Let $X$ be a Banach space and $A: X \to X$ a bounded linear operator. So, $A$ is the infinitesimal generator of an uniformly continuous semigroup $\{T(t)\}_{t\geq 0}$ on $X$. The proof, as presented in Pazy's book, is to define $T(t)=e^{At}$ and verify the conditions.

After, in the same book, in order to prove that there exists an unique semigroup whose generator is $A$, is proved the following:

Theorem: Let $T(t)$ and $S(t)$ be uniformly continuous semigrops of bounded linear operators. If $$\lim_{t\to 0^+}\left\|\frac{T(t)-I}{t}-A\right\|=0=\lim_{t\to 0^+}\left\|\frac{S(t)-I}{t}-A\right\|\tag{#}$$ then $T(t)=S(t)$ for $t\geq 0$.

Since the infinitesimal generator of a semigroup is an operator $A$ that satisfies

$$Ax=\lim_{t\to 0^+}\frac{T(t)x-x}{t},$$

why shouldn't we replace $(\#)$ by $(*)$?

$$\lim_{t\to 0^+}\left\|\frac{T(t)x-x}{t}-Ax\right\|_X=0=\lim_{t\to 0^+}\left\|\frac{S(t)x-x}{t}-Ax\right\|_X\;\forall x\in X\tag{*}$$

Are in this case $(\#)$ and $(*)$ equivalent?



1 Answer 1


The conditions are equivalent in the case that $A$ is bounded. Normally, though, where $A$ may be unbounded with a domain which is not all of $X$, you don't get uniform operator norm convergence because that would imply $A$ is bounded. The strong (vector) convergence is definitely implied by the uniform because $\|Ax\|\le \|A\|\|x\|$ for all $x\in X$ if $A$ is bounded. It's less obvious--but true--that strong convergence implies uniform convergence if the generator $A$ is bounded (this has to do with uniqueness of $C^{0}$ semigroups given the generator, and with the fact that you know how to construct one for a bounded $A$.)

This edit is to flesh out my comment that I made to you. Here's what I was saying you could prove with only minor modification to Pazy's argument:

Theorem: Let $X$ be a Banach space. Let $T : [0,\infty)\rightarrow\mathscr{L}(X)$ be a semigroup. Suppose that there exists $A\in\mathscr{L}(X)$ such that the following limits exist for all $x \in X$: $$ \lim_{t\downarrow 0}\left\|\frac{1}{t}\{T(t)x-x\}-Ax\right\|_{X}=0. $$ Then $T(t)=e^{tA}$. Therefore, for such $T$, one has $$ \lim_{t\downarrow 0}\left\|\frac{1}{t}\{T(t)x-x\}-Ax\right\|_{\mathscr{L}(X)}=0, $$ because the above holds for $e^{tA}$.

Proof: For each $x$, show that $S(t)x=e^{-tA}T(t)x$ has right derivative 0 for all $t \ge 0$. Conclude that $S(t)x=x$ for all $t \ge 0$, which means $e^{-tA}T(t)=I$ for all $t \ge 0$. Multiply on the left by $e^{tA}$ to conclude that $T(t)=e^{tA}$. $\Box$

  • $\begingroup$ I'm interested in the proof that "strong convergence implies uniform convergence if the generator $A$ is bounded". Could you help me with more details or any reference? $\endgroup$
    – Pedro
    Jan 14, 2014 at 23:04
  • $\begingroup$ @Pedro: If the derivative of $t \mapsto T(t)x$ exists for all $x$ at $t=0$ in the vector sense, and the limit is $Ax$ where $A$ is bounded, then $e^{-tA}$ is formed using power series, and $e^{-tA}T(t)x$ has derivative $0$ everywhere. So you conclude that $e^{tA}=T(t)$; so $T(t)$ inherits the stronger type of derivative from $e^{tA}$. $\endgroup$ Jan 14, 2014 at 23:18
  • $\begingroup$ I haven't understood why it proves that "strong convergence implies uniform". If I'm not wrong, your reasoning shows that every UCS is of the form $e^{At}$ (because your conlcusion is $T(t)=e^{At}$). However, in the Pazy's book, this conclusion is a corollary from theorem above so that (if we want follow Pazy's proof) we can't use it to show that "strong convergence implies uniform". $\endgroup$
    – Pedro
    Jan 15, 2014 at 13:25
  • $\begingroup$ @Pedro: I'll explain more about my last comment in an edit to the post $\endgroup$ Jan 15, 2014 at 16:55
  • $\begingroup$ Thanks for the help. One more comment: I noticed that in the proof of first theorem in Pazy's book is showed that if $T:=\{T(t)\}_{t\geq 0}$ is an UCS then there exists $\rho>0$ such that $$\lim_{t\to 0^+}\left\|\frac{T(t)-I}{t}-B_\rho\right\|=0.$$ where $$B_\rho=(T(\rho)-I)\left(\int_0^\rho T(s)ds\right)^{-1}$$ This implies that the IG (infinitesimal generator) of $T$ is $A=B_\rho$ (because uniform convergence implies strong). Hence, the IG of an UCS satisfies $$\lim_{t\to 0^+}\left\|\frac{T(t)-I}{t}-A\right\|=0.$$ This explains the Pazy's argument. $\endgroup$
    – Pedro
    Jan 16, 2014 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.