This is not a homework problem, so I am not sure whether this has a "good" answer or not. I came up with this question when I am now learning functional analysis and wonder whether my "freshman's intuition" for exponential works.

If $Q$ is some bounded linear operator on some Banach space(and maps to the same space), then since it is bounded, we can define the exponential operator $$P(t)=\exp(tQ)$$ for every $t>0$. It is again a bounded linear operator for every fixed $t$.

Question. What is a sufficient condition on $Q$ (if necessary, better!) for $\{P(t)\}$ to converge in operator norm to some operator as $t\rightarrow\infty$?

My intuition tells me there should be something of $Q$ being negative--at least non-positive, resulting in $P(t)\rightarrow 0$ or some other operator. If the spectrum of $Q$ is contained in $\{z\in\mathbb{C}\, :\, \operatorname{Re}(z)<0\}$, will this convergence happen? However, I failed to link this convergence to $Q$'s spectrum.

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    $\begingroup$ Use the spectral radius formula and the spectral mapping theorem (for the map $e^x$) $\endgroup$ – Pietro Majer Feb 23 at 8:03
  • $\begingroup$ en.wikipedia.org/wiki/Hille%E2%80%93Yosida_theorem $\endgroup$ – Steve Huntsman Feb 23 at 13:25
  • $\begingroup$ @SteveHuntsman: I'm not sure I see the relevance of the Hille-Yosida theorem here. Hille-Yosida characterises under which conditions an unbounded operator generates a $C_0$-semigroup; but for bounded operators (as considered by the OP) the generation property is trivial. On the other hand, the Hille-Yosida theorem does not tell us much about the long-time behaviour of the semigroup.. $\endgroup$ – Jochen Glueck Feb 23 at 14:28

Convergence to $0$ is simple:

Proposition 1. 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 2. 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 $Q$.

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).

  • $\begingroup$ More details of the second is highly welcomed! actually I am new to functional analysis and don't know what is mean ergodic projection and how it works here...or maybe some suggested materials for me to work out it by myself? $\endgroup$ – MikeG Feb 23 at 15:02
  • $\begingroup$ @MikeG: I've added more details to the proof, as well as several remarks and references after the proof. $\endgroup$ – Jochen Glueck Feb 23 at 17:56
  • $\begingroup$ Thanks a lot!!! $\endgroup$ – MikeG Feb 23 at 22:47

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