I'm studying some introductory material on the $C^0$ semigroup of solution operators S(t). I'm reading through a textbook and in one place, they state that $\bigcup\limits_{0\leq t\leq t_0}S(t) B$ is clearly bounded if $B$ is bounded, and also clearly closed if $B$ is compact. However, I'm not finding this obvious. I think all we need is the continuity of $S$ in both time and with respect to the initial data and the boundedness/closedness of B, but I'm not sure exactly how to flesh out the details. Any help would be appreciated. Thanks!

  • $\begingroup$ For reference, this is in Robinson's book (where he is taking closed neighborhoods of an absorbing set $B$) on Infinite-Dimensional Dynamical Systems pg 264 (and yes, I checked the errata as well). $\endgroup$ – mathishard.butweloveit Oct 21 '18 at 2:47
  • $\begingroup$ I think it may actually require $B$ to be compact... $\endgroup$ – mathishard.butweloveit Oct 24 '18 at 18:16

Partial answer: boundedness in the case where $\{S(t)\}_{t\geq 0}$ is a semigroup of linear operators.

Since $B$ is bounded, there exists $C>0$ such that $$\|b\|\leq C,\quad\forall\ b\in B.$$

Take $x\in\bigcup\limits_{0\leq t\leq t_0}S(t) B$. Then there exist $t_x\in[0,t_0]$ and $b_x\in B$ such that $$x=S(t_x)b_x.$$

From the joint continuity of $S$ (see equation (10.2) in the said book), there exists a constant $M$ (which depends only on $t_0$ and $C$) such that $$\|S(t)b_x\|\leq M,\quad\forall \ t\in[0,t_0].$$


$$\|x\|=\|S(t_x)b_x\|\leq M,\quad\forall \ x\in \bigcup\limits_{0\leq t\leq t_0}S(t) B$$ which shows that the union is bounded.

Partial answer 2: closedness in the case where $B\subset\mathbb R^n$ (as in the book).

Take $x\in\overline{\bigcup\limits_{0\leq t\leq t_0}S(t) B}$. Then there exists a sequence $(x_n)$ in $\bigcup\limits_{0\leq t\leq t_0}S(t) B$ such that $$x_n\overset{n\to\infty}{\longrightarrow} x\tag{1}.$$

For each $n$, there exist $t_n\in[0,t_0]$ and $b_n\in B$ such that $x_n=S(t_n)b_n$. From the Bolzano-Weierstrass theorem:

  • $(t_n)$ has a subsequence $(t_{n'})$ such that $t_{n'}\overset{n'\to\infty}{\longrightarrow} t^*\in [0,t_0]$.

  • $(b_{n'})$ has a subsequence $(b_{n''})$ such that $b_{n''}\overset{n''\to\infty}{\longrightarrow} b^*\in B$.

Then $$x_{n''}=S(t_{n''})b_{n''}\overset{n''\to\infty}{\longrightarrow} S(t^*)b^*\tag{2}$$ because $$\|S(t_{n''})b_{n''}-S(t^*)b^*\|\leq \|S(t_{n''})b_{n''}-S(t_{n''})b^*\|+\|S(t_{n''})b^*-S(t^*)b^*\|.$$

From $(1)$ and $(2)$, $$x=S(t^*)b^*\in \bigcup\limits_{0\leq t\leq t_0}S(t) B.$$

  • $\begingroup$ Great thanks! I will take this idea and see if I can apply it to the closedness portion. $\endgroup$ – mathishard.butweloveit Oct 21 '18 at 19:18
  • $\begingroup$ Ok, I do have a question. I thought I followed the second Inequality, but 10.2 in the book deals with a difference of solutions at time t with different initial data' you can't necessarily guarantee that you can find an initial condition such that at some time t the solution is identically 0. $\endgroup$ – mathishard.butweloveit Oct 22 '18 at 17:10
  • $\begingroup$ @mathishard.butweloveit Isn't 10.2 valid for $v_0=0$? $\endgroup$ – Pedro Oct 22 '18 at 17:20
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    $\begingroup$ @mathishard.butweloveit You are right. I edited the title of my post. $\endgroup$ – Pedro Oct 22 '18 at 18:06
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    $\begingroup$ @mathishard.butweloveit I thought $C^0$ was $C_0$, sory. It seems that the argument in your third comment is correct. $\endgroup$ – Pedro Oct 22 '18 at 18:13

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