0
$\begingroup$

I'm working through a problem, and it states "Show that if $X$ is bounded and $X \supset Y$, then $\omega(X) \supset \omega(Y)$. Deduce that if $Y$ is an absorbing set then $\omega(X) = \omega(Y)$." (In this case, we are assuming $S(t)$ is a semigroup of solution operators form a $C^0$ semigroup, i.e. that are NOT linear in space.)

First question: if we use the definition $\omega(X) = \{z : \exists t_n \to \infty, x_n \in X, S(t_n)x_n \to z\}$, it seems trivial that $\omega(Y) \subset \omega(X)$. I don't know why it is required that $X$ be bounded. Shouldn't it simply follow from the definition?

Additionally, it says deduce that if $Y$ is an absorbing set, then the $\omega$-limit sets are equal. If $Y$ is absorbing, then so is $X$. We can maybe use the other definition here, where $\omega(X) = \bigcap\limits_{t \geq 0} \overline{\bigcup\limits_{s\geq t} S(s)X}$, which since $Y$ is absorbing we can rewrite $\omega(X) = \bigcap\limits_{t \geq 0} \overline{\Big(\bigcup\limits_{t_0 \geq s \geq t} S(s)X \cup Y\Big)}$ and $\omega(Y) = \bigcap\limits_{t \geq 0} \overline{\Big(\bigcup\limits_{t_0 \geq s \geq t} S(s)Y \cup Y\Big)}$, for an appropriately chosen $t_0$. I don't see how we could use this to make a simple deduction, so I am not sure what I am missing. Any help would be much appreciated.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.