On the measure of nonconvexity (MNC)

Question

I'm actually working on the measure of nonconvexity and its application. Especially, the Eisenfeld–Lakshmikantham MNC defined - in a Banach space - by:

$$\alpha(A)=\sup_{b\in\operatorname{conv}(A)} \inf_{a\in A}|| b-a \|<\infty $$ Alternatively, if $H(X, Y)$ denotes the Hausdorff distance between two subsets $X$ and $Y$, $$\alpha(A)=H(A, \operatorname{conv}(A))$$

where $\operatorname{conv}(A)$ is the convex hull of $A$.

The interesting thing about the E-L MNC is that $\alpha(A)=0 \Longrightarrow \overline{A} \text{ is convex}.$

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I'm looking for a sequence of non-convex sets $(A_n)_n\subset E$ - which somehow - verify a "contraction" condition, that is, $$\alpha(A_{n+1})\leq \lambda\,\alpha(A_{n})\text{ where } \lambda \in (0,1).$$

Answer 1

This is a simple problem and I am voting to close it. Take any non-convex set $A$ with $\alpha(A)<\infty$ and define $A_n=\lambda^n A$ (dilation with the factor $\lambda^n$. Then $\alpha(A_n)=\lambda^n\alpha(A)$, because the distances are scaling by that factor and hence $\alpha(A_{n+1})=\lambda \alpha(A_n)$.