Is a set “trapped” between two convex sets, convex

convex-analysisconvex-geometrygeneral-topology

Let $A, C$ be two strictly-convex sets in $\mathbb{R}^n$. Let $B$ be a set such that:

$$A\subseteq B\subseteq C.$$

Is $B$ strictly-convex too? Not necessarily. For example, $A$ and $C$ can be to concentric discs, and $B$ can be a crazy amoeba-shaped figure trapped between them:

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But what if $A$ and $C$ only differ in their boundary? Consider two cases:

  • $A$ is the interior of some set and $C$ is the closure of the same set. Then, as shown by Anonymous below, $B$ might be arbirary

  • $A$ is the interior of $C$ and $C$ is the closure of $A$, so that $B$ contains $A$ plus some part of the boundary. In this case, is $B$ always strictly-convex?

Best Answer

Let $C$ be closed and strictly convex with $int \ C \ne \emptyset$. Let $B$ be such that $$ int \ C \subset B \subset C. $$ The interior of a set is an open set, so $int\ C \subset int \ B$.

Then $B$ is strictly convex: let $x,y\in B \subset C$. Then $\lambda x + (1-\lambda)y \in int \ C\subset int \ B$ for $\lambda \in (0,1)$.

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