I'm actually working on the measure of nonconvexity and its application. Especially, the Eisenfeld–Lakshmikantham MNC defined - in a Banach space - by:
$$\alpha(A)=\sup_{b\in\overline{\operatorname{conv}}(A)} \inf_{a\in A}|| b-a \|<\infty $$ Alternatively, if $H(X, Y)$ denotes the Hausdorff distance between two subsets $X$ and $Y$, $$\alpha(A)=H(A, \overline{\operatorname{conv}}(A))$$
where $\operatorname{conv}(A)$ is the convex hull of $A$.
The interesting thing about the E-L MNC is that $\alpha(A)=0 \Longrightarrow \overline{A} \text{ is convex}.$
Note that if a function $f$ is $\lambda$-norm-Lipschitzian, then the measure of (weak) noncompactness of De Blasi ($\beta$) is $\lambda$-Lipschitzian, that is $$\beta(f(X))\leq \lambda \beta(X)$$
Note that
$$ \beta(X)=\inf \left\{\varepsilon>0: \exists \text { weakly compact } Y \text { with } X \subset Y+\varepsilon B_E\right\} \text {, } $$
Is there a similar sufficient condition to have the $\lambda$-Lipschitzian property for the measure of nonconvexity.
$$\alpha(f(X))\leq \lambda \alpha(X)$$
Unfortunately, the $\lambda$-norm-Lipschitzianity of $f$ is not enough.
I think, at some stage, we need that $\overline{\operatorname{Conv}}(f(X))\subseteq f(\overline{\operatorname{Conv}}(X))$
Edit: After many discussions with colleagues, it's hard to impose a contraction condition with respect to $\alpha$. It's a very restrictive condition.
Now I'm looking for an example of an operator $T:C\subset E\to C$ and a decreasing sequence $\left\{X_{n}\right\}_{n \in \mathbb{N}}$ and $T$-invariant subsets of $C$, such that
$$\alpha(\left\{X_{n}\right\}_{n \in \mathbb{N}})\underset{n\to \infty}{\rightarrow}0.$$
