigroup is holomorphic.

This is not correct. The one-dimensional semigroup $(e^{it})_{t \ge 0}$ is a counterexample.

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  • $\begingroup$ A. Bobrwoski proved the converse under the assumption that semigroup is holomorphic and bounded. $\endgroup$ – Mark May 26 '20 at 22:40
  • $\begingroup$ @Mark: Yes. Please beware, though, of the following subtle terminology issue. There are two different meanings of the notion "holomorphic and bounded semigroup": (i) a semigroup that it bounded merely on the positive real axis, and holomorphic on a sector, or (ii) a semigroup that is holomorphic and bounded on a sector. This ambiguity occurs very often, and it easily causes misunderstandings. Proposition 10 in Adam Bobrowski's paper that you're referring to uses the second meaning of the notion. [to be continued] $\endgroup$ – Jochen Glueck May 27 '20 at 6:16
  • $\begingroup$ [continuation] For semigroups that are "bounded and holomorphic" merely in the weaker sense (i), this result does not hold, as the little counterexample in my answer shows. The main point is as follows: each of the properties (i) or (ii) implies that the spectrum of the generator is contained in a left sector. But in (i) the vertex of the sector need not be the point $0$, while (ii) always implies that one can choose the vertex to be $0$. Consequently, (ii) implies that the spectrum $\sigma(A)$ intersects $i\mathbb{R}$ only in $0$. [to be continued] $\endgroup$ – Jochen Glueck May 27 '20 at 6:21
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    $\begingroup$ [continuation] Hence, the result in Bobrowski's paper can be considered as a special case of the following Tauberian theorem: assume that $A$ generates a bounded $C_0$-semigroup $(T_t)_{t \ge 0}$, that the spectrum $\sigma(A)$ intersects the imaginary axis at most in $0$, and that the Abel limit $\lim_{\lambda \downarrow 0} \lambda R(\lambda,A)$ exists strongly. Then $T_t$ converges strongly as $t \to \infty$ (this result is, in turn, a special of the so-called ABLV theorem). $\endgroup$ – Jochen Glueck May 27 '20 at 6:27

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