I can't understand the following example of measure of non-compactness, which was given in a research article.
Definition: A nonnegative 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 $\alpha\in K$.
- $\phi(M\cup N)=\max\{\phi(M),\phi(N)\}$
- $\phi(M+N)\leq \phi(M)+\phi(N)$
- $\phi(\lambda M)=|\lambda|\phi(M)$
- $\phi(M)< \phi(N)$ for $M\subseteq N$
- $\phi([0, 1] ·M) = (M)$
- $\phi(\overline{co}M) = (M)$.
An additional property of a Sadovskij functional is
$\phi(M) = 0$ if and only if M is precompact,
which we call the regularity of . A regular Sadovskij functional is called measure of noncompactness.
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 $$\phi(M) := \lim \sup_{mesD\rightarrow 0} \sup_{u\in M}||\chi_Du||,\quad (M\subset X)$$ 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 H¨older 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.