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Let $M\subset \mathbb{R}^2$ be a finite set of points, $\operatorname{C}(M)$ the convex hull of M and $$\operatorname{diam}(M) = \sup_{x,y\in M}\|x-y\|_2$$ be the diameter of $M$

What I want to show now is, that it holds $$\operatorname{diam}(M) = \operatorname{diam}(\operatorname{C}(M))$$

Because $$M\subseteq\operatorname{C}(M)$$ we obtain $$\operatorname{diam}(M) \le\operatorname{diam}(\operatorname{C}(M))$$ but how to proof that $$\operatorname{diam}(M) \ge \operatorname{diam}(\operatorname{C}(M))$$

I suppose it should be possible to construct a contradiction assuming $\operatorname{diam}(M) <\operatorname{diam}(\operatorname{C}(M))$ but i do not see how at this moment.

user127.0.0.1
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2 Answers2

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Hint:

Prove this for a triangle and then use the fact that for every point of $C(M)$ there is a triangle that contains it, there are many ways to go from there.

I hope this helps ;-)

dtldarek
  • 37,381
3

Another method:

  • $C(M)$ is compact, so there exist $x_1,x_2 \in C(M)$ such that $d(x_1,x_2)=\mathrm{diam}(C(M))$.
  • Let $f : x \mapsto d(x_1,x)$. Because $f$ is convex, $f((1-t)x+ty) \leq \max(f(x),f(y))$.
  • Deduce that $x_2 \in \partial M$ and then that $x_2 \in M$.
  • With a symmetric argument, you have $x_1 \in M$.

Therefore, $\mathrm{diam}(C(M)) =d(x_1,x_2)\leq \mathrm{diam}(M)$.

Seirios
  • 33,157