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$\newcommand{\f}{\phi}$$\newcommand{\ep}{\varepsilon}$$\newcommand{\R}{\mathbb R}$

Suppose $(\f_t)_{t\ge0}$ is an abstract dynamical system in a Banach space $(X,\|\mathord\cdot\|)$. Let $C(x,\ep)$ denote the closed ball with radius $r$ centered at $x\in X$.

Recall that an abstract dynamical system $(\phi_t)_{t\ge0}$ on $X$ is a collections of maps $\f_t: X\to X$ such that $\f_0$ is the identity map on $X$ and $\f_t\circ\f_{s}=\f_{t+s}$ for all $t,s\ge0$.

Assume the following:

Suppose that for all $t\ge0$ there are maps $f_t,g_t:X\to X$ such that $$\f_t(x)=f_t(x)+g_t(x).\tag{1}$$

Additionally, suppose that for all $r\ge0$ there is a $T_r\ge0$ and a map $h=h_{T_r}:[0,\infty)\to[0,\infty)$ where $$\lim_{t\to\infty}h(t)=0\tag{2}$$ and$$\overline{f_t[C(0,r)]}\text{ is compact whenever }t>T_r\tag{3}$$(the overline is the closure) and for all $t\ge0$ and all $x\in C(0,r)$: $$\|g_t(x)\|\le h(t) \tag4$$

Then $$\lim_{t\to\infty}\alpha(\f_t[A])=0$$for all bounded sets $A\subset X$, where $\alpha$ is the Kuratowki measure of noncompactness.

I have been told that this a well-known result from the theory of abstract dynamical systems, but I can't find a proof. Is there someone who knows how to prove this statement or knows a nice reference (preferably a book)?

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Recall that $\alpha$ is monotone and subadditive: $\alpha(B+C)\le \alpha(B)+\alpha(C)$. [Reference: Nonlinear Functional Analysis by Deimling). Using (1), we find that $\alpha(\phi_t(A))\le \alpha(f_t(A))+\alpha(g_t(A))$. Pick $r$ such that $A\subset C(0,r)$. By (4) and (2) we have $\alpha(g_t(A))\le 2h(t)\to 0$ as $t\to\infty$. Finally, $\alpha(f_t(A))=0$ by (3) when $t>T_r$.

Unless I'm mistaken, the semigroup property isn't needed.

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Thank you for your help. I understand your reasoning, bu I do not see why (4) and (2) imply $\alpha(g_t(A))\le 2\cdot h(t)$. – Gifty May 22 '12 at 20:43
@Gifty For any bounded set $B$ we have $\alpha(B)\le \mathrm{diam} \, B$ by the definition of $\alpha$ (cover $B$ by itself). Here $B=g_t(A)$, and since this set is contained in the closed ball of radius $h(t)$, its diameter does not exceed $2h(t)$. – user31373 May 22 '12 at 20:45

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