The Frechet mean of $k$ elements $x_1, \ldots, x_k \in S = \{ x \in \mathbb{R}^n,\, \| x \| = 1 \}$ is defined as the $\text{arg}\underset{\|x\|=1}{\text{min}} \sum_{i=1}^n d^2(x_i,x)$. Where $d(x_i,x)= \text{arc }\text{cos}(<x_i,x>)$.
Is it true that if $k<n$ and the points are not colinears, there is a Frechet mean that belongs to the subspace generated by the $k$ points? This means that exists $x^\star \in \text{arg}\underset{\|x\|=1}{\text{min}} \sum_{i=1}^n d^2(x_i,x) $ such that $x^\star \in <x_1,\ldots, x_k>$.
If so how could I prove it? Does it hold for generalizations of Frechet mean, such as Frechet median? or a cost function $\sum_{i=1}^n \rho(d(x_i,x))$ where $\rho:[0,\infty)\to \mathbb{R}$ is a strictly increasing function.
It is true, here is a sketch of the idea:
First, if $ a $ and $ b $ are vectors defining a subspace $ V $, then moving $ b $ out of $ V $ increases $ d(a,b) $. It is a straightforward calculation: choose a coordinate system so that $ a = (1,\cdots,0)$ and $ b = (b_1,b_2,\epsilon,\cdots,0) $, where $ \epsilon $ parameterizes a deformation out of the plane $ V $. The angle between $ a $ and $ b $ is $ \cos^{-1} \left( \frac{b_1}{\sqrt{b_1^2 + b_2^2 + \epsilon^2} } \right) $, which increases with $ \epsilon $.
So, for your case, suppose $ x_i $ define a vector space $ V $. Let $ x $ be the Frechet mean on the $ (k-1) $-sphere $ S_{k-1} \subset V $ defined by the $ x_i $. Then by the above argument, deforming $ x $ outside of $ V $ will strictly increase distance to each $ x_i $, so that $ x $ is still at least a local minima.