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I am working on an article based mainly on the notion of Measure of non-compactness, to study a particular type of fixed point theorems.

Let $\mathcal M $ to be the family of all nonempty bounded subsets of $E$.

Definition:

The function $\nu : \mathcal M \rightarrow \mathbb R^+$ is said to be a measure of noncompactnees if:

1)$\nu (A)=\nu (\bar{A)}, \forall A \in \mathcal M$.

2)If $A_n \in \mathcal M$, and $A_{n+1}\subset A_{n}$ and if $\lim_{n \to +\infty} \nu (A_n)=0$, then: $A_{\infty}= \bigcap_{n=1}^{+\infty}A_n\neq \emptyset $.

3)If $A\in \mathcal M,$ closed and $\nu(A)=0$ Then, $A$ is compact.

Now, let $x\in BC(\mathbb R^+)$ (EDIT: $BC(\mathbb R^+)$ furnished with the standard supremum norm), i.e $x:\mathbb R ^+\rightarrow \mathbb{R}$ bounded, continuous.

Let $$\omega(x,r)=\sup\{|x(t)-x(s)| \colon t,s \in \mathbb R ^+ , \ |t-s|<r\}$$

(called modulus of continuity of $x$ ).

Let us fix $X$ a nonempty bounded subset of $BC(\mathbb R^+)$, and $$\begin{align*} \omega(X,r)&= \sup\{\omega(x,r), x\in X\}\\ \omega_0(X)&= \lim_{r\to 0} \omega(X,r)\} \end{align*} $$

I need to prove that $\mu$ is a measure of noncompactness, such that:

$$\mu(X)=\omega_0(X)+\lim_{t\to +\infty } sup\;diamX(t)$$

where $$diamX(t)=sup\{|x(t)-y(t)|:\: x,y \in X\}$$


Its Okay with 1) and 2) of the definiton, How can I get 3)? I guess we have to apply Arzelà–Ascoli somewhere..

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1 Answer 1

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$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}}$

Assume that the topology on $BC(\mathbb R^+)$ is induced by the $\sup$ norm and that $\lim\limits_{t\to +\infty } sup$ in the definition of $\mu(X)$ means $\limsup_{t\to\infty}$.

Let us now show that condition 3) in the definition of a measure of noncompactness holds. Accordingly, suppose that $X$ a nonempty closed bounded subset of $BC(\mathbb R^+)$ with $\mu(X)=0$. We need to show that then $X$ is compact.

Take any real $\ep>0$. Then the condition $\mu(X)=0$ implies that for some real $T>0$ one has \begin{equation} \sup\{|x(t)-y(t)|\colon x,y \in X, t\ge T\}<\ep. \tag{1} \end{equation} On the other hand, in view of the Arzelà--Ascoli theorem, the condition $\mu(X)=0$ implies that the set \begin{equation*} X_T:=\{x|_{[0,T]}\colon x\in X\} \end{equation*} of the restrictions $x|_{[0,T]}$ of the functions $x\in X$ to the interval $[0,T]$ is compact with respect to the $\sup$ norm on $[0,T]$. So, $X_T$ is totally bounded and thus has a finite $\ep$-net $\{x_1|_{[0,T]},\dots,x_k|_{[0,T]}\}$. Therefore and in view of (1), $\{x_1,\dots,x_k\}$ is a finite $\ep$-net for $X$. Thus, $X$ is totally bounded. Since $X$ is closed, it is compact, as desired.

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  • $\begingroup$ Great job @Iosif Pinelis Thank you! Firstly, your assumption ^^ is correct. Secondly, but still, I have a little problem to finish, I think that " $\{y,x_1,...,x_k\}$ is $\epsilon$-net for $X$ on all $\mathbb R ^+$" need more clarification. $\endgroup$
    – Motaka
    Apr 5, 2018 at 15:47
  • $\begingroup$ @Mokata : Thank you for your comment. I have fixed that; in fact, we don't need to fix a $y$. $\endgroup$ Apr 5, 2018 at 16:24

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