# Convergence of $\exp(tQ)$ in operator norm as $t\rightarrow\infty$

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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• Use the spectral radius formula and the spectral mapping theorem (for the map $e^x$) – Pietro Majer Feb 23 at 8:03
• en.wikipedia.org/wiki/Hille%E2%80%93Yosida_theorem – Steve Huntsman Feb 23 at 13:25
• @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.. – 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).

• 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? – MikeG Feb 23 at 15:02
• @MikeG: I've added more details to the proof, as well as several remarks and references after the proof. – Jochen Glueck Feb 23 at 17:56
• Thanks a lot!!! – MikeG Feb 23 at 22:47