TITLE: When does the center of a set of points lie in the convex hull?
QUESTION [5 upvotes]: Suppose we are given $n$ points $X_1,X_2\dots X_n$ in $d$-dimensional Euclidean space $\mathbb{R}^d$. I'm interested in understanding some properties of the ``center'' of $\{X_i\}_{i=1}^n$ w.r.t $L_p$ norm, for $p \geq 1$
$$
\min_{X} \sum_{i=1}^n\|X-X_i\|_p.
$$
For what values of $p$ does the minimizer of the above problem lie in the convex hull of $\{X_i\}_{i=1}^n$? 
PS: When $p=1$, the solution to the above problem is the coordinatewise median. When $p=2$ the minimizer is called the geometric median.

REPLY [5 votes]: Partial answer for $p=2$ (the easy case).
In all cases $f$ is convex and $\lim_{\|x\|_p \to \infty} f(x) = \infty$ so a minimiser exists.
Suppose $x \notin \operatorname{co} \{ x_k \}$. Then $f$ is differentiable at $x$ and
${\partial f(x) \over \partial x}= \sum_k {(x - x_k)^T \over \|x-x_k\|_2}$. If ${\partial f(x) \over \partial x} = 0$ then an easy computation shows that $x \in \operatorname{co} \{ x_k \}$, a contradiction. Hence $x$ cannot be a minimiser.
Partial counterexample for $p=1$.
Let the points be $(0,0), (2,0), (2,0), (-1, 1), (-1, 1), (-1, 1)$ (yes, some points are repeated). It is easy to see that the minimisers are the square $[-1,0]\times [0,1]$, but not all of these points lie in the convex hull.