Diameter of finite set of points is equal to diameter of its convex hull 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. 

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