Relation between symmetry about the origin and centroid of convex sets

Question

Given a convex set $C$ of points $x \in \mathbb{R}^d$.

Lets assume that $\forall x\in C \ (-x) \in C$, then is it possible that the centroid of C is the origin of $\mathbb{R}^d$ (the zero vector) ?

Please advise and thanks in advance.

Answer 1

We don't need convexity for this. However it is important that the centroid $c$ of $C$ is at all defined. Therefore let's assume that $C\subset {\mathbb R}^d$ is a bounded measurable set. Denote the volume element in ${\mathbb R}^d$ by ${\rm d}(x)$. Then the centroid $c$ of $C$ is given by $$c={\mathstrut\int_{\mathstrut C} x\>{\rm d}(x)\over\int_C^{\mathstrut} {\rm d}(x)}\ .\tag{1}$$

From $-C\subset C$ it follows by reflection of these sets that $C=-(-C)\subset -C$, so that in fact $-C=C$. The Jacobian of the reflection has constant absolute value $1$. We therefore have $$\int_C x\>{\rm d}(x)=\int_{-C} x\>{\rm d}(x)=\int_C(-y)\>{\rm d}(y)=-\int_C y\>{\rm d}(y)\ ,$$ which implies that the numerator in $(1)$ is $=0$, hence $c=0$.