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 $E$.


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..


1 Answer 1


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

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.

  • $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.