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

Question

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!

Answer 1

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