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Jochen Glueck
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Proposition 1. The following are equivalent:

Theorem 2. The following are equivalent:

If the equivalent assertions are satisfied, then the limit $\lim_{t \to \infty} e^{tQ}$ equals the spectral projection of $Q$ associated with the isolated spectral value $0$.

Sketch of proof. This also follows from classical spectral theory,"(i) $\Rightarrow$ (ii)" Let $P$ denote the limit operator; it commutes with $e^{tQ}$ for instance by splittingeach $t$. It's easy to check that the space intorange of $P$ consists precisely of the kernelfixed points of the operator semigroup $(e^{tQ})_{t \in [0,\infty)}$ and consequently, we can see that $P$ is a projection. Since the projection commutes with the semigroup, both the range and the kernel of $P$ are invariant under the mean ergodicsemigroup.

On the kernel of $P$, we have that $e^{tQ}$ converges in operator norm to $0$ as $t \to \infty$, so, according to Proposition 1, the restriction $Q|_{\ker P}$ has spectrum in the open left halfplane.

On the other hand, $e^{tQ}$ acts as the identity operator on the range $\operatorname{rg}P$. Since the space $\operatorname{rg}P$ is non-zero, the number $0$ is a spectral value of the restriction $Q|_{\operatorname{rg}P} = 0$, and a first order pole of its resolvent.

"(ii) $\Rightarrow$ (i)" Let $P$ denote the spectral projection of $Q$ associated with the semigroupisolated spectral value $(e^{tQ})_{t \ge 0}$$0$. Since $0$ is a first order pole of the resolvent, it follows that the restriction $Q|_{\operatorname{Rg}P}$ is the $0$ operator, so $e^{tQ}$ acts as the identity operator on $\operatorname{Rg}P$.

On the other hand, the restriction $Q|_{\ker P}$ has spectrum in the open left halfplane, so Proposition 1 tells us that $e^{tQ}$ norm converges to $0$ on $\ker P$ as $t \to \infty$.

So to sum up, on the whole space we have convergence of $e^{tQ}$ to $P$ as $t \to \infty$. $\square$

Remarks 3. (Ia) In the proof of the implication "(ii) $\Rightarrow$ (i)" we used several results about spectral projections and poles of the resolvents. These results can addbe found in various classical books about functional analysis; but they are a fewbit scattered through several books where they are, in my experience, not so easy to digest at first glance. I thus wrote a brief summary about spectral projections and properties of poles of the resolvent, along with detailed references to various books, in the Appendices A.1 - A.3 of my PhD thesis.

(b) Proposition 1 und Theorem 2 can be seen as very elementary special cases of the topic "long term behaviour of $C_0$-semigroups". The point is that, if we replace the bounded operator $Q$ with an unbounded closed operator that generates a so-called $C_0$-semigroup, the question whether one has operator norm convergence as $t \to \infty$ becomes suddenly much more details hereinvolved.

(c) The aforementioned topic becomes even subtler if requiredone replaces operator norm convergence with strong convergence. For this topology, even for bounded operators $Q$ the long-term behaviour of $e^{tQ}$ becomes quite non-trivial (and even more so for $C_0$-semigroups with unbounded generators).

(d) An introduction to the long-term behaviour of $\square$$C_0$-semigroups with many useful theorems can, for instance, be found in Chapter V of the book "Engel and Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)" (link to zbMATH).

Proposition. The following are equivalent:

Theorem. The following are equivalent:

Sketch of proof. This also follows from classical spectral theory, for instance by splitting the space into the kernel and the range of the mean ergodic projection of the semigroup $(e^{tQ})_{t \ge 0}$. (I can add a few more details here if required.) $\square$

Proposition 1. The following are equivalent:

Theorem 2. The following are equivalent:

If the equivalent assertions are satisfied, then the limit $\lim_{t \to \infty} e^{tQ}$ equals the spectral projection of $Q$ associated with the isolated spectral value $0$.

