$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Lemma Let $V$ be a normed vector space over the real numbers and let $K \subseteq V$ be the convex hull of a finite collection of points $v_0, v_1, \ldots , v_ n \in V$, given by the image of a continuous function:

\[ f: | \Delta ^ n | \rightarrow V \quad \quad (t_0, t_1, \ldots , t_ n) \mapsto t_0 v_0 + t_1 v_1 + \cdots + t_ n v_ n. \]

Let $\sigma $ be any $m$-simplex of the subdivision $\operatorname{Sd}( \Delta ^{n} )$, let $f_{\sigma }$ denote the composite map

\[ | \Delta ^{m} | \xrightarrow { | \sigma | } | \operatorname{Sd}( \Delta ^{n} ) | \simeq | \Delta ^{n} | \xrightarrow {f} V \]

(where the homeomorphism $| \operatorname{Sd}( \Delta ^{n} ) | \leq | \Delta ^ n |$ is supplied by Proposition, and let $K_0 \subseteq K$ be the image of $f_{\sigma }$. Then the diameters of $K_0$ and $K$ satisfy the inequality $\mathrm{diam}( K_0 ) \leq \frac{n}{n+1} \mathrm{diam}(K)$.

Proof. Let us denote the norm on the vector space $V$ by $| \bullet |_{V}$. Fix points $x,y \in | \Delta ^{m} |$; we wish to show that $| f_{\sigma }(x) - f_{\sigma }(y) |_{V} \leq \frac{n}{n+1} \mathrm{diam}(K)$. Note that, if we regard the point $x$ as fixed, then the function $y \mapsto | f_{\sigma }(x) - f_{\sigma }(y) |_{V}$ is convex, and therefore achieves its supremum at some vertex of $| \Delta ^{m} |$. We may therefore assume without loss of generality that $y$ is a vertex of $| \Delta ^{m} |$. Similarly, we may assume that $x$ is a vertex of $| \Delta ^{m} |$. We may also assume that $x \neq y$ (otherwise there is nothing to prove). Exchanging $x$ and $y$ if necessary, it follows that there exist disjoint nonempty subsets $A,B \subseteq \{ 0, 1, \ldots , n \} $ of cardinality $a = |A|$ and $b = |B|$ satisfying

\[ f_{\sigma }(x) = \sum _{i \in A} \frac{ v_ i}{a} \quad \quad f_{\sigma }(y) = \sum _{i \in A \cup B} \frac{ v_ i}{a+b}. \]

We then compute

\begin{eqnarray*} | f_{\sigma }(x) - f_{\sigma }(y) |_{V} & = & | \sum _{(i,j) \in A \times B} \frac{v_ i - v_ j}{a(a+b)} |_{V} \\ & \leq & \sum _{(i,j) \in A \times B} \frac{ | v_ i - v_ j |_{V} }{ a(a+b) } \\ & \leq & \sum _{(i,j) \in A \times B} \frac{ \mathrm{diam}(K) }{ a(a+b) } \\ & = & \frac{b}{a+b} \mathrm{diam}(K) \\ & \leq & \frac{n}{n+1} \mathrm{diam}(K). \end{eqnarray*}