The diameter of a convex hull. I want to prove the following statement:


  Given $A\subset \mathbb{R}^n$ let $C(A)$ be its convex hull. Prove that $\text{diam }(A)=\text{diam }(C(A))$.


I can suppose that $A$ is a bounded closed set and I know that if $x,y\in A$ are such that $d(x,y)=\text{diam }(A)$ then $x,y\in \partial A$. I tried proving that if $z,w\in \partial C(A)$ then $d(z,w)\leq d(x,y)$ but it is a little difficult to me using the fact $C(A)$ is the convex hull. 

Any hint?
Let $\ell = \mathrm{diam}\;A$. 

Since $C(A) \supseteq A$, it is trivial to see $\mathrm{diam}\;C(A) \ge \ell$. 

If $\mathrm{diam}\;C(A) &gt; \ell$, then one can find $p, q \in C(A)$ such that
$d(p,q) &gt; \ell$.

By Carathéodory's theorem, we can express $p$ as a convex linear combination of $P \le n+1$ points from $A$. More precisely, there exists
$$\begin{cases}
p_1, p_2, \ldots, p_P \in A\\
\lambda_1, \lambda_2, \ldots, \lambda_{P} \ge 0
\end{cases}
\quad\text{ with }\quad
\begin{cases}
\lambda_1 + \lambda_2 + \cdots + \lambda_P = 1\\
\lambda_1 p_1 + \lambda_2 p_2 + \cdots + \lambda_P p_P = p
\end{cases}
$$
Similarly, we can find $Q \le n + 1$ points in $A$ such that
$$
\begin{cases}
q_1, q_2, \ldots, q_Q \in A\\
\mu_1, \mu_2, \ldots, \mu_Q \ge 0
\end{cases}
\quad\text{ such that }\quad
\begin{cases}
\mu_1 + \mu_2 + \cdots + \mu_Q = 1\\
\mu_1 q_1 + \mu_2 q_2 + \cdots + \mu_Q q_Q = q
\end{cases}
$$
Notice as a function of $p$ and $q$, the distance $d(p,q) = |p-q|$ is a convex function in both of its arguments. This implies

$$d(p,q) = d\left(\sum_{i=1}^P \lambda_i p_i, \sum_{j=1}^Q \mu_j q_j \right)
\le \sum_{i=1}^P\sum_{j=1}^Q \lambda_i \mu_j d(p_i, q_j) \le \left(\sum_{i=1}^P\lambda_i\right)\left(\sum_{j=1}^Q \mu_j\right) \ell = \ell$$

Contradicts with our choice of $p, q$ which satisfy $d(p,q) &gt; \ell$. This implies the underlying assumption $\mathrm{diam}\;C(A) &gt; \ell$ is invalid. As a result, $\mathrm{diam}\;C(A) = \ell = \mathrm{diam}\;A$.

Note

If one adopt the definition that by convex hull, one mean the collection of all convex linear combinations of points from a set. We don't need Carathéodory's theorem at all. The argument above works as long as $P, Q$ exist and are finite.
Let $x,y\in C(A)$. Then by Carathéodory's theorem, there exists $p,q\le n+1$ and $x_1,\dots,x_{p},y_1,\dots,y_{q}\in A$ such that $x=\sum_{i=1}^{p}\lambda_{i}x_{i} $, $y=\sum_{j=1}^{q}\gamma_{j}y_{j}$, $\sum_{j=1}^{q}\gamma_{j}=1=\sum_{i=1}^{p}\lambda_{i}$ and $\lambda_{i}\ge0\ (i=1,\dots,p)$, $\gamma_{j}\ge0\ (j=1,\dots,q)$. Therefore
$$\|x-y\| = \left\|\sum_{i=1}^{p}\lambda_{i}x_{i}-y\right\|=\left\|\sum_{i=1}^{p}\lambda_{i}(x_{i}-y)\right\|\le\sum_{i=1}^{p}\lambda_{i}\|x_{i}-y\| \\ = \sum_{i=1}^{p}\lambda_{i}\left\|x_{i}-\sum_{j=1}^{q}\gamma_{j}y_{j}\right\| = \sum_{i=1}^{p}\lambda_{i}\left\|\sum_{j=1}^{q}\gamma_{j}(x_{i}-y_{j})\right\| \le \sum_{i=1}^{p}\lambda_{i}\sum_{j=1}^{q}\gamma_{j}‖x_{i}-y_{j}‖ \le \left(\sum_{i=1}^{p}\lambda_{i}\right)\left(\sum_{j=1}^{q}\gamma_{j}\right) \operatorname{diam}A = \operatorname{diam}A\text.\tag1\label1$$
Now by taking supremum over $x,y\in C(A)$, from \eqref{1} we obtain $\operatorname{diam}{C(A)}\le\operatorname{diam}A$. Since $A\subseteq C(A)$ we get $\operatorname{diam}{C(A)}\ge\operatorname{diam}A$. Consequently, $\operatorname{diam}{C(A)}=\operatorname{diam}A$.