# Boundedness of Union in Time of Semigroup Acting on a Bounded Set

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!

• 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). – mathishard.butweloveit Oct 21 '18 at 2:47
• I think it may actually require $B$ to be compact... – 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 $$\|S(t)b_x\|\leq M,\quad\forall \ t\in[0,t_0].$$

Therefore,

$$\|x\|=\|S(t_x)b_x\|\leq M,\quad\forall \ x\in \bigcup\limits_{0\leq t\leq t_0}S(t) B$$ 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 $$(2)$$, $$x=S(t^*)b^*\in \bigcup\limits_{0\leq t\leq t_0}S(t) B.$$

• Great thanks! I will take this idea and see if I can apply it to the closedness portion. – mathishard.butweloveit Oct 21 '18 at 19:18
• 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. – mathishard.butweloveit Oct 22 '18 at 17:10
• @mathishard.butweloveit Isn't 10.2 valid for $v_0=0$? – Pedro Oct 22 '18 at 17:20
• @mathishard.butweloveit You are right. I edited the title of my post. – Pedro Oct 22 '18 at 18:06
• @mathishard.butweloveit I thought $C^0$ was $C_0$, sory. It seems that the argument in your third comment is correct. – Pedro Oct 22 '18 at 18:13