Sketch of proof. "(i) $\Rightarrow$ (ii)" Let $P$ denote the limit operator; it commutes with $e^{tQ}$ for each $t$. It's easy to check that the range of $P$ consists precisely of the fixed points of the operator semigroup $(e^{tQ})_{t \in [0,\infty)}$ and consequently, we can see that $P$ is a projection. Since the projection commutes with the semigroup, both the range and the kernel of $P$ are invariant under the semigroup.

On the kernel of $P$, we have that $e^{tQ}$ converges in operator norm to $0$ as $t \to \infty$, so, according to Proposition 1, the restriction $Q|_{\ker P}$ has spectrum in the open left halfplane.

On the other hand, $e^{tQ}$ acts as the identity operator on the range $\operatorname{rg}P$. Since the space $\operatorname{rg}P$ is non-zero, the number $0$ is a spectral value of the restriction $Q|_{\operatorname{rg}P} = 0$, and a first order pole of its resolvent.

"(ii) $\Rightarrow$ (i)" Let $P$ denote the spectral projection of $Q$ associated with the isolated spectral value $0$. Since $0$ is a first order pole of the resolvent, it follows that the restriction $Q|_{\operatorname{Rg}P}$ is the $0$ operator, so $e^{tQ}$ acts as the identity operator on $\operatorname{Rg}P$.

On the other hand, the restriction $Q|_{\ker P}$ has spectrum in the open left halfplane, so Proposition 1 tells us that $e^{tQ}$ norm converges to $0$ on $\ker P$ as $t \to \infty$.

So to sum up, on the whole space we have convergence of $e^{tQ}$ to $P$ as $t \to \infty$. $\square$

Remarks 3. (a) In the proof of the implication "(ii) $\Rightarrow$ (i)" we used several results about spectral projections and poles of the resolvents. These results can be found in various classical books about functional analysis; but they are a bit scattered through several books where they are, in my experience, not so easy to digest at first glance. I thus wrote a brief summary about spectral projections and properties of poles of the resolvent, along with detailed references to various books, in the Appendices A.1 - A.3 of my PhD thesis.

(b) Proposition 1 und Theorem 2 can be seen as very elementary special cases of the topic "long term behaviour of $C_0$-semigroups". The point is that, if we replace the bounded operator $Q$ with an unbounded closed operator that generates a so-called $C_0$-semigroup, the question whether one has operator norm convergence as $t \to \infty$ becomes suddenly much more involved.

(c) The aforementioned topic becomes even subtler if one replaces operator norm convergence with strong convergence. For this topology, even for bounded operators $Q$ the long-term behaviour of $e^{tQ}$ becomes quite non-trivial (and even more so for $C_0$-semigroups with unbounded generators).

(d) An introduction to the long-term behaviour of $C_0$-semigroups with many useful theorems can, for instance, be found in Chapter V of the book "Engel and Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)" (link to zbMATH).

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Jochen Glueck
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Convergence to $0$ is simple:

Proposition. The following are equivalent:

(i) The operator $e^{tQ}$ converges to $0$ with respect to the operator norm as $t \to \infty$.

(ii) The spectrum of $Q$ is contained in the open left halfplane $\{\lambda \in \mathbb{C}: \, \operatorname{Re} \lambda < 0\}$.

Sketch of proof. As mentioned by Pietro Majer in the comments, this follows from the spectral mapping theorem for the operator exponential function, along with the spectral radius formula. $\square$

Convergence to a non-zero operator is a bit more involved. Here are the details:

Theorem. The following are equivalent:

(i) The operator $e^{tQ}$ converges with respect to the operator norm to a non-zero operator as $t \to \infty$.

(ii) The spectrum of $Q$ is contained in the set $\{\lambda \in \mathbb{C}: \, \operatorname{Re} \lambda < 0\} \cup \{0\}$, and $0$ is an isolated point in the spectrum and a first order pole of the resolvent of $Q$.

Sketch of proof. This also follows from classical spectral theory, for instance by splitting the space into the kernel and the range of the mean ergodic projection of the semigroup $(e^{tQ})_{t \ge 0}$. (I can add a few more details here if required.) $\square$