# Example of measure of non-compactness

I can't understand the following example of measure of non-compactness, which was given in this article.

Definition: A non-negative function $\phi$ defined on the bounded subsets of $X$ will be called Sadovskij functional if it satisfies the following for bounded subsets $M,N\subset X$ and $\alpha\in K$.

• $\phi(M\cup N)=\max\{\phi(M),\phi(N)\}$
• $\phi(M+N)\leq \phi(M)+\phi(N)$
• $\phi(\lambda M)=|\lambda|\phi(M)$
• $\phi(M) \leq \phi(N)$ for $M\subseteq N$
• $\phi([0, 1] ·M) = (M)$
• $\phi(\overline{co}M) = (M)$.

A Sadovskij functional is called measure of noncompactness if it satisfies

$\phi(M) = 0$ if and only if $M\subset X$ is pre-compact

.

Example: Let $X = L^p[0, 1]$ $(1 \leq p < \infty)$ be the Lebesgue space of all (classes of) p-integrable real functions on $[0, 1]$ with the usual norm, and denote by $\chi_D$ the characteristic function of a measurable subset $D\subset[0,1]$.

Can someone kindly explain me the following;

-How can we prove that $$\phi(M) := \limsup_{\textrm{mes}(D)\rightarrow 0} ~~\sup_{u\in M}||\chi_Du||,\quad (M\subset X)$$ is a Sadovskij functional?

-How is it not a measure of non-compactness?

-How to show that any set $M$ which is bounded in $L^q[0, 1]$ for some $q > p$ satisfies $\phi(M) = 0$ in $X$, by the Hölder inequality. How such a set need not be pre-compact in $X$?

Rather I'd be very thankful if someone can give very easy and simple examples of measure of non-compactness.

P.S. I couldn't even understand what $\limsup_{\textrm{mes}(D)\rightarrow 0}$ means.

• I asked this question on math.stackexchange first but nobody was able to answer it and that's why I posted it here. Commented Sep 2, 2015 at 8:43
• $\phi(M)$ is defined as the $\limsup$, as the measure of $D$ goes to zero, of $\sup_{u\in M}||\chi_D u||$, that is $\lim_{\varepsilon\to 0}\ \sup_{mes(D)\le\varepsilon}\ \sup_{u\in M}||\chi_D u||$ Commented Sep 2, 2015 at 14:00

-How is it not a measure of non-compactness?

The set of Rademacher functions $R:=\{r_n(t) : n \in N\}$ is bounded but not pre-compact (the sequence $(r_n)$ is equivalent to the unit vector basis of $\ell_2$) in $X$, but $\phi(R)=0$ because $|r_n(t)|\leq 1$.

-How to show that any set $M$ which is bounded in $L_q[0,1]$ for some $q>p$ satisfies $\phi(M)=0$ in $X$, by the Hölder inequality. How such a set need not be pre-compact in $X$?

Note that $\phi(M)>0$ in $X$ implies the existence of a constant $C>0$, a sequence $(f_n)$ in $M$, and a disjoint sequence of measurable subsets of $[0,1]$ such that $\|f_n\chi_{A_n}\|_p\geq C$. However, if $(f_n)$ is bounded in $L_q[0,1]$, the Hölder inequality implies $\|f_n\chi_{A_n}\|_p\to 0$ as $n\to\infty$. Note also that $R$ is bounded but not precompact in each $L_q[0,1]$.

For arguments and results of this kind, see the book Albiac, Fernando; Kalton, Nigel J. Topics in Banach space theory. Graduate Texts in Mathematics, 233. Springer, New York, 2006.

Example of a measure of non-compactness on $L^p[0,1]$ : $\phi(M):=\limsup_{h\to 0}\sup_{u\in M}(\int_0^{1}|u(x+h)-u(x)|^p)^{1/p}$. It satisfies the axioms of Sadovskij functionals (for $1\le p<\infty$, due to the convexity of the norm) and $M$ is precompact in $L^p$ iff $\phi(M)=0$ : that's Kolmogorov's criterion.

Another example, on $C(K)$ if $K$ is compact metric, with the sup-norm : $\phi(M):=\limsup_{h\to 0}\sup_{u\in M,\ d(x,y)\le h}|u(x)-u(y)|$, that vanishes if and only if $M$ is equicontinuous (i.e. precompact, by Ascoli).

• Thank you for your effort. Can you provide me some reference for these? @JeanDuchon Commented Sep 3, 2015 at 6:29