This is my last question in this area.
Let $p\geq 2$ and $X=\{v_1,\ldots,v_m\}$. Further let $x\in\mathbb R^n, r>0$ such that $\|x-v_i\|_p\leq r$ for all $i\in \{1,\ldots,m\}$.
Is it true, that for all $y\in\operatorname{conv} X$ there is a point $v_i\in X$ with $\|y-v_i\|_p\leq r$?
I know, it's true for $p=2$ (Covering of a polytope by balls with midpoints at the vertices) and it's false for $p=1$ (Let $K$ be an $o$-symmetric body, that covers $X$. Does it imply $\operatorname{conv}(X)\subset X+K$?)
I tried to prove it myself for $p=4$, but it's nasty, because we have no Hilbert-space. Here is, what i've done so far:
If the assumpptions from above hold, let $y=\sum_{i=1}^m\lambda_iv_i$. $$ \begin{align*} \| y-v_i\|^4 &=\|y-x\|^4 + 4\sum_{j=1}^n(y_j-x_j)^3(x_j-v_{i,j})+ 6\sum_{j=1}^n(y_j-x_j)^2(x_j-v_{i,j})^2\\ &\quad + 4\sum_{j=1}^n(y_j-x_j)^3(x_j-v_{i,j}) + \|x-v_i\|^4 \end{align*} $$ Multiplying by $\lambda_i$ and summing up yields $$\begin{align*} \sum_{i=1}^m\lambda_i\| y-v_i\|^4 &=\|y-x\|^4 + 4\sum_{j=1}^n(y_j-x_j)^3(x_j-\sum_{i=1}^mv_{i,j})+ 6\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)^2(x_j-v_{i,j})^2\\ &\quad + 4\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)(x_j-v_{i,j})^3 + \sum_{i=1}^m\lambda_i\|x-v_i\|^4\\ &=\|y-x\|^4 + 4\sum_{j=1}^n(y_j-x_j)^3(x_j-y_j)+ 6\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)^2(x_j-v_{i,j})^2\\ &\quad + 4\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)(x_j-v_{i,j})^3 + \sum_{i=1}^m\lambda_i\|x-v_i\|^4\\ &=\|y-x\|^4 - 4\sum_{j=1}^n(y_j-x_j)^3(y_j-x_j)+ 6\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)^2(x_j-v_{i,j})^2\\ &\quad + 4\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)(x_j-v_{i,j})^3 + \sum_{i=1}^m\lambda_i\|x-v_i\|^4\\ &=\|y-x\|^4 - 4\|y-x\|+ 6\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)^2(x_j-v_{i,j})^2\\ &\quad + 4\sum_{i=1}^m\sum_{j=1}^n\lambda_i(y_j-x_j)(x_j-v_{i,j})^3 + \sum_{i=1}^m\lambda_i\|x-v_i\|^4 \end{align*} $$ Hence, we get $$ \begin{align*} \sum_{i=1}^m\lambda_i\| y-v_i\|^4 +3\|y-x\|^4 &= 6\sum_{i=1}^m\sum_{j=1}^n\lambda_i(x_j-y_j)^2(x_j-v_{i,j})^2- 4\sum_{i=1}^m\sum_{j=1}^n\lambda_i(x_j-y_j)(x_j-v_{i,j})^3\\ &\quad + \sum_{i=1}^m\lambda_i\|x-v_i\|^4\\ &= \sum_{i=1}^m\sum_{j=1}^n\lambda_i(x_j-y_j)(x_j-v_{i,j})^2(2v_{i,j}+x_j-3y_j) + \sum_{i=1}^m\lambda_i\|x-v_i\|^4. \end{align*} $$ It is left to show that $$\sum_{i=1}^m\sum_{j=1}^n\lambda_i(x_j-y_j)(x_j-v_{i,j})^2(2v_{i,j}+x_j-3y_j)\leq 3\|y-x\|^4.$$
Maybe there is an easier way by the monotonicity of the $p$-norms?
Wlog we can assume $x=0$.
Here is a counterexample to $p=\infty$: Take $n=3$, $$ X = \left\{ \pmatrix{1\\1\\1}, \ \pmatrix{ 1\\-1\\-1}, \ \pmatrix{ -1 \\ 1\\-1}\right\} = \{ v_1 \dots v_3\}, $$ $$ y =\frac13(v_1+v_2+v_3) = \pmatrix{1/3\\1/3\\-1/3}. $$ Then $X$ is contained in the $\infty$-unit ball, but $\|y-v_i\|_\infty=\frac43$.
Actually, this selection of vectors gives a counterexample for a large range of $p$'s. Computing $p$-norms, we get $\|v_i\|_p^p = 3$ and $\|y-v_i\|_p^p = \frac{2\cdot 2^p+4^p}{3^p}$. According to wolframalpha, $\frac{2\cdot 2^p+4^p}{3^p} > 3$ for $p>3.08...$. Hence, the claim is false for this range of $p$.
I would suspect that taking $n\to\infty$ in the following pattern would produce counterexamples for all $p>2$: $v_1$ vector of all ones, $v_i$ for $i=2\dots n$: entries $n$ and $i-1$ set to $-1$, all others to $+1$. $y=\frac1n\sum_{i=1}^nv_i$.