Let $x_0,\ldots,x_{d+1}\in\mathbb{R}^d$, then can there be some $y\in\mathrm{int\!hull\!}_i x_i$ that is a unique convex combination of $x_i$?

Question

Fix some points $x_0,\ldots,x_{d+1}\in\mathbb{R}^d$ such that the convex hull $C:=\operatorname{hull}_{0\le i\le d+1} x_i$ has nonempty interior, and let $y$ be in that interior. Can we say definitively that there exists more that one $\mu$ in the simplex $$\Delta^{d+1}:=\{(\mu_0,\ldots,\mu_{d+1})\in\mathbb{R}^{d+2}\ |\ \mu_i\ge0\ \forall\ i,\ \sum_i\mu_i=1\}$$ such that $\sum_i\mu_i x_i=y$?

Answer 1

Here's a proof . . .

Let $S=\{x_0,...,x_{d+1}\}$.

Let $D$ be the interior of $C$.

Each point of $p\in C$ is in at least one closed simplex whose vertices are $d+1$ distinct points of $S$, hence $p$ can be represented in at least one way as a convex combination of $x_0,...,x_{d+1}$, with at least one of the convex weights equal to zero.

Let $E$ be the set of points in $\mathbb{R}^d$ which can represented as a strict convex combination of the $d+2$ elements of $S$.

Then $E$ is a convex open subset of $C$.

But for any $p\in C$, there are elements of $E$ which are arbitrarily close to $p$.

It follows that $E=D$.

Hence each $y\in D$ can be represented as a convex combination of the $d+2$ elements of $S$ in at least two ways; one way with all positive convex weights, and another way with at least one convex weight equal to zero.

Here's a little more detail with regard to the claim $E=D$.

Clearly, $E\subseteq D$.

For the reverse inclusion, let $y\in D$.

Let $p_0,...,p_n$ be the vertices of a regular $n$-simplex $P\subset D$, centered at $y$.

Then $y$ is in the interior of $P$, hence we can choose $q_0,...,q_n\in E$, with $q_0,...,q_n$ close enough to $p_0,...,p_n$, respectively, so that $y$ is in the interior of the simplex $Q$ with vertices $q_0,...,q_n$.

Thus, $y$ is a strict convex of $q_0,...,q_n$.

Since $y$ is a strict convex of $q_0,...,q_n$, and each of $q_0,...,q_n$ is a strict convex combination of $x_0,...,x_{n+1}$, it follows that $y$ is a strict convex combination of $x_0,...,x_{n+1}$, hence $y\in E$.

Thus, we get $D\subseteq E$.