A proof related to diameter of a simplex S

Question

Question: Prove that the diameter $\mathcal p(S)$ of a simplex $\mathcal S$ equals the greatest Eucledian distance between two vectors in the simplex.

My opinion: We all know what every vector in the $k$-dimensional simplex can be represented as a unique convex combination of $k+1$ affine independent vectors. Let $\mathcal S$$=<<x^0,x^1,...,x^k>>$ be $k$-dimensional simplex. Moreover, the definition of $\mathcal p(S) = max ||x^i-x^j||$ where $0 \le i \le j \le k$

I do not know how to proceed. Please help. I thank you very much.

Answer 1

Assuming that the simplex $S$ is the convex hull of $k+1$ affinely independent points $x^0,\dots,x^k$ in some $\Bbb R^n$, the diameter should be defined as $d=\text{diam}(S)=\sup\{||x-y||; x,y\in S\}$, and we want to show that $d=\max\{||x^i-x^j||; 0\le i<j\le k\}$.
We can do this by showing that, given points $x,y\in S$, we can find some vertex $x^j$ such that $ ||x-y||\le||x-x^j||$. Write $y$ as a convex combination $y=\sum_i t^ix^i$. Then $$\begin{align} ||x-y|| &= \textstyle ||\sum t^i x-\sum t^ix^i||\\ &= \textstyle ||\sum_i t^i(x-x^i)|| \\ &\le \textstyle \sum_i t^i||x-x^i|| \\ &\le \textstyle \sum_i t^i\max\{||x-x^j||; 0\le j\le k\} \\ & = \max\{||x-x^j||; 0\le j\le k\} \end{align}$$ We can now apply the same result letting $x^i$ take the role of $x$ and letting $x$ take the role of $y$.

Answer 2

Here is another (related) argument. Roughly speaking, a convex function achieves its maximum over a convex set at the extreme points of the set. In other words, the maximum of the convex function over a set $S$ is the same as its maximum over the set of extreme points (assuming enough regularity).

Let $S$ be a simplex with extreme points $E = \{x^i\}$. Then, $S$ is the (closed) convex hull of $E$. The function $x \mapsto \| x - y\|$ is a (continuous) convex function. Hence the maximum of this function over $S$ is the same as its maximum over $E$. Then, \begin{align*} \max_{y \in S} \max_{x \in S} \| x - y\| &= \max_{y \in S} \max_{i} \| x^i - y\| \\ &= \max_{i} \max_{y \in S} \| x^i - y\| \\ &= \max_{i} \max_{j} \| x^i - x^j\| \end{align*} where in the last line we applied the same idea to the function $y \mapsto \|x^i - y\|$.