Let $m$ be a positive integer and let $E(m)$ be the $m$-dimensional Euclidean space with its standard metric. For any positive integer $n > m$, let $P(1),P(2),\ldots,P(n)$ be a finite set of points of $E(m)$ whose convex hull $H$ has a non-empty interior (with respect to $E(m)$). Let $p$ be any interior point of $H$. Does there always exist a positive real number $e$ such that the following statement is true?
If $Q(1),Q(2),\ldots,Q(n)$ is any finite set of points of $E(m)$ for which the Euclidean distance between $P(i)$ and $Q(i)$ with $i \in \{ 1,2,\ldots,n \}$ is never greater than $e$, then the point $p$ belongs to the convex hull of $Q(1),Q(2),\dots,Q(n)$.
If such a number as $e$ always exists, then it would, of course, have to depend on $H$ and $p$.
Yes, this is true.
Since the convex hull of the $P(i)$ is of non-empty interior, I can choose $m+1$ points among the $P(i)$ that form an affine basis of $E(m)$. Up to reindexation, I assume that these points are $P(1),\dots,P(m+1)$ and I show that if $Q(1),\dots,Q(m+1)$ are close to $P(1),\dots,P(m+1)$ then $p$ is still in the convex hull of the $Q(i)$. (I don't need the other points).
I claim that if the $Q(i)$ are close to the $P(i)$, then the $Q(i)$ give an affine basis of $E(m)$. Indeed, this is the affine analogue of the well-known linear fact that if $n$ vectors are close to a basis of a $n$-dimensional vector space, then they still give a basis (use a determinant).
Hence, I can write the $P(j)$ as barycenters in the $Q(i)$. Since $p$ is a barycenter in the $P(j)$, I can write it as a barycenter in the $P(i)$ by associativity of barycenters. The only thing to check is that this barycenter has weights in $(0,1)$ if the $Q(i)$ are sufficiently close of the $P(i)$.
Say that $P(j)$ is a barycenter of the $Q(i)$ with weights $\omega_{ij}(Q_i)$. Then, you can show that $\omega_{ij}(Q_i)$ depends continuously on $Q_i$ (this is essentially solving a linear system). Now, $p$ is a barycenter of the $Q(i)$ (for $Q(i)$ close to $P(i))$ with weights $\nu_i(Q_i)$ and $\nu_i(Q_i)$ depends also continuously on the $Q_i$ (it is given by an explicit formula involving the $\omega_{ij}$ and the weights of $p$ relative to the $P_j$). Since for $Q_i = P_i$, $\nu_i(P_i)$ is in $(0,1)$, this concludes the proof.