Let $X$ be an infinite-dimensional Banach space and $C\subseteq X$ be a bounded closed convex subset. Let $\{z\_i\}\_{i\in\mathbb{N}}$ be a sequence of linearly independent points in $C$ and for each $n\in\mathbb{N}$, define $V\_{\leq n} = \operatorname{conv}\{z\_i\}\_{i\leq n}$ and $V\_{\geq n} = \operatorname{conv}\{z\_i\}\_{i\geq n}$. For each $z\in V\_{\geq 1}$, let $N(z)$ denote the least positive integer such that $z\notin V\_{\geq N(z)}$. For each $n\in\mathbb{N}$, define:


$$
d\_n: C\rightarrow [0, \infty), \hspace{0.3cm} c\mapsto \inf\big\{ \|c-z\|: z\in V\_{\geq n} \big\}
$$


and clearly each $d\_n$ is continuous.


**My question**: if, for a fixed $m\in\mathbb{N}$, there exists $\lambda>0$ and $M\in\mathbb{N}$ such that $d\_M(z\_i) > \lambda$ for each $i\leq m$, is it true that for each $z\in V\_{\leq m}$, there exists $n\geq \max(N(z), M)$ such that $d\_n(z) > \lambda$?


What inspires this question is that, given $m\in\mathbb{N}$ and $\lambda>0$ such that $d\_M(z\_i)>\lambda$ for each $i\leq m$, I need to find $M'\in\mathbb{N}$ such that:


$$
\inf\big\{ \|z-z'\|: z\in V\_{\leq m}, z'\in V\_{\geq M'}\big\} > 0
$$


Notice that each $d\_n$ is continuous. If my guess is true, then by the fact that $V\_{\leq m}$ is compact the inequality above will be true. Any hints, as well as other methods to prove the inequality above without answering my question, will be appreciated.



Distance between convex hulls in a bounded closed convex set
$\newcommand\la\lambda$The answer is no. E.g., suppose that $X=\ell^\infty$, $z\_1=e\_1$, $z\_2=-e\_1+e\_2$, and $z\_k=e\_2/2+e\_k/k$ for $k\ge3$, where $(e\_1,e\_2,\dots)$ is the standard basis of $\ell^\infty$. Let $C$ be the unit ball in $\ell^\infty$. Then $C$ is a bounded closed convex subset of $X$ and the $z\_k$'s are linearly independent points in $C$.


Let $m=2$, $M=3$, $\la=1/2$, and $z:=(z\_1+z\_2)/2=e\_2/2$. Then $d\_M(z\_i)=1>\la$ for each $i\le m$. However, for each $n\ge M$ we have $d\_n(z)=0\not>\la$. $\quad\Box$


