Distance of a point from centroid of simplex

Question

I have a full dimensional simplex $S \subset \mathbb{R}^n$ created by vectors $v_1,v_2,\ldots,v_{n+1}$. Let $v_c=\frac{\sum_{i=1}^{i=n+1} v_i}{n+1}$ be its centroid. Let $v \in S$ be some vector. Can we say $\|v-v_c\|_2 \leq \text{max} \{\|v_i-v_c\|_2\}$ for $i=1,2,\ldots n+1$ ?

It is trivial for $v=v_i$ or $v=v_c$. But what about in general ?

Answer 1

As pointed out in another answer, the answer is YES. Following is an algebraic way seeing this.

Since $v \in S$, $v$ is a convex linear combination of $v_i$. There are $n+1$ non-negative numbers $\lambda_1, \cdots\lambda_{n+1}$ such that $$ v = \sum_i \lambda_i v_i\quad\text{ and }\quad \sum_{i} \lambda_i= 1 $$ These are the barycentric coordinates of $v$ with respect to simplex $S$. Let

• $R_i = \| v_i - v_c\|$ be the distance between $v$ and $v_c$.
• $R_{\rm max} = \max\limits_{i} R_i$, the largest distance between $v_c$ and any vertex.
• $\ell_{ij} = \|v_i - v_j\|$ be the edge lengths of the $n$-simplex.
• $\ell_{\rm max} = \max\limits_{i<j} \ell_{ij}$ and $\ell_{\rm min} = \min\limits_{i<j} \ell_{ij}$ be the largest and smallest edge lengths.

In terms of the barycentric coordinates $\lambda_i$, we have $$\|v - v_c\|^2 = \left\|\sum_i\lambda_i(v_i-v_c)\right\|^2 = \sum_{i,j} \lambda_i\lambda_j(v_i-v_c)\cdot(v_j-v_c) = \frac12\sum_{i,j} \lambda_i\lambda_j ( R_i^2 + R_j^2 - \ell_{ij}^2) $$ This leads to $$\|v - v_c\|^2 = \sum_{i}\lambda_i R_i^2 - \sum_{i < j}\lambda_i\lambda_j \ell_{ij}^2 \quad\implies\quad \|v - v_c\| \le \left(\sum_i\lambda_i R_i^2\right)^{\frac12} \le R_{\rm max}$$

Notice we can express $R_i$ solely in terms of the edge lengths $\ell_{ij}$.

For any two points $u, v \in S$, let $\mu_i, \nu_i$ be their barycentric coordinates with respect to $S$.
In general, we have

$$\| u - v \|^2 = - \sum_{i<j} \ell_{ij}^2(\mu_i - \nu_i)(\mu_j - \nu_j)$$

For the special case $u = v_k$ and $v = v_c$, we have

$$\begin{align} R_k^2 = \| v_k - v_c \|^2 &= - \sum_{i<j} \ell_{ij}^2\left(\delta_{ik} - \frac{1}{n+1}\right)\left(\delta_{jk} - \frac{1}{n+1}\right)\\ &= \frac{1}{n+1}\sum_{i\ne k}\ell_{ik}^2 - \frac{1}{(n+1)^2}\sum_{i<j}\ell_{ij}^2 \end{align} $$ This leads to

$$R_{\rm max}^2 = \max_{k} R_k^2 \le \frac{1}{n+1}(n\ell_{\rm max}^2) - \frac{1}{(n+1)^2}\left(\frac{(n+1)n}{2}\ell_{\rm min}^2\right) = \frac{n}{2(n+1)}(2\ell_{\rm max}^2 - \ell_{\rm min}^2)$$

As a result, we have

$$\max_{v \in S}\| v - v_c \| =R_{\rm max} \le \sqrt{\frac{n}{2(n+1)}(2\ell_{\rm max}^2 - \ell_{\rm min}^2)}$$

In the special case when $S$ is regular, $\ell_{\rm max} = \ell_{\rm min} = \ell$. Above formula becomes $$R_{\rm max}^{\rm regular} = \sqrt{\frac{n}{2(n+1)}}\ell$$

Answer 2

Yes. You can easily verify that the function $v\to |\!|v-v_c|\!|$ defined on $S$ is convex, and so it attains its maximum on one (or more) of the vertices $v_i$. In fact, you can say more: for every $u\in S$ one has: $$\forall v\in S\quad |\!|v-u|\!|\leq\max_{1\leq i\leq n+1}|\!|v_i-u|\!|$$