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?



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 '14 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 $e^{tA}$. $\endgroup$ – DisintegratingByParts Jan 14 '14 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 '14 at 13:25
  • $\begingroup$ @Pedro: I'll explain more about my last comment in an edit to the post $\endgroup$ – DisintegratingByParts Jan 15 '14 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 '14 at 18:50

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